1. ## Non-Homogenous Heat Equation

Consider the following non-homogenous heat equation on $\displaystyle 0 \leq x \leq \pi$

$\displaystyle u_t = k u_{xx} - 1$ with $\displaystyle u(x,0) = 0, u(0,t) = 0, u(\pi, t) = 0$

Find a solution of the form

$\displaystyle \displaystyle \sum_1^{\infty} b_n(t) \phi_n (x)$

where $\displaystyle \phi_n(x)$ are the eigenfunctions of an appropriate homogenous problem, and find explicit expressions for $\displaystyle b_n(t)$

2. So I think

$\displaystyle \phi_n(x) = \sin \frac{n \pi x}{L}$

so I find solutions in the form

$\displaystyle \displaystyle u(x,t) = \sum_1^{\infty} b_n (t) \ \sin \frac{n \pi x}{L}$

Am I on the right track? Is the eigenfunction correct?

3. You are on the right track. You'll need a Fourier sine series for -1 and then sub in your expansion and equate coefficients.

4. Originally Posted by JoernE
So I think

$\displaystyle \phi_n(x) = \sin \frac{n \pi x}{L}$

so I find solutions in the form

$\displaystyle \displaystyle u(x,t) = \sum_1^{\infty} b_n (t) \ \sin \frac{n \pi x}{L}$

Am I on the right track? Is the eigenfunction correct?
Okay, ignore this last post. I have a different method....

Let $\displaystyle u = w(x,t) + v(x)$

Then

$\displaystyle \displaystyle \frac{\partial u}{\partial t} = \frac{\partial w}{\partial t}$ and $\displaystyle \displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 v}{\partial x^2}$

Sub these back into the PDE to obtain

$\displaystyle \displaystyle \frac{\partial w}{\partial t} = k \frac{\partial^2 w}{\partial x^2} + k \frac{\partial^2 v}{\partial x^2} - 1$

Choose $\displaystyle \displaystyle k \frac{\partial^2 v}{\partial x^2} - 1 = 0$

So

$\displaystyle \displaystyle V'' = \frac{1}{k}, \ V(0) = V(\pi) = 0$

$\displaystyle \displaystyle V' = \frac{1}{k} + A$

$\displaystyle \displaystyle V = \frac{1}{k} + Ax + B$

$\displaystyle \displaystyle V(0) = \frac{1}{k} + B = 0 \ \Rightarrow \ B = - \frac{1}{k}$

$\displaystyle \displaystyle V(\pi) = \frac{1}{k} + A \pi + B \ \Rightarrow \ A = -\frac{1}{k \pi} - \frac{B}{\pi}$

$\displaystyle \displaystyle V(x) = 0$??

Also since $\displaystyle \displaystyle k \frac{\partial^2 v}{\partial x^2} - 1 = 0$ we have

$\displaystyle \displaystyle \frac{\partial w}{\partial t} = k \frac{\partial^2 w}{\partial x^2}$ which is homogeneous.

$\displaystyle \displaystyle u(0, t) = v(0) + w(0, t) = 0$

$\displaystyle \displaystyle \Rightarrow \ w(0,t) = 0$

$\displaystyle \displaystyle u(x, 0) = v(x) + w(x,0) = 0$

$\displaystyle \displaystyle \Rightarrow \ w(x,0) = 0$

$\displaystyle \displaystyle u(\pi, t) = v(\pi) + w(\pi, t) = 0$

$\displaystyle \displaystyle \Rightarrow \ w(\pi, t) = 0$

Then we have

$\displaystyle \displaystyle w_t = k w_{xx}$ with $\displaystyle \displaystyle w(x,o) = 0, w(0,t) = 0, w(\pi, t) = 0$.

Since this is homogenous, we can solve by letting

$\displaystyle \displaystyle w = XT$

$\displaystyle \displaystyle \frac{T'}{kT} = \frac{X''}{X} = - \lambda$

$\displaystyle \displaystyle X = A \cos(\sqrt{\lambda}t) + B \sin(\sqrt{\lambda}t)$

$\displaystyle \displaystyle X(0) = A = 0$

but now I get a little lost.....I think I know where to go, just not exactly how to get there. Already have calculated that $\displaystyle v(x) = 0$, which will lead to what I need to prove (I think).

5. Originally Posted by JoernE

$\displaystyle \displaystyle X = A \cos(\sqrt{\lambda}t) + B \sin(\sqrt{\lambda}t)$
This should be $\displaystyle \displaystyle X = A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x)$

so

$\displaystyle \displaystyle X(\pi) = B \sin (\sqrt{\lambda} \pi) = 0$

$\displaystyle \displaystyle \ \sqrt{\lambda} \pi = \frac{n \pi }{2}$

$\displaystyle \displaystyle \ \sqrt{\lambda} = \frac{n}{2}$

$\displaystyle \displaystyle \ \lambda_n = \big( \frac{n}{2} \big)^2$ for $\displaystyle n = 1,2,...$

So

$\displaystyle \displaystyle \phi_n(x) = \sin \big(\frac{n \pi x}{2} \big)$

and so

$\displaystyle \displaystyle u(x,t) = v(x) + \sum_1^{\infty} b_n \sin \big(\frac{n \pi}{2} \big)$

$\displaystyle \displaystyle \ = \displaystyle \sum_1^{\infty} b_n(t) \phi_n (x)$

and I think

$\displaystyle \displaystyle \ b_n = \frac{1}{\pi} \int_0^{\pi} \sin(\sqrt{\lambda} x) dx$

Could someone please tell me if this is now correct?

Thanks