Solution. \begin {align*} u(x) &= (1 - x)\int _0^xy^2\,dy + x\int _x^1(1 - y)y\,dy \\ &= \frac {(1 - x)x^3}3 + x\Bigl (\frac {1 - x^2}2 - \frac {1 - x^3}3\Bigr ) = \frac {x(1 - x^2)}6 . \end {align*}

Check: \(-u'' = -\frac 16(-6x) = x\), and \(u(0) = u(1) = 0\).

Solution. By the Poisson integral, \(u = \frac y\pi \int _{-\infty }^0\frac {dt}{(x - t)^2 + y^2} = \frac 1\pi \bigl (\frac \pi 2 - \arctan \frac xy\bigr ) = \frac \theta \pi \), where \(\theta \in (0, \pi )\) is the polar angle. It is harmonic, being the imaginary part of \(\frac 1\pi \log z\), and tends to \(0\) on \(\theta = 0\) and \(1\) on \(\theta = \pi \).

Solution. For \(x = (x_1, x_2)\) put images \(-1\) at \((-x_1, x_2)\) and \((x_1, -x_2)\) and \(+1\) at \((-x_1, -x_2)\): \[ G(x, y) = \Phi (y - x) - \Phi \bigl (y - (-x_1, x_2)\bigr ) - \Phi \bigl (y - (x_1, -x_2)\bigr ) + \Phi (y + x) . \] On \(y_1 = 0\) the first two terms cancel and the last two cancel, by reflection in that axis; likewise on \(y_2 = 0\). All images lie outside the quarter-plane.

Solution. The potential above the plane is \(\Phi (y - x) - \Phi (y - x^*)\) with \(\Phi = 1/4\pi |y|\). The charge density is minus the normal derivative of the potential at the plane (the normal field there), and \(\partial _{y_3}\) of the potential at \(y_3 = 0\) is \(2\cdot \frac {h}{4\pi (r^2 + h^2)^{3/2}}\), so \(\sigma = -h/2\pi (r^2 + h^2)^{3/2}\). Its total is \(-\int _0^\infty \frac {hr\,dr}{(r^2 + h^2)^{3/2}} = -1\). The force on the charge is the attraction of its image at distance \(2h\), \(1/4\pi (2h)^2 = 1/16\pi h^2\) in these units, towards the plane.

Solution. \begin {align*} u &= \int _0^\infty \bigl (\Phi (x - y, t) - \Phi (x + y, t)\bigr )\,dy \\ &= \frac 12\Bigl (1 + \operatorname {erf}\frac x{\sqrt {4\kappa t}}\Bigr ) - \frac 12\Bigl (1 - \operatorname {erf}\frac x{\sqrt {4\kappa t}}\Bigr ) = \operatorname {erf}\frac x{\sqrt {4\kappa t}} . \end {align*}

The cold end penetrates to the diffusion length \(\sqrt {4\kappa t}\).

Solution. \(u(0) = \int G(0, y)\,dy = -\frac 1{2\pi }\int _0^1\log r\cdot 2\pi r\,dr = -\int _0^1r\log r\,dr = \frac 14\), and \((1 - 0)/4 = \frac 14\); also \(-\Delta \frac {1 - r^2}4 = -\frac 14\bigl (-2 - 2\bigr ) = 1\).