Is Miura transformation invertible?

Question

It is known that the KdV equation \begin{align} \frac{\partial u}{\partial t}+b u \frac{\partial u}{\partial x}-\frac{1}{2}\frac{\partial^3 u}{\partial x^3}=0\,,\quad b\in\mathbb{R}\,, \end{align} under the Miura transformation \begin{align} v=\frac{3}{4b}\left(\frac{1}{4}u^2+\frac{\partial u}{\partial x}\right)\,, \end{align} maps to equation \begin{align} \frac{3}{4b}\left(\frac{1}{2}u+\frac{\partial}{\partial x}\right)\left(\frac{\partial u}{\partial t}+\frac{3}{16}u^2\frac{\partial u}{\partial x}-\frac{1}{2}\frac{\partial^3u}{\partial x^3}\right)=0\,. \end{align} A solution to this system is that the MKdV equation, namely \begin{align} \frac{\partial u}{\partial t}+\frac{3}{16}u^2\frac{\partial u}{\partial x}-\frac{1}{2}\frac{\partial^3u}{\partial x^3}=0\,, \end{align} must be fulfilled. Is Miura transformation $v=\frac{3}{4b}\left(\frac{1}{4}u^2+\frac{\partial u}{\partial x}\right)$ invertible, i.e., can we obtain KdV from MKdV?

Answer 1

Not a complete answer - just some thoughts.

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The Miura transformation is the mapping $$u\mapsto \frac{3}{4b}\left(\frac{u^2}{4}+\frac{\partial u}{\partial x}\right)=M[u]$$ To ask if the Miura transformation is invertible is to ask whether there is at most one solution $u$ to the equation $$M[u]=v$$ This presents us with the "PDE" (really an ODE in disguise) $$u^2/4+\partial_xu=\frac{4b}{3}v$$

Write this as $$u'(x)=\frac{4b}{3}v(x)-\frac{{u(x)}^2}{4}\equiv F(x,u(x))$$ With of course $$F(x,y):= \frac{4b}{3}v(x)-\frac{y^2}{4}$$ The question is, are solutions of this ODE necessarily unique given an appropriate initial condition $u(0)=u_0$?

The Picard–Lindelöf theorem tells us that a solution to this equation exists and is unique in some open interval provided that $F$ is continuous in its first argument and Lipschitz continuous in its second argument.

$F$ is continuous in its first argument as long as $v$ is continuous, which I think we are safe to just assume. However, $F$ is not Lipschitz continuous in its second argument , at least not on all of $\mathbb R$, and therefore, uniqueness of solutions cannot be guaranteed, again, at least not on all of $\mathbb R$.

With that being said, I haven't yet found a counterexample. But you could probably construct one if you tried hard enough.