Update: One of the numerical example (example 1) has incorrect numerical results. This blog has been edited. The results posted are now correct. Apologizes. - Eddie
1/4/2015
Introduction
In this blog we will explore one of the ways to solve the differential equation y' = f(x,y) by numerical methods. One of the common methods is the Runge-Kutta 4th Order Method, probably the most common methods.
With the initial condition (x_n, y_n) and step size h, the next point (x_n+1, y_n+1) is calculated by the following equations:
x_n+1 = x_n + h
y_n+1 = y_n + (k1 + 2*k2 + 2*k3 + k4)/6 with
k1 = h * f(x_n, y_n)
k2 = h * f(x_n + h/2, y_n + k1/2)
k3 = h * f(x_n + h/2, y_n + k2/2)
k4 = h * f(x_n + h, y_n + k3)
(Almost) obviously use either a calculator or mathematical software for this process.
Examples:
1. dy/dx = -0.23 * (y + 1)
Initial Condition: (0, 24)
We want y1 when x1 = 1 and y2 when x2 = 2.
Observe that h = 1 with x0 = 0 and y0 = 24. Note the lack of x term in f(x,y).
Then: x1 = 0 + 1 = 1
k1 ≈ -5.75
k2 ≈ -5.08875
k3 ≈ -5.16479375
k4 ≈ -4.562097438
y1 ≈ 18.86346918
x2 = 1 + 1 = 2
k1 ≈ -4.568597911
k2 ≈ -4.043209151
k3 ≈ -4.103628858
k4 ≈ -3.624763273
y2 ≈ 14.78229631
So our points are:
(0, 24)
(1, 18.86346918)
(2, 14.78229631)
2. y' = y - x with I.C. y(0) = 2
Find y(0.1) and y(0.2).
Note h = 0.1 with x0 = 0, y0 = 2, and f(x,y) = -x + y.
x1 = 0.1
k1 = 0.1 * f(0, 2) = 0.2
k2 = 0.1 * f(0.5 * 0.1, 2 + 0.5 * 0.2) = 0.205
k3 = 0.1 * f(0.5 * 0.1, 2 + 0.5 * 0.205) = 0.20525
k4 = 0.1 * f(0.1, 2 + 0.20525) = 0.210525
y1 = 2 + (0.2 + 2 * 0.205 + 2 * 0.20525 + 0.210525)/6 ≈ 2.205171
x2 = 0.2
k1 ≈ 0.210517
k2 ≈ 0.216043
k3 ≈ 0.216319
k4 ≈ 0.222149
y2 ≈ 2.421203
So our points are:
(0, 2)
(0.1, 2.205171)
(0.2, 2.421203)
Talk to you all next time!
Eddie
This blog is property of Edward Shore. 2013
In this blog we will explore one of the ways to solve the differential equation y' = f(x,y) by numerical methods. One of the common methods is the Runge-Kutta 4th Order Method, probably the most common methods.
With the initial condition (x_n, y_n) and step size h, the next point (x_n+1, y_n+1) is calculated by the following equations:
x_n+1 = x_n + h
y_n+1 = y_n + (k1 + 2*k2 + 2*k3 + k4)/6 with
k1 = h * f(x_n, y_n)
k2 = h * f(x_n + h/2, y_n + k1/2)
k3 = h * f(x_n + h/2, y_n + k2/2)
k4 = h * f(x_n + h, y_n + k3)
(Almost) obviously use either a calculator or mathematical software for this process.
Examples:
1. dy/dx = -0.23 * (y + 1)
Initial Condition: (0, 24)
We want y1 when x1 = 1 and y2 when x2 = 2.
Observe that h = 1 with x0 = 0 and y0 = 24. Note the lack of x term in f(x,y).
Then: x1 = 0 + 1 = 1
k1 ≈ -5.75
k2 ≈ -5.08875
k3 ≈ -5.16479375
k4 ≈ -4.562097438
y1 ≈ 18.86346918
x2 = 1 + 1 = 2
k1 ≈ -4.568597911
k2 ≈ -4.043209151
k3 ≈ -4.103628858
k4 ≈ -3.624763273
y2 ≈ 14.78229631
So our points are:
(0, 24)
(1, 18.86346918)
(2, 14.78229631)
2. y' = y - x with I.C. y(0) = 2
Find y(0.1) and y(0.2).
Note h = 0.1 with x0 = 0, y0 = 2, and f(x,y) = -x + y.
x1 = 0.1
k1 = 0.1 * f(0, 2) = 0.2
k2 = 0.1 * f(0.5 * 0.1, 2 + 0.5 * 0.2) = 0.205
k3 = 0.1 * f(0.5 * 0.1, 2 + 0.5 * 0.205) = 0.20525
k4 = 0.1 * f(0.1, 2 + 0.20525) = 0.210525
y1 = 2 + (0.2 + 2 * 0.205 + 2 * 0.20525 + 0.210525)/6 ≈ 2.205171
x2 = 0.2
k1 ≈ 0.210517
k2 ≈ 0.216043
k3 ≈ 0.216319
k4 ≈ 0.222149
y2 ≈ 2.421203
So our points are:
(0, 2)
(0.1, 2.205171)
(0.2, 2.421203)
Talk to you all next time!
Eddie
This blog is property of Edward Shore. 2013
