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

For $I=(-a,a)$, the first-order Sobolev space isThe weak derivative $v=u^{\prime }$ is unique as an equivalence class modulo equality almost everywhere. We use the standard Sobolev norm $\Vert u{\Vert }_{H^1(I)}^2=\Vert u{\Vert }_{L^2(I)}^2+\Vert u^{\prime }{\Vert }_{L^2(I)}^2$. Pointwise values of an arbitrary representative have no intrinsic meaning; the one-dimensional Sobolev representative supplies the continuous representative used later.

Reflect $u$ across the two endpoints of $I$ into a slightly larger interval, giving an $H^1$ extension, and multiply by a smooth cutoff function equal to one near $\overline{I}$. Convolve the resulting function on $\mathbb{R}$ with a mollifier. The convolutions $u_{\epsilon }$ are smooth; differentiation commutes with convolution of the weak derivative. Continuity of translations in $L^2$ proves both $u_{\epsilon }\to u$ and $u_{\epsilon }^{\prime }\to u^{\prime }$ in $L^2(I)$, hence convergence in the Sobolev norm.

To obtain convergence almost everywhere, select a sequence with $\Vert u_n-u{\Vert }_2\le 2^{-n}$. Then $\int {\sum }_n|u_n-u{|}^2<\infty$, so the summands tend to zero almost everywhere. Thus a smooth approximating sequence converges both in $H^1$ and almost everywhere. This is density of smooth functions in a Sobolev space; smooth functions need not vanish at the endpoints.

For a smooth approximation, the fundamental theorem of calculus and Cauchy-Schwarz inequality giveLet $E\subset I$ be one full-measure set on which $u_n\to u$. Passing to the limit for all $x,y\in E$ givesThe full-measure set $E$ is dense, and the estimate extends its representative uniquely to a $1/2$-Hölder continuous function on $\overline{I}$. Its classical Hölder seminorm satisfies the same bound. Changing the original representative on a null set does not affect the essential supremum in the assertion.

With the standard Sobolev norm, the printed constant one is not valid for arbitrary $a$. The constant function $u=1$ has supremum norm one and $H^1$ norm $\sqrt{2a}$, which is smaller when $a<1/2$.

For the continuous one-dimensional Sobolev representative, let $\bar{u}=(2a{)}^{-1}{\int }_{I}u$. The Cauchy-Schwarz inequality gives $|\bar{u}|\le (2a{)}^{-1/2}\Vert u{\Vert }_2$. Using the preceding Hölder seminorm estimate,Consequently the corrected interval Sobolev supremum estimate isOne may first prove this for smooth approximations and pass on the full-measure convergence set. Averaging differences also handles complex-valued functions without assuming that a function takes its complex average at some point. At the fixed interval used below the constant is absolute, which suffices for the intended interior estimate.

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 $R$.

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.