# Diffusion equation problem

Heat conduction is described by the diffusion equation $$u_t = k u_{xx},$$ where $u(x,t)$ is the temperature at $x$ at time $t$ and $k$ is some constant. A homogeneous thin pipe occupying the region $0\leq x \leq \ell$ is completely insulated (laterally and on the ends). Its initial temperature is $f(x)$. After a sufficiently long time, the temperature of the object eventually reaches a steady (or equilibrium) state. We call this the steady state temperature. Find a formula for the steady state temperature in terms of $f$.

Hints:

• The steady state temperature does not change in time. What does this mean for $u_t$?
• The amount of heat is constant. None is lost or gained.

I'm not sure what the question is trying to ask. Any help would be appreciated!

• The statement of the problem looks incomplete. Presumably, there should be boundary conditions. If those boundary conditions are constant, say $u(0)=u_0$ and $u(\ell) = u_{\ell}$, then the temperature should change uniformly from $u_0$ to $u_{\ell}$ as we move along the bar. – Mark McClure Jan 25 '15 at 18:21
• The boundaries are insulated, which means $\dfrac{\partial u}{\partial x} =0$ at $x=0$ and $x=\ell$. – Robert Israel Jan 25 '15 at 18:22
• So I'm supposed to solve the diffusion equation? – mike russel Jan 25 '15 at 18:23
• Hint: The total amount of heat at time $t$ can be taken to be $\int_0^\ell u(x,t)\; dx$. – Robert Israel Jan 25 '15 at 18:24
• Thanks alot. Now i get it! – mike russel Jan 25 '15 at 19:42

The first hint states that the temperature does not change in steady state. In symbols, $u_t=0$. Thus, the equation $u_t=ku_{xx}$ becomes $u_{xx}=0$. This is an ODE that we can solve by simply integrating twice to get $u(x) = ax+b$ for some constants $a$ and $b$. Now, as the boundaries are insulated, we must have $u_x(0)=u_x(\ell)=0$. Since $u_x=a$, we must have $a=0$ so $u$ is the constant function $u(x)=b$.
The last piece of the puzzle is, just what constant should we use for the solution. That is where the second hint comes in that, due to the insulation, the amount of heat in the bar is constant. But, the amount of heat can be expressed as the integral of $u$ over the bar. Setting the total heat from the initial state $f(x)$ equal to the total heat from the steady state, $u(x)=b$, we have $$\int_0^{\ell} f(x) \, dx = \int_0^{\ell} b \, dx = b\ell.$$ Thus, $$b=\frac{1}{\ell}\int_0^{\ell} f(x) \, dx,$$ which just the average temperature in the bar.
Here's an example of the diffusive process. Suppose that we're working with a bar of length one with initial temperature profile $f(x)=x^2(2-x)^2$, which looks like so:
The bar is cooler on the left and warmer on the right. The average temperature in the bar is $$\int_0^2 x^2(2-x^2)\,dx = \frac{8}{15}.$$ The heat in the bar will spread out as it diffuses, but can't escape from the ends. As a result, the left side will heat up towards the average temperature of $8/15$ while the right hand side cools down to the same temperature. The evolution might look like so: