# PDE Separation of Variables with Nonzero Boundary Conditions

1. Nov 24, 2012

### jtleafs33

1. The problem statement, all variables and given/known data
Solve the diffusion equation:

$u_{xx}-\alpha^2 u_{t}=0$

With the boundary and initial conditions:

$u(0,t)=u_{0}$
$u(L,t)=u_{L}$
$u(x,0=\phi(x)$

3. The attempt at a solution

I want to solve using separation of variables...
I start by assuming a solution of the form:

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

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

$u_{x}=X'T$
$u_{xx}=X''T$
$u_{t}=XT'$

Substituting back into the PDE,

$X''T-\alpha^2 XT'=0$
$X''T=\alpha^2 XT$

$\frac{X''}{X}=\alpha^2 \frac{T'}{T}$

Let each side be equal to a constant, $-\lambda^2$
So now we have two ODE's:

$X''+\lambda^2 X=0$
$T'+(\frac{\lambda}{\alpha})^2 T=0$

These ODE's are easy to solve:

$T(t)=Ae^{-(\frac{\lambda}{\alpha})^2 t}$
$X(x)=Bsin(\lambda x)+Ccos(\lambda x)$

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

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

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.

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

This is where I'm confused.
To satisfy the B.C., my intuition says $B=u_{0}$ and $e^{-(\frac{\lambda}{\alpha})^2 t}$ must be a multiple of $\pi$.
Even if this is correct, I'm not sure how to express it as a constraint on $\lambda$.

2. Nov 24, 2012

### LCKurtz

Before you separate variables, make the substitution $u(x,t) = v(x,t) +\psi(x)$. Substitute that into your problem:$$v_{xx} + \psi''(x) + v_t = 0$$ $$v(0,t) + \psi(0) = u_0$$ $$v(L,t) + \psi(L) = u_L$$ $$v(x,0) + \psi(x) = \phi(x)$$
Now if you let $\psi''(x)=0,\ \psi(0)=u_0,\ \psi(L)=u_L$, you can solve for $\psi(x)$ and what is left is a system in $v(x)$ you can separate variables on. Your boundary conditions will be homogeneous and the last equation will be $v(x,0)=\phi(x) - \psi(x)$. Your final answer will be $u(x,t) = v(x,t)+\psi(x)$.

3. Nov 24, 2012

### jtleafs33

LCKurtz,

I can't find anything like that in my textbook. Did you do that so the solution would satisfy both the nonhomogenous and homogenous parts? I suppose you added the extra function of x because the nonhomogenous boundary conditions only depend on x?

4. Nov 24, 2012

### LCKurtz

I don't know about your text, but it is a standard method. You let the $\psi(x)$ take care of the nonhomogeneous part. So did you try it on your problem?

5. Nov 24, 2012

### jtleafs33

I will be trying it soon.

6. Nov 25, 2012

### jtleafs33

Got it! It was trivial after I did the substitution you mentioned. Thanks again!