Questions:

How can I describe radioactive decay using a first-order ODE?
What are initial conditions and why are they important?

Objectives:

Map between physical notation for a particular problem and the more general notation for all differential equations
Solve a linear, first order, separable ODE using integration
Understand the physical importance of initial conditions

Radioactive decay can be modelled a linear, first-order ODE

As our first example of an ODE we will model radioactive decay using a differential equation.

We know that the decay rate is proportional to the number of atoms present. Mathematically, this relationship can be expressed as:

\begin{equation} \frac{d N}{d t} = -\lambda N \end{equation}

Note that we could choose to use different variables, for example:

\begin{equation} \frac{d y}{d x} = cy \end{equation}

However we try to use variables connected to the context of the problem. For example $N$ for the Number of atoms.

For example, if we know that 10% of atoms will decay per second we could write:

\begin{equation} \frac{d N}{d t} = -0.1 N \end{equation}

where $N$ is the number of atoms and $t$ is time measured in seconds.

This equation is linear and first-order.

physical notation	generic notation
number of atoms $N$	dependent variable $y$
time $t$	independent variable $x$
decay rate $\frac{dN}{dt}$	differential $\frac{dy}{dx}$
constant of proportionality $\lambda=0.1 $	parameter $c$

The equation for radioactive decay is separable and has an analytic solution

The radioactive decay equation is separable. For example,

\begin{equation} \frac{d N}{d t} = -\lambda N \end{equation}

Can be seperated as

\begin{equation} \frac{dN}{N} = -\lambda dt. \end{equation}

We can then integrate each side:

\begin{equation} \ln N = -\lambda t + const. \end{equation}

and solve for N:

\begin{equation} N = e^{-\lambda t}e^{\textrm{const.}} \end{equation}

Note: Remember that $\int \frac{1}{x} dx = \ln x + \textrm{const.}$

To model a physical system an initial value has to be provided

At the beginning (when $t=0$):

\begin{equation} N = e^{-\lambda t}e^{\textrm{const.}} = e^{0}e^{\textrm{const.}} = e^{\textrm{const.}} \end{equation}

So we can identify $e^{\textrm{const.}}$ as the amount of radioactive material that was present in the beginning. We denote this starting amount as $N_0$.

Substituting this back into Equation 4, the final solution can be more meaningfully written as:

\begin{equation} N = N_0 e^{-\lambda t} \end{equation}

We now have not just one solution, but a whole class of solutions that are dependent on the initial amount of radioactive material $N_0$.

Remember that not all mathematical solutions make physical sense. To model a physical system, this initial value (also known as initial condition) has to be provided alongside the constant of proportionality $\lambda$.

ODEs can have initial values or boundary values

ODEs have either initial values or boundary values. For example, using Newton's second laws we could calculate the distance $x$ an object travels under the influence of gravity over time $t$

\begin{equation} \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -g \end{equation}

An initial value problem would be where we know the starting position and velocity. A boundary value problem would be where we specify the position of the ball at times $t=t_0$ and $t=t_1$.

In this course we will only study ODEs with initial values. ODEs with boundary values are more difficult to solve, but you can related materials listed under External resources.

The number of initial conditions depends on the order of the differential equation

Our radioactive decay example is a first-order ODE and so we only had to provide a single initial condition. For second-order ODEs (such as acceleration under gravity) we need to provide two initial/boundary conditions, for third-order ODEs we would need to provide three, and so on.

Keypoints:

Radioactive decay can be modelled a linear, first-order ODE
The equation for radioactive decay is separable and has an analytic solution
To model a physical system an initial value has to be provided
The number of initial conditions depends on the order of the differential equation

Do the quick-test.

Back to Modelling with Ordinary Differential Equations.