Docstoc

engineering_math

Document Sample
engineering_math Powered By Docstoc
					          ‫‪WWW.ESUD83.MIHANBLOG.COM‬‬
                                                 ‫ﻣﻌﺮﰲ ﻣﻌﺎﺩﻻﺕ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻭ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ:‬

    ‫ﺍﮔﺮ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺷﺎﻣﻞ ﺟﻨﺪ ﻣﺘﻐﲑ ﻭ ﻣﺸﺘﻘﺎﺕ ﺁ‪‬ﺎ ﺑﺎﺷﺪ، ﺁﻥ ﻣﻌﺎﺩﻟﻪ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻧﺎﻣﻴﺪﻩ ﻣـﻲ‬
    ‫ﺷﻮﺩ.ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﺳﻴﻞ ﺑﺎ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺍﺳﺖ ﻛﻪ ﺷﺎﻣﻞ ﻳﻚ ﺗﺎﺑﻊ ﻭ ﻣـﺘﻐﲑ ﻫـﺎﻱ ﺁﻥ ﻭ‬
    ‫ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﻣﺮﺑﻮﻁ ﻣﻲ ﺑﺎﺷﺪ.ﻓﺮﻡ ﻛﻠﻲ ﻣﻌﺎﺩﻻﺕ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺑﺎ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﺑﻪ ﺻﻮﺭﺕ ﺭﺍﺑﻄﻪ‬



                                                                                                          ‫)١(ﻣﻲ ﺑﺎﺷﺪ:‬
                                                                                              ‫ﺭﺍﺑﻄﻪ )١(: 0 = )......,‬
                                  ‫)‪( n‬‬                              ‫)‪( m‬‬           ‫) ‪(i + j‬‬
     ‫‪G(u, ux , uxx ,.......,ux‬‬           ‫‪, u y , u yy ,......,u y‬‬          ‫‪, uxy‬‬
                  ‫‪∂y‬‬        ‫‪∂u‬‬                                   ‫‪∂ 2u‬‬
      ‫‪u‬‬       ‫=‬                ‫‪= u‬‬                 ‫‪u‬‬        ‫=‬                                               ‫ﻛﻪ ﺩﺭ ﺁﻥ‬
                  ‫‪∂x‬‬        ‫‪∂y‬‬                                  ‫‪∂x∂y‬‬
          ‫‪x‬‬                                  ‫‪y‬‬         ‫‪xy‬‬




      ‫0 = ) ‪⎧u xxx ( x, y ) + u yyy ( x, y‬‬                                                                      ‫ﻣﺜﺎﻝ:‬
      ‫⎪‬
      ‫⎨‬         ‫‪∂ 3u ∂ 3u‬‬
      ‫⎪‬               ‫+‬        ‫0=‬
      ‫⎩‬         ‫3 ‪∂x 3 ∂y‬‬

    ‫ﳎﻤﻮﻋﻪ ﻣﻌﺎﺩﻻﺕ ‪ PDE‬ﻳﻚ ﳎﻤﻮﻋﻪ ﻣﻌﺎﺩﻻﺕ ﻧﺎﳏﺪﻭﺩ ﻣﻲ ﺑﺎﺷﺪ.ﺍﺯﺑﲔ ﺍﻳﻦ ﳎﻤﻮﻋﻪ ﺗﻨﻬﺎ ﲞﺶ ﻛﻮﭼﻜﻲ‬
                                                      ‫ﺩﺍﺭﺍﻱ ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ﺧﺎﺹ ﻣﻲ ﺑﺎﺷﺪ.‬
       ‫ﻏﺎﻟﺐ ﻣﺴﺎﺋﻠﻲ ﻛﻪ ﺩﺭ ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ‪ PDE‬ﻣﻄﺮﺡ ﻣﻲ ﺷﻮﻧﺪ ﺩﺍﺭﺍﻱ ﻓﺮﻣﻬﺎﻱ ﳏﺎﺳﺒﺎﰐ )ﺭﺍﺑﻄﻪ )٢((‬




‫١‬
                                              ‫ﻣﻲ ﺑﺎﺷﻨﺪ.ﻛﻪ ﺩﺭ ﺁﻥ ‪ A , B , ….., G‬ﺿﺮﺍﻳﺐ ﻣﻌﺎﺩﻟﻪ ‪PDE‬ﻫﺴﺘﻨﺪ.‬
    ‫ﺭﺍﺑﻄﻪ )٢(:0 = ‪Au xx ( x, y ) + Bu xy + Cu yy ( x, y ) + Du x ( x, y ) + Eu y ( x, y ) + Fu ( x, y ) + G‬‬
                                      ‫ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﻓﺮﻡ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﻣﻌﺎﺩﻟﻪ ﺩﺭﺟﻪ ٢ ﺍﺯ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﻣﻲ ﻧﺎﻣﻨﺪ.‬
            ‫ﭼﻨﺎﻧﭽﻪ ‪ A , B , ….., G‬ﺗﻮﺍﺑﻌﻲ ﺍﺯ ‪ x , y‬ﺑﺎﺷﻨﺪ، ﻣﻌﺎﺩﻟﻪ ﻱ ﺩﺭﺟﻪ ٢ ﺧﻄﻲ ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ ﻭﱄ ﭼﻨﺎﻧﭽﻪ‬
                                        ‫‪ A , B , ….., G‬ﺗﺎﺑﻌﻲ ﺍﺯ ‪ U‬ﺑﺎﺷﺪ ،ﻣﻌﺎﺩﻟﻪ ﻱ ﻏﲑ ﺧﻄﻲ ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ.‬
              ‫0 = ‪u xx + (sin u ) uy‬‬                                        ‫ﻣﺜﺎﻝ: ﻣﻌﺎﺩﻟﻪ ﻏﲑ ﺧﻄﻲ ﺩﺭﺟﻪ ﺩﻭ‬

             ‫01 = ) ‪zu xx ( x, y ) + xy 2 u yy ( x, y ) + e x u ( x, y‬‬
                                                                        ‫ﻣﺜﺎﻝ: ﻣﻌﺎﺩﻟﻪ ﺧﻄﻲ ﺩﺭﺟﻪ ﺩﻭ‬
                              ‫ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺑﺎ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﺧﻄﻲ ﺩﺭﺟﻪ ٢ ﺩﺭ ﺣﻮﺯﻩ ﻱ ) ‪u ( x, y, z‬‬
          ‫ﻣﻌﺎﺩﻻﺕ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺑﺎ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﺭﻭﻱ ﺧﻂ ، ﺻﻔﺤﻪ ﻭ ﻳﺎ ﻓﻀﺎ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﻧﺪ.ﺩﺭ ﺑﺴﻴﺎﺭﻱ‬
          ‫ﺍﺯ ﻛﺎﺭﺑﺮﺩﻫﺎ ،ﻣﻘﺎﺩﳝﺘﻐﲑ ‪U‬ﺭ ﺩﺭ ﻣﺮﺯ ﻧﺎﺣﻴﻪ ﻳﺎ ﻓﻀﺎﻳﺎ ﺭﻭﻱ ﺧﻂ ﻣﻌﻠﻮﻡ ﻣﻲ ﺑﺎﺷﺪ.ﻧﻴﺰﺩﺭ ﺑﺮﺧﻲ ﻛﺎﺭﺑﺮﺩﻫﺎ‬
          ‫ﻣﻘﺪﺍﺭ ﺗﺎﺑﻊ ‪ U‬ﺩﺭ ﻧﻘﻄﻪ ﺷﺮﻭﻉ ﺯﻣﺎﻥ ﻭ ﻣﻜﺎﻥ ﻣﻌﻠﻮﻡ ﻣﻲ ﺑﺎﺷﺪ.ﻣﺴـﺎﺋﻞ ﮔـﺮﻭﻩ ﺍﻭﻝ ‪Boundary value‬‬

          ‫‪) Problems‬ﻣﺴﺎﺋﻞ ﺑﺎ ﻣﻘﺎﺩﺭ ﻣﺮﺯﻱ( ﻭ ﻣﺴﺎﺋﻞ ﮔﺮﻭﻩ ﺩﻭﻡ ﺭﺍ ‪ Initional Value Problems‬ﻣﻴﻨﺎﻣﻴﻢ.ﺣﺎﻟـﺖ‬
          ‫ﺗﺮﻛﻴﱯ ﺍﺯ ﻣﻘﺎﺩﻳﺮ ﻣﺮﺯﻱ ﻭ ﺍﻭﻟﻴﻪ ﻧﻴﺰ ﻭﺟﻮﺩ ﺩﺍﺭﺩ. ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ،ﺧﻄﻲ ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ ﻭﻗـﱵ‬
                                                  ‫ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻭ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻳﺎ ﻣﻘﺎﺩﻳﺮﻣﺮﺯﻱ ﺁﻥ ﺧﻄﻲ ﺑﺎﺷﺪ.‬
          ‫ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ‪ PDE‬ﳘﮕﻦ ﻳﺎ ‪ Homogenus‬ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ ﻭﻗﱵ ﻣﻌﺎﺩﻟﻪ ﻭ ﺷﺮﺍﻳﻂ ﺁﻥ ﳘﮕﻦ ﺑﺎﺷﻨﺪ‬
          ‫ﺑﺮﺍﻱ ﺭﻭﺵ ﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺩﺭﺱ ﺍﺭﺍﺋﻪ ﻣﻲ ﺷﻮﺩ ،ﺧﻄﻲ ﻭ ﳘﮕﻦ ﺑﻮﺩﻥ ﻳﻚ ﻣﻌﺎﺩﻟـﻪ , ﺩﺭ ﻭﺟـﻮﺩ‬
          ‫ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﻧﻘﺶ ﺍﺳﺎﺳﻲ ﺩﺍﺭﺩ.ﺑﺮﺍﻱ ﺗﻮﺿﻴﺢ ﺍﻳﻨﻜﻪ ﺑﻪ ﭼﻪ ﺩﻟﻴﻞ ﺑﻪ ﺣﻞ ﻣﻌﺎﺩﻻﺕ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋـﻲ‬
          ‫ﻧﻴﺎﺯﻣﻨﺪﱘ . ﻣﺜﺎﳍﺎﺋﻲ ﺭﺍ ﺩﺭ ﻓﻴﺰﻳﻚ ﻣﻄﺮﺡ ﻣﻲ ﻛﻨﻴﻢ ﻛﻪ ﺣﻞ ﺩﻗﻴﻖ ﺁ‪‬ﺎ ﺑﻪ ﺣﻞ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ‪ PDE‬ﻧﻴﺎﺯﻣﻨﺪ‬
          ‫ﺍﺳﺖ.ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ ﺑﻴﺸﺘﺮ ﻛﺎﺭﺑﺮﺩ ﺍﻳﻦ ﺩﺭﺱ ﺑﺮﺍﻱ ﺣﻞ ﻣﺴﺎﺋﻠﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺯﻳﺮ ﳕﻮﻧﻪ ﻫﺎﺋﻲ ﺍﺯﺁﻥ‬
                                                                                    ‫ﺁﻭﺭﺩﻩ ﺷﺪﻩ ﺍﺳﺖ.‬
                                                                       ‫ﺣﻞ ﻣﺴﺎﹶﻟﻪ ﺳﻴﻢ ﻧﻮﺳﺎﻥ ﻛﻨﻨﺪﻩ :‬
          ‫ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ ﺳﻴﻤﻲ ﺑﻪ ﻃﻮﻝ ‪ L‬ﺩﺭ ﺩﻭ ﻧﻘﻄﻪ ‪ Fix‬ﺷﺪﻩ ﺑﺎﺷﺪ .ﺳﻴﻢ ﺭﺍ ﺑﺎﺭﻳﻚ ﻭ ﺩﺍﺭﺍﻱ ﭼﮕﺎﱄ ﺟﺮﻡ‬
                                                                      ‫ﻃﻮﱄ ﻳﻜﻨﻮﺍﺧﺖ ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ.‬
                                                     ‫) (‬
                                             ‫ﺳﻴﻢ ﺩﺍﺭﺍﻱ ﺣﺮﻛﺖ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ‪ y‬ﺑﺎ ﺳﺮﻋﺖ ‪ v x+∆x.t‬ﺍﺳﺖ .‬

                                                                  ‫ﺍﺯ ﺁﳒﺎ ﻛﻪ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ‪ x‬ﺣﺮﻛﺖ ﻧﺪﺍﺭﱘ ﺑﻨﺎﺑﺮﺍﻳﻦ:‬
               ‫‪H (v, t ) = H ( x + ∆x, t ) → H ( x, t ) = cte‬‬      ‫‪∀t‬‬    ‫‪0< x<l‬‬
                        ‫) ‪v ( x + ∆x , t‬‬                 ‫) ‪− v ( x, t‬‬
              ‫= ‪tan α‬‬                          ‫= ‪tan β‬‬
                              ‫‪H‬‬                            ‫‪−H‬‬



‫٢‬
       ‫) ‪tan α = y x ( x + ∆x, t‬‬                          ‫) ‪tan β = y x ( x,−t‬‬
                           ‫ﺑﺎﻗﺮﺍﺭ ﺩﺍﺩﻥ ﻣﻌﺎﺩﻟﻪ ﻧﲑﻭ ‪ ∆F = ma‬ﺑﺮﺍﻱ ﻳﻚ ﺍﳌﺎﻥ ﻃﻮﻝ ﺳﻴﻢ ﻭ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ‪y‬ﺩﺍﺭﱘ‬
        ‫) ‪⇒ v( x + ∆x, t ) − v( x, t ) = δ .∆s. ytt ( x, t‬‬
              ‫ﻛﻤﺎﻥ ﺍﳚﺎﺩ ﺷﺪﻩ ﺗﻮﺳﻂ ﺍﳌﺎﻥ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﺪﻩ ﻣﻲ ﺑﺎﺷﺪ ﻭ ﺩﺭ ﺻﻮﺭﰐ ﻛﻪ‬                    ‫‪∆s‬‬   ‫ﻛﻪ ﺩﺭ ﺍﻥ‬
                                     ‫) ‪v( x + ∆x, t ) − v( x, t‬‬
               ‫⇒ ‪∆x = ∆s‬‬                                        ‫) ‪= δ ytt ( x, t‬‬
                                              ‫‪∆x‬‬
             ‫⎞ ) ‪⎛ y ( x + ∆x, t ) − y x ( x, t‬‬
            ‫‪H⎜ x‬‬                              ‫) ‪⎟ = δ y tt ( x, t‬‬
             ‫⎝‬           ‫‪∆x‬‬                   ‫⎠‬

        ‫‪Hy XX ( x, t ) = δ y tt (x, t ) ⇒ H‬‬                                                      ‫ﺭﺍﺑﻄﻪ )٤(:‬
                                                      ‫) ‪δ y xx ( x, t ) = y tt (x, t‬‬
                          ‫ﻣﻘﺎﺩﻳﺮ ﺛﺎﺑﱵ ﻫﺴﺘﻨﺪ ‪δ‬ﺟﺮﻡ ﻭﺍﺣﺪ ﻃﻮﻝ ﺳﻴﻢ ﻭ ‪H‬ﻛﺸﺶ ﺳﻴﻢ ﻣﻲ ﺑﺎﺷﺪ.‬
                                                             ‫:‬                                      ‫‪δ ,H‬‬


                                          ‫ﻧﺸﺎﻥ ﺩﺍﺩﱘ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺣﺎﻛﻢ ﺑﺮ ﺳﻴﻢ ﻧﻮﺳﺎﻥ ﻛﻨﻨﺪﻩ ﺑﻪ ﻓﺮﻡ:‬
         ‫‪ytt ( x, t ) = H‬‬
                     ‫) ‪δ y xx ( x, t‬‬
    ‫ﻛﻪ ﺩﺭ ﺁﻥ )‪ y(x,t‬ﺍﳓﺮﺍﻑ ﺳﻴﻢ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ‪ y‬ﻭ ﺩﺭ ﻧﻘﻄﻪ ‪ x‬ﻭ ﺯﻣﺎﻥ ‪ t‬ﺑﺎﺷﺪ ﭼﻨﺎﳒﻪ ﻧﲑﻭﻳﻲ ﺑﺮ ﺳﻴﻢ‬
                                                          ‫ﻭﺍﺭﺩ ﳕﺎﺋﻴﻢ ﻣﻌﺎﺩﻟﻪ ﻧﲑﻭ ﺑﺼﻮﺭﺕ ﺭﺍﺑﻄﻪ )٥( ﺧﻮﺍﻫﺪ ﺑﻮﺩ:‬
     ‫‪δ ∆xy tt ( x, t ) = H ( y x ( x + ∆x ), t ) − y x ( x, t ) + F∆x‬‬                         ‫ﺭﺍﺑﻄﻪ )٥(:‬
                                                          ‫ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﺍﺯ ﺟﺎﻳﮕﺰﻳﻦ ﻛﺮﺩﻥ ﺭﺍﺑﻄﻪ )٣(ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ.‬


                      ‫‪⎛ y ( x + ∆x, t ) − y x ( x, t ) ⎞ f‬‬                                           ‫ﺑﻨﺎﺑﺮﺍﻳﻦ:‬
     ‫‪ytt ( x, t ) = H ⎜ x‬‬                              ‫‪⎟+ δ‬‬
                     ‫⎝‪δ‬‬           ‫‪∆x‬‬                   ‫⎠‬



                                                                                                     ‫ﺑﻨﺎﺑﺮﺍﻳﻦ:‬
       ‫‪y tt (x, y ) = H‬‬       ‫‪y XX ( x, t ) + f‬‬
                          ‫‪δ‬‬                       ‫‪δ‬‬



                                                                     ‫ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﱳ ﻣﺴﺎﹶﻟﻪ ﻭﺯﻥ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ:‬
      ‫‪δ ∆xy tt ( x, t ) = H ( y x ( x + ∆x, t ) − y x ( x, t )) − g ∆x‬‬

     ‫‪y tt ( x, y ) = H‬‬
                         ‫‪δ y xx ( x, t ) − g‬‬
    ‫ﺑﺮﺍﻱ ﺣﺎﻟﱵ ﻛﻪ ﺳﻴﻢ ﺩﺭ ﳏﻴﻂ ﺩﺍﺭﺍﻱ ﺍﺻﻄﻜﺎﻙ ﺑﺎﺷﺪ; ﻣﻲ ﺩﺍﻧﻴﻢ ﻛﻪ ﻧﲑﻭﻱ ﺍﺻﻄﻜﺎﻙ ﺍﺟﺴﺎﻡ ﻣﺘﺤـﺮﻙ‬
                      ‫ﺩﺭ ﻫﻮﺍ، ﮔﺎﺯﻫﺎ ﻭ ﺳﻴﺎﻻﺕ ﺑﺎ ﺳﺮﻋﺖ ﺣﺮﻛﺖ ﻫﻢ ﻣﺘﻨﺎﺳﺐ ﺍﺳﺖ.ﺭﺍﺑﻄﻪ)٦(‬



‫٣‬
                               ‫) ‪f == −by t ( x, t‬‬                ‫‪: b‬ﺿﺮﻳﺐ ﺍﺻﻄﻜﺎﻙ‬           ‫ﺭﺍﺑﻄﻪ)٦(:‬
                                           ‫) ‪y tt ( x, t ) = H y xx ( x, t ) − by t ( x, t‬‬    ‫ﺑﻨﺎﺑﺮﺍﻳﻦ:‬
                                                            ‫‪δ‬‬
    ‫ﻣﻌﺎﺩﻻﺕ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺑﺪﻭﻥ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻳﺎ ﻣﺮﺯﻱ ﺑﻪ ﺻﻮﺭﺕ ﻛﺎﻣﻞ ﻗﺎﺑﻞ ﺣﻞ ﻧﻴﺴﺘﻨﺪ ﺑﻨﺎﺑﺮﺍﻳﻦ ﻭﻗـﱵ‬
                                     ‫ﻛﺎﻣﻞ ﺍﺳﺖ ﻛﻪ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻳﺎ ﻣﺮﺯﻱ ﺁﻥ ﻧﻴﺰ ﻣﺸﺨﺺ ﺑﺎﺷﺪ.‬
                                                                  ‫ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺳﻴﻢ ﻧﻮﺳﺎﻥ ﻛﻨﻨﺪﻩ:‬
                         ‫ﺣﺎﻟﱵ ﻛﻪ ﺳﻴﻢ ﺩﺍﺭﺍﻱ ﺍﳓﺮﺍﻑ ﺍﻭﻟﻴﻪ ﺍﺳﺖ ﻭ ﲢﺖ ﺍﳓﺮﺍﻑ ﺍﻭﻟﻴﻪ ﺭﻫﺎ ﻣﻲ ﺷﻮﺩ.‬
              ‫0 = ) ‪y (0, t‬‬
              ‫0 = ) ‪y (l , t‬‬
     ‫‪y ( x ,0 ) = f ( x ) 0 < x < l‬‬
                                                                                         ‫ﺁﻧﮕﺎﻩ ﺩﺍﺭﱘ:‬
                                                                ‫ﻛﻪ ﺩﺭ ﺁﻥ )‪ f(x‬ﻣﻨﺤﲏ ﺍﳓﺮﺍﻑ ﺍﻭﻟﻴﻪ ﺍﺳﺖ.‬



                                   ‫ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺳﺮﻱ ﺩﻭﻡ:ﻭﻗﱵ ﻛﻪ ﺳﻴﻢ ﲢﺖ ﻳﻚ ﺿﺮﺑﻪ ﺩﺍﺭﺍﻱ ﺳﺮﻋﺖ ﺍﻭﻟﻴﻪ‬
                                                                                    ‫ﻣﻲ ﺷﻮﺩ:‬
     ‫0 = ) ‪y (0, t‬‬
     ‫0 = ) ‪y (l , t‬‬
     ‫0 = ) 0,‪y ( x‬‬
     ‫) ‪y t ( x,0 ) = g ( x‬‬

                                                     ‫ﻛﻪ ﺩﺭ ﺍﻥ )‪ g(x‬ﺳﺮﻋﺖ ﻧﻘﻄﻪ ‪ x‬ﺩﺭ ﳊﻈﻪ ﺻﻔﺮ ﻣﻲ ﺑﺎﺷﺪ.‬

                          ‫ﻣﺪﻟﺴﺎﺯﻳﻬﺎﻱ ﻣﺮﺑﻮﻁ ﺑﻪ ﻣﺴﺎﻳﻞ ﻧﻮﺳﺎﻥ ﺑﺎ ﻳﻜﻲ ﺍﺯ ﺩﻭ ﺻﻮﺭﺕ ﻓﻮﻕ ﺍﳒﺎﻡ ﻣﻲ ﺷﻮﺩ.‬
                                                                      ‫ﻣﺴﺎﹶﻟﻪ ﻣﻴﻠﻪ ﻧﻮﺳﺎﻥ ﻛﻨﻨﺪﻩ‬
    ‫ﻳﻚ ﻣﻴﻠﻪ ﻗﺎﺑﻞ ﺍﺭﲡﺎﻉ ﻛﻪ ﲢﺖ ﻳﻚ ﻧﲑﻭ ﻗﺮﺍﺭ ﺩﺍﺭﺩ ﺩﺭ ﻧﻈﺮ ﺑﮕﲑﻳﺪ.)‪y(x,t‬ﺭﺍ ﺗﺎﺑﻌﻲ ﺩﺭ ﻧﻈﺮﺑﮕﲑﻳﺪﻛﻪ‬
    ‫ﻣﻴﺰﺍﻥ ﺟﺎﲜﺎﻳﻲ ﻧﻘﻄﻪ ‪ X‬ﺭﺍ ﺩﺭ ﺯﻣﺎﻥ ‪ t‬ﻧﺴﺒﺖ ﺑﻪ ﻣﻜﺎﻥ ﺍﻭﻟﻴﻪ ﳏﺎﺳﺒﻪ ﻣﻲ ﻛﻨﺪ.ﻣﻴﺰﺍﻥ ﺟﺎﲜﺎﻳﻲ‬
    ‫‪ X + ∆X‬ﻣﻲ ﺑﺎﺷﺪ.ﻧﲑﻭﻱ ﻭﺍﺭﺩﻩ ﺑﺮ ﻣﻴﻠﻪ‬            ‫ﺑﺮﺍﺑﺮ ﺑﺎ ﻣﻜﺎﻥ ﺍﻭﻟﻴﻪ ﻧﻘﻄﻪ‬      ‫ﻧﻘﻄﻪ ‪X + ∆X‬‬
                               ‫)‪ F(X,t‬ﻣﺪ ﻧﻈﺮ ﻣﻲ ﮔﲑﱘ ﻛﻪ ﻧﲑﻭﻱ ﻭﺍﺭﺩ ﺑﺮ ﻭﺍﺣﺪ ﻃﻮﻝ ﻣﻲ ﺑﺎﺷﺪ.‬




‫٤‬
    ‫ﺍﺯ ﺭﻭﻱ ﻗﺎﻧﻮﻥ ﻫﻮﻙ ﻭ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺭﻭﺍﺑﻂ ﺗﻨﺶ ﻣﻴﺰﺍﻥ ﺗﻐﻴﲑ ﻃﻮﻝ ﺑﺎ ﺭﺍﺑﻄﻪ )‪(v‬ﻣﺸﺨﺺ ﻣـﻲ ﺷـﻮﺩ‬
      ‫) ‪F = y ( x + ∆x ) − y ( x, t‬‬                                    ‫ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ:‬
                                                                                       ‫‪: A‬ﺳﻄﺢ ﻣﻘﻄﻊ ﻣﻴﻠﻪ‬
      ‫= ) ‪F ( x, t‬‬
                     ‫‪E‬‬
                        ‫)) ‪( y(x + ∆x, t ) − y(x, t‬‬                      ‫‪:B‬ﻣﺪﻭﻝ ﺍﻻﺳﺘﻴﺴﻴﺘﻪ)ﻳﺎﻧﮓ(ﺭﺍﺑﻄﻪ )‪(v‬‬
                     ‫‪∆x‬‬
      ‫) ‪F ( x, t ) = AEy x ( x, t‬‬
                                 ‫ﻣﻘﺪﺍﺭ ﻧﲑﻭ ﺑﻪ ﺻﻮﺭﰐ ﺍﺳﺖ ﻛﻪ ﻣﻴﻠﻪ ﺍﺯ ﺣﺎﻟﺖ ﺍﻻﺳﺘﻴﻚ ﺧﺎﺭﺝ ﻧﺸﻮﺩ.‬
                                                                                                     ‫ﭘﺲ:‬
      ‫) ‪F ( x, t ) = − Ey x ( x, t‬‬                                                        ‫ﺑﺮﺍﻱ ﻧﻘﻄﻪ ‪x‬ﺩﺍﺭﱘ‬
      ‫) ‪F ( x + ∆x, t ) = AEy x ( x + ∆x.t‬‬
                                      ‫ﻣﻌﺎﺩﻟﻪ ﻧﲑﻭ ﺑﺮﺍﻱ ﻳﻚ ﺍﳌﺎﻥ ﻃﻮﱄ ﺍﺯﻣﻴﻠﻪ ﺑﻪ ﺻﻮﺭﺕ ﺭﺍﺑﻄﻪ )٨( ﺍﺳﺖ:‬
      ‫) ‪F ( x + ∆x, t ) − F ( x, t ) = δA∆xy tt ( x, t‬‬
                                                                       ‫‪:A‬ﺳﻄﺢ ﻣﻘﻄﻊ ﺭﺍﺑﻄﻪ )٨(‬
                            ‫⎞ ) ‪AE ⎛ y ( x + ∆x, t ) − y ( x, t‬‬                                    ‫∆ :ﻃﻮﻝ‬
      ‫= ) ‪⇒ y tt ( x, t‬‬        ‫⎜‬                              ‫⎟‬
                            ‫⎝ ‪δA‬‬           ‫‪∆x‬‬                 ‫⎠‬
                                                                                         ‫‪δ‬ﺟﺮﻡ ﻭﺍﺣﺪ ﻃﻮﻝ‬
          ‫= ) ‪y tt ( x, t‬‬       ‫) ‪y xx ( x, t‬‬
                            ‫‪E‬‬
                                                                         ‫ﺭﺍﺑﻄﻪ )٩(:ﻣﻌﺎﺩﻟﻪ ﻣﻴﻠﻪ ﻧﻮﺳﺎﻥ ﻛﻨﻨﺪﻩ‬
                            ‫‪δ‬‬


      ‫‪Z tt ( x, y , t ) = H‬‬       ‫(‬
                                ‫) ‪δ Z xx ( x, y, t ) + Z yy ( x, y, t‬‬
                                                                     ‫)‬
                                                         ‫ﺑﺮﺍﻱ ﻳﻚ ﺻﻔﺤﻪ ﻧﻮﺳﺎﻥ ﻛﻨﻨﺪﻩ ﺩﺍﺭﱘ:‬
                                        ‫ﻛﻪ ﺩﺭ ﺁﻥ ‪ Ztt‬ﻣﻴﺰﺍﻥ ﺍﳓﺮﺍﻑ ﻧﻘﻄﻪ ‪x‬ﻭ‪y‬ﺩﺭ ﳊﻈﻪ ‪ t‬ﻣﻴﺒﺎﺷﺪ.‬
                                                                       ‫ﻣﺴﺎﹶﻟﻪ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ:‬
    ‫ﺑﺮﺍﻱ ﻳﻚ ﺣﺠﻢ ‪ u(x,y,z,t) R‬ﺭﺍ ﺩﺭ ﺩﻣﺎﻱ ﻧﻘﻄﻪ ‪ x,y‬ﺩﺭ ﳊﻈﻪ ‪ t‬ﺩﺭ ﻧﻈﺮﺑﮕﲑﻳﺪ.ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺑـﺎ‬
    ‫ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻗﻮﺍﻧﲔ ﺗﺮﻣﻮ ﺩﻳﻨﺎﻣﻴﻚ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺑﺮﺍﻱ ﻫﺮ ﻧﻘﻄﻪ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯﺯﻣﺎﻥ ﺑﺪﺳﺖ‬
                                                                                      ‫ﺍﻭﺭﱘ.‬
    ‫ﳘﺎﻧﻄﻮﺭ ﻛﻪ ﻣﻴﺪﺍﻧﻴﻢ ﺷﺎﺭ ﺣﺮﺍﺭﰐ ﻛﻪ ﺍﺯ ﻳﻚ ﺍﳌﺎﻥ ﺻﻔﺤﻪ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﻣﺸﺨﺺ ﻣـﻲ‬
      ‫‪φ = −K‬‬
              ‫‪du‬‬                 ‫ﺷﻮﺩ.ﻛﻪ ‪n‬ﺑﺮﺩﺍﺭ ﻋﻤﻮﺩ ﺑﺮ ﺳﻄﺢ ﺍﳌﺎﻥ ﺻﻔﺤﻪ ﻣﻮﺭﺩ ﻧﻈﺮ ﻣﻲ ﺑﺎﺷﺪ‬
              ‫‪dn‬‬
                                     ‫‪:n‬ﺑﺮﺩﺍﺭ ﻋﻤﻮﺩ ﺑﺮ ﺳﻄﺢ‬        ‫‪:k‬ﺭﺳﺎﻧﺎﻳﻲ‬         ‫‪: φ‬ﺷﺎﺭ‬
           ‫ﻣﺜﻼ ﺣﺮﺍﺭﺕ ﺍﻧﺘﻘﺎﱄ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﳏﻮﺭﻫﺎﻱ ‪x‬ﻭ‪y‬ﻭ‪z‬ﺑﻪ ﺻﻮﺭﺕ ﺭﻭﺍﺑﻂ )٠١( ﺑﺪﺳﺖ ﻣﻲ ﺍﻳﺪ :‬      ‫ﹰ‬
                  ‫‪du‬‬
      ‫‪φ1 = − K‬‬
                  ‫‪dx‬‬
‫٥‬
                                                         ‫ﻭ ﺣﺮﺍﺭﺕ ﺍﻧﺘﻘﺎﱄ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﳏﻮﺭ ‪ :y‬ﺭﻭﺍﺑﻂ )٠١(‬
                ‫‪du‬‬
    ‫‪φ 2 = −K‬‬
                ‫‪dy‬‬
                ‫‪du‬‬
    ‫‪φ3 = −K‬‬
                ‫‪dt‬‬
                                              ‫ﺟﺮﻳﺎﻥ ﺣﺮﺍﺭﰐ ‪ y‬ﺭﺍ ﺑﻪ ﺑﻪ ﺻﻮﺭﺕ ﲨﻊ ﺷﺎﺭﻫﺎ ﻣﻌﺮﰲ ﻣﻲ ﻛﻨﻴﻢ.‬
                ‫∧‬            ‫∧‬            ‫∧‬
    ‫‪J = φ1 I + φ 2 J + φ 3 K‬‬
            ‫⎞ ∧ ‪⎛ ∂u ∧ ∂u ∧ ∂u‬‬
     ‫+ ‪J = −k ⎜ I‬‬
            ‫‪⎜ ∂x‬‬      ‫+‪J‬‬   ‫⎟‪K‬‬
            ‫⎝‬      ‫‪∂y‬‬    ‫⎟ ‪∂z‬‬‫⎠‬

     ‫) ‪J = −k (∆u‬‬
                                          ‫ﺑﻪ ﺍﻳﻦ ﺻﻮﺭﺕ ‪ φ‬ﺩﺭ ﻫﺮ ﺭﺍﺳﺘﺎ ﺑﻪ ﺻﻮﺭﺕ ‪ φ = J.n‬ﺑﺪﺳﺖ ﻣﻲ ﺍﻳﺪ‬
                                                           ‫.‬      ‫ﻣﺜﺎﻝ :ﺷﺎﺭ ﻋﺒﻮﺭﻱ ﺍﺯﺻﻔﺤﻪ ‪: yz‬‬
                        ‫‪du‬‬
    ‫‪φ=j‬‬         ‫‪j=−k‬‬
                        ‫‪dx‬‬
          ‫ﺍﮔﺮ1 ‪ I‬ﺑﺮﺍﺑﺮ ﻛﻞ ﺗﻐﻴﲑﺍﺕ ﺣﺮﺍﺭﰐ ﺟﺴﻢ ﻣﻮﺭﺩ ﻧﻈﺮ ﺑﮕﲑﱘ، 1 ‪ I‬ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﻣﺸﺨﺺ ﻣﻲ ﺷﻮﺩ:‬
     ‫‪I 1 = ∫ ∫ ∫ R δudv‬‬
                    ‫0‬
                                        ‫ﺷﺎﺭ ﺣﺮﺍﺭﰐ )ﻭ ﰊ ﺣﺮﺍﺭﰐ(‬
         ‫ﺍﺯ ﻃﺮﰲ ﺷﺎﺭﻱ ﻛﻪ ﺍﺯ ﺳﻄﺢ ﻧﺎﺣﻴﻪ ﺧﺎﺭﺝ ﻣﻲ ﺷﻮﺩ ﺑﺎﻋﺚ ﺗﻐﻴﲑ ﺍﻧﺮﮊﻱ ﺣﺮﺍﺭﰐ ﻧﺎﺣﻴﻪ ﻣﻴﮕﺮﺩﺩ.‬
                             ‫ﻛﻞ ﺷﺎﺭ ﺣﺮﺍﺭﰐ ﺧﺎﺭﺝ ﺷﺪﻩ ﺍﺯﻧﺎﺣﻴﻪ ﺍﻧﺘﮕﺮﺍﻝ ﺯﻳﺮ ﻣﺸﺨﺺ ﻣﻲ ﺷﻮﺩ:‬
     ‫‪I1 = ∫ ∫ R J .dA‬‬

