§3.6.1 Solution to the Dirichlet Problem on a Disk

3.6.1 Solution to the Dirichlet Problem on a Disk

A fundamental problem in the theory of partial differential equations is to find a function $𝑢$ that is continuous on the closed disk $\overline{𝐷 (0, 𝑅)}$ , harmonic on the open disk $𝐷 (0, 𝑅)$ , and identically equal to a given boundary function on $𝜕 𝐷 (0, 𝑅)$ . This is known as the Dirichlet problem (for Laplace’s equation) on a disk.

Theorem 3.6.1.1.

For a continuous function $𝜑 \in 𝐶 (𝜕 𝐷 (0, 𝑅))$ , the unique real-valued solution $𝑢 \in 𝐶 (\overline{𝐷 (0, 𝑅)})$ that solves

\begin{matrix} \nabla^{2} 𝑢 (𝑧) = 0 \forall 𝑧 \in 𝐷 (0, 𝑅), \\ 𝑢 (𝑧) = 𝜑 (𝑧) \forall 𝑧 \in 𝜕 𝐷 (0, 𝑅) \end{matrix}

is given by the Poisson integral formula:

𝑢 (𝑧) = \int_{0}^{2 π} 𝜑 (𝜁) 𝑃 (𝜁, 𝑧) d 𝜏, (3.6.1 . 1)

where

𝜁 = 𝑅 e^{i 𝜏}

Figure 3.5:

𝜁

𝜉

, and

𝑧

when

| 𝜗 - 𝜏 | > 𝛿

, with distances marked. The use of

+

and

-

denote a value more or less (respectively) than the preceding value.

Proof.

Since

𝑃 (𝜁, 𝑧) = \frac{1}{4 π} (\frac{𝜁 + 𝑧}{𝜁 - 𝑧} + \frac{\overline{𝜁} + \overline{𝑧}}{\overline{𝜁} - \overline{𝑧}}),

from (2.2.1.4) $(2.2.1.7) (2.2.1.4)$ , we have that $\nabla_{𝑧}^{2} 𝑃 (𝜁, 𝑧) = 4 \frac{𝜕^{2} 𝑃 (𝜁, 𝑧)}{𝜕 𝑧 𝜕 \overline{𝑧}} = 0$ (since each term is independent of either $𝑧$ or $\overline{𝑧}$ ). Moreover, by Theorem 1.2.8 $Theorem 1.2.8 Theorem 1.2.8$ , (3.6.1.1) $(3.6.1.11) (3.6.1.1)$ gives that

\nabla^{2} 𝑢 (𝑧) = \nabla^{2} \int_{0}^{2 π} 𝜑 (𝜁) 𝑃 (𝜁, 𝑧) d 𝜏 = \int_{0}^{2 π} \nabla^{2} [𝜑 (𝜁) 𝑃 (𝜁, 𝑧)] d 𝜏 = 0 .

Our goal is to show that for fixed $𝜉 = 𝑅 e^{i 𝜗} \in 𝜕 𝐷 (0, 𝑅)$ ,

lim_{\begin{matrix} 𝑧 \to 𝜉 \\ 𝑧 \in 𝐷 (0, 𝑅) \end{matrix}} 𝑢 (𝑧) = 𝜑 (𝜉) . (3.6.1 . 2)

Let $\frac{𝑅}{2} < 𝜌 < 𝑅$ and $𝑧 = 𝜌 e^{i 𝜃}$ . Then with $𝜁 = 𝑅 e^{i 𝜏}$ ,

| 𝜑 (𝜉) - 𝑢 (𝑧) | = | 𝜑 (𝑅 e^{i 𝜗}) - 𝑢 (𝜌 e^{i 𝜃}) | = | 𝜑 (𝑅 e^{i 𝜗}) - \int_{0}^{2 π} 𝑃 (𝜁, 𝑧) 𝜑 (𝜁) d 𝜏 | .

For a constant harmonic function identically equal to $1$ , we get $\int_{0}^{2 π} 𝑃 (𝜁, 𝑧) d 𝜏 = 1$ from (3.6.5) $(3.6.5) (3.6.5)$ . Hence,

| 𝜑 (𝜉) - 𝑢 (𝑧) | = | \int_{0}^{2 π} 𝑃 (𝜁, 𝑧) (𝜑 (𝑅 e^{i 𝜗}) - 𝜑 (𝜁)) d 𝜏 | .

By the continuity of $𝜑$ , $\forall 𝜀 > 0$ , $\exists 𝛿 > 0$ such that $\forall | 𝜗 - 𝜏 | < 𝛿 < \frac{π}{2}$ , we have that $| 𝜑 (𝑅 e^{i 𝜗}) - 𝜑 (𝜁) | < 𝜀$ . Therefore,

\begin{aligned} | 𝜑 (𝜉) - 𝑢 (𝑧) | & = | (\int_{| 𝜗 - 𝜏 | < 𝛿} + \int_{| 𝜗 - 𝜏 | > 𝛿}) 𝑃 (𝜁, 𝑧) (𝜑 (𝑅 e^{i 𝜗}) - 𝜑 (𝜁)) d 𝜏 | \\ = | 𝐼_{1} + 𝐼_{2} | \leq | 𝐼_{1} | + | 𝐼_{2} | . \end{aligned}

Since the Poisson kernel is non-negative,

| 𝐼_{1} | < \int_{| 𝜗 - 𝜏 | < 𝛿} 𝜀 𝑃 (𝜁, 𝑧) d 𝜏 < 𝜀 \int_{0}^{2 π} 𝑃 (𝜁, 𝑧) d 𝜏 = 𝜀 .

