1.4. Applications of ODEs

1.4.1. Kinematics

Example 1.11

Fig. 1.1 A free falling object and the forces acting on it.

The motion of a free falling object with mass \(m\) is given by

(1.16)\[ m\diff{v}{t} = m g - \gamma v \]

where \(v\) is its speed, \(g\) the gravitational acceleration, \(\gamma\) the air friction coefficient.

Find \(v(t)\) given that \(m=0.025\)kg, \(\gamma=0.007\)kg/s, \(g=9.8\) m/s² and \(v(0)=0\)m/s.

Solution

The given equation describing the physics is a first-order linear non-homogeneous equation, we will find its analytical solution then use a numerical method to solve the problem.

Re-arrange Eq. (1.16) as

\[ \diff{v}{t}+\frac{\gamma}{m} v = g, \]

we can decompose the equations into homogeneous and non-homogenous parts

\[\begin{split} \begin{aligned} v'_{h} + \frac{\gamma}{m} v_h & = 0 \\ v'_{p} + \frac{\gamma}{m} v_p & = g \end{aligned} \end{split}\]

The homogeneous solution is

\[ v_h(t)=C e^{-\frac{\gamma}{m} t} \]

where \(C\) is a constant to be determined, and the non-homogeneous solution is

\[ v_p(t)= \frac{mg}{\gamma}, \]

so the overall solution is

\[ v(t)=v_h(t)+v_p(t)=C e^{-\frac{\gamma}{m} t}+\frac{mg}{\gamma}, \]

By using the initial condition \(v(0)=0\), we can find

\[ C=-\frac{mg}{\gamma} \]

and

\[ v(t)=\frac{mg}{\gamma} \biggl(1-e^{-\frac{\gamma}{m} t}\biggr) = \frac{245}{7}\biggl(1-e^{-\frac{7}{25} t}\biggr). \]

1.4.2. Heat Transfer – Forensic Science

Example 1.12

A forensic scientist is called to the scene of a murder. The temperature of the corpse is found to be \(75\)^oF and one hour later the temperature has dropped to \(70\)^oF. If the temperature of the room in which the body was discovered is a constant \(68\)^oF, how long before the first temperature reading was taken did the murder occur? Assume that the body obeys Newton’s Law of Cooling,

\[ \diff{T}{t}=\beta(T-T_R), \]

where \(T\) is the temperature of the corpse, \(\beta\) is a constant, and \(T_R\) is room temperature. A normal body temperature for adults is generally around \(98.6\)^oF (\(37\)^oC).

Note: This will be a coursework question!

1.4.3. Electric Circuits

Example 1.13

Fig. 1.2 A simple RLC series circuit.

The current, \(I\), in a RLC circuit satisfies

\[ L \diff[2]{I}{t} + R \diff{I}{t} + \frac{1}{C} I = \diff{E}{t}. \]

Supposing \(L=1 \), \(R=5 \), \(C=\frac{1}{6}\) and \(E(t)=-10\cos t\), solve this differential equation given that \(I(0) = I'(0) = 0\) (a passive circuit).

Solution

By substituting the known conditions, the given differential equation can be simplified as

\[ \diff[2]{I}{t} + 5 \diff{I}{t} + 6 I = 10 \sin t. \]

Method 1 (Solution Decomposition)

Decompose the equation into homogeneous and non-homogenous parts

\[\begin{align*} I''_h + 5 I'_h + 6 I_h & = 0 \\ I''_p + 5 I'_p + 6 I_p & = 10 \sin t \end{align*}\]

For the homogeneous equation, its characteristic equation

\[ z^2 + 5z +6=0 \]

has two distinct roots \(z_1 = -2\) and \(z_2=-3\), so the homogeneous solution is

\[ I_h = C_1 e^{-2t} + C_2 e^{-3t}. \]

The right hand side of the non-homogeneous is a sinusoidal function, we guess the particular solution is given by

\[ I_p = A \cos t + B \sin t, \]

thus its first and second derivatives are

\[\begin{align*} I'_p & = -A \sin t + B \cos t, \\ I''_p & = -A \cos t - B \sin t. \end{align*}\]

Therefore, non-homogeneous equation is

\[\begin{align*} & I''_p + 5 I'_p + 6 I_p \\ = & (-A \cos t - B \sin t) + 5(-A \sin t + B \cos t) +6 (A \cos t + B \sin t) \\ = & (5B+5A) \cos t + (5B-5A) \sin t \\ = & 10 \sin t. \end{align*}\]

This requires two conditions

\[\begin{align*} 5B+5A & = 0, \\ 5B-5A & = 10, \end{align*}\]

so we get \(A=-1\) and \(B=1\).

Therefore, the particular solution is

\[ I_p = -\cos t + \sin t. \]

The overall solution is

\[ I = I_h + I_p = C_1 e^{-2t} + C_2 e^{-3t} -\cos t + \sin t, \]

and its derivative is

\[ I'= -2 C_1 e^{-2t} -3 C_2 e^{-3t} + \sin t + \cos t. \]

Using the initial conditions \(I(0)=I'(0)=0\),

\[\begin{align*} I(0) & = C_1 + C_2 -1 =0, \\ I'(0) & = -2 C_1 - 3 C_2 +1 =0, \end{align*}\]

we get \(C_1=2\) and \(C_2 =-1\), so the solution is

\[ I(t) = 2 e^{-2t} - e^{-3t} - \cos t + \sin t. \]

Method 2 (Laplace Transform)