     ‫= ‪I 2 = ∫ ∫ ∫ R (divj )dv‬‬      ‫‪∫∫∫ ∇ ( J ) dV‬‬
                                      ‫‪R‬‬                                  ‫ﺑﻨﺎﺑﺮﺍﻳﻦ:)ﺍﺯ ﺭﻳﺎﺿﻴﺎﺕ ﺩﺍﺭﱘ(:‬
     ‫‪I 2 = ∫ ∫ ∫ R − k∇.Jdv = ∫∫∫ − k∇ 2 udV‬‬

                                          ‫ﺍﺯ ﺁﳒﺎﻳﻲ ﻛﻪ ﺍﻳﻦ ﺭﻭﺍﺑﻂ ﺑﺮﺍﻱ ﻫﺮ ﺣﺠﻢ ﺑﺮﻗﺮﺍﺭ ﻣﻲ ﺑﺎﺷﺪ ﺑﻨﺎﺑﺮﺍﻳﻦ‬
     ‫2 ‪I1 = I‬‬                    ‫ﺑﺎﻣﻼﺣﻈﻪ ﻓﻴﺰﻳﻚ ﻣﺴﺎﹶﻟﻪ ﺩﺍﺭﱘ )ﻭﻗﱵ ﺩﺭ ﺩﺍﺧﻞ ﻧﺎﺣﻴﻪ ﻣﻨﺒﻊ ﺣﺮﺍﺭﰐ ﻧﺪﺍﺭﱘ( :‬
     ‫0 = ‪I 1 − I 2 = 0 ⇒ ∫ ∫ ∫ R (δ (u t ) − ku (vu ))dv‬‬
    ‫ﭼﻮﻥ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺑﺮﺍﻱ ﻫﺮ ﻧﺎﺣﻴﻪ ﻱ ‪ R‬ﺑﺮﻗﺮﺍﺭ ﺍﺳﺖ ﻭ ‪ U‬ﻛﻤﻴﱵ ﭘﻴﻮﺳﺘﻪ ﺍﺳﺖ ، ﺑﻨﺎﺑﺮﺍﻳﻦ ﻻﺯﻡ ﺍﺳﺖ .‬
     ‫‪δ u t = ku (v u ) = 0 ⇒ ∆ U‬‬                     ‫‪t‬‬   ‫‪= k∇ 2u‬‬

                                   ‫∧ ‪∂u ∧ ∂u ∧ ∂u‬‬
     ‫= ‪∇ u (x , y , t , z ) u‬‬
                        ‫‪ν‬‬             ‫+ ‪i‬‬    ‫+ ‪j‬‬    ‫‪k‬‬
                                   ‫‪∂x‬‬     ‫‪∂y‬‬     ‫‪∂z‬‬



‫٦‬
                 ‫∧‬          ‫∧‬             ‫∧‬
      ‫‪u1 = M 1 i + M 2 j + M 3 k‬‬
                                                             ‫3 ‪∂M 1 ∂M 2 ∂M‬‬
              ‫⎛⎞∧ ∂ ∧ ∂ ∧ ∂ ⎛‬          ‫∧‬       ‫∧‬       ‫∧‬
                                                         ‫= ⎞‬     ‫+‬    ‫+‬
      ‫‪UM‬‬      ‫⎟ ‪⎜ ∂x ∂y j + ∂z k ⎟.⎜ M 1 i + M 2 j + M 3 k‬‬
            ‫+‪= ⎜ i‬‬             ‫⎝⎟‬                             ‫‪∂x‬‬   ‫‪∂y‬‬   ‫‪∂z‬‬
              ‫⎝‬                ‫⎠‬                         ‫⎠‬


                                                                       ‫ﺑﺎ ﺗﺮﻛﻴﺐ ﻣﻮﺍﺭﺩ ﻓﻮﻕ ‪µ = Uu‬ﺩﺍﺭﱘ:‬
        ‫‪⎛ ∂ 2u ∂ 2u ∂ 2u‬‬             ‫⎞‬
       ‫+ 2 ⎜‪K‬‬
        ‫‪⎜ ∂x‬‬       ‫+‬                 ‫) ‪⎟ = δ u t (x , y , z , t‬‬
                                     ‫⎟‬
        ‫⎝‬      ‫2 ‪∂y 2 ∂z‬‬             ‫⎠‬                                                           ‫ﭘﺲ:‬
      ‫‪∂ 2 u (x, y , z , t ) ∂ 2 (x, y , z , t ) ∂ 2 u (x, t , z , t ) ∂ ∂ u‬‬
                           ‫+‬                   ‫+‬                     ‫=‬      ‫) ‪(x , y , z , t‬‬
             ‫2 ‪∂x‬‬                ‫2 ‪∂y‬‬                  ‫2 ‪∂z‬‬            ‫‪k ∂t‬‬

                                                                                   ‫ﻣﻌﺎﺩﻟﻪ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ‬
                                ‫‪δ‬‬
      ‫= ‪u xx + u yy + u zz‬‬           ‫‪ut‬‬
                                 ‫‪k‬‬

    ‫ﻫﺪﻑ ﺍﺯ ﲝﺚ ﻫﺎﻱ ﺍﺭﺍﺋﻪ ﺷﺪﻩ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻧﺸﺎﻥ ﺩﻫﻴﻢ ﻣﺴﺎﺋﻞ ﻭ ﺭﻓﺘﺎﺭﻫﺎﻳﻲ ﺩﺭ ﻓﻴﺰﻳﻚ ﻭ ﻭﺍﻗﻌﻴـﺖ‬
    ‫ﻭﺟﻮﺩ ﺩﺍﺭﻧﺪ ﻛﻪ ﻣﻲ ﺗﻮﺍﻥ ﺁ‪‬ﺎ ﺭﺍ ﺑﻪ ﻓﺮﻡ ﻣﻌﺎﺩﻻﺕ ‪ PDE‬ﳕﺎﻳﺶ ﺩﺍﺩﻭ ﺣﻞ ﺁ‪‬ﺎ ﺍﺣﺘﻴﺎﺝ ﺑﻪ ﺭﻭﺷﻬﺎﻱ ‪PDE‬‬
                                                                                                ‫دارد ‪E‬‬
    ‫‪P D E‬‬      ‫ﺩﺭﻣﺴﺎﹰﻟﻪ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ ﺩﻳﺪﱘ ﻛﻪ ﺭﺍﺑﻄﻪ ﻛﻠﻲ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ ﺩﺭ ﻫﺮ ﻧﺎﺣﻴﻪ ﺑﻮﺳﻴﻠﻪ ﺭﺍﺑﻄﻪ‬
                 ‫∂‬                    ‫∂‬      ‫∂‬       ‫∂‬                  ‫ﺯﻳﺮ ﻣﺸﺨﺺ ﻣﻲ ﺷﺪ.‬
      ‫= ‪? ∇ 2u‬‬        ‫‪ut‬‬        ‫2‬
                                 ‫= ∇‬             ‫+‬          ‫+‬
                  ‫‪k‬‬                       ‫2 ‪∂x‬‬       ‫2 ‪∂y‬‬       ‫2 ‪∂x‬‬

                                                  ‫ﺑﺮﺭﺳﻲ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ :‬
                                ‫ﭼﻨﺎﻧﭽﻪ ﺍﺯ ﻳﻚ ﻣﺮﺯ ﻧﺎﺣﻴﻪ ﺷﺎﺭ ﺛﺎﺑﺖ ﺣﺮﺍﺭﰐ ﻭﺍﺭﺩ ﺷﻮﺩ، ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ:‬
          ‫‪du‬‬
      ‫‪k‬‬      ‫0‪= φ‬‬
          ‫‪dn‬‬
       ‫‪du‬‬
          ‫0 =‬                   ‫ﺍﮔﺮ ﻣﺰ ﻋﺎﻳﻖ ﺑﺎﺷﺪ ﺩﺍﺭﺍﻱ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺍﺳﺖ ﻛﻪ:‬
       ‫‪dx‬‬
                                                       ‫ﺍﮔﺮ ﻋﺎﻳﻖ ﺑﺎﺷﺪ ﺷﺮﻁ ﻣﺮﺯﻱ ﺻﻔﺮ ﺍﺳﺖ‬
               ‫ﭼﻨﺎﻧﭽﻪ ﻣﺮﺯ ﺑﻪ ﻳﻚ ﺩﻣﺎﻱ ﺛﺎﺑﺖ ﻣﺘﺼﻞ ﺑﺎﺷﺪ ﺁﻧﮕﺎﻩ ﺍﻧﺘﻘﺎﻝ ﺭﺍﺭﺕ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﻣﻲ ﺑﺎﺷﺪ:‬

                      ‫ﺍﮔﺮ ﳏﻴﻂ ﺭﺳﺎﻧﺎ ﺑﺎﺷﺪ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ ﺑﻪ ﺻﻮﺭﺕ ﺭﺳﺎﻧﺎﻳﻲ ﻣﻲ ﺑﺎﺷﺪ ﻭ ﺩﺍﺭﱘ‬
                                                        ‫) ‪= h (T 0 − u‬‬
                                                     ‫‪du‬‬
                                                 ‫‪k‬‬
                                                     ‫‪dn‬‬
‫٧‬
    ‫-ﻳﻚ ﻭﺭﻕ ﻓﻠﺰﻱ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﲑﱘ .ﻣﻄﻠﻮﺏ ﺍﺳﺖ ﻣﻌﺎﺩﻻﺕ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﰐ ﺭﻭﻱ ﻫﺮ‬
                                                                ‫ﻧﻘﻄﻪ ﺍﺯ ﻭﺭﻕ:‬
                  ‫‪δ ∂u‬‬             ‫‪∂ 2u‬‬   ‫‪∂ 2 u δ ∂u‬‬
      ‫= ‪∇ u‬‬
         ‫2‬
                         ‫⇒‬              ‫+‬      ‫=‬
                  ‫‪k ∂t‬‬             ‫2 ‪∂x‬‬   ‫2 ‪∂y‬‬   ‫‪k ∂t‬‬

                                                                      ‫ﻣﻌﺎﺩﻟﻪ ﺑﺮﺍﻱ ﻫﺮ ﻧﻘﻄﻪ ﺍﺯ ﻧﻮﺍﺭ ﻓﻠﺰﻱ‬

      ‫ﺣﺎﻝ ﺑﺎﻳﺪ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺭﺍ ﺍﻋﻤﺎﻝ ﻛﻨﻴﻢ.ﺑﺮﺍﻱ ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﻧﻴﺎﺯ ﺑﻪ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺩﺍﺭﱘ.)ﺭﻭﻱ ﻭﺭﻕ(‬
                                         ‫0 = ) ‪u x (x , o , t‬‬
                          ‫‪du‬‬
             ‫‪ k‬ﻣﺮﺯ١ )١‬         ‫‪=o‬‬                              ‫‪0 < x <l‬‬
                                   ‫‪dx‬‬
                    ‫,‪o< x<l‬‬        ‫0=‪y‬‬


                                                    ‫0 ‪Ku x (o , y , t ) = φ‬‬
                                   ‫‪du‬‬
             ‫)٢‬     ‫ﻣﺮﺯ٢‬       ‫‪k‬‬      ‫0‪= φ‬‬                                          ‫0>‪y‬‬
                                   ‫‪dx‬‬
                      ‫,0 = ‪x‬‬   ‫0>‪y‬‬


                                      ‫) ‪= h (o, u‬‬      ‫) ‪KUx (0, y , t ) = − h (u‬‬       ‫)‪y > o‬‬
                                   ‫‪du‬‬
             ‫ﻣﺮﺯ ٣ )٣‬          ‫‪k‬‬
                                   ‫‪dx‬‬
                   ‫,‪x = e‬‬   ‫0>‪y‬‬


                                  ‫ﺩﻣﺎﻱ ﺍﻭﻟﻴﻪ ﺻﻔﺤﻪ ﺭﺍ ﻧﻴﺰ ﺑﺼﻮﺭﺕ )‪ u(x,y,0)=f(x‬ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﲑﱘ‬
                                                              ‫ﻣﻌﺎىﻠﻪ ﻧﻮﺳﺎﻥ 2 ‪∇ 2 u = ∂ 2 u ∂t‬‬
    ‫ﻣﻌﺎﺩﻻﺕ ﺑﺮ ﻓﺮﻡ ) ‪ ∇ 2 u = δ (u, x, y, z, t‬ﺭﺍ ﻣﻌﺎﺩﻻﺕ ﻻﭘﻼﺱ ﻣﻲ ﮔﻮﺋﻴﻢ ﻭ ‪u‬ﺗﺎﺑﻊ ﲢﺖ ﻻﭘﻼﺳـﲔ ﺭﺍ‬
                                                                           ‫ﺗﺎﺑﻊ ﻫﺎﺭﻣﻮﻧﻴﻚ‬
             ‫ﻣﻲ ﻧﺎﻣﻨﺪ.ﻛﺎﺭﺑﺮﺩ ﺍﻳﻦ ﻣﻌﺎﺩﻻﺕ ﺩﺭ ﻣﺒﺎﺣﺚ ﺳﻴﺎﻻﺕ ، ﺗﺮﻣﻮﺩﻳﻨﺎﻣﻴﻚ ، ﺍﻟﻜﺘﺮﻭ ﻣﻐﻨﺎﻃﻴﺲ ،ﻣﻮﺝ ،‬
                                                   ‫ﺁﻛﻮﺳﺘﻴﻚ ، ﲞﻮﻡ ،ﺍﺭﺗﻌﺎﺷﺎﺕ ، ﺑﻴﻮﻟﻮﮊﻱ ﻭ ..‬
        ‫ﺍﮔﺮ 1‪ u 2 , u‬ﺩﻭ ﺗﺎﺑﻊ ﻭ 1‪c 2 , c‬ﺩﻭ ﺿﺮﻳﺐ ﺩﺭ ‪ R‬ﺑﺎﺷﻨﺪ ، ﺁﻧﮕﺎﻩ 2 ‪c1u1 − c 2 u‬ﺭﺍ ﺗﺮﻛﻴﺐ ﺧﻄﻲ‬
                                                                             ‫1‪ u‬و 2‪ u‬ﻣﻲ ﻧﺎﻣﻴﻢ.‬
      ‫} 2 ‪e{c1u1 + c2 u 2 } = c1e{u1 } + c2 e{u‬‬               ‫-‪ e‬ﺭﺍ ﻳﻚ ﺍﭘﺮﺍﺗﻮﺭ ﺧﻄﻲ ﻣﻲ ﻧﺎﻣﻴﻢ ﻭﻗﱵ‬
                                                        ‫ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﺍﻧﺘﮕﺮﺍﻝ ﮔﲑ ﻳﻚ ﺍﭘﺮﺍﺗﻮﺭ ﺧﻄﻲ ﺍﺳﺖ:‬
      ‫= ‪e{u} = ∫ udt → e{c1u1 + c 2 u 2 } = ∫ (c1u1 + c 2 u 2 )du‬‬


‫٨‬
     ‫‪∫ c1u1 du + ∫ c 2 u 2 du = c1 ∫ u1 du + c 2 ∫ u 2 du‬‬
                                                                        ‫ﺧﻮﺍﺹ ﺍﭘﺮﺍﺗﻮﺭ ﺧﻄﻲ:‬
         ‫0 = }0{‪e‬‬

      ‫‪⎧ n‬‬                 ‫⎫‬
                            ‫{‪= ∑ c k e‬‬              ‫}‬
                                ‫‪n‬‬
     ‫‪e⎨ ∑ cku‬‬         ‫‪k‬‬   ‫⎬‬           ‫‪u‬‬         ‫‪k‬‬
      ‫1= ‪⎩ k‬‬              ‫⎭‬   ‫1= ‪k‬‬


                                             ‫ﺍﮔﺮ ‪ M‬ﻧﻴﺰ ﻳﻚ ﺍﭘﺮﺍﺗﻮﺭ ﺧﻄﻲ ﺑﺎﺷﺪ، ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ:‬
     ‫}‪e{M {u}} = M {e{u}} = eM {u‬‬
     ‫}‪(e + M ){u} = e{u}+ M {u‬‬
                                                                         ‫ﹰ‬
                                            ‫ﳘﺎﻧﻄﻮﺭ ﻛﻪ ﻗﺒﻼ ﲝﺚ ﺷﺪ، ﻣﻌﺎﺩﻻﺕ ﻣﻮﺭﺩ ﺗﻮﺟﻪ ﻓﺮﻡ:‬
      ‫‪∂ 2u‬‬  ‫‪∂ 2u‬‬   ‫‪∂ 2u‬‬ ‫‪∂u‬‬  ‫‪∂u‬‬
     ‫‪A 2 +B‬‬      ‫‪+c 2 +D +E‬‬    ‫) ‪+ Fu ( x, y ) = G ( x, y‬‬
      ‫‪∂x‬‬    ‫‪∂xy‬‬    ‫‪∂y‬‬   ‫‪∂x‬‬  ‫‪∂y‬‬                                                   ‫ﻣﻲ ﺑﺎﺷﺪ.‬
     ‫‪e‬‬   ‫0=‪ G‬ﺍﺳﺖ ، ﻣﺒﺎﱐ ﺣﻞ ﻣﻌﺎﺩﻻﺕ ﻣﻲ ﺑﺎﺷﺪ.ﺣﺎﻝ ﺍﭘﺮﺍﺗﻮﺭ‬     ‫ﺍﺯ ﺑﲔ ،ﺍﻳﻦ ﻣﻌﺎﺩﻻﺕ ﳘﮕﻦ ﻛﻪ ﺩﺭ ﺁﻥ‬
                                                              ‫ﺭﺍ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ:‬
             ‫2∂‬    ‫2∂‬    ‫‪∂ 2u‬‬  ‫∂‬   ‫∂‬
         ‫‪e= A 2 +B‬‬     ‫‪+C 2 + D + E + F‬‬
             ‫‪∂x‬‬    ‫‪∂xy‬‬   ‫‪∂y‬‬    ‫‪∂x‬‬  ‫‪∂y‬‬

                                                    ‫ﺣﺎﻝ ﺑﻴﺎﻥ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ ﺑﺎ ﺍﭘﺮﺍﺗﻮﺭ ‪ e‬ﺑﺼﻮﺭﺕ‬
         ‫0 = }‪e{u‬‬           ‫ﻣﻲ ﺑﺎﺷﺪ.)‪ u‬ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻭﺍﻟﺘﺎﮊ، ﺩﻣﺎ،ﻣﻜﺎﻥ ،ﭘﺘﺎﻧﺴﻴﻞ،ﻓﺸﺎﺭ ﻳﻚ ﺳﻴﺎﻝ (‬


      ‫ﺣﺎﻝ ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻏﲑ ﳘﮕﻦ ﺑﻪ ﺻﻮﺭﺕ ‪ e{u} = G‬ﺧﻮﺍﻫﺪ ﺑﻮﺩﻭ ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ‬
     ‫0 = } 1‪e{u‬‬    ‫0 = } 2 ‪e{u‬‬                 ‫ﺩﺍﺩ ﭼﻨﺎﻧﭽﻪ 2 ‪u1 , u‬ﺟﻮﺍ‪‬ﺎ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ ﺑﺎﺷﺪ:ﻳﻌﲏ‬
                                ‫ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ 0 = } 2‪c1e{u1} + c2 e{u2 } = e{c1u1 + c2u‬‬
    ‫ﻧﺘﻴﺠﻪ:ﺍﮔﺮ 2 ‪ u1 , u‬ﺟﻮﺍ‪‬ﺎﻱ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ ﺑﺎﺷﻨﺪ،ﻫﺮ ﺗﺮﻛﻴﺐ ﺧﻄﻲ ﺁ‪‬ﺎ ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ ﺍﺳﺖ.ﻛﺎ‬
                               ‫-ﰲ ﺁﻥ ﺩﺭ ﻳﻚ ﺿﺮﺏ ﺷﻮﺩ.ﰊ ‪‬ﺎﻳﺖ ﺟﻮﺍﺏ ﺑﺪﺳﺖ ﺧﻮﺍﻫﺪ ﺁﻣﺪ.‬
                    ‫ﺍﮔﺮ 1‪ u‬ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ ﺑﺎﺷﺪ ﻭ 2 ‪ u‬ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﻧﺎﳘﮕﻦ ، ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ.‬
     ‫0 = } 1‪e{u‬‬        ‫‪e{u 2 } = G‬‬
                ‫ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﻫﺮ ﺗﺎﺑﻊ ‪ u‬ﻛﻪ 2 ‪ u = cu1 + u‬ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﻏﲑ ﳘﮕﻦ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.‬
             ‫‪e{cu1 + u 2 } = ce{u1 } + e{u 2 } = G‬‬
    ‫) ‪u t ( x , t ) = Ku xx ( x , t‬‬                ‫ﻣﺜﺎﻝ:ﻣﺴﺎﹰﻟﻪ ﺳﻴﻢ ﻣﺮﺗﻌﺶ ﺯﻳﺮ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﺑﮕﲑﻳﺪ‬


‫٩‬
     ‫0 = ) ‪u x (o, t‬‬                                                 ‫ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺍﻳﻦ ﺍﺳﺖ:‬
     ‫0 = ) ‪u x (c, t‬‬


                                                           ‫ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ ﺑﻪ ﻓﺮﻡ‬
                                                ‫⎞ ‪⎛ n 2π 2 k‬‬      ‫‪nπx‬‬
     ‫1 = ) ‪u 0 ( x, t‬‬         ‫− ⎜‪U n ( x, t ) = exp‬‬
                                                ‫⎜‬         ‫‪t ⎟ cos‬‬
                                                            ‫⎟‬                   ‫......,2,1 = ‪n‬‬
                        ‫2‬                            ‫2‪c‬‬
                                                ‫⎝‬           ‫⎠‬      ‫‪c‬‬

                      ‫⎞ ‪⎛ − n 2π 2 k‬‬    ‫⎞ ‪⎛ − n 2π 2 k‬‬      ‫‪nπx‬‬
       ‫⎜ = ) ‪u t ( x, t‬‬            ‫⎜‪⎟ exp‬‬           ‫‪t ⎟ cos‬‬
                      ‫⎟ 2‪⎜ c‬‬
                      ‫⎝‬            ‫⎠‬
                                        ‫2‪⎜ c‬‬
                                        ‫⎝‬
                                                      ‫⎟‬
                                                      ‫⎠‬      ‫‪c‬‬              ‫‪t‬‬   ‫ﻣﺸﺘﻖ ﺗﺎﺑﻊ ﻧﺴﺒﺖ ﺑﻪ‬

               ‫2 ‪⎛ − n 2π‬‬   ‫⎞‬     ‫⎞ 2 ‪⎛ − n 2π‬‬  ‫‪nπx‬‬
       ‫‪u xx‬‬   ‫⎜=‬
               ‫2‪⎜ c‬‬         ‫⎜ ‪⎟ exp‬‬
                            ‫⎟‬     ‫‪⎜ c 2 kt ⎟ cos c‬‬
                                             ‫⎟‬                               ‫ﺗﻔﺎﻭﺗﺸﺎﻥ ﺩﺭ ‪ k‬ﺍﺳﺖ‬
               ‫⎝‬            ‫⎠‬     ‫⎝‬          ‫⎠‬

                                                                          ‫ﺳﭙﺲ ) ‪ku xx = u ( x, t‬‬
       ‫‪u = c o u o + c1u1 + ..... + c n u n‬‬
                                                                           ‫ﻣﻲ ﺗﻮﺍﻥ ﻧﺘﻴﺠﻪ ﮔﺮﻓﺖ ﻛﻪ‬
                                                                   ‫ﻧﻴﺰ ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﻓﻮﻕ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.‬
        ‫‪Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G‬‬                                ‫ﻣﻌﺎﺩﻻﺕ ﺑﻪ ﻓﺮﻡ‬
                                          ‫ﺧﻄﻲ ﺩﺭﺟﻪ ٢ ﻫﺴﺘﻨﺪ ﻭﻗﱵ ‪ A‬ﺗﺎ ‪ F‬ﺗﺎﺑﻌﻲ ﺍﺯ ‪ x , y , t‬ﺑﺎﺷﻨﺪ‬
        ‫ﭼﻨﺎﻧﭽﻪ ﺿﺮﺍﻳﺐ ‪ F-A‬ﻋﺪﺩ ﺛﺎﺑﱵ ﺑﺎﺷﻨﺪ،ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﻣﻌﺎﺩﻟﻪ ﺧﻄﻲ ﺩﺭﺟﻪ ٢ﺑﺎ ﺿﺮﻳﺐ ﺛﺎﺑﺖ ﻣﻲ ﻧﺎﻣﻴﻢ.‬
     ‫ﺣﺎﻝ ﺍﮔﺮ 0 > ‪B 2 − 4 Ac‬ﻣﻌﺎﺩﻟﻪ ﻫﺬﻟﻮﱄ ﻳﺎ ‪ H yperbolic‬ﮔﻮﺋﻴﻢ ﻭ ﺍﮔﺮ 0 = ‪ B 2 − 4 Ac‬ﺑﺎﺷﺪ ﻣﻌﺎﺩﻟﻪ‬
               ‫ﺭﺍ ﺑﻴﻀﻮﻱ ﻳﺎ ‪ Ecllipite‬ﮔﻮﻳﻨﺪ ﻭﺍﮔﺮ 0 > ‪B 2 − 4 Ac‬ﺑﺎﺷﺪ ﺳﻬﻤﻲ ﮔﻮﺋﻴﻢ.)‪.(Parabolic‬‬
                                                                       ‫ﺍﻧﻮﺍﻉ ﻣﻌﺎﺩﻻﺕ ﻻﭘﻼﺱ:‬
         ‫ﻣﺴﺎﹶﻟﻪ ﻻﭘﻼﺱ ﺭﺍ ﻣﺴﺎﹶﻟﻪ ﺩﻳﺮﻳﺸﻠﻪ ﻳﺎ ﻧﻮﻉ ﺍﻭﻝ ﮔﻮﻳﻨﺪ ﺍﮔﺮ ﺩﺭ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻣﻔﺪﺍﺭ ‪ u‬ﺭﻭﻱ ﻣﺮﺯ ﺩﺍﻧﺴﻪ‬
                                                                                     ‫ﺷﻮﺩ.‬
      ‫ﺭﻭﻱ ﲞﺸﻲ ﺍﺯ ﻣﺮﺯ ﺩﺍﺩﻩ ﺷﺪﻩ ﺑﺎﺷﺪ.‬       ‫ﻣﺴﹶﻟﻪ ﻻﭘﻼﺱ ﻧﻮﻉ ﺭﺍ ﻧﻴﻮﻣﻦ ﮔﻮﻳﻨﺪﺍﮔﺮ ﻣﻘﺪﺍﺭ ﻣﺸﺘﻖ ‪u‬‬‫ﺎ‬
               ‫ﻣﺴﺎﹰﻟﻪ ﻱ ﻻﭘﻼﺱ ﺭﺍ ﻧﻮﻉ ﺳﻮﻡ ﮔﻮﻳﻨﺪ ﺍﮔﺮ ﺭﻭﻱ ﻣﺮﺯ ﻣﻘﺪﺍﺭ ‪ Ku + du‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺑﺎﺷﺪ.‬
          ‫ﻣﺴﺎﹶﻟﻪ ﻻﭘﻼﺱ ﺭﺍ ﻧﻮﻉ ﭼﻬﺎﺭﻡ )ﻛﻮﺷﻲ( ﮔﻮﻳﻨﺪ ﺍﮔﺮ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺷﺎﻣﻞ ﺗﺮﻣﻬﺎﻱ ‪ u tt , u t‬ﺑﺎﺷﺪ ﻭ‬
                                      ‫‪dn‬‬

                                                ‫ﻣﻘﺪﺍﺭ ‪ u‬ﻳﺎ ‪ u t‬ﺩﺭ ﳊﻈﻪ ﺻﻔﺮ ﺩﺍﻧﺴﺘﻪ ﺷﺪﻩ ﺑﺎﺷﺪ.‬
                                                                                   ‫ﺑﻴﺎﻥ ﳐﺘﺼﺎ‪‬ﺎ:‬
                                                                            ‫ﻛﺎﺭﺗﺮﻳﻦ ﺑﻪ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ‬
       ‫‪⎧ x = p cos φ‬‬                 ‫2‪⎧ p = x2 + y‬‬
       ‫⎪‬                             ‫⎪‬
                                     ‫⎪‬
       ‫‪⎨ y = p sin φ‬‬                              ‫⎞ ‪⎛y‬‬
                                     ‫⎟ ‪⎨φ = Arc tan ⎜ x‬‬
       ‫‪⎪ z=z‬‬                         ‫⎪‬            ‫⎝‬   ‫⎠‬
       ‫⎩‬                             ‫⎪‬      ‫‪z= z‬‬
‫٠١‬                                   ‫⎩‬
                                                                            ‫ﻛﺮﻭﻱ‬         ‫ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ‬


     r=       p2 + z
                                θ = Arc tan⎛ p z ⎞
                                           ⎜     ⎟
                                           ⎝      ⎠

      φ =φ

     ⎧ x = rSin θ Cos φ
     ⎪
     ⎨ y = rSin θ Sin φ                                                        ‫ﺩﻛﺎﺭﰐ‬        ‫ﻛﺮﻭﻱ‬
     ⎪ z = rCos θ
     ⎩

     ⎧ p = rSin θ
     ⎪
     ⎨ φ =φ                                                                 ‫ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ‬      ‫ﻛﺮﻭﻱ‬
     ⎪ z = rCos θ
     ⎩

     r = x2 + y2 + z2

     φ = ArcTan y x                                                            ‫ﻛﺮﻭﻱ‬        ‫ﻛﺎﺭﺗﺰﻳﻦ‬
               ⎛ x2 + y2 ⎞
     z = ArcTan⎜         ⎟
               ⎜    z    ⎟
               ⎝         ⎠



                     ∂ 2u ∂ 2u ∂ 2u                    : ‫ﺑﻴﺎﻥ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺩﺭ ﳐﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ‬
     ∇ u=0
         2
                         +    +     =0
                     ∂x 2 ∂y 2 ∂z 2
                                               .‫ ﺧﻮﺍﻫﺪ ﺑﻮﺩ‬u ( p, φ , z ) ‫ ﺑﺼﻮﺭﺕ‬u ‫ﺩﺭ ﺍﻳﻨﺼﻮﺭﺕ ﺗﺎﺑﻊ‬
                                                                                           :‫ﺣﻞ‬
     ∂u ∂u ∂p   ∂u ∂φ   ∂u ∂z
       =  .   +   .   +   .
     ∂x ∂p ∂x ∂φ ∂x     ∂z ∂x

         ∂u      x           ∂u ⎛ − y ⎞ ∂u x Sinφ ∂u                ∂u Sinφ ∂u
     =
         ∂p
                         +       ⎜ x + y 2 ⎟ = ∂p . p − p ∂φ = Cosφ ∂p − p ∂φ ∗
                                .⎜ 2
                             ∂φ ⎝          ⎟
               x2 + y2                     ⎠
                                                            :‫ ﺩﺭ ﻋﺒﺎﺭﺕ * ﻗﺮﺍﺭ ﻣﻲ ﺩﻫﻴﻢ‬u ‫ﺑﻪ ﺟﺎﻱ‬
     ∂ 2 u ∂ ⎛ ∂u ⎞
          = ⎜ ⎟
     ∂x 2 ∂x ⎝ ∂x ⎠


١١
     ∂ 2u         ⎛ ∂ ⎛ ∂u ⎞ ⎞ Sin φ ⎛ ∂ ⎛ ∂u ⎞ ⎞
          = Cos φ ⎜ ⎜ ⎟ ⎟ −
                  ⎜ ∂p ∂x ⎟          ⎜   ⎜ ⎟⎟
     ∂x           ⎝ ⎝ ⎠⎠         p ⎜ ∂φ ⎝ ∂x ⎠ ⎟
                                     ⎝          ⎠

     ∂ 2 `u      ⎛ ∂ ⎛    ∂u Sinφ ∂u ⎞ ⎞ Sinφ ⎛ ∂    ∂u Sinφ ∂u ⎞
            = Cos⎜ ⎜ Cocφ
                      ⎜      −       ⎟⎟ −
                                     ⎟⎟       ⎜ Cosφ
                                              ⎜ ∂φ      −       ⎟
     ∂x   2      ⎜ ∂p
                 ⎝    ⎝   ∂p   p ∂φ ⎠ ⎠    p ⎝       ∂φ   p ∂φ ⎟⎠

      ∂ 2u       ⎛       ∂ 2 u Sin φ ∂u Sin ℑ ∂ 2 u ⎞
                 ⎜ Cos φ
     ⇒ 2 = Cos φ ⎜            +        −            ⎟−
      ∂x         ⎝       ∂p 2   p 2 ∂φ    p ∂φ∂p ⎟  ⎠
     Sinφ     ⎛        ∂u         ∂ 2 u Cosφ ∂u Sinφ ∂ 2 u ⎞
              ⎜ − Sinφ
              ⎜           + Cosφ       −       −           ⎟
       p      ⎝        ∂p        ∂u∂φ    p ∂φ     p ∂φ 2 ⎟ ⎠

                           ∂ 2 u CosφSinφ ∂u CosφSinφ ∂ u u Sin 2φ ∂u
                   = Cos φ    2
                                +            −             +          −
                           ∂p       p2    ∂φ    p     ∂φ∂p    p ∂p


                   Sin φCos φ ∂ 2 u   Sin φCos φ ∂u Sin 2 φ ∂ 2 u
                                    +               +             ⇒
                        p     ∂φ ∂p       p2     ∂φ   p ∂φ 2

          ∂ 2 p 2SinφCosφ ∂u 2CosφSinφ ∂ 2u Sin2φ ∂u Sin2φ ∂ 2 u
     Cos φ 2 +
          2
                             −              +       + 2
          ∂p        p2    ∂φ     p     ∂φ∂p   p ∂p    p ∂φ 2

                                                               ∂ 2u
                                                                      :‫ﳏﺎﺳﺒﻪ ﻱ‬
                                                               ∂y 2
     ∂u        ∂u Cosφ ∂u  ∂ 2u          ⎛ ∂u ⎞ Cosφ ∂ ⎛ ∂u ⎞
        = Sinφ    +       ⇒ 2 = Sinφ ∂ ⎜ ⎟ +           ⎜ ⎟
     ∂y        ∂p   p ∂φ   ∂y         ∂p ⎜ ∂y ⎟
                                         ⎝ ⎠     p ∂φ ⎜ ∂y ⎟
                                                       ⎝ ⎠

