How do I prove that the BVP $y'' = y\sin\left(x\right), y\left(0\right) = a, y\left(1\right) = b$ has a unique solution? I want to prove that the BVP $$ y'' = y\sin\left(x\right), y\left(0\right) = a, y\left(1\right) = b $$ has a unique solution.
I know how to prove uniqueness:
Suppose there are two different solutions $y_1,y_2$. We look at $z= y_1 - y_2$.
$z\left(0\right) = z\left(1\right) = 0$. So by Rolle's theorem $z$ has a positive maximum or a negative minimum. W.L.O.G $z$ has a positive maximum at point $x_0$.
Then on one side $z''\left(x_0\right) < 0$.
On the other side $z''\left(x_0\right) = z\sin\left(x_0\right) > 0$.
Now I need to prove existence of a solution. I think that I need to start by looking at the IVP
$$ {y_q}'' = {y_q}\sin\left(x\right), {y_q}\left(0\right) = a, {y_q}'\left(0\right) = q $$
I know that it has a solution, so I need to show that for a big q, ${y_q} > b$ and for a small q, $y_q < b$ but I don't know how to prove that.
 A: Let $y_q:[0,1] \to \mathbb{R}$ be the solution of your IVP and let $\|\cdot\|_\infty$ denote the maximum norm on $[0,1]$. We have
$$
(\ast) \quad y_q(x)=a+qx+ \int_0^x \int_0^t y_q(s)\sin(s) dsdt \quad (x \in [0,1]),
$$
thus
$$
|y_q(x)| \le |a| + |qx| + \int_0^x \int_0^t \sin(s)|y_q(s)| dsdt  
$$
$$
\le |a| + |q| + \int_0^1 \int_0^t \sin(s) dsdt \|y_q\|_\infty= |a| + |q| +(1-\sin(1))\|y_q\|_\infty \quad (x \in [0,1]).
$$
We get $\|y_q\|_\infty \le |a| + |q| +(1-\sin(1))\|y_q\|_\infty$, thus
$$
\|y_q\|_\infty \le \frac{|a| + |q|}{\sin(1)}.
$$
This a-priori inequality with $(\ast)$ yields
$$
|y_q(1)-q| \le |a|+ \int_0^1 \int_0^t \sin(s) dsdt \|y_q\|_\infty \le
|a|+ (1-\sin(1))\frac{|a| + |q|}{\sin(1)}=|a|+ c(|a| + |q|)
$$
with
$$
c:=\frac{1-\sin(1)}{\sin(1)} < 1.
$$
For $q \ge 0$ we obtain
$$
y_q(1) \ge (1-c)q - (1+c)|a| \to \infty \quad (q \to \infty),
$$
and for $q\le 0$ we get
$$
y_q(1) \le (1-c)q + (1+c)|a|\to -\infty \quad (q \to -\infty).
$$
Since $q \mapsto y_q(1)$ is continuous it has to be surjective. So $y_q(1)$ can reach any $b$ with a suitable shooting angle $q$.
