I'm having some trouble showing the following:
Let $u\subset\mathbb{R}^n$ be open and bounded and assume that $u\in C^{1,2}(U_T)$. Consider the partial differential operator $$Lu:=u_t-\Delta u+b\cdot\nabla u+cu\qquad\text{on $U_T$},$$ with $b\in\mathbb{R}^n$ and $c\in C(\overline{U_T})$. Show that if $Lu\leq0$ and $c\geq0$ on $U_T$ then $$\max_{\overline{U_T}}u\leq\max_{\Gamma_T}u^+,$$ where $u^+=\max\{u,0\}$ is the positive part of $u$.
And where $U_T:=U\times (0,T]$, and $\Gamma_T:=\overline{U_T}- U=(\partial U\times [0,T])\cup(U\times\{0\})$.
I can see this is quite similar to the maximum principles in Chapter 7 in Evans' book, and this can probably be proven in a similar fashion. Altough I'm having some trouble. Here's what I have thusfar:
Assume $Lu>0$ in $U_T$, and assume $u$ takes a negative minimum at $(x_0,t_0)\in U_T$, then $u_t(x_0,t_0)=0$, and $D^2u(x_0,t_0)\leq0$ an in particular $u_{x_ix_i}\leq0$.With this, $Lu$ also become non-positive, thus a contradiction with our assumption. Then when taking $Lu\geq0$ I don't know how to continue. I tried showing the problem using a similar technique as in Theorem 9 in section 7.1.4. of Evans' book. (also, why does it hold such a negative minimum exists? It's been a while since I've covered this and can't find an argument, although it is used in many other proof as well.)
Any hints are very much welcome! (I'm self-studying PDE using Evans' book (Chapter 2 and section 3.2) and Lecture Notes provided by my University, which is where this Problem is taken from)
First, I think you're missing the assumption that $u\in C(\overline{U_T})$, which is essential. Then due to compactness of $\overline{U_T}$ we infer that a maximum (and a minimum) in $\overline{U_T}$ exists.
Next, if $Lu \leq 0$, you consider $u_\varepsilon(x,t) = u(x,t) - \varepsilon t$ for $\varepsilon > 0$, for which it holds that $$ Lu_\varepsilon = Lu -\varepsilon - \varepsilon t c < 0,$$ since by assumption $c\geq 0$ and $Lu \leq 0$. If $u$ attains a positive maximum in $U_T$, then so does $u_\varepsilon$ if $\varepsilon$ is sufficiently small, and you get a contraction with the first step.