Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 5

Write $a=s+ib$ with real $s,b$, and first translate the real integration variable by $s$. It remains to integrate $e^{-z^2}$ along the horizontal line $\mathrm{Im}z=b$.

For $b>0$, apply the Cauchy integral theorem to the rectangle with vertices $-R,R,R+ib,-R+ib$. The integrand is an entire function, sois the negative of the two vertical-edge integrals. On either vertical edge, $z=\pm R+iy$ andTheir combined absolute value is at most $2b{e}^{-R^2+b^2}$ and tends to zero. The rectangle below the real axis gives the same estimate with $|b|$ when $b<0$; $b=0$ needs no deformation. Absolute convergence follows from $|e^{-(x+ib{)}^2}|=e^{b^2}{e}^{-x^2}$. Taking $R\to \infty$ proves the Gaussian contour translation identity

Take $\lambda >0$ and $t>0$, the positive diffusivity requirement for the forward heat kernel. The PDF does not explicitly specify the sign of $\lambda$; the stated real Gaussian function formula requires this assumption. For a direct calculation, initially assume sufficiently smooth and rapidly decreasing data, so that the Fourier transform, its inverse and the following integrations are justified. The resulting heat-kernel convolution extends to wider classes of initial data.

Using the stated Fourier transform convention and integration by parts,The transformed ordinary differential equation has solution $\tilde{u}(k,t)=e^{-\lambda k^{2}t}{\tilde{u}}_0(k)$. Fourier inversion then givesComplete the square:Part (a), after scaling by $\sqrt{\lambda t}$, evaluates the remaining Gaussian integral even though its shift is imaginary. ThereforeThe heat kernel has integral one and is an approximate identity as $t\downarrow 0$, which recovers the initial condition. Mere decay at spatial infinity does not by itself imply every Fourier integral is absolutely convergent; the regularity assumption above supplies a rigorous setting for the derivation.

Put $s=\lambda t>0$. In the heat-kernel convolution, complete the real square asThe Gaussian integral therefore givesAt $t=0$ the same formula gives $\beta =\alpha$. Equivalently, ${\beta }^{-2}={\alpha }^{-2}+4\lambda t$: the variance of this normalized Gaussian function grows from $1/(2{\alpha }^2)$ to $1/(2{\alpha }^2)+2\lambda t$.

The initial profiles have unit integral and concentrate at zero as $\alpha \to \infty$. Precisely, for every smooth compactly supported test function $\phi$,Thus the limit is the Dirac delta distribution, rather than an ordinary function: pointwise the profiles tend to zero away from the origin and diverge at the origin. At every fixed $t>0$,The limiting evolved profile is the heat kernel for a unit point source.