 Solution

The clamped energy space isTwice applying integration by parts, with the boundary terms killed by the clamped conditions, gives the symmetric bilinear formConsequentlyfor every nonzero $u\in \mathcal{H}$: equality forces $u^{\prime \prime }=0$ almost everywhere, and the clamped boundary values then force $u=0$. Hence $L$ is symmetric and positive definite.

Strictly under the stated assumption $p\in L^2$ rather than $L^{\infty }$, the first integral need not be finite for every $u\in H_0^2$. The literal energy domain is therefore $\{u\in H_0^2:\sqrt{p}{u}^{\prime \prime }\in L^2\}$; under the usual coefficient assumption $0<p_0\le p\le p_1<\infty$, it is exactly $H_0^2$ and the form is coercive there.