Runge-Kutta RK4 Method

As we know, Taylor’s series is a numerical method used for solving differential equations and is limited by the work to be done in finding the derivatives of the higher-order. To overcome this, we can use a new category of numerical methods called Runge-Kutta methods to solve differential equations. These will give us higher accuracy without performing more calculations. These methods coordinate with the solution of Taylor’s series up to the term in h_r, where r varies from method to method, representing the order of that method. One of the most significant advantages of Runge-Kutaa formulae is that it requires the function’s values at some specified points.

Before learning about the Runge-Kutta RK4 method, let’s have a look at the formulas of the first, second and third-order Runge-Kutta methods.

Consider an ordinary differential equation of the form dy/dx = f(x, y) with initial condition y(x₀) = y₀. For this, we can define the formulas for Runge-Kutta methods as follows.

1st Order Runge-Kutta method

y₁ = y₀ + hf(x₀, y₀) = y₀ + hy’₀ {since y’ = f(x, y)}

This formula is same as the Euler’s method

2nd Order Runge-Kutta method

y₁ = y₀ + (½) (k₁ + k₂)

Here,

k₁ = hf(x₀, y₀)

k₂ = hf(x₀ + h, y₀ + k₁)

3rd Order Runge-Kutta method

y₁ = y₀ + (⅙) (k₁ + 4k₂ + k₃)

Here,

k₁ = hf(x₀, y₀)

k₂ = hf[x₀ + (½)h, y₀ + (½)k₁]

k₃ = hf(x₀ + h, y₀ + k¹) such that k¹ = hf(x₀ + h, y₀ + k₁)

What is Fourth Order RK Method?

The most commonly used Runge Kutta method to find the solution of a differential equation is the RK4 method, i.e., the fourth-order Runge-Kutta method. The Runge-Kutta method provides the approximate value of y for a given point x. Only the first order ODEs can be solved using the Runge Kutta RK4 method.

Runge-Kutta Fourth Order Method Formula

The formula for the fourth-order Runge-Kutta method is given by:

y₁ = y₀ + (⅙) (k₁ + 2k₂ + 2k₃ + k₄)

Here,

k₁ = hf(x₀, y₀)

k₂ = hf[x₀ + (½)h, y₀ + (½)k₁]

k₃ = hf[x₀ + (½)h, y₀ + (½)k₂]

k₄ = hf(x₀ + h, y₀ + k₃)

Runge-Kutta RK4 Method Solved Examples

Example 1:

Consider an ordinary differential equation dy/dx = x² + y², y(1) = 1.2. Find y(1.05) using the fourth order Runge-Kutta method.

Solution:

Given,

dy/dx = x² + y², y(1) = 1.2

So, f(x, y) = x² + y²

x0 = 1 and y0 = 1.2

Also, h = 0.05

Let us calculate the values of k₁, k₂, k₃ and k₄.

k₁ = hf(x₀, y₀)

= (0.05) [x₀² + y₀²]

= (0.05) [(1)² + (1.2)²]

= (0.05) (1 + 1.44)

= (0.05)(2.44)

= 0.122

k₂ = hf[x₀ + (½)h, y₀ + (½)k₁]

= (0.05) [f(1 + 0.025, 1.2 + 0.061)] {since h/2 = 0.05/2 = 0.025 and k₁/2 = 0.122/2 = 0.061}

= (0.05) [f(1.025, 1.261)]

= (0.05) [(1.025)² + (1.261)²]

= (0.05) (1.051 + 1.590)

= (0.05)(2.641)

= 0.1320

k₃ = hf[x₀ + (½)h, y₀ + (½)k₂]

= (0.05) [f(1 + 0.025, 1.2 + 0.066)] {since h/2 = 0.05/2 = 0.025 and k₂/2 = 0.132/2 = 0.066}

= (0.05) [f(1.025, 1.266)]

= (0.05) [(1.025)² + (1.266)²]

= (0.05) (1.051 + 1.602)

= (0.05)(2.653)

= 0.1326

k₄ = hf(x₀ + h, y₀ + k₃)

= (0.05) [f(1 + 0.05, 1.2 + 0.1326)]

= (0.05) [f(1.05, 1.3326)]

= (0.05) [(1.05)² + (1.3326)²]

= (0.05) (1.1025 + 1.7758)

= (0.05)(2.8783)

= 0.1439

By RK4 method, we have;

y₁ = y₀ + (⅙) (k₁ + 2k₂ + 2k₃ + k₄)

y₁ = y(1.05) = y₀ + (⅙) (k₁ + 2k₂ + 2k₃ + k₄)

By substituting the values of y₀, k₁, k₂, k₃ and k₄, we get;

y(1.05) = 1.2 + (⅙) [0.122 + 2(0.1320) + 2(0.1326) + 0.1439]

= 1.2 + (⅙) (0.122 + 0.264 + 0.2652 + 0.1439)

= 1.2 + (⅙) (0.7951)

= 1.2 + 0.1325

= 1.3325

Example 2:

Find the value of k₁ by Runge-Kutta method of fourth order if dy/dx = 2x + 3y² and y(0.1) = 1.1165, h = 0.1.

Solution:

Given,

dy/dx = 2x + 3y² and y(0.1) = 1.1165, h = 0.1

So, f(x, y) = 2x + 3y²

x₀ = 0.1, y₀ = 1.1165

By Runge-Kutta method of fourth order , we have

k₁ = hf(x₀, y₀)

= (0.1) f(0.1, 1.1165)

= (0.1) [2(0.1) + 3(1.1165)²]

= (0.1) [0.2 + 3(1.2465)]

= (0.1)(0.2 + 3.7395)

= (0.1)(3.9395)

= 0.39395

Runge-Kutta RK4 Method Problems

1. Using the Runge-Kutta method of order 4, find y(0.2) if dy/dx = (y – x)/(y + x), y(0) = 1 and h = 0.2.
2. Find the value of y(0.3) from the differential equation dy/dx = 3e^x + 2y; y(0) = 0, h = 0.3 by the fourth order Runge-Kutta method.
3. Using RK4 method to find y(0.2) and y(0.4) if dy/dx = 1 + y + x²; y(0) = 0.5