Consider the Allen-Cahn equation $$ u_t=u_{xx}+f(u), x\in\mathbb{R}\qquad (*) $$ with $f(u)=u(1-u^2)$ and initial condition $u(0,x)=u_0(x)\in\text{BC}^0(\mathbb{R})$, the bounded uniformly continuous functions on the real line.
Now, it is said that $(*)$ is a gradient system in $L^2$ with respect to the energy $$ V(u):=\int_{\mathbb{R}}(\frac{1}{2}u_x^2+F(u))\, dx, $$ where $F(s)=\int_0^sf(s)$.
I would like to verify that.
My computations for the first variation $\delta V(u)$ give that $$ \langle \delta V(u),v\rangle_{L^2}=\int_{\mathbb{R}}u_xv_x+F'(u)v\, dx $$
Now, I think, I have to integrate by parts, i.e. $$ \int_{\mathbb{R}}u_xv_x\, dx= [u_xv]_{-\infty}^{+\infty}-\int_{\mathbb{R}}u_{xx}v\, dx. $$
Is it possible to use now $u_0(x)\in\text{BC}^0(\mathbb{R})$ to show that on the right hand side for the first summand we have $$ [u_xv]_{-\infty}^{+\infty}=0? $$
If yes, I get $$ \langle \delta V(u),v\rangle_{L^2}=\int_{\mathbb{R}}u_xv_x+F'(u)v\, dx=\int_{\mathbb{R}}(-u_{xx}+F'(u))v\, dx=\langle -u_{xx}+F'(u),v\rangle_{L^2}, $$ hence $\delta V(u)=-u_{xx}+F'(u)$.
Moreover, what is $F'(u)$? Is this $F'(u)=f(u)$ by fundamental theorem of calculus?
Please help me. :)
1) I think that in the definition of space $BC^0(\mathbb{R})$, superscript $0$ means that functions from $BC^0(\mathbb{R})$ have to vanish on infinity, that is $u(x) \to 0$ as $x \to \pm\infty$. (Or even have to have compact supports.) Moreover, you know that $v \in L^2(\mathbb{R})$, which implies that $v$ cannot tend to infinity as $x \to \infty$. Therefore, $[u_xv]_{-\infty}^{+\infty} = 0$, indeed.
2) Yes, $F'(u) = f(u)$ by fundamental theorem of calculus and the fact that $f(0) = 0$.