HeatWave by abhishekgoel137

VIEWS: 3 PAGES: 75

									PARTIAL DIFFERENTIAL
EQUATIONS AND BOUND
ARY VALUE PROBLEMS
VIBRATING STRINGS AN
D HEAT CONDUCTION
THE VIBRATING STRING:
Suppose that a flexible string is pulled taut on
the x-axis and fastened at two points, which w
e denote by x = 0 and x = L. The string is then
drawn aside into a certain curve y = f(x) in the
xy plane and then released from rest.
Problem: To find the vertical displacement yx
(,t) from the equilibrium position of the string
at the point (distance x from the origin) at time
t.
We make certain assumptions:
The subsequent motion of the string is entir
ely transverse (i.e. along the xy plane). (Thu
s the displacement of the string from the eq
uilibrium position is a function        y=y
(x, t).)

We look at a small piece of the string whic
h in the equilibrium position has length x
. If m is the (uniform) density of
the string, the mass of the piece is m x. By
Newton’s second law of motion, the transver
se force F acting on the piece is given by

                    ¶2 y
            F = m Dx 2
                    ¶t

Since the string is flexible, the tension T at
any point is directed along the tangent (see
the figure) and has T sin  as its y- compon
ent.
                 T sin 
                               T




We now assume that the motion of the string i
s due strictly to the tension T in it. (That is the
gravitational force is negligible.)
Hence F =  (T sin  )   (T tan  ) (as
                                      ¶y
is small) and so =  (T tan  ) = D (T    )
                                       ¶x
 Thus we get            ¶y         ¶ y
                                    2
                   D (T    ) = m Dx 2
                        ¶x         ¶t
Dividing by x and letting x  0, we get

        ¶     ¶y    ¶ y     2
           (T    )=m 2
        ¶x    ¶x    ¶t
Or as T is constant (the same for all x), we get
            ¶ y2
                   ¶ y  2
          T      =m 2
            ¶x 2
                   ¶t
             T
 Putting a =2

             M,

We get the one-dimensional wave equation:
                  ¶ y ¶ y
                     2    2
                a2
                       = 2
                  ¶x 2
                        ¶t
which is usually written as
                ¶ y
                 2
                       1 ¶ y
                          2
                     = 2 2
                ¶x 2
                      a ¶t
It is proved in Physics that y(x,t) satisfies the ‘o
ne-dimensional wave equation’
               2 y 2 y
           a 2
                     2
               x 2
                     t
where a is a constant (depending on the mass o
f the string and the tension of the string). We th
us want to solve the boundary value-problem

                y  y
                  2      2
                                (1)
             a2
                     2
               x 2
                     t
subject to the boundary conditions
  y (0, t )  y ( L, t )  0 for all t > 0  (2)
and the initial conditions
       y
           0 for t = 0              (3)
       t
     y(x,0) = f(x), 0 ≤ x ≤ L      (4)
We find the solution by the method of ‘sep
aration of variables’.
That is we let y(x,t) = u(x) v(t)
(a function of x  a function of t )
Substituting for y in the equation (1), we get

                                  
      a u  x ) v (t )  u ( x ) v  t )
        2
           (                        (

 or           
            u ( x )       
                       1 v ( t )
                      2                (5)
            u ( x)    a v (t )
where the ‘dashes’ denote derivative w.r.t. the c
oncerned variable. The L.H.S of (5) is indepen
dent of t and the R.H.S is independent of x and
hence both are independent
of both x and t and hence equals a pure consta
nt, say, - .
                           
             u  x ) 1 v t )
                (             (
Thus, we get u ( x )  a 2 v (t )  

 y (0, t )  y ( L , t )  0   for all t > 0
  implies u(0) = u(L) =0.
  So we have to solve for u
             
           u   u  0
with boundary conditions u(0) = u(L) =0.
This eigenvalue problem has nonttrivial soluti
ons only when
          np 2 2
       l = 2 (n =1,2,...)
           L
Corresponding solutions are
              np x
    u n = sin      ( n = 1, 2, ...)
               L
and non-zero multiples of them.
 y
     0 for t = 0 implies v  0 )  0
                            (
 t
Now for each n, v satisfies the d.e.
                  n p a
                   2   2   2
           v ¢¢ +     2
                        v = 0
                    L
with the initial condition v  0 )  0
                             (

                np at           np at
Thus v = c1 cos       + c 2 sin
                  L               L
        np a        np at      np a     np at
 v¢ = -      c1 sin       + c2      cos
         L            L         L         L
 v0)  0  c2  0
  (

