Euler's method

Differential equations: Direction field

Euler's method

If an explicit first-order dynamic system has no exact solution (in terms of known functions), or if a first orientation of the behavior of the system is desired, the geometric approach with the direction comes into the picture. By following the line elements of the direction field, we find an approximation of the solution curve with given initial condition by means of line segments.

Euler's method

Consider the dynamic system of the form

$\frac{\dd y}{\dd t}=\varphi(t,y)$ where $\varphi$ is a continuous function of two variables.

Let $\delta$ be a small positive number. A numerical approach to the solution curve of the dynamic system with initial condition $y(t_0)=\eta_0$ is given by the iterative formula

$\eta_{k+1}=\eta_k+\varphi(t_k,\eta_k)\cdot \delta$ where the points $t_k$ are chosen so each pair of subsequent points is at distance $\delta$ , so

$t_k=t_0+k\cdot \delta$ for $k=0,1,2,\ldots,n$ .

This approach is called Euler's method and the number $\delta$ is called the step size.

We begin with $\eta_0=y_0$ at time $t_0$ and determine the approximation $\eta_1$ of $y(t_1)$ by using the tangent at the point $\rv{t_0,\eta_0}$ as an approximation to the graph of the solution $y$ the segment $[t_0,t_1]$ :

$\eta_1=\eta_0+\varphi(t_0,\eta_0)\cdot \delta$ Starting with the known point $\rv{t_1,\eta_1}$ we approximate the next function value $y(t_2)$ by

$\eta_2=\eta_1+\varphi(t_1,\eta_1)\cdot \delta$ And so on. This approximation process leads to the general formula for $\eta_{k+1}$ .

In the figure below the direction field of the dynamic system

$\frac{\dd y}{\dd t}=-0.7y$ is drawn together with the approximation of the solution curve, calculated according to Euler's method with step size $\delta =1$ and initial condition $y(0)=6$ . As the step size $\delta$ becomes smaller, the graph of the calculated function values becomes smoother and the graph resembles an exact solution.

In this specific case, the formula $\eta_{k+1}=\eta_k+\varphi(t_k,\eta_k)\cdot \delta$ is the same as

$\eta_{k+1}=\eta_k-0.7\cdot \eta_k\cdot \delta=\eta_k\cdot 0.3\cdot \delta=\eta_0\cdot(0.3\cdot \delta)^{k+1}=6\cdot(0.3\cdot \delta)^{k+1}$ The exact solution to this initial value problem is $y(t)=6\e^{-0.7t}$

The method is based on the formula of the tangent line to the curve passing through the point $\rv{t_k,\eta_k}$ on the solution curve:

$y=\eta_k+\varphi(t_k,\eta_k)\cdot (x-t_k)$

Using step size $\delta$ we arrive at the point $\rv{t_{k+1},\eta_{k+1}}$ on this line, where $t_{k+1}=t_k+\delta$ so

$\eta_{k+1} = \eta_k+\varphi(t_k,\eta_k)\cdot \delta$

If, in the Euler method, a negative time step $\delta$ is used, the corresponding values of points close to the integral curve can be calculated just as well. Thus, for integral curves passing through a certain point, the Euler method is suitable for calculating approximations toward both the past and the future.

In the figure below, the integral curves of the ODE

$\frac{\dd y}{\dd t}=3\e^{-t}-\tfrac{1}{2}y,$ are drawn through the points $\rv{2,0},\;\rv{1,1},\;\rv{0,2}$

voorwaarte and terugwaarte Euler Method

The green integral curve has the general form of the concentration gradient of an orally administered drug. Later we will see that this pharmacokinetic model is also exactly solvable and that solution curves have this form.

Numerous algorithms for the numerical solution of differential equations exist; these are beyond the scope of this course. The most important message is that there are such numerical solution methods and these are useful for gaining insight into solutions of the ODE.

In order for you to experience Euler's method an interactive version is presented below.

Enlarge the number $s$ of steps in order to see a version with smaller step size, and enter different right hand sides of the differential equation in the input field behind " $\dd y/\dd t =$ ". Press the update button to see the result.

New example