PDE Separation of Variables with Nonzero Boundary Conditions

[jtleafs33]

Homework Statement

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)

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.

 

[LCKurtz]

Homework Statement

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)

The Attempt at a Solution

I want to solve using separation of variables...

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 $u(x,t) = v(x,t)+\psi(x)$.

 

[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?

 

[LCKurtz]

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?

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?

 

[jtleafs33]

I will be trying it soon.

 

[jtleafs33]

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