Past exam of the methods of applied functional analysis course of the University of Oxford / 2010 / B / 6

Using the dual pairing between $X^{\ast }$ and $X$, a monotone operator satisfiesA nonlinear coercive operator with respect to $u_0$ satisfiesThe directional derivative is ${\partial }_{v}F(u)={\lim }_{t\to 0}[F(u+tv)-F(u)]/t$, when this finite limit exists through admissible points. At a boundary point of $M$ only an inward direction may be admissible; then one uses the corresponding one-sided limit.

If $A:X\to X^{\ast }$ is monotone and $s<t$, apply its defining inequality at $u+tv$ and $u+sv$. It becomes $(t-s)[\phi (t)-\phi (s)]\ge 0$, so $\phi$ is nondecreasing. Conversely choose $u=y$, $v=x-y$. Nondecrease from zero to one gives $⟨A(x)-A(y),x-y⟩\ge 0$. This proves the line criterion for a monotone operator.

The graph area functional is finite on $H^1$, since $\sqrt{1+|\nabla u{|}^2}\le 1+|\nabla u|$ and $\Omega$ has finite volume. For $p=\nabla u$, $q=\nabla v$, the difference quotient of $\sqrt{1+|p+tq{|}^2}$ is bounded in absolute value by $|q|$. This is integrable by the Cauchy-Schwarz inequality. The dominated convergence theorem therefore givesThis defines a bounded linear functional of $v$, since $|⟨A(u),v⟩|\le |\Omega {|}^{1/2}\Vert \nabla v{\Vert }_2$.

For $a(p)=p/\sqrt{1+|p{|}^2}$ the Jacobian matrix isThe inequality follows from the Cauchy-Schwarz inequality. Integrating this derivative along the segment between any two gradients shows $(a(p)-a(r))\cdot (p-r)\ge 0$. Integrating over $\Omega$ proves that $A$ is a monotone operator.

It is not coercive on $H^1(\Omega )$. The constant functions $u_k\equiv k$ satisfy $\Vert u_k{\Vert }_{H^1}=k|\Omega {|}^{1/2}\to \infty$ but $A(u_k)=0$. Thus $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

The operator is well-defined in every dimension: $|\cos w|\le 1$ givesFor fixed $w$, its monotonicity condition isThus $\cos w\ge 0$ almost everywhere is sufficient. It is also necessary, by the positivity criterion for a scalar gradient form. Indeed, if the bounded coefficient $c=\cos w$ is negative on a set of positive measure, choose a Lebesgue point $x_0$ with $c(x_0)<0$ and a nonconstant compactly supported smooth $\psi$. Set $z_{\epsilon }(x)=\epsilon \psi ((x-x_0)/\epsilon )$, supported inside $\Omega$. Thencontradicting monotonicity. Therefore $\boxed{A(w,\cdot )\text{ is monotone exactly when }\cos w\ge 0\text{ almost everywhere}}$. In particular it is not monotone for every $w$: the admissible constant $w\equiv \pi$ makes it the negative Dirichlet form.

In dimension one, the Sobolev embedding theorem gives $H^1(\Omega )\hookrightarrow L^{\infty }(\Omega )$. Thereforeand, for fixed $w$,Hence the operator is well-defined and monotone in one dimension.

In every dimension $n\ge 2$ it is not well-defined on all of $H^1\times H^1$. We exhibit $w=u=v\in H^1$ for which the purported pairing $\int w^2|\nabla w{|}^2$ is infinite. Choose a ball compactly contained in $\Omega$, put $r=|x-x_0|$, and multiply each following radial expression by a smooth cutoff function that equals one near $r=0$.

For $n\ge 3$, set $w=r^{-\alpha }$ with $\alpha =(n-2)/3$. The local Sobolev norm is finite becausewhereassince $n-3-4\alpha =(-n-1)/3<-1$.

For $n=2$, take $w=(\log (1/r){)}^{1/3}$ on a sufficiently small ball. With $s=\log (1/r)$, the gradient integral is proportional to ${\int }^{\infty }{s}^{-4/3}ds<\infty$, while the weighted integral is proportional to ${\int }^{\infty }{s}^{-2/3}ds=\infty$. Its $L^2$ integral is finite because the area factor contributes exponential decay in $s$. These singular functions have the indicated weak derivatives; deleting a small central ball and passing to the limit in integration by parts leaves no boundary term.

Thus $w^2\nabla u\cdot \nabla v$ need not be integrable in dimensions at least two. This is the squared Sobolev gradient counterexample. With the additional restriction $w\in L^{\infty }$, the form is well-defined and monotone in any dimension.