Higher Order Differential Equations by yurtgc548

VIEWS: 1 PAGES: 54

									      Derivative Boundary Value
               Problems
            d2F                         F
               2  f ( x , F, F ' )  
Consider:
            dx                          4
      Derivative Boundary Value
               Problems
             d2F                         F
                2  f ( x , F, F ' )  
Consider:
             dx                          4

with B.C.’s: F (0)  F0  0 and F ( )  Fn  1
                                            
      Derivative Boundary Value
               Problems
             d2F                         F
                2  f ( x , F, F ' )  
Consider:
             dx                          4

with B.C.’s: F (0)  F0  0 and F ( )  Fn  1
                                            

 Note that in this case, we have derivative
 boundary conditions.
       Derivative Boundary Value
                Problems

Let us solve this using finite differences.
Using the finite difference approximations:
       Derivative Boundary Value
                Problems

Let us solve this using finite differences.
Using the finite difference approximations:

        d2F        Fi 1  2 Fi  Fi 1    Fi
          2                 2          
        dx x  x           h               4
                i
       Derivative Boundary Value
                Problems

Let us solve this using finite differences.
Using the finite difference approximations:

        d2F        Fi 1  2 Fi  Fi 1    Fi
          2                 2          
        dx x  x           h               4
                i




 Now if we attempt to set up the difference
 equations we obtain:
      Derivative Boundary Value
               Problems

Our Finite Difference Analog is:
      Derivative Boundary Value
               Problems

Our Finite Difference Analog is:

                          h2 
              Fi 1   2   Fi  Fi 1  0
                          4
      Derivative Boundary Value
               Problems

Our Finite Difference Analog is:

                          h2 
              Fi 1   2   Fi  Fi 1  0
                          4


 Evaluating this equation at i = 1 through 3,
 we obtain:
     Derivative Boundary Value
              Problems

For i = 1:   F0  2  1    2  F  F  0
                  
                      4   4      1
                                       2
     Derivative Boundary Value
              Problems

For i = 1:   F0  2  1    2  F  F  0
                  
                      4   4      1
                                       2




For i = 2    F  2  1    2  F  F  0
              1  
                     4   4      2
                                      3
     Derivative Boundary Value
              Problems

For i = 1:   F0  2  1    2  F  F  0
                  
                      4   4      1
                                       2




For i = 2    F  2  1    2  F  F  0
              1  
                     4   4      2
                                      3




For i = 3    F2  2  1    2  F  F  0
                  
                      4   4      3
                                       4
      Derivative Boundary Value
               Problems
These equations become:
F0 -1.84579F1 + F2                    = 0
      F1 - 1.84579 F2 + F3            = 0
                 F2 - 1.84579 F3 + F4 = 0
Notice that we now have 3 equations with 5
unknowns!
But we also know:
      Derivative Boundary Value
               Problems
These equations become:
F0 -1.84579F1 + F2                    = 0
      F1 - 1.84579 F2 + F3            = 0
                 F2 - 1.84579 F3 + F4 = 0
Notice that we now have 3 equations with 5
unknowns!
But we also know:               Fn 1  Fn 1
                          Fn 
                                     2h
      Derivative Boundary Value
               Problems
If we use our knowledge of the derivative at x0
and xn, i.e., F’0 = 0 and F’4 = 1:
      Derivative Boundary Value
               Problems
If we use our knowledge of the derivative at x0
and xn, i.e., F’0 = 0 and F’4 = 1:

              F  F1
     F0  1  1
               2 / 4
      Derivative Boundary Value
               Problems
If we use our knowledge of the derivative at x0
and xn, i.e., F’0 = 0 and F’4 = 1:

              F  F1                F5  F3
     F0  1  1            F4  1 
                             
               2 / 4                2 / 4
      Derivative Boundary Value
               Problems
If we use our knowledge of the derivative at x0
and xn, i.e., F’0 = 0 and F’4 = 1:

              F  F1                F5  F3
     F0  1  1            F4  1 
                             
               2 / 4                2 / 4
Notice, now we have two new equations,
but we have also introduced two new
unknowns!
We need two more equations!
      Derivative Boundary Value
               Problems
Two that we can use are :
      Derivative Boundary Value
               Problems
Two that we can use are :

