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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 4

Let
$$
g(x,t)=\frac1{\sqrt{4\pi Dt}}
\exp\left(-\frac{x^2}{4Dt}\right),
\qquad t>0.
$$
Convolution with the initial <Dirac delta function> translates the <heat kernel>, while part (b) propagates the impulsive source from time one. Hence
$$
\boxed{
\theta_c(x,t)=g(x-2\sqrt D,t)
-A H(t-1)g(x+2\sqrt D,t-1)}.
$$
At $(x,t)=(0,2)$ this becomes
$$
\theta_c(0,2)
=\frac{e^{-1/2}}{\sqrt{8\pi D}}
-A\frac{e^{-1}}{\sqrt{4\pi D}}.
$$
It vanishes exactly when
$$
A=\frac{e^{-1/2}}{\sqrt{8\pi D}}
\frac{\sqrt{4\pi D}}{e^{-1}}
=\boxed{\sqrt{\frac e2}}.
$$
This is the <cancellation of two heat-kernel impulses>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 14A / c / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 14A / c / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 14A / c / Solution

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