📂Partial Differential Equations

Maximum Principle of Harmonic Functions

Theorem¹

Let $\Omega \subset \mathbb{R}^n$ be open and bounded. Also, suppose $u : \Omega \to \mathbb{R}$ is equal to $u \in C^2(\Omega) \cap C(\bar \Omega)$ and satisfies the Laplace equation. Then, the following holds:

• (i) Maximum Principle

  $$\max \limits_{\bar \Omega} u = \max \limits_{\partial \Omega} u \quad \left( \mathrm{or} \ \ \min \limits_{\bar \Omega} u= \min \limits_{\partial \Omega} u \right)$$

• (ii) Strong Maximum Principle

  If $\Omega$ is a connected space and $$u(x_{0})=\max \limits_{\bar \Omega} u \left( \mathrm{or}\ \ u(x_{0})=\min \limits_{\bar \Omega} \right)$$ there exists a $x_{0} \in \Omega$ satisfying this, then $u$ is a constant function in $\Omega$ as follows: $$u(x) = M = \max_{\bar{\Omega}} u,\quad x\in \Omega$$

(i) The maximum (minimum) values on the boundary and the maximum (minimum) values including the boundary are equal.

Note that in (ii), the part where the function value of $u$ is constant applies to $\Omega$, but not to the boundary $\partial \Omega$.

Proof

(ii)

Let’s suppose $M=u(x_{0})=\max \limits_{\bar \Omega} u$. And let $V \subset \Omega$ be defined as follows.

$$V := \left\{ x\in\Omega | u(x)=M \right\}$$

Then, since it is $x_{0} \in V$, it is $V \ne \varnothing$. Also, since $\left\{ M \right\}^c=(-\infty, M)\cup (M,\infty)$ is open, $\left\{ M \right\}$ is closed, therefore, $u^{-1}(\left\{M\right\})=V$ is closed in $\Omega$. Now, for any $y \in V$, let’s set $d_{y}$ as follows.

$$d_{y} :=\mathrm{dist} (y,\partial \Omega)=\inf \limits_{ x \in \Omega} |y-x|>0$$

Then, according to the mean value theorem for harmonic functions, the following holds for $0<r<d_{y}$.

$$u(y)=-\!\!\!\!\!\! \int _{B(y,r)} u(x)dx \le M$$

Since the maximum value is $M$, the average must be less than or equal to $M$. In fact, since $y \in V$, the equality holds as follows.

$$u(y)=-\!\!\!\!\!\! \int _{B(y,r)} u(x)dx = M$$

For the average to be $M$ where the maximum value is $M$, for all $0<r<d_{y}$, $B(y,r)$ is $u=M$. Because if there’s even one place where the function value is less than $M$, the average must be less than $M$. Therefore, $V$ is open in $\Omega$ and by definition, the following holds.

$$B(y,r) \subset V \quad \forall r\in(0,d_{y})$$

Therefore, $V$ is both open and closed in $\Omega$. The necessary and sufficient condition for $\Omega$ to be a connected space is that a subset of $\Omega$ that is both open and closed is either $\emptyset$ or $\Omega$, and since it is $V \ne \emptyset$, it is $V=\Omega$. Thus, the following holds.

$$M=u(x) \quad \forall x \in \Omega$$

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Similarly, a proof can be made for the minimum value.

(i)

Since $u$ is continuous in the compact set $\bar \Omega$, a maximum (minimum) value $x_{0} \in \bar{\Omega}$ exists.

$$u(x_{0})=\max \limits_{\bar \Omega} u \quad \left( \mathrm{or} \ \ u(x_{0})=\min \limits_{\bar \Omega} u\right)$$

Now, let’s consider the cases where $x_{0}$ is on the boundary and inside.

• case 1. $x_{0} \in \partial \Omega$

  Trivially, the following holds.

  $$\max \limits_{\partial \Omega} = \max \limits_{\bar \Omega}$$

• Case 2. $x_{0} \in \Omega$

  Let’s say $\Omega_{0} \subset \Omega$ is a connected component that includes $x_{0}$. That is, $\Omega_{0}$ is $\mathrm{countable\ union\ oped\ set}$ that constitutes $\Omega$. Then $\Omega_{0}$ is open and connected. And $\partial \Omega_{0} \subset \partial \Omega$ holds. Applying (ii) to $\Omega_{0}$, the following is obtained.

  $$u=u(x_{0})=\max \limits_{\bar \Omega_{0}} u$$

  Then, by Part 1., the following holds.

  $$u(x_{0})=\max \limits_{\partial \Omega_{0}} u \le \max \limits_{\partial \Omega} u \le \max \limits_{\bar \Omega}u=u(x_{0})$$

  Thus, the following is obtained.

  $$\max \limits_{\partial \Omega} u = \max \limits_{\bar \Omega}u$$

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Similarly, a proof can be made for the minimum value.