1.7: Mathematical Modelling with First Order Dynamic Equations

Mathematical modelling of a system in an application area of interest (e.g., physics, biology, engineering, economics, ...) is the translation of our knowledge or beliefs of that system into the language of mathematics. It typically involves writing down an equation (or set of equations) that describes the behaviour of the system.

The majority of real-world problems are far too complex to model in their entirety. The process of developing a mathematical model therefore involves making a number of simplifications along the way. We accept that all models are limited and, to some extent, wrong; but if the modelling process is conducted with intelligence and care, models are still undeniably useful. They enable us to understand complex phenomena, make predictions about the future and test the impact of changes to a system.

Typically, mathematical models can be classified in a number of ways, e.g., empirical vs. mechanistic. Empirical models are often data-driven and use statistical methods to infer relations between different variables in a system; mechanistic models use prior knowledge and understanding of the mechanism that causes change in a system.

In many applications, we are given the value of a quantity at the present time (for example, the temperature of coffee in a cup, the number of people infected with a virus, the concentration of carbon dioxide in the atmosphere) and we wish to predict its value in the future. To do this, we must know how quickly the quantity is changing. Mathematically, the rate of change of this quantity is its derivative. If we can write down an equation relating this derivative to some knowledge of the mechanism that causes the change, we can calculate how the temperature changes, how the number of infected people changes, how the concentration of carbon dioxide changes. This gives rise to a differential equation—which is a mechanistic model of the quantity of interest.

Modelling in Geometry

We will find a curve in the \(t, y\)-plane that passes through the point \((t_0, y_0)\in \mathbb{T}\times \mathbb{T}_1\) and whose tangent at a point \((t, y)\) has slope

\[ \frac{\sigma(t)+t}{(\sigma_1(y))^2+y\sigma_1(y)+y^2}.\nonumber\]

Since the slope of the tangent line is \(\frac{\Delta_1 y}{\Delta t}\), we have

\[\frac{\Delta_1 y}{\Delta t}=\frac{\sigma(t)+t}{(\sigma_1(y))^2+y\sigma_1(y)+y^2}.\]

Because the curve passes through the point \((t_0, y_0)\), we have the initial condition

\[ y(t_0)=y_0.\]

Thus, we get the following initial value problem

\[\begin{eqnarray*}((\sigma_1(y))^2+y\sigma_1(y)+y^2)\Delta_1 y&=&(\sigma(t)+t) \Delta t,\\ \\y(t_0)&=& y_0.\end{eqnarray*}\nonumber\]

The obtained equation is a separable equation. For its general solution, we find

\[ \int((\sigma_1(y))^2+y\sigma_1(y)+y^2)\Delta_1y=\int(\sigma(t)+t)\Delta t+c,\quad (t, y)\in \mathbb{T}\times \mathbb{T}_1.\nonumber\]

Here \(c\) is a constant of integration that will be determined below. Hence,

\[ y^3+t^2=c,\quad (t, y)\in \mathbb{T}\times \mathbb{T}_1\nonumber\].

Now, we will find the cconstant \9c\) using the initial condition. We get

\[y_0^3=t_0^2+c\nonumber\]

or

\[c=y_0^3-t_0^2.\nonumber\]

Therefore

\[y^3=t^2+y_0^3-t_0^2,\quad (t, y)\in \mathbb{T}\times \mathbb{T}_1,\nonumber\]

is the desired curve.

Bacteria Growth

Let \(N(t)\) denote the number of bacteria growning on a rate of nutrients at time \(t\). Suppose that we know the initial value of \(N\) at time \(t=t_0\). At this time, the value of \(N\) is \(N_0\), or

\[ N(t_0)=N_0.\nonumber\]

We want to know at what time the number of bacteria becomes \(10N_0\).

To build a mathematical model for this problem, let us first describe how the number of bacteria changes with time using mathematics. Suppose that over the time interval \((t, t+\sigma(t))\), the number of bacteria increases by an amount

\[\delta N=N(\sigma(t))-N(t)\nonumber\]

which we write as

\[N(\sigma(t))-N(t)=k(N, t)(\sigma(t)-t).\nonumber\]

Then

\[\frac{N(\sigma(t))-N(t)}{\sigma(t)-t}=k(N, t).\nonumber\]

Thus, we get the following dynamic equation

\[N^\Delta(t)=k(N, t).\nonumber\]

To make a progress, we need some information on the function \(k(N, t)\). Suppose by doing experiments, we observe that the rate of change of \(N\) is proportional of \(N\). This means that if there are, say, twice as many bacteria, then \N\) will grow twice as rapidly. Mathematically,

\[k(N, t)=lN,\nonumber\]

where \(l\) is a constant we can measure from experiments. Therefore, the mathematical problem for our bacteria growth problem is

\[N^\Delta(t)=lN(t),\quad t\in \mathbb{T}.\nonumber\]

Note that

\[N(t)=e_l(t, t_0),\quad t\in \mathbb{T},\nonumber\]

is a solution of the obtained equation.

