Past exam of the methods of applied functional analysis course of the University of Oxford / 2010 / A

Here uniform ellipticity means that there is a constant $\lambda >0$, independent of $x$, such thatThe assumption that the entries are essentially bounded functions also gives an upper bound $|a(x)\xi |\le \Lambda |\xi |$ with a finite constant $\Lambda$. For this symmetric divergence-form operator, the lower bound makes its energy control the squared Euclidean norm of the gradient.

The space of test functions is $\boxed{C_0^{\infty }(\Omega )=\{\phi \in C^{\infty }(\Omega ):\mathrm{supp}\phi \text{ is a compact subset of }\Omega \}}$. The Sobolev space iswhere $D_{j}u$ denotes a weak derivative, and its Sobolev norm is $\Vert u{\Vert }_{H^1}^2=\Vert u{\Vert }_2^2+{\sum }_j\Vert D_{j}u{\Vert }_2^2$. The zero-boundary Sobolev space is $\boxed{H_0^1(\Omega )={\overline{C_0^{\infty }(\Omega )}}^{H^1}}$. This closure defines the boundary condition even when the boundary has insufficient regularity for a classical boundary value.

A weak solution is a function $u\in H_0^1(\Omega )$ satisfying the weak formulationMultiplication by a test function and integration by parts give this identity from the classical equation. Conversely it is the equation in the sense of distributions, with the zero boundary condition encoded by the zero-boundary Sobolev space. Testing only against $C_0^{\infty }(\Omega )$ is equivalent, by its density in $H_0^1$ and continuity of both sides.

Write $a(v,w)={\int }_{\Omega }a\nabla v\cdot \nabla w$ and $A[v]=a(v,v)$. The admissible set is the closed affine subspace $H_g^1=g+H_0^1$. Symmetry of the coefficient matrix givesIf $u$ minimizes the energy, its first variation vanishes: differentiating at $t=0$ gives $a(u,\phi )=0$. Hence $u$ is a weak solution.

Conversely, if $u$ is a weak solution, every admissible $v$ has $v-u\in H_0^1$ and thereforeby uniform ellipticity. This is the Dirichlet principle for the stated symmetric energy.

We prove existence directly through quadratic energy minimization by the parallelogram law. Let $m={\inf }_{v\in H_g^1}A[v]$. The set contains $g$, and uniform ellipticity gives $0\le m\le A[g]<\infty$. Choose a minimizing sequence $v_j$ with $A[v_j]\to m$. Since $(v_j+v_k)/2\in H_g^1$, the parallelogram law givesThe differences belong to $H_0^1$, so uniform ellipticity and the Poincaré inequality implyThus $(v_j)$ is a Cauchy sequence. Completeness of $H_0^1$, applied to $v_j-g$, produces $u\in H_g^1$ with $v_j\to u$ in $H^1$. The bounded coefficient matrix makes $a$ a bounded bilinear form, so $A[v_j]\to A[u]$. Consequently $A[u]=m$, and part (i) makes $u$ a weak solution.

For uniqueness, the difference $w$ of two weak solutions lies in $H_0^1$ and satisfies $a(w,w)=0$. Uniform ellipticity gives $\nabla w=0$ almost everywhere, and the Poincaré inequality gives $w=0$. The argument proves existence and uniqueness without assuming the Lax-Milgram theorem.

For one coordinate, integration by parts gives ${\int }_{\Omega }({\partial }_{j}u)\phi =-{\int }_{\Omega }u{\partial }_j\phi$: the test function vanishes outside a compact subset of $\Omega$, so there is no boundary contribution. Repeat this operation once for each of the $|\alpha |$ coordinate differentiations in the multi-index $\alpha$. Each operation contributes a factor $-1$, givingAll intermediate functions are continuous on the relevant compact support, since $u\in C^k$ and $|\alpha |\le k$.

The function $v$ is the weak derivative $D^{\alpha }u$ ifThe integrals exist by local integrability and the compact support of the test function. This is the definition of a distributional derivative represented by a locally integrable function.

The fundamental lemma of the calculus of variations says that if $h\in L_{\mathrm{loc}}^1(\Omega )$ and $\int h\phi =0$ for every $\phi \in C_0^{\infty }(\Omega )$, then $h=0$ almost everywhere.

If $v_1,v_2$ both represent $D^{\alpha }u$, subtracting their defining identities gives $\int (v_1-v_2)\phi =0$ for every test function. The lemma gives $v_1=v_2$ almost everywhere, proving uniqueness. If $u\in C^1$, part (i) gives exactly the defining identity with $v={\partial }_{j}u$. Since this classical derivative is locally integrable, it is the unique weak derivative $D_{j}u$, up to equality almost everywhere.

