 Solution

Use the zero-boundary Sobolev space $V=H_0^1(0,1)$ with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$ makes this a complete Hilbert space norm. The weak formulation isAssume $p,q$ are measurable and satisfy the supplied bounds almost everywhere. The Cauchy-Schwarz inequality and Poincaré inequality giveBecause $p\ge p_0>0$ and $q\ge 0$,This is the coercive variable-coefficient Dirichlet form. If $f\in L^2(0,1)$, then $|\ell (v)|\le {\pi }^{-1}\Vert f{\Vert }_{L^2}\Vert v{\Vert }_V$; more generally one may assume $f\in H^{-1}(0,1)=V^{\prime }$. Some such source regularity is necessary to invoke the theorem. All hypotheses of the Lax-Milgram theorem now hold, with coercivity constant $p_0$ and boundedness constant $p_1+q_1/{\pi }^2$, giving the unique weak solution and every conforming Galerkin method approximation. This argument does not require a pointwise derivative of $p$.