Exponential Growth, Decay, and Newton Cooling

Definition With $t$ as the independent variable and $Q' = dQ/dt$, a quantity $Q$ has a rate of change proportional to itself if

for a constant $k$. The constant $k$ has units of inverse time, and wherever $Q \ne 0$ it equals the relative rate $Q'/Q$. Every solution is defined for all $t$ and has the form

where $Q_0 = Q(0)$. No solution is missed, because $(e^{-kt}Q)' = e^{-kt}(Q' - kQ) = 0$, so $e^{-kt}Q(t)$ is constant on any interval where $Q$ solves the equation.

Definition Let $Q' = kQ$ with $Q_0 > 0$. If $k > 0$, the solution $Q_0e^{kt}$ increases without bound, which is exponential growth. If $k < 0$, it decreases toward $0$ as $t \to \infty$ but stays positive at every finite time, which is exponential decay. If $k = 0$, it is constant. In every case, for all $t$ and every time step $\Delta t$,

so equal time steps multiply the amount by equal factors. On the phase line of $Q' = kQ$, the equilibrium $Q = 0$ is unstable for $k > 0$ and asymptotically stable for $k < 0$.

Definition For decay write $k = -\lambda$ with $\lambda > 0$, so $Q' = -\lambda Q$ and $Q(t) = Q_0e^{-\lambda t}$ with $Q_0 > 0$. The half-life $t_{1/2}$ is the time in which the amount falls to half its value. It solves $e^{-\lambda t_{1/2}} = \tfrac{1}{2}$, so

The same duration halves the amount from any starting time, since $Q(t + t_{1/2}) = \tfrac{1}{2}Q(t)$ for every $t$. After $n$ half-lives the amount is $Q_0/2^n$, and in general $Q(t) = Q_0\,2^{-t/t_{1/2}}$. For growth with $k > 0$ the doubling time $(\ln 2)/k$ plays the same role.

Definition For $Q' = -\lambda Q$ with $\lambda > 0$, the time constant is $\tau = 1/\lambda$, measured in units of time, so $Q(t) = Q_0e^{-t/\tau}$. Over each time constant the amount is divided by $e$,

so about $36.8$ percent remains and about $63.2$ percent is lost. The two time scales are related by $t_{1/2} = \tau\ln 2 \approx 0.693\,\tau$, so the time constant is about $1.443$ half-lives. The tangent line to the decay curve at any time reaches zero exactly one time constant later. For growth with $k > 0$, the same name is used for $1/k$, the time in which the amount grows by the factor $e$.

Definition Let $T(t)$ be the temperature of an object at time $t$, with $T' = dT/dt$, and let $A$ be the constant ambient temperature of its surroundings. Newton’s law of cooling is the model

where $k > 0$ is a constant with units of inverse time. If $T > A$ the rate is negative and the object cools, and if $T < A$ the rate is positive and it warms. The difference $D = T - A$ satisfies $D' = -kD$, so with $T_0 = T(0)$

The model assumes that the object has nearly the same temperature throughout, that heat flows at a rate proportional to the temperature difference, and that the surroundings stay at $A$.

Definition The ambient temperature is the temperature of the surroundings. Newton’s formula assumes it is a constant $A$, which is reasonable when the surroundings are large enough that the object does not noticeably heat or cool them. Then the equilibrium $T = A$ is asymptotically stable, and every solution tends to it as $t \to \infty$. If the ambient temperature changes with time, $A = A(t)$ with $A$ differentiable, the difference $D = T - A$ satisfies

so the constant-ambient formula no longer solves the equation unless $A'$ is zero. The equation $T' + kT = kA(t)$ is still linear, and the integrating factor $e^{kt}$ solves it.