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Complex Analysis - Harmonic Functions

  • #1
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Homework Statement



Show ln(az) where a is a real number and z = x + iy is harmonic everywhere except z = 0.

Homework Equations



z = x + iy = rcos(θ) + irsin(θ) = re^iθ
z = u(x,y) + iv(x,y)

Cauchy Riemann test for analyticity:
∂u/∂x = ∂v/∂y
∂u/∂y = -∂v/∂x

The Attempt at a Solution



ln(az) = ln (rcos(θ) + irsin(θ)) = ln(rcos(θ+2nπ) + irsin(θ+2nπ))

= ln(a*re^iθ) = ln(ar) + i(θ+2nπ) <- this is multivalued, not harmonic.

How do I show it is harmonic?
 

Answers and Replies

  • #2
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Either rewrite ln(az) using x and y or the C-R relations using r and θ (this is the easier option) and proceed to check that they hold. The multivaluedness is in a constant which vanishes when you take derivatives.
 
  • #3
901
3
Either rewrite ln(az) using x and y or the C-R relations using r and θ (this is the easier option) and proceed to check that they hold. The multivaluedness is in a constant which vanishes when you take derivatives.
writing it in terms of r, theta and using CR, that shows it is analytic. how do i show it is harmonic? the e^iθ, after you take the ln, will no longer be harmonic!
 
  • #4
gabbagabbahey
Homework Helper
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writing it in terms of r, theta and using CR, that shows it is analytic. how do i show it is harmonic? the e^iθ, after you take the ln, will no longer be harmonic!
What is the definition of a harmonic function? Harmonic and periodic are not the same thing.
 
  • #5
Dick
Science Advisor
Homework Helper
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writing it in terms of r, theta and using CR, that shows it is analytic. how do i show it is harmonic? the e^iθ, after you take the ln, will no longer be harmonic!
Why do you think θ isn't harmonic? It's basically just arctan(y/x). That's perfectly fine harmonic function.
 
  • #6
901
3
Why do you think θ isn't harmonic? It's basically just arctan(y/x). That's perfectly fine harmonic function.
sorry, perhaps I was confused, but I think the definition of a harmonic function should be one such that for f(x), ∂^2f/∂x^2 = kf(x). Or as wikipedia says:

In mathematics, mathematical physics and the theory of stochastic processes, a harmonic function is a twice continuously differentiable function f : U → R (where U is an open subset of Rn) which satisfies Laplace's equation

http://en.wikipedia.org/wiki/Harmonic_function

That's true for sine: ∂^2/∂x^2 (sin x) = -1*sin x
Also true for cosine: ∂^2/∂x^2 (cos x) = -1*cos x

However it is not true for arctan.

for u = arctan(y/x), ∂^2u/∂x^2 =/= k*u. thus it is not a harmonic function.

∂^2u/∂x^2 (arctan(y/x) = -(2 x y)/(x^2+y^2)^2 instead.

So that's why I'm confused. Yes, harmonic doesn't necessarily mean periodic, e^x is an example, but I'm having a hard time finding a function from ln(az) that can satisfy the 2nd derivative rule.
 
  • #7
gabbagabbahey
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sorry, perhaps I was confused, but I think the definition of a harmonic function should be one such that for f(x), ∂^2f/∂x^2 = kf(x). Or as wikipedia says:

In mathematics, mathematical physics and the theory of stochastic processes, a harmonic function is a twice continuously differentiable function f : U → R (where U is an open subset of Rn) which satisfies Laplace's equation

http://en.wikipedia.org/wiki/Harmonic_function

That's true for sine: ∂^2/∂x^2 (sin x) = -1*sin x
Also true for cosine: ∂^2/∂x^2 (cos x) = -1*cos x
I think you need to look up Laplace's equation, because it is not ∂^2f/∂x^2 = kf(x).

To quote Tai L Chow's Mathematical Methods For Physicists,

Mathematical Methods For Physicists said:
If [itex]f(z)=u(x,y)+iv(x,y)[/itex] is analytic in some region of the [itex]z[/itex] plane, then at every point in the region the Cauchy-Riemann conditions are satisfied:

[tex]\frac{ \partial u }{ \partial x } = \frac{ \partial v }{ \partial y } \quad \text{and} \quad \frac{ \partial u }{ \partial y } = -\frac{ \partial v }{ \partial x }[/tex]

and therefor:

[tex]\frac{ \partial^2 u }{ \partial x^2 } = \frac{ \partial^2 v }{ \partial x \partial y } \quad \text{and} \quad \frac{ \partial^2 u }{ \partial y^2 } = -\frac{ \partial^2 v }{ \partial y \partial x }[/tex]

provided these second derivatives exist. In fact,one can show that if [itex]f(z)[/itex] is analytic in some region [itex]R[/itex], all its derivative exist and are continuous in [itex]R[/itex]. Equating the two cross terms, we obtain

[tex]\frac{ \partial^2 u }{ \partial x^2 } + \frac{ \partial^2 u }{ \partial y^2 } = 0[/tex]

throughout the region [itex]R[/itex].

Similarly, by differentiating the first of the Cauchy-Riemann equations with respect to [itex]y[/itex], the second with respect to [itex]x[/itex], and subtracting we obtain

[tex]\frac{ \partial^2 v }{ \partial x^2 } + \frac{ \partial^2 v }{ \partial y^2 } = 0[/tex]

Equations (6.12a) and (6.12b) are Laplace's partial differential equations in two independent variables [itex]x[/itex] and [itex]y[/itex]. Any function that has continuous partial derivatives of second order and that satisfies Laplace's equation is called a harmonic function.

We have shown that if [itex]f(z)=u(x,y)+iv(x,y)[/itex] is analytic, then both [itex]u[/itex] and [itex]v[/itex] are harmonic functions. They are called conjugate harmonic functions.
So, when you are being asked to show that [itex]\ln(az)[/itex] is harmonic, you are really being asked to show that the two-variable functions describing its real and imaginary parts are harmonic. From the above quote, you should see that this is equivalent to showing that [itex]\ln(az)[/itex] is analytic over the specified region (which is where the Cauchy Riemann equations come in, as clamtrox suggested).

Your textbook likely says something similar. Whenever you don't know how to tackle a problem, it is usually a good idea to open up your textbook and make sure that you understand the definitions of the relevant terms in the problem statement. Textbooks (good ones, anyways) are useful references to have.
 
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