Best Approximate Solution of Heat Equation (Diffusive Logistic Equation)

Question

If $u(x,t) \ge 0$ in the domain $ (0 \times 1) \times (0,\infty)$, find a function that caps the value of $u(x,t)$ in the region $(0 \times 1) \times (0,T)$. $u(x,t)$ is a solution of the following system: \begin{cases} u_t = u_{xx} + u(1-u), & (0 \times 1) \times (0,\infty) \\ u(0,t) = 0 = u(1,t), & t \in (0,\infty)\\ u(x,0) = sin(\pi x), & x \in (0,1) \end{cases} My attempt: I notice that without the term $u(1-u)$, the solution is simply a heat equation. So $u(x,t) \le ||u(x,0)|| = 1$. Adding in the effect of the term $u(1-u)$, which is positive on the interval $(0,1)$, then the following approximation can be made on $(0 \times 1) \times (0,T)$, $$u(x,t) \le 1 + ||u(1-u)||T$$

This is not a very strong approximation. So I am looking for something stronger. I appreciate any input. Thank you.

Answer 1

If you are allowed to assume that the function is two times differentiable with continuous second derivatives then the problem can be completed as follows.

Consider any finite value of $T > 0$ and concentrate on the restriction $u_T$ of $u$ to the rectangle \begin{equation} R_T = [0,1] \times [0,T]. \end{equation} By continuity, $u$ will assume its maximum value on this rectangle. There are now three distinct possibilities:

1. The maximum is assumed somewhere on the parabolic boundary. On the initial line \begin{equation} u(x,0)=\sin(\pi x) \end{equation} is bounded by 1 and equality is achieved at $x=\frac{1}{2}$. Since the boundary conditions are homogeneous, the maximum is not assumed there on the lines $x=0$ or $x=1$.

2. The maximum is assumed in the interior of $R_T$, i.e the rectangle \begin{equation} R'_T = (0,1) \times (0,T). \end{equation} In this case \begin{equation} u_t = u_x = 0, \quad \text{(critical point)} \end{equation} and \begin{equation} u_{xx} \leq 0 \quad \text{(maximum)}. \end{equation} It follows from the differential equation that $u(1-u) \geq 0$ which implies that $u \in [0,1]$.

3. The maximum is assumed on the line $t=T$. In this case \begin{equation} u_t \ge 0 \end{equation} or there would be an even larger value inside $R'_T$ (Taylor expansion). Moreover, by considering the restriction of $u$ to the line $t=T$ we deduce that \begin{equation} u_x = 0 \quad \text{and} \quad u_{xx} \leq 0. \end{equation} By rewriting the differential equation as \begin{equation} u_t - u_{xx} = u(1-u) \end{equation} we conclude that the right hand side must be nonnegative, because the left hand side is nonnegative. It follows again that $u \in [0,1]$.

Since $u$ assumes the value $1$ on the initial line, we conclude that $u \leq 1$ is the best possible upper bound for $u$ across the entire domain.

I hope this helps.