Heat equation with sine initial conditions

[mousemouse]

Hi,

I'm new here.

given the pde:

u(t) = Uxx - U
0<x<t
0<t<inf

B.C.
u(0,t) = 0
u(1,t) = 0

i.c.
u(x,0) = sin(pi*x) + 0.5(sin(3*pi*x))
when 0<x<1

can anyone help me with the solution?

 

[HallsofIvy]

When you say "u(t)" do you mean U_t?

If so, you should be able to use standard "separation of variables". Assume we can write U(x,t)= X(x)T(t). Then the equation is XT'= TX"+ XT. Dividng through by XT,
T'/T= (X"+ X)/X. Since the left side is a function of t only and the right side a function of x only, they must each be equal to the same constant:

T'/T= $\alpha$ and (X"+ X)/X= $alpha$.

(X"+ X)/X= $\alpha$ gives X"+ X= $\alpha$X or X"+ (1-$\alpha$)X= 0. If 1-$\alpha$> 0, that gives exponential solutions which cannot satisfy the boundary conditions. If 1- $\alpha$ = 0, that gives a linear functions which cannot satisfy the boundary conditions. In order to satisfy the boundary conditions, 1- $\alpha$ must equal a negative multiple of $\pi$: $1- \alpha= -n\pi$ so $\alpha= 1+ n\pi$. Put that into the equation for T and solve. The solution to the original problem is the sum, over n, of those solutions.