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This question concerns the Korteweg-de Vries equation.

It is known that the transform $F=f^2+f_x$ transforms $$F_t-6FF_x+F_{xxx}=0$$ into $$f_t-6f^2f_x+f_{xxx}=0$$ where $F=F(x,t), f=f(x,t)$

However, I have read that the converse does not work. In other words, given $F$, it is not necessarily true that $f$ implicitly given by $F=f^2+f_x$ is a solution to the second differential equation. Is there an example to illustrate this?

Thank you.

Anyone?

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1 Answer 1

Consider the time independent case.

Then $F \equiv C$ is a solution to the equation, since $F_x = F_{xxx} = 0$.

A time-independent transformation gives $$ C = f^2 + f_x \implies f_x = C - f^2 \tag{1}$$ So that $$ f_{xx} = -2 f f_x = -2C f + 2f^3 $$ and $$ f_{xxx} = -2C f_x + 6f^2 f_x = -2C^2 + 8C f^2 - 6f^4 $$

On the other hand we have $$ 6f^2 f_x = 6f^2 C - 6f^4 $$ so $$ f_t - 6f^2 f_x + f_{xxx} = -2C f_x = -2C^2 + 2C f^2 $$ which given the ODE (1) can only vanish identically if either

  1. $C = 0$, or
  2. $f^2 \equiv C$.

So in the case $C < 0$, since we cannot have it being a perfect square ($f$ being real valued) we conclude that $f$ cannot solve the given equation.


To get a real concrete function: let $C = -1$, by direct integration we see that $f = \cot (x)$ satisfies $f_x = -1 - f^2$ and so corresponds to the solution $F \equiv -1$ of the KdV equation, but $$ f_t - 6 f^2 f_x + f_{xxx} = -2 \csc(x) \neq 0 $$

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