3.8 - Interpreting differential-equation models

3.8 - Interpreting differential-equation models

Two models can agree with the same starting value and initial rate, yet make different predictions later. In this final part of 3.8, you will build simple rate models, solve and interpret them, and judge where their predictions remain meaningful. The context will supply the assumptions; no specialist knowledge is needed.

From a situation to a model

If two models fit the same first measurement, must they predict the same future? Consider a message spreading through a fixed group of 120120 people. Let NN be the number who have heard it, treating NN as continuous, and let tt be time in hours. A model must say what controls the rate at which new people hear it.

One possibility is that the rate depends on both the number who can pass it on and the number still available to hear it. Another is that only the number remaining matters. These are different assumptions, even if their constants can be chosen to fit the same data.

Differential-equation model

A differential-equation model expresses an assumed rate rule in terms of variables and derivatives. Define the variables and units, translate the rule, then use the data to evaluate its constants. A solution describes the consequences of those assumptions.

The number who have not heard the message is 120−N120-N, not 120120. This connects the changing quantity to the fixed total.

Worked example 1: a product of two changing numbers

Question. The rate at which people hear a message is proportional to the product of the number who have heard it and the number who have not. In a fixed group of 120120, initially 2020 have heard it and the rate is 1010 people per hour. Form the differential equation for NN.

Plan. Translate the product rule with an unknown constant, then use the initial measurement.

dNdt=kN(120−N),10=k(20)(100).\frac{\mathrm dN}{\mathrm dt}=kN(120-N), \qquad10=k(20)(100).

Therefore k=1/200k=1/200.

Answer. dNdt=N(120−N)200\displaystyle\frac{\mathrm dN}{\mathrm dt}=\frac{N(120-N)}{200}, with tt in hours. The constant has units person−1 h−1\mathrm{person^{-1}\,h^{-1}}.

Check. Substituting N=20N=20 gives 20×100/200=1020\times100/200=10. The rate is positive for 0<N<1200<N<120, and zero at N=0N=0 or N=120N=120, as the assumed product requires.

Keep the same group, starting number and measured rate. Change only the rate assumption by removing the factor representing people who have already heard the message.

Worked example 2: the number remaining

Question. Instead, suppose the rate is proportional only to the number who have not heard the message. Again N=20N=20 initially and the rate is 1010 people per hour. Form the new equation for a fixed group of 120120.

Plan. Use 120−N120-N as the single proportional factor and evaluate a new constant.

dNdt=c(120−N),10=c(100)⟹c=0.1.\frac{\mathrm dN}{\mathrm dt}=c(120-N), \qquad10=c(100)\quad\Longrightarrow\quad c=0.1.

Answer. dN/dt=0.1(120−N)\mathrm dN/\mathrm dt=0.1(120-N), where cc has units h−1\mathrm{h^{-1}}.

Check. The initial rate is again 1010. At N=0N=0, however, this model gives rate 1212, whereas the product model gives zero. They express different mechanisms.

Both models assume that people remain in the group and that hearing the message is permanent. They also smooth a count into a continuous variable: a predicted N=35.6N=35.6 is an approximation, not a claim that a fraction of a person literally heard it.

To establish a requested equation, start with the stated rate relationship and calculate its constant. Substituting measurements into an already supplied equation checks consistency, but does not derive that equation from the assumptions.

Watch out

Do not replace a stated product by a sum: “proportional to NN and 120−N120-N multiplied together” gives kN(120−N)kN(120-N), not k[N+(120−N)]k[N+(120-N)]. The sum would be constant and describe a different rate rule.

Form the product model with a new total.

Interpreting the solution

Solving a model gives an expression, but interpreting it means connecting every part of that expression back to the situation. Check the initial value, the time interval being modelled, the units and any long-term limit. Here the clock starts when 2020 people have heard the message, so the intended time domain is t≥0t\ge0.

Worked example 3: solving and interpreting the product model

Question. Solve dN/dt=N(120−N)/200\mathrm dN/\mathrm dt=N(120-N)/200, given N=20N=20 at t=0t=0, and interpret its long-term prediction.

Plan. Separate variables, use partial fractions, then apply the initial condition. The relevant branch has 0<N<1200<N<120.

The constant solutions N=0N=0 and N=120N=120 do not fit the initial value. For the required branch,

dNN(120−N)=dt2001120(1N+1120−N)=1N(120−N)ln⁡N120−N=0.6t+C\begin{aligned} \frac{\mathrm dN}{N(120-N)}&=\frac{\mathrm dt}{200}\\ \frac1{120}\left(\frac1N+\frac1{120-N}\right)&=\frac1{N(120-N)}\\ \ln\frac{N}{120-N}&=0.6t+C \end{aligned}

At t=0t=0, C=ln⁡(20/100)=−ln⁡5C=\ln(20/100)=-\ln5. Exponentiating and rearranging gives

N120−N=15e0.6t,N=1201+5e−0.6t.\frac{N}{120-N}=\frac15e^{0.6t}, \qquad N=\frac{120}{1+5e^{-0.6t}}.

Answer. N=120/(1+5e−0.6t)N=120/(1+5e^{-0.6t}), for t≥0t\ge0. It increases from 2020 towards 120120. The model predicts that the whole group is approached as a limit; NN is less than 120120 at every finite time.

Check. At t=0t=0, N=120/6=20N=120/6=20. Let u=5e−0.6tu=5e^{-0.6t}. Then N′=72u/(1+u)2N'=72u/(1+u)^2, while N(120−N)/200=72u/(1+u)2N(120-N)/200=72u/(1+u)^2, verifying the equation.

Now solve the alternative rate law using the same initial condition. This changes the predicted curve even though the starting value and initial gradient match.

Worked example 4: solving and interpreting the remaining-number model

Question. Solve dN/dt=0.1(120−N)\mathrm dN/\mathrm dt=0.1(120-N), given N=20N=20 at t=0t=0, and interpret its long-term prediction.

Plan. Separate and integrate the reciprocal of 120−N120-N, then determine the constant.

∫1120−N dN=∫0.1 dt−ln⁡∣120−N∣=0.1t+C120−N=Ae−0.1t\begin{aligned} \int\frac1{120-N}\,\mathrm dN&=\int0.1\,\mathrm dt\\ -\ln|120-N|&=0.1t+C\\ 120-N&=Ae^{-0.1t} \end{aligned}

The initial value gives A=100A=100, so

N=120−100e−0.1t.N=120-100e^{-0.1t}.

Answer. N=120−100e−0.1tN=120-100e^{-0.1t}, for t≥0t\ge0. It also starts at 2020 and tends to 120120, remaining below 120120 at finite times.

Check. At t=0t=0, N=20N=20 and N′=10N'=10. For every tt, N′=10e−0.1t=0.1(120−N)N'=10e^{-0.1t}=0.1(120-N). The excluded constant N=120N=120 solves the equation but does not satisfy the initial condition.

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