Consider the PDE $$ u_t=u_{xx}+\cos u-1+\mu.\qquad (1) $$ with $u\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}, (t,x)\mapsto u(t,x)$.
Does this PDE have a variational structure, i.e. can it be written as $$ u_t=-\delta \mathcal{E}(u) $$ for some integral functional $\mathcal{E}(u)$ and its first variation $\delta\mathcal{E}(u):=\lim_{\varepsilon\to 0}\frac{E(u+\varepsilon v)-E(u)}{\varepsilon}$? And which solution space is a good idea?
- Maybe solutions $u(t,\cdot)\in L^2(\mathbb{R})$ what was my first idea which I tried with the functional $$ \mathcal{E}(u):=\int_{\mathbb{R}}\frac{1}{2}(u_x)^2+F(u)\, dx,\qquad\text{ where }F'(u)=-\cos u+1-\mu. $$ This gave me $$ \langle \delta\mathcal{E}(u), v\rangle_{L^2}=\int_{\mathbb{R}}u_xv_x+F'(u)v\, dx. $$
Integrating by parts, I get $$ \int_{\mathbb{R}}u_xv_x+F'(u)v\, dx=[u_x(s)v(s)]_{-\infty}^{+\infty}-\int_{\mathbb{R}}(-u_{xx}+F'(u))v\, dx. $$ Now, I have to write this as some $L^2$-scalar product term but for this, I would need that the first summand on the right hand side vanishes, i.e. $$ u_x(+\infty)v(+\infty)-u_x(-\infty)v(-\infty)=0\qquad (2) $$ since then only the integral on the right hand side would remain and this is nothing else but $\langle -u_{xx}+F'(u),v\rangle_{L^2}$. Hence I would get $$ \delta\mathcal{E}(u)=-u_{xx}+F'(u) $$ and, consequently, $u_{t}=-\delta\mathcal{E}(u)$.
Unfortunately, I think (2) does not have to be true or to make sense in general for $u(t,\cdot),v(t,\cdot)\in L^2(\mathbb{R})$.
So, I think I have to use another space as $L^2(\mathbb{R})$.
Do you have any idea?