5. Boundary Value Problems#

The ODEs that we have encountered so far are initial value problems where we know the solution of the ODE at the lower boundary of the \(t\) domain. Another type of problem are Boundary Value Problems (BVP) which is where the solution at both boundaries of the \(t\) domain are known. Here we will be looking at solving two-point boundary value problems based on second-order ODEs.

Definition 5.1 (General two-point boundary value problem)

A two-point boundary value problem is a second-order ODE where the solutions at the lower and upper boundaries of the domain are known

(5.1)#\[y'' = f(t,y),\qquad t\in [t_{\min}, t_{\max}],\qquad y(t_{\min}) = \alpha ,\qquad y(t_{\max}) = \beta,\]

for some known values \(\alpha\) and \(\beta\).

5.1. Existence and uniqueness of solutions to boundary value problems#

If an initial value problem has a solution it will be unique for a particular initial value. This is not true for boundary value problems which can have a unique solution, no solution or an infinite number of solutions. Consider the following boundary value problem

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

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

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

so 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

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

Now consider the boundary value problem

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

and substituting the boundary values 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 consider the boundary value problem

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

and substituting the boundary values we have

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

Here we have \(c_1 = 1\) and \(c_2 = 2\) 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

(5.2)#\[ y'' = p(t)y' + q(t)y + r(t), \qquad t \in [t_{\min},t_{\max}], \qquad y(t_{\min}) = \alpha , \qquad y(t_{\max}) = \beta , \]

where \(p(t)\), \(q(t)\) and \(r(t)\) are some functions of \(t\) then it has a unique solution if the following are satisfied

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

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

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 (click to show)

Comparing this boundary value problem to equation (5.2) 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 \(q(t)\geq \) for all \(t\in [0,1]\) so this boundary value problem has a unique solution.