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May 03, 2020
Ajith B.P

First Order Ordinary Differential Equations

Differential equations are one of the most important mathematical tools used in producing models for physical and biological processes. Analytic solutions are available in some cases but numerical methods are employed for solving most of the problems encountered in science and engineering. Here we explore some of the popular methods for solving the initial value problems involving first-order ordinary differential equations(ODE). The focus will be on solving simple examples using Python programming language. The Python basics required for understanding the examples is given here

Consider the differential equation,

:math:\frac{dy}{dx}=f(x,y) with an initial value :math:y(x_0) = y_0

The derivative of the function and the value of the function at some initial value are known. The objective is to find out the value of the function for other values of x, starting from the given initial point. The underlying idea of any routine for solving the initial value problem is to rewrite the dy and dx as finite steps :math:\Delta x and :math:\Delta x, to convert the differential equation to an algebraic equation.

:math:y(x_{i+1}) = y_i + \Delta y = y_i + \frac{\Delta y}{\Delta x} \times {\Delta x} = y_i + \frac{dy}{dx} \times {\Delta x}

In the limit of making :math:\Delta x very small, a good approximation to the underlying differential equation is achieved. The process can be explained better with some examples. We choose a known function so that the error in the result can be easily checked.

euler-sine.py

import numpy as np
import matplotlib.pyplot as plt

def f1(x,y):		# derivative of the function to be evaluated
	return np.cos(x)

N = 4			# number of points
h = np. pi/(N-1)	# stepsize. Note that the Number of steps = Number of points -1 

xa = np.zeros(N)   	# numpy array of store the variable values
ya = np.zeros(N)	# numpy array to store results

xa[0] = 0.0
ya[0] = 0.0

def euler(x, y, fxy, h):
	return y + h * fxy(x,y)   # Euler method
	
for i in range(N-1):
	ya[i+1] = euler(xa[i], ya[i], f1, h)
	xa[i+1] = xa[i] + h

print (xa[-1], ya[-1],  'Err = ', ya[-1] - np.sin(xa[-1]))   # the last point and error

plt.plot(xa, ya, 'x-')
plt.plot(xa, np.sin(xa), '+-')
plt.show()

The output of the code is shown above. We have deliberately selected a very large stepsize to show the error. The lower graph, marked with dots, are the correct values. The points marked with crosses are the outputs of the calculation. The reason for error is very clear. The Euler method which advances a solution from :math:x_i to :math:x_{i+1} is not symmetric, it advances the solution through an interval h, but uses derivative information only at the beginning of that interval.

Better results are obtained if we take trial step to the midpoint of the interval and use the value of both x and y at that point to compute the real step across the whole interval. This is called the second-order Runge-Kutta or the midpoint method. To learn about the theory refer to Runge-Kutta methods_ , Second order Runge-Kutta methods_ or any other good books on numerical methods.

Runge-Kutta methods Second order Runge-Kutta methods

rk2-sine.py

import numpy as np
import matplotlib.pyplot as plt

def f1(x,y):		# derivative of the function to be evaluated
	return np.cos(x)

N = 4				# number of points
h = np.pi/(N-1) 	# stepsize. Note that the Number of steps = Number of points -1 

xa = np.zeros(N)   	# numpy array of store the variable values
ya = np.zeros(N)	# numpy array to store results

xa[0] = 0.0
ya[0] = 0.0

def rk2(x, y, fxy, h):   # x, y , f(x,y)
	k1 = fxy(x, y)
	k2 = fxy(x + h, h*k1)
	return y + h * ( k1/2 + k2/2)

for i in range(N-1):
	ya[i+1] = rk2(xa[i], ya[i], f1, h)
	xa[i+1] = xa[i] + h

print (xa[-1], ya[-1],  'Err = ', ya[-1] - np.sin(xa[-1]))
	 
plt.plot(xa, ya, 'x-')
plt.plot(xa, np.sin(xa), 'o-')
plt.show()

This procedure can be further extended to higher orders. The fourth order Runge-Kutta method is the most popular one and is commonly referred as the Runge-Kutta method. In each step the derivative is evaluated four times as shown in figure below. Once at the initial point, twice at trial midpoints, and once at a trial endpoint. Every trial evaluation uses the value of the function from the previous trial point, ie. :math:k_2 is evaluated using :math:k_1 and not using :math:y_i. From these derivatives the final function value is calculated.

rk4-sine.py

import numpy as np
import matplotlib.pyplot as plt

def f1(x,y):		# derivative of the function to be evaluated
	return np.cos(x)

N = 4				# number of points
h = np.pi/(N-1)   	# stepsize. Note that the Number of steps = Number of points -1 

xa = np.zeros(N)   	# numpy array of store the variable values
ya = np.zeros(N)	# numpy array to store results

xa[0] = 0.0
ya[0] = 0.0

def rk4(x, y, fxy, h):   # x, y , f(x,y)
	k1 = h * fxy(x, y)
	k2 = h * fxy(x + h/2.0, y+k1/2)
	k3 = h * fxy(x + h/2.0, y+k2/2)
	k4 = h * fxy(x + h, y+k3)
	return  y + ( k1/6 + k2/3 + k3/3 + k4/6 )

for i in range(N-1):
	ya[i+1] = rk4(xa[i], ya[i], f1, h)
	xa[i+1] = xa[i] + h

print (xa[-1], ya[-1],  'Err = ', ya[-1] - np.sin(xa[-1]))

plt.plot(xa, ya, 'x-')
plt.plot(xa, np.sin(xa), 'o-')
plt.show()

The reduction is error is clearly visible. We have used very coarse steps to highlight the error. You may change the value of ā€˜Nā€™ in the code and rerun. Outputs for N=100 are shown below. Also change the code to remove the marking of points.

As we explained earlier, the error is due to the fact that the value of the derivative changes within the interval but the calculations assume the derivative at the starting point or obtained by some averaging. This can be demonstrated by choosing a function representing a straight line, where derivative is constant, as shown in the example below.