Partial differential equation help

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SUMMARY

The discussion focuses on solving the partial differential equation (PDE) given by u(t) = Uxx - U with boundary conditions u(0,t) = 0 and u(1,t) = 0, and initial condition u(x,0) = sin(pi*x) + 0.5(sin(3*pi*x)). The method of separation of variables is employed, leading to the equations T'/T = α and (X'' + X)/X = α. The analysis concludes that to satisfy the boundary conditions, α must equal 1 + nπ, where n is a negative integer. The final solution is expressed as a sum over n of the derived solutions.

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

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= [itex]\alpha[/itex] and (X"+ X)/X= [itex]alpha[/itex].

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

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