Solution. \((r^k)'' + \frac 1r(r^k)' - \frac {k^2}{r^2}r^k = (k(k - 1) + k - k^2)r^{k - 2} = 0\). As the real part of \(z^3 = (x + iy)^3\), \(r^3\cos 3\theta = x^3 - 3xy^2\); check: \(\Delta (x^3 - 3xy^2) = 6x - 6x = 0\).

Solution. \(\cos ^2\theta = \frac 12 + \frac 12\cos 2\theta \), so \(u = \frac 12 + \frac 12r^2\cos 2\theta = \frac 12 + \frac 12(x^2 - y^2)\). On \(r = \frac 12\), \(u = \frac 12 + \frac 18\cos 2\theta \), with maximum \(\frac 58\), less than the boundary maximum \(1\), as the maximum principle requires.

Solution. Separating, \(u = \sum b_k\sin kx\,\sinh ky\), and matching at \(y = 1\) gives only \(k = 1\): \(u = \sin x\,\sinh y/\sinh 1\).

Solution. Let \(|u| \leq M\) and take points \(x\), \(y\). By the mean value property over balls of radius \(R\), \[ |u(x) - u(y)| = \frac 1{|B_R|}\Bigl |\int _{B_R(x)}u - \int _{B_R(y)}u\Bigr | \leq \frac {M\,|B_R(x)\,\triangle \,B_R(y)|}{|B_R|} , \] where \(\triangle \) is the symmetric difference. That set lies in the shell \(R - |x - y| \leq |z - x| \leq R + |x - y|\), whose volume divided by \(|B_R|\) is \(O(|x - y|/R)\). Letting \(R \to \infty \) gives \(u(x) = u(y)\).

Solution. By Harnack, \(\frac {1 - 1/2}{1 + 1/2} = \frac 13 \leq u \leq 3\). The Poisson kernel itself, \(u = 2\pi P(r, \theta - \varphi _0)\), the solution with all the boundary heat concentrated at the point \(\varphi _0\), is non-negative with \(u(0) = 1\), and at \(r = \frac 12\) equals \(3\) at \(\theta = \varphi _0\) and \(\frac 13\) opposite. Smooth non-negative boundary data concentrating at \(\varphi _0\) approach these values.

Solution. From the series, \(u_r(1, \theta ) = \sum k(a_k\cos k\theta + b_k\sin k\theta )\) has no constant term, so \(g\) must have mean \(0\) (this is the compatibility condition \(\int _{\partial \Omega }\partial _\nu u = \int _\Omega \Delta u = 0\)). For \(g = \cos \theta + \sin 2\theta \): \(a_1 = 1\), \(2b_2 = 1\), so \(u = C + r\cos \theta + \frac 12r^2\sin 2\theta = C + x + xy\), with \(C\) arbitrary.