How can I prove that $\|u \|$ is bounded in this?

Let $\Omega \subseteq \mathbb{R}^{n}$ be a bounded open set with boundary $\partial \Omega$. Let $\phi \in L^{2}(\Omega) .$ Consider the initial boundary value problem for $u=u(x, t)$,

$$u_{t}=\Delta u-\left\{\int_{\Omega} u^{2} \mathrm{d} x\right\} u, \quad x \in \Omega, t>0$$ $$u=0 ;x \in \partial \Omega, t>0$$ $$u(x, 0)=\phi(x), \quad x \in \Omega$$ Prove that

$$\frac{1}{2} \int_{\Omega}|\nabla u|^{2} d x \leqslant \frac{1}{2} \int_{\Omega}|\nabla \varphi|^{2} d x+\frac{1}{4}\left(\int_{\Omega} \varphi^{2}(x) d x\right)^{2}<+\infty$$

I took the steps that the researcher mentioned in the following article Ball 1977 This is my attempt. Multiplying by $u_t$ and integrating over $\Omega$

$$\int_{\Omega} u_{t} u_{t} d x=\int_{\Omega} u_{t} \Delta u d x-\int_{\Omega}\left(u_{t} \cdot u \int_{\Omega} u^{2} d x\right) d x$$ $$\int_{\Omega} u_t^{2} d x=\int_{\Omega} u_t \Delta u d x-\int_{\Omega}\left(u_t u \int_{\Omega} u^{2} d x\right) d x$$ $$\left\|u_{t}\right\|_{L^{2}}^{2}=\int_{\Omega} u_{t} \Delta{u} dx- \int_{\Omega} u_t u\|u\|_{L^{2}}^{2} d x$$ $$\| u_{t}\left\|_{L^{2}}^{2}=\int_{\Omega} u_{t} \Delta u d x-\right\| u \|_{L_{2}^{2}}^{2} \cdot \int_{\Omega} \frac{\partial}{\partial t}\left(\frac{1}{2} u^{2}\right) d x$$ $$\|u_{t}\left\|_{L^{2}}^{2}=\int_{\Omega} u_t \Delta u d x-\frac{1}{2}\right\| u\left\|^{2} \cdot \frac{d}{d t}\right\| u \|^{2}$$ $$\left\|u_{t}\right\|^{2}=\int_{\Omega} u_t \Delta u d x-\frac{\partial}{\partial t}\left[\|u\|^{2}\right]^{2}$$ $$\|u_t \|^{2}=\int_{\partial \Omega} u_{t} \frac{\partial u}{\partial n} d s-\int_{\Omega} \nabla u \nabla u_{t} d x-\frac{\partial}{\partial t}\left[\|u\|^{2}\right]^{2}.$$ $$\|u_t \|^{2}=-\int_{\Omega} \nabla u \nabla u_{t} d x-\frac{\partial}{\partial t}\left[\|u\|^{2}\right]^{2}.$$ $$\|u_t \|^{2}=-\frac{1}{2} \frac{\partial}{\partial t}\| \nabla u\|^{2}-\frac{1}{4}\frac{\partial}{\partial t}\left[\|u\|^{2}\right]^{2}$$

Can you help me to continue the proof

  • $\begingroup$ Now integrate from $0$ to $t$, and throw away $\|u_t\|^2 \ge0$. $\endgroup$
    – daw
    Jan 15, 2020 at 8:10
  • $\begingroup$ Could you do that, please @daw $\endgroup$
    – alaloush
    Jan 15, 2020 at 8:39

1 Answer 1


Integrating from $0$ to $t$ gives $$ 0 \le \int_0^t \|u_t(s)\|^2 ds = -\frac12 \left(\|\nabla u(t)\|^2 - \|\nabla \phi\|^2\right) - \frac14 \left( \|u(t)\|^4 - \|\phi\|^4\right), $$ where all norms are $L^2(\Omega)$-type. Then we get $$ \frac12 \|\nabla u(t)\|^2 + \frac14 \|u(t)\|^4 \le \frac12 \|\nabla \phi\|^2+ \frac 14 \|\phi\|^4, $$ which implies the claim.

  • $\begingroup$ think you so much @daw $\endgroup$
    – alaloush
    Jan 15, 2020 at 13:12

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