inverse of differential operator Define $A:\{u\in C^2[0,1];u(0)=0,u'(0)=0\}\to C[0,1]$, $A(u)=u''+au'+bu$ where $a,b\in\mathbb{R}$. Can we compute the inverted operator $A^{-1}$ ? Thank You in advance for Your answers.
 A: We first find the Green's function, which satisfies $g'' + ag' + bg = \delta(t-\tau)$ and the same initial conditions $g'(0) = g(0) = 0$. This looks like it might be easy to solve using a Laplace transform, which requires $g$ be defined on $[0, \infty)$ rather than just $[0,1]$. Not a problem; we'll just restrict the domain later.
The Laplace transform becomes
\begin{align*}
\mathcal{L}[g](s) = \frac{e^{-s\tau}}{s^2 + as + b}
\end{align*}
Now we can use table of Laplace transforms to find the inverse Laplace transform. We can use identity (27) of the linked pdf to see that if $F(s) = \frac{1}{s^2 + as + b}$, then $g(t, \tau) = \theta(t - \tau)f(t - \tau)$, where $\theta$ is the Heaviside step function.
Now we write
\begin{align*}
F(s) = \frac{1}{(s+a/2)^2 + b - a^2/4}
\end{align*}
and let $\omega^2 := b - a^2/4$. Then
\begin{align*}
\mathcal{L}^{-1}[F](t) = \frac{1}{\omega} \exp(-at/2)\sin(\omega t)
\end{align*}
and hence $g(t|\tau) = \theta(t-\tau) \exp(-a(t-\tau)/2)\sin(\omega (t-\tau))/\omega$
\begin{align*}
A^{-1}f(t) = \int_{0}^{1} \theta(t-\tau) \frac{\exp(-a(t-\tau)/2)\sin(\omega (t-\tau))}{\omega} f(\tau) \, \mathrm{d}\tau
\end{align*}
If $b < a^2/4$ then $\omega \in i\mathbb{R}$, so then it's more convenient to express this in terms of the $\sinh$ function, but nonetheless this should be ok.
