Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 7 / 2

Use the conventional Laplace operator $\Delta ={\partial }_{x_1}^2+{\partial }_{x_2}^2$. The fundamental solution of the Laplace equation in two dimensions is $G(x)=(2\pi {)}^{-1}\log |x|$, with $\Delta G={\delta }_0$. To check its sign, integrate the outward normal derivative around a circle: ${\int }_{\partial B_{\epsilon }}{\partial }_n\log |x|ds=2\pi$. Applying integration by parts against a smooth compactly supported test function and shrinking the circle gives the distributional identity.

The printed stream function has $\varphi =-G\ast \omega$. For each multi-index $\beta$ with $|\beta |\le 2$, transfer derivatives to the compactly supported $C^2$ function:The logarithmic singularity is locally integrable. On a bounded region in $x$, the convolution uses one fixed bounded region in the integration variable; splitting off a small disk around the singularity proves continuity of these derivatives. Thus $\varphi \in C^2$, and convolution of the distributional identity givesThe asserted plus sign is false for the printed kernel and the conventional Laplacian. Any nonzero smooth compactly supported $\omega$ is a counterexample. To obtain $\Delta \varphi =\omega$, replace the kernel's minus sign by a plus sign, or explicitly adopt the negative Laplace operator. Below, retain the printed kernel and velocity convention. With that convention, $\mathrm{curl}u=\Delta \varphi =-\omega$, so $\omega$ is the negative of conventional scalar vorticity. This sign does not change the transport or flow arguments.

Take canonical coordinates $q=x_1$ and $p=x_2$. The Hamiltonian system isIn the standard Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$, its Hamiltonian isAn arbitrary function of $t$ can be added without affecting the characteristic curves. The Hamiltonian generally depends on time through $\omega$ and need not be conserved along a trajectory.

Let $S_{s,t}(y)$ be the position at time $t$ of the characteristic curve satisfying $X(s)=y$ and $\dot{X}=u(t,X)$. Smooth uniqueness implies $S_{t,s}=S_{s,t}^{-1}$. Differentiating in the initial point gives the variational equationThe Liouville formula for a fundamental matrix, or the derivative formula for a determinant, therefore givesBecause $u=(-{\varphi }_{x_2},{\varphi }_{x_1})$, the mixed second derivatives cancel and $\mathrm{div}u=0$. The initial Jacobian determinant is one, soThus this Hamiltonian flow preserves planar Lebesgue measure. This calculation assumes smoothness as in the part; a rough-flow conclusion later must be justified by approximation rather than by differentiating a merely continuous velocity.

The planar vorticity velocity kernel has magnitude $(2\pi |z|{)}^{-1}$. Split its defining integral into $|x-y|<R$ and $|x-y|\ge R$ for any $R>0$. The singular part is absolutely integrable in two dimensions:In particular $R=1$ provesWhen both norms are nonzero, optimization in $R$ also gives $\Vert U[\Omega ]{\Vert }_{\infty }\le \sqrt{2/\pi }\Vert \Omega {\Vert }_1^{1/2}\Vert \Omega {\Vert }_{\infty }^{1/2}$. If either norm is zero, $\Omega =0$ almost everywhere. Absolute convergence supplies a well-defined velocity at every $x$, with these uniform bounds.

For $0<r<1$, $r\log r<0$. The printed upper bound is negative whenever $\Omega \ne 0$, so it cannot bound a nonnegative difference. The correct small-distance modulus is $r\log (1/r)$, and the radius in the hint must also be positive.

Write $h=|x_1-x_2|\le e^{-1}$, $x_3=(x_1+x_2)/2$, and $M=h\log (1/h)$. Then $h\le M\le e^{-1}$. Denote the planar vorticity velocity kernel by $K$. It satisfies $|K(z)|\le C/|z|$ and $|DK(z)|\le C/|z{|}^2$. Split the integral at radii $M$ and $1$ about $x_3$.

On $|y-x_3|\ge 1$, every point of the segment joining $x_1$ and $x_2$ remains at least $|y-x_3|-h/2\ge (1-h/2)|y-x_3|$ from $y$. The mean value theorem bounds the kernel difference by $Ch/|y-x_3{|}^2$, giving a contribution at most $Ch\Vert \Omega {\Vert }_1$.

On $M\le |y-x_3|\le 1$, the same estimate holds because $h/2\le M/2$. Polar integration givesHere $\log (1/M)\le \log (1/h)$ because $M\ge h$.

