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# Fourth Order Runge-Kutta Method in Python

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29 Jul 2014CPOL2 min read
Implementation of the fourth order Runge-Kutta method in Python for solving n-dimensional ordinary differential equations

## Introduction

The Python code presented here is for the fourth order Runge-Kutta method in n-dimensions. The Runge-Kutta method is a mathematical algorithm used to solve systems of ordinary differential equations (ODEs). The general form of these equations is as follows:

\Large\begin{aligned} \dot{x}&=f(t, x) \\ x(t_{0})&=x_{0} \end{aligned}

Where x is either a scalar or vector. The fourth order Runge-Kutta method is given by:

\Large\begin{aligned} x_{i+1}&=x_{i}+(k_{1}+2(k_{2}+k_{3})+k_{4})/6 \\ t_{i+1}&=t_{i}+h \end{aligned}

where h > 0 is a step size parameter, i=1, 2, 3,... and:

\Large\begin{aligned} k_{1}&=f(t_{i}, x_{i})h \\ k_{2}&=f(t_{i}+\frac{h}{2}, x_{i}+\frac{k_{1}}{2})h \\ k_{3}&=f(t_{i}+\frac{h}{2}, x_{i}+\frac{k_{2}}{2})h \\ k_{4}&=f(t_{i}+h, x_{i}+k_{3})h \end{aligned}

The Runge-Kutta method offers greater accuracy than the method of multiplying each function in the ODEs by a step size parameter and adding the results to the current values in x.

## Implementation

It is common practise to eliminate t with a suitable substitution such as:

$\Large c=\omega t$

Hence:

$\Large\dot{c}=\omega$

Presented here are two techniques for implementing the fourth order Runge-Kutta method. Here is a general Python function for the method in n-dimensions implemented using arrays (technique 1):

Python
def rKN(x, fx, n, hs):
k1 = []
k2 = []
k3 = []
k4 = []
xk = []
for i in range(n):
k1.append(fx[i](x)*hs)
for i in range(n):
xk.append(x[i] + k1[i]*0.5)
for i in range(n):
k2.append(fx[i](xk)*hs)
for i in range(n):
xk[i] = x[i] + k2[i]*0.5
for i in range(n):
k3.append(fx[i](xk)*hs)
for i in range(n):
xk[i] = x[i] + k3[i]
for i in range(n):
k4.append(fx[i](xk)*hs)
for i in range(n):
x[i] = x[i] + (k1[i] + 2*(k2[i] + k3[i]) + k4[i])/6
return x

Both x and fx are arrays, the latter is an array of functions, and n is the number of dimensions. The function definitions correspond with the ODEs being solved. Here is the ODE and an example usage of the rKN function for the forced Van der Pol oscillator:

$\Large\ddot{y}=\mu(1-y^2)\dot{y}-y+Asin(\omega t)$

Let:

$\Large x=(\dot{y},y,\omega t)$
Python
def fa1(x):
return 0.9*(1 - x[1]*x[1])*x[0] - x[1] + math.sin(x[2])

def fb1(x):
return x[0]

def fc1(x):
return 0.5

def VDP1():
f = [fa1, fb1, fc1]
x = [1, 1, 0]
hs = 0.05
for i in range(20000):
x = rKN(x, f, 3, hs)

In the above functions μ=0.9, A=1 and ω=0.5. Here is a general Python function for the fourth order Runge-Kutta method in 3-dimensions (technique 2):

Python
def rK3(a, b, c, fa, fb, fc, hs):
a1 = fa(a, b, c)*hs
b1 = fb(a, b, c)*hs
c1 = fc(a, b, c)*hs
ak = a + a1*0.5
bk = b + b1*0.5
ck = c + c1*0.5
a2 = fa(ak, bk, ck)*hs
b2 = fb(ak, bk, ck)*hs
c2 = fc(ak, bk, ck)*hs
ak = a + a2*0.5
bk = b + b2*0.5
ck = c + c2*0.5
a3 = fa(ak, bk, ck)*hs
b3 = fb(ak, bk, ck)*hs
c3 = fc(ak, bk, ck)*hs
ak = a + a3
bk = b + b3
ck = c + c3
a4 = fa(ak, bk, ck)*hs
b4 = fb(ak, bk, ck)*hs
c4 = fc(ak, bk, ck)*hs
a = a + (a1 + 2*(a2 + a3) + a4)/6
b = b + (b1 + 2*(b2 + b3) + b4)/6
c = c + (c1 + 2*(c2 + c3) + c4)/6
return a, b, c

