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Home / Find the Exact Solution to a Fourth Order Pde

Find the Exact Solution to a Fourth Order Pde

Given following inputs,

  • An ordinary differential equation that defines value of dy/dx in the form x and y.
  • Initial value of y, i.e., y(0)

Thus we are given below.
\frac{\mathrm{dy} }{\mathrm{d} x} = f(x, y),y(0)= y_o
The task is to find value of unknown function y at a given point x.
The Runge-Kutta method finds approximate value of y for a given x. Only first order ordinary differential equations can be solved by using the Runge Kutta 4th order method.
Below is the formula used to compute next value yn+1 from previous value yn. The value of n are 0, 1, 2, 3, ….(x – x0)/h. Here h is step height and xn+1 = x0 + h
. Lower step size means more accuracy.
K_1 = hf(x_n, y_n)\\ K_2 = hf(x_n+\frac{h}{2}, y_n+\frac{k_1}{2})\\ K_3 = hf(x_n+\frac{h}{2}, y_n+\frac{k_2}{2})\\ K_4 = hf(x_n+h, y_n+k_3)\\ y_{n+1} = y_n + k_1/6 + k_2/3 + k_3/3 + k_4/6 + O(h^{5})

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The formula basically computes next value yn+1 using current yn plus weighted average of four increments.



  • k1 is the increment based on the slope at the beginning of the interval, using y
  • k2 is the increment based on the slope at the midpoint of the interval, using y + hk1/2.
  • k3 is again the increment based on the slope at the midpoint, using using y + hk2/2.
  • k4 is the increment based on the slope at the end of the interval, using y + hk3.

The method is a fourth-order method, meaning that the local truncation error is on the order of O(h5), while the total accumulated error is order O(h4).
Source: https://en.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods

Below is implementation for the above formula.

C

#include<stdio.h>

float dydx( float x, float y)

{

return ((x - y)/2);

}

float rungeKutta( float x0, float y0, float x, float h)

{

int n = ( int )((x - x0) / h);

float k1, k2, k3, k4, k5;

float y = y0;

for ( int i=1; i<=n; i++)

{

k1 = h*dydx(x0, y);

k2 = h*dydx(x0 + 0.5*h, y + 0.5*k1);

k3 = h*dydx(x0 + 0.5*h, y + 0.5*k2);

k4 = h*dydx(x0 + h, y + k3);

y = y + (1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4);;

x0 = x0 + h;

}

return y;

}

int main()

{

float x0 = 0, y = 1, x = 2, h = 0.2;

printf ( "\nThe value of y at x is : %f" ,

rungeKutta(x0, y, x, h));

return 0;

}

Java

import java.io.*;

class differential

{

double dydx( double x, double y)

{

return ((x - y) / 2 );

}

double rungeKutta( double x0, double y0, double x, double h)

{

differential d1 = new differential();

int n = ( int )((x - x0) / h);

double k1, k2, k3, k4, k5;

double y = y0;

for ( int i = 1 ; i <= n; i++)

{

k1 = h * (d1.dydx(x0, y));

k2 = h * (d1.dydx(x0 + 0.5 * h, y + 0.5 * k1));

k3 = h * (d1.dydx(x0 + 0.5 * h, y + 0.5 * k2));

k4 = h * (d1.dydx(x0 + h, y + k3));

y = y + ( 1.0 / 6.0 ) * (k1 + 2 * k2 + 2 * k3 + k4);

x0 = x0 + h;

}

return y;

}

public static void main(String args[])

{

differential d2 = new differential();

double x0 = 0 , y = 1 , x = 2 , h = 0.2 ;

System.out.println( "\nThe value of y at x is : "

+ d2.rungeKutta(x0, y, x, h));

}

}

Python

def dydx(x, y):

return ((x - y) / 2 )

def rungeKutta(x0, y0, x, h):

n = ( int )((x - x0) / h)

y = y0

for i in range ( 1 , n + 1 ):

"Apply Runge Kutta Formulas to find next value of y"

k1 = h * dydx(x0, y)

k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1)

k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * k2)

k4 = h * dydx(x0 + h, y + k3)

y = y + ( 1.0 / 6.0 ) * (k1 + 2 * k2 + 2 * k3 + k4)

x0 = x0 + h

return y

x0 = 0

y = 1

x = 2

h = 0.2

print 'The value of y at x is:' , rungeKutta(x0, y, x, h)

C#

using System;

class GFG {

static double dydx( double x, double y)

{

return ((x - y) / 2);

}

static double rungeKutta( double x0,

double y0, double x, double h)

{

int n = ( int )((x - x0) / h);

double k1, k2, k3, k4;

double y = y0;

for ( int i = 1; i <= n; i++)

{

k1 = h * (dydx(x0, y));

k2 = h * (dydx(x0 + 0.5 * h,

y + 0.5 * k1));

k3 = h * (dydx(x0 + 0.5 * h,

y + 0.5 * k2));

k4 = h * (dydx(x0 + h, y + k3));

y = y + (1.0 / 6.0) * (k1 + 2

* k2 + 2 * k3 + k4);

x0 = x0 + h;

}

return y;

}

public static void Main()

{

double x0 = 0, y = 1, x = 2, h = 0.2;

Console.WriteLine( "\nThe value of y"

+ " at x is : "

+ rungeKutta(x0, y, x, h));

}

}

PHP

<?php

function dydx( $x , $y )

{

return (( $x - $y ) / 2);

}

function rungeKutta( $x0 , $y0 , $x , $h )

{

$n = (( $x - $x0 ) / $h );

$k1 ; $k2 ; $k3 ; $k4 ; $k5 ;

$y = $y0 ;

for ( $i = 1; $i <= $n ; $i ++)

{

$k1 = $h * dydx( $x0 , $y );

$k2 = $h * dydx( $x0 + 0.5 * $h ,

$y + 0.5 * $k1 );

$k3 = $h * dydx( $x0 + 0.5 * $h ,

$y + 0.5 * $k2 );

$k4 = $h * dydx( $x0 + $h , $y + $k3 );

$y = $y + (1.0 / 6.0) * ( $k1 + 2 *

$k2 + 2 * $k3 + $k4 );;

$x0 = $x0 + $h ;

}

return $y ;

}

$x0 = 0;

$y = 1;

$x = 2;

$h = 0.2;

echo "The value of y at x is : " ,

rungeKutta( $x0 , $y , $x , $h );

?>

Javascript

<script>

function dydx(x, y)

{

return ((x - y) / 2);

}

function rungeKutta(x0, y0, x, h)

{

let n = parseInt((x - x0) / h, 10);

let k1, k2, k3, k4, k5;

let y = y0;

for (let i = 1; i <= n; i++)

{

k1 = h * dydx(x0, y);

k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1);

k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * k2);

k4 = h * dydx(x0 + h, y + k3);

y = y + (1 / 6) * (k1 + 2 * k2 +

2 * k3 + k4);;

x0 = x0 + h;

}

return y.toFixed(6);

}

let x0 = 0, y = 1, x = 2, h = 0.2;

document.write( "The value of y at x is : " +

rungeKutta(x0, y, x, h));

</script>

Output: 

The value of y at x is : 1.103639

Time Complexity of above solution is O(n) where n is (x-x0)/h.
Some useful resources for detailed examples and more explanation.
http://w3.gazi.edu.tr/~balbasi/mws_gen_ode_txt_runge4th.pdf
https://www.youtube.com/watch?v=kUcc8vAgoQ0

This article is contributed by Arpit Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Find the Exact Solution to a Fourth Order Pde

Source: https://www.geeksforgeeks.org/runge-kutta-4th-order-method-solve-differential-equation/