Differential Equations - Partial: Solving The Heat Equation - i

Okay, it is finally time to completely solve a partial differential equation. In the previous section we applied separation of variables to several partial differential equations and reduced the problem down to needing to solve two ordinary differential equations. In this section we will now solve those ordinary differential equations and use the results to get a solution to the partial differential equation. We will be concentrating on the heat equation in this section and will do the wave equation and Laplace’s equation in later sections.

The first problem that we’re going to look at will be the temperature distribution in a bar with zero temperature boundaries. We are going to do the work in a couple of steps so we can take our time and see how everything works.

The first thing that we need to do is find a solution that will satisfy the partial differential equation and the boundary conditions. At this point we will not worry about the initial condition. The solution we’ll get first will not satisfy the vast majority of initial conditions but as we’ll see it can be used to find a solution that will satisfy a sufficiently nice initial condition.

Example 1 Find a solution to the following partial differential equation that will also satisfy the boundary conditions.$$\begin{array}{cc}& \frac{\partial u}{\partial t}=k\frac{{\partial }^{2}u}{\partial x^2}\\ & u\left(x,0\right)=f\left(x\right)u\left(0,t\right)=0u\left(L,t\right)=0\end{array}$$

Okay the first thing we technically need to do here is apply separation of variables. Even though we did that in the previous section let’s recap here what we did.
First, we assume that the solution will take the form,
$$u\left(x,t\right)=\phi \left(x\right)G\left(t\right)$$and we plug this into the partial differential equation and boundary conditions. We separate the equation to get a function of only on one side and a function of only $x$ on the other side and then introduce a separation constant. This leaves us with two ordinary differential equations.
We did all of this in Example 1 of the previous section and the two ordinary differential equations are,
$$\begin{array}{cc}\frac{dG}{dt}=-k\lambda G& \frac{d^2\phi }{d{x}^2}+\lambda \phi =0\\ & \phi \left(0\right)=0\phi \left(L\right)=0\end{array}$$The time dependent equation can really be solved at any time, but since we don’t know what $\lambda$ is yet let’s hold off on that one. Also note that in many problems only the boundary value problem can be solved at this point so don’t always expect to be able to solve either one at this point.
The spatial equation is a boundary value problem and we know from our work in the previous chapter that it will only have non-trivial solutions (which we want) for certain values of $\lambda$, which we’ll recall are called eigenvalues. Once we have those we can determine the non-trivial solutions for each $\lambda$, i.e. eigenfunctions.
Now, we actually solved the spatial problem,
$$\begin{array}{cc}& \frac{d^2\phi }{d{x}^2}+\lambda \phi =0\\ & \phi \left(0\right)=0\phi \left(L\right)=0\end{array}$$in Example 1 of the Eigenvalues and Eigen functions section of the previous chapter for $L=2\pi$. So, because we’ve solved this once for a specific $L$ and the work is not all that much different for a general $L$ we’re not going to be putting in a lot of explanation here and if you need a reminder on how something works or why we did something go back to Example 1 from the Eigenvalues and Eigen functions section for a reminder.
We’ve got three cases to deal with so let’s get going.
$\underset{–}{\lambda >0}$
In this case we know the solution to the differential equation is,
$$\phi \left(x\right)=c_1\cos \left(\sqrt{\lambda }x\right)+c_2\sin \left(\sqrt{\lambda }x\right)$$Applying the first boundary condition gives,
$$0=\phi \left(0\right)=c_1$$Now applying the second boundary condition, and using the above result of course, gives,
$$0=\phi \left(L\right)=c_2\sin \left(L\sqrt{\lambda }\right)$$Now, we are after non-trivial solutions and so this means we must have,
$$\sin \left(L\sqrt{\lambda }\right)=0\Rightarrow L\sqrt{\lambda }=n\pi n=1,2,3,\dots$$The positive eigenvalues and their corresponding eigenfunctions of this boundary value problem are then,
$${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2{\phi }_n\left(x\right)=\sin \left(\frac{n\pi x}{L}\right)n=1,2,3,\dots$$Note that we don’t need the $c_2$ in the eigen function as it will just get absorbed into another constant that we’ll be picking up later on.
$\underset{–}{\lambda =0}$
The solution to the differential equation in this case is,
$$\phi \left(x\right)=c_1+c_{2}x$$Applying the boundary conditions gives,
$$0=\phi \left(0\right)=c_{1}0=\phi \left(L\right)=c_{2}L\Rightarrow c_2=0$$So, in this case the only solution is the trivial solution and so $\lambda =0$ is not an eigenvalue for this boundary value problem.
$\underset{–}{\lambda <0}$
Here the solution to the differential equation is,
$$\phi \left(x\right)=c_1\cosh \left(\sqrt{-\lambda }x\right)+c_2\sinh \left(\sqrt{-\lambda }x\right)$$Applying the first boundary condition gives,
$$0=\phi \left(0\right)=c_1$$and applying the second gives,
$$0=\phi \left(L\right)=c_2\sinh \left(L\sqrt{-\lambda }\right)$$So, we are assuming $\lambda <0$ and so $L\sqrt{-\lambda }\ne 0$ and this means $\sinh \left(L\sqrt{-\lambda }\right)\ne 0$. We therefore we must have $c_2=0$and so we can only get the trivial solution in this case.
Therefore, there will be no negative eigenvalues for this boundary value problem.
The complete list of eigenvalues and eigenfunctions for this problem are then,
$${\lambda }_n={\left(\frac{n\pi }{L}\right)}^2{\phi }_n\left(x\right)=\sin \left(\frac{n\pi x}{L}\right)n=1,2,3,\dots$$Now let’s solve the time differential equation,
$$\frac{dG}{dt}=-k{\lambda }_{n}G$$and note that even though we now know $\lambda$ we’re not going to plug it in quite yet to keep the mess to a minimum. We will however now use ${\lambda }_n$ to remind us that we actually have an infinite number of possible values here.
This is a simple linear (and separable for that matter) 1^st order differential equation and so we’ll let you verify that the solution is,
$$G\left(t\right)=c{e}^{-k{\lambda }_{n}t}=c{e}^{-k{\left(\frac{n\pi }{L}\right)}^{2}t}$$Okay, now that we’ve gotten both of the ordinary differential equations solved we can finally write down a solution. Note however that we have in fact found infinitely many solutions since there are infinitely many solutions (i.e. eigenfunctions) to the spatial problem.
Our product solution are then,
$$u_n\left(x,t\right)=B_n\sin \left(\frac{n\pi x}{L}\right)e^{-k{\left(\frac{n\pi }{L}\right)}^{2}t}n=1,2,3,\dots$$We’ve denoted the product solution $u_n$ to acknowledge that each value of $n$ will yield a different solution. Also note that we’ve changed the $c$ in the solution to the time problem to $B_n$ to denote the fact that it will probably be different for each value of $n$ as well and because had we kept the $c_2$ with the eigen function we’d have absorbed it into the $c$ to get a single constant in our solution.

So, there we have it. The function above will satisfy the heat equation and the boundary condition of zero temperature on the ends of the bar.

The problem with this solution is that it simply will not satisfy almost every possible initial condition we could possibly want to use. That does not mean however, that there aren’t at least a few that it will satisfy as the next example illustrates.