2D heat equation bounday conditions for different intervals

[Gregg]

Homework Statement

I have boundary conditions on my heat equation $\dot{T}(x,t) = T''(x,t)$

$T(0,t) = T(L,t),$

$\frac{\partial T(0,t)}{\partial x} = \frac{\partial T(L,t)}{\partial x}$

Then

at $T= 0$

$T(x,0) = 1$ for $0<x<L/4$
$T(x,0) = 0$ for $L/4<x<L$

Homework Equations

None.

The Attempt at a Solution

I have done the usual separation of variables, then getting two ODEs equal to a constant. I've called it $\alpha^2$ I have eliminated the possibility that $\alpha^2 = 0.$ Will this always happen?

Now do I need to consider $\alpha^2 < 0$ and $\alpha^2 >0$ rather than just $\alpha^2\ne 0$?

I find the fact that the boundary conditions run over an interval to be off putting. Am I to use the differential boundary conditions instead for now? I have deduced that:

$T(x,t) = c_1(e^{\alpha x} - e^{-\alpha x})c_2 e^{\alpha^2 t}$

Of course if $T_i(x,t)$ is a solution then $T(x,t) = \displaystyle \sum_i T_i(x,t)$

I have found that

$T(x,t) = \sum_{n=1}^{\infty} C_n \sin ({n \pi x \over l}) e^{n^2 \pi^2 t \over l^2}$

So I need to find the Fourier coefficients of that function that is constant on the two different intervals. This is where I get stuck.

$C_n = \frac{1}{l} \int_0^{2l} f(x) \sin ({n \pi x \over l}) dx = \frac{1}{l} \int_0^{l/4} 0\cdot dx + \int_{l/4}^{l} \sin ({n \pi x \over l}) dx+ \frac{1}{l} \int_{l}^{5l/4} 0\cdot dx + \int_{5l/4}^{2l} \sin ({n \pi x \over l}) dx$

I don't think this the correct form for the coefficient:

$C_n = \frac{1}{n \pi} \left[ -(-1)^n + 1 +0 + \cos({4 n \pi \over 5})\right]$

So is it just integration over $[0,l]$ ?

This would give

$T(x,t) = \sum_{n=1}^{\infty} 2 \sin ({(2n-1) \pi x \over l}) e^{n^2 \pi^2 t \over l^2}$

 

[Dustinsfl]

I don't think your solution is correct. I don't like using T for the solution so I will write it as u(x,t)

Let $u(x,t)=\varphi(x)T(t)$. Then $\frac{\varphi''}{\varphi}=\frac{T'}{T}=-\lambda$.

So we have $\varphi''+\lambda\varphi = 0\Rightarrow \varphi(x)=A\cos x\sqrt{\lambda} + B\sin x\sqrt{\lambda}$ and $T(t)=\exp(-\lambda t)$.

Next we have to solve $\varphi$ for
$\varphi_1(0) = 1\quad\quad\varphi_2'(0) = 0$
$\varphi_1(0) = 0\quad\quad\varphi_2'(0) = 1$

$\varphi(x) = A\cos x\sqrt{\lambda} + B\frac{\sin x\sqrt{\lambda}}{\sqrt{\lambda}}$

Using your initial conditions, we get
$$
A(1 - \cos L\sqrt{\lambda}) = B\frac{\sin L\sqrt{\lambda}}{\sqrt{\lambda}} \quad \text{and} \quad B(1 - \cos L\sqrt{\lambda}) = -A\sqrt{\lambda}\sin L\sqrt{\lambda}
$$
I solved for B and obtained
$$
B = \frac{-A\sqrt{\lambda}\sin L\sqrt{\lambda}}{1 - \cos L\sqrt{\lambda}}.
$$
From this, I can substitute and get
$$
A\left[(1 - \cos L\sqrt{\lambda})^2 + \sin^2 L\sqrt{\lambda}\right] = 0
$$
That means $\lambda_n = \frac{4\pi^2n^2}{L^2}$.
This leads to
$$
u(x,t) = \sum_{n=1}^{\infty}a_n\left(\cos \frac{2x\pi n}{L} + \frac{L}{2\pi n}\sin\frac{2x\pi n}{L}\right)\exp\left(-\frac{4t\pi^2 n^2}{L^2}\right)
$$