Show that $L$ is formally self-adjoint.

Question

Consider the differential operator $$L=e^xD^2+e^xD,\;\;D=\frac{d}{dx},\;0\leq x\leq1,$$ $$u^\prime(0)=0,\;\;\; u(1)=0.$$ Show that $L$ is formally self-adjoint.

I just don't really know how to start this. I'm not quite sure how "formally self-adjoint" is different from just "self-adjoint". Any help/hints would be welcome.

Answer 1

Let $f,g\in dom(D^2)$, recall that $f'(0)=g'(0)=0$ and $f(1)=g(1)=0$.

$$\begin{align*} \langle Lf,g \rangle &= \int\limits_0^1(e^xf_{xx}+e^xf_x)gdx \\ &=\int\limits_0^1gd(e^xf_x) \\ &=\overbrace{ge^xf_x\Big|_0^1}^{=0} - \int\limits_0^1e^xf_xg_xdx\\ &=- \int\limits_0^1e^xf_xg_xdx\\ &=- \int\limits_0^1e^xg_xdf\\ &=\overbrace{g_xe^xf\Big|_0^1}^{=0} + \int\limits_0^1f(g_xe^x)'dx\\ &=\int\limits_0^1f(e^xg_x+e^xg_{xx})dx = \langle f, Lg \rangle, \end{align*}$$

conclude.