     ∂ 2u      2 ∂ u
                  2
                      2 SinφCosφ ∂ 2 u Cos 2φ ∂ 2 u Cos 2φ ∂u
          = Sin φ 2 +                 +            +          :‫ﺎﻳﺖ‬ ‫ﺩﺭ‬
     ∂y 2        ∂p        p     ∂p∂φ   p 2 ∂φ 2      p ∂p
                              2 sin φCosφ ∂u
                        −
                                   p2     ∂φ

     ∂ 2 u ∂ 2 ( x, y , z )
          =                       ∇ 2u = 0
     ∂z 2       ∂z 2

                   ∂ 2u ∂ 2u ∂ 2u
     ∇ 2u = 0 →        +    +     =0⇒
                   ∂x 2 ∂y 2 ∂z 2

     ∂ 2 u 1 ∂u    1 2 ∂ u + ∂ u =0
                        2     2
          +      +
     ∂p 2   p ∂p    p ∂φ 2 ∂z 2

١٢
                                                                         ‫ﺑﻴﺎﻥ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺩﺭ ﳐﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ:‬

         ‫‪∂ 2u‬‬    ‫‪∂u ∂ 2 p 2 ∂ 2 u‬‬
       ‫‪P‬‬   ‫2‬
              ‫‪+p‬‬   ‫+‬     ‫‪p‬‬        ‫0=‬
         ‫2 ‪∂p‬‬    ‫2 ‪∂p ∂φ‬‬    ‫2 ‪∂z‬‬

                                                                                ‫ﺍﮔﺮ ﻣﺴﺎﹶﻟﻪ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ‪ Z‬ﺛﺎﺑﺖ ﺑﺎﺷﺪ:‬
         ‫‪∂ u‬‬    ‫2‬
                 ‫‪∂u ∂ p 2 ∂ u‬‬                    ‫2‬       ‫2‬
                                                                              ‫‪∂ 2u‬‬
       ‫‪P‬‬   ‫2‬
              ‫‪+p‬‬   ‫+‬     ‫‪p‬‬      ‫0=‬                                          ‫‪p‬‬      ‫0=‬
         ‫‪∂p‬‬ ‫2‬
                 ‫2 ‪∂p ∂φ‬‬   ‫2 ‪∂z‬‬                                               ‫2 ‪∂z‬‬

                                                             ‫ﻣﻌﺎﺩﻟﻪ ﺯﻳﺮ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺩﺭﳐﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺍﺳﺖ:‬
                ‫‪∂u‬‬      ‫‪∂u ∂ u‬‬               ‫2‬                     ‫∞<‪o<x‬‬
       ‫2‪x‬‬            ‫‪+x‬‬   ‫+‬     ‫0=‬
               ‫‪∂x‬‬ ‫+2‬
                        ‫2 ‪∂x ∂y‬‬                                    ‫‪0 < y < 2π‬‬

                         ‫ﲤﺮﻳﻦ:ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺭﺍ ﺑﻪ ﻣﻌﺎﺩﻟﻪ ﻛﺮﻭﻱ ﺗﺒﺪﻳﻞ ﳕﺎﺋﻴﺪ.)ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺩﺭ ﻓﺮﻡ ﻛﺮﻭﻱ(‬

                             ‫‪∂ 2 u 2 ∂u‬‬    ‫1‬   ‫‪∂ 2u‬‬   ‫‪1 ∂ 2 u Cosθ ∂u‬‬                                   ‫ﺩﺭ ‪‬ﺎﻳﺖ:‬
       ‫⇒ 0 = ‪∇ 2u‬‬                 ‫+‬     ‫2 2 +‬       ‫2 +‬      ‫2 +‬      ‫0=‬
                             ‫2 ‪∂r 2 r ∂r r Sin θ ∂φ 2 r ∂φ‬‬     ‫‪r‬‬   ‫‪∂θ‬‬

                                                                                                ‫ﺭﻭﺵ ﺩﻳﮕﺮ ﳕﺎﻳﺶ:‬
                                             ‫1‬             ‫1‬
       ‫1=‪∇ u‬‬
           ‫2‬
                         ‫‪(ru )rr‬‬         ‫2 +‬      ‫2 + ‪u φφ‬‬      ‫‪(uθSin θ )θ‬‬
                     ‫‪r‬‬                    ‫‪r Sin θ‬‬
                                               ‫2‬
                                                        ‫‪r Sin θ‬‬

       ‫‪(u + ru r ) → (u r‬‬                    ‫) ‪+ u r + ru rr‬‬

                                                                                                ‫ﺭﻭﺵ ﺩﻳﮕﺮ ﳕﺎﻳﺶ:‬
       ‫= ‪∇ 2u‬‬
                    ‫2 1‬
                         ‫(‬
                       ‫‪r ur‬‬          ‫)‬   ‫+‬
                                                ‫1‬
                                                     ‫2 + ‪uφφ‬‬
                                                             ‫1‬
                                                                 ‫‪(uθ Sinθ )θ‬‬
                    ‫2‪r‬‬                       ‫‪r Sin θ‬‬
                                                  ‫2‬  ‫2‬
                                                          ‫‪r Sinθ‬‬
                                     ‫‪r‬‬
                                                                                      ‫ﺍﮔﺮ ﻣﺴﺎﹶﻟﻪ ﻣﺴﺘﻘﻞ ﺍﺯ ‪φ‬ﺑﺎﺷﺪ:‬

       ‫‪r‬‬
           ‫2 1‬
                ‫(‬
              ‫‪r ur‬‬       ‫)‬       ‫+‬
                                        ‫1‬
                                            ‫‪(uθ Sinθ )θ = ∇ 2 u‬‬
           ‫2‪r‬‬                        ‫‪r Sinθ‬‬
                                      ‫2‬
                             ‫‪r‬‬




                                                                           ‫ﻓﺼﻞ ﺩﻭﻡ:ﺣﻞ ﻣﻌﺎﺩﻟﻪ ‪∇ 2 u = G‬‬
     ‫ﺭﻭﺵ ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﺑﻪ ﻭﺳﻴﻠﻪ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ ﻭ ﺑﺪﺳﺖ ﺁﻭﺭﺩﻥ ﻳﻚ ﺟﻮﺍﺏ ﺍﺧﺘﺼﺎﺻﻲ ﺑﺮﺍﻱ ﻣﺴﺎﻟﻪ ﻣـﻲ‬
     ‫ﺑﺎﺷﺪ.ﻳﻚ ﺭﻭﺵ ﻛﺎﺭﺑﺮﺩﻱ ﺑﺮﺍﻱ ﺣﻞ ﻣﻌﺎﺩﻻﺕ ﻻﭘﻼﺱ ،ﺭﻭﺵ ﺟﺪﺍﺳﺎﺯﻱ ﻣﺘﻐﲑﻫﺎ ﻣﻴﺒﺎﺷﺪ. ‪(Sepration‬‬
                                                                                                     ‫)‪of Vauiable‬‬



‫٣١‬
                                                          ‫ﺑﺮﻱ ﺗﻮﺿﻴﺢ ﺍﻳﻦ ﺭﻭﺵ ﺍﺯ ﻳﻚ ﻣﺜﺎﻝ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﻛﻨﻴﻢ:‬
       ‫) ‪Ku xx ( x, t ) = u t ( x, t‬‬                ‫0> ‪0< x<c t‬‬



                                                       ‫) ‪⎧ Ku xx ( x, t ) = u t ( x, t‬‬
                                                       ‫⎪‬     ‫0 = ) ‪u x (0, t‬‬
                                                       ‫⎪‬
                ‫)‪u ( x,0) = f ( x‬‬                      ‫⎨‬
                                                             ‫0 = ) ‪u x ( c, t‬‬
                                                                                                       ‫ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ‬
                                                       ‫⎪‬
                                                       ‫)‪⎪ u ( x,0) = f ( x‬‬
                                                       ‫⎩‬
                                                                  ‫ﺩﺭ ‪‬ﺎﻳﺖ ﺑﻪ ﻣﺘﻮﺳﻂ ﻳﻌﲏ )‪ f(x‬ﺧﻮﺍﻫﻴﻢ ﺭﺳﻴﺪ.‬
     ‫ﺟﻮﺍﺏ ﻫﺎﻱ ﻣﻌﺎﺩﻟﻪ ﻓﻮﻕ ﺑﺎﺷﻨﺪ ‪u = ∑ c k u k‬ﻧﻴﺰ ﺟﻮﺍﰊ ﺑـﺮﺍﻱ‬                         ‫1 ‪u n ......... u‬‬   ‫ﻗﻀﻴﻪ ١(:ﺍﮔﺮ‬
                                                                                                        ‫ﻣﺴﺎﹶﻟﻪ ﺍﺳﺖ.‬
     ‫ﻗﻀﻴﻪ ٢(:ﺍﮔﺮ ‪ u‬ﭘﻴﻮﺳﺘﻪ ﻭ ﳘﮕﺮﺍﻱ ﻳﻜﻨﻮﺍﺧﺖ ﺑﺎﺷﺪ ﺳﺮﻱ ﺟﻮﺍﺏ ﺑﻪ ﺍﺯﺍﻱ ∞ = ‪ n‬ﳘﮕﺮﺍ ﻭﺍﻫﺪ ﺑﻮﺩ.‬
                                                ‫ﻗﻀﻴﻪ ٣(:ﺟﻮﺍﰊ ﻛﻪ ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ ﻳﻜﺘﺎﺳﺖ.‬
                                                                              ‫) ‪ u ( x, t‬ﺭﺍ ﻣﻲ ﺧﻮﺍﻫﻴﻢ.‬
       ‫) ‪u ( x, t ) = X ( x )T (t‬‬



                                                                                          ‫ﺭﻭﺵ ﺟﺪﺍﺳﺎﺯﻱ ﻣﺘﻐﲑﻫﺎ‬

                                            ‫) ‪KX ′′( x ) T ′(t‬‬
      ‫→ ) ‪KX ′′( x )T (t ) = X ( x )T ′(t‬‬             ‫=‬                              ‫ﺗﺎﺑﻌﻲ ﺍﺯ ﻛﺎﻥ =ﺗﺎﺑﻌﻲ ﺍﺯ ﺯﻣﺎﻥ‬
                                             ‫) ‪X (x‬‬     ‫) ‪T (t‬‬


                                        ‫ﺍﻳﻦ ﺗﺴﺎﻭﻱ ﺩﺭ ﺻﻮﺭﰐ ﺑﺮﻗﺮﺍﺭ ﺍﺳﺖ ﻛﻪ ﻧﺴﺒﺖ ﻛﺴﺮﻫﺎ ﺛﺎﺑﺖ ﺑﺎﺷﺪ .‬
       ‫) ‪X ′′( x ) 1 T ′(t‬‬
                ‫=‬          ‫→‪=λ‬‬                        ‫ﻋﺪﺩ ﺛﺎﺑﺖ‬
       ‫) ‪X ( x ) K T (t‬‬

       ‫) ‪X ′( x‬‬           ‫) ‪T ′(t‬‬
                ‫‪=λ‬‬                ‫‪=λ‬‬
       ‫) ‪X (x‬‬             ‫) ‪KT (t‬‬

       ‫0 = ) ‪X ′′( x ) − λX ( x‬‬                                                                         ‫ﻣﻌﺎﺩﻟﻪ 1‬



       ‫0 = ) ‪T ′(t ) − λKT (t‬‬                                              ‫ﻣﻌﺎﺩﻟﻪ٢‬
     ‫ﺑﺎ ﺣﻞ ﻣﻌﺎﺩﻻﺕ ١ﻭ٢ ﺟﻮﺍﺏ ﺑﺪﺳﺖ ﺧﻮﺍﻫﺪ ﺁﻣﺪ.ﺑﻪ ﻣﻌﺎﺩﻻﺕ ٢،١ ﻣﻌﺎﺩﻻﺕ ﺍﺷﺘﺮﻭﻡ ﻟﻴﻮﻭﻳﻞ ﮔﻮﻳﻨﺪ.‬
                                                               ‫ﳓﻮﻱ ﺍﻋﻤﺎﻝ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ:‬


‫٤١‬
         1       X ′(o )T (t ) = X ′(c )T (t ) = 0

       ⇒ T (t ) ≠ 0, X ′(o ) = o1 X ′(c ) = 0



                                                                        T ′(t )
       ⎧ X ′( x ) − λx( x ) = 0           T ′(t ) − kλt (t ) = 0 ⇒              = kλ ⇒ CosT (t ) = kλt + CosA
       ⎪                                                                T (t )
       ⎨       X (o ) = 0
       ⎪      X ′(c ) = 0                                   T (t )         T (t )
       ⎩                                       ⇒ Cos               = kλt ⇒        = e kλt ⇒ T (t ) = Ae kλt
                                                             A              A


                                                               :‫:ﺳﻪ ﺣﺎﻟﺖ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﲑﱘ‬             ‫ﳓﻮﻱ ﺣﻞ ﻣﻌﺎﺩﻟﻪ‬

       1)λ > 0         X ′′( x ) − λX ( x) = 0 ⇒ X ( x ) = a1e         λx
                                                                            + a2 e −   λx
                                                                                            λ>0


       ‫ﻓﺮض‬        λ = a 2 ⇒ X ′′( x ) = a 2 × ( x ) ⇒ X ( x ) = Ae ax + Be − ax



       X ( x ) = Aae ax − Bae − ax

       ⎧ X ′(o ) = 0 ⇒ A − B = 0
       ⎨                                                     ⇒ A = 0, B = 0
       ⎩ X ′(c ) = 0 ⇒ Ae − Be = 0
                         ac   − ac




                                .‫ ﺟﻮﺍﰊ ﺍﳚﺎﺩ ﳕﻲ ﻛﻨﺪ‬λ > 0 ‫ ﺻﻔﺮ ﺍﺳﺖ ﭘﺲ‬u (x,t) ‫ ﺻﻔﺮ ﺑﺎﺷﺪ‬X(x) ‫ﺍﮔﺮ‬
      2) λ = 0        X ′′( x ) = 0 ⇒ X ( x ) = Ax + B

                        X ′(o ) = 0 ⇒ A = 0

                        X ′(c ) = 0 ⇒ A = 0

                                                                                                      ‫ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ‬
       X (x ) = B            T (t ) = A              u ( x, t ) = X ( x )T (t ) − A.B

       u ( x, o ) = f ( x ) ⇒ AB = f ( x )


                                       .‫ ﻭﺍﺿﺢ ﺍﺳﺖ‬u (x, t ) = f ‫ ﺛﺎﺑﺖ ﺑﺎﺷﺪ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ‬f                 (x )   ‫ﭘﺲ ﺍﮔﺮ‬
     3)λ < 0              X ′′( x ) = −a 2 X ( x ) ⇒ X ( x ) = ACosax + BSinax
      λ = −a 2

                          X ′(o ) = 0 ⇒ − AaSinac + aBCosax = 0                               B=0
                                                                                  x=0

١٥
                                                                                                   nπ
        X ′(o ) = 0 ⇒ − AaSinac = 0 ⇒ Sinac = 0 → ac = nπ ⇒ a =
                                                                                                    c
                                            ⎛ nπ ⎞
                                X (x ) = Cos⎜   x⎟                       n = 0,1,......
                                            ⎝ c ⎠
                                                                  ⎛ n 2π 2       ⎞
                                T (t ) = e kλt = e − ka t = e − k ⎜ 2
                                                       2

                                                                  ⎜ c            ⎟t
                                                                                 ⎟          n = 0,1,...
                                                                  ⎝              ⎠



     ‫ﻭ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺍﺵ ﺻﺪﻕ ﻣﻲ ﻛﻨﺪ.ﭘﺲ ﺳﺮﻱ ﺯﻳـﺮ ﻧﻴـﺰ ﺩﺭ‬ku xx = u t ‫ ﻫﺎ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻱ‬u n ‫ﲤﺎﻣﻲ‬
            ∞                               :‫ﻣﻌﺎﺩﻟﻪ ﻱ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺻﺪﻕ ﻣﻲ ﻛﻨﺪ‬
        u = ∑ a n u n ( x, t )
                  n =1
                                                               n 2π 2       nπx
                     u ( x, t ) = X ( x )T (t ) = e − k            2
                                                                      + Cos
                                                                 c           c

                            − kn 2π 2    ⎛ nπx ⎞
           u n ( x, t ) = e      2
                                      Cos⎜     ⎟                            n = 0,1,.....
                               c         ⎝ c ⎠
                                  − n 2π 2
                          ∞
                                                        nπ x
        u ( x, t ) = ∑ a n e
                                             kt
                                    c2
                                                  Cos
                         n =1                            c
                                                                                :‫ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﻳﺪ‬a n ‫ﭘﺲ ﻧﻘﻂ ﻛﺎﰲ ﺿﺮﺍﻳﺐ‬
                                                                  u ( x,0 ) = f ( x ) ‫ﳏﺎﺳﺒﻪ ﺿﺮﺍﻳﺐ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺷﺮﻁ ﻣﺮﺯﻱ‬
                          ∞
                                         nπx
        u ( x,0 ) = ∑ a n Cos                = f (x )
                         n =1             c

                                                                                                             nηx
                                                                      .‫ ﳘﮕﺮﺍ ﺑﺎﺷﺪ‬f (x) ‫ ∑ﺑﺎﻳﺴﱵ ﺑﻪ‬a n Cos c ‫ﺳﺮﻱ‬
                                                                               :‫ ﺑﻪ ﻣﻮﺍﺭﺩ ﺯﻳﺮ ﺗﻮﺟﻪ ﺑﺸﻮﺩ‬a n ‫ﺑﺮﺍﻱ ﳏﺎﺳﺒﻪ‬
        c
                     nπx       c     nπ x                          ⎧o  n = 1.2,........
        ∫ Cos
        0
                      c
                         dx =
                              nπ
                                 Sin
                                      c
                                          =0                       ⎨
                                                                   ⎩ c      n=0

                     nπx     mπx          ⎛             πx              nπ                            ⎞
        c                                                 c

        ∫ Cos            Cos     dx = 1 ∫ ⎜ Cos (n + m ) + Cos (n − m )                               ⎟dx
                      c       c        2 ⎝               c               c                            ⎠
        0                               0




        ⇒         ‫ ﻃﺒﻴﻌﻲ ﺍﻧﺪ‬n,mn,m ‫ ﻓﺮﺽ ﻛﻨﻴﻢ‬n = 1,2,......                            ⎧
                                                                                      ⎪0       ∀m, n m ≠ n
                                                                                      ⎨ c
                                                                 m = 1,2,.....        ⎪         n=0
                                                                                      ⎩   2
       ∞
                         nπx                         nπx           2πx                       nπx
      ∑ a Cos n              = f ( x ) = a 0 + a1Cos     + a 2 Cos     + ......... + a n Cos     = f (x )
      n =0                c                           c             c                         c


١٦
                                                :‫ ﺑﮕﲑﱘ .ﺩﺭﺍﻳﻨﺼﻮﺭﺕ‬c ‫ﻛﺎﰲ ﺍﺯ ﺩﻭ ﻃﺮﻑ ﺍﻧﺘﮕﺮﺍﻝ ﺍﺯ 0 ﺗﺎ‬                                       ‫ﺑﺮﺍﻱ ﳏﺎﺳﺒﻪ‬
         c                    c                                                    c
                                            dx + ....... + ....... = ∫ f ( x )dx
                                         nx
     a 0 ∫ dx + a1 ∫ Cos
         0                    0
                                         c                           0


               c                                               c
     a0 c = ∫      f ( x )dx ⇒ a 0 = 1                             f ( x )dx
                                                          c∫
               0                                               0


                            .‫)ﺿﺮﻳﺐ ﺧﻮﺩﺵ(ﺿﺮﺏ ﻛﺮﺩﻩ ﻭ ﺍﻧﺘﮕﺮﺍﻝ ﻣﻲ ﮔﲑﱘ‬                                               Cos
                                                                                                                      nx
                                                                                                                         ‫ در‬a1         ‫ﺑﺮﺍﻱ ﳏﺎﺳﺒﻪ‬
                                                                                                                      c
        c                                   c
              nx            nx  nx
     a 0 ∫ Cos dx + a1 ∫ Cos Cos dx + ...........
         0
              c        0
                            c   c

                                                      ηx
                      f ( x )dx = ai c = ∫ f ( x )cCos dx
                   nx
     = ∫ Cos
                   c                  2                c

                                                                                                                                   :‫ﺑﻪ ﳘﲔ ﺗﺮﺗﻴﺐ‬
                                            2ηx
                   c
                        f ( x )Cos
              c∫
     a1 = 2                                     dx
                   0
                                             c

                                            nη x
                   c
                 f ( x )Cos
              c∫
     an = 2                                      dx                ∀n = 1,2,........
                   0
                                             c

             Ku xx = u t                                  ‫ﻣﺴﺂَﻟﻪ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ‬                     u x (0, t ) = 0            u x (c, t ) = 0

                                                                                                                 :‫ﺩﻣﺎﻱ ﺍﻭﻟﻴﻪ ﻧﻘﻄﻪ‬
                                                                                                                u ( x ,0 ) = ?

      f ( x ) = 10 x                        c = 10 x                             .‫ ﺗﺎﺑﻌﻲ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺑﺎﺷﺪ‬f ( x ) ‫ﻣﺜﺎﻝ:ﻓﺮﺽ ﻛﻨﻴﺪ ﻛﻪ‬
                                                                                            .‫ﺟﻮﺍﺏ ﻫﺮ ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﻳﺪ‬
                                      n 2π 2
                                  −
                                                          nηx
                        ∞                      kt
                                       c2
     u ( x, t ) = ∑ a n e                           Cos
                       n =0                                c
             ‫ﺣﺠﻢ‬              ‫واﺣﺪ‬                   ‫ﺣﺮارﺗﻲ‬     ‫ﻇﺮﻓﻴﺖ‬
     k=
                               ‫ﺣﺠﻢ‬                  ‫واﺣﺪ‬    ‫ﺟﺮم‬

                                                                                            nπx
                        C                                                 c
     a 0 = 1 ∫ 10 x dx                                                         f ( x )Cos
                                                                      10 ∫
                                                       an = 2                                   dx
            10                                                                               c
                        0                                                 0



                                                                                        nπx
                                                                         10

                                                                      10 ∫
     a 0 = 50‫ﻣﺘﻮﺳﻂ‬                                     an = 2              10 xCos          dx
                                                                          0
                                                                                         c

                         c
                                        nπ            ⎛ u10     nπx                                   10
                                                                                                           10
                                                                                                                   nπx ⎞
      an = 2             ∫ xCos            x dx = 2 ⎜       Sin                              10
                                                                                                  −        ∫ Sin      dx ⎟
                   10                   10         10 ⎜ nπ       c
                                                                                             0
                                                                                                      nπ            c    ⎟
                         0                            ⎝                                                    0             ⎠
      ⎧ ‫زوج‬    an = 0
      ⎪
      ⎨‫ﻓﺮد‬         400
            an = − 2 2
                                                          ‫ﻳﺎ‬          an =
                                                                                200
                                                                                nπ
                                                                                 2 2
                                                                                       (
                                                                                     (− 1)n − 1        )
      ⎪
      ⎩            nπ
١٧
                    ‫2 ⎞ 01 ⎛ ⎛‬     ‫‪nπx‬‬                ‫⎞‬
       ‫2 = ‪an‬‬       ‫⎜‬                            ‫01‬   ‫)1 − ‪⎟ ⇒ a n = 200 (Cosnπ‬‬
                 ‫⎟ ‪10 ⎜ ⎜ nπ‬‬
                               ‫‪Cos‬‬
                                   ‫01‬
                                                 ‫0‬
                                                      ‫⎟‬         ‫2 ‪n 2π‬‬
                    ‫⎝⎝‬     ‫⎠‬                          ‫⎠‬

                                           ‫(‬
                                          ‫‪− n 2π 2 k‬‬
                                                       ‫)‬   ‫‪nπ x‬‬
                               ‫∞‬
                              ‫002‬
      ‫‪u ( x, t ) = 50 + ∑ 2 2 (− 1) − 1 e‬‬
                                   ‫‪n‬‬
                                                     ‫‪t Cos‬‬
                        ‫‪n =1 n π‬‬             ‫01‬            ‫01‬


                                                               ‫)‪f (x‬‬       ‫ﺟﻮﺍﺏ ﻣﺴﺎﹶﻟﻪ ﺍﻧﺘﻘﺎﱄ ﺣﺮﺍﺭﺕ ﺑﺎ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ‬

         ‫ﺭﻭﺵ ‪ Sepration od Voualley‬ﺘﺮﻳﻦ ﺭﻭﺵ ﺑﺮﺍﻱ ﺣﻞ ﻣﻌﺎﺩﻻﺕ ﺑﺎ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﻣﻲ ﺑﺎﺷﺪ.‬
     ‫ﺑﺎ ﺭﻭﺵ ﻓﻮﻕ ﺍﻟﺬﻛﺮ ، ﺟﻮﺍﰊ ﺑﺮﺍﻱ ﻣﻌﺪﻻ ﻻﭘﻼﺱ ﺍﺭﺍﺋﻪ ﳕﻮﺩﱘ .ﺳﻮﺍﱄ ﻛﻪ ﻣﻄﺮﺡ ﺍﺳﺖ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ‬
                                              ‫ﺁﻳﺎ ﺟﻮﺍﺏ ﺑﺪﺳﺖ ﺁﻣﺪﻩ ﻳﻜﺘﺎﺳﺖ ﻳﺎ ﺧﲑ ؟‬
                      ‫ﻓﺼﻞ ٥١ ﻛﺘﺎﺏ ﭼﺮﭼﻴﻞ ،ﺍﺛﺒﺎﺕ ﻳﻜﺘﺎﻳﻲ ﺟﻮﺍﺏ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺑﻪ ﺍﻳﻦ ﺭﻭﺵ ﺍﺳﺖ.‬
                                                                            ‫ﺭﻭﺵ ﻫﺎﻱ ﺩﻳﮕﺮ ﺣﻞ ﻣﻌﺎﺩﻻﺕ ‪:P D E‬‬
                     ‫ﹰ‬
     ‫ﺑﺮﺧﻲ ﺍﺯ ﻣﺴﺎﺋﻞ ‪ PDE‬ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎ ﺭﻭﺵ ﻫﺎﻱ ﻣﺒﺘﻜﺮﺍﻧﻪ ﺣﻞ ﳕﻮﺩ.ﺍﻳﻦ ﺭﻭﺵ ﻫﺎ ﺻﺮﻓﺎ ﺑﺮﺍﻱ ﺣـﺎﻻ‬
                                                                     ‫ﺧﺎﺹ ﺑﺮﻗﺮﺍﺭﻧﺪ.‬
       ‫0 = ) ‪u xx ( x, y‬‬ ‫2 ‪u (0, y ) = y‬‬ ‫1 = ) ‪u (1, y‬‬
            ‫∞ < ‪0 < x <1 − ∞ < y‬‬                                                   ‫ﻣﺜﺎﻝ:‬


       ‫) ‪u xx ( x, y ) = 0 ⇒ u x ( x, y ) = φ ( y‬‬                  ‫‪ ⇒ u( x, y ) = φ ( y )x + h y‬ﻭﱄ ﳑﻜﻦ ﺍﺳﺖ ﺛﺎﺑﺖ ﺑﺎﺷﺪ‬

       ‫2 ‪u (0, y ) = y 2 ⇒ h( y ) = y‬‬                      ‫)0 × ) ‪(φ ( y‬‬

       ‫2 ‪u (1, y ) = φ ( y ) + h( y ) = 1 ⇒ φ ( y ) = 1 − y‬‬

       ‫2 ‪u ( x, y ) = (1 − y 2 )x + y‬‬



       ‫) ‪a 2 y xx ( x, t ) = y tt ( x, t‬‬                                                                      ‫ﺭﻭﺵ ﺩﻳﮕﺮ‬
                                                                                                        ‫ﺑﺮﺍﻱ ﻣﺜﺎﻝ:‬
        ‫) ‪y ( x ,0 ) = h ( x‬‬               ‫∞<‪−∞<x‬‬
          ‫0 = ) ‪y t ( x, o‬‬                     ‫0> ‪t‬‬

       ‫‪⎧u = x + at‬‬
       ‫⎨‬                                                         ‫ﺣﻞ:ﺑﺮﺍﻱ ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﺍﺯ ﺗﺒﺪﻳﻞ ﺯﻳﺮ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﻛﻨﻴﻢ:‬
       ‫‪⎩ v = x − at‬‬



‫٨١‬
                                    ‫( ﻧﻮﺷﺖ‬     x,t)     ‫ﻳﺎ‬   (v,u)   ‫ﻳﻌﲏ ﻣﻐﲑﻫﺎ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺮﺣﺴﺐ‬

     ∂y ∂y ∂u ∂y ∂u
       =  .  + .                           y ( x, t ) → y (u , v )
     ∂t ∂u ∂t ∂v ∂t
     ∂y      ∂u             ∂y    ∂y
        .a +    .(− a ) = a    −a
     ∂u      ∂v             ∂u    ∂v

     ∂2 y    ∂ ⎛ ∂y ⎞ ∂ 2 y  ∂ ⎛ ∂y    ∂y ⎞  ∂ ⎛ ∂y    ∂y ⎞
          . = ⎜ ⎟ ⇒ 2 = a ⎜a        − a ⎟ − a ⎜a     −a ⎟
     ∂t 2
             ∂t ⎝ ∂t ⎠ ∂t   ∂u ⎝ ∂u    ∂v ⎠  ∂v ⎝ ∂u   ∂v ⎠

             ∂2 y     ∂2 y    ∂2 y      ∂2 y     ∂2 y
     ⇒            = a2 2 − a2      − a2      + a2 2
             ∂t 2     ∂u      ∂u∂v      ∂u∂v     ∂u

               ∂2 y         ∂2      ∂2 y
     ⇒ a2           − 2a 2      + a2 2
               ∂u 2        ∂u∂v     ∂v

     ∂ 2 y ∂ ⎛ ∂y ⎞
          = ⎜ ⎟
     ∂x 2 ∂x ⎝ ∂x ⎠

     ∂y ∂y ∂u ∂y ∂v ∂y ∂y
       = .   + . =    +
     ∂x ∂u ∂x ∂v ∂x ∂u ∂v

     ∂ 2 y ∂ ⎛ ∂y ∂y ⎞ ∂ ⎛ ∂y ∂y ⎞ ∂ 2 y ∂2 y ∂2 y
          = ⎜ + ⎟+        ⎜   + ⎟= 2 +2      +
     ∂x 2 ∂x ⎝ ∂x ∂v ⎠ ∂u ⎝ ∂u ∂v ⎠ ∂u   ∂u∂v ∂u 2

                                                                       PDE   ‫ﺑﺎ ﺟﺎﮔﺬﺍﺭﻱ ﺩﺭ ﻣﻌﺎﺩﻟﻪ‬
       ⎛ ∂2 y ∂2 y ∂2 y ⎞  2⎛∂ y
                               2
                                   ∂2 y ∂2 y ⎞
     a ⎜ 2 +
         2
       ⎜ ∂u       +     ⎟=a ⎜ 2 −2     +     ⎟
       ⎝      ∂u∂v ∂u 2 ⎟
                        ⎠
                            ⎜ ∂u
                            ⎝      ∂u∂v ∂u 2 ⎟
                                             ⎠

         ∂2 y     ∂2 y    ⎧ y (u , v ) = h(u )
     4        =0⇒      =0⇒⎨ n
         ∂u∂v     ∂u∂v    ⎩ y v (u , v ) = q (v )


                                                               :‫ﻳﻚ ﺗﺎﺑﻊ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺍﺳﺖ‬

     g (u , v ) = f (u ) + g (v )   f ′(u ) → 0




                                                                             ‫ﭘﺲ ﺟﻮﺍﺏ ﺑﻪ ﺻﻮﺭﺕ‬
     y (u , v ) = f (u ) + g (v )




١٩
                                                                               ‫‪x + at = u‬‬      ‫ﺍﺳﺖ ﻭ ﺟﺎﻳﮕﺬﺍﺭﻱ‬
                                                                               ‫‪x − at = v‬‬

                                                                                            ‫ﺍﺯ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺩﺍﺭﱘ :‬
            ‫) ‪y ( x, t ) = f ( x, at ) + g ( x − at‬‬

            ‫) ‪af ′( x ) − ag ( x ) = 0 ⇒ f ′( x ) = g ′( x‬‬


                                         ‫ﻳﺎ ‪f ( x ) = g ( x ) + k‬‬         ‫‪c‬‬

            ‫1 = ) ‪2 f ( x ) = h( x ) + c ⇒ f ( x‬‬           ‫) ‪(h( x ) + c‬‬
                                                       ‫2‬
            ‫1 = ) ‪2 g ( x ) = h( x ) − c ⇒ g ( x‬‬           ‫) ‪(h(x ) − c‬‬
                                                       ‫2‬


                                                                       ‫ﺳﺎﻳﺮ ﺭﻭﺵ ﻫﺎﻱ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺪﻩ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ :‬
                                                ‫٣(ﺗﺒﺪﻳﻞ ﺍﻧﺘﮕﺮﺍﱄ ﻓﻮﺭﻳﻪ‬                           ‫١(ﺗﺒﺪﻳﻞ ﻻﭘﻼﺱ‬
‫ﭘﺲ‬   ‫)) ‪y ( x, t ) = f ( x, at ) + g ( x − at ) = 1 (h(a + at ) + h( x − at‬‬
                                                ‫2‬
                                                 ‫٤(ﻧﮕﺎﺷﺖ ﻛﺎﻧﺘﻮﺭﻣﺎﻝ‬                               ‫٢(ﺗﺒﺪﻳﻞ ﻓﻮﺭﻳﻪ‬
                                                                            ‫ﻣﺜﺎﻝ:ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺯﻳﺮ ﺭﺍ ﺣﻞ ﻛﻨﻴﺪ؟‬
                ‫) ‪y tt ( x, t ) = a y xx ( x, t‬‬
                                 ‫2‬



            ‫0 = ) ‪y (o, t ) = y (c, t‬‬             ‫1‬

            ‫) ‪y ( x,0) = f ( x‬‬                 ‫0 = )0,‪y t ( x‬‬              ‫3‬


             ‫‪ a‬ﺿﺮﻳﱯ ﺍﺳﺖ ﻛﻪ ﺑﻪ‬               ‫‪, c‬‬   ‫ﳘﺎﻧﻄﻮﺭ ﻛﻪ ﻣﻲ ﺩﺍﻧﻴﻢ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻧﻮﺳﺎﻥ ﺗﺎﻣﺴﺮ ﻧﻘﺶ ﺍﺳﺖ.ﻃﻮﻝ ﺗﺎﺭ‬
                                                                    ‫ﺟﻨﺲ ﺗﺎﺭ ﻭ ﺟﺮﻡ ﻭﺍﺣﺪ ﻃﻮﻝ ﺑﺴﺘﮕﻲ ﺩﺍﺭﺩ.‬
                                                                                 ‫ﺭﻭﺵ ﺣﻞ:ﺟﺪﺍﺳﺎﺭﻱ ﻣﺘﻐﲑﻫﺎ‬
            ‫) ‪y ( x, t ) = X ( x )T (t‬‬
                                                                         ‫2‪⎧ λ‬‬
                                                    ‫⎪ ) ‪X ′′( x ) 1 T ′′(t‬‬
            ‫⇒ ) ‪X ( x )T ′′(t ) = a 2 X ′′( x )T (t‬‬          ‫=‬          ‫0 ⎨=‬
                                                    ‫2 ⎪ ) ‪X ( x ) a 2 T (t‬‬
                                                                         ‫‪⎩− λ‬‬