We extract the answer to the question we ask. Let \(t_1\) be the time at which we have

\[N=10N_0.\nonumber\]

Then

\[e_l(t_1, t_0)=10N_0.\nonumber\]

The Newton Law of Cooling

Sir Isaac Newton FRS: 1643–1727, English physicist, mathematician, astronomer, natural philosopher, alchemist and theologian.

The Newton law of cooling states that if an object is hotter than the ambient temperature, then the rate of change of the object’s temperature is proportional to the temperature difference between the object and its surrounding. Mathematically, we write

\[\begin{eqnarray*}Y^\Delta(t)&=& k(Y^\sigma-A),\quad t\in \mathbb{T},\quad t\geq t_0,\\ \\Y(t_0)&=& Y_0,\end{eqnarray*}\nonumber\]

where \(Y(t)\) is the object temperature, \(A\) is the ambient temperature (obviously it is a constant), \(t\) is time and \(k\) is a positive constant. This is a first order dynamic equation

\[\label{*} Y^\Delta(t)+kY^\sigma(t)=k A,\quad t\in \mathbb{T},\quad t\geq t_0.\]

So, the integrating factor is

\[R(t)= e_k(t, t_0),\quad t\in \mathbb{T},\quad t\geq t_0.\nonumber\]

We multiply the equation \eqref{*} by \(e_k(t, t_0)\) and we get

\[Y^\Delta(t)e_k(t, t_0)+ke_k(t, t_0)Y^\sigma(t)= k A e_k(t, t_0),\quad t\in \mathbb{T},\quad t\geq t_0,\nonumber\]

or

\[ (Y e_k(\cdot, t_0))^\Delta(t)= kA e_k(t, t_0),\quad t\in \mathbb{T},\quad t\geq t_0.\nonumber\]

By the last expression, we get

\[Y(t)e_k(t, t_0)= Y_0+kA \int\limits_{t_0}^t e_k(s, t_0)\Delta s\nonumber\]

\[= Y_0+A e_k(s, t_0)\bigg|_{s=t_0}^{s=t}\nonumber\]

\[= Y_0+A e_k(t, t_0)-A,\quad t\in \mathbb{T},\quad t\geq t_0.\nonumber\]

Hence,

\[Y(t)= (Y_0-A) e_{\ominus k}(t, t_0)+A,\quad t\in \mathbb{T},\quad t\geq t_0.\nonumber\]

Suppose that \(\mu(t)k<1\), \(t\in \mathbb{T}, t\geq t_0\). Then

\[e_{\ominus k}(t, t_0)= e^{\int\limits_{t_0}^t \frac{1}{\mu(s)}\log(1+\mu(s)(\ominus k)(s))\Delta s}\nonumber\]

\[= e^{\int\limits_{t_0}^t \frac{1}{\mu(s)}\log\left(\frac{1}{1+k\mu(s)}\right)\Delta s}\nonumber\]

\[= e^{-\int\limits_{t_0}^t \frac{1}{\mu(s)} \log(1+k \mu(s))\Delta s}\nonumber\]

\[\leq e^{-\int\limits_{t_0}^t \frac{1}{mu(s)}\left(\mu(s) k-\frac{(\mu(s))^2k^2}{2}\right)\Delta s}\nonumber\]

\[= e^{-\int\limits_{t_0}^t \left(k- \frac{\mu(s) k^2}{2}\right)\Delta s}\nonumber\]

\[= e^{-\int\limits_{t_0}^t k\left(1-\frac{\mu(s)k}{2}\right)\Delta s}\nonumber\]

\[\leq e^{-\frac{k}{2}(t-t_0)}\nonumber\]

\[ \to 0\quad \mbox{as}\quad t\to \infty.\nonumber\]

Thus, when \(\mu(t)k<1\), \(t\in \mathbb{T},\quad t\geq t_0\), we have that \(Y(t)\to A\), \(t\to \infty\).