(chap:pre:4)= # Applications of ODEs ## Kinematics ::::{prf:example} ```{figure} /images/00/ball.svg --- height: 250px name: Free fall --- A free falling object and the forces acting on it. ``` The motion of a free falling object with mass $m$ is given by $$ m\diff{v}{t} = m g - \gamma v $$(eq:ode:freefall) 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. :::{admonition} Solution :class: solution, dropdown 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. {eq}`eq:ode:freefall` as $$ \diff{v}{t}+\frac{\gamma}{m} v = g, $$ we can decompose the equations into homogeneous and non-homogenous parts $$ \begin{aligned} v'_{h} + \frac{\gamma}{m} v_h & = 0 \\ v'_{p} + \frac{\gamma}{m} v_p & = g \end{aligned} $$ 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 $$ v(t)=\frac{mg}{\gamma} \biggl(1-e^{-\frac{\gamma}{m} t}\biggr) = \frac{245}{7}\biggl(1-e^{-\frac{7}{25} t}\biggr). $$ ::: :::: ## Heat Transfer -- Forensic Science ::::{prf:example} 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!** :::: ## Electric Circuits ::::::{prf:example} ```{figure} /images/00/circuit.svg --- height: 300px name: RCL circuit --- 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). :::::{admonition} Solution :class: solution, dropdown By substituting the known conditions, the given differential equation can be simplified as $$ \diff[2]{I}{t} + 5 \diff{I}{t} + 6 I = 10 \sin t. $$ ::::{tab-set} ```{tab-item} 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 $$ I(t) = 2 e^{-2t} - e^{-3t} - \cos t + \sin t. $$ ``` ```{tab-item} 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*} ``` :::: ::::: :::::: ## Chemical Reactions ::::{prf:example} 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. ```{figure} /images/00/chemical.svg --- height: 80px name: Chemical reactions --- 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 $$ M + M \biggl( \frac{\beta e^{-\alpha t}-\alpha e^{-\beta t}}{\alpha-\beta} \biggr). $$ :::{admonition} Solution :class: solution, dropdown From the given question, we can form a system of ODEs $$ \begin{aligned} \diff{A}{t} & = - \alpha A \\ \diff{B}{t} & = \alpha A - \beta B \\ \diff{C}{t} & = \beta B \end{aligned} $$ 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 $$ \begin{aligned} C(t) & = \int \frac{\alpha \beta M}{\alpha - \beta} ( e^{-\beta t} - e^{-\alpha t}) \dt \\ & = \frac{\alpha \beta M}{\alpha - \beta} \biggl( \frac{e^{-\beta t}}{-\beta}- \frac{e^{-\alpha t}}{-\alpha}\biggr) + K_2 \\ & = \frac{M}{\alpha - \beta} (\beta e^{-\alpha t} - \alpha e^{-\beta t}) + K_2 \\ \end{aligned} $$ By using the initial condition $c(0)=0$, we can find $K_2=M$ and so the solution is $$ C(t) = M + \frac{M}{\alpha - \beta} (\beta e^{-\alpha t} - \alpha e^{-\beta t}) $$ ::: :::: ## Population Dynamics :::::{prf:example} 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. ::::{admonition} Solution :class: solution, dropdown 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 ```{figure} /images/00/population.svg --- height: 300px name: fish population --- The population of fish in a lake. ``` :::: ::::: ## Economics ::::{prf:example} 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)*](https://www.intechopen.com/chapters/76374) for more examples of ODEs in dynamic economics.