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The divergence is ${\partial }_x{u}_x+{\partial }_y{u}_y=1-1=0$, while the scalar vorticity is ${\partial }_x{u}_y-{\partial }_y{u}_x=0$. The flow is therefore incompressible flow and irrotational flow. Integrating the streamfunction equations givesThe time dependence must be retained: no time-independent function $\psi (x,y)$ represents this unsteady field for every $t$. At $t=0$ the streamlines are $xy=\mathrm{constant}$, with the coordinate axes as separatrices. Their arrows point towards increasing $|x|$ and decreasing $|y|$, as shown below.

A pathline solves the material ODEs, givingFor $x_0\ne 0$, elimination of time yieldsIf $x_0=0$, the pathline stays on the $y$ axis and the parametric expression remains valid; it cannot in general be written as a single-valued $y=f(x)$ along that vertical trajectory. For $x_0=y_0=1$, the path is $y=\log x-1+2/x$: it initially descends, reaches its minimum $(2,\log 2)$, then rises. Unlike a frozen streamline, it records the time-varying flow.

The steady Euler equations for an inviscid fluid with no body force are $(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla p/\rho$. Use the convective acceleration identityto obtainFor the question's streamfunction convention, $\mathbf{u}=({\psi }_y,-{\psi }_x,0)$, so $\mathbf{u}\times \mathbf{\omega }=-\omega \nabla \psi$. Constant $\omega$ therefore gives the Bernoulli function for planar constant-vorticity flowon the connected fluid region.

For circular streamlines, write $\mathbf{u}=w(r)\hat{\mathbf{\varphi }}$. The cylindrical curl gives $(rw{)}^{\prime }/r=\omega$, so $w=\omega r/2+C_1/r$. The outer wall condition gives $C_1=-2\omega a^2$. Choose the positive azimuthal orientation so that $w(a)=V$, as implicit in the requested sign for $\omega$. ThenReversing the circulation reverses $\omega$ but leaves the speed and pressure difference unchanged. For this uniform-vorticity circular flow in an annulus, one convenient streamfunction is $\psi =V(r^2/2-4a^2\log (r/a))/(3a)+C_2$.

The radial component of the Euler equations for an inviscid fluid is $p_r=\rho w^2/r$. Thus pressure increases outwards, andTherefore

For the linear planar flow with strain and rotation, ${\partial }_{x}u+{\partial }_{y}v=ϵ-ϵ=0$, proving incompressibility. The Cartesian streamfunction convention gives $u={\psi }_y$, $v=-{\psi }_x$, and direct integration givesup to an irrelevant additive constant. Along a streamline, $\mathbf{u}\cdot \nabla \psi ={\psi }_y{\psi }_x-{\psi }_x{\psi }_y=0$. Sincethe stated quadratic expression is constant along each streamline. The three cases below distinguish strain, rotation and their degenerate shear limit.

Here $u=v=x-y$. The streamlines are $x-y=\text{constant}$, straight lines parallel to $(1,1)$. The entire line $x=y$ is stationary. On the $x-y>0$ side the motion is northeast, and on the $x-y<0$ side it is southwest. In orthonormal coordinates $p=(x+y)/\sqrt{2}$, $q=(x-y)/\sqrt{2}$, the velocities are $\dot{p}=2q$, $\dot{q}=0$: this is simple shear flow. The right panel shows both moving families and the stationary line.

Here $u=-y$, $v=x$. The streamlines are $x^2+y^2=\text{constant}$. This is rigid rotation counterclockwise with angular velocity one; the origin is a stagnation point and all other trajectories are circles. The middle panel shows their orientation.

Here $u=x$, $v=-y$. The streamlines are $xy=\text{constant}$, rectangular hyperbolas, together with the coordinate axes as separatrices. Particles satisfy $x(t)=x(0)e^t$, $y(t)=y(0)e^{-t}$: the flow stretches along the $x$ axis and compresses along the $y$ axis, with a saddle stagnation point at the origin. The left panel shows the flow directions.

Choose $U$ and $Q$ as the speed and source strength in the same kinematic velocity convention. With $r=(x^2+y^2{)}^{1/2}$ and $\vartheta =\arg (x+iy)$, superposition of uniform potential flow and a two-dimensional point source giveswhere $(u,v)=\nabla \Phi =({\psi }_y,-{\psi }_x)$. The angular branch cut records the volume flux from the source; the logarithm's arbitrary reference length only changes an additive constant in the velocity potential.

