Rs t-e tzt p GoJ =O7,4 t at {116'ko9tft{ € Fouc-*OE; Soza.c r,vt l-r{ -u 4s DF |dGO1 tl={2 L ID1-D2...

Question

Rs t-e tzt p GoJ =O7,4 t at {116'ko9tft{ € Fouc-*OE; Soza.c r,vt l-r{ -u 4s DF |dGO1 tl={2 L ID1-D2 9'l

Robert · Accepted Answer

Ans. 1
So the given initial boundary problem, 
2 0 1
( ,0) (1 )
( ,0) 0 0 1
(0, ) (1, ) 0 0
tt xx
t
u C u x
u x x x
u x x
u t u t t
  
 
  
  
 
So, we use separation of the variable,  
Let, 
( , ) ( ) ( )u x t X x T t  
2
2
2
2
2 2
1 1
dT d X
X c T
dt dx
dT d X
dt Xc T dx


  
 
Separating the boundary conditions: 
(0, ) 0 (0) ( ) 0 (0) 0
( , ) 0 ( ) ( ) 0 ( ) 0
x
dX dX
u t T t
dx dx
u L t X L T t X L
    
    
 
Now, the solution toeigon valueproblem is, 
2
(2 1) (2 1)
, ( ) cos , 1,2,...
2 2
n n
n n x
X x n
L L
 

  
   
 
 
And, 
 
2
2 2 1
2
( )
n
c t
L
n nT t c e
 
   
   
Now, for n  
2
2 (2 1)
2 (2 1)( , ) cos
2
n
c t
L
n n
n x
u x t c e
L


 
  
 
 
  
 
 
Or,
2
2 (2 1)
2
1
(2 1)
( , ) cos
2
n
c t
L
n n
n
n x
u x t c e
L


    
 

 
  
 

Now, to solve the initial value constant apply the boundary conditions, 
2
2 (2 1)
2
1
(2 1)
( ) ( ,0) ( , ) cos
2
n
c t
L
n n
n
n x
f x u x u x t c e
L


    
 

 
    
 
  
So from earlier, the section 
(2 1)
cos
2
n x
L
 
 
 
is  lie in a orthogonalsection 0 1x    
So,

Rs t-er> tzt p G oJ =O 7,4 t at {116 'ko9t ft{ € Fouc -*O E/ ; Soza.c r, vt l-r{ -u 4 s DF |dG >O 1 tl ={2 L I D1- D2 9'l

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