Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 5A (Differential Equations)

Measure temperature relative to ambient by ${\theta }_i=T_i-T_{\infty }$. Newton's law of cooling models ordinary cooling by ${\theta }_i^{\prime }=-a{\theta }_i$, since heat loss is proportional to the excess temperature. The additional negative forcing represents the cooling from the milk, with the temperature drop or rate normalized to one in the chosen units.

The Dirac delta function has unit integral and is supported at the instant of addition. It therefore imposes a sudden unit drop for cup 1. The difference $H(t-1)-H(t-2)$ of Heaviside step functions is one during the addition interval and zero outside it, giving a unit cooling rate for cup 2. The second temperature is continuous; only its derivative changes at the switching times. The value assigned to a Heaviside step function exactly at its jump does not change the integrated solution. This is the impulse versus finite-duration cooling comparison.

Before time one, both forcing terms vanish. The separable differential equation ${\theta }_i^{\prime }=-a{\theta }_i$ gives ${\theta }_i(t)=C_i{e}^{-at}$. The initial values give $C_1=C_2=T_0-T_{\infty }$, and henceIn particular the common left-hand limit at time one is $T_{\infty }+(T_0-T_{\infty })e^{-a}$.

Integrate the first equation over $(1-\epsilon ,1+\epsilon )$. The integral of the ordinary cooling term tends to zero as $\epsilon \downarrow 0$, while the Dirac delta function integrates to one. Thus the jump condition for an impulse isAfter the jump there is no further forcing, so Newton's law of cooling givesEquivalently,The last term has jump minus one at time one and then decays at the same exponential rate as the unforced temperature.

On the active interval the excess temperature satisfies ${\theta }_2^{\prime }+a{\theta }_2=-1$. Multiplication by the integrating factor $e^{at}$ and integration from time one givesThere is no impulse in this equation, so continuity supplies ${\theta }_2(1)=(T_0-T_{\infty })e^{-a}$. ThereforeThe equivalent correction to the common unforced temperature is $-(1-e^{-a(t-1)})/a$, which vanishes continuously at the beginning of the interval.

At time two, continuity of the second temperature givesThe forcing then turns off, so Newton's law of cooling again gives exponential decay:Subtracting this from the first temperature and expressing both corrections with the same exponential factor givesBecause $e^a>1+a$ for $a>0$, this difference is positive. Thus the cup cooled instantaneously is warmer after the addition interval: its cooling occurred earlier and has undergone more subsequent exponential attenuation.

During the addition interval, subtracting the formulas in (ii) and (iii) cancels the common unforced temperature:This is zero precisely when $e^{-a(t-1)}=1/(1+a)$, soThe inequalities $0<\log (1+a)<a$ give $1<t^{\ast }<2$, so using the active-interval formula was justified. The difference has derivative $(1+a)e^{-a(t-1)}>0$, making this crossing unique within that interval. Immediately after time one it is minus one; after time two it is strictly positive by (iv). Hence there is no other equality at any finite time after time one. Both temperatures approach the same ambient limit, but that limiting equality is not an additional crossing.