- #1

jtleafs33

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## Homework Statement

Solve the diffusion equation:

[itex]u_{xx}-\alpha^2 u_{t}=0[/itex]

With the boundary and initial conditions:

[itex]u(0,t)=u_{0}[/itex]

[itex]u(L,t)=u_{L}[/itex]

[itex]u(x,0=\phi(x)[/itex]

## The Attempt at a Solution

I want to solve using separation of variables...

I start by assuming a solution of the form:

[itex]u(x,t)=X(x)T(t)[/itex]

Differentiating and dropping the notation so f(F) = F :

[itex]u_{x}=X'T[/itex]

[itex]u_{xx}=X''T[/itex]

[itex]u_{t}=XT'[/itex]

Substituting back into the PDE,

[itex]X''T-\alpha^2 XT'=0[/itex]

[itex]X''T=\alpha^2 XT[/itex]

[itex]\frac{X''}{X}=\alpha^2 \frac{T'}{T}[/itex]

Let each side be equal to a constant, [itex]-\lambda^2[/itex]

So now we have two ODE's:

[itex]X''+\lambda^2 X=0[/itex]

[itex]T'+(\frac{\lambda}{\alpha})^2 T=0[/itex]

These ODE's are easy to solve:

[itex]T(t)=Ae^{-(\frac{\lambda}{\alpha})^2 t}[/itex]

[itex]X(x)=Bsin(\lambda x)+Ccos(\lambda x)[/itex]

So now, let AB=A and AC=B so the PDE has the general solution:

[itex]u_{\lambda}(x,t)=e^{-(\frac{\lambda}{\alpha})^2 t}[Asin(\lambda x)+Bcos(\lambda x)][/itex]

So now, I want to match my boundary conditions. All the other PDE's I've solved have zero-valued boundary conditions, so I'm not exactly sure what to do here.

[itex]u(0,t)=u_{0}=e^{-(\frac{\lambda}{\alpha})^2 t}[Asin(0)+Bcos(0)][/itex]

[itex]u(0,t)=u_{0}=Be^{-(\frac{\lambda}{\alpha})^2 t}[/itex]

This is where I'm confused.

To satisfy the B.C., my intuition says [itex]B=u_{0}[/itex] and [itex]e^{-(\frac{\lambda}{\alpha})^2 t}[/itex] must be a multiple of [itex]\pi[/itex].

Even if this is correct, I'm not sure how to express it as a constraint on [itex]\lambda[/itex].