Solution. Write \(s = 1 + 4\kappa t\). Then \(u_t = 4\kappa \bigl (-\frac 1{2s} + \frac {x^2}{s^2}\bigr )u\), and \(u_x = -\frac {2x}su\), \(u_{xx} = \bigl (-\frac 2s + \frac {4x^2}{s^2}\bigr )u\), so \(u_t = \kappa u_{xx}\). The data \(e^{-x^2}\) are \(\sqrt \pi \) times a normal density of variance \(\frac 12\); convolving with \(\Phi \), of variance \(2\kappa t\), gives \(\sqrt \pi \) times a normal density of variance \(\frac 12 + 2\kappa t = s/2\), which is \(u\).

Solution. The lower bound is the minimum principle, since \(u \geq 0\) on the parabolic boundary. For the upper bound, \(v = \pi ^{-1}e^{-\pi ^2t}\sin \pi x\) solves the heat equation and vanishes at the ends, so by comparison it suffices that \(x(1 - x) \leq \pi ^{-1}\sin \pi x\) on \([0, 1]\). Both sides are symmetric about \(\frac 12\); the difference \(d(x) = \pi ^{-1}\sin \pi x - x(1 - x)\) has \(d(0) = 0\) and \(d'(x) = \cos \pi x - (1 - 2x) \geq 0\) on \([0, \frac 12]\), because the concave function \(\cos \pi x\) lies above its chord \(1 - 2x\) there. So \(d \geq 0\).

Solution. For the difference \(w\) of two solutions, \(\frac {d}{dt}\int w^2 = 2\kappa \int w\Delta w = 2\kappa \int _{\partial \Omega }w\,\partial _\nu w\,dS - 2\kappa \int |\nabla w|^2 = -2\kappa \int |\nabla w|^2 \leq 0\), and \(\int w^2 = 0\) at \(t = 0\), so \(w = 0\). Also \(\frac {d}{dt}\int u = \kappa \int \Delta u = \kappa \int _{\partial \Omega }\partial _\nu u\,dS = 0\): an insulated body keeps its heat, and tends to the uniform temperature equal to its initial mean.

Solution. \(q = \sqrt {\omega /2\kappa }\) with \(\omega = 2\pi /86400 = 7.27 \times 10^{-5}\) s\(^{-1}\), so \(q = \sqrt {72.7} = 8.53\) m\(^{-1}\) and the depth is \(\ln 10/q = 0.27\) m. The phase lag there is \(qx = \ln 10 = 2.30\) radians, a time lag of \(2.30/\omega = 31\,700\) s, about \(8.8\) hours: the afternoon maximum arrives at \(27\) cm around midnight.

Solution. Setting \(\sqrt {4\kappa t} = L\) gives \(t = L^2/4\kappa \): for \(1\) m, \(t = 1/(4.8 \times 10^{-5}) = 20\,800\) s, about \(6\) hours; for \(1\) cm, \(10^{-4}\) of that, about \(2\) seconds. The quadratic dependence on \(L\) is why a thin pan heats quickly and a thick casting slowly.

Solution. The first is the scaling invariance; the second follows by differentiating the equation; the third is \(\frac {d}{d\lambda }u(\lambda x, \lambda ^2t)\) at \(\lambda = 1\), a limit of solutions. Applying the second to \(\Phi \) gives \(\Phi _x = -\frac {x}{2\kappa t}\Phi \), a constant multiple of \(xt^{-3/2}e^{-x^2/4\kappa t}\). It is the solution on the half-line \(x > 0\) with \(u(0, t) = 0\) produced by a dipole, a source and an equal sink at the origin, and it is the derivative of the kernel along the direction of the dipole.