Section 1.15 Boundary Value Problems
Consider the 2nd order ODE \(x''=f(t,x,x')\text{.}\) By the Picard theorem we know that, if \(f(t,x,x')\) is continuous, then a unique solution to the IVP \(x(t_0)=x_0\text{,}\) \(x'(t_0)=v_0\) 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*}
x''=f(t,x,x')\,,\;x(t_0)=x_0\,,\;x(t_1)=x_1.
\end{equation*}
Is there some general condition on \(f\) under which a unique solution exists for every choice of \(t_0,t_1,x_0,x_1\text{?}\) Simple examples show that this is not the case. Consider, for instance, the ODE
\begin{equation*}
x''=-x.
\end{equation*}
Recall that this ODE models harmonic oscillations with period \(2\pi\text{,}\) so it is easy to find boundary values that are compatible and others that are not compatible with this type of motion.
Example 1. Consider first tha BVP
\begin{equation*}
x''=-x\,,\;x(0)=0\,,\;x(\pi)=0.
\end{equation*}
The general solution of the ODE 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 now
\begin{equation*}
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 3. Consider finally
\begin{equation*}
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.
Remark. Some "not so general" statement does exist. For instance, we have the following: Theorem 1.15.1.
Suppose that \(f:\bR^3\to\bR\) and its partial derivatives \(f_2\) and \(f_3\) are continuous in \(R=[a,b]\times\bR^2\) and that:- \(f_2|_R>0\text{;}\)
- \(\sup_R|f_3|<\infty\text{.}\)
\begin{equation*}
x''=f(t,x,x')\,,\;x(t_0)=x_0\,,\;x(t_1)=x_1
\end{equation*}
has a unique solution for each \(x_0,x_1\text{.}\)