Mc31 - Forming differential equation models

Mc31 - Forming differential equation models

Build first-order differential equations from information about rates of change. You will fit constants to observations, choose the right derivative, identify equilibrium states and assess assumptions in population, temperature, flow and demand models.

Expressing a rate as a differential equation

Does a larger population necessarily grow faster? Suppose a model gives the net rate 20−0.1N20-0.1N individuals per year when the population is NN. At N=80N=80, the model gives an increase of 1212 per year. At N=250N=250, it gives a decrease of 55 per year. The quantity is positive in both cases, but its rate changes sign. What population would stay constant, and what assumption is hidden in treating individuals as a smooth curve?

You already know

The derivative of a quantity with respect to time is its instantaneous rate of change. A negative derivative means the quantity decreases. An initial condition supplies a value at a stated input, rather than a rule for all rates.

Write the rate rule as an equation with a derivative. The notation tells us precisely what changes and what it changes with respect to.

First-order differential equation

A differential equation connects an unknown function with its derivatives. It is first-order if the highest derivative present is a first derivative. For example, dN/dt=20−0.1N\mathrm dN/\mathrm dt=20-0.1N is first-order. The equation N=20−0.1tN=20-0.1t instead states a particular linear function, with constant derivative −0.1-0.1.

You can use a differential equation at a known state without yet finding the entire function.

Worked example 1: a population and its rate

Question. A population has net rate of change 20−0.1N20-0.1N individuals per year. Initially N=80N=80. State the differential equation, the initial condition and the initial rate.

Plan. Put the net rate after the derivative, then use the initial population only to evaluate that rate.

dNdt=20−0.1Nt in yearsN(0)=80initial populationN′(0)=20−0.1(80)=12initial rate\begin{aligned} \frac{\mathrm dN}{\mathrm dt}&=20-0.1N && t\text{ in years} \\ N(0)&=80 && \text{initial population} \\ N'(0)&=20-0.1(80)=12 && \text{initial rate} \end{aligned}

Why this step? The value 8080 identifies one state; replacing every NN in the model by 8080 would incorrectly make the rate constant.

Answer. dN/dt=20−0.1N\mathrm dN/\mathrm dt=20-0.1N, N(0)=80N(0)=80, with initial increase 1212 individuals per year.

Check. Input 2020 exceeds the loss contribution 88 by 1212.

Change only the initial population to 250250. The same differential equation describes a different initial trend.

Worked example 2: the same law predicts decline

Question. The same rate rule 20−0.1N20-0.1N applies, but initially N=250N=250. State the initial condition and initial rate, and interpret its sign.

Plan. Keep the rate law and substitute the changed population.

N(0)=250new initial stateN′(0)=20−0.1(250)evaluate the same rule=−5net rate\begin{aligned} N(0)&=250 && \text{new initial state} \\ N'(0)&=20-0.1(250) && \text{evaluate the same rule} \\ &=-5 && \text{net rate} \end{aligned}

Why this step? A positive population does not force a positive derivative: the loss contribution is now larger than the input.

Answer. With N(0)=250N(0)=250, the initial rate is −5-5 individuals per year, meaning an initial decrease of 55 per year.

Check. The loss contribution is 2525, exceeding the input 2020 by 55.

A pure curve problem uses the same translation. “The gradient is twice the yy-coordinate minus the xx-coordinate” means dy/dx=2y−x\mathrm dy/\mathrm dx=2y-x. At (3,4)(3,4), it gives gradient 55; y(3)=4y(3)=4 is the separate point condition. Neither model requires us to solve the differential equation merely to find that one rate.

Watch out

Wrong: identify N(0)=80N(0)=80 with N′(0)=80N'(0)=80. Right: distinguish an initial population from an initial rate, and include their different units. Between two times, a change divided by elapsed time is an average rate, not automatically either endpoint's derivative.

Use another initial population in this diagnostic calculation.