                   2  1  2  F  F  0
 For i = 0   F1 
                   
                       4
                          4  0 1
                              
      Derivative Boundary Value
               Problems
Two that we can use are :

                   2  1  2  F  F  0
 For i = 0   F1 
                   
                       4
                          4  0 1
                              

                  2  1  2  F  F  0
 For i = 4   F3 
                  
                      4
                         4  4 5
                             
      Derivative Boundary Value
               Problems
Two that we can use are :

                   2  1  2  F  F  0
 For i = 0   F1 
                   
                       4
                          4  0 1
                              

                  2  1  2  F  F  0
 For i = 4   F3 
                  
                      4
                         4  4 5
                             
This gives us two more equations, and we have
the same number of unknowns.
      Derivative Boundary Value
               Problems
So now our equations become:
F-1 + -1.84579F0 + F1                   = 0
- F-1                + F1               = 0
-1.84579F1 +         F2                 = 0
      F1 - 1.84579 F2 +      F3         = 0
                F2 - 1.84579 F3 + F4 = 0
                 -F3           + F5 = 1.5708
                 F3 - 1.84579 F4 + F5 = 0
Notice that we now have 7 eqns with 7 unkns!
       Derivative Boundary Value
                Problems
Or: [A][F] = [b]
where:
    Derivative Boundary Value
             Problems
       1   0    1    0    0    0   0
       1   a1   1    0    0    0   0
       0   1    a1   1    0    0   0
A      0   0    1    a1   1    0   0
       0   0    0    1    a1   1   0
       0   0    0    0    1    0   1
       0   0    0    0    1    a1 1
     Derivative Boundary Value
              Problems

where
-a1 = -1.84579
and :
     Derivative Boundary Value
              Problems
                     0
where
                     0
-a1 = -1.84579
and :                0
                     0
                 b
                     0
                     
                     2
                     0
     Derivative Boundary Value
              Problems
                     0        1.88253
where
                     0        2.03981
-a1 = -1.84579
and :                0
                              1.88253
                     0
                 b       F=   1.43494
                     0
                              0.76606
                     
                     2        0.02095
                     0        0.80474
       Derivative Boundary Value
                Problems
Comparing to exact solution of F = -2cos(x/2):
                           2


            F
                j          0

                      x
                       j
             2. cos        2
                      2


                           4
                               2   0           2   4
                                       x
                                           j
     Derivative Boundary Value
                                 xj
              Problems    2. cos
                                     2
                   1.88253
Compared to                  1.414
the exact          2.03981
solution:                      2
                   1.88253
                             1.848
              F=   1.43494   1.414
                   0.76606   0.765
                   0.02095     0
                   0.80474   0.765
            Nonlinear Problems
              2
             d F      dF
                2  F dx  e  0
Consider:                   x
             dx
            Nonlinear Problems
              2
             d F      dF
                2  F dx  e  0
Consider:                   x
             dx
with B.C.’s: F(0)  F0  1 and F(1)  Fn  1
            Nonlinear Problems
              2
             d F      dF
                2  F dx  e  0
Consider:                   x
             dx
with B.C.’s: F(0)  F0  1 and F(1)  Fn  1

 Note that in this case, we have a nonlinear
 D.E. (because of the second term)
           Nonlinear Problems

Let us solve this using finite differences.
Using the finite difference approximations:
              Nonlinear Problems

Let us solve this using finite differences.
Using the finite difference approximations:
          2
        d F          Fi 1  2 Fi  Fi 1
                   
        dx 2 x  x           h2
                 i
              Nonlinear Problems

Let us solve this using finite differences.
Using the finite difference approximations:
          2
        d F          Fi 1  2 Fi  Fi 1
                   
        dx 2 x  x           h2
                 i


        dF          Fi 1  Fi 1
                  
        dx x  xi        2h
           Nonlinear Problems

Our Finite Difference Analog is:
            Nonlinear Problems

Our Finite Difference Analog is:
                         h                           2 xi
   Fi 1  2 Fi  Fi 1  ( Fi Fi 1  Fi Fi 1 )  h e
                         2
            Nonlinear Problems

