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The boundary of a steady Kida vortex must be material. Put $E=x^2/b^2+y^2/a^2$. Requiring $\mathbf{u}\cdot \nabla E=0$ on the ellipse givesThe relative vorticity is ${\partial }_x{u}_y-{\partial }_y{u}_x=-\beta -\alpha =-S+{\zeta }_0$, so the supplied vorticity relation impliesCombining these independent conditions yieldsFor the usual positive orbital shear both are positive, the flow circulates clockwise and its streamlines have the same aspect ratio as the patch. The velocity coefficients in this question are unrelated to the alpha viscosity parameter in Question 1.

Write $B=2\Omega -S$, $A(t)=c_s^2-2\pi G{\Sigma }_0/k(t)$, and $w=i{k}_x{\tilde{v}}_y-i{k}_y{\tilde{v}}_x$. Differentiation, including ${\dot{k}}_x=S{k}_y$, givesThe two pressure/self-gravity cross terms cancel. Continuity gives ${\dot{\tilde{\Sigma }}}^{\prime }=-i{\Sigma }_0(k_x{\tilde{v}}_x+k_y{\tilde{v}}_y)$, so $\dot{w}=B{\dot{\tilde{\Sigma }}}^{\prime }/{\Sigma }_0$. ConsequentlyThe conserved amplitude is linearized vortensity: the perturbation of the absolute vertical vorticity divided by surface density,A two-dimensional inviscid barotropic fluid materially conserves this potential vorticity. Both pressure enthalpy and gravitational forces are gradients, so they create no baroclinic vorticity source. Since $q_0$ is uniform, its linear perturbation obeys $D_0{q}^{\prime }=0$, explaining why a shearing wave has constant vortensity amplitude.

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.

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. Arzela-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.