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

For a real weak solution $u\in H_{\mathrm{loc}}^1(B_1)$, $\int A\nabla u\cdot \nabla \varphi =0$ for compactly supported $H^1$ tests. The symmetric coefficient bounds give $A\xi \cdot \xi \ge \lambda |\xi {|}^2$ and $|A\xi |\le \Lambda |\xi |$. Test with $\varphi ={\zeta }^{2}u$ and apply the Cauchy-Schwarz inequality:Thus $E\le 4R^{2}I$, including $E=0$. As $\zeta =1$ on $B_{r_1}$ and has support in $B_{r_2}$,This Caccioppoli inequality is valid for smooth solutions and, by density of admissible tests, for locally $H^1$ weak solutions with bounded measurable coefficients.

Take $\alpha =1+1/\ell >1$. Then $2\alpha <p^{\ast }$ in every $\ell \ge 2$, including $p^{\ast }=\infty$ for $\ell =2$. The supplied Sobolev embedding theorem, by the fixed scaling from $B_{1/2}$ to $B_1$, gives $\Vert w{\Vert }_{L^{2\alpha }(B_1)}\le C_{\ell }\Vert w{\Vert }_{H^1(B_1)}$.

Choose $0\le \zeta \le 1$, equal to one on $B_{r_1}$, supported in $B_{r_2}$, with $|\nabla \zeta |\le C/(r_2-r_1)$. Extend $w=\zeta u$ by zero to $B_1$. The Caccioppoli inequality before discarding the cutoff weight givesAlso $\Vert w{\Vert }_2\le \Vert u{\Vert }_{L^2(B_{r_2})}$. Since $R>1$ and $r_2-r_1\le 1$, both terms are controlled by $CR(r_2-r_1{)}^{-1}\Vert u{\Vert }_2$. Applying the Sobolev embedding theorem to $w$ givesUsing a fixed ambient ball prevents the Sobolev constant from acquiring an unintended dependence on either radius.

The original PDF assumes $u\ge 0$ at the beginning of this question; the TeX conversion omits that hypothesis. First replace a smooth nonnegative solution by $v=u+\epsilon >0$, which solves the same homogeneous divergence-form elliptic equation. For $w=v^{q/2}$, the chain rule gives the favorable termin the distributional sense. Thus $w$ is a weak subsolution. Testing its subsolution inequality with ${\zeta }^{2}w$ yields the same Caccioppoli inequality as before. More precisely, testing the equation for $v$ with ${\zeta }^2{v}^{q-1}$ givesIndeed the main term is at least $(q-1)\lambda \int {\zeta }^2{v}^{q-2}|\nabla v{|}^2$, and the cross term is bounded by twice $\Lambda$ times the square roots of this weighted gradient integral and $\int v^q|\nabla \zeta {|}^2$. Let $\epsilon \downarrow 0$ to obtain the stated power energy estimate.

Apply the fixed-domain Sobolev embedding theorem to $\zeta u^{q/2}$. The crucial constant stays uniform before taking the $2/q$ power:where $D_{\ell }\ge 1$ is independent of $q$. This is uniform power gain for elliptic solutions. The exponent is $q\alpha$, not the TeX conversion's $q^{\alpha }$. The PDF's weaker prefactor $C_{\ell }$ outside the $2/q$ power follows from this stronger estimate, but that weaker form alone cannot be iterated infinitely.

For completeness, the estimate also extends to nonnegative locally $H^1$ weak solutions without assuming they are already bounded. For $s\ge 0$ define $F_M(s)=s^{q/2}$ up to $M$ and continue it linearly with slope $(q/2)M^{q/2-1}$ above $M$; set ${\Psi }_M(s)={\int }_0^s{F}_M^{\prime }(r{)}^{2}dr$. Both have bounded derivatives for fixed $M$. Direct calculation gives ${\Psi }_M/F_M^{\prime }\le F_M$ and $F_M(s)\le (q/2)s^{q/2}$. Test the weak equation with ${\zeta }^2{\Psi }_M(u)$, using smooth approximations to these Sobolev chain rule functions if needed. The resulting weighted energy bound is $\int {\zeta }^2|\nabla F_M(u){|}^2\le 4R^2\int F_M(u{)}^2|\nabla \zeta {|}^2$. If $u\in L^q(B_{r_2})$, dominated convergence on the right and the Fatou lemma on the Sobolev left give the boxed gain as $M\to \infty$, with the same uniform constant. Starting with $q=2$, this justifies every subsequent iteration step.

Set $q_i=2{\alpha }^i$, $r_i=1/2+2^{-i-1}$ and $N_i=\Vert u{\Vert }_{L^{q_i}(B_{r_i})}$. Then $r_0=1$, $r_i-r_{i+1}=2^{-i-2}$, and the sharper uniform power gain for elliptic solutions givesThe Moser product on geometric radii converges becauseThus every $N_i$ is at most $B=(D_{\ell }R{)}^{S_0}{2}^{S_1}\Vert u{\Vert }_{L^2(B_1)}$. Since $B_{1/2}\subset B_{r_i}$, its $L^{q_i}$ norms are also bounded by $B$. If $u>B+\delta$ on a set of positive measure $m$ in $B_{1/2}$, these norms are at least $(B+\delta )m^{1/q_i}$, tending to $B+\delta$, a contradiction. HenceThis is the required Moser iteration estimate, with a finite constant depending only on $\ell$ and $R$. The weak-test truncation argument in the preceding solution makes the conclusion valid for locally $H^1$ weak solutions with $u\in L^2(B_1)$ and measurable uniformly elliptic coefficients. No differentiability of $A$ or unproved smooth approximation of solutions is required.