Differential Equations - Fourier Series: Boundary Value Problems - i

Before we start off this section we need to make it very clear that we are only going to scratch the surface of the topic of boundary value problems. There is enough material in the topic of boundary value problems that we could devote a whole class to it. The intent of this section is to give a brief (and we mean very brief) look at the idea of boundary value problems and to give enough information to allow us to do some basic partial differential equations in the next chapter.

Now, with that out of the way, the first thing that we need to do is to define just what we mean by a boundary value problem (BVP for short). With initial value problems we had a differential equation and we specified the value of the solution and an appropriate number of derivatives at the same point (collectively called initial conditions). For instance, for a second order differential equation the initial conditions are,

$$y\left(t_0\right)=y_0{y}^{\prime }\left(t_0\right)=y_0^{\prime }$$

With boundary value problems we will have a differential equation and we will specify the function and/or derivatives at different points, which we’ll call boundary values. For second order differential equations, which will be looking at pretty much exclusively here, any of the following can, and will, be used for boundary conditions.

$$\begin{array}{}y\left(x_0\right)=y_{0}y\left(x_1\right)=y_1& \text{(1)}\end{array}$$$$\begin{array}{}{y}^{\prime }\left(x_0\right)=y_0{y}^{\prime }\left(x_1\right)=y_1& \text{(2)}\end{array}$$$$\begin{array}{}{y}^{\prime }\left(x_0\right)=y_{0}y\left(x_1\right)=y_1& \text{(3)}\end{array}$$$$\begin{array}{}y\left(x_0\right)=y_0{y}^{\prime }\left(x_1\right)=y_1& \text{(4)}\end{array}$$

As mentioned above we’ll be looking pretty much exclusively at second order differential equations. We will also be restricting ourselves down to linear differential equations. So, for the purposes of our discussion here we’ll be looking almost exclusively at differential equations in the form,

$$\begin{array}{}{y}^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y=g\left(x\right)& \text{(5)}\end{array}$$

along with one of the sets of boundary conditions given in $\text{(1)}$ – $\text{(4)}$. We will, on occasion, look at some different boundary conditions but the differential equation will always be on that can be written in this form.

As we’ll soon see much of what we know about initial value problems will not hold here. We can, of course, solve $\text{(5)}$provided the coefficients are constant and for a few cases in which they aren’t. None of that will change. The changes (and perhaps the problems) arise when we move from initial conditions to boundary conditions.

One of the first changes is a definition that we saw all the time in the earlier chapters. In the earlier chapters we said that a differential equation was homogeneous if $g\left(x\right)=0$ for all $x$. Here we will say that a boundary value problem is homogeneous if in addition to $g\left(x\right)=0$ we also have $y_0=0$ and $y_1=0$(regardless of the boundary conditions we use). If any of these are not zero we will call the BVP nonhomogeneous.

It is important to now remember that when we say homogeneous (or nonhomogeneous) we are saying something not only about the differential equation itself but also about the boundary conditions as well.

The biggest change that we’re going to see here comes when we go to solve the boundary value problem. When solving linear initial value problems a unique solution will be guaranteed under very mild conditions. We only looked at this idea for first order IVP’s but the idea does extend to higher order IVP’s. In that section we saw that all we needed to guarantee a unique solution was some basic continuity conditions. With boundary value problems we will often have no solution or infinitely many solutions even for very nice differential equations that would yield a unique solution if we had initial conditions instead of boundary conditions.

Before we get into solving some of these let’s next address the question of why we’re even talking about these in the first place. As we’ll see in the next chapter in the process of solving some partial differential equations we will run into boundary value problems that will need to be solved as well. In fact, a large part of the solution process there will be in dealing with the solution to the BVP. In these cases, the boundary conditions will represent things like the temperature at either end of a bar, or the heat flow into/out of either end of a bar. Or maybe they will represent the location of ends of a vibrating string. So, the boundary conditions there will really be conditions on the boundary of some process.

So, with some of basic stuff out of the way let’s find some solutions to a few boundary value problems. Note as well that there really isn’t anything new here yet. We know how to solve the differential equation and we know how to find the constants by applying the conditions. The only difference is that here we’ll be applying boundary conditions instead of initial conditions.

Example 1 Solve the following BVP.$$y^{\prime \prime }+4y=0y\left(0\right)=-2y\left(\frac{\pi }{4}\right)=10$$

Okay, this is a simple differential equation to solve and so we’ll leave it to you to verify that the general solution to this is,
$$y\left(x\right)=c_1\cos \left(2x\right)+c_2\sin \left(2x\right)$$Now all that we need to do is apply the boundary conditions.
$$\begin{array}{cc}-2& =y\left(0\right)=c_1\\ 10& =y\left(\frac{\pi }{4}\right)=c_2\end{array}$$The solution is then,
$$y\left(x\right)=-2\cos \left(2x\right)+10\sin \left(2x\right)$$

We mentioned above that some boundary value problems can have no solutions or infinite solutions we had better do a couple of examples of those as well here. This next set of examples will also show just how small of a change to the BVP it takes to move into these other possibilities.

Example 2 Solve the following BVP.$$y^{\prime \prime }+4y=0y\left(0\right)=-2y\left(2\pi \right)=-2$$

We’re working with the same differential equation as the first example so we still have,
$$y\left(x\right)=c_1\cos \left(2x\right)+c_2\sin \left(2x\right)$$Upon applying the boundary conditions we get,
$$\begin{array}{cc}-2& =y\left(0\right)=c_1\\ -2& =y\left(2\pi \right)=c_1\end{array}$$So, in this case, unlike previous example, both boundary conditions tell us that we have to have $c_1=-2$ and neither one of them tell us anything about $c_2$. Remember however that all we’re asking for is a solution to the differential equation that satisfies the two given boundary conditions and the following function will do that,

$$y\left(x\right)=-2\cos \left(2x\right)+c_2\sin \left(2x\right)$$In other words, regardless of the value of $c_2$ we get a solution and so, in this case we get infinitely many solutions to the boundary value problem.