I'm trying understand the proof of the Weak Maximum Principle for Parabolic Equations in this presentation (the proof starts on the slide $11$) for the operator

$$Lu \equiv \sum_{i,j = 1}^n a_{ij}(x,t) \frac{\partial^2 u}{\partial x_i \partial x_j} + \sum_{i = 1}^n b_i(x,t) \frac{\partial u}{\partial x_i} + c(x,t)u - \frac{\partial u}{\partial t}$$

when $c \leq 0$.

My doubts are

1) Why $D^2u \leq 0$ on the slide $11$? I know that the operator $Lu$ is evaluated on a maximum point, but how $D^2u$ appears in $Lu$ instead of $\sum_{i,j = 1}^n a_{ij}(x,t) \frac{\partial^2 u}{\partial x_i \partial x_j}$?

2) How exactly appears the inequality on slide $12$ which involves $||b_1||_{\infty}||$ and $||c||_{\infty}$?

3) Why $\sup_{\overline{\Omega_T}} u_{\varepsilon} \leq \max_{\overline{\Omega_T}} u_{\varepsilon}^+$ and not $\sup_{\overline{\Omega_T}} u_{\varepsilon} = \max_{\overline{\Omega_T}} u_{\varepsilon}^+$?

Thanks in advance!


2) I understood how appears the inequality. The term $\alpha^2 \lambda$ appears by definition of parabolic operator $L$, the terms $- ||b_1||_{\infty}$ and $- ||c||_{\infty}$ appear because $|b_1(x,t)| \leq ||b_1||_{\infty}$ and $|c(x,t)| \leq ||c||_{\infty}$ for every $(x,t) \in \Omega_T$ by definition of the uniform norm and, by the inequalities involving modulus, we have

$$- ||b_1||_{\infty} \leq b_1(x,t) \leq ||b_1||_{\infty} \hspace{1cm} \text{and} \hspace{1cm} - ||c||_{\infty} \leq c(x,t) \leq ||c||_{\infty}, \forall (x,t) \in \Omega_T.$$

3) I understood why $\sup_{\overline{\Omega_T}} u_{\varepsilon} \leq \max_{\overline{\Omega_T}} u_{\varepsilon}^+$. I saw now that $u = u^+ - u^-$ and $u^- = - \min \{ u, 0 \}$, then the inequality is clear now for me.

  • $\begingroup$ I think that's just the fact that at a maximum the hessian is negative semi-definite. $\endgroup$
    – Jose27
    Dec 21, 2017 at 23:24
  • $\begingroup$ @Jose27, yes, but my doubt is how exactly the hessian is related to $\sum_{i,j = 1}^n a_{ij}(x,t) \frac{\partial^2 u}{\partial x_i \partial x_j}$? $\endgroup$
    – George
    Dec 22, 2017 at 11:13

1 Answer 1


The sum $\sum_{i,j=1}^n a_{ij}(x,t)\frac{\partial^2 u}{\partial x_i\partial x_j}$ is exactly the trace of product $AH$ where $H$ is the hessian of $u$ and $A$ is the matrice with coefficients $(a_{ij})_{ij}$. Since $A$ is positive definite and $H$ is negative semidefinite, the trace $AH$ is non positive.

  • $\begingroup$ Sorry for not seeing your answer before, I didn't see any warnings in my inbox about your answer, thanks for the help! :) $\endgroup$
    – George
    Mar 4, 2019 at 19:16

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