[proofplan] We differentiate the energy along the smooth solution and identify the first variation of the gradient term by [integration by parts](/theorems/210). The boundary hypothesis removes the boundary contribution, leaving the bulk factor $-\Delta u+u^3-u$. The Allen-Cahn equation rewrites this factor as $-\partial_t u$, so the energy derivative is the negative $L^2(\Omega)$ norm squared of $\partial_t u$. [/proofplan] [step:Differentiate the potential and gradient parts of the energy] Define the potential function $W:\mathbb{R}\to\mathbb{R}$ by \begin{align*} W(s)=\frac{1}{4}(s^2-1)^2. \end{align*} Then $W$ is smooth and \begin{align*} W'(s)=s^3-s. \end{align*} Since $u\in C^1([0,T];C^2(\overline{\Omega}))$ and $\Omega$ is bounded, differentiation under the integral sign gives, for each $t\in(0,T)$, \begin{align*} \frac{d}{dt}E[u(t)]=\int_\Omega \left(\nabla u(t,x)\cdot \nabla \partial_t u(t,x)+(u(t,x)^3-u(t,x))\partial_t u(t,x)\right)\,d\mathcal{L}^n(x). \end{align*} [guided] We first isolate the two pieces of the energy. The potential function is the smooth map $W:\mathbb{R}\to\mathbb{R}$ defined by \begin{align*} W(s)=\frac{1}{4}(s^2-1)^2. \end{align*} Differentiating this scalar function gives \begin{align*} W'(s)=\frac{1}{4}\cdot 2(s^2-1)\cdot 2s=s^3-s. \end{align*} Now fix $t\in(0,T)$. Because $u\in C^1([0,T];C^2(\overline{\Omega}))$ and $\Omega$ is bounded, the functions appearing in the energy and their time derivatives are continuous on the compact set $\overline{\Omega}$. Thus the usual differentiation-under-the-integral rule applies to the [Lebesgue integral](/page/Lebesgue%20Integral) over $\Omega$. For the gradient part, the chain rule gives \begin{align*} \frac{\partial}{\partial t}\left(\frac{1}{2}|\nabla u(t,x)|^2\right)=\nabla u(t,x)\cdot \nabla \partial_t u(t,x). \end{align*} For the potential part, the chain rule and the computation of $W'$ give \begin{align*} \frac{\partial}{\partial t}W(u(t,x))=(u(t,x)^3-u(t,x))\partial_t u(t,x). \end{align*} Combining these two identities inside the integral yields \begin{align*} \frac{d}{dt}E[u(t)]=\int_\Omega \left(\nabla u(t,x)\cdot \nabla \partial_t u(t,x)+(u(t,x)^3-u(t,x))\partial_t u(t,x)\right)\,d\mathcal{L}^n(x). \end{align*} [/guided] [/step] [step:Integrate the gradient term by parts and use the boundary hypothesis] For the fixed time $t\in(0,T)$, [integration by parts](/theorems/2098) on the smooth bounded domain $\Omega$ gives \begin{align*} \int_\Omega \nabla u(t,x)\cdot \nabla \partial_t u(t,x)\,d\mathcal{L}^n(x)=\int_{\partial\Omega}\partial_\nu u(t,x)\partial_t u(t,x)\,d\mathcal{H}^{n-1}(x)-\int_\Omega \Delta u(t,x)\partial_t u(t,x)\,d\mathcal{L}^n(x). \end{align*} The boundary integral is zero by hypothesis. Therefore \begin{align*} \frac{d}{dt}E[u(t)]=\int_\Omega \left(-\Delta u(t,x)+u(t,x)^3-u(t,x)\right)\partial_t u(t,x)\,d\mathcal{L}^n(x). \end{align*} [/step] [step:Substitute the Allen-Cahn equation to obtain dissipation] The Allen-Cahn equation says \begin{align*} \partial_t u(t,x)-\Delta u(t,x)=u(t,x)-u(t,x)^3. \end{align*} Rearranging pointwise on $(0,T)\times\Omega$ gives \begin{align*} -\Delta u(t,x)+u(t,x)^3-u(t,x)=-\partial_t u(t,x). \end{align*} Substituting this identity into the previous formula yields \begin{align*} \frac{d}{dt}E[u(t)]=-\int_\Omega |\partial_t u(t,x)|^2\,d\mathcal{L}^n(x). \end{align*} Since the integrand $|\partial_t u(t,x)|^2$ is nonnegative for every $x\in\Omega$, the integral is nonnegative, and hence \begin{align*} \frac{d}{dt}E[u(t)]\leq 0. \end{align*} This proves the asserted energy dissipation identity. [/step]