Our Finite Difference Analog is:
                         h                           2 xi
   Fi 1  2 Fi  Fi 1  ( Fi Fi 1  Fi Fi 1 )  h e
                         2

 Evaluating this equation with h = 0.2 and
 thus at i = 1 through 4, we obtain:
               Nonlinear Problems
For i = 1:
             1  2 F  F2  0.1( F F2  F F0 )  0.04e 0.2
                    1             1      1
               Nonlinear Problems
For i = 1:
             1  2 F  F2  0.1( F F2  F F0 )  0.04e 0.2
                    1             1      1
For i = 2
            F  2 F2  F3  0.1( F2 F3  F2 F )  0.04e
             1                               1
                                                          0. 4
                Nonlinear Problems
For i = 1:
              1  2 F  F2  0.1( F F2  F F0 )  0.04e 0.2
                     1             1      1
For i = 2
            F  2 F2  F3  0.1( F2 F3  F2 F )  0.04e
             1                               1
                                                          0. 4

For i = 3
            F2  2 F3  F4  0.1( F3 F4  F3 F2 )  0.04e 0.6
                Nonlinear Problems
For i = 1:
              1  2 F  F2  0.1( F F2  F F0 )  0.04e 0.2
                     1             1      1
For i = 2
            F  2 F2  F3  0.1( F2 F3  F2 F )  0.04e
             1                               1
                                                          0. 4

For i = 3
            F2  2 F3  F4  0.1( F3 F4  F3 F2 )  0.04e 0.6

For i = 4
             F3  2 F4  1  0.1(  F4  F4 F3 )  0.04e 0.8
          Nonlinear Problems

Using MathCad to solve these four
simultaneous non-linear equations,
the results are :
          Nonlinear Problems

Using MathCad to solve these four
simultaneous non-linear equations,
the results are :
                      1.000
                      0.49454
             [F] =   -0.01211
                     -0.45794
                     -0.79503
                     -1.000
           Nonlinear Problems
Graphically:             1


                       0.5

               F         0
                   i

                       0.5

                         1
                             0   0.5   1
                                 x
                                  i
   Two Dimensional Steady-State..

Consider:
            T T
              2        2
                       0
            x 2
                   y 2



            Laplace Equation
    Two Dimensional Steady-State..

Analytical:
      4To     
                   1                y
T 
              2n  1 sin( 2n  1) a
              n 0

                                        b
sinh( b  1)( 2n  1)   cos ec ( 2n  1)
                      a                   a
   Two Dimensional Steady-State..

Consider: System of grid-points (i,j)
           and its four immediate neighbors
                                      i,j+1


                         I-1,j        i,j        i+1,j
                                            Dx
                                 Dy
                                      i,j-1
  Two Dimensional Steady-State..

            T T
              2        2
                       0
            x 2
                   y 2

Difference equation:
    Ti  1, j  2Ti , j  Ti  1, j
                                      
                 Dx  2

    Ti , j  1  2Ti , j  Ti , j  1
                                      0
                 Dy  2
Two Dimensional Steady-State..

If Dx = Dy:


           Ti  1, j  Ti  1, j  Ti , j  1  Ti , j  1
Ti , j 
                                 4
    Two Dimensional Steady-State..
Example:
What is the temperature at P            0
                                 100        0
P= (100 + 0 + 0 + 0)/4
 = 25                            100        0

not very accurate…grid is too coarse!
                                 100        0
                                        0
   Two Dimensional Steady-State..

Apply the equation to each interior node:
                                       0       0
                             100                   0
Ta = (Tb + 0 + 100 + Tc)/4
                                   a       b
Tb = (0 + 0 + Ta + Td)/4     100                   0
                             100   c       d
Tc = (Td + Ta + 0 + 100)/4                         0
Td = (0 + Tb + Tc + 0)/4     100                   0
                                       0       0

      Four equations --> four unknown!
Two Dimensional Steady-State..

 Gauss-Seidel Iteration:
 a. replace all “=“ by “<--”
 b. make starting guesses
 c. compute new approximations
   for Ta, Tb, Tc, Td
 d. repeat step c until converge.

								
To top