                          ‫) ‪X ′′( x‬‬
                                    ‫) ‪= λ 2 ⇒ x ′′( X ) = λ 2 X ( X‬‬
                          ‫)‪X (x‬‬
             ‫اﻟﻒ‬
                          ‫) ‪T ′′(t‬‬
                                    ‫) ‪= λ 2 → T ′′(t ) = a 2 λ 2 T (t‬‬
                           ‫) ‪T (t‬‬



‫٠٢‬
       ⎧ X (t ) = Ae λx + Be − λx
       ⎨
       ⎩T (t ) = Ce + De
                   aλt      − aλt



       1 ⇒ X (0 ) = 0 ⇒ A + B = 0                          ⇒ A = 0, B = 0

        X (c ) = 0 ⇒ Ae λc + Be − λt = 0                                ‫غ. ق. ق‬
                 X ′′( x ) = 0 ⎧ X ( x ) = Ax + B
       ‫)ب‬λ = 0 ⇒               ⎨
                 T ′′(t ) = 0 ⎩ T (t ) = Ct + D

                                  ⎧ X (0) = 0 ⇒ B = 0
       ‫ﻣﺮزي‬            ‫ﺷﺮاﻳﻂ‬      ⎨
                                  ⎩ X (c ) = 0 ⇒ A = 0
                        ⎧ X ′′( x ) = −λ 2 X ( x ) ⇒ X ( x ) = B Sinλx + A Cosλx
       ‫) ج‬λ < 0         ⎨
                        ⎩T ′′(t ) = −a λ T (t ) ⇒ T (t ) = C Sinaλ + + D Cosλt
                                       2 2



       1 → X (0 ) = 0 ⇒ A = 0                     nλ
                                         X n ( x ) = Sin       x
                                                           c
                                                                   nπ
       X (c ) = 0 ⇒ BSinλc = 0 ⇒ λc = nπ ⇒ λ =                              ∀ n = 0,1,.....
                                                                    c
                                                                                          :‫ﺎﻳﺖ ﺟﻮﺍﺏ ﺩﺍﺭﱘ‬ ‫ﭘﺲ ﰊ‬

                         anπ            anπ
       Tn (t ) = CSin          + + DCos     +∗
       3:      X ( x )T ′(0c) = 0        c

       ∗ = 0 ⇒ Tn′ (0) = 0 ⇒ C
                                                               ‫ﺑﺎﻳﺪ ﺻﻔﺮﺑﺎﺷﺪ‬
                               anπC
                  Tn(t ) = Cos
                                 T
                                           nπx                                    aπx
       ‫ﺑﻨﺎﺑﺮاﻳﻦ‬                : X n ( x ) = Sin           , Tn (t ) = Cos            t
                                            c                                      c
                                nπx     aπx
       ‫ﭘﺲ‬     Yn ( x, t ) = Sin     Cos     t
                                 c       c
                                       ‫ﭘﺲ‬
                                                           ∞
                                                                  . ‫ﺍﻳﻦ ﺗﺮﻛﻴﺐ ﺧﻄﻲ ﺟﻮﺍﺏ ﺍﺳﺖ‬
                                              y ( x, t ) = ∑ bn Sin
                                                                        nπx
                                                                            Cos
                                                                                aπx
                                                                                    t
         ∞
                    nπx                                                  c       c
                                                                  ‫ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﱘ‬bn ‫ﻓﻘﻂ ﺑﺎﻳﺪ‬
       ∑1 b n Cos c
                                           n =1

        n=
     ‫ ﻻﺯﻡ ﺍﺳﺖ ﻛﻪ ﺿـﺮﺍﻳﺐ ﳏﺎﺳـﺒﻪ ﺷـﻮﻧﺪ.ﺍﺯ ﺷـﺮﻁ ﺍﻭﻟﻴـﻪ‬y(x,t) ‫ﺑﺮﻧﺎﻣﻪ ﺍﻱ ﳏﺎﺳﺒﻪ ﺍﻱ ﺩﻗﻴﻖ‬
      (‫ ﺻﻔﺮ ﻣﻲ ﮔﺰﺍﺭﱘ‬t ‫ ﺍﳓﺮﺍﻑ ﺍﻭﻟﻴﻪ ﺗﺎﺭﺍﺳﺖ.)ﺑﻪ ﺟﺎﻱ‬f(x) ‫ ﺍﺳﻨﻔﺎﺩﻩ ﻣﻲ ﻛﻨﻴﻢ.ﻛﻪ‬g(x,0)=f(x)
                   ∞
                               nπx
      y ( x,0) = ∑ bn Sin          = f (x )
                  n =1          c


             nπx             nπx
                                                                                      (          )     (         )
        c                                      c

        ∫ Sin c dx = − nπ Cos c ∫ = − nπ (Cosnπ − 1) = nπ (−1) + 1 = nπ 1 − (−1)
                        c              c                c     n+1     c         n

        0                        0
٢١
       c
                       nπx     mπx      ⎧0
                                        ⎪   n≠m
       ∫ Sin            c
                           Sin
                                c
                                   dx = ⎨ c
                                        ⎪ 2 n=m
       0                                ⎩
                                   :‫ﻭ ﮔﺮﻓﱳ ﺍﻧﺘﮕﺮﺍﻝ ﺩﺍﺭﱘ‬            Sin
                                                                         nπx   ‫ ∑ ﺩﺭ‬bn Sin nπx = f (x ) ‫ﺑﺎ ﺿﺮﺏ ﺭﺍﺑﻄﻪ‬
                                                                          c                   c
           ∞
                                nπx     m πx                                                      mπx
                   c                                                            c

       ∑∫               b n Sin     Sin      dx =                              ∫    f ( x )Sin        dx
       n =1 o                    c       c                                      0
                                                                                                   c

       c b = f ( x )Sin mπx ⇒ b = 2              mπx
                            c                                       c
                                       f (x )Sin
        2 m ∫                       c∫
                               m                     dx
                         c  0
                                                  c                 0

     ⎧        f ( x ) = 0.2 x             0 < x < 10                                           :‫ ﺗﺎﺑﻊ ﺭﻭﺑﺮﻭ ﺑﺎﺷﺪ‬f(x) ‫ﻣﺜﺎﻝ‬
     ⎨
     ⎩ f ( x ) = −0.2 x + 4                  10 < x < 20
              ⎛    ⎛                                                  nπx ⎞ ⎞
              ⎜ 2 ⎜ − x 20 Cos nπx
                                                               10
                                                            20
                                                                          dx ⎟ ⎟
                                                            nπ ∫
     =2                                              10
                                                          +       Cos
           20 ⎜ 10 ⎜    nπ      20
                                                     0
                                                                       20    ⎟⎟
              ⎝    ⎝                                            0            ⎠⎠


                                       nπ x         ⎡10
                                                 2 ⎢ 2 x Sin nπx dx + − 2 + 4 Sin nπx dx  (            )
                            20                                       20
                            f ( x )Sin
                       20 ∫                       20 ∫ 10            ∫ 20
       bn = 2                               dx =
                          0
                                        20          ⎣o        20     10
                                                                                   20



        ⎛⎛                                                                        nπx ⎞ ⎞ ⎞ ⎞
               (     ⎛ − 20     cπx
                                     )
                                                                      20
                                                                   20 ⎛ − 2
      + ⎜⎜ − 2 x + 4 ⎜      Cos                             20
                                                                 +    ∫  ⎜    Cos    dx ⎟ ⎟ ⎟ ⎟
        ⎜⎜    20     ⎜ nπ        20
                                                            0
                                                                   nπ 10 ⎝ 10      c    ⎠⎟⎟⎟
        ⎝⎝           ⎝                                                                    ⎠⎠⎠

                                nπ x            nπx
                       20                       20
                          −2
      ........ +                         ∫
                       ∫ 20 xSin 20 dx + 10 4Sin 20 dx
                       10



                                   nπx                     nπx ⎞ 20
                       20
                                            4 × 20 ⎛
               20 ∫
      =2                    xSin       dx +        ⎜ − Cos     ⎟ 10 = .........
                       10
                                    20        nπ ⎝          20 ⎠

                                                                                         :‫ﻓﺼﻞ ﺳﻮﻡ ﺳﺮﻱ ﻫﺎﻱ ﻓﻮﺭﻳﻪ‬
                                                                                              :‫ﻣﻌﺮﰲ ﺗﻮﺍﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ‬
       ، ‫ﺑﻪ ﺟﺰ ﭼﻨﺪ ﻧﻘﻄﻪ ﺍﻱ ﳏﺪﻭﺩ ﻗﺎﺑﻞ ﴰﺎﺭﺵ ﭘﻴﻮﺳﺘﻪ ﺑﺎﺷﺪ،ﺑﻪ ﺁﻥ ﺗﺎﺑﻊ‬                               [a,b]‫ﻓﺎﺻﻠﻪ‬   ‫ ﺩﺭ‬f(x) ‫ﺍﮔﺮ ﺗﺎﺑﻊ‬



٢٢
                                                                                 ‫ﺗﺎﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﻣﻲ ﮔﻮﻳﻨﺪ.‬
                                                                             ‫ﻧﻘﺎﻁ ﻧﺎﭘﻴﻮﺳﺘﮕﻲ ‪ x1 , x 2 ,.........x n‬ﻛﻪ‬
       ‫‪a < x1 < x 2 ................... < x n < b‬‬

                                                                              ‫ﻧﻴﺰ ﻻﺯﻡ ﺍﺳﺖ ﺗﺎ ﺣﺪﻭﺩ ﺯﻳﺮ ﻣﻮﺟﻮﺩ ﺑﺎﺷﻨﺪ‬
            ‫) ( ) ( ) () (‬
       ‫‪f x1− f x1+ , f x 2 , f x 2 ,........, f x n , f x n‬‬
                         ‫−‬       ‫+‬                ‫−‬       ‫+‬
                                                                ‫) ( ) (‬
                                                                                                                 ‫ﻣﺜﺎﻝ:‬
                  ‫5.0− < ‪⎧− 1 − 1Mx‬‬
                  ‫⎪‬
        ‫5. 0 < ‪f ( x ) = ⎨ a − 0 .5 < x‬‬
                  ‫1 < ‪⎪ 1 0 .5 ≤ x‬‬
                  ‫⎩‬


     ‫ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳـﻒ ﻣـﻲ‬               ‫) ‪( c, d ) ⊂ ( a , b‬‬   ‫ﺍﻧﺘﮕﺮﺍﻝ ﺗﺎﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺭﺍ ﺩﺭ ﻓﺎﺻﻠﻪ )‪ (c,d‬ﻛﻪ‬
                                                                                                   ‫ﺷﻮﺩ:‬
       ‫‪d‬‬                        ‫‪xM‬‬                    ‫1+ ‪x M‬‬                                  ‫‪d‬‬

       ‫‪∫ f (x )dx = ∫ f (x )dx + ∫ f (x )dx + ......... + ∫ f (x )dx‬‬
        ‫‪c‬‬                         ‫‪c‬‬                    ‫‪xM‬‬                                    ‫1− ‪x r‬‬

                                         ‫ﺗﻮﺍﺑﻊ ﭘﻴﻮﺳﺘﻪ ﺣﺎﻟﺖ ﺧﺎﺻﻲ ﺍﺯ ﺗﻮﺍﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺍﻧﺪ.‬
     ‫ﻓﻮﺭﻳﻪ ﺑﺮﺍﻱ ﺗﻘﺮﻳﺐ ﺗﻮﺍﺑﻊ ﺩﺭ ﻳﻚ ﺑﺎﺯﻩ ﺍﺯ ﺳﺮﻱ ﻫﺎﻱ ﺳﻴﻨﻮﺳﻲ ﻭ ﻛﺴﻴﻨﻮﺳﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ ﻛﻪ ﻣﻲ ﺗﻮﺍﻥ‬
     ‫ﺑﺎ ﺗﻘﺮﻳﺐ ﻣﻨﺎﺳﺐ ﺍﺯ ﻫﺮ ﺩﻗﺖ ﺩﳋﻮﺍﻩ ﻛﻪ ﲞﻮﺍﻫﻴﻢ ﺗـﺎﺑﻊ )‪ f (x‬ﺍﻱ ﺭﺍ ﻛـﻪ ﻗﻄﻌـﻪ ﭘﻴﻮﺳـﺘﻪ‬
     ‫ﺍﺳﺖ،)ﺩﺭ ﻓﺎﺻﻠﻪ ]‪، [a,b‬ﺗﻮﺳﻂ ﺳﺮﻱ ﻫﺎﻱ ﺳﻴﻨﻮﺳﻲ ﻭ ﻛﺴﻴﻨﻮﺳﻲ ﺗﻘﺮﻳﺐ ﺯﺩ ﻭ ﺩﻭ ﺳﺮﻱ ﺍﺭﺍﺋﻪ‬
                     ‫ﻛﺮﺩ ﻛﻪ ﺑﻪ ﺳﺮﻱ ﻫﺎﻱ ﺳﻴﻨﻮﺳﻲ ﻓﻮﺭﻳﻪ ﻭ ﻛﻮﺳﻴﻨﻮﺳﻲ ﻓﻮﺭﻳﻪ ﻣﻌﺮﻭﻑ ﻫﺴﺘﻨﺪ.‬
                          ‫ﺳﺮﻱ ﻛﺴﻴﻨﻮﺳﻲ ﻓﻮﺭﻳﻪ: ﭘﻴﺸﻨﻬﺎﺩ ﻓﻮﺭﻳﻪ ﺑﺮﺍﻱ ﺑﺴﻂ ﻛﺴﻴﻨﻮﺳﻲ ﺑﺼﻮﺭﺕ ﺭﻭﺑﺮﻭ ﺍﺳﺖ‬
                                 ‫∞‬
                                            ‫‪nπx‬‬
                              ‫‪+ ∑ a n Cos‬‬
                    ‫‪ao‬‬
        ‫= )‪f ( x‬‬
                          ‫2‬     ‫1= ‪n‬‬         ‫‪c‬‬
                  ‫ﻗﻄﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺩﺭ ﻓﺎﺻﻠﻪ 0 ﺗﺎ‪ C‬ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ.‬                ‫)‪f(x‬‬   ‫ﺍﺯ ﺍﻳﻦ ﺳﺮﻱ ﺑﺮﺍﻱ ﺗﻘﺮﻳﺐ ﺗﻮﺍﺑﻊ‬
                                       ‫ﺭﺍ ﺗﻮﺳﻂ ﺳﺮﻱ ﻛﺴﻴﻨﻮﺳﻲ ﺗﻘﺮﻳﺐ ﺑﺰﻧﻴﺪ.‬               ‫‪0 < x <η‬‬        ‫ﻣﺜﺎﻝ:ﺗﺎﺑﻊ )‪ f(x‬ﻛﻪ‬
                                ‫∞‬
                                            ‫‪nπx‬‬
        ‫≈ ) ‪f (x‬‬              ‫‪+ ∑ a n Cos‬‬
                   ‫‪ao‬‬
                          ‫2‬     ‫1= ‪n‬‬         ‫‪π‬‬

                                                                                     ‫ﺩﺍﺭﱘ:‬      ‫ﺿﺮﻳﺐ ‪a n‬‬    ‫ﺑﺮﺍﻱ ﳏﺎﺳﺒﻪ‬
                   ‫∞‬                                      ‫‪η‬‬       ‫∞‬      ‫‪η‬‬              ‫‪η‬‬
                 ‫⇒ ) ‪+ ∑ a n Cosnx = f ( x‬‬                ‫‪∫ dx + ∑ a n ∫ Cos nx dx = ∫ f (x )dx‬‬
        ‫‪ao‬‬                                       ‫‪ao‬‬
             ‫2‬     ‫1= ‪n‬‬
                                                      ‫2‬           ‫1= ‪n‬‬
                                                          ‫0‬              ‫0‬              ‫0‬




‫٣٢‬
                                      η                                 η
                     × π + o = ∫ f ( x )dx ⇒ ao = 2
                                                                      π ∫ f ( x )dx
          ao
     =
                 2
                 η                    o              η                  o                η
                                          ∞

                 ∫ Cos mx dx + ∑ a ∫ Cos nx Cos mx dx =                                  ∫ f (x )Cos mx dx
     ao
             2                            n =1
                                                 n
                 o                                   o                                   o


             ∞   ⎡η                                    η
              1 ⎢ (Cos (n − m )x − Cos (nn + m )x )dx = f ( x )Cos mx dx
         ∑ a n 2 ⎢∫                                    ∫
         n =1    ⎣o                                    o


             ∞   ⎡η                                    η
              1 ⎢ (Cos (n − m )x − Cos (nn + m )x )dx = f ( x )Cos mx dx
         ∑ a n 2 ⎢∫                                    ∫                                                               ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﺭ ﻛﻞ‬
         n =1    ⎣o                                    o

                                                                                             η
                                                                                an = 2
                                                                                         π ∫ f ( x )Cos nx dx
                                                                                             o


                                                                                                            :‫ﺑﺴﻂ ﺳﺮﻱ ﺳﻴﻨﻮﺳﻲ‬
     ‫ ﺭﺍ ﺗﻮﺳﻂ‬f(x) ‫,0 [ ﺑﺎﺷﺪ ﻭ ﺑﺼﻮﺭﺕ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺑﺎﺷﺪ .ﻣﻲ ﺗﻮﺍﻥ‬c] ‫ ﻳﻚ ﺗﺎﺑﻊ ﺩﺭ ﻓﺎﺻﻠﻪ‬f(x) ‫ﺍﮔﺮ‬
                ∞
                         nπx                             .‫ﻳﻚ ﺳﺮﻱ ﺳﻴﻨﻮﺳﻲ ﻓﻮﺭﻳﻪ ﺗﻘﺮﻳﺐ ﺯﺩ‬
      f (x ) ≅ ∑ bn Sin
                          n =1              c

                                                 .‫ﺎﻳﺖ ﻣﻴﻞ ﻛﻨﺪ ،ﺗﻘﺮﻳﺐ ﺗﺎ ﺣﺪﻭﺩ ﺩﳋﻮﺍﻩ ﺩﻗﻴﻖ ﻣﻲ ﺷﻮﺩ‬ ‫ ﺑﻪ ﰊ‬n ‫ﭼﻨﺎﻧﭽﻪ‬
                          ∞
                                          nπx
         f (x ) ≅ ∑ bn Sin
                          n =1             c
                                                                                                           :‫ ﻫﺎ‬bn ‫ﻣﻄﻠﻮﺏ ﺍﺳﺖ‬
                     nπx     mπx                  mπx
         c                                           c

      ∫ Sin              Sin     dx = ∫ f (x )Sin     dx                                          : ‫ ﺗﻮﺟﻪ ﺷﻮﺩ ﻛﻪ‬f(x) ‫ﺑﺮﺍﻱ ﳏﺎﺳﺒﻪ‬
         o
                      c       c       o
                                                   c
                                                    mπx  ‫ﻭ ﺍﻧﺘﮕﺮﺍﻝ ﺟﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ‬                     ‫ﭘﺲ ﺑﺎ ﺿﺮﻳﺐ ﻃﺮﻓﲔ ﺩﺭ ﺳﻴﻨﻮﺱ‬
                                                     c
                       mπx              2 mπx )Sin mπx dx
              c                            c
            = f ( x )x                       f(
           2 ∫
      ∑ bn ∫ Sin c Sin c c dx = ∫ f (x )Sin∫ c dx c
       ∞c
                nπ Sin bπxdx ⇒ bm =
            c                   c
     bm
                                         c
              o                            o
      n =1  o                   o


                                      :‫( ﺑﺎ ﺑﺴﻂ ﺳﻴﻨﻮﺳﻲ ﻓﻮﺭﻳﻪ ﺛﻘﺮﻳﺐ ﺑﺰﻧﻴﺪ‬o, π ) ‫ ﺭﺍ ﺩﺭ ﻓﺎﺻﻠﻪ‬f(x)=x ‫ﻣﺜﺎﻝ:ﺗﺎﺑﻊ‬

                              ∞
                                nπx              ∞
             f ( x ) ≅ ∑ bn Sin     ⇒ f ( x ) = ∑ bn Sin πx
                       n =1      c              n =1




                          π                                      π          π
                                                  −x
      bn = 2
                      π   ∫ xSin nx dx =
                          o
                                                  n
                                                     Cos nx ∫ + 1 ∫ Cos nx dx
                                                            o
                                                                 n
                                                                   o


          ⎛                        π
                                                                     ⎞ ⎛ (− 1)n +1 × π       ⎞      2(− 1)
                                                                                                           n +1
      = 2 ⎜ π (− 1) − o + 1 2 Sinx ∫                                 ⎟=⎜                     ⎟× 2 =
                   n +1
         π⎜ n              n                                         ⎟ ⎜                     ⎟ π
          ⎝                        o                                 ⎠ ⎝      n              ⎠         n


         f (x) = ∑
                          ∞
                              (− 1)n +1 × 2 Sin nx = 2SinX − Sin2 x + 2                      Sin3x............
                      n =1        n                                                      3
٢٤
        ‫ﺍﺳﺖ.ﻭ ﻧﻴﺰ ﺑﻪ ﺩﻟﻴﻞ‬                               ‫ﹰ‬
                                            ‫ﺍﻭﻻ ﺩﺍﺭﺍﻱ ﭘﺮﻳﻮﺭ‬                        ‫ﺩﺭ ﺧﺎﺭﺝ ﺍﺯﺍﻳﻦ ﻓﺎﺻﻪ ﺑﺴﻂ ﺳﻴﻨﻮﺳﻲ‬
                                                                    ‫∞‬

                                                    ‫‪ ∑ n‬ﻫﻢ ﻓﺮﺩ ﺍﺳﺖ.‬
                                                                        ‫‪b Sin nx‬‬
                                                                ‫1= ‪n‬‬
                                                                            ‫ﺍﻳﻨﻜﻪ ‪ Sin nx‬ﻓﺮﺩ ﺍﺳﺖ ﺑﻨﺎﺑﺮﺍﻳﻦ‬
                                                  ‫ﺷﻜﻞ ﺭﻭﺑﺮﻭ ﺑﺴﻂ ﺗﺎﺑﻊ ) ‪ f ( x‬ﻭ ﺧﻮﺩ ﺗﺎﺑﻊ ﺭﺍ ﻣﺸﺨﺺ ﻣﻲ ﻛﻨﺪ.‬
       ‫‪f (x) = x‬‬                  ‫2]‪[0,ππ‬‬                        ‫‪∑ b Sin nx‬‬
                                                                        ‫‪n‬‬



                         ‫ﺍﻟﺒﺘﻪ ﺍﮔﺮ ﳘﲔ ﺗﺎﺑﻊ ﺭﺍ ﺗﻮﺳﻂ ﻛﺴﻴﻨﻮﺳﻲ ﻓﻮﺭﻳﻪ ﺑﻴﺎﻥ ﻣﻲ ﻛﺮﺩﱘ ﺩﺍﺭﺍﻱ ﺧﻮﺍﺹ ﺯﻳﺮ ﺑﻮﺩ‬
       ‫= ) ‪f (x‬‬              ‫‪+ ∑ an Cosnx‬‬
                  ‫‪ao‬‬
                        ‫2‬
                                                                            ‫)ﻙ،ﻡ،ﻡ(‬     ‫ﭘﺮﻳﻮﺭ 2‪π‬‬   ‫ﺧﻮﺍﺹ:١-ﺑﺴﻂ ﺩﺍﺭﺍﻱ‬
                                   ‫٢-ﺑﺴﻂ ﺑﻪ ﺩﻟﻴﻞ ﺍﻳﻨﻜﻪ ﺍﺯ ﲨﻼﺕ ﻛﺴﻴﻨﻮﺳﻲ ﺗﺸﻜﻴﻞ ﺷﺪﻩ ﺯﻭﺝ ﺍﺳﺖ.‬

                                                               ‫ﻧﻜﺘﻪ:ﭘﺮﻳﻮﺩ ﻳﻚ ﺗﺎﺑﻊ ﺛﺎﺑﺖ ﻫﺮ ﻋﺪﺩﻱ ﻣﻲ ﺗﻮﺍﻧﺪ ﺑﺎﺷﺪ .‬

      ‫ﻓﺮﺽ ﻛﻨﻴﺪ ﻛﻪ ) ‪ f ( x‬ﻳﻚ ﺗﺎﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺩﺭ ﻓﺎﺻﻠﻪ ﻱ )‪(-c،c‬ﳘﺎﻧﻄﻮﺭ ﻛﻪ ﻣﻲ ﺩﺍﻧﻴﻢ ﻫﺮ ﺗﺎﺑﻊ‬
                                           ‫ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺼﻮﺭﺕ ﲨﻊ ﺩﺭ ﺗﺎﺑﻊ ﺯﻭﺝ ﻭ ﻓﺮﺩ ﻧﻮﺷﺖ.‬

        ‫) ‪f ( x ) = h( x ) + g ( x‬‬            ‫آﻪ‬         ‫, ) ‪g (− x ) = − g ( x‬‬       ‫) ‪h(− x ) = h( x‬‬



             ‫) ‪f ( x ) + f (− x‬‬                          ‫) ‪f ( x ) − f (− x‬‬
     ‫= ) ‪h( x‬‬                                 ‫= )‪g (x‬‬
            ‫ﻛﺴﻴﻨﻮﺳﻲ ﺗﻘﺮﻳﺐ‬
                      ‫2‬                  ‫ﺑﺎ ﺑﺴﻂ‬   ‫) ‪ h( x‬ﺭﺍ ﺩﺭﻓﺎﺻﻠﻪ )‪(٠،c‬‬
                                                                  ‫2‬           ‫ﺗﺎﺑﻊ ) ‪ h( x‬ﻳﻚ ﺗﺎﺑﻊ ﺯﻭﺝ ﺍﺳﺖ ﺑﻨﺎﺑﺮﺍﻳﻦ ﭼﻨﺎﻧﭽﻪ‬

                                                         ‫ﺷﺪ.‬   ‫ﺑﺰﻧﻴﻢ ،ﺗﺎﺑﻊ ) ‪ h( x‬ﺩﺭ ﻧﺎﺣﻴﻪ)‪ (-c،c‬ﺗﻘﺮﻳﺐ ﺯﺩﻩ ﺧﻮﺍﻫﺪ‬
                                                                                      ‫‪nπx‬‬
                                                                        ‫‪c‬‬
                                                                        ‫‪h( x )Cos‬‬
                                                                     ‫∫‪c‬‬
                             ‫‪nπx‬‬
                              ‫∞‬
                                                            ‫2 = ‪an‬‬
      ‫‪h( x ) = o + ∑ a n Cos‬‬
              ‫‪a‬‬                                                                           ‫‪dx‬‬
                ‫1= ‪2 n‬‬                                                  ‫‪o‬‬
                                                                                       ‫‪c‬‬
                              ‫‪c‬‬
       ‫) ‪f ( x ) = h( x ) + g ( x‬‬            ‫) ‪(− c, c‬‬
                        ‫ﭘﺲ‬        ‫∞‬
                                             ‫∞ ‪nπx‬‬         ‫‪nπx‬‬
         ‫= ) ‪f (x‬‬            ‫‪+ ∑ a n Cos‬‬        ‫‪+ ∑ bn Sin‬‬                            ‫) ‪x ∈ (− c, c‬‬
                    ‫‪ao‬‬
                         ‫2‬
                                                                              ‫آﻪ‬
                                  ‫1= ‪n‬‬        ‫‪c‬‬   ‫1= ‪n‬‬      ‫‪c‬‬
            ‫ﻛﻪ ﺍﻳﻦ ﺑﺴﻂ ﺗﺎﺑﻊ ﻓﻮﺭﻳﻪ ) ‪ f ( x‬ﺩﺭ ﻓﺎﺻﻠﻪ )‪ (-c،c‬ﻣﻲ ﺑﺎﺷﺪ.ﺣﺎﻝ ﺑﺎﻳﺪ ﺿﺮﺍﻳﺐ ﺭﺍ ﭘﻴﺪﺍ ﳕﺎﺋﻴﻢ.‬
                   ‫) ‪f ( x ) + f (− x‬‬     ‫‪nπx‬‬        ‫‪⎡c‬‬           ‫‪nπx‬‬                 ‫⎤ ‪nπx‬‬
                 ‫‪c‬‬                                                       ‫‪c‬‬
                                                        ‫‪f (x )Cos‬‬     ‫‪dxt f (− x )Cos‬‬
                  ‫∫‪c‬‬                                             ‫∫⎢ ‪c‬‬                        ‫∫‬
      ‫‪an‬‬   ‫2=‬                         ‫‪Cos‬‬     ‫1 = ‪dx‬‬                                     ‫‪dx‬‬                    ‫⎥‬
                    ‫‪o‬‬
                                  ‫2‬                 ‫‪c‬‬              ‫‪⎣o‬‬               ‫‪c‬‬        ‫‪o‬‬
                                                                                                           ‫‪c‬‬   ‫⎦‬
           ‫‪⎡c‬‬                                     ‫⎤‬
         ‫1 = ⎥ ‪1 ⎢ f (x )Cos nπx dx + f ( x )Cos nπx dx‬‬                ‫‪nπx‬‬
                               ‫‪o‬‬                         ‫‪c‬‬
                                                            ‫‪f ( x )Cos‬‬
          ‫∫ ‪c‬‬                  ‫‪∫c‬‬                       ‫∫‪c‬‬
       ‫=‬                                                                   ‫‪dx‬‬
           ‫‪⎣o‬‬           ‫‪c‬‬      ‫−‬
                                            ‫‪c‬‬     ‫⎦‬      ‫‪−c‬‬
                                                                        ‫‪c‬‬


‫٥٢‬
                                                                          .‫ −( ﺑﺎ ﺑﺴﻂ ﺯﻳﺮ ﻣﺸﺨﺺ ﻣﻲ ﺷﻮﺩ‬c, c ) ‫ ﺩﺭ ﻓﺎﺻﻠﻪ‬f (x) ‫ﺑﻨﺎﺑﺮﺍﻳﻦ‬
                                                         ∞
                                                                 ⎛                             nπx                            nπx ⎞
      f    (x ) =            a    0
                                       2
                                               +     ∑           ⎜ a      n   Cos                  + b n Sin                      ⎟
                                                      n =1       ⎝                              c                              c ⎠
                                                 nπx                                                                                 nπx
                             c                                                                                  c
     an = 1                  ∫        f ( x )Cos     dx                                    bn = 1               ∫       f ( x )Sin       dx
                         c                        c                                                     c                             c
                             −c                                                                             −c



                                                                                                x ←(−π,π) :‫ﻣﺜﺎﻝ: ﺗﺎﺑﻊ ﺯﻳﺮ ﺭﺍ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺑﺪﻫﻴﺪ‬
             ⎧ 0                                           − π            <       x < − π
             ⎪                                                                                      2
             ⎪
      f ( x )⎨ 1                                         − π                  <       x < π
                                                                      2                             2
             ⎪                                               π
             ⎪ 0                                                              <       x < π
             ⎩                                                        2


                                                     N
      f (x ) =                                  ∑ (a                                                                      )
                          a0
                                           +                              Cos nx + b n Sin nx
                                      2             n =1
                                                                  n


                             π                                                                          π
                                                                                                                                 nπx
     an = 1
                      π      ∫π f (x )Cos nx dx
                             −
                                                                                          bn = 1
                                                                                                    π   ∫π f (x )Sin
                                                                                                        −
                                                                                                                                     π
                                                                                                                                         dx

                                                                              π
            π                                                                     2
                                                                                           1 ⎛     nπ      nπ ⎞
     1
            ∫       1× Cos nx dx = 1                          Sin nx          ∫       =      ⎜ Sin    + Sin ⎟ = 1 × 2Sin nπ
      π     π
                                                         nπ                   π           nπ ⎝      2       2 ⎠  nπ         2
           −                                                                  −
                2                                                                 2

                                           π
                                                2
     ‫ﺣﺎل‬                 bn : 1
                                       π    ∫ Sin            nx dx = 0
                                           −π
                                                2




                                                                                                            π
                                                                                                                                                      ‫ﺑﻨﺎﺑﺮﺍﻳﻦ‬
                             ∞
     f ( x ) = 1 + ∑ ⎛ 2 Sin nη ⎞Cos nx
                                                                                                                2
                        ⎜ nπ   2⎟                                                                           ∫1dx = 1                     ‫ ﺭﺍ ﳏﺎﺳﺒﻪ ﻣﻲ ﻛﻨﻴﻢ‬a o
                   n =1 ⎝       ⎠
                                                                                                            π
                                                                                                            −
                                                                                                                    2                       = ‫ﻳﺎ ﻣﻘﺪﺍﺭ ﻣﺘﻮﺳﻂ‬

               ‫ ﺑﺴﻂ‬f (x) ‫ −( ﺑﺎﺷﺪ ﻭ‬π , π ) ‫ ﻳﻚ ﺗﺎﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺩﺭ ﻓﺎﺻﻠﻪ ﻱ‬f (x) ‫ﻗﻀﻴﻪ:ﺍﮔﺮ ﺗﺎﺑﻊ‬
                                                                .‫ ﺑﺎﺷﺪ،ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ‬f (x) ‫ﻓﻮﺭﻳﻪ‬
                          ∞
                     a0
     f (x) =            + ∑ (a n Cos nx dx + bn Sin nx )
                     2 n =1
                     π                                                                     π
                1                                                                     1
     an =            ∫    f ( x )Cosnx dx                        , bn =                    ∫ f (x )Sin nx dx
                π    −π
                                                                                      π    −π




٢٦
     ‫ﺣﺎﻝ ﭼﻨﺎﻧﭽﻪ ﻳﻚ ﻧﻘﻄﻪ ﻧﺎﭘﻴﻮﺳﺘﮕﻲ ﺗﺎﺑﻊ ) ‪ f (x‬ﺑﺎﺷﺪ ﻭ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻦ ﻛﻪ ﺗﺎﺑﻊ ) ‪ f (x‬ﭘﻴﻮﺳﺘﻪ ﺍﺳﺖ.‬
      ‫∑ = )‪f (x‬‬
               ‫∞‬
                   ‫)1 −(2‬
                          ‫1+ ‪n‬‬
                               ‫‪Sin nx‬‬
                                                  ‫ﻣﻘﺪﺍﺭ ) ‪ f (x‬ﺑﻮﺳﻴﻠﻪ ﺭﺍﺑﻂ ﺯﻳﺮ ﻣﺸﺨﺺ ﻣﻲ ﺷﻮﺩ:‬                        ‫‪o‬‬


