Consider the partial differential equation
$$u_{tt}-\omega\cdot u_{xx}=u^{2r+1}-u,\; (x,t) \in \mathbb{R}\times \mathbb{R}_+$$
where $w>0$ and $r \in \mathbb{N}$ are fixeds.
What is the method to follow to find a conserved quantity for this equation?
Best Answer
Multiplying the equation by $u_t$ and integrating by parts over $\mathbb{R}$ we have: \begin{eqnarray} \int_{\mathbb{R}}u_{tt}u_t \, dx &=& \frac{1}{2} \frac{d}{dt} \|u_t\|_{L^2(\mathbb{R})}^{2}\\ -\int_{\mathbb{R}}\omega u_{xx}u_t\, dx&=& \omega \int_{\mathbb{R}} u_x u_{xt} \, dx\\ &=& \frac{\omega}{2} \frac{d}{dt} \|u_x\|_{L^2(\mathbb{R})}^{2}\\ \int_{\mathbb{R}} u^{2r+1}u_t\,dx&=&\int_{\mathbb{R}} \frac{1}{2r+2}\frac{d}{dt} u^{2r+2}\,dx\\ &=&\frac{1}{2r+2}\frac{d}{dt}\int_{\mathbb{R}}u^{2r+2}\,dx\\ \int_{\mathbb{R}}uu_t\,dx&=&\frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}}u^2\,dx \end{eqnarray} Therefore, defining $$E(t)=\frac{1}{2}\|u_t\|_{L^2(\mathbb{R})}^{2}+\frac{\omega}{2}\|u_x\|_{L^2(\mathbb{R})}^{2}+\frac{1}{2}\|u\|_{L^2(\mathbb{R})}^{2}-\frac{1}{2r+2}\int_{\mathbb{R}}u^{2r+2}\,dx$$ it follows that $E'(t)=0$