The stagnation point lies on the negative axis, where $u=U+Q/(2\pi x)=0$. Thus the limiting upstream reach of source fluid isThe stagnation point is approached asymptotically, rather than reached in finite time by a parcel on the upstream axis. The dividing streamlines through it have $\psi =\pm Q/2$ when the angle is taken in $(-\pi ,\pi )$. Far downstream $\vartheta \to 0$, so these streamlines approach $y=\pm Q/(2U)$. Consequently the source fluid occupies the Rankine half-body between them, with downstream width $2b=Q/U$. In the upper half-plane its boundary can also be parametrized by $y=b(1-\vartheta /\pi )$, $x=y\cot \vartheta$, $0<\vartheta <\pi$.

For arrival at the downstream axis, the direct streamline $y=0$, $x>0$ has speed $U+Q/(2\pi x)$. Its transit time from the ideal point source isThe first source fluid at a downstream cross-section lies on this symmetry axis: for positive $x$, the horizontal source contribution $Qx/[2\pi (x^2+y^2)]$ is maximal at $y=0$, and off-axis paths cannot advance through positive $x$ more quickly. A far-away parcel passing the inlet line at the same release time has transit time $x/U$. The time advance is thereforeThis is an advance, not the source parcel's absolute transit time. If a source parcel is released at time $t_0$ while the distant comparison parcel crosses $x=0$ at time zero, the advance relative to that parcel is $\tau -t_0$; the printed formula uses simultaneous releases. If $U,Q$ instead denote Darcy velocity and bulk-area source flux in a medium of uniform porosity $\varphi$, their ratio still gives the same $b$, but both parcel times acquire a factor $\varphi$. No porosity is specified here, so the displayed time formula fixes the kinematic convention.

Parametrize a streamline by $\mathbf{r}(s)$ with $d\mathbf{r}/ds=\mathbf{u}(\mathbf{r}(s))$. The chain rule givesTherefore $H$ is constant along every streamline. A different regular parametrization merely multiplies this derivative by a scalar and gives the same conclusion. For time-dependent flow, the claim concerns streamlines at a fixed time, rather than Lagrangian trajectories.

In steady flow, the Euler equations for an inviscid fluid and part (a) giveTaking the dot product with $\mathbf{u}$ yields $\mathbf{u}\cdot \nabla H=0$, because a cross product is perpendicular to its first factor. Part (b) therefore gives Bernoulli's quantity along streamlines:For rotational flow these constants can differ between streamlines, since $\mathbf{u}\times \mathbf{\omega }$ need not vanish and therefore $\nabla H$ need not be zero. Special rotational flows can still have one global constant when that cross product vanishes; the general conclusion is only constancy along individual streamlines.

For $\mathbf{u}=\alpha y{\mathbf{e}}_x-\beta x{\mathbf{e}}_y$, the velocity divergence is zero and $\mathbf{u}\cdot \nabla \mathbf{u}=-\alpha \beta (x{\mathbf{e}}_x+y{\mathbf{e}}_y)$. Substituting into the steady Euler equations yieldsIntegration givesThe steady enthalpy equation also requires $\mathbf{u}\cdot \nabla Q=0$. Its $xy$ coefficient is $\alpha (3{\Omega }^2+\alpha \beta -{\beta }^2)$, so, since $\alpha >0$,For a bounded patch with positive mass density and a vacuum fluid free boundary, choose $Q_0>0$ and require both quadratic coefficients to confine the fluid. With $\Omega >0$ and $\alpha ,\beta >0$, this meansThe lower bound is imposed by $\alpha >0$; the upper bound by bounded positive $Q$. At $\beta =2\Omega$, $Q$ is constant and has no nontrivial finite vacuum boundary; $\beta \le \sqrt{3}\Omega$ violates or degenerates the stipulated elliptical flow.

For a polytropic elliptical patch in a shearing sheet, the physical fluid free boundary is $p=\rho =Q=0$. It is the ellipsewhere $a_x^2=2Q_0/[\beta (2\Omega -\beta )]$ and $a_y^2=2Q_0/[\alpha (2\Omega -\beta )]$. In the interior $\rho =[Q/((n+1)K){]}^n$. The boundary is a material streamline because $\mathbf{u}\cdot \nabla (\beta x^2+\alpha y^2)=0$. Since $\alpha <\beta$, it is elongated in the azimuthal $y$ direction.

This is a pressure-supported fluid patch in the supplied tidal model; no additional planetary self-gravity potential was assumed.