Answer to Question #236286 in Chemical Engineering for Lokika

Question #236286

2) Show that v(x,y)=x²-y²-y is harmonic function.Find it's conjugate harmonic function u(x,y) and corresponding analytic function f(z).

Explain the problem with step by Step process?


1
Expert's answer
2021-09-27T01:45:41-0400

Harmonic functions appear regularly and play a fundamental role in math, physics and engineering. In this topic we’ll learn the definition, some key properties and their tight connection to complex analysis. The key connection to 18.04 is that both the real and imaginary parts of analytic functions are harmonic. We will see that this is a simple consequence of the Cauchy-Riemann equations. In the next topic we will look at some applications to hydrodynamics.


The connection between analytic and harmonic functions is very strong. In many respects it mirrors the connection between ez and sine and cosine.

Let z = x + iy and write f (z) = u(x, y) + iv(x, y).

If f (z) = u(x, y) + iv(x, y) is analytic on a region A then both u and v are

harmonic functions on A.

. This is a simple consequence of the Cauchy-Riemann equations. Since ux = vy


Since we know an analytic function is infinitely differentiable we know u and v have the required two continuous partial derivatives. This also ensures that the mixed partials agree, i.e. vxy = vyx.

To complete the tight connection between analytic and harmonic functions we show that any harmonic function is the real part of an analytic function.

If u(x,y) is harmonic on a simply connected region A, then u is the real part of an analytic function f (z) = u(x, y) + iv(x, y).

This is similar to our proof that an analytic function has an antiderivative. First we come up with a candidate for f(z) and then show it has the properties we need. Here are the details broken down into steps.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS