Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 5 / 3

For this Neumann Poisson problem, interpret the forcing in $L^2(U)$, as is automatic if it is smooth up to the boundary. Literal interior smoothness alone does not ensure the integrals or bounded functionals required in this question: for example, $f(x)=x^{-2}$ on $(0,1)$ is interior smooth but even $\int f\cdot 1$ diverges. Classical regularity in the converse is likewise understood up to the boundary.

Suppose the weak solution is smooth on $\overline{U}$. Testing against compactly supported test functions gives $-\Delta u=f$ in distributions and hence pointwise. Now the weak identity and Green's first identity implyfor every smooth $v$ on $\overline{U}$. Every smooth boundary function has such an extension, so ${\partial }_{n}u=0$ on $\partial U$. This proves both the interior equation and the boundary condition.

Conversely, for $u\in C^2(\overline{U})$ satisfying the equation and zero normal derivative, Green's first identity gives the weak identity for all smooth $v$ on $\overline{U}$. The density of smooth functions in a Sobolev space and the Cauchy-Schwarz inequality extend it continuously to every $v\in H^1(U)$. Thus the classical solution is a weak solution, and the smooth weak solution is classical.

Subtract the two weak solution identities and test with their difference $w\in H^1(U)$. ThenA Sobolev function with zero weak gradient is constant on each connected component. One justification is to mollify locally: each mollification has zero gradient and is constant on its ball, and overlaps identify the constants; taking limits gives the original assertion. Since $U$ is connected, $w$ is one constant on $U$.

Conversely, adding a constant changes neither the weak derivative nor the weak identity. The solution is unique up to an additive constant.

There is a normalization error in the PDF. With the printed unnormalized integral, testing a constant function equal to one would give a left side $|U|(1-|U|{)}^2$ and a right side zero. Thus that formulation fails whenever $|U|\ne 1$.

Use instead the average $v_U=|U{|}^{-1}{\int }_{U}v$. The Poincare-Wirtinger inequality, also called the Neumann-Poincare inequality, isIf no $C_P$ exists, subtract the average and normalize a violating sequence to obtain $w_j\in H^1(U)$ with ${\int }_U{w}_j=0$, $\Vert w_j{\Vert }_2=1$, and $\Vert \nabla w_j{\Vert }_2\to 0$. This sequence is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrachov compactness theorem supplies a subsequence converging strongly in $L^2(U)$ to $w$.

For every compactly supported test function $\phi$, integration by parts and these convergences give $\int w{\partial }_i\phi =0$. Thus $w$ has zero weak gradient. The Sobolev function with zero weak gradient result and connectedness make $w$ constant. Strong $L^2$ convergence preserves its zero integral, so $w=0$. It also preserves its norm one, a contradiction. This proves the correctly normalized Neumann-Poincare inequality.

Work in the mean-zero Sobolev spaceThe integral is a continuous functional on $H^1(U)$, so $H$ is closed. The Neumann-Poincare inequality shows that $\Vert v{\Vert }_a=\Vert \nabla v{\Vert }_2$ is equivalent to the usual $H^1$ norm on $H$, making $a$ a complete inner product there.

For $f\in L^2(U)$ the functional $L(v)={\int }_{U}fv$ satisfiesThe Riesz representation theorem, or the Lax-Milgram theorem, gives a unique $u\in H$ with $a(u,v)=L(v)$ for all $v\in H$. To recover every $H^1(U)$ test, write $v=(v-v_U)+v_U$. The constant contributes zero to $a$ and contributes $v_U{\int }_{U}f=0$ to $L$. Thus the same equality holds for all $v\in H^1(U)$.

A weak solution exists whenever ${\int }_{U}f=0$; fixing its average to zero makes it unique. Moreover $\Vert \nabla u{\Vert }_2\le \sqrt{C_P}\Vert f{\Vert }_2$ and the Neumann-Poincare inequality also controls $\Vert u{\Vert }_2$.

The constant function $v=1$ is an admissible $H^1(U)$ test for the Neumann Poisson problem. Its weak gradient is zero, so the weak identity immediately givesThis is necessary, and the preceding Hilbert-space construction proves sufficiency for $f\in L^2(U)$. It is the balance condition corresponding to zero total boundary flux.

The clamped second-order Sobolev space is $H_0^2(U)={\overline{C_c^{\infty }(U)}}^{H^2}$. On a smooth bounded domain the Sobolev trace theorem characterizes it by zero value and zero normal derivative on the boundary. In particular, its whole first-order boundary jet is zero, since tangential derivatives of the zero trace also vanish. As above, assume $f\in L^2(U)$ and classical regularity up to the boundary.

For a smooth weak solution, compactly supported test functions and two integrations by parts giveso ${\Delta }^{2}u=f$ pointwise. Membership in $H_0^2(U)$ supplies $u={\partial }_{n}u=0$ on $\partial U$. Thus it is a classical solution of the clamped biharmonic problem.

Conversely, a $C^4(\overline{U})$ classical solution with these traces belongs to $H_0^2(U)$. For every compactly supported test function, two integrations by parts give ${\int }_U\Delta u\Delta v={\int }_{U}fv$. Both sides are continuous for the $H^2$ norm, so the defining density of $C_c^{\infty }(U)$ in $H_0^2(U)$ extends this equality to every required test. The two notions agree under the stated smoothness.

The difference $w$ of two weak solutions lies in the clamped second-order Sobolev space. Testing with $w$ yields ${\int }_U(\Delta w{)}^2=0$, so $\Delta w=0$. Since $w\in H_0^1(U)$, integration by parts givesThe Poincare inequality for zero boundary values now implies $\Vert w{\Vert }_2=0$. The clamped biharmonic problem has at most one weak solution. Unlike the Neumann Poisson problem, no additive constant is allowed by these boundary traces.

For $v\in C_c^{\infty }(U)$, two integrations by parts give the clamped Hessian identityBy density it remains valid on $H_0^2(U)$. Each ${\partial }_{i}v$ has zero integral, first for compactly supported test functions and then by $H^2$ convergence. Applying the Neumann-Poincare inequality to each ${\partial }_{i}v$ givesThe zero-boundary Poincare inequality also gives $\Vert v{\Vert }_2^2\le C_D\Vert \nabla v{\Vert }_2^2$. Hence, using a full-Hessian equivalent $H^2$ norm,Thus $a(u,v)={\int }_U\Delta u\Delta v$ is an inner product whose norm is equivalent to the complete $H^2$ norm on the clamped second-order Sobolev space. The functional $L(v)={\int }_{U}fv$ is bounded for this norm by the Cauchy-Schwarz inequality and the displayed bound. Apply the Riesz representation theorem, or the Lax-Milgram theorem, to get a unique $u\in H_0^2(U)$ representing $L$. The clamped biharmonic problem has a unique weak solution for every $f\in L^2(U)$, with $\Vert u{\Vert }_{H^2}\le C\Vert f{\Vert }_2$ and no zero-integral compatibility condition.