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 $\max_{\overline{\Omega}} u = \max_{\partial' \Omega} u$.


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.


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!


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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.