Growth and Decay with Differential Equations

An exponential model can be derived from an assumption about the rate of change. If change is proportional to the current amount, a differential equation yields exponential growth or decay. If change is proportional to a difference from equilibrium, the exponential applies to that difference instead.

The numerical scenarios below are stipulated teaching models. They illustrate derivation and interpretation, not predictions about a particular physical or biological system.

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Derive a constant proportional-rate law

Suppose

Q’=kQ,\qquad Q(0)=Q_0.

For a nonzero solution, separation gives $\ln|Q|=kt+C$ . Applying the initial condition yields

Q(t)=Q_0e^{kt}.

The equilibrium is included when $Q_0=0$ . For positive initial amount, positive $k$ gives growth and negative $k$ gives decay. The parameter has inverse-time units, making $kt$ dimensionless.

A continuous proportional rate of $0.05$ per year gives annual factor $e^{0.05}$ , not exactly $1.05$ . An annual factor of $1.05$ corresponds instead to $k=\ln(1.05)$ per year. Continuous rates and per-period percentage changes must not be conflated.

Recover a rate from two observations

Suppose $Q(0)=200$ and $Q(3)=250$ , with time in hours. Under the exponential assumption,

250=200e^{3k},\qquad k=\frac{\ln(5/4)}3\approx0.0743812\text{ h}^{-1}.

The resulting model recovers both observations exactly. They determine a parameter within that model; they do not establish that the rate remained constant between measurements or will remain constant afterwards.

Additional observations, residuals and knowledge of the mechanism help assess the assumption. Systematic residual patterns can indicate a missing baseline or changing rate. Verified evaluation of a formula is not empirical validation.

Derive doubling time and half-life

For $k>0$ , doubling requires $e^{kT}=2$ , giving

T_{\mathrm d}=\frac{\ln2}{k}.

The fitted model has doubling time $3\ln2/\ln(5/4)\approx9.31885$ hours. It is not twelve hours: repeated proportional growth is not repeated addition of fifty units.

For decay, write $Q’=-\lambda Q$ with $\lambda>0$ . Then

T_{1/2}=\frac{\ln2}{\lambda},\qquad Q(t)=Q_0\left(\frac12\right)^{t/T_{1/2}}.

An initial amount of eighty with a four-hour half-life leaves ten after twelve hours. A fixed half-life follows from the constant-rate model; it is not a property of every decreasing function.

Model the excess above a baseline

Let a numerical Celsius temperature reading $\theta$ satisfy

\theta’=-k(\theta-\theta_a),

with constant ambient reading $\theta_a$ and positive $k$ . Setting $u=\theta-\theta_a$ gives $u’=-ku$ , hence

\theta(t)=\theta_a+(\theta_0-\theta_a)e^{-kt}.

The exponentially decaying quantity is the temperature difference, not the Celsius reading itself. Celsius and kelvin temperature differences have the same numerical size, although their zero points differ.

With ambient reading twenty, initial reading eighty and a five-minute half-life for the excess,

\theta(t)=20+60\,2^{-t/5}.

At $0,5,10,15,20$ minutes, the readings are $80,50,35,27.5,23.75$ . They approach twenty, not zero.

Cooling towards a fixed baseline

The dashed line is the ambient reading.

Diagram explanation and exact data

The excess, not the Celsius reading itself, halves every five minutes.

Cooling towards a fixed baseline: exact reference data
Time (min)	Reading	Excess
0	80	60
5	50	30
10	35	15
15	27.5	7.5
20	23.75	3.75

Solve a threshold and interpret its timing

The crossing of twenty-five satisfies

20+60\,2^{-t/5}=25,\qquad t=5\frac{\ln12}{\ln2}\approx17.9248\text{ min}.

The model is strictly decreasing above ambient, so it stays at or below twenty-five afterwards. If observations occur only at whole minutes, the first scheduled reading meeting the threshold is at minute eighteen. Continuous crossing time and first sampled time are different quantities.

A threshold below ambient is never reached from above by this model. Taking a logarithm of a negative ratio would not produce a meaningful real crossing.

Derive mixing from an amount balance

A tank holds $100$ L, with equal inflow and outflow of $2$ L/min. Incoming concentration is $3$ g/L; initially none of the substance is present. Under perfect mixing,

Q’=6-\frac Q{50},\qquad Q(0)=0.

The equilibrium amount is three hundred grams. With $u=Q-300$ , the equation becomes $u’=-u/50$ , so

Q(t)=300(1-e^{-t/50}).

Its derivative is $6e^{-t/50}=6-Q/50$ , verifying the equation. The limiting concentration is $300/100=3$ g/L, matching the inflow.

If inflow and outflow differ, volume is no longer constant and the outflow concentration term changes. If mixing is incomplete, the outflow may not match the tank-wide average. The balance equation depends on those assumptions.

Recognise limits to unrestricted growth

A logistic model uses

Q’=rQ\left(1-\frac QK\right),\qquad r>0,\ K>0.

For initial amounts between zero and $K$ , growth slows towards the model equilibrium $K$ . This parameter is not a guarantee about a changing environment. The logistic-equation reference develops its assumptions and solutions.

Completing the square gives

rQ\left(1-\frac QK\right)=\frac{rK}{4}-\frac rK\left(Q-\frac K2\right)^2.

Thus the greatest amount-growth rate on $[0,K]$ occurs at $Q=K/2$ and equals $rK/4$ . The proportional rate $Q’/Q$ is a different quantity and decreases as the amount grows.

Allow the proportional rate to vary

For continuous $k(t)$ and initial value at $t_0$ ,

Q’=k(t)Q\quad\Longrightarrow\quad Q(t)=Q_0\exp\left(\int_{t_0}^t k(s)\,ds\right).

For piecewise-constant rates, multiply the interval growth factors or add their exponents. Averaging rates without respecting durations can be wrong. A new interval begins from the previous interval’s final amount unless an explicitly modelled jump occurs.

Practice questions

A quantity starts at ninety-six with a six-hour half-life. Find the predicted amount after eighteen hours.
What is halved every five minutes in the cooling example?
Explain the limiting amount and concentration in the mixing example.

Check your answers

Three half-lives leave $96(1/2)^3=12$ .
The excess above ambient twenty, not the Celsius reading itself. The excess changes from sixty to thirty to fifteen.
The exponential term tends to zero, so the limiting amount is three hundred grams. In one hundred litres, that is three grams per litre, matching inflow concentration.

Exponential Growth and Decay with Differential Equations