4.1: Introduction

Until this point we have solved initial value problems. For an initial value problem one has to solve a differential equation subject to conditions on the unknown function and its derivatives at one value of the independent variable. For example, for \(x = x(t)\) we could have the initial value problem

\[x'' + x = 2, \quad x(0) = 1, \quad x'(0) = 0 \label{4.1} \]

In the next chapters we will study boundary value problems and various tools for solving such problems. In this chapter we will motivate our interest in boundary value problems by looking into solving the one-dimensional heat equation, which is a partial differential equation. for the rest of the section, we will use this solution to show that in the background of our solution of boundary value problems is a structure based upon linear algebra and analysis leading to the study of inner product spaces. Though technically, we should be lead to Hilbert spaces, which are complete inner product spaces.

For an initial value problem one has to solve a differential equation subject to conditions on the unknown function or its derivatives at more than one value of the independent variable. As an example, we have a slight modification of the above problem: Find the solution \(x = x(t)\) for \(0 ≤ t ≤ 1\) that satisfies the problem

\[x'' + x = 2, \quad x(0) = 1, \quad x(1) = 0 \label{4.2} \]

Typically, initial value problems involve time dependent functions and boundary value problems are spatial. So, with an initial value problem one knows how a system evolves in terms of the differential equation and the state of the system at some fixed time. Then one seeks to determine the state of the system at a later time.

For boundary values problems, one knows how each point responds to its neighbors, but there are conditions that have to be satisfied at the endpoints. An example would be a horizontal beam supported at the ends, like a bridge.

The shape of the beam under the influence of gravity, or other forces, would lead to a differential equation and the boundary conditions at the beam ends would affect the solution of the problem. There are also a variety of other types of boundary conditions. In the case of a beam, one end could be fixed and the other end could be free to move. We will explore the effects of different boundary value conditions in our discussions and exercises.

Let’s solve the above boundary value problem. As with initial value problems, we need to find the general solution and then apply any conditions that we may have. This is a nonhomogeneous differential equation, so we have that the solution is a sum of a solution of the homogeneous equation and a particular solution of the nonhomogeneous equation, \(x(t) = x_h(t) + x_p(t)\). The solution of \(x′′ + x = 0\) is easily found as

\[x_{h}(t)=c_{1} \cos t+c_{2} \sin t. \nonumber \]

The particular solution is easily found using the Method of Undetermined Coefficients,

\[x_{p}(t)=2 \nonumber \]

Thus, the general solution is

\[x(t)=2+c_{1} \cos t+c_{2} \sin t \nonumber \]

We now apply the boundary conditions and see if there are values of \(c_{1}\) and \(c_{2}\) that yield a solution to our problem. The first condition, \(x(0)=0\), gives

\[0=2+c_{1} . \nonumber \]

Thus, \(c_{1}=-2\). Using this value for \(c_{1}\), the second condition, \(x(1)=1\), gives

\[0=2-2 \cos 1+c_{2} \sin 1 \nonumber \]

This yields

\[c_{2}=\dfrac{2(\cos 1-1)}{\sin 1} \nonumber \]

We have found that there is a solution to the boundary value problem and it is given by

\[x(t)=2\left(1-\cos t \dfrac{(\cos 1-1)}{\sin 1} \sin t\right) \nonumber \]

Boundary value problems arise in many physical systems, just as many of the initial values problems we have seen. We will see in the next section that boundary value problems for ordinary differential equations often appear in the solution of partial differential equations.