I have the following initial value problem: $$ \dfrac{\partial}{\partial t} u(x,t) - \frac{1}{2} \dfrac{\partial^2}{\partial x^2}u(x,t) = (u(x,t))^2$$ $$ u(x,0) = u_0(x)\in C^2(S^1)$$ and want to show that if $u,v \in C^2(S^1 \times [0, T])$ are two solutions, then $u = v$.
As a hint it said I should consider $\eta(t)=\int_{0}^{2\pi}(u(x,t)-v(x,t))^2 dx$
and we proved earlier that if $\dfrac{d}{dt}\eta(t) \le \eta(t)\phi(t)$ then $$\eta(t) \le \eta(0)e^{\int_{0}^{2\pi}\phi(s)ds}$$ I don't have much experience with nonlinear differential equations and I'm not sure how to proceed. Can you help me? Thanks in advance!
Consider $w=u-v$. Then $w$ satisfies $w_t-1/2w_x=(u+v)w$. Therefore we have $$ \partial_t w^2 =2ww_t=2w(w_x+(u+v)w)=(w^2)_x +2(u+v)w^2 . $$ Notice now that $$ \int_0^{2\pi} (w^2)_x dx=0 $$ so that, with your notation, $$ \eta'(t)=2\int_0^{2\pi}(u+v)w^2 dx \leq \phi(t) \eta(t), $$ where $\phi(t)=2\sup_{x\in [0,2\pi )} |u+v|$. Now apply Gronwall.