12. Application of Non Linear Equations

Dated: 15-11-2024

Logistic Equation

The logistic model is also called Verhulst-Pearl model.
Suppose \(a > 0\) is the constant average rate of birth and that death rate is proportional to the population \(P(t)\) at any time \(t\).
Thus if \(\frac 1 P \cdot \frac {dP}{dt}\) is the rate of growth per individual then

\[ \frac{dP}{dt} = P(a-bP) \]

The term \(-bP^2\) where \(b > 0\) can be interpreted as inhibition term.
When \(b = 0\) then the equation reduces to the one in exponential model.¹

Solution to Logistic Equation

For the constant solutions

\[P (a - bP) = 0\]

\[\implies P = \frac a b\]

For non constant solutions,

\[ \frac{dP}{P(a-bP)} = dt \]

Resolve into partial fractions,² we have

\[ \left[\frac{\frac 1 a}{P} + \frac{\frac b a}{a-bP}\right] dP = dt \]

\[ \frac{1}{a}\ln|P|-\frac{1}{a}\ln|a-bP|=t+C \]

\[ \ln\left|\frac{P}{a-bP}\right|=at+aC \]

\[ \frac{P}{a-bP} = C_{1}e^{at} \quad \text{where } C_1 = e^{aC} \]

\[ P(t)=\frac{aC_{1}e^{at}}{1+bC_{1}e^{at}}=\frac{aC_{1}}{bC_{1}+e^{-at}} \]

If we are given any initial value problem such that

\[P(0) = P_0, \quad P_0 \ne \frac a b\]

We get

\[ C_{1}=\frac{P_{0}}{a-bP_{0}} \]

Plugging it back in, we get

\[ P(t)=\frac{aP_{0}}{bP_{0}+(a-bP_{0})e^{-at}} \]

It is clear that there is limited growth as \(t \to \infty\).

\[ \lim_{t\rightarrow\infty}P(t)=\frac{aP_{0}}{bP_{0}}=\frac{a}{b} \]

Special Cases of Logistic Equation

Epidemic Spread

Suppose that one person infected from a contagious disease is introduced in a fixed population of \(n\) people.

Assumption

Rate of spread of disease \(\frac {dx}{dt}\) is proportional to number of infected people \(x(t)\) and number of uninfected people \(y(t)\).

\[\frac {dx}{dt} = kxy\]

\[x + y = n + 1\]

Therefore,

\[ \frac{dx}{dt}=kx(n+1-x), \quad x(0)=1 \]

This is used for

• Spread of information
• Impact of advertising

Modification of Logistic Equation

\[ \frac{dP}{dt}=P(a-b\ln P) \]

This equation is used in studies of

• Solid tumors
• Actuarial Predictions
• Growth of revenue from the sale of a commercial product
• Growth or decline of population

Example

A flu virus spreads on a college campus of 1000 students at a rate proportional to the product of infected (\(x\)) and uninfected students \((1000 - x)\).
If \(x(4) = 50\), determine the number of infected students after \(6\) days.

Solution

\[ \frac{dx}{dt}=kx(1000-x), \quad x(0)=1 \]

\[a = 1000k \text { and } b = k\]

Therefore, the solution is

\[ P(t)=\frac{aP_{0}}{bP_{0}+(a-bP_{0})e^{-at}} \]

\[ x(t)=\frac{1000k}{k+999ke^{-1000kt}}=\frac{1000}{1+999e^{-1000kt}} \]

Now using \(x(4) = 50\), we determine \(k\).

\[ 50=\frac{1000}{1+999e^{-4000k}} \]

\[ k=\frac{-1}{4000}\ln\frac{19}{999}=0.0009906 \]

\[ x(t)=\frac{1000}{1+999e^{-0.9906t}} \]

\[ x(6)=\frac{1000}{1+999e^{-5.9436}}=276 \text{ students} \]

Chemical Reactions

In a first-order chemical reaction, substance \(A\) decomposes into smaller molecules at a rate proportional to the remaining amount of \(A\).
Radioactive decay is an example of this type of reaction.
If \(X\) is the remaining amount of substance \(A\) at any time \(t\) then

\[\frac {dX}{dt} = kX\]

\(k < 0\) because \(X\) is decreasing.

