The temperature $T$ in a one-dimensional bar whose sides are perfectly insulated obeys the heat flow equation $$ \frac{\partial T}{\partial t} = \kappa \frac{\partial^2 T}{\partial x^2} $$ where $\kappa$ is a positive constant. The bar extends from $x=0$ to $x=L$ and is perfectly insulated at $x=L$. At $t < 0$ the temperature is $0^\circ$C throughout the bar and at $t=0$ the uninsulated end is placed in contact with a heat bath at $100^\circ$C. Show that the temperature of the bar at subsequent times is given by

$$ \frac{T}{100}=1-\sum_{n=0}^\infty \frac{4}{(2n+1) \pi} \sin\left(\frac{(2n+1)\pi x}{2L}\right) \exp \left[ - \kappa \left(\frac{(2n+1)\pi x}{2L}\right)^2 t\right] $$

Here is what I have so far, and where I am stuck.

Assume separable solution $T(x,t) = F(t)G(x)$ $$ \frac{F'}{F} = \frac{\kappa G''}{G} = -\lambda $$ So: $$ F(t)\propto e^{- \lambda t} $$ $$ G(x)\propto e^{\pm \sqrt{\frac{\lambda}{\kappa}} t} $$ Let's define $q \equiv \sqrt{\frac{\lambda}{\kappa}}$ for convenience, giving a general solution of $$ T(x,t)=T_0 + \sum_{q>0}e^{-\kappa q^2 t} \left[ A_q \sin q x + B_q \cos q x \right] $$ We have 4 boundary conditions to apply:

  1. $T(x,t) \to T_0$ as $T \to \infty$. This fixes the value of $T_0$ as $100^\circ$C.

  2. $T(0,t)=T_0$ for all $t>0$. This sets all $B_q = 0$.

  3. $\frac{\partial T}{\partial x} \Big|_{x=L}=0$. This sets $\cos (qL) = 0$, restricting $q$ to the values $q_n = \frac{\pi}{L}(\frac{1}{2} + n)$

So our solution now looks like

$$T(x,t) = T_0 + \sum_{n=0}^{\infty} A_n e^{-kq_n^2 t} \sin{(q_n x)}$$

  1. $T(0<x<L , t=0) = 0^\circ C$. Applying this to our solution gives: $$-T_0= \sum_{n=0}^{\infty} A_n \sin{(q_n x)}$$

And here's the problem: I can't get out an expression for $A_n$. I would expect to take the inner product of each side with some $\sin mx$ to knock out some terms, giving me an expression for $A_n$.

The problem is that $\sin q_n x = \sin \frac{\pi}{L}(\frac{1}{2} + n)$ which is no longer orthogonal to $\sin mx$ over [0,L].

Perhaps I am meant to take the inner product with $\sin q_m x$. I've used Wolfram alpha to find out that $\sin q_n x$ and $\sin q_m x$ are indeed orthogonal for $n \neq m$ over $[0,L]$, but I can't see why.


Perhaps I am meant to take the inner product with $\sin q_m x$.

Yes, that's the thing: you should take inner product of both sides with the eigenfunction whose coefficient you want to find. And here is why the eigenfunctions are orthogonal:

You are considering the eigenfunctions of $d^2/dx^2$ with boundary conditions $\phi(0)=0$, $\phi'(L)=0$. If $\phi$ and $\psi$ are two eigenfunction for distinct eigenvalues $\lambda$ and $\mu$, then $$ \lambda\int_0^L \phi(x)\psi(x)\,dx = \int_0^L \phi''(x)\psi(x)\,dx = (\phi'\psi) |_0^L -\int_0^L \phi'(x)\psi'(x)\,dx \tag{1} $$ and $$ \mu\int_0^L \phi(x)\psi(x)\,dx = \int_0^L \phi(x)\psi''(x)\,dx = (\phi\psi') |_0^L -\int_0^L \phi'(x)\psi'(x)\,dx \tag{2}$$ The boundary conditions ensure that in both (1) and (2), the boundary term is zero. Hence $$ \lambda\int_0^L \phi(x)\psi(x)\,dx = \mu\int_0^L \phi(x)\psi(x)\,dx $$ which implies $ \int_0^L \phi(x)\psi(x)\,dx=0$.

More generally, the above argument works whenever the boundary conditions are such that $$(\phi'\psi)|_0^L = (\phi\psi')|_0^L \tag{3}$$ holds for all pairs $\phi,\psi$ satisfying these conditions. Such boundary conditions are called symmetric, and they are useful precisely because the corresponding eigenfunctions are orthogonal. The Dirichlet, Neumann, and Robin boundary conditions are all symmetric, and so is any mix of them.

  • $\begingroup$ Thanks for your help, I think I understand it now! $\endgroup$ – Tarrare Jan 11 '15 at 0:47

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