By continuity of $𝜑$ on the compact set $𝜕 𝐷 (0, 𝑅)$ , by Theorem 1.2.26 $Theorem 1.2.26 Theorem 1.2.26$ , it is bounded and $𝑀 = {sup}_{| 𝜁 | = 𝑅} | 𝜑 (𝜁) |$ is finite. The Poisson kernel can be rewritten as

𝑃 (𝜁, 𝑧) = \frac{𝑅^{2} - 𝜌^{2}}{2 π {| 𝜁 - 𝑧 |}^{2}},

where $𝜁 = 𝑅 e^{i 𝜏}$ and $𝑧 = 𝜌 e^{i 𝜃}$ , with $| 𝜗 - 𝜏 | > 𝛿$ . Then $\exists 𝜂 > 0$ such that $\forall 𝑧$ with $| 𝜉 - 𝑧 | < 𝜂$ (small enough so that $| 𝜗 - 𝜃 | < \frac{𝛿}{2}$ ),

| 𝜃 - 𝜏 | > \frac{𝛿}{2} (3.6.1 . 3)

and

(𝜌 > \frac{𝑅}{2}) and 𝜂 \leq \frac{𝑅}{2} (3.6.1 . 4)

as in Figure 3.5 $Figure 3.5 Figure 3.5$ . Then,

{| 𝜁 - 𝑧 |}^{2} > 4 𝜌^{2} {sin (\frac{𝛿}{4})}^{2} > \frac{1}{2} 𝑅^{2} (1 - cos (\frac{𝛿}{2})) .

We aim to prove that $| 𝐼_{2} | < 𝜀$ . Since $| 𝜑 (𝑅 e^{i 𝜗}) - 𝜑 (𝜁) | < 2 𝑀$ , the condition is satisfied if

\int_{| 𝜗 - 𝜏 | > 𝛿} \frac{𝑅^{2} - 𝜌^{2}}{π 𝑅^{2} (1 - cos (\frac{𝛿}{2}))} d 𝜏 < 2 \frac{𝑅^{2} - 𝜌^{2}}{𝑅^{2} (1 - cos (\frac{𝛿}{2}))} < \frac{𝜀}{2 𝑀},

and from rearrangement, we can tighten the constraint with:

𝑅^{2} - 𝜌^{2} < \frac{𝜀}{4 𝑀} 𝑅^{2} (1 - cos (\frac{𝛿}{2})),

which follows in particular from

𝑅 - 𝜌 < \frac{𝜀}{8 𝑀} 𝑅 (1 - cos (\frac{𝛿}{2})) . (3.6.1 . 5)

From Figure 3.5 $Figure 3.5 Figure 3.5$ , it is evident that $𝑅 - 𝜌 < | 𝜉 - 𝑧 | < 𝜂$ . For (3.6.1.3) $(3.6.1.13) (3.6.1.3)$ to be true, we previously had that $| 𝜗 - 𝜃 | < \frac{𝛿}{2}$ . In other words

{| 𝜉 - 𝑧 |}^{2} < 𝑅^{2} + 𝜌^{2} - 2 𝑅 𝜌 cos (\frac{𝛿}{2}) .

Obviously, this is satisfied if ${| 𝜉 - 𝑧 |}^{2} < \frac{𝑅^{2}}{2} (1 - cos (\frac{𝛿}{2})) < 2 𝜌^{2} (1 - cos (\frac{𝛿}{2}))$ . This can be rearranged into

| 𝜉 - 𝑧 | < 𝑅 \sqrt{\frac{1 - cos (\frac{𝛿}{2})}{2}} = 𝑅 sin (\frac{𝛿}{4}) .

Therefore, we can choose

𝜂 = min (\frac{𝜀}{8 𝑀} 𝑅 (1 - cos (\frac{𝛿}{2})), 𝑅 sin (\frac{𝛿}{4}), \frac{𝑅}{2},) > 0,

under which (3.6.1.3) $(3.6.1.13) (3.6.1.3)$ , (3.6.1.4) $(3.6.1.14) (3.6.1.4)$ , and (3.6.1.5) $(3.6.1.15) (3.6.1.5)$ are satisfied.

Hence, $\forall 𝜀 > 0$ , $\exists 𝜂 > 0$ such that $\forall 𝑧$ with $0 < | 𝜉 - 𝑧 | < 𝜂$ , we have $| 𝜑 (𝜉) - 𝑢 (𝑧) | < 2 𝜀$ . Then (3.6.1.2) $(3.6.1.12) (3.6.1.2)$ follows.

We will now show that $𝑢 (𝑧)$ is unique. Assume that $𝑣 \neq 𝑢$ on $\overline{𝐷 (0, 𝑅)}$ also solves the problem. Then $𝑢 - 𝑣$ is harmonic and vanishes on $𝜕 𝐷 (0, 𝑅)$ . By the Poisson Integral Formula ((3.6.5) $(3.6.5) (3.6.5)$ ),

𝑢 (𝑧) - 𝑣 (𝑧) = \int_{0}^{2 π} 𝑃 (𝜁, 𝑧) (𝑢 (𝜁) - 𝑣 (𝜁)) d 𝜏 = 0

for all $𝑧 \in 𝐷 (0, 𝑅)$ . Hence $𝑢 \equiv 𝑣$ , a contradiction. ⁠ $□$