Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 5A (Differential Equations)

Away from the impulse, the harmonic oscillator equation has a homogeneous solution. Before $x=\pi /2$, the initial conditions select $y=\cos x$. The jump condition for an impulse requires continuity of $y$ and a unit jump in $y^{\prime }$: a jump in $y$ would produce a derivative of the Dirac delta distribution, which is absent from the forcing, while integrating through the impulse gives $[y^{\prime }]=1$.

Just before the impulse, $y=0$ and $y^{\prime }=-1$. Just afterwards, $y=0$ and $y^{\prime }=0$, so uniqueness for the harmonic oscillator equation makes the subsequent solution identically zero. Equivalently the causal response is $H(x-\pi /2)\sin (x-\pi /2)$, with $H$ the Heaviside step function, and it cancels the original oscillation.

The solution isThe impulse cancellation of a harmonic oscillator produces a continuous graph with a corner at $(\pi /2,0)$, followed by a horizontal zero line.

With a positive cooling rate $k$, Newton's law of cooling gives $T^{\prime }=-k(T-T_0)$. The temperature excess $U=T-T_0$ satisfies the first-order linear differential equation $U^{\prime }=-kU$. Applying the initial condition to its exponential decay solution givesThe temperature approaches the ambient value $T_0$, rather than zero. The intended cooling situation has $T_0>0$ and $k>0$.

At the first half-temperature time, the solution of Newton's law of cooling givesSince $\alpha >2$, this time is positive and finite. The instantaneous temperature increase supplies the new initial condition $T(t_1+)=\alpha T_0/2+\beta$. Restart the same first-order linear differential equation at $t_1$:This is Newton cooling with an instantaneous temperature jump. The time translation in the exponential decay is essential: its prefactor is the excess immediately after the jump.

Each smooth segment follows Newton's law of cooling: $T^{\prime }=-k(T-T_0)<0$ and $T^{\prime \prime }=k^2(T-T_0)>0$. Thus the graph decreases and is convex on both sides of $t_1$. It starts at $\alpha T_0$, falls to $\alpha T_0/2$, jumps upward by $\beta$, and then decreases asymptotically to $T_0$.

The right-hand limit at the jump is above the original temperature, because $\beta >\alpha T_0/2$ implies $T(t_1+)>\alpha T_0$. The vertical dotted segment in the sketch denotes an instantaneous jump, not a continuous evolution through intermediate temperatures. The example uses $\alpha =4$, $\beta /T_0=3$; its qualitative shape holds for all the stipulated parameters.

Impose the target temperature on the post-jump exponential decay:Solving for the jump and eliminating $t_1$ using $e^{k{t}_1}=2(\alpha -1)/(\alpha -2)$ givesThis is the unique required temperature increase, and it satisfies the stipulated lower bound. Indeed $t_2>t_1$ impliesThus no additional restriction on the prescribed time is needed.