An integrable example on the whole real line is the tent functionIt is not differentiable at $-1,0,1$. For a test function $\phi$, split the integral into $(-1,0)$ and $(0,1)$ and use integration by parts. The endpoint terms at $-1,1$ vanish because $u=0$, and those at zero cancel because $u$ is continuous. The result is $\int u{\phi }^{\prime }=-\int (Du)\phi$. Thus the corners obstruct classical differentiability but not this weak derivative.

Choose a standard mollifier $\rho \in C_0^{\infty }({\mathbb{R}}^n)$ with $\rho \ge 0$, $\int \rho =1$ and support in the unit ball, and set ${\rho }_{\epsilon }(z)={\epsilon }^{-n}\rho (z/\epsilon )$. Define the convolutionThe smoothing properties of a mollifier give $u_{\epsilon }\in C^{\infty }$ and ${\partial }_j{u}_{\epsilon }={\rho }_{\epsilon }\ast D_{j}u$. Apply the chain rule to $t\mapsto u_{\epsilon }(y+t(x-y))$ and integrate its derivative by the fundamental theorem of calculus:

Write $g=(g_1,\dots ,g_n)$ for the continuous representatives of the weak derivatives. Integrating the preceding identity over $B=B(0,1)$ givesThe first term converges to ${\int }_{B}u$ because $u_{\epsilon }\to u$ in $L^1$. For fixed $x$, all segments in the second term lie in a compact set. Continuity of $g$ implies ${\rho }_{\epsilon }\ast g\to g$ uniformly on that set. Hence every $x$ has the finite limitThe same argument is uniform for $x$ in any fixed compact set: all the segments then lie in a common compact set. Thus $u_{\epsilon }\to U$ locally uniformly, so $U$ is continuous; the local $L^1$ convergence also gives $U=u$ almost everywhere.

Passing to the limit in the segment identity, and taking its endpoints to be $x$ and $x+h$, givesTherefore $U$ is differentiable with continuous gradient $g$: $\boxed{U\in C^1,U=u\text{ almost everywhere},\nabla U=g}$. This is the continuous weak gradient gives a C1 representative criterion.

The real Lax-Milgram theorem states: let $H$ be a Hilbert space and let $a:H\times H\to \mathbb{R}$ be a bounded bilinear form such that $a(v,v)\ge \alpha \Vert v{\Vert }_H^2$ for some $\alpha >0$ and every $v\in H$. For every bounded linear functional $\ell \in H^{\ast }$ there exists a unique $u\in H$ satisfyingSymmetry of $a$ is not required. The complex version replaces bilinearity by sesquilinearity and the lower bound by $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }_H^2$.

Introduce coordinates $s=x+y$, $t=x-y$, so $dxdy=\frac{1}{2}dsdt$. For a test function $\phi (x,y)=\Phi (s,t)$,almost everywhere. HenceThe first derivative integrates to zero in $s$, and the second to zero in $t$, using compact support. Thus $\boxed{\mathrm{div}V=0}$ as a weak divergence. Geometrically, the jump across $x+y=0$ is $(1,-1)$, perpendicular to its normal $(1,1)$; there is no distributional flux concentrated on the interface.

A weak solution belongs to the zero-boundary Sobolev space $H_0^1(\Omega )$ and satisfiesThe boundedness of $b$ and the Cauchy-Schwarz inequality make the drift term integrable. This weak formulation moves one derivative from the Laplacian onto the test function, while retaining the first-order drift term.

Give $H_0^1$ the equivalent Hilbert space norm $\Vert u{\Vert }_H=\Vert \nabla u{\Vert }_2$, using the Poincaré inequality. DefineThe Cauchy-Schwarz inequality gives $|a(u,v)|\le (1+C_P\Vert b{\Vert }_{\infty })\Vert u{\Vert }_H\Vert v{\Vert }_H$ and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_H$.

For $\phi \in C_0^{\infty }(\Omega )$, the definition of weak divergence, applied to ${\phi }^2/2$, givesThe left side is continuous in the $H_0^1$ norm. Approximate any $u\in H_0^1$ by test functions and pass to the limit on the left to obtain $\int (b\cdot \nabla u)u\ge 0$. This density argument is essential: local integrability of $\mathrm{div}b$ alone does not justify treating $(\mathrm{div}b)u^2$ as integrable for arbitrary $u\in H_0^1$.

Consequently $a(u,u)\ge \Vert u{\Vert }_H^2$, so $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique weak solution with $\boxed{\Vert \nabla u{\Vert }_2\le C_P\Vert f{\Vert }_2}$. The reusable estimate is coercivity from a drift with nonpositive weak divergence.