Module1: Numerical Solution of Ordinary Differential Equations. Lecture 6. Higher order Runge Kutta Methods

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1 Module1: Numerical Solution of Ordinary Differential Equations Lecture 6 Higher order Runge Kutta Methods Keywords: higher order methods, functional evaluations, accuracy

2 Higher order Runge Kutta Methods The local error of RK4 is O(h 5 ) while the global error is O(h 4 ). If the solution of IVP is to be obtained for t in the interval (0,1] with h=0.1 using RK4, then the solution is to be computed at 10 grid points t=0.1, 0.2,,1.0. At each grid point four functional evaluations are required. This way, the solution at t=1 requires 40 functional evaluations and the accuracy will be of order If Euler method is to be used then h= will yield the desired accuracy of Accordingly, the solution is to be computed at grid points to reach to t=1.this means Euler method requires10,000 functional evaluations (one for each grid point) to compute approximate solution at t=1. Evidently, RK4 is very efficient and saves lots of computational effort. We can further improve the efficiency by employing still higher order Runge-Kutta methods. The higher order methods, say of order 5 and 6 are developed on the same lines. These are more efficient as the higher accuracy is achieved with less computational effort as compared to lower order methods. Runge-Kutta method ( say of order 4) is applied for obtaining approximate solution y 1 at t=t 1 =t+h for IVP with chosen value of h. The value of h is then halved and the solution is again obtained at t 1. Now 7 functional evaluations are needed to compute y1. If the difference between two solutions is not substantial then the approximation is accepted. Otherwise, the iteration (halving h) is repeated again till the desired accuracy is achieved. Each halving h will require 4+7 functional evaluations. This way higher accuracy can be achieved with more computational effort. In another approach, higher performance with less computational effort is achieved when the Runge-Kutta methods of different orders are used to move from one grid point to the next point. One such method known as Runge-Kutta Fehlberg method is based on the formulae (1.19) given below. In this, two estimates are obtained for y k+1 using RK method of global error O(h 4 ) and O(h 5 ) with six functional evaluations.

3 K f( t, y ) 1 k k h h K2 f( tk, yk K1) 4 4 3h 3h 9h K3 f( tk, yk K1 K2) h 1932h 7200h 7296h K4 f( tk, yk K1 K2 K3) h 3680h 845h K5 f( tk h, yk K1 8hK2 K3 K4) h 8h 3544h 1859h 11h K6 f( tk, yk K1 2hK2 K3 K4 K5) h yˆ k 1 yk [ K1 K3 K4 K5] h y k 1 yk [ K1 K3 K4 K5 K6] Error y ˆ k 1 yk 1 [ K1 K3 K 4 K5 K6] h (1.19) Since the lower order method is of order four, the step size adjustment factor s can be computed as s T i 1 14 / ˆ Here, ε is the accuracy requirement and T is the truncation errort y y i 1 i 1 i 1. If the desired accuracy is not achieved the solution is iterated taking new value of h. Depending upon error requirement the step size h can be increased or decreased. The solution y k+1 of desired accuracy is obtained at t k+1 =t k +sh. The method is known as RKF45. To implement the method, the user specifies the allowable smallest step size h min, largest step size h max and the maximum allowable local truncation error ε. The following algorithm is used to solve IVP using RKF45 formulae with self adjusting variable step sizes:

4 Algorithm RKF45 [Step 1] set k=0, t=a=t 0, y=y k, h=h max, flag=1 [Step 2] while (flag==1) repeat steps 3-7 [Step 3] compute y, yˆ andr yˆ y k 1 k 1 k 1 k 1 [Step 4] compute s [Step 5] if s >h max then h=h max else if (s<h min ) exit else h=s [Step 6] if ( R ) flag=0 [Step 7] go to step 2 [Step 8] t=t+h, y=yk+1, k++; flag=1 [Step 9] if( t<=b) goto step 2 [Step 10] stop Example 1.8: Solve IVP using y 2ty 2, y( 0) Solution: the matlab code for solving the system is given as function dydt= rkf4(t,y) dydt=y.^2.2*t; [t,y]=ode45(@rkf4,[0 1], 1); %[0 1]-time span, 1-initial condition t y Table 1.6 Solution of example 1.8

5 Example 1.9: find the solution of IVP using higher order order Runge-Kutta method RKF45 given in (1.19) in the interval (0,2) taking h max=0.25, hmin=0.01 and accuracy as dy 2 y t 1; y( 0) 0. 5 dt Solution: The detailed solution of the problem is worked out in the excel sheet rkf45.xls h t y k1 k2 k3 k4 k5 Kk6 y Table 1.7 Details of Solution of example 1.9 tk+1 y5 exact tk+1 y5 exact Table 1.8 Comparison with exact Solution of example 1.9

6 Exercise Apply Eulers method to solve the initial value problems in the interval (0,1]: y 2y 3t, y( 0) 2 y 2ty, y( 0) x y, y( 0) y Take h=0.1 and compare with exact solutions. Reduce h=0.05 and again solve the two IVPs. 1.2 Apply modified Euler method to IVPs in Ex. 1.1 with h=0.1. Compare the effort and accuracy achieved in two methods. 1.3 Can Eulers method be applied to solve the following IVP in the interval (0,2) y () t 4 y 2, y( 0) Show that Eulers method and modified Eulers method fail to approximate the solution yt () 3/ 2 8t of the IVP / y () t 6y 13, y( 0) Solve the following IVP using Runge Kutta method of order two and four: y () t t 2 y 2, y() 1 0 at t 2using h=0.5 y () t y sin x, y( 0) 2 at t 1using h=0.1