A simple first order ODE¶
The simplest example of a first order ODE is
with the solution
where \(y\left( 0 \right)\) or \(y_{0}\), the value of \(y\left( t \right)\) at \(t = 0\), is the initial condition. The final steady state solution as \(t \rightarrow \infty\) is \(y\left( \left. \ t \right|_{\infty} \right) = y_{\infty} = \frac{b}{a}\) (see Figure 6). Note that \(t = \tau = \frac{1}{a}\) is called the time constant of the exponential decay, and that
At \(t = \tau\) , \(y\left( t \right)\) has therefore fallen to \(\frac{1}{e}\) (or about 37%) of the difference between the initial (\(y\left( 0 \right)\)) and final steady state ( \(y\left( \infty \right)\)) values[1].
Choosing parameters \(a = \tau = 1;b = 2\) and \(y\left( 0 \right) = 5\), the CellML Text for this model is
def model first_order_model as
def comp main as
var t: dimensionless {init: 0};
var y: dimensionless {init: 5};
var a: dimensionless {init: 1};
var b: dimensionless {init: 2};
ode(y,t)=-a*y+b;
enddef;
enddef;
The solution by OpenCOR is shown in Fig. 7(a) for these parameters (a decaying exponential) and in Fig. 7(b) for parameters \(a = 1;b = 5\) and \(y\left( 0 \right) = 2\) (an inverted decaying exponential). Note the simulation panel with Ending point=10, Point interval=0.1. Try putting \(a = - 1\).
These two solutions have the same exponential time constant (\(\tau = \frac{1}{a} = 1\)) but different initial and final (steady state) values.
The exponential decay curve shown on the left in Fig. 7 is a common feature of many models and in the case of radioactive decay (for example) is a statement that the rate of decay (\(- \frac{\text{dy}}{\text{dt}}\)) is proportional to the current amount of substance (\(y\)). This is illustrated on the NZ$100 note (should you be lucky enough to possess one), shown in Figure 8.
Footnotes