This function performs the same calculations as rKN, but specifically in 3-dimensions and with the loops unravelled. Numerics need to be passed to the parameters a, b and c. Functions, each taking three numerics, and each returning numerics need to be passed to the parameters fa, fb and fc. So x is represented here by [a, b, c], k1 is represented by [a1, b1, c1], k2 by [a2, b2, c2] and so on. The variables ak, bk and ck are temporary variables used to optimize the calculations. Here is an example usage of the rK3 function, again for the forced Van der Pol oscillator:

Python
def fa2(a, b, c):
return 0.9*(1 - b*b)*a - b + math.sin(c)

def fb2(a, b, c):
return a

def fc2(a, b, c):
return 0.5

def VDP2():
a, b, c, hs = 1, 1, 0, 0.05
for i in range(20000):
a, b, c = rK3(a, b, c, fa2, fb2, fc2, hs)

The program TestRK.py demonstrates that technique 2 is faster by a factor of about 3.

## Code Generation

The program GenRK.py can be used to generate the Python code (for technique 2) in any dimension up to n=26. Here the function rK3 was created using the generator program GenRK.py with n=3. The generator works using a number of for loops in a function called genRK. Below is the code for GenRK.py:

Python
def getChr(i):
return chr(i + 97)

def genRK(n):
print("# fourth order Runge-Kutta method in " + str(n) + " dimensions")
u = ""
v = ""
f = ""
for i in range(n):
c = getChr(i)
if i != 0:
u += ", "
v += ", "
f += ", "
u += c
v += c + "k"
f += "f" + c
print("def rK" + str(n) + "(" + u + ", " + f + ", hs" + "):")
for i in range(n):
c = getChr(i)
print("\t" + c + "1 = f" + c + "(" + u + ")*hs")
for i in range(n):
c = getChr(i)
print("\t" + c + "k = " + c + " + " + c + "1*0.5")
for i in range(n):
c = getChr(i)
print("\t" + c + "2 = f" + c + "(" + v + ")*hs")
for i in range(n):
c = getChr(i)
print("\t" + c + "k = " + c + " + " + c + "2*0.5")
for i in range(n):
c = getChr(i)
print("\t" + c + "3 = f" + c + "(" + v + ")*hs")
for i in range(n):
c = getChr(i)
print("\t" + c + "k = " + c + " + " + c + "3")
for i in range(n):
c = getChr(i)
print("\t" + c + "4 = f" + c + "(" + v + ")*hs")
for i in range(n):
c = getChr(i)
print("\t" + c + " = " + c + " + (" + c + "1 + 2*(" + c + "2 + " + c + "3) + " + c + "4)/6")
print("\treturn " + u)

## Conclusion

Whilst the first technique is easier to implement, it is somewhat slower than the second. Technique 2 becomes harder to implement in higher dimensions, though it is relatively easy to generate the code.

## About the Author

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## Comments and Discussions

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 Mr tyebillion Member 1475786227-Feb-20 21:19 Member 14757862 27-Feb-20 21:19
 thanks Member 128379927-Nov-16 16:45 Member 12837992 7-Nov-16 16:45
 Re: thanks tyebillion26-Jan-17 3:45 tyebillion 26-Jan-17 3:45
 help for line 9 in method 1 Member 1269267018-Aug-16 20:18 Member 12692670 18-Aug-16 20:18
 Re: help for line 9 in method 1 tyebillion19-Aug-16 10:44 tyebillion 19-Aug-16 10:44
 My vote of 5 Member 1131046914-Dec-14 14:58 Member 11310469 14-Dec-14 14:58
 My vote of 5 CatchExAs8-Jul-14 13:08 CatchExAs 8-Jul-14 13:08
 Re: My vote of 5 tyebillion9-Jul-14 1:45 tyebillion 9-Jul-14 1:45
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