Taking Laplace transforms of both sides:

\[ \Laplace{\diff[2]{I}{t} + 5 \diff{I}{t} + 6 I} = \Laplace{10 \sin t}, \]

we get

\[ \biggl( s^2 \bar{I}(s) - s I(0) - I'(0) \biggr) + 5 \biggl( s \bar{I}(s) - I(0) \biggr) + 6 \bar{I}(s) = \frac{10}{s^2+1}. \]

Substitute the initial conditions to obtain

\[ (s^2+5s+6)\bar{I}(s) = \frac{10}{s^2+1}. \]

Splitting into partial fractions gives

\[ \bar{I}(s) = \frac{2}{s+2} - \frac{1}{s+3} + \frac{1}{s^2+1} - \frac{s}{s^2+1}. \]

Take inverse Laplace transforms to get

\[\begin{align*} I(t) & = \invLaplace{\bar{I}(s)} \\ & = \invLaplace{\frac{2}{s+2} - \frac{1}{s+3} + \frac{1}{s^2+1} - \frac{s}{s^2+1}} \\ & = 2 e^{-2t} - e^{-3t} + \sin t - \cos t. \end{align*}\]

1.4.4. Chemical Reactions

Example 1.14

A chemical substance \(A\) changes into substance \(B\) at a rate \(\alpha\) times the amount of \(A\) present. Substance \(B\) changes into \(C\) at a rate \(\beta\) times the amount of \(B\) present.

Fig. 1.3 Chemical reactions of three substances.

If initially only substance \(A\) is present and its amount is \(M\), show that the amount of \(C\) present at time \(t\) is

\[ M + M \biggl( \frac{\beta e^{-\alpha t}-\alpha e^{-\beta t}}{\alpha-\beta} \biggr). \]

Solution

From the given question, we can form a system of ODEs

\[\begin{split} \begin{aligned} \diff{A}{t} & = - \alpha A \\ \diff{B}{t} & = \alpha A - \beta B \\ \diff{C}{t} & = \beta B \end{aligned} \end{split}\]

By using the initial condition \(A(0)=M\), we can find the solution for substance \(A\)

\[ A(t) = M e^{-\alpha t}. \]

The equation for substance \(B\) then is

\[ \diff{B}{t} + \beta B = M e^{-\alpha t}, \]

so its general solution is

\[ B(t) = K e^{-\beta t} - \frac{\alpha M}{\alpha - \beta} e^{-\alpha t}, \]

by using the initial condition \(B(0)=0\), we can find the solution for substance

\[ B(t) = \frac{\alpha M}{\alpha - \beta} ( e^{-\beta t} - e^{-\alpha t}). \]

Thus the equation for substance \(C\) is

\[ \diff{C}{t} = \beta B(t) = \frac{\alpha \beta M}{\alpha - \beta} ( e^{-\beta t} - e^{-\alpha t}), \]

so we can obtain the general solution by integrating the equation

\[\begin{split} \begin{aligned} C(t) & = \int \frac{\alpha \beta M}{\alpha - \beta} ( e^{-\beta t} - e^{-\alpha t}) \dt \\ & = \frac{\alpha \beta M}{\alpha - \beta} \biggl( \frac{e^{-\beta t}}{-\beta}- \frac{e^{-\alpha t}}{-\alpha}\biggr) + K_2 \\ & = \frac{M}{\alpha - \beta} (\beta e^{-\alpha t} - \alpha e^{-\beta t}) + K_2 \\ \end{aligned} \end{split}\]

By using the initial condition \(c(0)=0\), we can find \(K_2=M\) and so the solution is

\[ C(t) = M + \frac{M}{\alpha - \beta} (\beta e^{-\alpha t} - \alpha e^{-\beta t}) \]

1.4.5. Population Dynamics

Example 1.15

The population of a certain species of fish living in a large lake at time \(t\) can be modelled using Verhulst’s equation, otherwise known as the logistic equation,

\[ \diff{P}{t} = P (\beta - \delta P), \]

where \(P(t)\) is the population of fish measured in tens of thousands, and \(\beta\) and \(\delta\) are constants representing the birth and death rates of the fish living in the lake, respectively. Suppose that \(\beta=1\), \(\delta = 10^{-3}\), and the initial population is \(N=800\). Solve this initial value problem and interpret the results in physical terms.

Solution

Using the methods of separation of variables gives

\[ \int \frac{1}{P (\beta - \delta P)} \d P = \int \dt. \]

The solution to the integral on the left may be determined using partial fractions. The general solution is

\[ P(t) = \frac{\beta N e^{\beta t}}{\beta - \delta N + N \delta e^{\beta t}}. \]

Substituting the initial conditions, the solution is

\[ P(t) = \frac{800}{0.8+0.2 e^{-t}}. \]

Thus as time increases, the population of fish tends to a value of 1000. The solution curve is plotted in the following figure

Fig. 1.4 The population of fish in a lake.

1.4.6. Economics

Example 1.16

The Harrod-Domar model was developed to analyse business cycles originally but later was used to explain an economy’s growth rate through savings and capital productivity. Output, \(Y\), is a function of capital stock, \(K\), \(Y=F(K)\), and the marginal productivity,

\[ \diff{Y}{K}=c=\text{constant}. \]

The model postulates that the output growth rate is given by

\[ \frac{1}{Y} \diff{Y}{t} = sc - \delta, \]

where \(s\) is the savings rate, and \(\delta\) the capital depreciation rate. The straightforward solution,

\[ Y(t) = Y_0 e^{(sc-\delta)t}, \]

clearly demonstrates that increasing investment through savings and productivity boosts economic growth but does not take into account labour input and population size.

See Anastasios Tsoularis (2020) for more examples of ODEs in dynamic economics.