= Solution
The clamped energy space is
$$
\mathcal H=H_0^2(-1,1)
=\{u\in H^2(-1,1):u(\pm1)=u'(\pm1)=0\}.
$$
Twice applying <integration by parts>, with the boundary terms killed by the clamped conditions, gives the symmetric bilinear form
$$
a(u,v)=\int_{-1}^1
\left(pu''v''+qu'v'+ruv\right)dx.
$$
Consequently
$$
\langle Lu,u\rangle=a(u,u)
=\int_{-1}^1\left(p|u''|^2+q|u'|^2+r|u|^2\right)dx>0
$$
for every nonzero $u\in\mathcal H$: equality forces $u''=0$ 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 $\{u\in H_0^2:\sqrt p\,u''\in L^2\}$; under the usual coefficient assumption $0<p_0\leq p\leq p_1<\infty$, it is exactly $H_0^2$ and the form is coercive there.
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