Harmonic Function Values (Complex Analysis)

[EC92]

Homework Statement

Let [itex]u[/itex] be a continuous real-valued function in the closure of the unit disk [itex]\mathbb{D}[/itex] that is harmonic in [itex]\mathbb{D}[/itex]. Assume that the boundary values of [itex]u[/itex] are given by

[itex] u(e^{it}) = 5- 4 \cos t. [/itex]

Furthermore, let [itex]v[/itex] be a harmonic conjugate of [itex]u[/itex] in [itex]\mathbb{D}[/itex] such that [itex]v(0) = 1[/itex]. Find [itex]u(1/2)[/itex] and [itex]v(1/2)[/itex].

Homework Equations

Harmonic functions fulfill
[itex] u_{xx} + u_{yy} = 0[/itex].
Poisson's Integral Formula (for harmonic functions) tells us that
[itex] u(z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|e^{i\theta}-z|^2}u(e^{i\theta}) d\theta [/itex].
The harmonic conjugate of [itex]u[/itex] is given by the line integral
[itex] v(z) = \operatorname{Im} \int_0 ^z u_x(w) - iu_y(w) dw [/itex].
The harmonic conjugate is itself harmonic, so Poisson's formula applies to it as well.

The Attempt at a Solution

It's easy to find [itex]u(1/2)[/itex]: it is
[itex] \frac{1}{2\pi} \int_0^{2\pi} \frac{3/4}{5/4 - \cos\theta} (5-4\cos \theta) d\theta = \frac{1}{2\pi} \int_0 ^{2\pi} 3 d\theta = 3 [/itex].

However, I can't figure out how to compute [itex]v(1/2)[/itex]; the formulas just seem to give a hopelessly complicated mess.
Any ideas on how to proceed?
Thanks.

 

[mighty2000]

Hello, I am a scientist and I would like to help you with your problem. First, let's review the equations that we will need to use to solve this problem. As you mentioned, harmonic functions fulfill the equation u_{xx} + u_{yy} = 0. Additionally, we can use Poisson's Integral Formula to find u(z) in terms of its boundary values. This formula tells us that u(z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|e^{i\theta}-z|^2}u(e^{i\theta}) d\theta.

Now, let's focus on finding v(1/2). We know that v is a harmonic conjugate of u, so we can use the line integral formula v(z) = \operatorname{Im} \int_0 ^z u_x(w) - iu_y(w) dw. Using this formula, we can write v(1/2) as v(1/2) = \operatorname{Im} \int_0 ^{1/2} u_x(w) - iu_y(w) dw.

To simplify this integral, we can use the Cauchy-Riemann equations, which state that u_x = v_y and u_y = -v_x. Substituting these into our integral, we get v(1/2) = \operatorname{Im} \int_0 ^{1/2} v_y(w) + iv_x(w) dw.

Now, we can use the fact that v(0) = 1 to simplify this integral even further. Since v is harmonic, we can use Poisson's Integral Formula for v as well, which gives us v(z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|e^{i\theta}-z|^2}v(e^{i\theta}) d\theta. Substituting this into our integral, we get v(1/2) = \operatorname{Im} \int_0 ^{1/2} v_y(w) + iv_x(w) dw = \operatorname{Im} \int_0 ^{1/2} \frac{1-|w|^2}{