Numerical Analysis Lecture 38

Chapter 7 Ordinary Differential Equations

Introduction Taylor Series Euler Method Runge-Kutta Method Predictor Corrector Method

TAYLOR’S SERIES METHOD

We considered an initial value problem described by

We expanded y (t ) by Taylor’s series about the point t = t0 and obtain

Noting that f is an implicit function of y, we have

Similarly

EULER METHOD

Consider the differential equation of first order with the initial condition y(t0) = y0.

The value of y corresponding to t = t1

Similarly

Then, we obtained the solution in the form of a recurrence relation

MODIFIED EULER’S METHOD

The recurrence relation is the modified Euler’s method.

RUNGE – KUTTA METHODS

We considered the IVP We also defined and took the weighted average of k1 and k2 and added to yn to get yn+1

We obtained Implying

We considered two cases, Case I We choose W2 = 1/3, then W1 = 2/3 and

Case II: We considered W2 = ½, then W1 = ½ and Then

Finally, we obtain The expression can further be simplified to

Therefore, the expression for local truncation error is given by

The fourth-order R-K method was described as

where

Example Use the following second order Runge-Kutta method described by

where and and find the numerical solution of the initial value problem described as at x = 0.4 and taking h = 0.2.

Solution Here We calculate

Now, using the given R-K method, we get Now, taking x1 = 0.2, y1 = 1.24, we calculate

Again using the given R-K method, we obtain

Example Solve the following differential equation with the initial condition y(0) = 1, using fourth- order Runge-Kutta method from t = 0 to t = 0.4 taking h = 0.1

Solution The fourth-order Runge-Kutta method is described as (1) where

In this problem, As a first step, we calculate

Now, we compute from Therefore y(0.1) = y1=1.1103 In the second step, we have to find y2 = y(0.2)

We compute

From Equation (1), we see that Similarly we calculate,

Using equation (1), we compute Finally, we calculate

Using them in equation (1), we get which is the required result.

RUNGE – KUTTA METHOD FOR SOLVING A SYSTEM OF EQUATIONS

The fourth-order Runge-Kutta method can be extended to numerically solve the higher-order ordinary differential equations- linear or non-linear

For illustration, let us consider a second order ordinary differential equation of the form

Using the substitution this equation can be reduced to two first-order simultaneous differential equations, as given by

Now, we can directly write down the Runge-Kutta fourth-order formulae for solving the system. Let the initial conditions of the above system be given by

Then, we define

Now, using the initial conditions yn, pn and 4th -order R-K formula, we compute

This method can be extended on similar lines to solve system of n first order differential equations.

Example Solve the following equation Using 4th order Runge-Kutta method for x = 0.2, with the initial values y(0) = 1, y’(0)=0

Solution Let Then Thus, the given equation reduced to two first-order equations.

In the present problem, we are given x0 = 0, y0 = 1, p0 =yo’ =0 Taking h = 0.2, we compute

Now, y (0.2) = y1 is given by

Therefore, the required solution is

Numerical Analysis Lecture 38