Uniqueness solution 2nd order differential equation

Question

Let $q, r, f \in C[0,1], (\alpha, \beta) \in \mathbb{R}^2$ and show that the the inhomogenous lineair boundary value problem: $$y''(t) + q(t)y'(t) + r(t)y(t) = f(t) $$ $$y(0) = \alpha, y'(0) = \beta$$

has a unique solution $y \in C^2[0,1]$ and show that this $y$ depends continuously on $\alpha, \beta$ and $f$.

Let's define the operator $T_{q, r}: C^2[0,1] \rightarrow C[0,1]$ as follows (here $C^2[0,1]$ is endowed with the norm $ || f ||_{C^2} = || f ||_\infty + || f' ||_\infty +|| f'' ||_\infty)$: $$T_{q, r}(y) := y''(t) + q(t)y'(t) + r(t)y(t)$$

I've been given the hint that I need to apply the open mapping theorem on the linear operator $T$. It is not difficult to show that $T$ is bounded so all we need to show is that $T_{q,r}$ is surjective. My question (or questions, rather) is the following: How do I show that $T_{q,r}$ is surjective and what does applying the open mapping theorem actually accomplish here?

Thanks in advance

Answer 1

This is an initial value problem, not a boundary value problem. Both conditions are at $t=0$.

The existence and uniqueness of solution is a standard theorem, Picard-Lindelöf. You aren't going to get them from the abstract Banach space reformulation. My guess is that the open mapping theorem is meant as a tool to prove stability. I would apply it thus:

let $Y=\mathbb R^2\times C[0,1]$, equipped with some product norm. Define $T:C^2[0,1]\to Y$ so that $y$ is sent to $(y(0),y'(0), y''+qy'+r)$. This is a bounded operator. By the aforementioned existence-uniqueness result $T$ is a bijection. The open mapping theorem tells you $T$ has a bounded inverse $T^{-1}:Y\to C^2[0,1]$. This is precisely the stability statement you need.