Math Problems Vault

Problem 5.9.1

Let $u : \mathbb {R}^{2} \to \mathbb {R}$ be the function defined by $u(x,y) = x^{3} - 3xy^{2}$ . Show that $u$ is harmonic and find $v : \mathbb {R}^{2} \to \mathbb {R}$ such that the function $f : \mathbb {C} \to \mathbb {C}$ defined by

f(x+iy) = u(x,y) + iv(x,y)

is analytic.

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Problem 5.9.2

Let $f(z)$ be an analytic function defined for $\left|z\right| \leq 1$ and let

u(x,y) = \mathrm{Re}\left(f(z)\right), \qquad z = x+iy .

Prove that

\int _{C} \frac{\partial u}{\partial y} \, dx - \frac{\partial u}{\partial x} \, dy = 0

where $C$ is the unit circle, $x^{2}+y^{2} = 1$ .

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Problem 5.9.3

(1)(2)

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(1)

Let $f$ be a complex function which is analytic on an open set containing the disc $\left|z\right| \leq 1$ , and which is real valued on the unit circle. Prove that $f$ is constant.

(2)

Find a nonconstant function which is analytic at every point of the complex plane except for a single point on the unit circle $\left|z\right| = 1$ , and which is real valued at every other point of the unit circle.

Problem 5.9.4

Let $s$ be a real number, and let the function $u$ be defined in $\mathbb {C} \setminus (-\infty , 0]$ by

u(re^{i\theta }) = r^{s} \cos s\theta \qquad (r > 0, \ -\pi < \theta < \pi ) .

Prove that $u$ is a harmonic function.

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Problem 5.9.5

Let $u$ be a positive harmonic function on $\mathbb {R}^{2}$ ; that is,

\frac{\partial ^{2} u}{\partial x^{2}} + \frac{\partial ^{2} u}{\partial y^{2}} = 0 .

Show that $u$ is constant.

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Problem 5.9.6

Let $u$ be a real valued harmonic function in the complex plane such that

u(z) \leq a\left|\log \left|z\right|\right| + b

for all $z$ , where $a$ and $b$ are positive constants. Prove that $u$ is constant.

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Problem 5.9.7

Let $D = \left\{ z \in \mathbb {C} : \left|z\right| \leq 1\right\} \setminus \left\{ 1,-1\right\}$ . Find an explicit continuous function $f : D \to \mathbb {R}$ satisfying all the following conditions:

$f$ is harmonic on the interior of $D$ (the open unit disk),
$f(z) = 1$ when $\left|z\right| = 1$ and $\mathrm{Im}\left(z\right) > 0$ , and
$f(z) = -1$ when $\left|z\right| = 1$ and $\mathrm{Im}\left(z\right) < 0$ .

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