I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul

August 27, 2020

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 9-6 : Heat Equation with Non-Zero Temperature Boundaries

In this section we want to expand one of the cases from the previous section a little bit. In the previous section we look at the following heat problem.

\[\begin{align*}& \frac{{\partial u}}{{\partial t}} = k\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ & u\left( {x,0} \right) = f\left( x \right)\hspace{0.25in}u\left( {0,t} \right) = 0\hspace{0.25in}\,\,\,\,\,u\left( {L,t} \right) = 0\end{align*}\]Now, there is nothing inherently wrong with this problem, but the fact that we’re fixing the temperature on both ends at zero is a little unrealistic. The other two problems we looked at, insulated boundaries and the thin ring, are a little more realistic problems, but this one just isn’t all that realistic so we’d like to extend it a little.

What we’d like to do in this section is instead look at the following problem.

\[\begin{equation}\begin{aligned}& \frac{{\partial u}}{{\partial t}} = k\frac{{{\partial ^2}u}}{{\partial {x^2}}}\\ & u\left( {x,0} \right) = f\left( x \right)\hspace{0.25in}u\left( {0,t} \right) = {T_1}\hspace{0.25in}\,\,\,\,\,u\left( {L,t} \right) = {T_2}\end{aligned}\label{eq:eq1}\end{equation}\]In this case we’ll allow the boundaries to be any fixed temperature, \({T_1}\) or \({T_2}\). The problem here is that separation of variables will no longer work on this problem because the boundary conditions are no longer homogeneous. Recall that separation of variables will only work if both the partial differential equation and the boundary conditions are linear and homogeneous. So, we’re going to need to deal with the boundary conditions in some way before we actually try and solve this.

Luckily for us there is an easy way to deal with them. Let’s consider this problem a little bit. There are no sources to add/subtract heat energy anywhere in the bar. Also, our boundary conditions are fixed temperatures and so can’t change with time and we aren’t prescribing a heat flux on the boundaries to continually add/subtract heat energy. So, what this all means is that there will not ever be any forcing of heat energy into or out of the bar and so while some heat energy may well naturally flow into our out of the bar at the end points as the temperature changes eventually the temperature distribution in the bar should stabilize out and no longer depend on time.

Or, in other words it makes some sense that we should expect that as \(t \to \infty \) our temperature distribution, \(u\left( {x,t} \right)\) should behave as,

\[\mathop {\lim }\limits_{t \to \infty } u\left( {x,t} \right) = {u_E}\left( x \right)\]where \({u_E}\left( x \right)\) is called the **equilibrium temperature**. Note as well that is should still satisfy the heat equation and boundary conditions. It won’t satisfy the initial condition however because it is the temperature distribution as \(t \to \infty \) whereas the initial condition is at \(t = 0\). So, the equilibrium temperature distribution should satisfy,

This is a really easy 2^{nd} order ordinary differential equation to solve. If we integrate twice we get,

and applying the boundary conditions (we’ll leave this to you to verify) gives us,

\[{u_E}\left( x \right) = {T_1} + \frac{{{T_2} - {T_1}}}{L}x\]Okay, just what does this have to do with solving the problem given by \(\eqref{eq:eq1}\) above? We’ll let’s define the function,

\[\begin{equation}v\left( {x,t} \right) = u\left( {x,t} \right) - {u_E}\left( x \right)\label{eq:eq3}\end{equation}\]where \(u\left( {x,t} \right)\) is the solution to \(\eqref{eq:eq1}\) and \({u_E}\left( x \right)\) is the equilibrium temperature for \(\eqref{eq:eq1}\).

Now let’s rewrite this as,

\[u\left( {x,t} \right) = v\left( {x,t} \right) + {u_E}\left( x \right)\]and let’s take some derivatives.

\[\frac{{\partial u}}{{\partial t}} = \frac{{\partial v}}{{\partial t}} + \frac{{\partial {u_E}}}{{\partial t}} = \frac{{\partial v}}{{\partial t}}\hspace{0.25in}\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}} + \frac{{{\partial ^2}{u_E}}}{{\partial {x^2}}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}\]In both of these derivatives we used the fact that \({u_E}\left( x \right)\) is the equilibrium temperature and so is independent of time \(t\) and must satisfy the differential equation in \(\eqref{eq:eq2}\).

What this tells us is that both \(u\left( {x,t} \right)\) and \(v\left( {x,t} \right)\) must satisfy the same partial differential equation. Let’s see what the initial conditions and boundary conditions would need to be for \(v\left( {x,t} \right)\).

\[\begin{align*}v\left( {x,0} \right) & = u\left( {x,0} \right) - {u_E}\left( x \right) = f\left( x \right) - {u_E}\left( x \right)\\ v\left( {0,t} \right) & = u\left( {0,t} \right) - {u_E}\left( 0 \right) = {T_1} - {T_1} = 0\\ v\left( {L,t} \right) & = u\left( {L,t} \right) - {u_E}\left( L \right) = {T_2} - {T_2} = 0\end{align*}\]So, the initial condition just gets potentially messier, but the boundary conditions are now homogeneous! The partial differential equation that \(v\left( {x,t} \right)\) must satisfy is,

\[\begin{align*}\frac{{\partial v}}{{\partial t}} & = k\frac{{{\partial ^2}v}}{{\partial {x^2}}}\\ v\left( {x,0} \right) & = f\left( x \right) - {u_E}\left( x \right)\hspace{0.25in}v\left( {0,t} \right) = 0\hspace{0.25in}\,\,\,\,\,v\left( {L,t} \right) = 0\end{align*}\]We saw how to solve this in the previous section and so we the solution is,

\[v\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \]where the coefficients are given by,

\[{B_n} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{\left( {f\left( x \right) - {u_E}\left( x \right)} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,\hspace{0.25in}n = 1,2,3, \ldots \]The solution to \(\eqref{eq:eq1}\) is then,

\[\begin{align*}u\left( {x,t} \right) & = {u_E}\left( x \right) + v\left( {x,t} \right)\\ & = {T_1} + \frac{{{T_2} - {T_1}}}{L}x + \sum\limits_{n = 1}^\infty {{B_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - k{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}} \end{align*}\]and the coefficients are given above.