Hence               np a
         v = c1 cos      t
                     L
We shall take c1=1 and denote the correspo
nding solution (which depends on n ) by vn
. Thus                     np a
                     v n = cos       t
                                 L
  Thus for each n =1,2,3,…
                             np x     np a t
        y n = u n v n = s in      cos
                              L         L
is a solution to the given problem satisfying th
e first three conditions. So also any finite or in
finite lc of them. But no finite lc of them wou
ld satisfy the initial condition
     y(x,0) = f(x), 0 ≤ x ≤ L
So we consider their infinite lc so as to satisfy t
he above initial condition.
Thus look at the infinite linear combination
                       ¥
                                      np x     np a t
  y =   å   yn =      å
                      n =1
                             b n s in
                                       L
                                           cos
                                                 L

 Now t = 0 implies
                  ¥
                           np x
        f ( x ) = å bn sin      , 0£ x£ L
                  n =1      L
Thus bn= Fourier sine coefficient of f(x) in [0, L]
                              n x
                  L
              2
            
              L   f ( x ) sin L dx
                  0
Thus the solution to the wave equation satisf
ying the initial and boundary conditions is

                  ¥
                                 np x     np a t
    y( x,t) =   ån =1
                        b n s in
                                  L
                                      cos
                                            L
                                 0 ≤ x ≤ L, t  0
where
            L
        2                   np x
   bn =
        L   ò
            0
                f ( x ) sin
                             L
                                 d x , n = 1, 2, ...
Example: Solve the wave equation when

   f ( x ) = x ( L - x ), 0 £ x £ L
 Thus             2
                        L
                                    np x
              bn = ò x ( L - x )sin      dx
                  L0                 L
                                                                      L
  é           æ    np x ö              æ     np x ö     æ     np x öù
 2ê           ç cos L ÷                ç sin      ÷     ç cos      ÷ú
= ê x( L - x) ç -         - ( L - 2 x) ç - 2 L ÷ + (-2) ç 3 L ÷ú
 Lê               np ÷                 ç   n p2 ÷       ç   n p 3 ÷ú
              ç         ÷              ç          ÷     ç          ÷ú
  ê
  ë           è    L ø                 è    L ø
                                              2
                                                        è L   3
                                                                   øû 0
                         ì 8L2
   4L 2
                         ï 3 3 n = 1,3,..
 = 3 3 (1 - cos np )   = ín p
  np                     ï0    n = 2,4,..
                         î
Thus the solution to the given problem is

  y(x,t)
 8L 2 ¥
           1        (2n -1)p x    (2n -1)p at
= 3
 p
      å (2n -1)3 sin L cos L
      n=1
                          0 ≤ x ≤ L, t  0
Example: Solve the wave equation when

                             L
                 x      0x
       f ( x)               2
                         L
                L  x     xL
                        2
Thus
         L                                  
                nx                    nx 
          2              L
        2
    bn   sin
            x       dx   L  x ) sin
                           (              dx 
        L 0      L                      L
         
         
                         L
                         2
                                             
                                             
    éì    æ        np x   ö       np x   ü
                                          L/2

    êï
  2 êï    ç - cos L       ÷ s in L       ï
                                         ï
=    íx   ç               ÷+             ý
  L êï          np             n p
                                2   2
                                         ï
          ç               ÷
    êï    è      L        ø              ï0
    ëî                                   þ
                                  2
                                L

   ì        æ        np x   ö      np x       ü
                                               L
                                                   ù
   ï
   ï        ç - co s L      ÷ sin L           ï
                                              ï
                                                   ú
 + í(L - x) ç               ÷-                ý    ú
   ï              np            n 2p 2        ï    ú
            ç               ÷
   ï
   î        è      L        ø     L 2
                                              ïL/2 ú
                                              þ    û
    2 é    L  2
                    np     L     2
                                                  np
  =   ê - 2 np c o s 2 + n 2p        2
                                             s in
    L ë                                            2

            L 2
                     np x   L        2
                                                 np ù
         +      co s      + 2                sin    ú
           2 np       L    n p           2
                                                  2 û

               ì 0              n = 2, 4, 6, ...
  4L      np   ï
= 2 2 sin    = í 4L / n p 2   2
                                n = 1, 5, 9, ...
 n p       2   ï - 4 L / n 2p 2 n = 3, 7 ,1 1, ...
               î
Hence the solution to the wave equatio
n is
  y( x,t)

 4L ¥            1          (2n -1)p x     (2n -1)p at
= 2 å(-1) n+1
                        sin            cos
 p n=1        (2n -1) 2
                                L               L
Example: Solve the wave equation when

             ì x,        0£ x £p /4
             ï
   f ( x ) = í p / 4, p / 4 £ x £ 3p / 4
             ï p - x, 3p / 4 £ x £ p
             î
                  p
 Thus         2
         bn =
              p   ò f ( x) sin nx dx
                  0
 2é
    p /4                 3p / 4
                        p
= ê ò x sin nx dx + ò sin nx dx
 pë0               p /4
                        4
            p
      +     ò     (p - x)sin nx dx]
           p
           3 /4
 2 éì
                           p /4          3p / 4
           cos nx sin nx ü p ì cos nx ü
= ê í x( -       ) + 2 ý + í-         ý
 p êî
   ë         n       n þ0  4î    n þp / 4

                  cos nx sin nx ü ù
                                  p
       ì
     + í(p - x)(-       )- 2 ý ú
       î            n      n þ3p / 4 ú
                                     û
   2      np       3np
 = 2 (sin    + sin     )
  pn       4        4
Thus the solution to the given problem is

  y( x,t)

 2   ¥
       1  np     3np
= å 2 (sin + sin     )sin nx cos nat
 p n=1 n   4      4
D’Alembert’s solution of the wave equation:

Consider the wave equation  y  1  y
                               2         2


                           x 2
                                 a t
                                  2   2


We change the independent variables x, t to u,
v by substituting u = x + at , v = x - at
    y   y u   y v   y   y
                         
    x   u x   v x   u   v
        y
         2
                 y    y 
                 u    x   v 
       x 2
              x             
     y u  y v  y u  y v
      2            2        2         2
    2                  2
    u x vu x uv x v x
    y2
           u  y
               2       2
    2 2      2
    u    uv v
(assuming that all mixed partial derivates are
continuous). Similarly
      y  2
                2 y   u  y
                        2   2
            a  2 2
              2
                u        2
      t 2
                     uv v 
                              
Hence the wave equation becomes
   y
    2
              y
               2
                     or    y
                           2
2        2                  0
  u v      u v       uv

Integrating partially w.r.t u, we get
  y    = a constant independent of u
  v
        = a function of v = g(v) say
Again integrating partially w.r.t v, we get
 y   (v ) dv  a
      g                (where a is a constant
     v
                       independent of v)
    = G (v) + F (u)(say)

    = F (x + at) + G (x - at)
We have to chose F,G so as to satisfy the initi
al and boundary conditions. This is extremel
y difficult. Hence we have to adopt the Fouri
er Series method for solving the wave equati
on.
ONE-DIMENSIONAL
HEAT EQUATION
The heat equation:

We consider the flow of heat in a thin cylin
drical rod of cross-sectional area A whose l
ateral surface is perfectly insulated so that n
o heat flows through it. The word ‘thin’ me
ans that the temperature is uniform on any
cross- section and thus is a function of the t
ime t and its horizontal distance x from one
end, w(x,t).
We note the following physical principles:
• Heat flows in the direction of decreasing te
  mperature, that is from hot regions to cold re
  gions
• The rate at which heat flows across an area i
  s proportional to the area and to the rate of ch
  ange of temperature with respect to the dista
  nce in a direction perpendicular to the area. (
  This proportionality factor is denoted by k an
  d called the thermal conductivity of the subst
  ance.)
c) The quantity of heat gained or lost by a b
  ody when its temperature changes is propo
  rtional to the mass of the body and to the c
  hange of temperature. (This proportionalit
  y factor is denoted by c and called the spe
  cific heat of the substance.)
Using the above three principles, we can deri
ve the heat equation describing the temperatu
re w(x,t) as
                        2 w    1 w
                              2
                        x 2
                               a t
Here the constant a2 depends on three things:
,density of the rod, c, specific heat of the su
bstance and k, thermal conductivity of the subs
tance.
              k
 In fact a =
          2

