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Here both the divergence and vorticity vanish:The flow is incompressible and irrotational. With $\mathbf{u}=\nabla \varphi$ and $(u_x,u_y)=({\psi }_y,-{\psi }_x)$,The streamlines are the level sets $x^2-y^2=\text{constant}$. To determine the arrows, put $r=x+y$ and $s=x-y$. A particle satisfies $\dot{r}=-2r$, $\dot{s}=2s$. Thus the separatrix $y=x$ approaches the stagnation point at the origin, whereas $y=-x$ leaves it; the other streamlines are hyperbolas following those asymptotes. The origin itself is a stationary particle.

Write $\mathbf{g}=-g{\mathbf{e}}_z$. The vertical component of the balance gives hydrostatic pressure,Its horizontal gradient is ${\nabla }_{h}p=\rho g{\nabla }_{h}h$. The horizontal balance is consequentlyThus $\mathbf{u}\cdot {\nabla }_{h}h=0$: the velocity is tangent to every level set of $h$, so those contours are streamlines of the geostrophic flow.

Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|{\nabla }_{h}h|$, while $|\mathbf{u}|=(g/f)|{\nabla }_{h}h|$. The volumetric flow rate is thereforeFor dimensional analysis, $[g]=L{T}^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3{T}^{-1}$, exactly that of a volume flux.

Along a trajectory, the chain rule givesThus $\boxed{H=x^2+y^2-y^4}$ is a first integral, and every trajectory lies on one of its level sets.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Frechet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

A level set at a regular value is a smooth submanifold. By the regular level set theorem, its codimension is the dimension of the target. For determinant-one real symmetric matrices, $D(\det {)}_B(C)=\mathrm{tr}(B^{-1}C)$ is nonzero because $C=B$ gives $n$, so the constraint has codimension one.