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Suppose $U=\Omega \times (0,\infty),$ where $\Omega$ is a bounded domain in $\mathbb{R}^n$ and $u \in C_{2}^{1}(\overline{U})$ satisfies $$u_t=\Delta u - u^3~\textrm{in}~U,~u(x,t)=0~\textrm{on}~\partial \Omega \times (0,\infty).$$

Two questions:

(i).Construct an energy functional.

(ii).Using part (i) show that $u(x,t) \rightarrow 0$ as $t \rightarrow \infty.$

My approach: Following the method of Evan's (page 63), I was able to obtain the following:

Setting $e(t)=\int_{U} (u(x,t))^2~dx$ for $0 < T < \infty,$ $$\frac{de}{dt}= -2 \bigg\{ \int_\Omega |\nabla u(x,t)|^2~dx - \int_\Omega u(x,t)^4~dx \bigg\} \leq 0.$$

From which it follows that $e(t) \leq e(0)=0~(0 < T < \infty).$

Does this e(t) is the energy functional required by the problem ? If so is my answer is the correct explanation for part (i) ? How can I prove part (ii).

This seems to be a lengthy question/answer. But I appreciate any help in solving this.

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    $\begingroup$ Hint: try to obtain an upper bound on $\frac{de}{dt}$ in terms of $e$ itself. For this you need to compare $\int u^4$ with $e$, $\endgroup$ – Michał Miśkiewicz Dec 18 '16 at 11:40
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Note that $$ \left( \int_\Omega u(x,t)^2~dx \right)^2 \le |\Omega| \int_\Omega u(x,t)^4~dx. $$

Combining this with the energy inequality you obtained, we get \begin{align*} \frac{de}{dt} & = -2 \left( \int_\Omega |\nabla u(x,t)|^2~dx + \int_\Omega u(x,t)^4~dx \right) \le -c e^2, \end{align*} where $c = 2/|\Omega|$.

All we need to do is to apply some Gronwall-type inequality. Transforming $\frac{de}{dt} \le -ce^2$, \begin{align*} \frac{d}{dt} (e^{-1}) = -e^{-2} \frac{de}{dt} & \ge c, \\ e^{-1}(t) - e^{-1}(0) & \ge ct, \\ e(t) & \le \frac{e(0)}{1+ct \cdot e(0)}. \end{align*} The last inequality implies that $e(t) \to 0$ and in consequence $u(\cdot, t) \to 0$ in $L^2(\Omega)$ as $t \to \infty$.

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  • $\begingroup$ I don't understand the question. First, these both set are finite dimensional (of dimensions $n$ and $1$, respectively). Second, this has nothing to do with the problem. $\endgroup$ – Michał Miśkiewicz Dec 18 '16 at 16:52
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    $\begingroup$ Since $\Omega$ is a bounded domain, its measure is finite. Note that we're talking about $\Omega$, not $U = \Omega \times (0,\infty)$. And dimensions are irrelevant here. $\endgroup$ – Michał Miśkiewicz Dec 18 '16 at 19:45

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