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Let $G:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be smooth vector field over $\mathbb{R}^3$. For which vector fields $F:\mathbb{R}^3\rightarrow\mathbb{R}^3$ does the PDE

$$\dfrac{\partial}{\partial t}\textbf{v} + \textbf{Curl}(F(\textbf{v})) = \textbf{G}$$ have solutions $\textbf{v}(x,t):\mathbb{R}^3\times (0,\infty) \rightarrow\mathbb{R}^3$ for all time $t>0$, given smooth initial condition $$\textbf{v}(x,0) = \textbf{i}(x)?$$

It's not hard to see F has to be smooth. Are there any other implications we can make? If I give you $F$ can you find $\textbf{v}$ every time? Are the solutions unique?

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    $\begingroup$ The way you ask this question is unusual. Analysts prefer to choose a PDE (or a system of PDEs), the lanscape, and then solve it if possible, for every data $(G,i)$. $\endgroup$
    – Denis Serre
    Commented Jan 24, 2023 at 7:27
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    $\begingroup$ Do you want to assume also that $\mathbf{i}(x)$ is smooth? I can imagine that if $F$ is close to the identity, you might want some control on some function space norm of $\mathbf{i}(x)$, but for more general $\mathbf{F}$, I can't imagine how you could get any control on the growth of $\mathbf{v}(x,t)$. $\endgroup$
    – Ben McKay
    Commented Jan 24, 2023 at 13:28
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    $\begingroup$ @BenMcKay yes the initial condition should be smooth. $\endgroup$
    – MrPie
    Commented Jan 24, 2023 at 16:14
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    $\begingroup$ @DenisSerre yes of course but I am not doing that sorry that it is unusual. My point here is to try to see the constraint $F$ has on this type of system and see if there is universal technique to solve for arbitrary fields $F$ and $G$. I would prefer that $F$ doesn't really matter, but I cannot make that conclusion myself. $\endgroup$
    – MrPie
    Commented Jan 24, 2023 at 16:16

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This is an example of a first order system. A necessary condition for well-posedness of initial value problems is hyperbolicity. But even if F is the identity, the simplest case, this system is not hyperbolic.

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    $\begingroup$ @MichealRenardy when $F$ is the identity I believe this problem has been solved. $\endgroup$
    – MrPie
    Commented Jan 24, 2023 at 16:18
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    $\begingroup$ Take the case where F is the identity and G=0. We have $\partial v/\partial t=curl \,v$, which implies $\partial^2 v/\partial t^2=curl\, curl\, v=-\Delta\, v+\nabla(div\, v)$. If div v is zero initially, this persists, so we have $\partial^2 v/\partial t^2=-\Delta\, v$, which is an elliptic equation, and therefore ill-posed. $\endgroup$
    – Michael Renardy
    Commented Jan 24, 2023 at 22:00

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