The question is let $u(t, x, y)$ be a twice continuously differential solution of \begin{aligned} &u_t=\Delta u- u^3 \ \mbox{ in } \Omega \subset \mathbb{R}^2, \ t\geq 0\\ &u(0,x, y)=0 \ \mbox{ in } \Omega \\ &u(t, x, y)=0 \ \mbox{ in } \partial \Omega, \ t\geq 0 .\\ \end{aligned} Prove that $u(t, x, y) =0$ in $\Omega \times \left[0, T \right] $.
This looks like an appication of maxium principle of heat eqaution, but it is not homogeneous. Anyone can give a hint on how to aproach?Thanks!
The direct way is to apply the comparison principle to the parabolic operator $Lu=u_t-\Delta u+u^3.$ However, we can also derive it by the energy method.
In this answer, we use the following notation \begin{align*}x&=(x_1,x_2)\in\Omega,\\ \Omega_T&=\Omega\times(0,T],\\ u_t&=\partial_tu,\nabla u=(\partial_{x_1}u,\partial_{x_2}u),\\ \Delta u&=\sum_{i=1}^2\partial_{x_ix_i}u.\end{align*}
Suppose $u\in C^{2,1}(\Omega_T)\cap C^{1,0}(\bar\Omega_T)$ solves the heat equation above, and $2\|u\|_{L^\infty(\Omega_T)}^2\leq M.$ For any $\tau\in(0,T),$ multiplying by $u$ in the equation and integrating by parts, thanks to the initial and boundary conditions, we have\begin{align*}M\int_0^\tau\int_\Omega u^2dxdt&\geq\int_0^\tau\int_\Omega2u^4dxdt=\int_0^\tau\int_\Omega2uu_tdxdt-\int_0^\tau\int_\Omega2u\Delta udxdt\\ &=\int_\Omega u^2(x,\tau)dx+2\int_0^\tau\int_\Omega|\nabla u|^2dxdt.\end{align*}Now, we set $$F(\tau)=\int_0^\tau\int_\Omega u^2(x,t)dxdt,\tau\in[0,T],$$then $F$ is non-negative on $[0,T]$ and $F(0)=0.$ The preceding inequality shows that $$\forall\tau\in[0,T]:\frac{d}{d\tau}F(\tau)\leq MF(\tau),$$ which implies $F\equiv0.$ Thus we obtain $u(x,t)\equiv0$ in the whole $\bar\Omega_T.$