Solution. \(\int _{-\infty }^\infty e^{-a|x| - ix\xi }\,dx = \frac 1{a + i\xi } + \frac 1{a - i\xi } = \frac {2a}{a^2 + \xi ^2}\). By inversion, \(e^{-a|x|} = \frac 1{2\pi }\int \frac {2a}{a^2 + \xi ^2}e^{ix\xi }\,d\xi \); exchanging the names of \(x\) and \(\xi \) and replacing \(x\) by \(-x\), \(\int \frac {a}{\pi (a^2 + x^2)}e^{-ix\xi }\,dx = e^{-a|\xi |}\). Put \(a = y\).

Solution. \(\hat u_t = (-\kappa \xi ^2 + ic\xi )\hat u\), so \(\hat u = \hat ge^{-\kappa \xi ^2t}e^{ic\xi t}\). The factor \(e^{ic\xi t}\) is a shift by \(-ct\), so \(u(x, t) = (\Phi (\cdot , t) * g)(x + ct)\): diffusion as before, carried to the left at speed \(c\).

Solution. \(\hat f(\xi ) = \int _{-1}^1e^{-ix\xi }\,dx = \frac {2\sin \xi }\xi \), and \(\int |f|^2 = 2\). Plancherel gives \(\int \frac {4\sin ^2\xi }{\xi ^2}\,d\xi = 2\pi \cdot 2\), so the integral is \(\pi \).

Solution. Substituting \(e^{i(kx - \omega t)}\): \(-i\omega + (ik)^3 = -i\omega - ik^3 = 0\), so \(\omega = -k^3\). The phase velocity is \(\omega /k = -k^2\) and the group velocity \(d\omega /dk = -3k^2\). A packet with \(k = 2\) moves to the left at speed \(12\), three times as fast as its crests, which move at speed \(4\).

Solution. \(\bar u = A(s)e^{-x\sqrt {s/\kappa }}\) and \(\bar u_x(0, s) = -A\sqrt {s/\kappa } = -q/s\), so \(A = q\sqrt \kappa \,s^{-3/2}\) and \(\bar u(0, s) = q\sqrt \kappa \,s^{-3/2}\). Since \(\overline {t^{1/2}} = \Gamma (\frac 32)s^{-3/2} = \frac {\sqrt \pi }2s^{-3/2}\), the surface temperature is \(u(0, t) = \frac {2q\sqrt {\kappa t}}{\sqrt \pi }\): it rises like \(\sqrt t\), not linearly, because the heat spreads into a layer of growing thickness.

Solution. \(\hat u = \hat ge^{\kappa \xi ^2t}\). Take \(g_n\) with \(\hat g_n\) equal to \(\varepsilon \) on \(n \leq \xi \leq n + 1\) and \(0\) elsewhere; then \(\|g_n\|^2 = \varepsilon ^2/2\pi \) by Plancherel, while \(\|u_n(\cdot , t)\|^2 \geq \frac 1{2\pi }\varepsilon ^2e^{2\kappa n^2t} \to \infty \). Data of fixed small size produce arbitrarily large solutions at any \(t > 0\).