3
$\begingroup$

Consider the Kirchhoff equation, given by $$u_{tt}-\left(1+\int_{\mathbb{R}} u_x^2\;dx\right)u_{xx}+f(u)=0, (x,t) \in \mathbb{R}\times \mathbb{R}_+$$ where $f(u)=u-u^{2r+1}$, for $r \in \mathbb{N}$. How to find the conserved quantitie of this equation?

$\endgroup$

1 Answer 1

2
$\begingroup$

Some of them:

  • Multiply by $u_x$ and integrate, you get $$ \int u_{tt} u_x ~dx = 0 $$ so $$ \partial_t \int u_{t} u_x ~dx - \int u_t u_{tx} ~dx = 0 $$ the second term integrates to zero.
  • Multiply by $u_t$ and integrate by parts you get $$ \int u_{tt} u_t + (1 + \int u_x^2 ~dx) u_{xt} u_x + uu_t - u^{2r+1} u_t ~dx = 0 $$ This you can rewrite as $$ \partial_t \int \frac12 u_t^2 + \frac12 (1 + \frac12 \int u_x^2~dx) u_x^2 + \frac12 u^2 - \frac1{2r+2} u^{2r+2} ~dx = 0$$
$\endgroup$
3
  • $\begingroup$ Why multiplying by $u_x$ and integrante we get only $\int u_{tt}u_x \; dx=0$? And the rest of the terms? $\endgroup$
    – Guilherme
    Jun 10, 2020 at 13:23
  • 1
    $\begingroup$ Integration of a total derivative yields 0. $$\int u_{xx} u_x = \int \partial_x (u_x)^2 = 0$$ and similarly the $f(u)$ terms. $$ \int f(u) u_x = \int \partial_x F(u) = 0$$ where $F$ is the primitive to $f$. $\endgroup$ Jun 10, 2020 at 17:17
  • $\begingroup$ Very good. Thank you! $\endgroup$
    – Guilherme
    Jun 11, 2020 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.