proof of weak maximum principle (parabolic equation)

We had the following version of the weak maximum principle for parabolic differential equations of second order in our lecture:

Let $L$ be a parabolic differential operator of the form

$$(Lu)(x,t)= - \partial_t u(x,t) + \sum_{j,k=1}^n a_{j,k}(x,t) \partial_{x_j} \partial_{x_k} u(x,t) + \sum_{j=1}^n b_j(x,t) \partial_{x_j} u(x,t) + c(x,t)u(x,t)$$ for $x \in \Omega$, $t \in (0,T)$ with $\Omega \subseteq \mathbb{R}^n$ open, bounded and non-empty, $T >0$, $a_{j,k}, b_j, c \in C^0(\bar{Q_T})$, $Q_T= \Omega \times (0,T]$ and $\sum_T = (\Omega \times {0}) \cup (\partial \Omega \times [0,T])$.

The maximum principle now states that for $L, \Omega, Q_t, \sum_t$ given as above and $c(x,t)=0 \forall x \in Q_t$, we have for all $u \in C^2(Q_T) \cap C^0(\bar{Q_T})$:

$Lu(x,t) \geq 0 ~ \forall (x,t) \in Q_T$ implies ${max}_{(x,t)\in \bar{Q_T}} u(x,t)= {max}_{x \in \sum_T} u(x,t)$

My questions now concern two aspects in the proof of this. In the first case ($Lu(x,t)>0$), we showed that there is no $(x_0,t_0) \in \Omega \times (0,T)$ st $u(x_0,t_0)={max}_{(x,t)\in \bar{Q_T}} u(x,t)=:m$. We then argue that, for this case, it is sufficient to show that there's no $x_0 \in \Omega$ st $u(x_0, T)=m$. Question: Why is this enough and why do I plug in $T$ and focus on $x_0$?

In the second case ($Lu(x,t) \geq 0$), we defined $u_\epsilon(x,t)= u(x,t)+ \epsilon e^{-t} ~ \forall (x,t) \in \bar{Q_t}, \epsilon >0$ and said that $Lu_\epsilon(x,t)=Lu(x,t) - \epsilon \partial_t(e^{-t}) >0$ Question: Why is this strictly greater than 0?

Thank you very much for your help!

• For the 2nd question: because $\partial_t \exp(-t)=-\exp(-t)<0$. – Elsa Aug 16 '17 at 12:46
• For the 1st part: because if you show that there is no maximizer on $\Omega x T$ then a maximizer must be either in $\Omega x 0$ (because ruled out $t\in (0,T]$) or a point with $x$ on the boundary of $\Omega$, i.e. only $\Sigma _T$ remains. – Elsa Aug 16 '17 at 12:56