5. Boundary Value Problems

The ordinary differential equations (ODEs) considered so far have been initial value problems (IVPs), where the value of the solution and, for higher-order equations, its derivatives are specified at a single point in the domain. Another important class of problems are boundary value problems (BVPs), where conditions are specified at the endpoints of the interval. In this chapter, we consider two-point boundary value problems governed by second-order ODEs.

Definition 5.1 (General two-point boundary value problem)

A two-point boundary value problem is a second-order ODE together with boundary conditions specified at two distinct points of the domain

(5.1)\[ y' = f(t,y),\qquad t\in [t_{0}, t_{\max}],\qquad y(t_{0}) = a ,\qquad y(t_{\max}) = b, \]

for some known values \(a\) and \(b\).

5.1. Existence and uniqueness of solutions to boundary value problems

For well-posed linear initial value problems satisfying the conditions of the existence and uniqueness theorem, a solution is unique. Boundary value problems behave differently and may possess a unique solution, no solution, or infinitely many solutions.

To illustrate the different possibilities, consider the following examples.

\[ y'' + 4y = 0, \qquad y(0) = 1, \qquad y\left(\dfrac{\pi}{4}\right) = 2. \]

The general solution to the ODE \(y'' + 4y = 0\) is

(5.2)\[ y = c_1 \cos(2 t) + c_2 \sin(2 t). \]

Substituting the boundary values we have

\[\begin{split} \begin{align*} 1 &= c_1 \cos(0) + c_2 \sin(0) = c_1, \\ 2 &= c_1 \cos\left(2\left(\frac{\pi}{4}\right)\right) + c_2 \sin\left(2\left(\frac{\pi}{4}\right)\right) = c_2. \end{align*} \end{split}\]

Here we can determine the values of \(c_1\) and \(c_2\) so this boundary value problem has the unique solution

\[ \begin{align*} y = \cos(2t) + 2\sin(2t). \end{align*} \]

Changing the boundary conditions produces a very different outcome.

\[ \begin{align*} y'' + 4y = 0, \qquad y(0) = 1, \qquad y(\pi) = 1. \end{align*} \]

Substituting the boundary values into equation (5.2) we have

\[\begin{split} \begin{align*} 1 &= c_1 \cos(0) + c_2 \sin(0) = c_1, \\ 1 &= c_1 \cos(2\pi) + c_2 \sin(2\pi) = c_1. \end{align*} \end{split}\]

Here we have a solution for \(c_1\) but we cannot determine the value of \(c_2\) so the solution to the boundary value problem is

\[ \begin{align*} y = \cos(2t) + c_2 \sin(2t), \end{align*} \]

where \(c_2\) can be any value. So we have infinitely many solutions.

Finally, a slight modification of the upper boundary condition leads to an inconsistent system

\[ \begin{align*} y'' + 4y = 0, \qquad y(0) = 1, \qquad y(\pi) = 2. \end{align*} \]

and substituting the boundary values equation (5.2) we have

\[\begin{split} \begin{align*} 1 &= c_1 \cos(0) + c_2 \sin(0) = c_1, \\ 2 &= c_1 \cos(2\pi) + c_2 \sin(2\pi) = c_1. \end{align*} \end{split}\]

Here we have \(c_1 = 1\) and \(c_1 = 2\) which is a contradiction, so this boundary value problem does not have a solution.

Theorem 5.1 (Uniqueness of the solution to boundary value problems)

A linear boundary value problem of the form

(5.3)\[ y'' = p(t)y' + q(t)y + r(t), \qquad t \in [t_0,t_{\max}], \qquad y(t_0) = a , \qquad y(t_{\max}) = b , \]

where \(p(t)\), \(q(t)\) and \(r(t)\) are continuous functions on the interval. The boundary value problem has a unique solution provided that the following conditions are satisfied

• \(p(t)\), \(q(t)\) and \(r(t)\) are continuous on \([t_0, t_{\max}]\);

• \(q(t) > 0\) for all \(t\in [t_0,t_{\max}]\).

Example 5.1

Show that the following boundary value problem has a unique solution

\[ y'' = (t^3 +5)y + \sin (t), \qquad t \in [0,1], \qquad y(0) = 0, \qquad y(1) = 1. \]

Solution

Comparing this boundary value problem to equation (5.3) we have

\[\begin{split} \begin{align*} p(t) &= 0,\\ q(t) &= t^3 + 5,\\ r(t) &= \sin(t), \end{align*} \end{split}\]

which are all continuous on \([0,1]\) and \(q(t) > 0\) for all \(t\in [0,1]\) so this boundary value problem has a unique solution.