Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 13

Set $T=t+a>0$. The similarity solution uses $y=(x+b)T^{-1/2}$, whose partial derivatives at fixed $x$ or $t$ areApply the multivariable chain rule to $u=T^{-1/2}v(y)$:Substituting in the heat equation and cancelling the positive common factor $T^{-3/2}$ gives

For a Gaussian function, the required derivatives areSubstitute these into the reduced ordinary differential equation and compare the constant and quadratic coefficients:The constant terms give $c=1/(4\lambda )$, and this value also satisfies the quadratic terms. HenceA constant multiple is also a solution by linearity.

The heat equation is linear and homogeneous in $u$: the operator $L={\partial }_t-\lambda {\partial }_x^2$ satisfies $L(A{u}_1+B{u}_2)=AL{u}_1+BL{u}_2$. Thus the superposition principle applies to its solutions. Spatial translations are also allowed because the coefficients are constant.

Choose $a=1/(4\lambda )$ in part (b) and multiply by $\sqrt{a}$ to make the initial amplitude one. Taking the two shifts $b=1$ and $b=-1$ then givesAt $t=0$ this reproduces both initial peaks. Each term is an instance of Gaussian initial data for the heat equation, so their sum satisfies the equation. Its extension to negative times is a real smooth solution precisely on the connected intervalAt the lower endpoint the peaks narrow without bound and their heights diverge; there is no finite classical continuation there. For positive times each peak spreads and decreases in height while preserving its integral.