Question

Problem 3. Show that the solution of the partial differential equation (Laplace equation), Wxx(x,y) + Wyy(x, y) = 0, with the four boundary conditions: w(x,0) = 0, w(x, 1) = 0, w(0,y) = 0 and w(1, y) = 24 sin ny, can be obtained as w(x, y) = 2 sinh nx · sin ny.

Answer 1

Solution:-

Given that

The given function is,

......(1)

Wrr(x, y) + Wyyx, y) = - 0

W 2,0) = 0
,
W(2, 1) = 0

W(0,y) = 0
and
W(1,y) = 24 sin ny

Let be a solution of (1)

(6) A(x)x = (i =)

Then the function (1) becomes

x"y+y'r = 0

or,

J'y = -'T

or,  

2 y Y

Let   constant (say)

Y 1 Ꮖ

........(2)

I" - ur = 0

..........(3)

0 = hrl +,

So,

W 2,0) = 0

X (0)Y (0) = 0

Y(0) = 0

and  

W(1, 1) = X(C)Y(y) = 0

Y(1) = 0

where

, Since otherwise W = 0

XC) 70

which does not imply

W(1,y) = 24 sin ny

Now we are to solve the equation (3)

Case 1:-

Let

0

Then the solution of (2) is

Y y) = Ay) + B

where A, B are constants

Y(0) = 0 =B=0

Y (1) = 0 > A+B=0

A = 0

So, W = 0 which does not satisfy

W(1,y) = 24 sin ny

Case 2:-

Let (negative)  

=r

Here

170

Then the solution of (3) is

Y(y) = Aeyd + Be-y!

Y (0) = 0 > A+B=0

Y(1) = 0 = A + Be- 0

trivial solution

A=B=0

So, Y = 0

Then
WC, y) = 0
which does not satisfy
W(1,y) = 24 sin ny

Case 3:-

Let $\mu=\lambda^2$ (positive)

here

170

Then the solution of (3) is

Y(y) = Acos ly + B sin ly

or,

Y(0) = 0 A=0

Y(1) = 0 = B sin = 0

sin X=0

Since if B = 0

B=0

then Y = 0

and

WC, y) = 0

Which does not imply

W(1,y) = 24 sin ny

sin X=0

n = 1, 2, 3, ...

=nT

Hence

Yn(y) = Bn sin ny

Also, then

u=- -12 -n-H

then the equation (2)

reduces to, n = 1, 2, 3, ...

0 = X H U – X

So the solution is,

..........(4)

пат -NTI XnC) = Cne + Dne

Now,

W(0,y) = 0

XOY (y) = 0

..X(0) = 0

So,

gives
C + D = 0

$D_n=-C_n$

  

$X_n(x)=2C_n\sin h n \pi x$

Therefore

$\sum^\infty_{n=1}W_n(1,y)=24\sin \pi y=W(1,y)$

n = 1, 2, 3, ...

$E_n\sin h n \pi=2\int_{0}^{1}24\sin \pi y\ \sin n\pi ydy$

So, $W_n(x,y)=E_n\sin hn \pi x\ \sin n\pi y$

$W_n(x,y)=\sum^\infty_{n=1}(E_n\sin hn \pi x)\ \sin n\pi y$

hence proved

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