              ‫1= ‪n‬‬    ‫‪n‬‬
                                                              ‫ﻛﻪ ﺑﻪ ﻗﻀﻴﻪ ﻣﺘﻮﺳﻂ ﺟﺎﻡ ﻣﻮﺳﻮﻡ ﺍﺳﺖ.‬
                                                               ‫ﺍﻳﻦ ﻗﻀﻴﻪ ﺩﻭﱂ ﺭﻳﺎﺿﻲ ﺍﺛﺒﺎﺕ ﻣﻲ ﺷﻮﺩ.‬
                ‫(ﺑﺴﻂ ﺩﺍﺩﻩ ﻭ ﻗﻀﻴﻪ ﻣﺘﻮﺳﻂ ﺁﻥ ﺭﺍ ﺗﺴﺖ ﻛﻨﻴﺪ.‬                   ‫ﻓﺎﺻﻠﻪ ) ‪π , π‬‬   ‫= ) ‪ f (x‬ﺩﺭ‬       ‫‪x‬‬       ‫ﻣﺜﺎﻝ :ﺗﺎﺑﻊ‬
                                           ‫‪ao‬‬
      ‫‪f ( x) = x‬‬                ‫= )‪f ( x‬‬        ‫+1‬      ‫+‪an Cosnx‬‬       ‫‪bn Sinnx‬‬


                          ‫‪π‬‬                                                       ‫‪π‬‬
      ‫1 = ‪an‬‬              ‫‪∫π‬‬    ‫‪f ( x ) Cos nx dx‬‬                    ‫1 = ‪bn‬‬       ‫‪∫π f (x )Cos‬‬    ‫‪nx dx‬‬
                     ‫‪π‬‬                                                        ‫‪π‬‬
                          ‫−‬                                                       ‫−‬

                                                                             ‫0 = ‪an‬‬   ‫) ‪ f (x‬ﻳﻚ ﺗﺎﺑﻊ ﻓﺮﺩ ﺍﺳﺖ ﭘﺲ:‬
                          ‫‪π‬‬                                 ‫‪π‬‬                                          ‫‪π‬‬
                                                                                    ‫‪⎛− x‬‬                    ‫⎞‬
      ‫1 = ‪bn‬‬
                     ‫‪π‬‬    ‫∫‬
                         ‫‪−π‬‬
                               ‫2 = ‪x Sin x dx‬‬
                                                       ‫‪π‬‬    ‫∫‬
                                                            ‫‪o‬‬
                                                                ‫2 = ‪Sin nx dx‬‬       ‫⎜‬
                                                                                  ‫‪π ⎜ π Cosx‬‬
                                                                                    ‫⎝‬
                                                                                                       ‫∫‬
                                                                                                       ‫‪o‬‬
                                                                                                           ‫⎟+‬
                                                                                                            ‫⎟‬
                                                                                                            ‫⎠‬

                                                 ‫) 1 −( ‪− π‬‬                           ‫1+ ‪⎡π ⎤ 2 (− 1 )n‬‬
                ‫‪π‬‬                                               ‫‪n‬‬
      ‫1 +‬
            ‫‪π‬‬   ‫2 = ‪∫ Cos nx dx‬‬
                 ‫‪o‬‬
                                            ‫∫‪η‬‬          ‫‪π‬‬
                                                                    ‫1 +‪+o‬‬
                                                                            ‫2‪n‬‬
                                                                               ‫= ⎥ ⎢ ‪Sin nx‬‬
                                                                                      ‫⎦ ‪⎣o‬‬     ‫‪n‬‬
      ‫ﭘﺲ‬
                                        ‫) 1 −( 2‬
                                  ‫∞‬                  ‫1+ ‪n‬‬
                ‫≅ ) ‪f (x‬‬        ‫∑‬                           ‫‪Sin nx‬‬
                                 ‫1= ‪n‬‬        ‫‪n‬‬


                                        ‫) 1 −( 2‬
                                                     ‫1+ ‪n‬‬
            ‫= ) ‪f (x‬‬
                         ‫ﭘﺲ‬     ‫∞‬

                               ‫∑‬
                               ‫1= ‪n‬‬          ‫‪n‬‬
                                                            ‫‪Sin nx‬‬


            ‫(‬
      ‫......... + ‪= 2 Sinx − 1 Sin 2 x + 1 Sin 3 x − 1 Sin 4 x‬‬
                  ‫2‬           ‫3‬           ‫4‬
                                                                                                               ‫)‬
      ‫)‪ f (x) = f (x‬ﻭﻗﱵ ‪ x‬ﻧﻘﻄﻪ ﭘﻴﻮﺳﺘﮕﻲ ‪ f‬ﺑﺎﺷﺪ ﺩﺭ ﻏﲑ ﺍﻳﻨﺼﻮﺭﺕ ﺩﺭ ﻧﻘﻄﻪ ﻧﺎﭘﻴﻮﺳﺘﮕﻲ ﻣﺘﻮﺳﻂ ﺗﺎﺑﻊ‬
                                                                              ‫‪ f‬ﻣﻲ ﺷﻮﺩ.‬
      ‫0 = ) ‪F (− π‬‬            ‫0 = ) ‪F (π‬‬
                                                         ‫‪−π,π : f‬‬         ‫ﻧﻘﺎﻁ ﻧﺎﭘﻴﻮﺳﺘﮕﻲ‬
                                                   ‫ﻛﻪ ﺍﻳﻦ ﺑﺎ ﻣﺘﻮﺳﻂ ﻣﻘﺎﺩﻳﺮ ﺟﺎﻡ ﺑﺮﺍﺑﺮ ﺍﺳﺖ.‬




‫٧٢‬
                                                                                                                                         ‫ﺧﻮﺍﺹ ﺿﺮﺍﻳﺐ ﺳﺮﻱ ﻓﻮﺭﻳﻪ :‬
                                                                                                                                                   ‫‪N‬‬

                                                                                                          ‫ﺍﮔﺮ ‪ S N (x) = o 2 + ∑ an Cosnx‬ﺁﻧﮕﺎﻩ ﺩﺍﺭﱘ :‬
                                                                                                                                            ‫‪a‬‬
                              ‫‪π‬‬

                              ‫‪∫ S (x )Cos‬‬
                                                                                                                                       ‫1=‪n‬‬
                 ‫2‬                                                ‫‪nx = a n‬‬
                      ‫‪π‬‬                   ‫‪N‬‬
                                                                                                                                                       ‫‪N‬‬
                                                                                                                             ‫= ) ‪S N (x‬‬              ‫‪+ ∑ a n Cosnx‬‬
                              ‫‪o‬‬
                                                                                                                                            ‫0‪a‬‬
                                                                                                                                                 ‫2‬     ‫1= ‪n‬‬


                 ‫‪π‬‬
                                                                                                                   ‫ﻧﻴﺰ ﺍﻧﺘﮕﺮﺍﻝ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﳏﺎﺳﺒﻪ ﻣﻲ ﺷﻮﺩ:‬
     ‫2‬
                 ‫)) ‪∫ (S ( x‬‬
                                                        ‫2‬
                                                            ‫‪dx‬‬
         ‫‪π‬‬                        ‫‪N‬‬
                  ‫‪o‬‬

                 ‫‪π‬‬                                                                            ‫‪π‬‬                                              ‫‪π‬‬
                                      ‫2 = ‪+ ∑an Codnx S N (x)dx‬‬                                                ‫2 + ‪S N (x)dx‬‬           ‫‪∑an ∫ S N (x)Cosnx dx‬‬
                   ‫‪a‬‬                                                                              ‫0‪a‬‬
         ‫2‬
             ‫‪π‬‬   ‫‪∫( o‬‬
                 ‫‪o‬‬
                              ‫2‬
                                                ‫)‬
                                                                                             ‫∫‪π‬‬
                                                                                              ‫‪o‬‬
                                                                                                           ‫2‬                      ‫‪π‬‬
                                                                                                                                             ‫‪o‬‬


                                           ‫‪π‬‬
                         ‫2‬                                             ‫‪N‬‬
                                                                                                     ‫2‪⎛ a‬‬                          ‫‪N‬‬
                                                                                                                                          ‫⎞ 2‬
                                                                             ‫‪an × π‬‬
                        ‫‪ao‬‬
         ‫2 =‬                               ‫∫‬    ‫2 + ‪dx‬‬
                                                                 ‫∑‪π‬‬                                ‫+ ‪= ⎜ o‬‬                     ‫∑‬         ‫⎟ ‪an‬‬
                                                                              ‫2‬
                      ‫4 ‪π‬‬                                                                         ‫2 ⎜ 2‬                                     ‫⎟‬
                                           ‫‪o‬‬                          ‫1= ‪n‬‬                           ‫⎝‬                            ‫1= ‪n‬‬      ‫⎠‬
                                                ‫∞‬                                                                                      ‫ﺣﺎﻝ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺳﻴﻨﻮﺳﻲ ﺩﺍﺭﱘ:‬
         ‫‪S‬‬   ‫‪N‬‬        ‫∑ = ) ‪(x‬‬                              ‫‪b n Sin‬‬    ‫‪nx‬‬
                                               ‫1= ‪n‬‬



                                      ‫‪π‬‬
                          ‫2‬
                              ‫‪π‬‬       ‫∫‬‫0‬
                                               ‫‪S‬‬    ‫‪N‬‬   ‫‪( X ) Sinmxdx‬‬                        ‫‪= bm‬‬                                                                     ‫ﺩﺍﺭﱘ‬
                      ‫‪π‬‬                                               ‫‪N‬‬                  ‫‪π‬‬
         ‫2‬
             ‫‪π‬‬        ‫= ‪∫ [ S N ( X )] dx‬‬
                                    ‫2‬
                                                                      ‫∑‬
                                                                      ‫1= ‪n‬‬
                                                                             ‫2‬
                                                                                 ‫‪π‬‬       ‫‪∫S‬‬       ‫‪N‬‬       ‫‪( X ) b n Sinnxdx‬‬                                             ‫ﻟﺬﺍ‬
                      ‫0‬                                                                  ‫0‬
                 ‫‪N‬‬                                      ‫‪π‬‬                                                      ‫‪N‬‬
         ‫=‬   ‫∑‬   ‫1= ‪n‬‬
                          ‫2 × ‪bn‬‬
                                                ‫‪π‬‬       ‫∫‬    ‫= ‪S N ( X ) Sinnxdx‬‬                           ‫‪∑ (b‬‬
                                                                                                            ‫1= ‪n‬‬
                                                                                                                    ‫‪n‬‬   ‫2)‬
                                                        ‫0‬




                                               ‫‪π‬‬
                                                                                                                                                ‫ﺑﺮﺍﻱ ﻫﺮ ﺩﻭ ﺣﺎﻟﺖ ﺩﺍﺭﱘ :‬
             ‫2 = ‪E‬‬
                                      ‫‪π‬‬        ‫(∫‬
                                                ‫0‬
                                                            ‫‪f (x) − S‬‬            ‫‪N‬‬       ‫0 ≥ ‪( X )) 2 dx‬‬


                                                                                     ‫‪π‬‬                                        ‫‪π‬‬                                  ‫‪π‬‬
                                                                             ‫2‬
                                                                                 ‫‪π‬‬   ‫∫‬        ‫‪f‬‬       ‫2‬
                                                                                                          ‫‪(x )dx‬‬        ‫4‬
                                                                                                                            ‫∫‪η‬‬           ‫2 + ) ‪f ( x )S N ( x‬‬
                                                                                                                                                                ‫∫‪η‬‬   ‫‪(S N (x ))2 dx‬‬
                 ‫ﺑﻪ ﺍﻳﻦ ﺻﻮﺭﺕ ﻣﻲ ﻧﻮﻳﺴﻴﻢ‬                                               ‫0‬                                        ‫0‬                                  ‫0‬


                                ‫ﺍﺯ ﺁﳒﺎ ﻛﻪ ‪ f‬ﺗﺎﺑﻌﻲ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﻪ ﺍﺳﺖ ﻛﻪ ﺑﻪ ﺻﻮﺭﺕ ﺳﻨﻮﺳﻲ ﻳﺎ‬
          ‫ﻛﺴﻴﻨﻮﺳﻲ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ ﻭ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ ﺗﺎﺑﻊ )‪ f (x‬ﺩﺭ ﻓﺎﺻﻠﻪ )‪ (0,π‬ﳏﺪﻭﺩ‬
         ‫‪π‬‬
                                                           ‫ﺍﺳﺖ ، ﻟﺬﺍ ﻋﺒﺎﺭﺕ ﺍﺯ 0 ﺗﺎ ‪: η‬‬
         ‫‪∫ f (x )dx‬‬                        ‫ﳏﺎﺳﺒﻪ ﻣﻘﺪﺍﺭ ﻋﺒﺎﺭﺕ:‬
                          ‫2‬

         ‫0‬

            ‫‪π‬‬                                                         ‫)ﻗﻄﻌﻪ ﺍﻱ ﺍﺳﺖ ﳏﺪﻭﺩ(‬
         ‫2‬
                     ‫∫‬                ‫‪f ( x )S N (x )dx‬‬
             ‫‪π‬‬
                     ‫0‬

‫٨٢‬
                                                                                                                              ‫ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﻛﺴﻴﻨﻮﺳﻲ ﺩﺍﺭﱘ:‬
             ‫‪π‬‬                                                                                          ‫‪π‬‬                                   ‫‪π‬‬
                            ‫‪⎛a‬‬                                ‫⎞‬       ‫‪a‬‬                                                       ‫‪N‬‬
                       ‫+ ‪f (x)⎜ o‬‬                                                                                ‫+‪f (x)dx‬‬     ‫‪∑a‬‬                      ‫‪f (x)Cosnxd‬‬
         ‫∫‪π‬‬                                             ‫∫ 0 × 2 =‪anCosnxdx‬‬
                                                                                                                                        ‫∫‪π‬‬
     ‫2‬                                                        ‫⎟‬                                                                         ‫2‬
                            ‫2⎝‬                                ‫⎠‬     ‫2 ‪π‬‬                                                     ‫1=‪1 n‬‬
                                                                                                                                    ‫‪n‬‬
             ‫0‬                                                           ‫0‬                                                                  ‫0‬

                      ‫2‬                          ‫‪N‬‬
                     ‫0‪a‬‬
                   ‫=‬
                     ‫2‬
                        ‫+‬                       ‫∑‬
                                                ‫1= ‪n‬‬
                                                              ‫2‬
                                                             ‫‪an‬‬

                                                                                                                                ‫ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺳﻴﻨﻮﺳﻲ ﺩﺍﺭﱘ:‬
             ‫‪π‬‬
                             ‫‪⎛π‬‬                        ‫⎞‬                                    ‫‪N‬‬                    ‫‪π‬‬                              ‫‪N‬‬
                        ‫∫ ⎜)‪f (x‬‬                       ‫⎟‬                                    ‫2 ×‪∑ bn‬‬                   ‫‪f (x)Sinnxdx‬‬          ‫‪∑b‬‬
         ‫∫‪π‬‬                                                                                                  ‫∫‪π‬‬
     ‫2‬                                          ‫= ‪bn Sinnx dx‬‬                                                                     ‫=‬
                                                                                                                                                           ‫2‬
                             ‫⎜‬                         ‫⎟‬                                                                                               ‫‪n‬‬
             ‫0‬               ‫0⎝‬                        ‫⎠‬                                    ‫1=‪n‬‬                  ‫0‬                              ‫1=‪n‬‬



              ‫‪π‬‬                                       ‫2‬                        ‫‪π‬‬

                                                                              ‫‪∑ (a‬‬                      ‫)‬
                                                     ‫‪ao‬‬
                           ‫= ‪f 2 (x)dx‬‬                                                          ‫ﻗﻀﻴﻪ ﭘﺎﺭﺳﻮﺍﻝ ‪+ bn‬‬
         ‫∫‪π‬‬
     ‫2‬                                                  ‫+‬                                   ‫2‬      ‫2‬
                                                                                            ‫‪n‬‬
                                                     ‫2‬                         ‫1=‪n‬‬
                                                                                                       ‫ﺩﺭ ﺣﺎﻟﺖ ﻛﺴﻴﻨﻮﺳﻲ ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﻧﺎﻣﺴﺎﻭﻱ ﺩﺍﺭﱘ:‬
              ‫0‬
             ‫‪π‬‬
                                         ‫‪⎛a‬‬                                          ‫‪N‬‬          ‫⎞‬
     ‫2‬
             ‫∫‬             ‫+ ‪f 2 ( x )dx ≥ ⎜ o‬‬                                       ‫∑‬       ‫⎟ ‪an‬‬
                                                                                              ‫2‬
         ‫‪π‬‬                               ‫2 ⎜‬                                                    ‫⎟‬
              ‫0‬                          ‫⎝‬                                           ‫1= ‪n‬‬       ‫⎠‬
         ‫‪N‬‬                                          ‫‪π‬‬

     ‫∑‬
     ‫1= ‪n‬‬
                  ‫2 ≤ 2‪b n‬‬
                                            ‫‪π‬‬        ‫∫‬       ‫‪f‬‬       ‫2‬
                                                                         ‫‪( x ) dx‬‬
                                                     ‫0‬

                             ‫ﺍﻳﻦ ﻧﺎﻣﺴﺎﻭﻱ ﺑﻪ ﺍﺯﺍﺀ ﻫﺮ ‪ N‬ﺑﺮﻗﺮﺍﺭ ﺍﺳﺖ)ﻣﻦ ﲨﻠﻪ ﺯﻣﺎﱐ ﻛﻪ ‪ N‬ﺑﻪ ﰊ ‪‬ﺎﻳﺖ ﻣﻴﻞ ﻛﻨﺪ(‬
                                                                                               ‫ﭘﺲ:‬
                             ‫‪N‬‬                                   ‫‪π‬‬
     ‫0‪a‬‬
              ‫2‬
                       ‫+‬   ‫∑‬‫1= ‪n‬‬
                                   ‫‪a‬‬    ‫2‬
                                        ‫‪n‬‬       ‫2 ≥‬
                                                          ‫‪π‬‬      ‫∫‬       ‫‪f 2 ( x ) dx‬‬
                                                                 ‫0‬
                                                                                                                                        ‫2‬
                                                                                                                                      ‫‪ao‬‬
                                                    ‫ﺳﺮﻱ ﺑﺎ ﲨﻼﺕ ﻣﺜﺒﺖ ﻭ ﻣﻘﺪﺍﺭ ﳏﺪﻭﺩ ﺍﺳﺖ ﭘﺲ:‬                ‫ﺳﺮﻱ‬                         ‫2‪ 2 + ao‬ﻳﻚ‬
                                                                                                                                          ‫1= ‪n‬‬

     ‫0 = ‪Lim a n‬‬                                                                       ‫ﺑﺮﺍﻱ ﺣﺎﻟﺖ ﺳﻴﻨﻮﺳﻲ ﺩﺍﺭﱘ:‬
       ‫‪n‬‬
                  ‫‪π‬‬                                                      ‫‪N‬‬                        ‫‪N‬‬
     ‫2‬
                  ‫∫‬        ‫+ 2‪f ( x ) ≥ 2 × ( ∑ b n‬‬                                             ‫∑‬      ‫0 ≥ ) ‪an‬‬
                                                                                                        ‫2‬
          ‫‪π‬‬                                                          ‫1= ‪n‬‬                       ‫1= ‪n‬‬
                   ‫0‬



      ‫‪N‬‬                                     ‫‪π‬‬
                                                                                                                     ‫0 = ‪Lim b n‬‬
     ‫∑‬
     ‫1= ‪n‬‬
              ‫2 ≤ ‪b‬‬‫2‬
                   ‫‪n‬‬               ‫‪π‬‬        ‫∫‬    ‫‪f 2 ( x ) dx‬‬
                                                                                                                      ‫‪n‬‬
                                            ‫0‬

                            ‫ﭘﺲ ﳘﻮﺍﺭﻩ ﺩﻧﺒﺎﻟﻪ ﻫﺎﻱ ‪ an , bn‬ﺩﺭ ﺳﺮﻱ ﻓﻮﺭﻳﻪ ﺑﻪ ﲰﺖ ﻣﻴﻞ ﺧﻮﺍﻫﻨﺪ ﻛﺮﺩ.‬
                       ‫ﻗﺒﻼ ﮔﻔﺘﻴﻢ ﻛﻪ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺗﺎﺑﻊ )‪ f (x‬ﻛﻪ ﺩﺭ ﺑﺎﺯﻱ ) ‪ (− n, n‬ﺗﻌﺮﻳﻒ ﺷﺪﻩ ﺑﻮﺩ ﺑﺼﻮﺭﺕ:‬
                                                                                               ‫ﹰ‬
                                                         ‫∞‬
                  ‫= )‪f (x‬‬
                                   ‫0‪a‬‬
                                        ‫2‬
                                                ‫+‬    ‫‪∑ (a‬‬
                                                        ‫1= ‪n‬‬
                                                                         ‫‪m‬‬   ‫‪Cosx + b n Sinn‬‬             ‫‪x‬‬   ‫)‬
                                                                                                                      ‫ﺑﻮﺩ ﻭ ﺑﺎ ﺿﺮﺍﻳﺐ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﺪ:‬


‫٩٢‬
                        ‫1‬                                                                                                            ‫1‬
                                                 ‫‪( x )Cosdx‬‬                                                                                   ‫∫‬                ‫‪( x )Sinnxdx‬‬
                                     ‫‪n‬‬                                                                                                            ‫‪n‬‬
      ‫‪a‬‬   ‫‪n‬‬       ‫=‬
                        ‫‪n‬‬       ‫∫‬   ‫‪−n‬‬
                                            ‫‪f‬‬                                                                    ‫‪b‬‬       ‫‪n‬‬       ‫=‬
                                                                                                                                     ‫‪n‬‬            ‫‪− n‬‬
                                                                                                                                                          ‫‪f‬‬




     ‫ﻭ ﮔﻔﺘﻴﻢ ﻛﻪ ﺗﺎﺑﻊ )‪ f (x‬ﺑﺎﻳﺴﱵ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺑﺎﺷﺪ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺗﺎﺑﻊ )‪ (x‬ﻛﻪ ﺩﺭ‬
         ‫‪f‬‬

          ‫ﻧﻘﺎﻃﻲ ﻛﻪ )‪ f (x‬ﭘﻴﻮﺳﺘﻪ ﺍﺳﺖ ﺑﻪ )‪ f (x‬ﳘﮕﺮﺍ ﺍﺳﺖ ﻭ ﺩﺭ ﻧﻘﺎﻃﻲ ﻛﻪ ) ‪f (x‬ﻧﺎﭘﻴﻮﺳﺘﻪ ﺍﺳﺖ‬
                                    ‫ﺑﺎ ﻣﺘﻮﺳﻂ ﺟﺎﻡ ﺑﺮﺍﺑﺮ ﺍﺳﺖ .ﭘﺲ ﳘﻮﺍﺭﻩ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﳘﮕﺮﺍﺳﺖ.‬
      ‫0 = ‪lim ax‬‬            ‫0 = ‪, lim bx‬‬                        ‫ﻃﺒﻖ ﻗﻀﻴﻪ ﻧﺸﺎﻥ ﺩﺍﺭﱘ ﻛﻪ :‬
                                    ‫‪N‬‬                                      ‫‪R‬‬
      ‫2 0‪a‬‬
                                ‫∑‬                        ‫1 ≤‬               ‫∫‬
                                                 ‫2‬
                        ‫+‬                   ‫‪an‬‬                                      ‫‪f ( x ) Cos‬‬              ‫2‬
                                                                                                                 ‫≥ ‪dx‬‬                     ‫1‬
                   ‫2‬            ‫1= ‪n‬‬
                                                                      ‫‪R‬‬
                                                                           ‫‪−R‬‬
                                                                                                                                                                               ‫ﻭ ﻧﻴﺰ ﻧﺸﺎﻥ ﺩﺍﺭﱘ ﻛﻪ:‬
          ‫‪N‬‬                                               ‫‪R‬‬

      ‫∑‬
      ‫1= ‪n‬‬
                  ‫‪b‬‬    ‫‪n‬‬
                        ‫2‬
                                ‫1 ≤‬
                                                ‫‪R‬‬        ‫∫‬        ‫‪f‬‬   ‫2‬
                                                                           ‫‪Cos‬‬                  ‫‪( x ) dx‬‬
                                                                                                                                 ‫∞‬
                                                         ‫‪− R‬‬
                                                                                                                             ‫∑‬           ‫‪ax‬‬   ‫2‬
                                                                                                                                                   ‫‪+ bx‬‬        ‫2‬

                                                                                         ‫ﳘﮕﺮﺍ ﺍﺳﺖ:‬               ‫ﻗﻀﻴﻪ:ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺳﺮﻱ‬
                                                                                                                             ‫1= ‪n‬‬



                                                                                    ‫ﺍﺛﺒﺎﺕ:ﺍﮔﺮ ‪ b , a‬ﺿﺮﺍﻳﺐ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﻱ )‪ f (x‬ﺑﺎﺷﻨﺪ ﺩﺍﺭﱘ:‬                                             ‫‪n‬‬    ‫‪n‬‬

                                 ‫‪π‬‬                                                                                           ‫‪π‬‬
      ‫‪an‬‬          ‫1 =‬
                            ‫‪π‬‬    ‫‪∫ β (x )Cosnx‬‬
                                ‫‪−π‬‬
                                                                                        ‫,‬           ‫‪bn‬‬   ‫1 =‬
                                                                                                                     ‫‪π‬‬       ‫‪∫ β (x ) sin‬‬
                                                                                                                             ‫‪−π‬‬
                                                                                                                                                          ‫‪nxdx‬‬


                                                                                ‫‪β‬ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻢ:‬                                                  ‫‪n‬‬           ‫,‬        ‫‪αn‬‬       ‫ﺣﺎﻝ ﺿﺮﺍﻳﺐ‬
                                ‫‪π‬‬                                                                                            ‫‪π‬‬
     ‫‪α n = 1π‬‬                   ‫‪∫ f ′ (x )cos‬‬
                                ‫‪−π‬‬
                                                                  ‫‪nxdx‬‬                  ‫,‬           ‫‪β n = 1π‬‬                 ‫‪∫ f ′ (x )Sinnxdx‬‬
                                                                                                                             ‫‪−π‬‬

                  ‫‪α‬‬    ‫‪n‬‬        ‫‪= an × n‬‬                              ‫,‬             ‫‪β n = nb n‬‬                                        ‫ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﲑﻱ ﺟﺰﺀ ﺑﻪ ﺟﺰﺀ ﺩﺍﺭﱘ:‬

                                                                                                                 ‫∞‬

                            ‫‪S‬ﺭﺍ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ‬                                     ‫‪N‬‬           ‫∑ ﻣﻘﺪﺍﺭ‬      ‫1= ‪n‬‬
                                                                                                                             ‫2 ‪a m 2 + bn‬‬                 ‫ﺣﺎﻝ ﺑﺮﺍﻱ ﻧﺸﺎﻥ ﺩﺍﺩﻥ ﳘﮕﺮﺍﻳﻲ‬
                                                                                                         ‫2‬
                                                                                                                                                                           ‫ﻣﻲ ﻛﻨﻴﻢ.‬
                        ‫⎛‬               ‫‪N‬‬
                                                                                                  ‫⎞‬
      ‫‪S‬‬       ‫‪N‬‬       ‫⎜ =‬
                        ‫⎝‬
                                 ‫∑‬  ‫1= ‪n‬‬
                                                         ‫‪a‬‬    ‫‪n‬‬   ‫‪2 + b‬‬                 ‫‪n‬‬       ‫⎟ 2‬
                                                                                                  ‫⎠‬


                                                                                                    ‫ﺑﻪ ﺍﺯﺍﺀ ﻫﺮ ﻣﻘﺪﺍﺭ ‪ N‬ﳘﮕﺮﺍﺳﺖ:‬                                             ‫‪SN‬‬      ‫ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﻴﻢ ﻛﻪ‬
                                                                                                                                                                               ‫2‬
                                                                                            ‫2‬         ‫⎛‬                                                                ‫1‬
                                                                                                                                                                           ‫⎞‬
                     ‫⎛‬                                                                  ‫⎞‬             ‫⎜‬                  ‫‪⎛ α‬‬         ‫2‬
                                                                                                                                                      ‫‪β‬‬    ‫2‬
                                                                                                                                                                   ‫⎞‬       ‫⎟‬
                                ‫‪∑ (a‬‬                                       ‫)‬
                                    ‫‪N‬‬                                           ‫1‬                                ‫‪N‬‬                                                     ‫2‬
      ‫‪S‬‬    ‫‪N‬‬       ‫⎜ =‬
                     ‫⎜‬
                                                     ‫2‬
                                                     ‫‪n‬‬       ‫‪+ b‬‬      ‫‪n‬‬
                                                                       ‫2‬        ‫2‬       ‫⎟‬
                                                                                        ‫⎟‬           ‫⎜ =‬      ‫∑‬           ‫⎜‬
                                                                                                                         ‫‪⎜ n‬‬
                                                                                                                                     ‫‪n‬‬
                                                                                                                                     ‫2‬
                                                                                                                                          ‫+‬               ‫‪n‬‬
                                                                                                                                                          ‫2‬
                                                                                                                                                                   ‫⎟‬
                                                                                                                                                                   ‫⎟‬       ‫⎟‬       ‫=‬
                     ‫⎝‬          ‫1= ‪n‬‬                                                    ‫⎠‬             ‫⎜‬      ‫1= ‪n‬‬        ‫⎝‬                            ‫‪n‬‬            ‫⎠‬       ‫⎟‬
                                                                                                      ‫⎝‬                                                                    ‫⎠‬

                                                                                                         ‫2‬
                    ‫‪⎛ N‬‬                                                                              ‫⎞‬         ‫‪⎛ N‬‬      ‫‪⎞⎛ N‬‬
                  ‫2‪= ⎜ ∑ ( 1 2 (a n + b n‬‬                                                   ‫)‬                ‫‪< ⎜ ∑ 1 2 ⎟ ⎜ ∑ (α‬‬                                                              ‫⎞)‬
                                                                                                ‫1‬
      ‫‪S‬‬   ‫‪N‬‬         ‫⎜‬
                                 ‫2‬                                                              ‫2‬   ‫⎟)‬
                                                                                                     ‫⎟‬
                                                                                                                                                                           ‫2‬
                                                                                                                                                                           ‫‪n‬‬    ‫‪+ β‬‬      ‫2‬
                                                                                                                                                                                         ‫‪n‬‬    ‫⎟‬
                    ‫1= ‪⎝ n‬‬ ‫‪n‬‬                                                                         ‫⎠‬         ‫1= ‪⎝ n =1 n ⎠ ⎝ n‬‬                                                              ‫⎠‬
                                                                                                                                      ‫ﺣﺎﻝ ﺑﺎﻳﺪ ﺍﻳﻦ ﻧﺎﻣﺴﺎﻭﻱ ﺗﻮﺟﻴﻪ ﺷﻮﺩ.‬
                                                                                                                               ‫) 2‪(α n2 + β n‬‬
                                                                                                                             ‫‪N‬‬                           ‫‪N‬‬

                                                                                                                             ‫∑‬                          ‫2 ‪∑1 1 n‬‬
                                                                                                                                                        ‫=‪n‬‬
                                                                                                                             ‫1= ‪n‬‬

‫٠٣‬
                   ‫ﻫﻢ ﻃﺒﻖ ١ﻭ٢ ﳘﮕﺮﺍ ﻫﺴﺘﻨﺪ.ﺍﻟﺒﺘﻪ ‪‬ﺘﺮ ﺍﺳﺖ ﺑﮕﻮﺋﻴﻢ‬                                                                     ‫ﳘﮕﺮﺍﺳﺖ ﻭ‬             ‫ﺍﻣﺎ‬

                                                                                          ‫1‬

                                                ‫‪∑ (α‬ﳘﮕﺮﺍﺳﺖ.‬                               ‫ﳏﺪﻭﺩ ﻫﺴﺘﻨﺪ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﻧﺰ ﳏﺪﻭﺩ ﻣﻲ ﺑﺎﺷﺪ.ﺳﭙﺲ ﺳﺮﻱ )‬
                                                            ‫‪N‬‬                                 ‫2‬
                                                                           ‫2‬
                                                                           ‫‪n‬‬   ‫‪+ β‬‬   ‫2‬
                                                                                     ‫‪n‬‬
                                                           ‫1= ‪n‬‬


         ‫ﻗﻀﻴﻪ ﭘﺎﺭﺳﻮﺍﻝ:ﻃﺒﻖ ﺍﻳﻦ ﻗﻀﻴﻪ ﺍﻧﺮﮊﻱ ﻣﺘﻮﺳﻂ )‪ f (x‬ﺩﺭ ﻓﺎﺻﻠﻪ ) ‪ (− π , π‬ﺑﺮ ﺣﺴﺐ ﺿﺮﺍﻳﺐ ﺑﺴﻂ‬
              ‫‪π‬‬                   ‫‪π‬‬
                                           ‫⎛‬                          ‫ﻓﻮﺭﻳﻪ ﻗﺎﺑﻞ ﳏﺎﺳﺒﻪ ﺍﺳﺖ.‬
                                                                                       ‫⎞‬
                  ‫1‬
                      ‫‪π‬‬   ‫‪∫π‬‬           ‫‪f‬‬    ‫2‬
                                                ‫‪(x )dx‬‬   ‫1 =‬
                                                               ‫‪π‬‬      ‫‪∫π‬‬           ‫+ ‪f ( x )⎜ o‬‬
                                                                                            ‫‪a‬‬
                                                                                          ‫2 ⎜‬              ‫∑‬          ‫‪am‬‬      ‫‪Cosbx + bn‬‬       ‫‪sin nx ⎟dx‬‬
                                                                                                                                                      ‫⎟‬
                          ‫−‬                                         ‫−‬                     ‫⎝‬                ‫1= ‪n‬‬                                       ‫⎠‬
                                                                               ‫‪η‬‬
                                                                                                             ‫⎛‬  ‫‪η‬‬ ‫∞‬                 ‫⎞‬
                                       ‫1 (‪f ( x )dx + ∑ an‬‬                           ‫‪f (x )Cos nx dx) + ∑ bn‬‬ ‫1⎜‬    ‫⎟ ‪f ( x )Sin cx dx‬‬
         ‫0‪a‬‬
     ‫=‬
              ‫2‬
                   ‫1 ×‬
                              ‫∫‪η‬‬                                           ‫∫‪η‬‬                                ‫∫‪⎜ η‬‬                   ‫⎟‬
                                                                               ‫‪−η‬‬                       ‫1= ‪n‬‬ ‫⎝‬  ‫‪−η‬‬                  ‫⎠‬

