Let $u$ be sufficiently smooth and satisfy $$ \begin{cases} u_t − u_{xx}= f(x,t) & \{(x,t) : 0 < x < 1, 0 < t < 1\},\\ u(0,t) \ge 0, u(1,t) \ge 0 & 0 \le t \le 1\\ u(x, 0) \ge 0 & 0 \le x \le 1 \end{cases} $$
- Prove that $u \ge 0$ on $\{(x,t) : 0 < x < 1, 0 < t < 1\}$ if $f(x,t) \ge \epsilon$ in $\{(x,t) : 0 < x < 1, 0 < t < 1\},$ where $\epsilon > 0$.
- Prove that $u \ge 0$ on $\{(x,t) : 0 < x < 1, 0 < t < 1\}$ if $f(x,t) \ge 0$ in $\{(x,t) : 0 < x < 1, 0 < t < 1\}$.
My approach
For the equation given as the condition, I think that this is related to diffusion equation, but I need to assume 2 solution ( for example, a&b) and come up with its difference w, so I can rewrite the condition as the proper diffusion equation with boundary conditions. After this process, I thought this would be related to maximum principle, but I'm quite stuck since it's all about .
I want to know whether my approach is right, or is there another proper approach for this problem.
And, for these 2 questions, I get the point that the difference is that f(x,t) can be 0 in question #2, but I don't get its significance.
Would be thankful if you could come up with proper approach for each questions.
If your region is $D = \{ (x,t): 0<x<1, 0<t<1\}$ and its boundary is $\partial D$, then $D \cup \partial D$ is compact, hence the image under $u$ of this set is also compact, i.e. closed and bounded by Heine-Borel since we're dealing with a real-valued function. Thus, the function must attain its minimum somewhere in $D \cup \partial D$.
It suffices to show that this minimum cannot be attained on the line $t=1$ or in the interior. For the latter case, if it were in the interior, then we'd have $u_t = u_x = 0$ at this point, as well as $u_{xx} \geq 0$. This contradicts the heat equation since $f$ is strictly positive. If the minimum instead lies on $t=1$, then again $u_x=0$, $u_{xx} \geq 0$, but now $u_t \leq 0$ - otherwise this minimum would be attained vertically below this line. Again, a contradiction. So the minimum is attained on $x=0,$ $x=1,$ or $t=0$, thus the function is non-negative everywhere.
For the second part, define a new solution of the heat equation by: \begin{equation} v(x,t) = u(x,t) - \frac{\epsilon}{2}x^2. \end{equation} In this case, $v_t-v_{xx} = f(x,t)+\epsilon >0$, so the previous part applies to $v$, i.e. $v$ attains its minimum on $\partial D$ less $\{t=1\}$. So if the minimum value of $u$ on $\partial D$ less $\{t=1\}$ is $M$, then: \begin{equation} u \geq v \geq M - \frac{\epsilon}{2}, \end{equation} since the maximum value of $|x|$ is $1$. We may then take $\epsilon \to 0^+$ and conclude that the minimum value of $u$ is attained on $\partial D$ less $\{t=1\}$.