Let $\phi \in C^1(\mathbb{R})$ and periodic. We consider the problem $$u_t=u_{xx}, \ x\in \mathbb{R}, 0<t<\infty$$ with initial data $\phi$. I have computed a formula for a solution $u$. I want to show that there is no other solution ($C^1(\mathbb{R})$ and periodic).
For that we have to use the energy method, or not?
I have done the following:
Let $u_1, u_2$ be two solutions of the problem. Then $w=u_1-u_2$ solves the problem $$w_t=w_{xx}, \ x\in \mathbb{R}, 0<t<\infty\\ w(-L, t)=w(L, t) \\ w_x(-L, t)=w_x(L, t)$$
Then we have the following: $$w_t=w_{xx}\Rightarrow ww_t=ww_{xx} \Rightarrow \int_{-L}^Lww_t\, dx=\int_{-L}^Lww_{xx}\, dx \\ \Rightarrow \int_{-L}^L\frac{1}{2}\frac{\partial}{\partial{t}}\left (w^2\right )\, dx=ww_x\mid_{x=-L}^L-\int_{-L}^L\left (w_x^2\right )\, dx \\ \Rightarrow \frac{d}{dt}\left (\frac{1}{2}\int_{-L}^Lw^2\, dx\right )=w(L, t)w_x(L, t)-w(-L, t)w_x(-L, t)-\int_{-L}^Lw_x^2\, dx \\ \Rightarrow \frac{d}{dt}\left (\frac{1}{2}\int_{-L}^Lw^2\, dx\right )=-\int_{-L}^Lw_x^2\, dx\leq 0$$
Let $E(t)=\frac{1}{2}\int_{-L}^Lw^2\, dx$ then $E'(t)\leq 0$, i.e. the energy $E(t)$ is decreasing, so $E(t)\leq E(0)$.
Is everything correct so far? How could we continue? We need that $E(0)=0$, or not? But how do we get that?