             cr
We assume the boundary conditions
      w (0, t )  w ( L , t )  0
This means that the ends are maintained at zer
o temperature at all times.
We also assume that the rod has an initial temp
erature distribution f(x) at time t = 0
i.e. w(x,0) = f(x)    0<x<L
Again ‘separating’ the variables, we find that
w=uv
where u satisfies the eigenvalue problem
              
            u   u  0
     u(0)  u(L)  0        (1)
and that v satisfies first order d.e
      v    a 2 v        (2)
Thus for nontrivial solutions,
        n 2p 2
   ln = 2 ( n = 1, 2,...)
           L
Corresponding eigenfunctions are
                     np x
           u n = sin
                      L
                                           n 2p 2 a 2 t
                                      -
For each n=1,2,3,…    vn = e                       L2


                              np       2       2
is a solution of (2) with ln = 2
                               a
Hence for each n = 1,2,3,…
                                n 2p 2 a 2 t
                     np x     -
   wn = u n vn = sin      e         L2
                      L
is a solution of the heat equation.
Again we assume that the infinite linear combin
ion
                                         n 2p 2 a 2 t
       ¥             ¥
                            np x       -
 w = å b n w n = å b n s in      e           L2

     n =1        n =1        L
is also a solution.             
Putting t = 0, we get f ( x)   b sin nx
                                    n
                               n 1     L
Hence bn = nth Fourier sine coefficient of f(x)
                           n x
               L
            2
in [0, L]  f ( x ) sin        dx
            L 0              L
Problem

Solve the heat equation given that
w (0, t ) = w ( L , t ) = 0;
                       ì x    0 < x £ L/2
w ( x , 0) = f ( x ) = í
                       îL - x L / 2 £ x £ L
 Solution

The solution to the heat equation subject to the
given boundary and initial conditions are:
                                         n 2p 2 a 2 t
       ¥               ¥
                            np x       -
 w = å b n w n = å b n s in      e           L2

     n =1        n =1        L
 where bn = nth Fourier sine coefficient of f(x)
                               n x
                   L
               2
   in [0, L] 
               L   f ( x ) sin L dx
                   0
      L                                 
            nx                    nx 
                     L
    2 2
bn   sin
         x      dx   L  x ) sin
                       (              dx 
    L 0      L                      L
      
      
                     L
                     2
                                         
                                         


                  Continued on the next
                  slide
    éì    æ        np x   ö       np x   ü
                                          L/2

    êï
  2 êï    ç - cos L       ÷ s in L       ï
                                         ï
=    íx   ç               ÷+             ý
  L êï          np             n p
                                2   2
                                         ï
          ç               ÷
    êï    è      L        ø              ï0
    ëî                                   þ
                                  2
                                L

   ì        æ        np x   ö      np x       ü
                                               L
                                                   ù
   ï
   ï        ç - co s L      ÷ sin L           ï
                                              ï
                                                   ú
 + í(L - x) ç               ÷-                ý    ú
   ï              np            n 2p 2        ï    ú
            ç               ÷
   ï
   î        è      L        ø     L 2
                                              ïL/2 ú
                                              þ    û
    2 é    L  2
                    np     L     2
                                              np
  =   ê - 2 np c o s 2 + n 2p        2
                                         s in
    L ë                                        2

             L    2
                      np   L     2
                                             np ù
          +      co s    + 2             sin    ú
            2 np       2  n p        2
                                              2 û

               ì 0              n = 2, 4, 6, ...
  4L      np   ï
= 2 2 sin    = í 4L / n p 2   2
                                n = 1, 5, 9, ...
 n p       2   ï - 4 L / n 2p 2 n = 3, 7 ,1 1, ...
               î
The solution to the heat equation subject to the
given boundary and initial conditions are:

w( x , t ) =
                                      ( 2 n -1) 2 p 2 a 2t
  4L    ¥
                   (2 n - 1)p x     -
   2 å
              n +1
         ( -1) sin              e             L2
  p n =1                L
Steady state temperature
We assume that when t is large, w a function i
ndependent of t, w(x) say. This is called the ste
ady state temperature. And noting that in stead
y state       w
                   0
               t
           w d w
             2      2
we get           2 0
           x 2
                 dx
 Therefore w = c1 x + c 2
Suppose that the ends are maintained at temper
ature w1,w2 at all times. Hence
     x  0  w  w1  c2  w1
                          1
     x = L Þ w = w2 Þ c1 = ( w2 - w1 )
                          L
Hence the steady state temperature is
            1
    w  w1  ( w2  w1 ) x   0xL
            L
Problem
Find the solution of the one-dimensional heat e
quation satisfying the boundary and initial cond
itions
     w ( 0 , t )  100 , w ( L , t )  0 , for all t.

