Zero vorticity for ideal flow

The equation with $\nu=0$ can be written $$\partial_t \omega(t,x) = L_x\omega(t,x)\tag{0}$$ where $L_x$ is a first order differential operator in the variable $x$.

Heuristically, if $v$ in $L_x$ does not depend on $t$, the formal solution is $$\omega(t,x) =e^{tL_x}\omega(0,x). \tag{1}\:.$$

This solution is the correct and unique one provided one fixes a suitable space of solutions where to interpret the problem.

(For instance assuming that the solution is jointly smooth in both variables and analytic in $t$ for every given $x$, and that $v$ is smooth in $x$, (0) yields $$\frac{\partial^n \omega}{\partial t^n}= L_x^n \omega(t,x).$$ As a consequence of $t$-analyticity $$\omega(t,x)= \sum_{n=0}^{+\infty} \frac{t^n}{n!}\frac{\partial^n \omega}{\partial t^n}|_{t=0}= \sum_{n=0}^{+\infty} \frac{t^n}{n!} L_x^n\omega(0,x) =: e^{tL_x}\omega(0,x)\:. $$ However analyticity is an incredibly strong requirement and the same result can be found in some functional spaces using the semi-group theory.)

If the solution is (1), it is obvious true that $\omega(0,x)=0$ everywhere in $x$ then the solution is the trivial one.

If $v$ depends on time, a formal solution can be given in terms of $T$ ordered exponential. But here details on domains are crucial: $$\omega(t,x) = T[e^{\int_0^{t}L_{s,x} ds}]\omega(0,x).$$ At least formally we obtain the same result as before.

The fact that $\omega = \nabla \wedge v$ is not a problem because this identity can be imposed at the end of computations.

I observe that if the fluid is incompressible ($\nabla \cdot v =0$) and also using $\omega= \nabla \wedge v$ so that $\nabla \cdot \omega =0$, the form of the operator $L_{t,x}$ simplifies considerably: $$L_{t,x} \omega = \omega \cdot \nabla_x v(t,x) -v(t,x) \cdot \nabla \omega = \nabla \wedge (v(t,x)\wedge \omega).$$