In a second-order reaction, chemicals \(A\) and \(B\) react to form chemical \(C\) at a rate proportional to the product of their remaining concentrations.
If \(X\) is the amount of \(C\) formed at time \(t\), the unconverted amounts of \(A\) and \(B\) are \(\alpha - X\) and \(\beta - X\), respectively.
Hence the rate of formation of chemical \(C\) is

\[ \frac{dX}{dt}=k(\alpha-X)(\beta-X) \]

Example

A compound \(C\) forms when chemicals \(A\) and \(B\) react in a ratio of \(1:4\).
If \(30\) grams of \(C\) are formed in \(10\) minutes, and the reaction rate is proportional to the product of the remaining amounts of \(A\) and \(B\), determine the amount of \(C\) at any time, given initial amounts of \(50\) grams of \(A\) and \(32\) grams of \(B\).
Find \(C\) at \(15\) minutes and interpret as \(t \to \infty\).

Solution

If \(X(t)\) denotes the number of grams of chemical \(C\) present at time \(t\), then

\[X(0) = 0 \text{ and } X(10) = 30\]

Suppose there are \(2\) grams of compound \(C\) and we have used \(\alpha\) grams of \(A\) and \(b\) grams of \(B\) then

\[a + b = 2 \text{ and } b = 4a\]

Solving these equations, we get

\[a = 2 \cdot \frac 1 5\]

\[b = 2 \cdot \frac 4 5\]

If there were \(X\) amount of \(C\) then

\[a = \frac X 5 \text{ and } b = \frac 4 5 \cdot X\]

Therefore, the amount of \(A\) and \(B\) remaining at any time \(t\) are

\[50 - \frac X 5 \text{ and } 32 - \frac 4 5 \cdot X\]

Therefore,

\[ \frac{dX}{dt}=\lambda\left(50-\frac{X}{5}\right)\left(32-\frac{4}{5}X\right) \]

\[ \frac{dX}{dt}=k(250-X)(40-X), \quad k=\frac {4\lambda}{25} \]

Solving the equation

\[ \frac{dX}{(250-X)(40-X)}=kdt \]

\[ -\frac{1}{210}\frac{dX}{250-X}+\frac{1}{210}\frac{dX}{40-X}=kdt \]

\[ \ln\left|\frac{250-X}{40-X}\right|=210kt+c_{1} \]

\[ \frac{250-X}{40-X}=c_{2}e^{210kt} \quad \text{ where } \quad c_{2}=e^{c_{1}} \]

When \(t = 0\), \(X = 0\), so it follows at this point that \(c_2 = \frac {25} 4\).
Using \(X = 30\) at \(t = 10\), we find

\[ 210k=\frac{1}{10}\ln\frac{88}{25}=0.1258 \]

We will solve for \(X\)

\[ X(t)=1000\left(\frac{1-e^{-0.1258t}}{25-4e^{-0.1258t}}\right) \]

It is clear that \(e^{-0.1258t} \to 0\) as \(t \to \infty\).
Therefore, \(X \to 40\) as \(t \to \infty\).

t	X
10	30
15	34.78
20	37.25
25	38.54
30	39.22
35	39.59

\[ 50-\frac{1}{5}(40)=42 \text{ grams of chemical } A \]

\[ 32-\frac{4}{5}(40)=0 \text{ grams of chemical } B \]

Miscellaneous Applications

Application 1

The velocity \(v\) and a falling mass \(m\) subject to air resistance, is given by

\[ m\frac{dv}{dx}=mg-kv \]

Application 2

The rate at which a drug disseminates into bloodstream is governed by the differential equation

\[ \frac{dx}{dt}=A-Bx \]

\(x(t)\) describes the concentration of drug in the bloodstream at any time \(t\).

Application 3

The rate of memorization of a subject is given by

\[ \frac{dA}{dt}=k_{1}(M-A)-k_{2}A \]

\(A(t)\) is the amount of material memorized at any time \(t\).
\(M\) is the total amount.

References

Read more about notations and symbols.

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1. Read more about exponential model. ↩

2. Read more about partial fractions. ↩