Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 20

Write $\Psi (\eta ,\xi )=\psi (x,t)$. The inverse coordinate formulas $x=(\eta +\xi )/2$ and $t=(\eta -\xi )/(2c)$ are valid because $c>0$. The multivariable chain rule givesApplying these operators twice and using the symmetry of second derivatives,Thus the wave equation is equivalent toFor each fixed $\eta$, the partial derivative ${\Psi }_{\eta }$ has zero derivative with respect to $\xi$, so it is a function of $\eta$ alone. Taking an antiderivative in $\eta$ gives $\Psi =f(\eta )+g(\xi )$, where the integration constant may depend on $\xi$. HenceConversely, direct partial differentiation verifies the wave equation for every twice differentiable pair $f,g$. The two terms describe profiles moving left and right respectively at wave speed $c$ along the characteristic lines for the one-dimensional wave equation.

At $t=0$, the initial conditions implyDifferentiate the first relation and solve for the two derivatives:Fix a reference point $x_{\ast }$ and set $V(x)={\int }_{x_{\ast }}^{x}v(s)ds$. Taking antiderivatives givesThe constants must be opposite to preserve the first initial condition; the arbitrary constant of integration $K$ does not change $\psi$. Substitution yields D'Alembert's formula:The fundamental theorem of calculus verifies that its time derivative at zero is $v(x)$, and its value at zero is $u(x)$. The oriented integral also makes the formula valid for negative $t$.

An antiderivative of the initial velocity is $-2c{e}^{-s^2/2}$. Thus D'Alembert's formula givesThis is a positive Gaussian function moving right and a negative Gaussian function moving left. Their cancellation at $t=0$ gives the zero initial displacement, while direct differentiation gives ${\psi }_t(x,0)=2cx{e}^{-x^2/2}$.

For every $t>0$, the profile is an odd function of $x$, vanishes at $x=0$, is positive for $x>0$ and negative for $x<0$, and tends to zero at both spatial infinities. The signs follow from $(x+ct{)}^2-(x-ct{)}^2=4xct$. The positive and negative lobes move farther apart as $t$ increases; for large separation they approach the individual unit-height Gaussian functions centred at $x=\pm ct$. At finite separation the actual extrema lie slightly beyond these centres: writing $a=ct$, the positive global maximum occurs at the unique $x>a$ satisfying $(x-a)/(x+a)=e^{-2ax}$.