Translating proportional growth and decline

A proportional rate law has two decisions: which expression controls the rate, and which sign describes the change. The constant is then fitted to a measurement.

Signed proportional rate

For a positive quantity MM, a rate of decrease proportional to MnM^n gives

dMdt=−kMn,k>0.\frac{\mathrm dM}{\mathrm dt}=-kM^n,\qquad k>0.

A rate of increase uses the positive sign. A reciprocal law requires M≠0M\ne0; keep the physical domain M>0M>0 when MM is a positive mass.

Not in the formula sheet: learn it.

The same observation can calibrate different proposed laws. It does not prove that either law remains valid at other states.

Worked example 3: calibrating a square law

Question. A mass MM, in grams, decreases at a rate proportional to its square. It loses 44 grams per day when M=8M=8. Form the differential equation.

Plan. Write the square dependence with a negative derivative, then determine its coefficient.

dMdt=−kM2the proposed loss law−4=−64kuse M=8k=116calibrate\begin{aligned} \frac{\mathrm dM}{\mathrm dt}&=-kM^2 && \text{the proposed loss law} \\ -4&=-64k && \text{use }M=8 \\ k&=\frac1{16} && \text{calibrate} \end{aligned}

Why this step? The measured loss is positive, but the mass's signed rate of change is negative.

Answer. dM/dt=−M2/16\mathrm dM/\mathrm dt=-M^2/16, with tt in days. The coefficient has units g−1 day−1\mathrm{g^{-1}\,day^{-1}}.

Check. At M=8M=8, −M2/16=−4-M^2/16=-4 grams per day.

Change only the dependence from the square of mass to its reciprocal. Keep the measured state and loss rate.

Worked example 4: calibrating a reciprocal law

Question. A mass M>0M>0 decreases at a rate inversely proportional to MM, losing 44 grams per day when M=8M=8. Form its differential equation.

Plan. Put MM in the denominator and fit the new constant to the same observation.

dMdt=−kMreciprocal dependence−4=−k8measured state and ratek=32solve for the constant\begin{aligned} \frac{\mathrm dM}{\mathrm dt}&=-\frac{k}{M} && \text{reciprocal dependence} \\ -4&=-\frac{k}{8} && \text{measured state and rate} \\ k&=32 && \text{solve for the constant} \end{aligned}

Why this step? The new constant is not 1/161/16: it belongs to a different expression and has different units.

Answer. dM/dt=−32/M\mathrm dM/\mathrm dt=-32/M, for M>0M>0, with coefficient units g2 day−1\mathrm{g^2\,day^{-1}}.

Check. At M=8M=8, the rate is −4-4. At M=16M=16, it is −2-2, whereas the square law would give −16-16.

For comparison, a square-law sample losing 22 grams per day at mass 55 gives k=2/25k=2/25. Each observation must be paired with its own state. If instead a curve's gradient is inversely proportional to positive xx, with gradient 33 at x=2x=2, the equation is dy/dx=6/x\mathrm dy/\mathrm dx=6/x. The controlling variable in the words determines the denominator.

Watch out

Wrong: interpret N′=0.08NN'=0.08N as an exact 8%8\% increase over a complete year. Right: N′/N=0.08N'/N=0.08 is an instantaneous relative rate, when time is in years and N>0N>0. The quantity, and hence the rate, changes during the year.

The next reciprocal model uses different measurements. Keep the decline sign throughout the calibration.

Combining rates and preserving equilibrium states

A restoring rate changes sign on opposite sides of a preferred state. A cooling model can therefore also describe warming, without changing its equation.

Equilibrium state

An equilibrium state is a constant value satisfying the differential equation. Its derivative is zero. In a balance of input and output, equilibrium means equal rates; it does not mean that both processes stop.

For an object at temperature TT in fixed surroundings at temperature SS, a supplied restoring assumption says that the signed rate is opposite to T−ST-S.

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