NA Boundary Value Problems

Section 9.10 Boundary Value Problems

Consider the 2nd order ODE

$\begin{equation*} \ddot x=f(t,x,\dot x). \end{equation*}$

By the Picard theorem we know that, if

$f(t,x,\dot x)$ is continuous, then a unique solution to the IVP

$x(t_0)=x_0\text{,}$

$\dot x(t_0)=v_0$ exists for some time interval containing

$t_0$ for every choice of

$t_0,x_0,v_0\text{.}$

Now consider instead the BVP

$\begin{equation*} \ddot x=f(t,x,\dot x)\,,\;x(t_0)=x_0\,,\;x(t_f)=x_f. \end{equation*}$

Does a unique solution exists for every choice of

$t_0,t_f,x_0,x_f\text{?}$

The simple examples below show that this is not the case.

Example 1. Consider the BVP problem

$\begin{equation*} \ddot x=-x\,,\;x(0)=0\,,\;x(\pi)=0. \end{equation*}$

The general solution is

$\begin{equation*} x(t)=A\cos(t)+B\sin(t) \end{equation*}$

and, from

$x(0)=0\text{,}$ we know that

$A=0\text{.}$ Since

$\sin(\pi)=0\text{,}$ then every function

$x(t)=B\sin(t)$ solves the problem, namely there are infinitely many solutions!

Example 2. Consider

$\begin{equation*} \ddot x=-x\,,\;x(0)=0\,,\;x(\pi/2)=0. \end{equation*}$

Again from

$x(0)=0$ we get that

$x(t)=B\sin(t)$ but then

$x(\pi/2) = B\sin(\pi/2)=B$ and so there is in this case a unique solution

$x(t)=0\text{.}$

Example 2. Consider

$\begin{equation*} \ddot x=-x\,,\;x(0)=0\,,\;x(\pi)=1. \end{equation*}$

We already notice that if

$x(0)=0$ then

$x(\pi)=0\text{,}$ so in this case there is no solution to this problem.