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 $f(u)=uu^{2r+1}$, for $r \in \mathbb{N}$. How to find the conserved quantitie of this equation?
1 Answer
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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 $$ \partial_t \int \frac12 u_t^2 + \frac12 (1 + \frac12 \int u_x^2~dx) u_x^2 + \frac12 u^2  \frac1{2r+2} u^{2r+2} ~dx = 0$$

$\begingroup$ Why multiplying by $u_x$ and integrante we get only $\int u_{tt}u_x \; dx=0$? And the rest of the terms? $\endgroup$ Commented 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 $f$. $\endgroup$ Commented Jun 10, 2020 at 17:17
