Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 5

Differentiate the spatial Fourier transform under the integral, and integrate the second spatial derivative by parts twice. For sufficiently decaying temperature and first derivative, the boundary terms vanish:Hence the transformed heat equation is the ordinary differential equation

Solving the transformed ordinary differential equation with its initial condition gives $\tilde{\theta }(k,t)=e^{-k^{2}t}\tilde{\theta }(k,0)$. The Fourier inversion theorem therefore yieldsFor $t>0$ the Gaussian multiplier suppresses large spatial frequencies, which explains the smoothing produced by heat conduction.

The convolution theorem turns the product $e^{-k^{2}t}\tilde{\theta }(k,0)$ into a spatial convolution. Its kernel isComplete the square:After rescaling the variable, the supplied translated Gaussian integral givesThis is the heat-kernel solution for diffusivity one. Its kernel integrates to one, preserving the total temperature integral, and approaches a Dirac delta function as $t\downarrow 0$. Its variance is $2t$; the factor is appropriate to ${\theta }_t={\theta }_{xx}$, rather than the alternative convention with diffusivity $1/2$.

For the Gaussian initial data for the heat equation, substitute the initial profile into the heat-kernel solution. Completing the square in $y$ givesThe remaining Gaussian integral yieldsThe centre stays at $x_0$ and the squared width grows linearly with time.

For the interval pulse, the convolution reduces to $B{\int }_a^{b}G(x-y,t)dy$. Using $u=(y-x)/(2\sqrt{t})$ and the error function givesThis is the translated and scaled heat equation with interval-indicator initial data. At its jump points the $t\downarrow 0$ limits are $B/2$; away from them the initial data are recovered pointwise.

For fixed $a<x<b$ at late time, both error function arguments are small. Since $\mathrm{erf}u=2u/\sqrt{\pi }+O(u^3)$,The estimate is uniform for $x\in [a,b]$. Equivalently, on this fixed interval the Gaussian in the convolution is $1+O((b-a{)}^2/t)$.

The spreading distance is more apparent on the growing spatial scale. For the Gaussian pulse the exponent is order one when $|x-x_0|=O(\sqrt{t})$. For the interval pulse, set $x=c+\xi \sqrt{t}$ with $c=(a+b)/2$ and fixed $\xi$. Uniformly for bounded $\xi$, its convolution givesThus both nonzero pulses spread over a distance $O(\sqrt{t})$, while their heights are $O(t^{-1/2})$ and their total integrals remain constant. More quantitatively, the variances of their normalized profiles are $1/(2a_0)+2t$ and $(b-a{)}^2/12+2t$, respectively, confirming the same width law.