                               ‫‪η‬‬                                                      ‫‪η‬‬                                           ‫‪η‬‬
                                                 ‫+ ‪f (x)dx‬‬
                                                                                    ‫1=‪η f (x)Cosnxdxt+ n‬‬                      ‫‪η f (x)Sinnxdx‬‬
                                   ‫0‪a‬‬
               ‫1 =‬                                                      ‫1 ‪an‬‬                                               ‫1 ‪bn‬‬
                          ‫‪η‬‬                 ‫2‬
                                                                ‫1= ‪n‬‬
                               ‫‪η‬‬                                                      ‫‪η‬‬                                           ‫‪η‬‬



                          ‫2‬ ‫∞‬        ‫∞‬
                      ‫‪ao‬‬
              ‫=‬          ‫2 ‪+ ∑ an 2 + ∑ bn‬‬
                       ‫2‬   ‫1= ‪n‬‬     ‫1= ‪n‬‬

                                            ‫‪η‬‬                              ‫2‬
                          ‫1 ﭘﺲ‬             ‫) 2 ‪f ( x )dx = o + ∑ (an 2 + bn‬‬                                                           ‫)ﺍﺯ ﺧﻮﺍﺹ ﺑﺴﻂ ﻓﻮﺭﻳﻪ(‬
                                                            ‫∞‬
                                                      ‫‪a‬‬
                                       ‫∫‪η‬‬
                                                   ‫2‬

                                                       ‫2‬
                                                                                                                      ‫ﻣﺸﺘﻖ ﭘﺬﻳﺮﻱ ﻭ ﺍﻧﺘﮕﺮﺍﻝ ﺑﺴﻂ ﻓﻮﺭﻳﻪ:‬
                                        ‫‪−η‬‬                 ‫1= ‪n‬‬




                                           ‫ﺍﮔﺮ)‪ f(x‬ﺗﺎﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺩﺭ ﻓﺎﺻﻠﻪ ) ‪ (− η, η‬ﺑﺎﺷﺪ ﻭ ﺩﺍﺭﺍﻱ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺯﻳﺮ ﺑﺎﺷﺪ:‬
                                   ‫∞‬                            ‫∞‬
                          ‫‪+ ∑ a n Cos nx + ∑ bn Sin nx‬‬
              ‫0‪a‬‬
                      ‫2‬        ‫1= ‪n‬‬                            ‫1= ‪n‬‬

                                       ‫‪π‬‬                                                              ‫‪π‬‬
              ‫1 = ‪an‬‬
                               ‫‪π‬‬
                                   ‫‪−π‬‬
                                       ‫‪∫ f (x )Cos nx dx‬‬                       ‫1 = ‪, bn‬‬
                                                                                                  ‫‪π‬‬   ‫‪∫ f (x )Sin nx dx‬‬
                                                                                                      ‫‪−π‬‬



                                                                                                  ‫ﺁﻧﮕﺎﻩ ﺍﻧﺘﮕﺮﺍﻝ ﺗﺎﺑﻊ )‪ f (x‬ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﳏﺎﺳﺒﻪ ﻣﻲ ﺷﻮﺩ:‬

              ‫)) )1 −( ‪∫ f (x )du = 0 2 (x + π ) + ∑ 1π (a n Sin nx − bn (Cos nxt‬‬
              ‫‪π‬‬                                                            ‫∞‬
                          ‫‪a‬‬                                                   ‫1+ ‪n‬‬

              ‫‪−π‬‬                                                        ‫1= ‪n‬‬



                                                      ‫ﻳﻌﲏ ﺍﻧﺘﮕﺮﺍﻝ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﳘﻮﺍﺭﻩ ﻗﺎﺑﻞ ﳏﺎﺳﺒﻪ ﺍﺳﺖ.‬
          ‫ﺑﺮﺍﻱ ﻣﺸﺘﻖ ﭘﺬﻳﺮﻱ ﺑﺴﻞ ﻻﺯﻡ ﺍﺳﺖ ﻛﻪ )‪ f (x‬ﺩﺭ ﻓﺎﺻﻠﻪ ] ‪ [− π , π‬ﺗﻌﺮﻳﻒ ﺷﺪﻩ ﺑﺎﺷﺪ ﻭ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ:‬
            ‫) ‪f ′( x ) = f ( x‬‬

                                   ‫ﺁﻧﮕﺎﻩ ) ‪ f ′( x‬ﺩﺭ ﻧﻘﺎﻃﻲ ﻛﻪ ) ‪ f ′′( x‬ﻣﻮﺟﻮﺩ ﺍﺳﺖ ﺑﺎ ﻣﺸﻖ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺑﺮﺍﺑﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ:‬
          ‫) ‪f ′( x ) = ∑ (− n a n Sin nx + nbn Cos nx‬‬


‫١٣‬
       ‫) ‪y t = a 2 y xx ( x, t‬‬   ‫ﺩﺭ ﻓﺼﻞ ﻗﺒﻞ ﺣﻞ ﻣﻌﺎﺩﻻﺕ ﻻﭘﻼﺱ ﺑﻪ ﺭﻭﺵ ﺟﺪﺍﺳﺎﺯﻱ ﻣﺘﻐﲑﻫﺎ ﺁﻭﺭﺩﻩ ﺷﺪ‬
       ‫0 = ) ‪y (a, t ) = 0 → y (c, t‬‬          ‫) ‪y ( x, o ) = f ( x‬‬

                                                                   ‫ﻣﻌﺎﺩﻻﺕ ﺑﻪ ﻓﺮﻡ ﺯﻳﺮ ﺭﺍ ﺑﻪ ﺭﺍﺣﱵ ﻣﻲ ﺗﻮﺍﻥ ﺣﻞ ﻛﺮﺩ:‬
      ‫) ‪yt ( x, t ) = a 2uxx ( x, t‬‬

                                                        ‫ﺭﺍ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﲑﱘ:‬               ‫)‪f (x‬‬   ‫ﺑﺮﺍﻱ ﺣﻞ ﻣﻌﺎﺩﻻﺕ‬
     ‫) ‪u t ( x, t ) = a 2 u xx ( x, t‬‬

     ‫) ‪U ( x, t ) = u ( z , t ) + φ ( x‬‬

     ‫) ‪U t ( x, t ) = a 2 u xx ( x, t ) + h( x‬‬

     ‫) ‪u t ( x, t )a 2 (u xx ( x, t ) + φ ′′( x )) + h( x‬‬

     ‫1 − = ′′ ‪a 2φ ′′( x ) + h( x ) = 0 → φ‬‬             ‫) ‪h( x‬‬
                                                   ‫2‪a‬‬

                ‫1−‬
     ‫= ) ‪φ (x‬‬          ‫‪h(t )dtdv‬‬                      ‫) ‪u t ( x, t ) = a 2 u xx ( x, t‬‬
                ‫∫∫ 2 ‪a‬‬


                                                                           ‫ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺩﺭ ﺻﻔﺤﻪ )ﻧﻮﺳﺎﻥ ﺻﻔﺤﻪ(‬
      ‫]) +‪U tt (xyt) = a 2 [(Z xx )(xy +) + Z yy (xy‬‬

             ‫‪0< x < a‬‬           ‫‪0, < y <b‬‬        ‫0> +‬
                                                                                                             ‫ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ:‬
      ‫0 = ) ‪Z (0, y, t‬‬                      ‫0 = ) ‪Z (a, y, t‬‬                  ‫0 = ) ‪Z ( x,0, t‬‬
      ‫0 = ) ‪Z ( x, b, t‬‬                    ‫) ‪Z ( x, y,0) = f ( x, y‬‬
                                                                                                 ‫ﺭﻭﺵ :ﺟﺪﺍ ﺳﺎﺯﻱ ﻣﺘﻐﲑﻫﺎ:‬
      ‫) ‪Z ( x , y , t ) = X ( x ).Y ( y ).T ( t‬‬
                                                                      ‫′′ ‪T‬‬   ‫′′ ‪x‬‬   ‫′′ ‪y‬‬
       ‫‪XY T ′′ = a‬‬          ‫2‬
                                ‫‪( x ′′yt‬‬    ‫‪+ X Y ′′T‬‬       ‫⇒)‬         ‫2‬
                                                                           ‫=‬      ‫+‬
                                                                     ‫‪a T‬‬      ‫‪x‬‬      ‫‪y‬‬
                                                                                             ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﺍﺭﱘ:‬
       ‫′′ ‪T‬‬
      ‫‪a 2T‬‬
            ‫2‪= −λ‬‬                                                    ‫ﹰ‬                     ‫ﹰ‬
                                              ‫ﳘﺎﻧﻄﻮﺭ ﻛﻪ ﻗﺒﻼ ﮔﻔﺘﻴﻢ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺻﺮﻓﺎ ﺑﺮﺍﻱ 0 < ‪ λ‬ﺑﺮﻗﺮﺍﺭ ﺍﺳﺖ:‬

     ‫‪T ′′ = −a2λ2T‬‬
                                                                             ‫‪⎧+ η‬‬                    ‫2‬
      ‫) ‪X ′′ ( x‬‬   ‫′′ ‪y‬‬                                 ‫′′ ‪X‬‬   ‫′′ ‪− y‬‬        ‫⎪‬
                 ‫+‬      ‫→ 2‪= −λ‬‬                              ‫=‬        ‫0 ⎨ = 2‪− λ‬‬
       ‫) ‪x (x‬‬       ‫‪y‬‬                                   ‫‪X‬‬        ‫‪y‬‬           ‫‪⎪− η‬‬                    ‫2‬
                                                                             ‫⎩‬

‫٢٣‬
     ⎧ x (0 ) = 0               x (a ) = 0                                                   :‫ﭘﺲ ﺩﺍﺭﱘ‬
     ⎨
     ⎩ y (0 ) = 0               y (b ) = 0              ‫ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ‬
             :‫ﺣﻞ‬
     X ′′
          = −η 2 ⇒ X ′′( x ) = −η 2 X ⇒ X = BSinηx + ACosηx
     X
           X (a ) = 0 → A = 0

                                                                         nη
     X (a ) = 0 → o = BSin η a ⇒ η a = x η ⇒ η =
                                                                          a

                    (x )                      n η
     ⇒      X       n      =   Sin    =   (       ) x
                                               a

     y ′′                y ′′
     y
          − λ 2 = −η 2 ⇒
                         y
                              = λ 2 − η 2 ⇒ y ′′ = λ 2 − η 2 y       (        )

         .‫ ﻣﻨﻔﻲ ﺑﺎﺷﺪ‬λ2 − η 2 ‫ﭘﺲ ﺑﺮﺍﻱ ﺍﻳﻨﻜﻪ ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﺟﻮﺍﺏ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ،ﻣﻄﺎﺑﻖ ﺑﺎﻻ ﺑﺎﻳﺪ ﺿﺮﺍﻳﺐ‬
     y(0) = 0
     y(b) = 0
                (
     y ′′ = − n 2 λ 2 y    )
     y = c n 2 − λ2 y + DSin η 2 − λ2 y
                                                          2              2           2
                                   ⎛ mη ⎞     ⎛ nη ⎞  ⎛ mη ⎞
      η 2 − λ2 = mη b ⇒ η 2 − λ2 = ⎜    ⎟ ⇒λ =⎜
                                            2
                                                   ⎟ −⎜    ⎟
                                   ⎝ b ⎠      ⎝ a ⎠   ⎝ b ⎠

                        n2  m2
     ⇒ λ =η                − 2
                        a2  b


                ⎛ mη ⎞                                                       .‫ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﻨﺪ‬   η, λ   ‫ﭘﺲ‬
     γm   = Sin ⎜    ⎟y
                ⎝ b ⎠


                                     :‫2 − 3 = 3 ﻣﻄﻠﻮﺏ ﺍﺳﺖ‬λ
                                              Z                 Z2 = 1 − λ     z1 = 2λ ‫١-ﺍﮔﺮ‬
                                         3i    (‫ﺝ‬         1 + 3i (‫ﺏ‬                 −1 + i (‫ﺍﻟﻒ‬
                                                       :‫٢-ﻓﺮﻡ ﻗﻄﱯ ﺍﻋﺪﺍﺩ ﳐﺘﺼﺎﺕ ﺯﻳﺮ ﺭﺍ ﻧﺸﺎﻥ ﺩﻫﻴﺪ‬


٣٣
                                      ‫‪3i‬‬      ‫ﺝ(‬           ‫‪1 + 3i‬‬       ‫ﺏ(‬            ‫‪−1+ i‬‬   ‫ﺍﻟﻒ(‬
                                                  ‫‪i‬‬                 ‫‪π‬‬

               ‫ﻣﻄﻠﻮﺏ ﺍﺳﺖ:‬              ‫‪Z 3 = se‬‬   ‫‪u‬‬
                                                      ‫3 ‪, Z 2 = 2e‬‬      ‫‪, Z1 = 2eiπ‬‬      ‫٣-ﺍﮔﺮ‬
                   ‫ﺝ( 0 = 4 + 2 ‪Z 4 − 2 z‬‬                         ‫ﺏ(‬                 ‫ﺍﻟﻒ( 1‪Z‬‬
                                                        ‫2 ‪Z1 Z‬‬
                                                                                     ‫3‪Z‬‬
                                ‫٤-ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻗﺎﻧﻮﻥ ﺩﻭ ﻣﻘﺪﺍﺭ ‪ Cos3θ , Sin3θ‬ﺭﺍ ﺑﻴﺎﺑﻴﺪ‬
                                                        ‫٥-ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﺭﻭﺍﺑﻂ ﺯﻳﺮ ﺑﺮﻗﺮﺍﺭ ﺍﺳﺖ:‬
                                   ‫‪Im(iz ) = Re z‬‬    ‫ﺏ(‬                ‫ﺍﻟﻒ( ‪Re(iz ) = − Im z‬‬
                                   ‫‪Z + 3i = z − 3i‬‬    ‫ﺩ(‬                        ‫ﺝ( ‪iz = −i z‬‬
                                    ‫٦-ﳎﻤﻮﻋﻪ ﻧﻘﺎﻃﻲ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ ﻛﻪ ﺩﺭ ﺷﺮﺍﻳﻂ ﺯﻳﺮ ﺻﺪﻕ ﻛﻨﻨﺪ:‬
       ‫‪) Re(z − i ) = z‬ﺍﻟﻒ‬
        ‫‪) z − i = z + i‬ﺏ‬
                                     ‫٧-ﺍﮔﺮ 0 = 2 ‪ Z 1 Z‬ﺛﺎﺑﺖ ﻛﻨﻴﺪ ﺣﺪﺍﻗﻞ ﻳﻜﻲ ﺍﺯ ﺁ‪‬ﺎ ﺻﻔﺮ ﺍﺳﺖ.‬
                                                            ‫٨-ﺭﻳﺸﻪ ﻣﻌﺎﺩﻻﺕ ﺯﻳﺮ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ:‬
                                             ‫ﺏ( 0 = ‪ Z 3 + 3i‬ﺝ(‬              ‫ﺍﻟﻒ( 0 = 4 + 2 ‪Z‬‬
                                                   ‫2‬



               ‫٩-ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻫﺬﻟﻮﻱ 1 = 2 ‪ x 2 − y‬ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺼﻮﺕ 2 = 2− ‪ z 2 + z‬ﻧﻮﺷﺖ.‬
                                                     ‫٠١-ﺑﺎ ﺍﺳﺘﻘﺮﺍﺀ ﻓﺮﻣﻮﻝ ﭼﻨﺪ ﲨﻠﻪ ﺍﻱ ﻣﺘﻘﺎﺑﻞ‬


     ‫‪(1+ z)n =1+ n z + n(n −1) z2 + .....n(n −1)(n − 2) + .....(n − k +1) zx + .....+ zn‬‬
                   ‫‪i‬‬           ‫‪2i‬‬                          ‫‪ki‬‬

                                                  ‫ﺭﺍ ﻛﻪ ﺩﺭ ﺁﻥ ‪ n‬ﻋﺪﺩﻱ ﺻﺤﻴﺢ ﻭ ﻣﺜﺒﺖ ﺍﺳﺖ ﺑﻴﺎﺑﻴﺪ.‬
                                           ‫ﺭﺍ ﺑﻪ ﻃﺮﻳﻖ ﺑﺮﺩﺍﺭﻱ‬        ‫2 ‪z1 − z 2 , z 1 + z‬‬

                                                                                  ‫ﻣﺸﺨﺺ ﻛﻨﻴﺪ.‬
      ‫) 4 ,1( = 2 ‪) z 1 = = (− 3 ,1 ) z‬ﺍﻟﻒ‬
       ‫1‪) z1 = x1 + iy1 z2 = = x1−iy‬ﺏ‬
                                                                                   ‫٢١-ﻧﺸﺎﻥ ﺩﻫﻴﺪ:‬
          ‫()5 + ‪(2 z‬‬       ‫)‬
                       ‫5 + ‪2 − i = 3 2z‬‬


                                                       ‫‪Re z + Im z < 2 z‬‬           ‫٣١-ﺛﺎﺑﺖ ﻛﻨﻴﺪ:‬
                                                             ‫٤١-ﻳﻚ ﻣﻘﺪﺍﺭ ﺍﺯ ‪ argz‬ﺭﺍ ﺑﻴﺎﺑﻴﺪﻫﺮﮔﺎﻩ:‬
                                      ‫2−‬                                             ‫‪i‬‬
                               ‫=‪Z‬‬                                            ‫=‪Z‬‬
                                    ‫‪1 + 3i‬‬                                        ‫‪− 2 − 2i‬‬

‫٤٣‬
                                                                     ‫ﺏ(‬                                             ‫ﺍﻟﻒ(‬




                   ‫ﺍﺯ ﺧﻮﺍﺹ ﻣﺮﺑﻮﻁ ﺑﻪ ﻓﻀﺎﻱ ﺑﺮﺩﺍﺭﻱ ﺩﺍﺷﺘﻴﻢ ﻛﻪ ﺍﮔﺮ : ‪ g , f‬ﺩﻭ ﺑﺮﺩﺍﺭ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺑﺎﺷﻨﺪ‬
        ‫→‬                                  ‫→‬
        ‫) 3 ‪f = (a1 , a 2 , a‬‬              ‫) 3‪g : (b1 , b2 , b‬‬
                                                                                 ‫ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺧﻮﺍﺹ ﺯﻳﺮ ﺭﺍ ﺩﺍﺭﱘ:‬
       ‫→ →‬
                                                          ‫2‬
       ‫3‪f . g = a1b1 + a 2 b2 + a 3 b‬‬                 ‫‪f‬‬       ‫3 ‪= f . f = a12 + a 2 + a‬‬
                                                                                ‫2‬     ‫2‬




                                                               ‫(‬                                        ‫)‬
        ‫→ →‬        ‫→ →‬                         ‫→‬     ‫→‬                                                      ‫1‬
        ‫‪f . g = f . g .Cosθ‬‬                    ‫) 3‪f − g = (a1 − b1 ) + (a 2 − b2 ) + (a 3 − b‬‬
                                                                          ‫2‬               ‫2‬                     ‫2‬




                     ‫→ →‬
        ‫0 = ‪f + g ⇒ f .g‬‬                       ‫ﺑﺮدارﻧﺮﻣﺎل = ‪f = 1 ⇒ f‬‬

                                                   ‫ﺍﮔﺮ 1‪φ n ,........, φ 2 , φ‬ﳎﻤﻮﻋﻪ ﺑﺮﺩﺍﺭﻱ ﺍﺭﺗﻮﮔﻮﻧﺎﻝ ﺑﺎﺷﻨﺪ ﻳﻌﲏ:‬
                                                     ‫2‬
        ‫0 => ‪< φ i , φ J‬‬        ‫‪, < φ i φ i >= φ i‬‬
                                                                ‫‪φn‬‬              ‫‪φ φ‬‬
                                                                    ‫1 , 2 ,..........,‬
                    ‫ﺑﺮﺩﺍﺭﻫﺎﻱ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ ﻫﺴﺘﻨﺪ.‬             ‫‪ φ n‬ﳎﻤﻮﻋﻪ‬             ‫1‪φ 2 φ‬‬        ‫ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺑﺮﺩﺍﺭﻫﺎﻱ‬

                                                              ‫‪φ‬‬
      ‫ﺗﺮﺟﻴﺎ ﻻﺯﻡ ﺍﺳﺖ ﻛﻪ ﺑﺮﺩﺍﺭﻫﺎﻱ 1 ,........, ‪ φn‬ﺍﺯ ﻫﻢ ﻣﺴﺘﻘﻞ ﺑﺎﺷﻨﺪ .ﭼﻮﻥ ﺩﺭ ﻏﲑ ﺍﻳﻦ ﺻﻮﺭﺕ ﻳﻜﻲ‬
                                                                                      ‫ﹰ‬
                               ‫ﺭﺍ ﺑﺮ ﺣﺴﺐ ﺩﻳﮕﺮﻱ ﻣﻲ ﺗﻮﺍﻥ ﻧﻮﺷﺖ ﻭ ﺻﻔﺮ ﺩﺍﺧﻠﻲ ﺁ‪‬ﺎ ﺻﻔﺮ ﳔﻮﺍﻫﺪ ﺷﺪ.‬
     ‫ﻫﺮ ﺑﺮﺩﺍﺭ ﺩﺭ ﻓﻀﺎﻱ ﺍﳚﺎﺩ ﺷﺪﻩ ﺑﻮﺳﻴﻠﻪ ﺍﻳﻦ ﺑﺮﺩﺍﺭ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎ ﺗﺮﻛﻴﺐ ﺧﻄﻲ ﺍﺯ ﺍﻳﻦ ﺑﺮﺩﺍﺭﻫﺎ ﺑﺪﺳﺖ‬
       ‫‪f = c1φ1 + c 2φ 2 ,........., c nφ n‬‬                                     ‫ﺁﻭﺭﺩ.‬

                                                                              ‫ﳓﻮﻱ ﳏﺎﺳﺒﻪ ﺿﺮﺍﻳﺐ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺍﺳﺖ:‬
                                                     ‫1‬
       ‫= ‪C K < φ k φ k >=< f .φ k >⇒ C k‬‬                      ‫> ‪< f ,φk‬‬
                                                     ‫‪φk‬‬
                                                                   ‫ﺍﺳﺘﻨﺘﺎﺝ ﺑﺮﺍﻱ ﻓﻀﺎﻱ ﺗﻮﺍﺑﻊ:‬
     ‫ﺍﮔﺮ ‪ f , g‬ﺗﻮﺍﺑﻌﻲ ﺩﺭ ﻓﺎﺻﻠﻪ )‪ (a,b‬ﻭ ﺑﻪ ﻓﺮﻡ ﻗﻄﻌﻪ ﺍﻱ –ﭘﻴﻮﺳﺘﻪ ﺑﺎﺷﺪ)‪، (pwc‬ﭼﻨﺎﻧﭽﻪ ﻓﺎﺻﻠﻪ ﺭﺍ ﺑﻪ ‪∆x‬‬
                                                     ‫ﻫﺎﻱ ﻛﻮﭼﻚ ﺗﻘﺴﻴﻢ ﻛﻨﻴﻢ ،ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ:‬
        ‫) ‪f = ( f 1 , f 2 ,........, f n‬‬
        ‫) ‪g = ( g 1 , g 2 ,........, g n‬‬


‫٥٣‬
                                                         ‫ﺿﺮﺏ ﺑﺮﺩﺍﺭﻱ ﺭﺍ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ ﻛﻪ ‪:f , g‬‬
     ‫‪( f , g ) = ∑ f K g K ∆x K‬‬
                            ‫‪b‬‬
      ‫‪Lim( f , g ) = ∫ − f ( x )g ( x )dx‬‬
      ‫0 → ‪∆x‬‬
                            ‫‪a‬‬               ‫ﻛﻪ ﺍﻳﻦ ﺗﻌﺮﻳﻒ ﺿﺮﺏ ﺩﺍﺧﻠﻲ ﺑﺮﺍﻱ ﺗﻮﺍﺑﻊ ﻣﻲ ﺑﺎﺷﺪ.‬
      ‫ﻧﺎﺣﻴﻪ ‪ a ,b‬ﺭﺍ ﺿﺮﺏ ﺩﺍﺧﻠﻲ ﺭﻭﻱ ﺁﻥ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ‪ Fundamental interval‬ﻳﺎ ﺑﺎﺯﻩ ﻱ ﺍﺳﺎﺳﻲ‬
                                                                                ‫ﻣﻴﻨﺎﻣﻨﺪ.‬
                                            ‫ﺧﻮﺍﺹ ﺯﻳﺮ ﺑﺮﺍﻱ ﺿﺮﺏ ﺩﺍﺧﻠﻲ ﺗﻮﺍﺑﻊ ﺑﺮﻗﺮﺍﺭ ﺍﺳﺖ:‬
     ‫) ‪( f , g ) = (g , f‬‬
     ‫) ‪( f , g + h) = ( f , g ) + ( f , g‬‬
     ‫) ‪(cf , g ) = c( f , g‬‬
                            ‫‪⎛b‬‬                                ‫⎞‬
     ‫‪( f , f )= f ≥ 0 → f = ⎜∫ f‬‬                       ‫⎟ ‪( x )dx‬‬
                        ‫2‬                          ‫2‬
                            ‫⎜‬                                  ‫⎟‬
                            ‫‪⎝a‬‬                                ‫⎠‬
     ‫‪( f , g) = 0 ⇒ f + g‬‬
                    ‫‪b‬‬
      ‫‪f − g = ∫ ( f − g ) dx‬‬
                ‫2‬                   ‫2‬

                    ‫‪a‬‬



     ‫⇒1= )‪( f , g‬‬               ‫‪ f‬ﻧﺮﻣﺎﻝ ﻣﻲ ﺑﺎﺷﺪ‬

                                                                                                                     ‫ﻣﺜﺎﻝ:‬

      ‫) ‪< Ψi (x,Ψ j ( x )) >= 2 < Ψi Ψ j >= δ (i, j‬‬                    ‫‪0< x<c‬‬                 ‫‪Ψn ( x ) = Sin‬‬
                                                                                                               ‫‪nπx‬‬
                             ‫‪c‬‬
                                                                                                                ‫‪c‬‬
                                                                                            ‫2‬     ‫‪nπx‬‬
                                ‫ﺩﺳﺘﻪ ﺗﻮﺍﺑﻌﻲ ﺍﺯ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ ﻫﺴﺘﻨﺪ ﺯﻳﺮﺍ:‬              ‫= ‪φi‬‬        ‫‪Sin‬‬         ‫ﺣﺎﻝ ﺩﺳﺘﻪ ﺗﻮﺍﺑﻊ‬
                                                                                            ‫‪c‬‬      ‫‪c‬‬
                                                                                  ‫2‬     ‫‪nπx‬‬
                                        ‫1 = ) ‪φ 0 (x‬‬               ‫= ) ‪, φ n (x‬‬
          ‫ﺩﺳﺘﻪ ﺗﻮﺍﺑﻊ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ‬                          ‫‪c‬‬                         ‫‪c‬‬
                                                                                    ‫‪Cos‬‬
                                                                                         ‫‪c‬‬         ‫ﻣﺜﺎﻝ:ﺩﺳﺘﻪ ﺗﻮﺍﺑﻊ‬
                                                                                              ‫ﻫﺴﺘﻨﺪ. ‪0 < x < c‬‬
                                  ‫‪nπx‬‬                                    ‫‪nπx‬‬                       ‫ﻣﺜﺎﻝ:ﺩﺳﺘﻪ ﺗﻮﺍﺑﻊ‬
     ‫1 = ) ‪φ 0 (x‬‬           ‫‪Sin‬‬            ‫1 = 1− ‪, φ 2n‬‬             ‫‪Cos‬‬         ‫1 = 0‪, φ‬‬
                        ‫‪c‬‬          ‫‪c‬‬                               ‫‪c‬‬      ‫‪c‬‬                 ‫‪2c‬‬
                                                                            ‫ﺩﺳﺘﻪ ﺗﻮﺍﺑﻊ ﺍﺯ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ ﺩﺭ)‪ (-C,C‬ﻫﺴﺘﻨﺪ.‬
                                                              ‫ﺭﻭﺵ ﻛﻠﻲ ﻓﻮﺭﻳﻪ ﺑﺮﺍﻱ ﺑﺴﻂ ﺗﻮﺳﻂ ﺗﻮﺍﺑﻊ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ:‬
      ‫‪f = c 0φ 0 + c1φ1 + ........ + c k φ k + ........ + c nφ n‬‬
                                                                         ‫‪b‬‬
     ‫آﻪ‬        ‫> ‪C k =< f .φ k‬‬            ‫,.....,1,0 = ‪k‬‬       ‫‪C k = ∫ f ( x )φ k ( x )dx‬‬
                                                                         ‫‪a‬‬




‫٦٣‬
                                                                                                                     ‫ﻣﺜﺎﻝ ﺍﺯ ﳏﺎﺳﺒﻪ ﺿﺮﺍﻳﺐ:‬
                                 ‫1‬                    ‫1‬          ‫‪nπx‬‬      ‫1‬      ‫‪nπx‬‬
        ‫0‪f (x ) = C‬‬                      ‫1‪+ C‬‬             ‫‪Cos‬‬        ‫2‪+ C‬‬    ‫‪Sin‬‬     ‫.......... +‬
                                 ‫‪2c‬‬                   ‫2‬           ‫‪c‬‬        ‫2‬      ‫‪c‬‬

                             ‫‪nπ x‬‬                                                        ‫‪nπx‬‬
                                                                          ‫‪c‬‬
                                 ‫1‬   ‫1‬
       ‫.) ‪C k =< f ( x‬‬   ‫‪Cos‬‬      ‫=>‬                                      ‫‪∫ f (x )Cos‬‬        ‫‪dx‬‬
                       ‫‪c‬‬      ‫‪c‬‬       ‫‪c‬‬                                   ‫‪−c‬‬
                                                                                          ‫‪c‬‬
                                                                                         ‫‪nπx‬‬
                                                                          ‫‪c‬‬

                                                                          ‫‪∫ f (x )Cos‬‬
                                                          ‫‪ck‬‬
                                             ‫= ‪ak‬‬               ‫1=‬                           ‫‪dx‬‬           ‫ﻛﻪ ﳘﺎﻥ ﺿﺮﺍﻳﺐ ﺳﺮﻱ ﻓﻮﺭﻳﻪ ﺍﺳﺖ.‬
                                                           ‫‪c‬‬         ‫‪c‬‬                    ‫‪c‬‬
                                                                          ‫‪−c‬‬




                                                           ‫) ‪ φ n ( x ),....., φ 2 ( x ), φ1 ( x‬ﺗﻮﺍﺑﻊ ﺍﺯ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ ﺑﺎﺷﺪ،ﻳﻌﲏ:‬     ‫ﻗﻀﻴﻪ ﺍﮔﺮ‬
                                 ‫0⎧‬                             ‫‪k≠m‬‬
        ‫2 ⎨ => ) ‪< φ m ( x ).φ k ( x‬‬                                               ‫.....,2,1 = ‪k , m‬‬
                                 ‫1 = ‪⎩c k‬‬                       ‫‪k =m‬‬

                                     ‫) ‪φ n ( x ),.......φ1 ( x‬ﺑﺎﺷﺪ،‬                         ‫ﻭ ) ‪ φ (x‬ﻳﻚ ﺗﺮﻛﻴﺐ ﺧﻄﻲ ﺍﺯ ﺗﻮﺍﺑﻊ ﺍﺭﺗﻮﮔﻮﻧﺎﻝ‬
        ‫) ‪φ ( x ) = δ 1φ1 ( x ) + ....... + δ nφ n ( x‬‬

           ‫ﻭ ﻧﻴﺰ ﺍﮔﺮ )‪ f (x‬ﻳﻚ ﺗﺎﺑﻊ ﺩﺭ ﳘﺎﻥ ﺑﺎﺯﻩ ﻱ ﺗﻮﺍﺑﻊ ﺍﺭﺗﻮﮔﻮﻧﺎﻝ ﺑﺎﺷﺪ، ﺁﻧﮕﺎﻩ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ ‪‬ﺘﺮﻳﻦ‬
       ‫ﲣﻤﲔ ﺗﺎﺑﻊ )‪ f (x‬ﺗﻮﺳﻂ‪ φ k‬ﻫﺎ ﻛﻪ ‪ k=1,2,……..n‬ﺑﺎﺷﺪ ،ﺗﺎﺑﻌﻲ ﳘﺎﻧﻨﺪ ) ‪ φ ( x‬ﺍﺳﺖ ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ‬
                                                                              ‫ﺿﺮﺍﻳﺐ ﳘﺎﻥ ﺿﺮﺍﻳﺐ ﻓﻮﺭﻳﻪ ﺑﺎﺷﺪ.‬
     ‫ﺍﺛﺒﺎﺕ:ﺍﮔﺮ ﻓﺮﺽ ﻛﻨﻴﻢ 1 ‪δ n ,.......δ 2 , δ‬ﺍﻋﺪﺍﺩ ﺩﳋﻮﺍﻫﻲ ﺑﺎﺷﺪ ﻭ ) ‪φ (x‬ﺗﺎﺑﻊ ﲣﻤﲔ )‪ f f (x‬ﺑﺎﺷـﺪ ﻻﺯﻡ‬
        ‫) ‪f (x ) − φ (x‬‬                                                 ‫ﺍﺳﺖ ﺗﺎ ﺍﻧﺪﺍﺯﻩ ﺧﻄﺎ ﻣﻴﻨﻴﻤﻢ ﺷﻮﺩ‬
                                                     ‫ﺍﻳﻦ ﻣﻘﺪﺭﺍ ﺑﺎﻧﺎﻡ ﺧﻄﺎ ﻳﻌﲏ ) ‪ f ( x ) − φ ( x‬ﻧﺎﻣﻴﺪﻩ1 ﻣﻲ ﺷﻮﺩ.‬
                       ‫⎞ ‪ − φ ( x ) = ⎛ ∫ f ( x ) − φ ( x )2 dx‬ﭘﺲ‬
                                        ‫‪b‬‬                         ‫2‬
               ‫)‪E = f (x‬‬              ‫⎜‬
                                                      ‫⎜‬
                                                                ‫⎟‬
                                                                              ‫⎟‬
                                                      ‫‪⎝a‬‬                      ‫⎠‬
         ‫) ‪E 2 = f (x ) − φ (x‬‬                    ‫‪= ∫ ( f ( x )) − φ ( x ) dx‬‬
                                              ‫2‬                           ‫2‬

         ‫‪b‬‬                           ‫‪b‬‬                     ‫‪b‬‬

        ‫∫‬    ‫‪f‬‬   ‫2‬
                     ‫‪( x )dx − 2∫ f ( x )dx + ∫ φ 2 ( x )dx‬‬
         ‫‪a‬‬                           ‫‪a‬‬                     ‫‪a‬‬

                                                                                                      ‫2‬
                                                              ‫‪⎛ n‬‬              ‫⎞‬
                                 ‫‪b‬‬                ‫‪n‬‬                            ‫‪b‬‬
        ‫) ‪= f (x‬‬            ‫‪− 2∫ f ( x )∑ δ k φ k ( x )dx + ∫ ⎜ ∑ δ k φ k ( x )⎟ dx‬‬
                        ‫2‬