On $|y-x_3|<M$, use the sum of the two kernel magnitudes instead of their derivatives. This disk lies inside a disk of radius $M+h/2\le 3M/2$ about either $x_i$. Hence its contribution is bounded by $CM\Vert \Omega {\Vert }_{\infty }$. Combining the three regions yieldsAt coincident points the difference is zero. At larger distances use the boundedness from part (d). A convenient global log-Lipschitz modulus isgiving $|U[\Omega ](x)-U[\Omega ](y)|\le C(\Vert \Omega {\Vert }_1+\Vert \Omega {\Vert }_{\infty })\mu (|x-y|)$. This is positive, continuous and nondecreasing. It also proves continuity of $U[\Omega ]$.

Literally, the printed hypothesis with positive $C$ is impossible for pairs at distance less than one, even when $U$ is constant. Use the corrected log-Lipschitz modulus $\mu$ from part (e). Local existence for the continuous finite-dimensional vector field follows from the Peano existence theorem. Boundedness gives $|X(t)-X(s)|\le \Vert U{\Vert }_{\infty }|t-s|$. Thus a trajectory cannot escape to infinity in finite time; at a finite endpoint it has a limit, and local existence there extends it. Solutions exist globally.

For uniqueness, let $X,Y$ have the same initial point and set $\delta (t)=|X(t)-Y(t)|$. This absolutely continuous function satisfies, almost everywhere,This is the Osgood uniqueness criterion, rather than the usual linear Gronwall inequality, because the velocity need not be Lipschitz continuous. To see the argument directly, put $w_{\epsilon }=\delta +\epsilon$. As long as $w_{\epsilon }\le 1$, monotonicity of $\mu$ givesFor $z_{\epsilon }=1-\log w_{\epsilon }$, this becomes $z_{\epsilon }^{\prime }\ge -C{z}_{\epsilon }$. Integration givesFor any fixed finite interval, this upper bound tends uniformly to zero as $\epsilon \downarrow 0$. Taking $\epsilon$ small prevents a first exit through $w_{\epsilon }=1$, so the bound remains valid throughout that interval. Thus $\delta =0$. Backward time has the same proof. Every initial point has exactly one global trajectory.

The decisive fact is ${\int }_0^{1}dr/\mu (r)=\infty$. For different nearby initial points, the same comparison givesuntil the separation reaches one. This quantitative continuous dependence makes the flow a homeomorphism, although it need not be a smooth diffeomorphism.

For smooth vorticity, parts (b)--(c) and the transport equation give $\omega (t,S_{0,t}(y))={\omega }_{\mathrm{in}}(y)$ and measure preservation. The change of variables formula therefore conserves $\Vert \omega (t){\Vert }_1$ and $\Vert \omega (t){\Vert }_{\infty }$. Parts (d)--(e) give a uniformly bounded velocity and a uniform log-Lipschitz modulus; applying the time-dependent version of part (f) constructs unique global characteristic curves.

For a rough Yudovich characteristic flow, state the needed temporal hypothesis explicitly:The usual weak Euler solution class provides these conditions and the initial time trace. On $[0,T]$, put $A(t)=\Vert \omega (t){\Vert }_1+\Vert \omega (t){\Vert }_{\infty }$. The velocity bounds are $|u(t,x)|\le CA(t)$ and $|u(t,x)-u(t,y)|\le CA(t)\mu (|x-y|)$, with $A$ integrable on this interval.

For completeness, mollify $u$ spatially. The resulting smooth velocities have trajectories and obey the same speed and modulus bounds, up to one common constant. Their speeds have the integrable majorant $CA(t)$, giving uniform boundedness and equicontinuity of trajectories. Arzelà-Ascoli theorem supplies a uniformly convergent subsequence. The modulus estimate controls $u(t,X_{\epsilon })-u(t,X)$, while the spatial mollification error tends to zero with an integrable majorant. Thus the limit solvesReplace $Ct$ by $C{\int }_0^{t}A(s)ds$ in the Osgood uniqueness criterion proof. It gives uniqueness and prevents finite-time escape. The characteristics are globally uniquely defined for each given Euler solution in the usual Yudovich class. This constructs the flow for a fixed solution; it is not by itself the entire uniqueness proof for the nonlinear Euler equation.

Merely saying that both spatial norms are finite separately at each time does not state the local temporal bound or measurability used in this argument. The printed formulation must be interpreted in the usual solution class, or supplemented with these temporal conditions. Also, an initial datum here is a function of $x\in {\mathbb{R}}^2$; the extra time variable in the displayed initial-data space is extraneous.