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I am self-studying Evan's PDE, chapter 6.4, which talks about the maximum principle of elliptic equations.

Theorem (Weak maximum principle) Define $L := \sum_{i,j=1}^n a^{ij} u_{x_i x_j} + \sum_{i=1}^n b^i u_{x_i} + cu$ and let $U \subset \mathbb{R}^n$ be a bounded open set. Assume $u \in C^2(U) \cap C(\overline{U})$ and $c \equiv 0$ in $U$. If $Lu \leq 0$ in $U$, then $\max_U u = \max_{\partial U} u$.

Proof Suppose $Lu<0$ in $U$ first and suppose there exists some $x_0 \in U$ such that $u(x_0)$ attains the maximum. Then $Du(x_0) = 0$ and $D^2u(x_0)$ is non-positive definite at $x_0$.

My question: Why $D^2u(x_0) \leq 0$? My attempts are as follows:

The conclusion is true in 1-dimensional case. WLOG let $x_0 = 0$. Fix some $y \in \mathbb{R}^n$, define $f(t) := u(ty)$. Then $f'(t) = Du(ty) \cdot y = \sum_{i=1}^n u_{x_i}(ty)y_i$ and then $f''(t) = \sum_{i=1}^n y_i \nabla (u_{x_i}(ty)) \cdot y = y^T D^2u(ty) y$. Since $u\in C^2(U)$, sending $t \to 0$ shows that $D^2u(0) \leq 0$.

Is my above proof correct? Could anyone give me some hint?

Edit: with the help in comments, the above proof should be correct; and $u \in C^2$ is not necessary, instead, we could directly write $f''(0) = y^T D^2(0) y \leq 0$.

mathdoge
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  • How are you getting $y^T D^2u(ty)y \le 0$? – Calvin Khor Jul 10 '20 at 01:16
  • @CalvinKhor Expanding out $f''(t) = \sum_{i=1}^n y_i \sum_{j=1}^n u_{x_i x_j} (ty) y_j = y^T D^2u(ty) y$. $u$ attains maximum at $0$ means that $f$ also attains maximum at $0$, so $f''(0) \leq 0$. – mathdoge Jul 10 '20 at 05:04
  • @CalvinKhor Oh I see your point. Yes you're right, I cannot obtain $f''(t) \leq 0$ for all $t$; but I could get $y^T D^2u(0) y \leq 0$ by the fact that $u \in C^2$ right? – mathdoge Jul 10 '20 at 05:06
  • $f''(0)\le 0$ is correct, but your sentence in the post is $f''({\Huge t})\le 0$ – Calvin Khor Jul 10 '20 at 05:07
  • Rereading the proof after the above fix, I believe it is also correct – Calvin Khor Jul 10 '20 at 05:09
  • @CalvinKhor Thank you, yes, I will fix the post. BTW: your huge $t$ looks funny. – mathdoge Jul 10 '20 at 05:10
  • you're welcome. If its funny, then you will look at it :) its easier to type {\huge t}${\huge t}$ than it is to type {\color{red} t}${\color{red} t}$ which is what I used to do to emphasize. – Calvin Khor Jul 10 '20 at 05:12
  • Where in your proof did you use $u\in C^2$? I think twice differentiable is enough – Calvin Khor Jul 10 '20 at 05:16
  • @CalvinKhor I want to send $t \to 0$ to conclude that $0 \geq f''(0) = \lim_{t \to 0} f''(t) = \lim_{t \to 0} y^T D^2(ty) y = y^T D^2(0) y$, and I think I need $C^2$ to pass the limit. – mathdoge Jul 10 '20 at 13:59
  • But you don’t need to consider $f’’(t)$ at all, just directly use $f’’(0)$ – Calvin Khor Jul 10 '20 at 14:00
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    @CalvinKhor I think you're right, at this step we don't even need the continuity of the second derivative. In your proof, you only use twice differentiability as well. Maybe Evans intends to use the continuity in later of the proof. Thank you for notifying me of this! – mathdoge Jul 10 '20 at 14:09

1 Answers1

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Evans is appealing to a general result coming only from the fact that $u$ is sufficiently smooth (twice differentiable) and has a (local) maximum at the interior point $x_0$.

It is standard that $Du$ vanishes at the maxmimum $x_0$. Suppose that $D^2u(x_0)\not\le 0$. Then there would exist a unit vector $v$ such that $v^T D^2 u(x_0)v=C>0$ (as real numbers). Recall the second order Taylor expansion as $x\to x_0$, (Peano remainder is enough) $$ u(x) = u(x_0) + \underbrace{Du(x_0)\cdot (x-x_0)}_{=0} + (x-x_0)^T Du(x_0)(x-x_0) + o(|x-x_0|^2). $$ Choose $x-x_0 = \lambda v$ where $\lambda\ll1$. Then we see as $\lambda\to 0$, $$ u(x) = u(x_0) + \lambda^2 C + o(\lambda^2).$$ since the $o(\lambda^2)$ term is eventually smaller than $\lambda^2C/2$, we see that $u(x)>u(x_0)$, which contradicts the fact that $u(x_0)$ was a maximum.

Calvin Khor
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