Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2022 / Paper 1 / 10 / b

Write $k=2\pi /\lambda$. For $T=A{e}^{\alpha t}\cos (ky)$, direct partial differentiation givesHence $T_t=\kappa T_{yy}$ exactly when $\alpha =-\kappa k^2$, proving the claimed diffusion equation solution. At $t=0$, $e^0=1$, so $T(0,y)=A\cos (2\pi y/\lambda )$. The infinite spatial domain needs no end-boundary conditions. For positive thermal diffusivity the amplitude decreases, as described by the thermal diffusion rate of a Fourier mode.

Choose $t_1=1/(\kappa k^2)$ to make the sketch explicit. The three profiles areThey have the same wavelength, zero crossings and phase, with amplitudes $A$, $0.3679A$ and $0.1353A$. Diffusion reduces the temperature gradient and amplitude; it does not advect or change this individual mode’s wavelength.

The diffusion equation is linear, so superposition givesEach term separately satisfies the equation, and their sum recovers the stated initial data at $t=0$. Its e-folding times are ${\tau }_i={\lambda }_i^2/(4{\pi }^2\kappa )$, so ${\lambda }_1<{\lambda }_2$ implies ${\tau }_1<{\tau }_2$. The shorter-wavelength oscillation loses a given fraction of its amplitude first. It approaches zero exponentially rather than disappearing at a finite instant. This preferential damping explains smoothing of fine-scale temperature structure.

For an integrable initial profile $f(y)$, use its Fourier transform with the conventionThe transformed diffusion equation is ${\partial }_t\hat{T}=-\kappa k^2\hat{T}$, soEquivalently the heat-kernel convolution isThis also accommodates bounded piecewise-continuous initial profiles; it recovers $f$ at continuity points as $t\downarrow 0$. For periodic initial data use a Fourier series and multiply each mode by $e^{-\kappa k_r^{2}t}$; the constant mean mode remains unchanged. “Any function” needs the usual existence/integrability or growth conditions: an arbitrarily fast-growing initial function need not give a finite heat-kernel integral.