$u_t=u_{xx}-2u^3$: Non-negative Solution for Non-negative Initial Conditions

Question

I'm attempting to show the following IBVP has a non-negative solution for non-negative initial conditions:

Suppose $u$ is continuous and satisfies

\begin{cases} u_t=u_{xx}-2u^3 \\ u(a,t)=\psi_a(t) \\ u(b,t)=\psi_b(t) \\ u(x,0)=h(x) \end{cases}

Show if $\psi_a(t)$, $\psi_b(t)$, and $h(x)$ are $\ge0$, then $u(x,t)\ge 0$

I was unable to find an explicit solution and I'm not sure how else to approach this.

Any help is appreciated! Thank you!

Answer 1

Suppose $u$ is negative on some interval of $x$ for some amount of time.

Since $u$ is continuous, there exists a moment where $u<0$ (by assumption), $u_t<0$ ($u$ must decrease to satisfy the assumption), and $u_{xx}>0$ (after crossing the $x$-axis $u$ must go concave up to reach $\psi_b$).

This contradicts the PDE because the LHS$<0$ and the RHS$>0$