Solving du/dt = (d^2)u/d(x^2) with Boundary Conditions

[CarmineCortez]

Homework Statement

du/dt = (d^2)u/d(x^2), t>0, 0<x<1

u(0,t) = 0 = u(1,t) , t>0
u(x,0) = P(x), 0<x<1

P(x) = {0 , if abs(x-1/2) >epsilon/2
{u/epsilon, if abs(x-1/2) <= epsilon/2


i need to find u(1/2,1/pi^2)

Homework Equations

i have u(x,t) = SUM{ 2/(n*pi) *P(x)*(1-cos(n*pi))sin(n*pi*x)exp(-t*(n*pi)^2)}

The Attempt at a Solution

so when I try to get u(1/2,1/pi^2) i get 2U/e * 2/(pi*epsilon) * (exp(-2)-exp(-10)+exp(-26)-...)


I know the answer is u(1/2,1/Pi^2) = 2U/e * (sin(pi*epsilon/2)/(pi*epsilon/2))

 

[mighty2000]

but I don't know how to get there. Can someone please help me understand how to get to this answer?

First, let's rewrite the given equation as:

du/dt = u''(x)

where u''(x) is the second derivative of u with respect to x.

Next, we can use the method of separation of variables to solve this differential equation. We assume that u(x,t) can be written as a product of two functions, one depending only on x and the other depending only on t:

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

Substituting this into our equation, we get:

X''(x)T(t) = X(x)T'(t)

Dividing both sides by X(x)T(t), we get:

X''(x)/X(x) = T'(t)/T(t)

Since the left side of the equation depends only on x and the right side depends only on t, both sides must be constant. Let's call this constant -λ^2. This gives us two separate equations:

X''(x) + λ^2X(x) = 0

T'(t) + λ^2T(t) = 0

The solution to the second equation is:

T(t) = Ae^-λ^2t

where A is a constant of integration.

For the first equation, we need to solve for X(x). The general solution to this equation is:

X(x) = c1cos(λx) + c2sin(λx)

Applying the boundary conditions u(0,t) = 0 and u(1,t) = 0, we get:

X(0) = c1 = 0

X(1) = c2sin(λ) = 0

Since we want a non-trivial solution, c2 must be non-zero. This means that sin(λ) = 0, which gives us the values of λ as nπ, where n is a positive integer.

Therefore, our solution for X(x) is:

X(x) = c2sin(nπx)

Substituting this into our original equation, we get:

u(x,t) = T(t)X(x) = Ae^-n^2π^2t sin(nπx)

We can now use the initial condition u(x,0) = P(x) to solve for A. Since u(x,0) =