Question

(a) Prove the following version of Green's theorem: suppose that $\Omega$ is a bounded region, bounded by finitely many $C^2$ simple closed curves. If $u$ and $v$ are twice continuously differentiable on $\bar{\Omega}$ then $$ \int_{\partial \Omega}\left(u \frac{\partial v}{\partial \eta}-v \frac{\partial u}{\partial \eta}\right)|d z|=\int_{\Omega}(u \Delta v-v \Delta u) d x d y, $$ where $\eta(\zeta)$ is the unit normal at $\zeta \in \partial \Omega$ pointing out of the region $\Omega,|d z|$ is arclength measure on $\partial \Omega$ and $d x d y$ is area measure on $\Omega$. If you already know a proof of some version of Green's theorem, you may use it to derive this version. (b) Prove that (ii) implies (i) in Theorem 7.12 by directly verifying that the mean-value property holds. Hint: In $\mathbb{D} \backslash B(0, \varepsilon)$ set $v=\log |z|$, apply (a) and let $\varepsilon \rightarrow 0$. (c) Similarly, show that if $v$ is twice continuously differentiable and $\Delta v \geq 0$ then $v$ is subharmonic because the mean-value inequality holds. (d) If $g$ is differentiable on $\mathbb{R}$ and $g^{\prime \prime}>0$ then prove that $v\left(r e^{i t}\right)=g(\log r)$ is subharmonic.

   (a) Prove the following version of Green's theorem: suppose that $\Omega$ is a bounded region, bounded by finitely many $C^2$ simple closed curves. If $u$ and $v$ are twice continuously differentiable on $\bar{\Omega}$ then
$$
\int_{\partial \Omega}\left(u \frac{\partial v}{\partial \eta}-v \frac{\partial u}{\partial \eta}\right)|d z|=\int_{\Omega}(u \Delta v-v \Delta u) d x d y,
$$
where $\eta(\zeta)$ is the unit normal at $\zeta \in \partial \Omega$ pointing out of the region $\Omega,|d z|$ is arclength measure on $\partial \Omega$ and $d x d y$ is area measure on $\Omega$. If you already know a proof of some version of Green's theorem, you may use it to derive this version.
(b) Prove that (ii) implies (i) in Theorem 7.12 by directly verifying that the mean-value property holds. Hint: In $\mathbb{D} \backslash B(0, \varepsilon)$ set $v=\log |z|$, apply (a) and let $\varepsilon \rightarrow 0$.
(c) Similarly, show that if $v$ is twice continuously differentiable and $\Delta v \geq 0$ then $v$ is subharmonic because the mean-value inequality holds.
(d) If $g$ is differentiable on $\mathbb{R}$ and $g^{\prime \prime}>0$ then prove that $v\left(r e^{i t}\right)=g(\log r)$ is subharmonic.
Show more…
Complex Analysis
Complex Analysis
Donald E. Marshall 1st Edition
Chapter 7, Problem 5 ↓

Instant Answer

verified

Step 1

(a) To prove this version of Green's theorem, we can start by applying the standard version of Green's theorem to the functions $u$ and $v$.  Show more…

Show all steps

lock
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
(a) Prove the following version of Green's theorem: suppose that $\Omega$ is a bounded region, bounded by finitely many $C^2$ simple closed curves. If $u$ and $v$ are twice continuously differentiable on $\bar{\Omega}$ then $$ \int_{\partial \Omega}\left(u \frac{\partial v}{\partial \eta}-v \frac{\partial u}{\partial \eta}\right)|d z|=\int_{\Omega}(u \Delta v-v \Delta u) d x d y, $$ where $\eta(\zeta)$ is the unit normal at $\zeta \in \partial \Omega$ pointing out of the region $\Omega,|d z|$ is arclength measure on $\partial \Omega$ and $d x d y$ is area measure on $\Omega$. If you already know a proof of some version of Green's theorem, you may use it to derive this version. (b) Prove that (ii) implies (i) in Theorem 7.12 by directly verifying that the mean-value property holds. Hint: In $\mathbb{D} \backslash B(0, \varepsilon)$ set $v=\log |z|$, apply (a) and let $\varepsilon \rightarrow 0$. (c) Similarly, show that if $v$ is twice continuously differentiable and $\Delta v \geq 0$ then $v$ is subharmonic because the mean-value inequality holds. (d) If $g$ is differentiable on $\mathbb{R}$ and $g^{\prime \prime}>0$ then prove that $v\left(r e^{i t}\right)=g(\log r)$ is subharmonic.
Close icon
Play audio
Feedback
Powered by NumerAI
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever