Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 5 / 3 / 2

Write $R=\Lambda /\lambda$. The coefficient is measurable, so interpret the divergence-form elliptic operator weakly: $\int A{u}^{\prime }{\varphi }^{\prime }=0$ for compactly supported $H^1$ tests. Testing with $\varphi ={\zeta }^{2}u$ givesDivide if the first integral is positive; the zero case is immediate. Since $\zeta =1$ on the inner interval,This Caccioppoli inequality applies in particular to the smooth solutions in the question. No derivative of $A$ is taken.

Choose a fixed smooth cutoff function with $\Vert {\zeta }^{\prime }{\Vert }_{\infty }=C_0$ and set $J=(-1/2,1/2)$. The Caccioppoli inequality gives $\Vert u^{\prime }{\Vert }_{L^2(J)}\le 2R{C}_0\Vert u{\Vert }_{L^2(-1,1)}$. Since the length of $J$ is one, the corrected interval Sobolev supremum estimate yields $\Vert u{\Vert }_{L^{\infty }(J)}\le \Vert u{\Vert }_{L^2(J)}+\Vert u^{\prime }{\Vert }_{L^2(J)}$, and the Hölder seminorm is at most $\Vert u^{\prime }{\Vert }_{L^2(J)}$. ThereforeFor locally $H^1$ weak solutions with $u\in L^2(-1,1)$, density of smooth functions in a Sobolev space justifies the test ${\zeta }^{2}u$, and the previous representative estimates already apply to $H^1$. Mollifying a solution need not preserve the equation with the same measurable coefficient, so approximation is used for admissible tests and Sobolev estimates, not to assert that the mollified function solves the original equation. The required constant depends only on $R$.