Question

Problem # 3 [20 Points] Solve
PDE:   ut = uxx - u, 0 < x < 1, 0  < t < ∞

BCs:   u(0, t)=0
u(1, t)=0 0 < t < ∞

IC:   u(x, 0) = sin(πx), 0 ≤ x ≤   1
directly by separation of variables without making any preliminary trans- formation. Does your solution agree with the solution you would obtain if transformation

u(x, t)= e(caret)(-t) w(x, t)
were made in advance?

Answer 1

Let

u(z,t) = X(z)T(t)

Then $u_t=X(x)T'(t)~,~u_{xx}=X''(x)T(t)$

Using in the given equation we get

seperateing the variables XT(t)

Equate to some constant we get

X ( T(t) 2

Now as boundary conditions are

u(0,t0-u(1,t)

So ---------(3)

X(1) (0) = 0

Now the solution of (1) is  

X (z) C1 cos pr + C2 sin pr sin D.T

Using (3) we get $C_1=0$ and

C2 sin p 0

Thus the solution of (1) is

7t 71

Also, the solution of (2) is

$T_{n}(t)=B_ne^{-n^2\pi^2 t}$

Therefore the solution for the given PDE is

$u_{n}(x,t)=X_{n}(x)T_{n}(t)\\ \\~~~~ ~~~~~~~~~~~= [C_n\sin{n\pi{x}}]B_n e^{-n^2\pi^2 t} \\ \\~~~~ ~~~~~~~~~~~=A_n \sin{n\pi{x}}e^{-n^2\pi^2 t}$

Using Superimposition principle we get

$u(x,t)=\sum_{n=1}^{\infty} u_{n}(x,t)=\\ \\~~~~ ~~~~~~~~~~~=\sum_{n=1}^{\infty}A_n \sin{n\pi{x}}e^{-n^2\pi^2 t}$

As

a(2,0) = sin(TI)

Hence $A_1=1, A_2=A_3=\cdots=0$

Hence the required solution is $u(x,t)= \sin({ \pi{x}})~e^{- \pi^2 t}$