I have the following exercise problem: (Gewöhnliche Differentialgleichungen und dynamische Systeme by Mathias Wilke and Jan W. Prüss, Page 82, 8)
Look at the following boundary value problem: $$x''(t) + g(x(t)) = 0,x(0)=x(1)=0\tag1$$ with $g \in C^1(\mathbb{R})$. One can transform this ODE to an initial value problem: $$y' = f(y), \; y(0) = (0,a) $$ where $y=(y_1,y_2)=(x,x') \quad y'=(x',x'') \quad f(y_1,y_2) = (y_2,-g(y_1))$ to solve the problem by using the shooting method. So we search $a \in \mathbb{R}$.
The exercise is:
Prove the existence of the solution of $(1)$, by using the shooting method, if $\| g(x) \|\le M \; \forall x \in \mathbb{R}$.
At this point i am absolutely clueless. I tried using the continuity of a the starting value problem as function of its parameters. I tried to show that the starting value problem has a solution on $[0, \infty)$ for all $a \in \mathbb{R}$.
None of that produced some good results. Has anyone another idea where to start do or what my mistake could be.
Consider $a=\pm 3M$ and show that the resulting slope is always positive/negative and thus the second boundary values of opposite sign.
If you want it maximally tight, consider $a=\frac M2$. Then $y'\ge \frac M2(1-2t)$ and $y\ge\frac M2 t(1-t)$ and thus $y(1)\ge 0$. Similarly, $a=-\frac M2$ gives $y(1)\le 0$.