For $k>0$, consider the Camassa-Holm equation: $$ u_t-u_{txx}+2k u_x=-3uu_x+2u_xu_{xx}+uu_{xxx}, \quad (t,x)\in\mathbb{R}^2. $$ I've been trying to (formally) recover the second of its well-known conservation laws, that is to say, to prove that the following functional $$ F(u):=\int \big(u^3+uu_x^2+2k u^2\big)dx, $$ is conserved. In other words, considering smooth solutions decaying sufficiently fast at $\pm\infty$, proving that the time derivative of $F(u(t))$ is equivalently zero. It is very intriguing for me that I have tried by multiplying the equation by all sort of factors, like $u^2$, $u_x^2$,... and then integrating in $\mathbb{R}$, and performing several integration by parts, but it has been imposible for me to recover the conservation law. Does anyone know how to do it? It also intrigues me that the first two terms in the conservation law are cubic, but the last one is just quadratic, that makes me thinks that this is not just multiplying the equation (not at least once) and performing some integration by parts.


1 Answer 1


The following should work.

First, denote by $U(t,x) = \int_{-\infty}^x u(t,y) ~dy$.

If you write the equation as $$ u_t-u_{txx}+2k u_x=-3uu_x+u_xu_{xx}+(uu_{xx})_x$$ and take the primitive in $x$, you find $$ U_t - u_{tx} + 2k u + \frac32 u^2 - \frac12 (u_x)^2 - u u_{xx} = 0. $$ Now, multiply the entire equation by $u_t$ and integrate over $x$ over the entire $\mathbb{R}$.

  • The first two terms vanish since they are integration of a real function against its first derivative.
  • The final term you integrate by part in $x$.

This should yield you

$$ \int 2 k u u_t + \frac32 u^2 u_t + \frac12 (u_x)^2 u_t + u u_{tx} u_{x} ~dx = 0 $$

which gives

$$ \frac{d}{dt} \int k u^2 + \frac12 u^3 + \frac12 u (u_x)^2 ~dx = 0 $$


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