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I'm trying understand the proof of the Maximum Principle of Heat Equation given by John Fritz in his book "Partial Differential Equations - Third Edition" on page 175.

Before the theorem, he introduced the following sets:

$\Omega = \{ (x,t) \ | \ x \in \omega, 0 < t < T \}$, where $\omega \subset \mathbb{R}^n$ is an open bounded set and $T > 0$ is fixed.

$\partial' \Omega = \{ (x,t) \ | \ \text{either} \ x \in \partial \omega, 0 \leq t \leq T \ \text{or} \ x \in \omega, t = 0 \}$

$\partial'' \Omega = \{ (x,t) \ | \ x \in \omega, t = T \}$.

$\textbf{Theorem}$ Let $u$ be continuous in $\overline{\Omega}$ and $u_t$, $u_{x_ix_j}$ exist and be continuous in $\Omega$ and satisfy that $u_t - \triangle u \leq 0$, then $\max_{\overline{\Omega}} u = \max_{\partial' \Omega} u$.

$\textbf{Proof}:$

Let at first $u_t - \triangle u < 0$ in $\Omega$. Let $\Omega_{\epsilon}$ for $0 < \epsilon < T$ denote the set

$$\Omega_{\epsilon} = \{ (x,t) \ | \ x \in \omega, 0 < t < T - \epsilon \}.$$

Since $u \in C^0 \left( \overline{\Omega_{\epsilon}} \right)$ there exists a point $(x_0,t_0) \in \overline{\Omega_{\epsilon}}$ with $u(x_0, t_0) = \max_{\overline{\Omega_{\epsilon}}} u$ (the maximum here is in $\overline{\Omega_{\epsilon}}$)

If here $(x_0,t_0) \in \overline{\Omega_{\epsilon}}$ the necessary relations $u_t = 0$ and $\triangle u \leq 0$ would contradict $u_t - \triangle u < 0$. If $(x_0,t_0) \in \partial'' \Omega_{\epsilon}$ we would have $u_t \geq 0$ and $\triangle u \leq 0$ leading to same contradiction.

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The proof continues, but my doubt emerge here: "If $(x_0,t_0) \in \partial'' \Omega_{\epsilon}$ we would have $u_t \geq 0$ and $\triangle u \leq 0$ leading to same contradiction."

Well, $(x_0,t_0)$ is a maximum in $\overline{\Omega_{\epsilon}}$, then $u_t = 0$ and $\triangle u \leq 0$, so why $u_t \geq 0$?

Thanks in advance!

$\textbf{EDIT:}$

I tried understand why $u_t \geq 0$ and I thought that, fixing $x_0$ and expanding $u$ by Taylor's series in terms of $t$ and considering that $(x_0,t_0) \in \partial'' \Omega_{\epsilon}$, we have $t_0 = T - \epsilon$ and

$$u(x_0,t) \approx u(x_0,t_0) + u_t(x_0,t_0) \left( t - (T - \epsilon) \right)$$

Since $(x,t) \in \Omega_{\epsilon}$, $t < T - \epsilon$, then $t - (T - \epsilon) < 0$

Since $(x_0,t_0)$ is a maximum of $u$ in $\Omega_{\epsilon}$, we know that $u(x_0,t) - u(x_0,t_0) \leq 0$, therefore we have $u_t(x_0,t_0) \geq 0$.

Could anyone confirm if this is the reason for $u_t \geq 0$?

Thanks in advance!

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