                          px
     w ( x , 0) = 100 cos       0xL
                          2L
Solution

We write the temperature distribution as
           w( x, t ) = wS ( x) + wT ( x, t )

= steady state temperature + Transient tempe
rature.
We impose the boundary conditions              wS(0
) = 100, wS(L) = 0 on wS(x).
Thus
               1                 x
ws ( x)  100  (0 100) x  100   0  x  L
                                 1
               L                 L

We impose on wT(x,t) the boundary conditions
wT(0,t) = wT(L,t) = 0 for all t
and the initial condition
                        px
   wT ( x, 0) = 100 cos    - wS ( x)   0£ x£ L
                        2L
That is : wT(x, t) satisfies the boundary conditio
ns
             wT(0,t)=wT(L,t) = 0 for all t

 and the initial condition
                   px            x
wT ( x, 0) = 100 cos    - 100(1 - )   0£ x£ L
                     2L          L
                                          n 2p 2 a 2 t
Thus
                      ¥
                                 np x -
       w T ( x , t ) = å b n sin     e        L2

                       n =1       L
where
     2
         L
                       px               x       np x
bn =
     L   ò [1 0 0 co s 2 L - 1 0 0 (1 - L )] sin L d x
         0


         L
    2              px     np x
  =   ò 1 0 0 co s 2 L sin L d x
    L 0
               L
          2            x      np x
         - ò 1 0 0 (1 - ) sin      dx
          L 0          L       L
Now
 2
      L
                px     np x
   ò 1 0 0 co s 2 L sin L d x
 L 0


             ( 2 n  1)x ( 2 n  1)x 
          L
  100

   L      sin 2 L  sin
          0 
                                 L      
                                        
                                         dx
                                                    L
  100  2L           (2n 1)x 2L        (2n 1)x 
     (2n 1) cos 2L  (2n 1) cos 2L 
   L                                              x0
                                          
  200   cos( 2 n  1) 2 1  cos( 2 n  1) 2 
       1
                                           
         ( 2 n  1)          ( 2 n  1)     
                                             

   200 é 1             1 ù
 =     ê (2n + 1) + (2n - 1) ú
    p ë                      û
   800 n
           ( n  1,2,3,..)
     4n 1
        2
                              np x
          L
 And    2              x
          ò 1 0 0 (1 - L ) sin L d x
        L 0

                                             L
      ì                np x          np x   ü
      ï           co s          sin         ï
  200 ï     x            L )- 1         L   ï
=     í (1 - )( -                           ý
   L ï      L        np       L n 2p 2      ï
      ï
      î                L            L 2
                                            ï x=0
                                            þ

          200 L   200
        =       =
           L np   np
Thus
       w( x, t )
                                       n2p 2a2
         x 200 æ 4n¥
                            1 ö np x  - 2 t
 = 100(1- ) +
         L
                åç 4n2 -1 - n ÷ sin L e
              p n=1 è         ø
                                         L
 Problem
The temperatures at the ends x = 0 and x = 100
of a rod 100 cm in length with insulated sides a
re held at 0 and 100 degree centigrade respectiv
ely, until steady-state conditions prevail. Then a
t the instant t = 0, the temperatures of the two e
nds are interchanged. Find the resultant tempera
ture distribution of the rod as a function of x an
d t.
Solution

We note that the steady state temperature (w
hen the ends x = 0 and x = 100 of the rod 10
0 cm in length are held at 0 and 100 degree
centigrade respectively) is

  f ( x ) = x , 0 < x < 100
Now, the temperatures of the two ends are i
nterchanged.
And we have to find the new temperature di
stribution in the rod.

