Gill’s 4th Order Method to solve Differential Equations
Given the following inputs:
- An ordinary differential equation that defines the value of dy/dx in the form x and y.
- Initial value of y, i.e., y(0).
The task is to find the value of the unknown function y at a given point x, i.e. y(x).
Examples:
Input: x0 = 0, y = 3.0, x = 5.0, h = 0.2
Output: y(x) = 3.410426
Input: x0 = 0, y = 1, x = 3, h = 0.3
Output: y(x) = 1.669395
Approach:
Gill’s method is used to find an approximate value of y for a given x. Below is the formula used to compute the next value yn+1 from the previous value yn.
Therefore:
yn+1 = value of y at (x = n + 1) yn = value of y at (x = n) where 0 ? n ? (x - x0)/h h is step height xn+1 = x0 + h
The essential formula to compute the value of y(n+1):
The formula basically computes the next value yn+1 using current yn:
- K1 is the increment based on the slope at the beginning of the interval, using y.
- K2 is the increment based on the slope, using
- K3 is the increment based on the slope, using
- K4 is the increment based on the slope, using
The method is a fourth-order method, meaning that the local truncation error is of the order of O(h5).
Below is the implementation of the above approach:
C++
// C++ program to implement Gill's method #include <bits/stdc++.h> using namespace std; // A sample differential equation // "dy/dx = (x - y)/2" float dydx( float x, float y) { return (x - y) / 2; } // Finds value of y for a given x // using step size h and initial // value y0 at x0 float Gill( float x0, float y0, float x, float h) { // Count number of iterations // using step size or height h int n = ( int )((x - x0) / h); // Value of K_i float k1, k2, k3, k4; // Initial value of y(0) float y = y0; // Iterate for number of iteration for ( int i = 1; i <= n; i++) { // Apply Gill's Formulas to // find next value of y // Value of K1 k1 = h * dydx(x0, y); // Value of K2 k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1); // Value of K3 k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * (-1 + sqrt (2)) * k1 + k2 * (1 - 0.5 * sqrt (2))); // Value of K4 k4 = h * dydx(x0 + h, y - (0.5 * sqrt (2)) * k2 + k3 * (1 + 0.5 * sqrt (2))); // Find the next value of y(n+1) // using y(n) and values of K in // the above steps y = y + (1.0 / 6) * (k1 + (2 - sqrt (2)) * k2 + (2 + sqrt (2)) * k3 + k4); // Update next value of x x0 = x0 + h; } // Return the final value of dy/dx return y; } // Driver Code int main() { float x0 = 0, y = 3.0, x = 5.0, h = 0.2; printf ( "y(x) = %.6f" , Gill(x0, y, x, h)); return 0; } |
Java
// Java program to implement Gill's method class GFG{ // A sample differential equation // "dy/dx = (x - y)/2" static double dydx( double x, double y) { return (x - y) / 2 ; } // Finds value of y for a given x // using step size h and initial // value y0 at x0 static double Gill( double x0, double y0, double x, double h) { // Count number of iterations // using step size or height h int n = ( int )((x - x0) / h); // Value of K_i double k1, k2, k3, k4; // Initial value of y(0) double y = y0; // Iterate for number of iteration for ( int i = 1 ; i <= n; i++) { // Apply Gill's Formulas to // find next value of y // Value of K1 k1 = h * dydx(x0, y); // Value of K2 k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1); // Value of K3 k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * (- 1 + Math.sqrt( 2 )) * k1 + k2 * ( 1 - 0.5 * Math.sqrt( 2 ))); // Value of K4 k4 = h * dydx(x0 + h, y - ( 0.5 * Math.sqrt( 2 )) * k2 + k3 * ( 1 + 0.5 * Math.sqrt( 2 ))); // Find the next value of y(n+1) // using y(n) and values of K in // the above steps y = y + ( 1.0 / 6 ) * (k1 + ( 2 - Math.sqrt( 2 )) * k2 + ( 2 + Math.sqrt( 2 )) * k3 + k4); // Update next value of x x0 = x0 + h; } // Return the final value of dy/dx return y; } // Driver Code public static void main(String[] args) { double x0 = 0 , y = 3.0 , x = 5.0 , h = 0.2 ; System.out.printf( "y(x) = %.6f" , Gill(x0, y, x, h)); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program