                               ‫‪a‬‬        ‫1= ‪k‬‬                ‫1= ‪a ⎝ k‬‬           ‫⎠‬
                                              ‫‪n‬‬            ‫‪b‬‬                                  ‫‪b‬‬       ‫‪n‬‬          ‫‪n‬‬
         ‫‪= f ( x ) − 2 ∑ δ k ∫ f ( x ) k ( x )dx + ∫ ∑ ∑ δ k φ k ( x )δ lφ l ( x )dx‬‬
                             ‫2‬
                                     ‫‪φ‬‬
                                             ‫1= ‪k‬‬          ‫‪a‬‬                                  ‫1= ‪a k =1 k‬‬



     ‫ﺑﺮﺍﻱ ‪ Min‬ﻛﺮﺩﻥ 2 ‪ E‬ﺑﺎﻳﺪ ﲰﺖ ﺭﺍﺳﺖ ﺗﺎﺑﻊ ‪ Min‬ﺷﻮﺩ.ﺍﻧﭽﻪ ﻣﺘﻐﻴﲑ ﺍﺳﺖ ‪ δ k‬ﺍﺳﺖ .ﻳﻌـﲏ‬
      ‫) ‪ φ (x‬ﺳﺎﺧﺘﻪ ﺷﺪﻩ ﺑﺎ ﺿﺮﺍﻳﺐ ‪ δ x‬ﻭﻗﱵ ‪‬ﺘﺮﻳﻦ ﲣﻤﲔ ﺍﺳﺖ ﻛﻪ ﺿﺮﺍﻳﺐ ‪ δ k‬ﳘﺎﻥ ﺿﺮﺍﻳﺐ ﻓﻮﺭﻳﻪ‬


‫٧٣‬
                                                                      ‫(‬                           ‫)‬
                                         ‫‪n‬‬                       ‫‪nn‬‬                                        ‫‪n‬‬
        ‫‪⇒ E 2 ∑ δ k2 − 2δ k (δ k= ck ) +− ∑δ k2k2− 2ckδ k + ck − ∑ ck‬‬
        ‫‪f + = f + ∑ ck − f‬‬               ‫‪∑C‬‬
             ‫2‬               ‫2‬                             ‫22‬       ‫2‬                         ‫2‬

                                     ‫1= ‪k‬‬                         ‫=‪k‬‬
                                                                ‫1 1= ‪k‬‬                                    ‫1= ‪k‬‬
                                                                                                                        ‫ﺑﺎﺷﻨﺪ.‬

                                         ‫‪δ k = ck‬‬
                                   ‫‪n‬‬                ‫‪n‬‬
      ‫)1( 2‪= f (x ) − 2∑ δ K ck + ∑ δ k‬‬
                     ‫2‬

                                  ‫1= ‪k‬‬             ‫1= ‪k‬‬

                              ‫ﭘﺲ‬
      ‫> ) ‪ck =< f ( x )φk ( x‬‬

                                                     ‫‪n‬‬
                                                 ‫2‪− ∑ Ck‬‬
                                            ‫2‬
      ‫‪δ k = ck ⇒ E 2 = f‬‬
                                                    ‫1= ‪k‬‬

                                                                              ‫)ﺍﻧﺘﮕﺮﺍﻝ ﺍﺯ ﻳﻚ ﺗﺎﺑﻊ ﻣﺜﺒﺖ(ﭘﺲ:‬     ‫0 > 2‪E‬‬   ‫ﻭ ﭼﻮﻥ‬
                          ‫‪n‬‬
       ‫2‪f ( x ) ≥ ∑ Ck‬‬
                 ‫2‬
                                            ‫‪∀n ∈ N‬‬                         ‫)ﻟﺬﺍ ﻭﻗﱵ ﻛﻪ ∞ =‪(n‬‬
                         ‫1= ‪k‬‬


                                                                                 ‫ﺑﻪ ﻧﺎﻣﺴﺎﻭﻱ ﻓﻮﻕ ﻧﺎﻣﺴﺎﻭﻱ ﺑﺴﻞ ﮔﻮﻳﻨﺪ.‬
                                             ‫‪n‬‬
         ‫ﺑﻪ‬          ‫∑‬
      ‫2 ‪ ck‬ﺑﻪ ﺍﺯﺍﺀ ∞ → ‪ n‬ﲰﺖ‬                                   ‫ﻭﻗﱵ ∞ → ‪n‬ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ 0=‪ lim Cn‬ﺑﻪ ﻋﺒﺎﺭﰐ ﺳﺮﻱ‬
                                                                                         ‫) ‪ f ( x‬ﳘﮕﺮﺍ ﻣﻲ ﺷﻮﺩ. ﻳﻌﲏ:‬
                                            ‫1= ‪k‬‬                                                                   ‫2‬
                          ‫‪n‬‬
       ‫‪f (x ) = ∑ C‬‬
              ‫2‬                     ‫2‬
                                    ‫‪k‬‬
                         ‫1= ‪k‬‬


           ‫ﳎﻤﻮﻋﻪ }) ‪ {φ k ( x‬ﺭﻭﻱ )‪ (a,b‬ﺭﺍ ﻛﺎﻣﻞ ﮔﻮﺋﻴﻢ ﺍﮔﺮ ﺑﺮﺍﻱ ﻫﺮ ﺗﺎﺑﻊ ‪ f‬ﺑﺎ ﺩﺍﻣﻨﻪ )‪ (a,b‬ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ:‬
       ‫‪n‬‬

      ‫‪∑C‬‬
                              ‫2‬
             ‫2‬
             ‫‪k‬‬    ‫‪= f‬‬
      ‫1= ‪k‬‬
                                                                                                    ‫ﻳﻌﲏ ﺧﻄﺎ‬      ‫=0‬
                                             ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ﲣﻤﲔ ﺑﺎ ﳎﻤﻮﻋﻪ ﺍﻱ ﺍﺯ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ ﻛﺎﻣﻞ ﺑﻪ ﺧﻄﺎﻱ ﺻﻔﺮ ﻣﻲ ﺍﳒﺎﻣﺪ.‬
                                                                       ‫ﻣﺜﺎﻝ:ﳎﻤﻮﻋﻪ ﺍﻱ ﺗﻮﺍﺑﻊ ﺍﺭﺗﻮﻧﺮﻣﺎﻝ ﻫﻨﺪﺳﻲ:‬
                 ‫1‬                                   ‫1‬                               ‫1‬
     ‫= 0‪φ‬‬                ‫= ) ‪, φ 2 n −1 ( x‬‬                 ‫= ) ‪Cos nx , φ 2 n ( x‬‬       ‫‪Sin nx‬‬
              ‫‪2π‬‬                                        ‫‪π‬‬                            ‫‪π‬‬
                                                                                         ‫ﺍﻳﻦ ﺗﻮﺍﺑﻊ ﺭﻭﻱ ) ‪ (− π , π‬ﻛﺎﻣﻞ ﻫﺴﺘﻨﺪ.‬
                                                                                                    ‫‪ C‬ﻫﺎﻱ ﺯﻭﺝ:ﺿﺮﺍﻳﺐ ‪bn‬‬

                                                                                                    ‫‪ C‬ﻫﺎﻱ ﻓﺮﺩ :ﺿﺮﺍﻳﺐ ‪a n‬‬
                                                                                                   ‫ﺍﻧﻮﺍﻉ ﺩﻳﮕﺮ ﺍﺭﺗﻮﮔﻮﻧﺎﻟﻴﺴﱵ:‬
                                                          ‫) ‪ φ k ( x‬ﺩﺭ )‪ (a , b‬ﺑﺎ ﻭﺯﻥ )‪ p(x‬ﺍﺭﺗﻮﮔﻮﻧﺎﻝ ﺍﺳﺖ ﺍﮔﺮ‬      ‫ﮔﻮﺋﻴﻢ ﺗﻮﺍﺑﻊ‬
       ‫‪b‬‬
                                     ‫0⎧‬                                 ‫‪m≠k‬‬
       ‫∫‬ ‫⎨ = ‪P( x ) k ( x )φ m ( x )dx‬‬
              ‫‪φ‬‬
       ‫‪a‬‬                             ‫‪⎩cte‬‬                               ‫‪m=k‬‬




‫٨٣‬
                   ‫ﹰ‬
     ‫ﺍﻳﻦ ﺍﺭﺗﻮﮔﻮﻧﺎﻳﺴﱵ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﻣﻌﺎﺩﻝ ﺑﺎ ﺍﺭﺗﻮﮔﻮﻧﺎﻟﻴﺴﱵ ﺗﻮﺍﺑﻊ ) ‪ p( x )φ k ( x‬ﮔﺮﻓﺖ.ﻣﺜﻼ ﺗﻮﺍﺑﻊ ﺑﺴﻞ ﺑﻪ‬
        ‫ﺍﻳﻦ ﺻﻮﺭﺕ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﻧﺪ ﻭ ﺗﻮﺍﺑﻌﻲ ﻛﻪ ﺩﺍﺭﺍﻱ ﺩﻭﻣﺘﻐﻴﲑ ﻫﺴﺘﻨﺪ ﻧﻴﺰ ﻣﻲ ﺗﻮﺍﻥ ﲢﺖ ﺗﻮﺍﺑﻊ ﭘﺎﻳﻪ‬
            ‫) ‪φ mn ( x, y‬ﺑﺴﻂ ﺩﺍﺩ ﺑﻪ ﮔﻮﻧﻪ ﺍﻱ ﻛﻪ ﺿﺮﺏ ﺩﺍﺧﻠﻲ ﻣﺮﺑﻮﻁ ﺑﻪ ﺻﻮﺭﺕ ﺳﻄﺤﻲ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﺷﻮﺩ.‬

                                                                 ‫0 ⎧‬        ‫‪m ≠ e, n ≠ k‬‬
       ‫∫∫‬       ‫) ‪φ mn ( x, y‬‬           ‫⎨ = ‪φ ek ( x, y )dxdy‬‬
            ‫‪R‬‬                                                       ‫2‬
                                                                 ‫‪⎩c mn‬‬    ‫‪m = e, n = k‬‬


                                                          ‫ﺣﺎﻝ ﻫﺮ ﺗﺎﺑﻊ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺮ ﺣﺴﺐ ﺗﻮﺍﺑﻊ ) ‪ φ mn ( x, y‬ﺑﺴﻂ ﺩﺍﺩ.‬
                                                                      ‫ﲣﻤﲔ ﻭﻗﱵ ‪‬ﻴﻨﻪ ﺍﺳﺖ ﻛﻪ ﺿﺮﺍﻳﺐ ﳘﺎﻥ ﻓﻮﺭﻳﻪ ﺑﺎﺷﻨﺪ.‬
       ‫‪ω = u + iv‬‬                                               ‫ﺑﺮﺍﻱ ﺗﻮﺍﺑﻊ ﳐﺘﻠﻒ ﻧﻴﺰ ﻣﻲ ﺗﻮﺍﻥ ﺍﺭﺗﻮﮔﻮﻧﺎﻟﻴﺴﱵ ﺗﻌﺮﻳﻒ ﳕﻮﺩ.‬
        ‫‪b‬‬                          ‫‪b‬‬                ‫‪b‬‬

        ‫‪∫ ω (t )dt =∫ u(t )dt + i ∫ v(t )dx‬‬
        ‫‪a‬‬                          ‫‪a‬‬                ‫‪a‬‬


                 ‫) ‪ω (t ) = u (t ) − iv(t‬‬


                                                                         ‫ﺣﺎﻝ ﺿﺮﺏ ﺩﺍﺧﻠﻲ ﺭﺍ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ:‬
                                                                       ‫0 ⎧‬        ‫‪m≠k‬‬
                                                                       ‫⎪‬
                                              ‫‪b‬‬
      ‫) ‪< ω m (t )., ω k (t ) >= ∫ ω m (t )ω k (t‬‬
                                              ‫‪a‬‬
                                                                  ‫2 ⎨ = ‪dt‬‬
                                                                           ‫(‬              ‫)‬
                                                                         ‫‪u (t ) + v m (t ) dt‬‬
                                                                       ‫‪⎪∫ m‬‬
                                                                                    ‫2‬

                                                                       ‫⎩‬
                                                                                     ‫ﻛﻪ ﺁﻥ ﺭﺍ ﺿﺮﺏ ﺩﺍﺧﻠﻲ ﻫﺮﻣﺘﻴﻚ ﮔﻮﻳﻨﺪ.‬
                 ‫ﺍﻧﺘﮕﺮﺍﻝ ﻓﻮﺭﻳﻪ:ﺗﺎﲝﺎﻝ ﺩﺭ ﻣﻮﺭﺩ ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺑﺮﺍﻱ ﺗﻮﺍﺑﻊ ﭘﺮﻳﻮﺩﻳﻚ ﺩﺭ ﻓﺎﺻﻠﻪ )2,2-(ﺻﺤﺒﺖ‬
                                ‫ﻛﺮﺩﻩ ﺍﱘ.ﻧﺸﺎﻥ ﺩﺍﺩﱘ ﻛﻪ ﲣﻤﲔ ‪‬ﻴﻨﻪ ﺍﻱ ﺑﺮﺍﻱ ﺗﻮﺍﺑﻊ ﺩﺭ ﻓﻀﺎﻱ ﻫﻨﺪﺳﻲ ﻣﻮﭼﻮﺩ ﺍﺳﺖ.‬

                                       ‫∞‬
                                                        ‫‪nηx‬‬          ‫‪nηx‬‬
        ‫= )‪f (x‬‬                    ‫‪+ ∑ a n Cos‬‬
                      ‫0‪a‬‬
                                                            ‫‪+ bn Sin‬‬
                               ‫2‬       ‫1= ‪n‬‬              ‫‪c‬‬            ‫‪c‬‬
                           ‫‪c‬‬
                                                                                 ‫‪nηx‬‬
                         ‫‪f ( x )dx‬‬                                  ‫‪f ( x )Cos‬‬
                      ‫∫‪c‬‬                                         ‫∫‪c‬‬
        ‫1 = 0‪a‬‬                                          ‫1 = ‪an‬‬                       ‫‪dx‬‬
                           ‫‪−c‬‬
                                                                                  ‫‪c‬‬

                                              ‫‪nη x‬‬
                      ‫‪c‬‬
      ‫1 = ‪bn‬‬
                  ‫‪c‬‬   ‫‪∫ f (x )Sin‬‬
                      ‫‪−c‬‬
                                               ‫‪c‬‬
                                                   ‫‪dx‬‬



                                                                                                  ‫ﺑﺎ ﺟﺎﻳﮕﺬﺍﺭﻱ ﺩﺍﺭﱘ ﻛﻪ:‬
                         ‫∞‬
                             ‫⎡1‬           ‫‪nπs‬‬     ‫‪nπx‬‬                    ‫‪nπs‬‬     ‫⎤ ‪nπx‬‬
         ‫‪c‬‬                        ‫‪c‬‬                        ‫‪c‬‬
     ‫1‬
        ‫∫‬   ‫‪f ( s )ds + ∑ ⎢ ∫ f ( s ) cos‬‬     ‫‪cos‬‬     ‫‪ds + ∫ f ( s ) sin‬‬     ‫‪sin‬‬    ‫⎥ ‪ds‬‬
     ‫‪2c − c‬‬             ‫‪n =1 c ⎣ − c‬‬       ‫‪c‬‬       ‫‪c‬‬       ‫‪−c‬‬
                                                                          ‫‪c‬‬       ‫‪c‬‬    ‫⎦‬


‫٩٣‬
                                                 ‫∞‬
                                                                      ‫‪nη‬‬
                      ‫‪c‬‬                              ‫‪c‬‬
                           ‫1 + ‪f (s )ds‬‬                   ‫‪f (s )Cos‬‬      ‫‪(s − x )dx‬‬
               ‫∫ ‪2c‬‬                           ‫∫∑‪c‬‬
       ‫1=‬
                      ‫‪−c‬‬                       ‫1= ‪n‬‬  ‫‪−c‬‬
                                                                       ‫‪c‬‬

                                                                          ‫ﻛﻪ ﳕﺎﻳﺶ ﺟﺪﻳﺪﻱ ﺑﺮﺍﻱ ﺳﺮﻱ ﻓﻮﺭﻳﻪ ﺍﺳﺖ.‬
                                        ‫ﺣﺎﻝ ﭼﻨﺎﻧﭽﻪ )‪ f(x‬ﻳﻚ ﺗﺎﺑﻊ ﺍﺯ ) ∞ ,∞ −( ﺑﺎﺷﺪ ﻭ ﻣﻄﻠﻘﺎ ﺍﻧﺘﮕﺮﺍﻝ ﭘﺬﻳﺮ ﺑﺎﺷﺪ ﻳﻌﲏ:‬
                                                               ‫ﹰ‬
       ‫∞‬
                                                                 ‫ﳏﺪﻭﺩ ﺑﺎﺷﺪ:‬
       ‫‪∫ f (s ) ds‬‬
      ‫∞−‬

                                    ‫ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺑﺒﻴﻨﻴﻢ ﺩﺭ ﺷﺮﺍﻳﻂ ﻓﻮﻕ ،ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺑﻪ ﭼﻪ ﻓﺮﻡ ﺟﺪﻳﺪﻱ ﺗﺒﺪﻳﻞ ﻣﻲ ﺷﻮﺩ.‬
       ‫‪∆α = η‬‬                                                               ‫ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ :‬
                       ‫‪c‬‬
                                               ‫ﲢﺖ ﺷﺮﺍﻳﻂ ﺁﻭﺭﺩﻩ ﺷﺪﻩ ،ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺗﺒﺪﻳﻞ ﻣﻲ ﺷﻮﺩ:‬
               ‫∞+ ‪n‬‬                                                         ‫∞‬         ‫∞+‬
                           ‫1 = ‪f (s )Cos(n∆α ( x − s ))ds‬‬
           ‫∫ ∑‪c‬‬                                                           ‫‪η ∑ ∆α ∫ f (s )Cos(n∆α (x − s ))ds‬‬
       ‫1‬
            ‫1= ‪n‬‬   ‫∞−‬                                                       ‫1= ‪n‬‬      ‫∞−‬

               ‫∞‬

             ‫‪η ∑ f (n∆α )∆x‬‬
          ‫1‬
               ‫1= ‪n‬‬

                              ‫∞‬                            ‫∞ ∞‬
      ‫1‬        ‫1 =‬
                          ‫‪η‬‬   ‫1 = ‪∫ F (α , x )dα‬‬
                               ‫‪o‬‬
                                                          ‫∫‪η‬‬
                                                           ‫∞− 0‬
                                                               ‫‪∫ f (s )Cos(α (x − s ))dsdα‬‬
       ‫∞‬                                  ‫‪x‬‬                           ‫∞‬

      ‫∑‬       ‫) ‪f (n∆α )∆x = ∫ f ( x )dx = f ( x ) = ∫ ( A(α )Cos αx + B(α )Sin αx‬‬
      ‫1= ‪n‬‬                                ‫‪o‬‬                           ‫0‬




     ‫ﺑﺴﻂ ﻓﻮﺭﻳﻪ ﺑﻪ ﻓﺮﻡ ﻓﻮﻕ ﻛﻪ ﺑﺎ ﻧﺎﻡ ﺍﻧﺘﮕﺮﺍﻝ ﻓﻮﺭﻳﻪ ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ ﻣﻴﻞ‬                                       ‫∞→‪c‬‬   ‫ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﺭ ﺣﺎﻟﱵ ﻛﻪ‬
                              ‫∞ ∞‬                                                                                        ‫ﻣﻲ ﻛﻨﺪ:‬
     ‫1 = )‪f (x‬‬                     ‫‪∫ f (s )Cos α (s − x )dsdx‬‬
                      ‫∫‪η‬‬                                                  ‫∞<‪−∞< x‬‬
                              ‫∞− 0‬
                          ‫1‬        ‫∞+‬                                             ‫1‬       ‫∞+‬
       ‫= ) ‪A(α‬‬
                       ‫‪π‬‬      ‫∫‬‫∞−‬
                                        ‫‪f (u ) cosαudu‬‬                 ‫= ) ‪B (α‬‬
                                                                                  ‫‪π‬‬   ‫∫‬
                                                                                      ‫∞−‬
                                                                                               ‫‪sin αudu‬‬
                                               ‫) ‪B (α ), A(α‬‬          ‫‪ f ( x ) = e −2 x‬ﻣﻄﻠﻮﺏ ﺍﺳﺖ ﳏﺎﺳﺒﻪ ﺿﺮﺍﻳﺐ‬                  ‫ﻣﺜﺎﻝ:‬
                                   ‫∞‬                                                  ‫∞‬
       ‫1 = ) ‪A(α‬‬                   ‫‪∫ f (s )Cos αsds‬‬                   ‫1 = ) ‪B(α‬‬       ‫‪∫ f (s )Sin αsds‬‬
                              ‫‪η‬‬                                                   ‫‪η‬‬
                                   ‫∞−‬                                                 ‫∞−‬


                                                 ‫‪nηx‬‬          ‫‪nηx‬‬
       ‫‪bn n 4 < M → 4bn Sin‬‬                          ‫‪→ bn Sin‬‬     ‫‪≤ bn‬‬
                                                  ‫‪c‬‬            ‫‪c‬‬
                                       ‫∞‬
                               ‫‪M‬‬            ‫1‬
      ‫≤ ‪≤ ∑ bn‬‬                   ‫4‬
                                   ‫4 ∑‪= M‬‬
                               ‫‪n‬‬      ‫‪n =1 n‬‬




‫٠٤‬
                                                                                                            :‫ ﺭﺍ ﺑﻨﻮﻳﺴﻴﺪ‬e − x ‫ﺍﻧﺘﮕﺮﺍﻝ ﻓﻮﺭﻳﻪ‬
           ‫:ﺭﻭﺵ ﺣﻞ ﺩﻭﻡ‬f ( x ) = ∫ ( A(α )Cosx + B(α )Sin αx )dx
                                ∞                                                        ∞
      A(α ) = 1                     ∫ f (x )Cos αx dx              , B(α ) = 1           ∫ f (x )Sin αxdx
                           η                                                     η
                                −∞                                                   −∞

                                                                                             ∞
      f (x) = e                                                        B(α ) = 1
                                                                                     η ∫ e Sin αx dx
                           −x                                                             −x
                                               −∞<x<∞
                                                                                             −∞
                   ∞                                                             ∞                                                  ∞
     f ( x ) = ∫ ( A(α )Cos αx + B(α )Sin αx )dx Re ∫ e − x e iαx                                          B (α ) = 0, A(α ) = 2
                                                                                                                                   η ∫ e Cos αxdx
                                                                                                                                        −x

                   0                                                             o                                                  0



      A(α ) = 1
                           η∫e
                                        −x
                                              Cos αx dx




     = 2 × 1
               η                [
                           η α − α A(α ) × 2
                             1  1         η
                                                               ]

     ⇒ A(α ) = 2
                                    η( α
                                      1       2            )
                                                (1 − A(α )) ⇒ ηα ( A(α )) = 1 − A(α )η 2
                                                                2


                                                                            2

                                        2
     ⇒ A(α ) =
                            η (α 2 + 1)
                   ∞
                                    2
               =∫
          −x
                                              Cosαx dx ⇒
                       η (α 2 + 1)
     e
                   0


                           Cos               5 x                   R
      2
                   ∫
                                                                           − 5
                                                 dx    =               e
               R           1 +               x 2                   2

          Cos5α                         π
     ∫ 1+α         2
                           dα =
                                        2
                                             e−5 → α = x

                                                                                                      ∞
                                                                           .‫ ∫ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﻳﺪ‬Sin 2π x dx ‫ﲤﺮﻳﻦ :ﻣﻘﺪﺍﺭ ﺍﻧﺘﮕﺮﺍﻝ‬
                                                                                            1+ x      0
                                                                                                                2


                                                                                                    :‫ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺍﺳﺖ ﻛﻪ‬
              ⎧− e − x                         x>0                                       π
     f (x ) = ⎨    −x
                                                               ‫ﺟﻮاب‬                  =            e − 2π
              ⎩− e                             x<0                                           2

                   Cos 5 x      R −5
      2
          R    ∫   1+ x 2
                           dx =
                                2
                                  e → α = x

                       ∞                                                                           :‫ﻣﺴﺎﹶﻟﻪ ﺍﻧﺘﻘﺎﻝ ﺣﺮﺍﺭﺕ ﺯﻳﺮ ﺭﺍ ﺣﻞ ﻛﻨﻴﺪ‬
     f ( x0 ) = ∫ A(α )Cosαx0 + B(α )Sinαx0 dα
                       0



٤١
     ‫) ‪u t ( x, t ) = ku xx ( x, t‬‬
     ‫0 = ) ‪u(0, t‬‬               ‫)‪u(x, o) = F (x‬‬                ‫0>‪x‬‬           ‫0> ‪t‬‬
     ‫ﳘﺎﻧﻄﻮﺭ ﻛﻪ ﻣﺸﺎﻫﺪﻩ ﻣﻲ ﺷﻮﺩ ﺣﻮﺯﻩ ﺗﻐﻴﲑﺍﺕ ‪ x‬ﺗﺎ ﰊ ‪‬ﺎﻳﺖ ﺍﺳﺖ. ﺑﻨﺎﺑﺮﺍﻳﻦ ﺭﻭﺵ ﺣﻞ ﺍﺯ ﺗﺒﺪﻳﻞ ﺩﺭ‬
                                  ‫ﺳﺮﻱ ﻓﻮﺭﻳﻪ ﻛﻪ ﳘﺎﻥ ﺍﻧﺘﮕﺮﺍﻝ ﻓﻮﺭﻳﻪ ﻣﻲ ﺑﺎﺷﺪ ﺑﺪﺳﺖ ﺧﻮﺍﻫﺪ ﺁﻣﺪ.‬


                                                                            ‫2‪⎧ x‬‬
                                               ‫= ) ‪) T (t‬‬    ‫) ‪X ′′ ( x‬‬
                                                   ‫′‬                        ‫⎪‬
     ‫→ ‪u ( x , t )X ( x )T (t‬‬                                             ‫غ. ق. ق ⇒ 0 ⎨ =‬
                                                 ‫) ‪kt (t‬‬     ‫) ‪X (x‬‬         ‫2 ‪⎪− λ‬‬
                                                                            ‫⎩‬
     ‫0 = ) 0( ‪U ( x1 ) = 0 → X‬‬
      ‫‪X‬‬       ‫= ) ‪(x‬‬       ‫(‪A cos‬‬         ‫(‪λ x ) + B sin‬‬         ‫)‪λx‬‬

      ‫‪X ( x ) = sin λx‬‬

     ‫‪T (t ) = e − λ zkt‬‬

                           ‫∞‬                                         ‫∞‬
     ‫‪U‬‬       ‫∫ = ) ‪(x , t‬‬      ‫‪(Bλe‬‬     ‫, ‪− kλ‬‬
                                             ‫+‬   ‫⇒ ‪Sin λ x ) dx‬‬      ‫‪∫ B (λ )sin‬‬    ‫) ‪λ xd λ = f ( x‬‬   ‫ﺍﻧﺘﮕﺮﺍﻝ ﻓﻮﺭﻳﻪ‬
                           ‫0‬                                         ‫∞−‬


                                    ‫∞‬
      ‫2 = ) ‪B (λ‬‬                    ‫∫‬      ‫‪f‬‬     ‫‪( x ) sin‬‬   ‫‪λ xdx‬‬
                                ‫‪R‬‬
                                    ‫∞ −‬

                                                                                ‫ﻣﻌﺮﰲ ﺗﻮﺍﺑﻊ ﺑﺴﻞ ﻭ ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ﺁﻬﻧﺎ:‬
                                                                     ‫ﺩﺭﻓﺮﻡ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺩﺍﺭﱘ :‬
                       ‫0 = ‪+ (λp 2 − v 2 )y‬‬
            ‫2 ‪du‬‬    ‫‪du‬‬
      ‫‪∆u= p‬‬
         ‫2‬
               ‫2‬
                 ‫‪+p‬‬    ‫2‬

            ‫‪dp‬‬      ‫‪dp‬‬
                                                 ‫= ‪ x‬ﺗﻐﻴﲑ ﻣﺘﻐﻴﲑ ﺩﻫﻴﻢ ﺩﺍﺭﱘ:‬     ‫‪λp‬‬     ‫ﭼﻨﺎﻥ ﭼﻪ ﺩﺭ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺑﻪ ﺟﺎﻱ‬
      ‫‪x × ∂ u . ∂x‬‬
           ‫2‬       ‫2‬

       ‫2 ‪λ ∂p 2 ∂p‬‬

      ‫‪∂ 2 u ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ∂x ⎞ ex‬‬
     ‫= 2 ⇒‬     ‫=⎟ ⎜‬      ‫.⎟ . ⎜‬
      ‫‪∂p‬‬    ‫‪∂p ⎜ ∂p ⎟ ∂p ⎜ ∂x ∂p ⎟ ep‬‬
               ‫⎠ ⎝‬       ‫⎝‬       ‫⎠‬


                                                       ‫ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺑﻪ ﻧﺎﻡ ﻣﻌﺎﺩﻟﻪ ﺑﺴﻞ ﻣﻲ ﺷﻨﺎﺳﻴﻢ:‬
          ‫⎛ ∂‬    ‫⎞ ‪∂u‬‬     ‫‪∂ 2u‬‬
     ‫=‬       ‫‪⎜ λ‬‬    ‫2 ‪⎟ λ =λ‬‬
          ‫⎝ ‪∂x‬‬   ‫⎠ ‪∂x‬‬     ‫‪∂x‬‬

                                     ‫0 = ‪+ (x 2 − v 2 )y‬‬
             ‫2‪x‬‬        ‫‪d2y x‬‬      ‫‪dy‬‬
     ‫⇒‬            ‫‪.λ‬‬        ‫‪+ . λ‬‬
              ‫‪λ‬‬        ‫‪dx‬‬ ‫2‬
                             ‫‪λ‬‬    ‫‪dx‬‬

‫٢٤‬
                                                                      ‫ﺭﻭﺵ ﺣﻞ ﺍﻳﻦ ﮔﻮﻧﻪ ﻣﻌﺎﺩﻻﺕ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺳﺮﻳﻬﺎ ﺑﻮﺩ.‬
                                                                                       ‫ﺳﺮﻱ ﭘﻴﺸﻨﻬﺎﺩﻱ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺍﺳﺖ:‬
                                    ‫∞‬
           ‫‪y(x ) = x r ∑ ak x k‬‬
                                        ‫(‬            ‫)‬
           ‫2‬
        ‫‪2 d y‬‬      ‫‪dy‬‬
     ‫‪⇒x‬‬      ‫2‬
               ‫0 = ‪+ x k+ x2 − v2 y‬‬
                       ‫0=‬
          ‫‪dx‬‬       ‫‪dx‬‬


      ‫ﻋﻠﺖ ﭘﻴﺸﻨﻬﺎﺩ ﺍﻳﻦ ﺍﻳﻦ ﺟﻮﺍﺏ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻧﻘﻄﻪ ﻱ 0 = ‪ x‬ﻳﻚ ﻧﻘﻄﻪ ﺗﻜﲔ ﻣﻨﻈﻢ ﺍﺳﺖ‬
                                                 ‫2‪x2 − v‬‬                        ‫ﭼﻮﻥ :‬
      ‫2‬          ‫2‬    ‫2‬
                               ‫(‬
     ‫0 − ‪x y ′′ + xy ′ + x − v y‬‬        ‫1‬   ‫)‬                     ‫+ ′′ ‪y‬‬           ‫+ ′‪y‬‬              ‫0=‪y‬‬
                                                                           ‫‪x‬‬                    ‫2‪x‬‬

                                                                      ‫ﻛﻪ ﻣﺸﺎﻫﺪﻩ ﻣﻲ ﺷﻮﺩ ﻧﻘﻄﻪ 0=‪ x‬ﻳﻚ ﻧﻘﻄﻪ ﺗﻜﲔ ﺍﺳﺖ.‬
               ‫∞‬                                                                      ‫∞‬
     ‫‪x‬‬    ‫2‬
              ‫‪∑ (r + k )(r + k − 1)a‬‬
              ‫0= ‪k‬‬
                                                         ‫‪k‬‬   ‫‪x‬‬   ‫2− ‪r +k‬‬
                                                                           ‫)1− ‪+ x∑ (r + k )a k x (r + k‬‬
                                                                                     ‫0= ‪k‬‬


      ‫(‬                   ‫‪)∑ (r + k )(r + k − 1)a‬‬
                           ‫∞‬
     ‫2‪t x2 + v‬‬                                               ‫‪k‬‬   ‫0 = ‪x r +k‬‬
                          ‫0= ‪k‬‬



                                                                                          ‫,.......,2,1 = ‪n‬‬        ‫‪v=n‬‬    ‫ﺩﺭ ﺣﺎﻟﱵ ﻛﻪ‬

     ‫‪∑ ((r + k )(r + k − 1)av x‬‬                  ‫‪r +k‬‬
                                                                                                              ‫)‬
                                                             ‫0 = 2 + ‪+ (r + k )akx r + k − n 2 akx r + k + ∑ akx r + k‬‬



     ‫0 = ) ‪∑ x ((r + k )(r + k − 1) + (r + k ) − n ak + akx‬‬
      ‫∞‬
               ‫‪r +k‬‬                                                    ‫2‬                    ‫2‬

     ‫0= ‪k‬‬




               ‫ﺩﺭ ‪‬ﺎﻳﺖ‬
     ‫0 = ‪: (r − n)(r + n)a0 + (r − n +1)(r + n +1)a1x + ∑((r − n + k )(r + n + k )ak + ak − 2)xk‬‬

     ‫0‪⇒ a‬‬                 ‫ﻳﺎ‬       ‫0 = )‪(r + n)(r + n‬‬

     ‫0 = )1 + ‪(r − n + 1)(r − n‬‬                 ‫ﻳﺎ‬               ‫0 = 1‪a‬‬

                                         ‫1 −‬
     ‫‪∗ ⇒ a‬‬                ‫=‬                                                ‫‪a‬‬        ‫2 −‬
                      ‫‪k‬‬
                               ‫‪(r − n + k )(r + n + k‬‬                  ‫)‬       ‫‪k‬‬



