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My problem is the following:


Let $u$ be a continuous real-valued function in the closure of the unit disk $\mathbb{D}$ that is harmonic in $\mathbb{D}$. Assume that the boundary values of $u$ are given by $ u(e^{it}) = 5- 4 \cos t. $ Furthermore, let $v$ be a harmonic conjugate of $u$ in $\mathbb{D}$ such that $v(0) = 1$. Find $u(1/2)$ and $v(1/2)$.


It's easy to find $u(1/2)$ using the Poisson integral formula: $u(z) = \frac{1}{2\pi} \int_0 ^{2\pi} \frac{1-|z|^2}{|e^{i\theta}-z|^2} u(e^{i\theta}) d\theta$ yields $u(1/2) = \frac{1}{2 \pi} \int_0^{2\pi} \frac{3/4}{5/4-\cos \theta} (5-4\cos \theta) d\theta = \frac{1}{2\pi} \int_0^{2\pi} 3 d\theta = 3.$

I get stuck trying to find the value for $v$. I know that $ 1 = v(0) = \frac{1}{2\pi} \int_0^{2\pi} v(e^{i\theta}) d\theta.$ Also, $ v(1/2) = \frac{1}{2\pi} \int_0^{2\pi} \frac{3/4}{5/4 - \cos \theta} v(e^{i\theta}) d\theta = \frac{1}{2\pi} \int_0^{2\pi} \frac{3v(e^{i\theta})}{u(e^{i\theta})} d\theta, $ and in general, the harmonic conjugate is given by the line integral $ v(z) = \mathcal{Im} \int_0^z f'(w) dw + C.$ I don't know how to proceed with this information to find the $v$-value.

2 Answers 2

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HINT Look at $f(z) = 5 -4z + ic$, where $c \in \mathbb{R}$.

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    @ec92 The most general approach is to guess the right underlying analytic function and then find out the harmonic conjugate :).2012-06-27
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Note that $u(z)=5-4\Re(z)$ on $\partial D$ and that the function $z\mapsto5-4\Re(z)$ is linear hence harmonic everywhere. The function $f:z\mapsto5-4z$ is obviously holomorphic and $f=u+\mathrm iw$ with $w:z\mapsto-4\Im(z)$, hence, for every real number $c$, $v=w+c$ is a harmonic conjugate of $u$. The condition that $v(0)=1$ imposes $c=1$, hence $v(z)=1-4\Im(z)$, $u(1/2)=3$ and $v(1/2)=-1$.