Can Euler be better than Runge-Kutta for some functions?

Question

I'm trying to solve exercises from Steven Strogatz's Non-Linear Dynamics and Chaos. In exercise 2.8.3, 2.8.4, and 2.8.5, one is expected to implement the Euler method, improved Euler method, and Runge-Kutta (4th order) method respectively for the initial value problem dx/dt = -x; x(0) = 1 to find x(1).

Analytically, the answer is 1/e. And I was finding the error obtained in each method. Much to my surprise, I was getting lesser error in Euler than in Improved Euler and Runge-Kutta!

My code looks like this. Sorry for the shabbiness.

from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt

to = 0
xo = 1
tf = 1

deltaT = np.zeros([5])
errorE = np.zeros([5])
errorIE = np.zeros([5])
errorRK = np.zeros([5])


for j in range(0,5):
  n = pow(10,j)
  deltat = (tf - to)/(n)

  print ("delta t is",deltat)

  deltaT[j] = deltat

  t = np.linspace(to,tf,n)
  xE = np.zeros([n])
  xIE = np.zeros([n])
  xRK = np.zeros([n])

  xE[0] = xo
  xIE[0] = xo
  xRK[0] = xo

  for i in range (1,n):
    #Regular Euler
    xE[i] = deltat*(-xE[i-1]) + xE[i-1]

    #Improved Euler
    IEintermediate = deltat*(-xIE[i-1]) + xIE[i-1]
    xIE[i] = xIE[i-1] - deltat*(xIE[i-1] + IEintermediate)/2 

    #Runge-Kutta fourth order
    k1 = -deltat*xRK[i-1]
    k2 = -deltat*(xRK[i-1] + k1/2)
    k3 = -deltat*(xRK[i-1] + k2/2)
    k4 = -deltat*(xRK[i-1] + k3)

    xRK[i] = xRK[i-1] + (k1 + 2*k2 + 2*k3 + k4)/6

    print (deltat,xE[i],xIE[i],xRK[i])

  errorE[j] = np.exp(-1) - xE[n-1]
  errorIE[j] = np.exp(-1) - xIE[n-1]
  errorRK[j] = np.exp(-1) - xRK[n-1]

The errors :

For delT = 1.0

Euler error is -0.6321205588285577
I.Euler error is -0.6321205588285577
RK error is -0.6321205588285577

For delT = 0.1

Euler error -0.019541047828557645
I.Euler error -0.039348166379443716
RK error -0.03869055002863331

For delT = 0.01

Euler -0.0018501964782845493
I.Euler -0.003703427083890265
RK -0.0036972498815148747

For delT = 0.001

Euler -0.0001840470877806366
I.Euler -0.00036812480143849635
RK-0.00036806344222467535

For delT = 0.0001

Euler -1.839504510836587e-05
I.Euler -3.67903967520844e-05
RK -3.678978357835039e-05

Is this legit? If not, why is this happening?

Lutz Lehmann Lutz Lehmann · Accepted Answer · 2018-10-31T08:28:53

You are doing only n-1 integration steps of step size h=1/n, thus you compute

exp(-(n-1)/n)=1/e*exp(1/n)

which has the approximate value

1/e + 1/e*1/n

The reported error values are exactly that, -h/e, which is first order and thus gets visibly distorted by the order 1 Euler method. The Euler value is, more exactly

(1-1/n)^(n-1) = exp((n-1)*(-1/n-1/(2n^2)+O(1/n^3))
              = 1/e*exp(1/(2n)+..)
              = 1/e + h/(2e) + ...

If you adapt the code to make the extra step to reach time 1, you will get a correct error picture.

delta t is  1.0
Euler          0.0             0.367879441171
imp. Euler     0.5            -0.132120558829
Runge-Kutta 4  0.375          -0.00712055882856

delta t is  0.1
Euler          0.3486784401    0.0192010010714
imp. Euler     0.368540984834 -0.00066154366211
Runge-Kutta 4  0.367879774412 -3.33241056083e-07

delta t is  0.01
Euler          0.366032341273  0.00184709989821
imp. Euler     0.367885618716 -6.17754474969e-06
Runge-Kutta 4  0.367879441202 -3.09130498977e-11

delta t is  0.001
Euler          0.367695424771  0.000184016400479
imp. Euler     0.367879502531 -6.13592486265e-08
Runge-Kutta 4  0.367879441171 -4.05231403988e-15

delta t is  0.0001
Euler          0.367861046433  1.83947385133e-05
imp. Euler     0.367879441785 -6.13176398545e-10
Runge-Kutta 4  0.367879441171 -2.6645352591e-15

Can Euler be better than Runge-Kutta for some functions?

1 Answers