                                                                       ‫1−‬
              ‫‪ ∗ ⇒ a‬ﻣﻲ ﺗﻮﺍﻥ ﲤﺎﻡ ﺭﻭﺍﺑﻂ‬                    ‫=‬                           ‫2− ‪a‬‬                      ‫ﺩﺭ ﻧﺘﻴﺠﻪ ﺗﻮﺳﻂ ﺭﺍﺑﻄﻪ ﺑﺎﺯﮔﺸﱵ‬
                                                     ‫‪k‬‬
                                                             ‫‪(r − n + k )(r + n + k ) k‬‬
                                                                                                             ‫ﺳﺮﻱ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﺩ.ﭼﻮﻥ:‬
                                             ‫1−‬
     ‫= ‪r = n ⇒ ak‬‬                                          ‫2− ‪a‬‬
                                   ‫‪(r − n + k )(r + n + k ) k‬‬
‫٣٤‬
                                                                                   ‫ﻛﻪ ﺑﺮﺍﻱ ‪ k‬ﻫﺎﻱ ﺯﻭﺝ ﻋﺪﺩ ﻏﲑ ﺻﻔﺮ ﻣﻲ ﺑﺎﺷﺪ.‬
      ‫0 = 1 + ‪a1 , a3, a2m‬‬                                                                                  ‫ﺍﻣﺎ 0 ≠ 0 ‪ a‬ﻭ‬
                      ‫1−‬                             ‫1‬                                             ‫ﺩﺭ ‪‬ﺎﻳﺖ ﻓﺮﻡ ﻛﻠﻲ ﺟﻮﺍﺏ:‬
     ‫= 2‪a‬‬                     ‫= 4‪a0 , a‬‬
                 ‫) 2 + ‪2 (2 n‬‬            ‫) 2 + ‪4 (2 n + 2 )(2 )(2 n‬‬
     ‫‪a 2k‬‬       ‫=‬
                                 ‫‪(− 1 )k‬‬
                   ‫‪ki (n + 1 )(n + 2 )(n + k )2 2 k‬‬

                  ‫ﺑﺮﺍﻱ ﺍﻳﻨﻜﻪ ﺑﺘﻮﺍﻥ ﻓﺮﻡ ﺳﺎﺩﻩ ﺷﺪﻩ ﺍﻱ ﺑﺮﺍﻱ ﺟﻮﺍﺏ ﭘﻴﺸﻨﻬﺎﺩ ﻛﺮﺩ ،ﻣﻘﺪﺍﺭ 0‪a‬ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ‬
      ‫0‪a‬‬        ‫=‬
                    ‫1‬
                                          ‫‪a‬‬   ‫=‬
                                                       ‫) 1 −(‬                 ‫ﺍﻧﺘﺨﺎﺏ ﻣﻲ ﻛﻨﻴﻢ:‬
                                                                                          ‫‪k‬‬


                                                                       ‫2 ) ‪k ! (n + k‬‬
                                                                 ‫‪2 k‬‬                            ‫‪n + 2 k‬‬
                    ‫‪n!2n‬‬                                                           ‫!‬

                                        ‫∞‬
                                              ‫‪(− 1 )k‬‬                     ‫) 2 ‪(x‬‬      ‫‪n + 2 k‬‬
                                                                                                          ‫ﭘﺲ ﺟﻮﺍﺏ ﺑﻪ ﻓﺮﻡ ﺯﻳﺮ ﺍﺳﺖ :‬
                                       ‫!) ‪∑ 0 k ! (n + k‬‬
                                       ‫= ‪k‬‬




                                                                                                  ‫ﺑﻪ ﻣﻘﺪﺍﺭ ﺍﻳﻦ ﺟﻮﺍﺏ )‪ yn (x‬ﮔﻮﻳﻨﺪ.‬
                                         ‫‪(− 1 )k‬‬
                                  ‫‪∑ 0 k ! (n + k )! (x‬‬                               ‫)‬
                                   ‫∞‬                                                   ‫‪n + 2 k‬‬
       ‫‪y‬‬        ‫‪n‬‬   ‫= ) ‪(x‬‬                                                         ‫2‬
                                  ‫= ‪k‬‬


                                                                                    ‫)‪ yn (x‬ﺭﺍ ﺗﺎﺑﻊ ﻣﺮﺗﺒﻪ ‪ n‬ﺍﻡ ﺑﺴﻞ ﻣﻲ ﻧﺎﻣﻨﺪ.‬
                                                                               ‫ﺍﻳﻦ ﺗﺎﺑﻊ ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻣﺮﺗﺒﻪ ‪ n‬ﻣﻲ ﺑﺎﺷﺪ.‬
                                        ‫(‬
       ‫) ‪x 2 y ′′ + x y ′ + x 2 − n 2 y = 0 → y = y n ( x‬‬        ‫)‬
                                                                                          ‫ﺑﺮﺍﻱ ﻣﺜﺎﻝ:ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﺑﺴﻞ ﻣﺮﺗﺒﻪ ﺻﻔﺮ:‬

      ‫= ) ‪x 2 y ′′ + x y ′ + x 2 y = 0 → y = y 0 ( x‬‬
                                                                           ‫∞‬
                                                                             ‫‪(− 1)k‬‬      ‫) 2 ‪(x‬‬    ‫‪2k‬‬
                                                                          ‫2)! ‪∑0 (k‬‬
                                                                          ‫=‪k‬‬


                                                                                              ‫ﺷﻜﻞ ) ‪y (x‬ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﻲ ﺑﺎﺷﺪ :‬    ‫0‬



                                                                                                     ‫ﺍﻧﺪﺍﺧﱳ ﺩﺭ ﺁﺏ ﳕﺎﺩ 0‪ y‬ﺍﺳﺖ.‬
       ‫‪y‬‬    ‫0‬   ‫= ) ‪(x‬‬      ‫− 1‬   ‫) 2 ‪(x‬‬    ‫2‬
                                                ‫+‬
                                                      ‫1‬
                                                    ‫) ! 2(‬   ‫2‬
                                                                 ‫) 2 ‪(x‬‬‫4‬
                                                                               ‫+‬   ‫) 2 ‪( 1 3 ! ) (x‬‬
                                                                                          ‫2‬               ‫6‬
                                                                                                              ‫..........‬   ‫.‬




                           ‫ﺍﺳﺖ.‬                 ‫ﺩﺭ ﻫﺮﻧﻘﻄﻪ ﺗﺎﺑﻊ‬                                                             ‫ﻗﻀﻴﻪ ﺟﻮﺍﺏ ﻣﻌﺎﺩﻟﻪ ﳘﮕﻦ‬


                                                                                                                           ‫ﺧﺎﺻﻴﺖ ﺗﺎﺑﻊ ﺑﺴﻞ:‬




‫٤٤‬
                                                                                                                                                :‫ﺧﻮﺍﺹ ﺗﻮﺍﺑﻊ ﺑﺴﻞ‬
                                                                                                        (1                            )y
                                                                                               ∞
      1 ) Cos          ( xSin            θ       )=            y      o   (x ) + ∑                               +   (−      1   )n        n   ( x )Cos    φ
                                                                                           n =1


                                                                       (1                            )y
                                                              ∞
     2 ) Sin          ( xSin         θ       )= ∑                             −   (−       1   )n            n   ( x )Sin φ
                                                             n =1

                                                                              R

      ‫ﭘﺲ ﺩﺍﺭﱘ‬                            n (x ) =                             ∫            ( xSin φ )Sin (2 x                         − 1 )φ d φ
                                 y2                                1              Sin
                                                                          R
                                                                              0


                                                                                       R

      ٣)‫ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ‬                              y   n   (x ) =               1
                                                                                  R   ∫    Cos          (n φ         − xSin           φ )d φ
                                                                                       0

                                                                                                                                                   ‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ‬
                                             − 1k
                                                                                  ( )
                                 ∞                                                             n+2k
        y n (x ) =              ∑                      x
                                k=0      K ! (x + k )!   2

                                                                                 − 1k
                                                                                                          ( )
                                                                  ∞
                           (x ) =
                                                                                                                       2 k
        x   − n
                  y    n
                                             1
                                                 2       n    ∑k = 0        k ! (n + k )!
                                                                                          x
                                                                                            2



                       x yn ( x) = −nx                                    yn ( x) + x
                                                                                                    ∞
                                                                                                         (− 1)k (n + 2x) .(x )n−2k −1
        4) d
                  dx
                           −n                                 − n −1                           −n
                                                                                                    ∑ k!(n + k )! 2
                                                                                                    k =0
                                                                                                                            2

        − x − n yn+1 ( x )                   ‫ﭘﺲ‬                   y′(0) = − y1 ( x )




                                                                                                                      .‫ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﻲ ﺑﺎﺷﺪ‬y0(x) ‫ﺗﺎﺑﻊ‬
                  2      x                 x4   x4      1     x6       1 1
     Y0 ( x) =      [(log − ν ) J 0 ( x)] + 2 − 2 2 (1 + ) + 2 2 2 (1 + + ) + ....
                  π      3                 x   x 4      2   x 46       2 3



                                      R (v + 1 )
                  R (V          )=
                                          v
       V → −1t                       V → −1t
       V 3  ( 2 )=              n
                                     2
                                          = min


                                                                                                                                                :‫ﺧﻮﺍﺹ ﺗﻮﺍﺑﻊ ﺑﺴﻞ‬
                                                                                                        (1                            )y
                                                                                               ∞
      1 ) Cos          ( xSin            θ       )=            y      o   (x ) + ∑                               +   (−      1   )n        n   ( x )Cos    φ
                                                                                           n =1


                                                                       (1                            )y
                                                              ∞
     2 ) Sin          ( xSin         θ       )= ∑                             −   (−       1   )n            n   ( x )Sin φ
                                                             n =1



٤٥
                                                                  R

     ‫ﭘﺲ ﺩﺍﺭﱘ‬                         n (x ) =                     ∫           ( xSin φ )Sin (2 x        − 1 )φ d φ
                                y2                    1               Sin
                                                             R
                                                                  0


                                                                          R

     ٣)‫ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ‬                      y   n   (x ) =            1
                                                                      R   ∫   Cos       (n φ   − xSin   φ )d φ
                                                                          0

                                                                                                                     ‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ‬
                                         − 1k
                                                                  ( )
                               ∞                                              n+2k
       y n (x ) =              ∑                   x
                               k=0   K ! (x + k )!   2

                                                                   − 1k
                                                                                         ( )
                                                      ∞
                          (x ) = 1
                                                                                                 2 k
       x   − n
                 y    n
                                        2       n   ∑
                                                    k = 0     k ! (n + k )!
                                                                            x
                                                                              2



                      x yn ( x) = −nx                        yn ( x) + x
                                                                                    ∞
                                                                                        (− 1)k (n + 2x) .(x )n−2k −1
       4) d
                 dx
                          −n                        − n −1                    −n
                                                                                   ∑ k!(n + k )! 2
                                                                                   k =0
                                                                                                           2

       − x − n yn+1 ( x )             ‫ﭘﺲ‬             y′(0) = − y1 ( x )
                                                                                         :‫ﺭﻭﺍﺑﻂ ﺯﻳﺮ ﺍﺯ ﺧﻮﺍﺹ ﺗﻮﺍﺑﻊ ﺑﺴﻞ ﻣﻲ ﺑﺎﺷﻨﺪ‬

      1 ) xy n ( x ) = ny      ( x ) − xy n + 1 ( x )
                                        n

       2 ) xy n ( x ) = − ny n ( x ) + xy n −1 ( x )
       xy n + 1 ( x ) = 2 ny n ( x ) − xy n + 1

                                                             : ‫ﺑﺮﺭﺳﻲ ﺻﻔﺮﻫﺎﻱ ﺗﺎﺑﻊ ﺑﺴﻞ‬
     .‫ ﺩﺍﺭﺍﻱ ﺭﻳﺸﻪ ﻫﺎﻱ ﻧﺎﳏﺪﻭﺩ ﺍﺳﺖ )ﺍﺯ ﻧﻈﺮ ﺗﻌﺪﺍﺩ(ﻛﻪ ﺩﺭ ﻣﻜﺎﻥ ﻫﺎﻱ ﻣﺸﺨﺺ ﻫﺴﺘﻨﺪ‬yn(x) ‫ﺗﺎﺑﻊ‬
        y n ( x ) = 0 → ‫{ = ﺭﻳﺸﻪ ﻫﺎ‬α n 0 ,α n1 ,....,α A}}

      φ a ‫ → ﻩ‬y n (α ni ) = 0
                                                      .‫ﺍﺯ ﺍﻧﺪﺍﺧﱳ ﺳﻨﮓ ﺩﺭ ﺁﺏ ﻧﺎﺷﻲ ﻣﻲ ﺷﺪ‬                      yo    ‫ﺩﻳﺪﱘ ﻛﻪ ﺷﻜﻞ ﺗﺎﺑﻊ‬

                    f ( x) − f (− x)
                      c
                                         Nηx        ⎡c                                   ⎤
                                                  1 ⎢ f ( x)Sin Nηx dx − f ( x)Sin Nηx dx⎥
                                                                        c

                 c∫                                c ∫                  ∫
       bn = 2                        Sin     dx =
                  o
                            2             c         ⎣o           c      o
                                                                                    c    ⎦

       x = −x                               dx = − dx

                                     N ηx
       c

       ∫    f    ( x )Sin                           (−       dx       )
       o
                                       c


٤٦
                                         ‫ﹰ‬
     ‫ﺩﺭ ﺧﺎﺭﺝ ﺍﺯﺍﻳﻦ ﻓﺎﺻﻪ ﺑﺴﻂ ﺳﻴﻨﻮﺳﻲ ‪ ∑ bn Sin nx‬ﺍﻭﻻ ﺩﺍﺭﺍﻱ ﭘﺮﻳﻮﺩ ‪ 2π‬ﺍﺳﺖ.ﻭ ﻓﺮﺽ ﻛﻨﻴﺪ ﻛﻪ‬
     ‫) ‪ f ( x‬ﻳﻚ ﺗﺎﺑﻊ ﻗﻄﻌﻪ ﺍﻱ ﭘﻴﻮﺳﺘﻪ ﺩﺭ ﻓﺎﺻﻠﻪ ﻱ )٢،٢-(ﳘﺎﻧﻄﻮﺭ ﻛﻪ ﻣﻲ ﺩﺍﻧﻴﻢ ﻫﺮ ﺗﺎﺑﻊ ﺭﺍ ﻣـﻲ ﺗـﻮﺍﻥ‬
                                                ‫ﺑﺼﻮﺭﺕ ﲨﻊ ﺩﺭ ﺗﺎﺑﻊ ﺯﻭﺝ ﻭ ﻓﺮﺩ ﻧﻮﺷﺖ.‬
        ‫) ‪f ( x ) = h( x ) + g ( x‬‬             ‫آﻪ‬         ‫, ) ‪g (− x ) = − g ( x‬‬      ‫) ‪h(− x ) = h( x‬‬


                   ‫) ‪f ( x ) + f (− x‬‬                        ‫) ‪f ( x ) − f (− x‬‬
       ‫= ) ‪h( x‬‬                                    ‫= )‪g (x‬‬
                            ‫2‬                                         ‫2‬

              ‫ﺗﺎﺑﻊ ) ‪ h( x‬ﻳﻚ ﺗﺎﺑﻊ ﺯﻭﺝ ﺍﺳﺖ ﺑﻨﺎﺑﺮﺍﻳﻦ ﭼﻨﺎﻧﭽﻪ ) ‪ h( x‬ﺭﺍ ﺩﺭ ﻓﺎﺻﻠﻪ )٠،٢( ﺑﺎ ﺑﺴﻂ ﻛﺴﻴﻨﻮﺳﻲ ﺗﻘﺮﻳﺐ‬

            ‫ﺑﺰﻧﻴﻢ ،ﺗﺎﺑﻊ ) ‪ h( x‬ﺩﺭ ﻧﺎﺣﻴﻪ)٢،٢-( ﺗﻘﺮﻳﺐ ﺯﺩﻩ ﺧﻮﺍﻫﺪ ﺷﺪ.) ) ‪ h(x‬ﻭ ﻛﻴﻨﻮﺱ ﻫﺮ ﺩﻭ ﺯﻭﺝ ﺍﻧﺪ(.‬
                                    ‫∞‬
                                             ‫‪nηx‬‬                         ‫‪c‬‬
                                                                                     ‫‪nη x‬‬
       ‫= ) ‪h( x‬‬                ‫‪+ ∑ a n Cos‬‬
                   ‫‪ao‬‬
                                                                         ‫‪h( x )Cos‬‬
                                                                      ‫∫‪c‬‬
                           ‫‪η‬‬                                 ‫2 = ‪an‬‬                       ‫‪dx‬‬
                                    ‫1= ‪n‬‬      ‫‪c‬‬                          ‫‪o‬‬
                                                                                      ‫‪c‬‬
        ‫) ‪f ( x ) = h( x ) + g ( x‬‬            ‫) ‪(− c, c‬‬
                                                                                                             ‫ﭘﺲ‬
                              ‫∞ ‪nηx‬‬  ‫∞‬
                                            ‫‪nηx‬‬
        ‫‪f ( x ) = o + ∑ anCos‬‬    ‫‪+ ∑ bn Sin‬‬                                        ‫) ‪x ∈ (− c, c‬‬
                 ‫‪a‬‬
                   ‫1= ‪2 n‬‬
                                                                             ‫آﻪ‬
                               ‫‪c‬‬   ‫1= ‪n‬‬      ‫‪c‬‬
           ‫ﻛﻪ ﺍﻳﻦ ﺑﺴﻂ ﺗﺎﺑﻊ ﻓﻮﺭﻳﻪ ) ‪ f ( x‬ﺩﺭ ﻓﺎﺻﻠﻪ )٢،٢-( ﻣﻲ ﺑﺎﺷﺪ.ﺣﺎﻝ ﺑﺎﻳﺪ ﺿﺮﺍﻳﺐ ﺭﺍ ﭘﻴﺪﺍ ﳕﺎﺋﻴﻢ.‬
              ‫‪η‬‬
        ‫2‬
                   ‫⎡ ‪f ( ( )Sinmxdx nηx = bm‬‬
          ‫) ‪η 2 `SNx ) +x f (− x‬‬              ‫⎥ ‪1 ⎢ f ( x )Cos nηx dx + f (− x )Cos nηx dx‬‬
                                                                                          ‫⎤‬
                 ‫‪c‬‬                                ‫‪c‬‬                     ‫‪c‬‬
       ‫∫ 0 = ‪an‬‬
                                                ‫∫ ‪c‬‬                    ‫∫‬
                                 ‫‪Cos‬‬     ‫= ‪dx‬‬
              ‫‪c‬‬                         ‫2‬           ‫‪c‬‬                                ‫‪c‬‬                   ‫‪c‬‬
                       ‫‪o‬‬                                            ‫‪⎣o‬‬                         ‫‪o‬‬             ‫⎦‬




                     ‫ﻓﺮﺽ ﻛﻨﻴﺪ ﻧﺎﺣﻴﻪ ‪ C‬ﻧﺎﺣﻴﻪ ﺍﻱ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺑﺎﺷﺪ ﻭ ) ‪ β (Z‬ﺑﻐﲑﺍﺯ ﻧﻮﺍﺣﻲ ﻣﺸﺨﺺ‬
            ‫0 = ‪β (Z )dZ‬‬                               ‫ﺷﺪﻩ ﲢﻠﻴﻠﻲ ﺑﺎﺷﺪ ﺁﻧﮕﺎﻩ ﺍﻧﺘﮕﺮﺍﻝ ) ‪β (Z‬‬
       ‫∫‬‫‪B‬‬


               ‫‪dZ‬‬
       ‫∫‬‫‪C‬‬          ‫(‬
             ‫‪Z Z2 +q‬‬
               ‫2‬
                     ‫‪=φ‬‬
                                         ‫)‬
                                                                                   ‫ﺭﻭﻱ ﻧﺎﺣﻴﻪ ‪ D‬ﻓﻮﻕ ﺑﺮﺍﺑﺮ ﺻﻔﺮ ﺍﺳﺖ.‬

                                                                ‫ﮔﻮﺋﻴﻢ ﺍﻧﺘﮕﺮﺍﻝ ‪∫ β (Z )dZ‬ﻣﺴﺘﻘﻞ ﺍﺯ ﻣﻴﺴﺮ ﺍﺳﺖ ﺍﮔﺮ‬
                                ‫‪b‬‬

       ‫)) ‪∫ β (Z )dZ = ∫ β (Z )dZ = F (Z (b )) − F (Z (a‬‬
        ‫‪C‬‬
                                ‫‪a‬‬




‫٧٤‬
     ‫ﺍﻟﺒﺘﻪ ﻣﻲ ﺗﻮﺍﻥ ﮔﻔﺖ ﻳﻚ ﻧﺎﻡ ) ‪ β (Z‬ﻭﻟﺪﺍﻱ ﺍﻧﺘﮕﺮﺍﻝ ﻣﺴﺘﻘﻞ ﺍﺯ ﻣﺴﲑ ﺍﺳﺖ ﺍﮔﺮ ﺑﻨﺎﻡ ﻭﻟﺪﺍﻱ ﺗﺎﺑﻊ ﺍﻭﻟﻴﻪ‬
                               ‫3‬
                                                                                ‫ﺑﺎﺷﺪ. ﻣﺜﺎﻝ:‬
       ‫‪β (Z )cZ 2 → F (Z )c‬‬
                             ‫‪Z‬‬
                                                  ‫3‬
      ‫‪1+ i‬‬
                              ‫1‬
         ‫= ‪∫Z Z‬‬
           ‫2‬
                                ‫+ 3) ‪(1 + i‬‬
         ‫0‬
                              ‫3‬
         ‫‪2i‬‬                      ‫‪2i‬‬

       ‫‪∫2i Z = log z −∫2i= log 2i − log(− 2i ) = tui‬‬
          ‫‪dZ‬‬
      ‫−‬




                                                                                            ‫ﺍﻧﺘﮕﺮﺍﻝ ﻛﻮﺷﻲ‬
                 ‫ﺍﮔﺮ ) ‪ β (Z‬ﻳﻚ ﺗﺎﺑﻊ ﲢﻠﻴﻠﻲ ﺩﺭ ‪ C‬ﺑﺎﺷﺪ ﻭ ﺗﻨﻬﺎ ﺩﺭ ﻧﻘﻄﻪ ﺩﺭ ‪ C‬ﲡﻠﻴﻠﻲ ﻧﺒﺎﺷﺪ ﺁﻧﮕﺎﻩ‬
       ‫‪β (Z )dz = 2πi‬‬
     ‫0‪∫ Z − Z‬‬               ‫) 0 ‪β (Z‬‬
     ‫‪c‬‬                                                                              ‫ﻣﺜﺎﻝ:‬
             ‫‪β (Z )dz‬‬
     ‫‪∫ Z −Z‬‬                    ‫‪= 2πi‬‬              ‫) 0 ‪β (Z‬‬
     ‫‪c‬‬                   ‫0‬

                       ‫‪Z dz‬‬                           ‫‪i‬‬      ‫‪− 2π‬‬
     ‫= 1 + ‪∫ (q − Z )(Z − i ) = 2πi q‬‬
     ‫‪c‬‬
                                                              ‫01‬



     ‫‪∫ z β (Z )dZ = Z ∫ β (Z )dZ‬‬
     ‫‪c‬‬
              ‫0‬                       ‫0‬
                                          ‫‪c‬‬



     ‫‪∫ (β (Z + g t ))dZ = ∫ β (Z )aZ + ∫ g (Z )dZ‬‬
     ‫‪c‬‬                                        ‫‪c‬‬                ‫‪c‬‬




                   ‫∫ ≤ ‪∫ β (Z )dx‬‬
                   ‫‪c‬‬                      ‫‪a‬‬
                                                  ‫‪β (Z (t )Z ′) dt‬‬



          ‫‪Ic ∫ Z 2 dZ‬‬                                                                           ‫ﻣﺜﺎﻝ:‬
                  ‫1‪C‬‬




                                                                     ‫ﺑﺼﻮﺭﺕ ﻫﺎﻱ ﺯﻳﺮ‬   ‫ﺗﺎ‬   ‫؟ ﺍﻧﺘﮕﺮﺍﻝ ﺍﺯ‬
                   ‫11 2‬
      ‫= 1‪I‬‬          ‫‪+ i‬‬
                   ‫3 3‬
                   ‫11 2‬
      ‫= 2‪I‬‬          ‫‪+ i‬‬
                   ‫3 3‬




‫٨٤‬
       Ic ∫ Z 2 dZ
           C
                                                                                                                       ‫ﻃﺒﻖ ﻗﻀﻴﻪ ﻛﻮﺷﻲ‬

                                               ⎛ ∂Q            ∂R ⎞
       ∫ Pdx + Qdy = ∫ ∫ ⎜ ∂X
       C
                         ⎜
                         ⎝             R
                                                           −      ⎟dxdy
                                                               ∂y ⎟
                                                                  ⎠

       ∫ β (Z )dδ = ∫ Ndx − Vdy + i ∫ Xdx + udy
       C                           C


           ⎛ − ∂u ∂n ⎞           ⎛ ∂u ∂x ⎞
       = ∫∫⎜
           ⎜ ∂y  − ⎟dxdy + i ∫ ∫ ⎜
                                 ⎜ ∂y − ∂y ⎟dxdy
         R ⎝
                  ∂x ⎟
                     ⎠         R⎝
                                           ⎟
                                           ⎠

                                                                ‫ ﲢﻠﻴﻠﻲ ﺑﺎﺷﺪ‬β (Z ) ‫ ﻧﺎﺣﻴﻪ ﺑﺴﺘﻪ ﺑﺎﺷﺪ ﻭ‬C ‫ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ ﺍﮔﺮ‬
        ux = vy
       Xy = −V X

     ‫ﺣـﺪ ﺭﻭ ﻭ‬β (Z ) ‫ ﺍﻱ ﻛﻪ ﺣـﺪ‬C ‫ ﲢﻠﻴﻠﻲ ﺑﺎﺷﺪ ﺍﻧﺘﮕﺮﺍﻝ ﺁﻥ ﺭﻭﻱ ﻫﺮ ﻧﺎﺣﻴﻪ‬β (Z ) ‫ﻟﺬﺍ ﻃﺒﻖ ﻗﻀﻴﻪ ﺍﮔﺮ‬
                                                                             C ‫ﺩﺭﻭﻥ‬
                                                              ‫ﲢﻠﻴﻠﻲ ﺑﺎﺷﺪ ﺑﺮﺍﺑﺮ ﺑﺎ ﺻﻔﺮ ﺍﺳﺖ‬
       ω (t ) = X (t ) + iy (t )                                                                            :‫ﻣﻔﻬﻮﻡ ﺍﻧﺘﮕﺮﺍﻝ‬

       b                   b                           b

       ∫ ω (t )d + ∫ X (t )dt + o∫ y(t )dt
       a                   a                           a



                                                                                                                       :‫ﻣﺜﺎﻝ‬
               ∫θ
                    i 2t
                           dt = 3              + i
                                           2           4



                                                                                                                      :‫ﺧﻮﺍﺹ‬
                               b                       b

                    Re ∫ ∫ ω (t )dt = ∫ Re Z (t ) dt
                               a                       a


                                                       b

                    ∫ Zω (t )dt = Z ∫ ω (t )           a




                    r0 eθ 0 = ∫ ω (t )dt ⇒ ∫ e −uθ 0 ω (t ) dt ‫ﻳﺎ‬  (           )                     (            )
                                                                                           ro c ∫ Re e −1θ 0 ω (t ) dt ‫!؟‬

                     b                             b

                     ∫ X (t )dt
                     a
                                               ≤   ∫ ω (t ) dt
                                                   a

٤٩
     ‫‪Line Integral‬‬                                                                                      ‫ﺍﻧﻮﺍﻉ ﺍﻧﺘﮕﺮﺍﻝ ١(‬
                                        ‫2‪z‬‬

     ‫‪∫ β (t )dt‬‬
     ‫‪C‬‬
                         ‫ﻳﺎ‬             ‫‪∫ β (t )dt‬‬
                                        ‫1‪Z‬‬

                                                                       ‫‪ Z‬ﺭﻭﻱ ﻳﻚ ﺧﻂ ﺣﺮﻛﺖ ﻣﻲ ﳕﺎﻳﺪ.‬            ‫ﺑﺎ ﻓﺮﺽ ﺍﻳﻨﻜﻪ‬

     ‫) ‪Z = Z (t‬‬
      ‫‪b‬‬                       ‫1‪Z‬‬

     ‫‪∫ β (Z )dt = ∫ (u / x(t ) + iv(x(t ), y(t )))(dX ′ + 1dy ′)dt‬‬
      ‫‪a‬‬                       ‫1‪Z‬‬

          ‫1‪Z / t‬‬

     ‫=‬         ‫‪∫ (Ux ′ − Vy ′) + i(uy ′ + Vx ′)dt‬‬
          ‫0‪Z / t‬‬
          ‫1‪t‬‬                                ‫1‪t‬‬

     ‫‪= ∫ (Ux ′ + Vy ′)dt + ∫ (Uy ′ + Vx ′)dt‬‬
          ‫0‪t‬‬                                ‫0‪t‬‬




      ‫‪∫ β (Z )dZ − ∫ β (Z )dZ‬‬
     ‫‪−C‬‬                       ‫‪C‬‬


                                                                                                     ‫ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﻟﮋﺍﻧﺪﺭ:‬
                    ‫ﻣﻌﺎﺩﻟﻪ 0 = ‪ (1 − x )y ′′ − 2 x y ′ + y‬ﺭﺍ ﺑﺎ ﻧﺎﻡ ﻣﻌﺎﺩﻟﻪ ﺛﺮﺍﻧﺪﺭ ﻣﻲ ﺷﻨﺎﺳﻴﻢ.‬
                                                             ‫2‬


                          ‫ﺑﺎ ﺣﻞ ﻣﻌﺎﺩﻟﻪ ﻓﻮﻕ ﺑﺎ ﺭﻭﺵ ﺳﺮﻱ ﻫﺎ ﺩﺍﺭﱘ .ﺍﻳﻦ ﺑﻨﺎﻡ ﺣﺪ ﻓﺎﺻﻠﻪ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ‬
                           ‫∞‬
                                                                ‫ﺗﻮﺍ‪‬ﺎﻱ ﻣﻌﺎﺩﻟﻪ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻫﺴﺘﻨﺪ‬
      ‫= 1‪y‬‬         ‫+ ‪a‬‬
                     ‫0‬        ‫∑‬
                              ‫‪a h × 2h‬‬
                              ‫1− ‪h‬‬
                                                 ‫2‬

                                        ‫∞‬
      ‫‪y‬‬   ‫2‬     ‫+ ‪= a1x‬‬                ‫∑‬
                                       ‫1− ‪h‬‬
                                                     ‫‪a‬‬   ‫1+ ‪2 h‬‬   ‫1 + ‪× 2h‬‬

          ‫ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ ﻛﻪ ﺩﺭ ﺁﻥ )1 + ‪λ = n(n‬‬                                ‫‪Pn‬‬   ‫ﺑﻨﺎﻡ ﺛﺮﺍﻧﺪﺭ ﺍﺯ ﻣﺮﺗﺒﻪ ‪ n‬ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ) ‪( x‬‬

                ‫1‬
     ‫‪Pn ( x ) − n‬‬
                                  ‫‪(− 1 )x (2 n − 2 h )! X‬‬
               ‫2‬
                              ‫) ‪∑ h! (n − 2 h )h (n + x‬‬                              ‫‪n−2h‬‬
                                                                                                ‫....,1, 0 = ‪n‬‬
                                               ‫•‬

                   ‫1 = )‪P0 ( x‬‬
                   ‫‪P ( x) = X‬‬
                    ‫1‬


     ‫= ) ‪P 2 (x‬‬
                      ‫1‬
                      ‫2‬
                         ‫(‬
                        ‫‪3x‬‬         ‫2‬
                                       ‫1−‬        ‫)‬



‫٠٥‬
                   1
     P3 ( x ) =      (3 S − 3 )
                   2

     P4       (x ) =    1
                        2
                                 (3 s   × 4 − 30 x            2
                                                                  + 3        )
                                                                                 Pn (− x ) = (− 1)Pn ( x ) ‫ﺧﻮﺍﺹ ﺩﺍﺭﱘ ﻛﻪ‬
                ‫ﺎ ﺑﻪ ﺻﻮﺭﺕ‬‫(ﺟﻮﺍ‬i − x g )y n − 3 × y ′ + n (n = 1 ) y = 0 ‫!؟ﺍﮔﺮ ﺣﻞ ﻣﻌﺎﺩﻟﻪ‬
                                                                     .‫ﺗﻮﺍﺑﻊ ﻟﮋﺍﻧﺪﺭ ﻣﻲ ﺑﺎﺷﺪ‬
      1

     ∫        P m ( x ) + P n ( x )d × =                  ∫   mn                       m ≠ n
     −1


                             P n (x )
     φ        (x ) =                   →         {φ       ( x )}
                             P n (x )
          n                                           n
                                     1


     P n (x ) =
                        1 1    d n
                                                 (x   2
                                                              − 1   ) n

                        2n n1 d × n

     P n + 1 ( x ) − xp             n   =
                                             n
                                                   D      n −1
                                                                  u     n
                                                                             ,     u = x    (   2
                                                                                                    −1   )
                                            2 n!
                                             n




     P n′+ 1 ( x ) − P ′n − 1( x ) = (2 n + 1 )Pn ( x )

                                    2                                       2n + 1
     1 Pn ( x )1 =                       → φ n (x ) =                              Pn ( x )
                                  2n + 1                                      2

                       ∞                                                           1
                                                                  2n + 1
     β (x ) =          ∫         A n Pn ( x ) ⇒ A n =                    ∫ β ( x )Pn ( x )dx
                       n=0
                                                                    2 −1

                   .‫ﻛﺎﺭﺑﺮﺩ ﺛﺮﺍﻧﺪﺭ ﺩﺭﺣﻞ ﻣﺴﺎﺋﻞ ﻣﻌﺎﺩﻻﺕ ﺑﺎ ﻣﺸﺘﻘﺎﺕ ﺟﺰﺋﻲ ﺩﺭ ﳐﺘﺼﺎﺕ ﻛﺮﻭﻱ ﻣﻲ ﺑﺎﺷﺪ‬
         ∂2
     r        (ru) + 1 ∂ ⎛ Sinθ ∂v ⎞ = 0
                            ⎜      ⎟                                  0 < r < c , 0 < ∂ <π
         ∂r 2
                    Sinθ ∂θ ⎝   ∂θ ⎠

     V (c , ∂ ) = F (θ       )

                 ‫ﺑﺎﺟﺪﺍﺳﺎﺯﻱ ﺑﻪ ﻣﻌﺎﺩﻟﻪ‬                                                   ∞
                                                                                                    ⎛ r ⎞
                                                                                                             n

                                                   V          (r , θ ) = ∑                  A   n   ⎜   ⎟        Pn   (Cos θ )
                    .‫ﺛﺮﺍﻧﺪﺭ ﻣﻲ ﺭﺳﻴﻢ‬                                                 n = 0           ⎝ c ⎠


           ⎡c                                      ⎤
       1 = ⎢ f ( x )Cos nηx dx + f ( x )Cos nηx dx ⎥ = 1                nηx
                                 o                        c

            ∫                   ∫c                           f ( x )Cos
                                                         c∫
     =                                                                      dx
        c
           ⎣o            c      −
                                             c     ⎦      −c
                                                                         c


٥١

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:55
posted:10/3/2011
language:Persian
pages:51