- Let u be a real-valued function defined on the unit disc D. Suppose that \(u\) is twice continuously diferentiable and harmonic, that is,
\[ \triangle u (x, y) = 0 \]
for all \(( x , y ) \in \mathbb { D }\)
- Prove that there exists a holomorphic function f on the unit disc such that
\[ \operatorname{Re} (f) = u. \]
Also show that the imaginary part of f is uniquely defined up to an additive (real) constant. [Hint: From the previous chapter we would have \(f ^ { \prime } ( z ) =\) \(2 \partial u / \partial z\) . Therefore, let \(g ( z ) = 2 \partial u / \partial z\) and prove that \(g\) is holomorphic. Why can one find \(F\) with \(F ^ { \prime } = g \ ?\) Prove that \(\mathrm { R e } ( F )\) difers from u by a real constant.]
- Deduce from this result, and from Exercise 11, the Poisson integral representation formula from the Cauchy integral formula: If u is harmonic in the unit disc and continuous on its closure, then if \(z = r e ^ { i \theta }\) one has
\[ u (z) = \frac {1}{2 \pi} \int_ {0} ^ {2 \pi} P _ {r} (\theta - \varphi) u (\varphi) d \varphi \]
where \(P _ { r } ( \gamma )\) is the Poisson kernel for the unit disc given by
\[ P _ {r} (\gamma) = \frac {1 - r ^ {2}}{1 - 2 r \cos \gamma + r ^ {2}}. \]
Solution
1. (a) There is a holomorphic \(f\) on \(\mathbb D\) with \(\Re f = u\), and \(\Im f\) is unique up to an additive real constant.
1.1. \(g = 2\frac{\partial u}{\partial z} = u_x - i u_y\) is holomorphic on \(\mathbb D\).
Since \(u\) is \(C^2\), \(g\) is \(C^1\) and \(\frac{\partial g}{\partial \bar z} = 2\frac{\partial^2 u}{\partial \bar z \partial z} = \frac{1}{2}\Delta u = 0\).
1.2. There is a holomorphic \(f\) on \(\mathbb D\) with \(\Re f = u\).
A holomorphic function on a disc has a primitive, so there is \(F\) with \(F' = g\). Then \(\partial_x\Re F = \Re F' = \Re g = u_x\) and \(\partial_y\Re F = \Re(iF') = \Re(ig) = u_y\), so \(\Re F - u\) has zero gradient on the connected set \(\mathbb D\) and equals a real constant \(c\). Put \(f = F - c\).
1.3. Q.E.D.
Step 2 gives existence. If \(\Re f_1 = \Re f_2 = u\), then \(f_1 - f_2\) is holomorphic with zero real part, hence constant by E-SS1.EX-13, so \(f_1 - f_2 = ic\) with \(c\) real.
2. (b) If \(u\) is harmonic in \(\mathbb D\) and continuous on \(\overline{\mathbb D}\), then for \(z=re^{i\theta}\), \(0\le r<1\), \(u(z) = \frac{1}{2\pi}\int_0^{2\pi} P_r(\theta - \varphi) u(e^{i\varphi})\, d\varphi\).
2.1. For \(r<\rho<1\), \(u(re^{i\theta}) = \frac{1}{2\pi}\int_0^{2\pi} \frac{\rho^2-r^2}{\rho^2-2\rho r\cos(\theta-\varphi)+r^2}\, u(\rho e^{i\varphi})\, d\varphi\).
By step 1, \(u = \Re f\) with \(f\) holomorphic on \(\mathbb D\). Part (a) of E-SS2.EX-11 with \(R=\rho\) expresses \(f(z)\) as the integral of \(f(\rho e^{i\varphi})\) against the real kernel \(\Re\frac{\rho e^{i\varphi}+z}{\rho e^{i\varphi}-z}\); taking real parts gives the same formula for \(u\), and part (b) of E-SS2.EX-11 evaluates the kernel.
2.2. Q.E.D.
2.1. Since \(u\) is uniformly continuous on \(\overline{\mathbb D}\), \(u(\rho e^{i\varphi})\to u(e^{i\varphi})\) uniformly in \(\varphi\), and the kernels converge uniformly in \(\varphi\) to \(P_r(\theta-\varphi)\) because \(\rho^2-2\rho r\cos\gamma+r^2\ge(\rho-r)^2\) stays bounded away from \(0\). Hence the integrals converge to \(\frac{1}{2\pi}\int_0^{2\pi}P_r(\theta-\varphi)u(e^{i\varphi})\,d\varphi\).