Hence we have to solve the heat equation wi
th the boundary and initial conditions

w (0, t ) = 100, w (100, t ) = 0, for all t.

  w ( x , 0) = x ,   0xL
We write the temperature distribution as
      w( x, t ) = wS ( x) + wT ( x, t )

= (new) steady state temperature + Transient
temperature.
We impose the boundary conditions         wS(0
) = 100, wS(100) = 0 on wS(x).
Thus
                  1
 ws ( x) = 100 +     (0 - 100) x = 100 - x 0 £ x £ 100
                 100

We impose on wT(x,t) the boundary conditions
wT(0,t) = wT(100,t) = 0 for all t
 and the initial condition
       wT ( x,0) = x - wS ( x)   0 £ x £ 100
i.e.    wT ( x,0) = 2 x - 100    0 £ x £ 100
That is : wT(x, t) satisfies the boundary conditio
ns
             wT(0,t)=wT(100,t) = 0 for all t

 and the initial condition
        wT ( x,0) = 2 x - 100   0 £ x £ 100
                                              n 2p 2 a 2 t
Thus
                        ¥
                                   np x     -
        w T ( x , t ) = å b n s in      e       1002

                        n =1       100
                 100
where b n = 2                              np x
           100   ò0
                       [ 2 x - 1 0 0 ] sin
                                           100
                                                dx


         é            æ      np x           ö
      2  ê            ç co s 1 0 0          ÷
   =     ê(2 x - 100) ç -                   ÷
     100 ê            ç np / 1 0 0          ÷
         ê
         ë            è                     ø
                                             100
                     æ      np x          öù
                     ç  sin               ÷ú
                 + 2 ç 2 2 100 2          ÷ú
                     ç n p /100           ÷ú
                     è                    øú0
                                           û
      200                              400
    =     (1 - co s np )             =     , n odd
      np                               np

Hence the desired temperature distribution is:
 w( x, t ) = wS ( x) + wT ( x, t )

  = 100 - x
                                              (2 n-1)2 p 2a2t
    400 ¥   1       (2n -1)p x              -
  +    å (2n -1) sin 100 e
     p n=1
                                                   1002
END OF WAVE AND HE
AT EQUATIONS
TWO-DIMENSIONAL H
EAT EQUATION
- LAPLACE’S EQUATIO
N
The two-dimensional heat equation:
A thin rectangular plate has its boundary kep
t at temperatures 0,0,0 and f(x) at all times. F
ind the steady temperature distribution in the
plate.
Thus we have to solve the Laplace’s equation
         w  w
          2       2
              2 0
        x 2
              y
subject to the boundary conditions
  w ( 0, y )  w ( L , y )  0   0 yM

  w( x, M ) = 0, w( x,0) = f ( x) 0 £ x £ L
We again solve by the method of separation of
variables. We assume w(x,y) = X(x) Y(y)
Substituting and separating the variables we get
    X   Y
             (a constant)
    X     Y
Thus X satisfies the boundary value problem
      X ¢¢ + l X = 0, X(0) = X(L) = 0
We know nontrivial solutions exist when
              n p2   2
     l = ln =   2
                  ( n = 1, 2, 3, ...)
               L
and the corresponding eigenfunction is
                   nx
         X n  sin
                    L
For each n, Y satisfies the BV problem
     
   Y  n Y  0 , Y ( M )  0
               ny          ny
Now Y  B sinh      C cosh
                L            L
                      np M          np M
 Y ( M ) = 0 Þ B sinh      + C cosh      =0
                        L             L
                     np M              np M
Taking    B = - cosh      , C = + sinh
                       L                 L

we find              np ( M - y )
            Y = sinh
                          L
Thus for each n,
                  np x      np ( M - y )
      wn = bn sin      sinh
                   L             L
is a solution satisfying the first three boundary
conditions. Hence
               ¥
                    np x      np ( M - y )
w = å wn = å bn sin      sinh
           n =1      L             L
is also a solution.
Putting y = 0, we get
           
                       nM     nx
    f ( x)   bn sinh     sin
             n 1       L       L
              nM
Hence bn sinh
               L
 = nth Fourier sine coefficient of f(x) in [0, L]
     L
 2              np x
= ò f ( x ) sin      dx ( n = 1, 2,3,...)
 L0              L

Thus we have found out the steady state temp
erature distribution.


          THE END
BEST OF LUCK IN T
HE COMPRE

								
To top