In this post, we shall see the maximum principles for subharmonic ( ${\Longleftrightarrow}$ the Laplacian is nonnegative) ${{\mathcal C}^2}$ functions( ${\Longleftrightarrow}$ twice differentiable with continuous second derivative). As an application, we shall see the uniqueness of a solution to the Dirichlet problem. The Dirichlet problem is:

Suppose ${\Omega}$ is an open bounded subset of ${{\mathbb R}^n}$. Let ${\phi}$ be a continuous function defined on ${\partial \Omega}$, the boundary of ${\Omega}$. Does there exist a function ${f\in {\mathcal C}^2(\Omega) \cap {\mathcal C}(\overline\Omega)}$ that solves $\displaystyle \begin{gathered} \triangle f = 0 \qquad \text{ in } \Omega, \\ f = \phi \qquad \text{ on } \partial \Omega. \end{gathered} \ \ \ \ \ (1)$

Historically, the Dirichlet problem has been important to solve PDEs that arise naturally in Physics and engineering. But as mathematicians, let us not worry about that. For us, it is important to know when a solution to the DP exists and when its unique. Curiously, a sufficient condition on the existence of a solution is how ${\Omega}$ looks! For example, if ${\Omega = B(0,r)}$ then ${f}$ exists! In the following part, we shall show using the “Maximum Principle” that if an ${f}$ exists, then it must be unique.

The Maximum Principle

Theorem 1 (Strong form) Suppose ${\Omega }$ is an open subset of ${{\mathbb R}^n}$ and ${f \in {\mathcal C}^2(\Omega), \triangle f >0}$ in ${\Omega}$. Then ${f}$ cannot attain a maximum in ${\Omega}$.

Idea of proof:

By contradiction: Suppose ${f}$ attains its maximum at ${x_0\in \Omega}$. Find an ${r>0}$ such that ${B(x_0, r)\subseteq \Omega}$ and let ${\phi_i : (-r,r) \rightarrow {\mathbb R}}$ be ${\phi_i(t) = f(x_0 + te_i)}$. Show that ${\phi_i \in {\mathcal C}^2}$ and ${\phi_i''(0) \leq 0}$. Add over all ${i = 1, \cdots, n}$ to get ${\triangle f(x_0) \leq 0}$, a contradiction. $\square$

In the weak form, we assume the function to be subharmonic, i.e., ${\triangle f \geq 0}$.

Theorem 2 ( Weak form) Suppose ${\Omega}$ is an open bounded subset of ${{\mathbb R}^n}$. (So now, both ${\partial \Omega}$ and ${\overline\Omega}$ are compact sets). If ${\triangle f \geq 0}$ in ${\Omega}$, then $\displaystyle \max_{\overline \Omega} f = \max_{\partial \Omega} f.$

Idea of proof: We use the ${\varepsilon}$ trick.

Define $\displaystyle g_\varepsilon(x) = f(x) + \varepsilon \frac{|x|^2}{2n}, \qquad x \in \overline \Omega.$
Show that ${\triangle g_\varepsilon\geq \varepsilon > 0}$, and use the strong form of the maximum principle. $\square$

Proof of uniqueness in the DP:

Suppose that ${f_1}$ and ${f_2}$ both solve the DP equation (1). Define ${f = f_1 - f_2}$. Then, ${\triangle f = 0}$ in ${\Omega}$ and ${f=0}$ on ${\partial \Omega}$. Hence by the weak form of the maximum principle, ${f\leq 0}$ in ${\Omega}$. Since we could have taken ${f}$ to be ${f_2-f_1}$ instead, it follows that ${f = 0}$, by which ${f_1 = f_2}$ in ${\Omega}$ and uniqueness has been established. $\square$