Section 9.9 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.