to implement Gill's method from math import sqrt # A sample differential equation # "dy/dx = (x - y)/2" def dydx(x, y): return (x - y) / 2 # Finds value of y for a given x # using step size h and initial # value y0 at x0 def Gill(x0, y0, x, h): # Count number of iterations # using step size or height h n = ((x - x0) / h) # Initial value of y(0) y = y0 # Iterate for number of iteration for i in range ( 1 , int (n + 1 ), 1 ): # Apply Gill's Formulas to # find next value of y # Value of K1 k1 = h * dydx(x0, y) # Value of K2 k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1) # Value of K3 k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * ( - 1 + sqrt( 2 )) * k1 + k2 * ( 1 - 0.5 * sqrt( 2 ))) # Value of K4 k4 = h * dydx(x0 + h, y - ( 0.5 * sqrt( 2 )) * k2 + k3 * ( 1 + 0.5 * sqrt( 2 ))) # Find the next value of y(n+1) # using y(n) and values of K in # the above steps y = y + ( 1 / 6 ) * (k1 + ( 2 - sqrt( 2 )) * k2 + ( 2 + sqrt( 2 )) * k3 + k4) # Update next value of x x0 = x0 + h # Return the final value of dy/dx return y # Driver Code if __name__ = = '__main__' : x0 = 0 y = 3.0 x = 5.0 h = 0.2 print ( "y(x) =" , round (Gill(x0, y, x, h), 6 )) # This code is contributed by Surendra_Gangwar |
C#
// C# program to implement Gill's method using System; class GFG{ // A sample differential equation // "dy/dx = (x - y)/2" static double dydx( double x, double y) { return (x - y) / 2; } // Finds value of y for a given x // using step size h and initial // value y0 at x0 static double Gill( double x0, double y0, double x, double h) { // Count number of iterations // using step size or height h int n = ( int )((x - x0) / h); // Value of K_i double k1, k2, k3, k4; // Initial value of y(0) double y = y0; // Iterate for number of iteration for ( int i = 1; i <= n; i++) { // Apply Gill's Formulas to // find next value of y // Value of K1 k1 = h * dydx(x0, y); // Value of K2 k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1); // Value of K3 k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * (-1 + Math.Sqrt(2)) * k1 + k2 * (1 - 0.5 * Math.Sqrt(2))); // Value of K4 k4 = h * dydx(x0 + h, y - (0.5 * Math.Sqrt(2)) * k2 + k3 * (1 + 0.5 * Math.Sqrt(2))); // Find the next value of y(n+1) // using y(n) and values of K in // the above steps y = y + (1.0 / 6) * (k1 + (2 - Math.Sqrt(2)) * k2 + (2 + Math.Sqrt(2)) * k3 + k4); // Update next value of x x0 = x0 + h; } // Return the final value of dy/dx return y; } // Driver Code public static void Main(String[] args) { double x0 = 0, y = 3.0, x = 5.0, h = 0.2; Console.Write( "y(x) = {0:F6}" , Gill(x0, y, x, h)); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program to implement Gill's method // A sample differential equation // "dy/dx = (x - y)/2" function dydx(x, y) { return (x - y) / 2; } // Finds value of y for a given x // using step size h and initial // value y0 at x0 function Gill(x0, y0, x, h) { // Count number of iterations // using step size or height h let n = ((x - x0) / h); // Value of K_i let k1, k2, k3, k4; // Initial value of y(0) let y = y0; // Iterate for number of iteration for (let i = 1; i <= n; i++) { // Apply Gill's Formulas to // find next value of y // Value of K1 k1 = h * dydx(x0, y); // Value of K2 k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1); // Value of K3 k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * (-1 + Math.sqrt(2)) * k1 + k2 * (1 - 0.5 * Math.sqrt(2))); // Value of K4 k4 = h * dydx(x0 + h, y - (0.5 * Math.sqrt(2)) * k2 + k3 * (1 + 0.5 * Math.sqrt(2))); // Find the next value of y(n+1) // using y(n) and values of K in // the above steps y = y + (1.0 / 6) * (k1 + (2 - Math.sqrt(2)) * k2 + (2 + Math.sqrt(2)) * k3 + k4); // Update next value of x x0 = x0 + h; } // Return the final value of dy/dx return y; } // Driver Code let x0 = 0, y = 3.0, x = 5.0, h = 0.2; document.write( "y(x) = " , Gill(x0, y, x, h).toFixed(6)); </script> |
y(x) = 3.410426
Time Complexity: O(n3/2)
Auxiliary Space: O(1)
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