Section 7.4 Runge-Kutta Methods
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1 Section 7.4 Runge-Kutta Methods Key terms: Taylor methods Taylor series Runge-Kutta; methods linear combinations of function values at intermediate points Alternatives to second order Taylor methods Fourth order Runge-Kutta methods; very popular
2 Taylor methods require derivatives of f(t, y), which makes them difficult to use effectively. This drawback is so severe that they are seldom used in practice. However, there is another approach that is very effective. Runge-Kutta methods are related to Taylor methods in the following way. Instead of using derivatives of f(t, y), Runge-Kutta methods use evaluations of the f(t, y) at alternative pairs of points (t, y) that are not restricted to discrete points with t = t 0, t 1 = t 0 +h, t 2 = t 0 + 2h, etc. These techniques were developed around 1900 by the German mathematicians C. Runge and M.W. Kutta. The determination of these alternative pairs requires certain parameters to be selected so that we match the Taylor expressions and maintain the same power of stepsize h in the error expression. The advantage of this approach is that we use only evaluations of f not its derivatives. Thus the cost of using Runge-Kutta methods is FUNCTION EVALUATIONS instead of DERIVATIVE EVALUATIONS. Hence the Runge-Kutta methods are simple to use and have error properties corresponding to the Taylor methods upon which they are based.
3 The development of R-K methods expresses the difference between y(t n+1 ) and y(t n ) as a linear combination of function evaluations of f(t, y(t)). Then coefficients in such linear combinations are determined so that terms of Taylor expansions are matched. The general approach is to construct a difference equation of the form where the coefficients a j are constants and the k j are expressions involving stepsize h times evaluations of function f. That is, where the coefficients a j, α j, and β js are chosen to match terms of the Taylor expansion of y(t). (Note that k j can depend on values of previous k s and the value t n + α j h t n+1.)
4 Example: Construct a R-K method of the form We use Taylor expansions for the corresponding expression involving the exact solution y(t), which looks like then following. (We replace w s with y(t).) We choose parameters a 1, a 2, α, and β to match terms of the expansion Recall that When we do this (via ugly algebra & calculus) we get THREE EQUATIONS in FOUR UNKNOWNS (nonlinear system).
5 When these equations are satisfied then the R-K method has local truncation error O(h 2 ). That is, it will in effect act like a second order Taylor method, but with a (somewhat) different error expression. Since we have a nonlinear system of 3 equations in 4 unknowns we expect many solutions.
6 Some particular choices for the parameters a 1, a 2, α, and β: (Warning: the names vary from book to book. The names used here correspond to those used in MATLAB software and are not the same as those used in the text by Bradie.) Midpoint Method Bradie name,modified Euler method Modified Euler Method Bradie name,heun method Heun s Method Bradie name, Optimal RK2 method
7 Higher order R-K methods are constructed in a similar way just using more terms of the difference equation with and of course matching more terms of the Taylor expansion of y(t+h). The algebra & calculus becomes very tedious. One of the most popular R-K methods is a 4th order method given by the following: Note that the k-values must be computed in order.
8 Exact solution y(t) = 2 e 1 (-t) In a graph the sets of data look coincident. The main computational effort in RK-methods is evaluating f. There should be some payoff to using a method like RK4. We hope that such higher order methods will let us use larger step sizes h and maintain the same accuracy as lower order methods with smaller step sizes.
9 RK4 should give more accurate answers than Euler s method with about a stepsize about 4 times larger. RK4 should give more accurate answers than RK2 with about a stepsize about 2 times larger. To get methods which are as accurate as RK-methods, but with fewer evaluations of f, we need to use multistep methods. (Discussed in the next section.)
10 If you had to use one of the Runge-Kutta methods by hand (calculator) care must be used to get the substitutions correct. We illustrate this below. Example: The RK2 Midpoint Method is given by This is what you substitute for t. This is what you substitute for y. is evaluated where t is and y is
11 A comparison of behaviors:
12 MATLAB command names: eulersys rkmidpt rkmodeuler rkheun rk4th All of these routines are designed to handles systems of ODEs: The system has the form X' = F(t,X) X(t0) = X0 <-- initial condition where X is column vector and F is a vector function. The independent variable is t and the components of F(X) are functions of the variables t and x(1), x(2),..., x(k), for k<=9. Before using any of these read the help files carefully. The names used by the m-files are different than names appearing in the text.
13 Example: Use rk4th on IVP y = y t 2 + 1, 0 t 2, y(0) = 0.5 with stepsize h = 0.1. Use command >>rk4th What you see on the MATLAB screen that requests input: Enter size (<= 9) of the system: k = 1 Enter f1 = x(1)-t^2+1 You can see a table of output by performing several more steps: Your system is X' = x(1)-t^2+1 1 Accept current functions in f. 2 Re-enter functions. 0 Quit! Enter your choice ---> 1 Enter the initial condition X(t0) = [x1 x2...xk] Enter value of t0: t0 = 0 Enter 1 initial values in the form [a b.. etc]: X(t0) = 0.5 Enter right end of solution interval: b = 2 Enter the stepsize: h = 0.1 The solution at t = b is W = e+00 Input in red output in blue. t w The true solution is, y = 2t -0.5e t + t so you could consturct the error at each t value and graph the error. B&F 5.4 Examp 4
14 How do you approximate the solution of an IVP whose DE is more than first order? Answer: Convert the higher order DE to a system of first order DEs. Example: Use rk4th to approximate the solution of the IVP Linear second order nonhomogenous DE rk4th INPUT 2 3 t y'' 2ty' 2y t ln(t), 1 t 2, y( 1) 1, y'( 1) 0, with h The true solution is y(t) t t ln(t) t. u 1(t) y(t) Let. u (t) y'(t) 2 u ' y' u 1 2 Now differentiate to get 2 2. u 2 ' y'' u2 u 2 1 t ln(t) t t u 1 x(1); u 2 x(2) in rk4th Enter size (<= 9) of the system: k = 2 Enter f1 = x(2) Enter f2 = (2/t)*x(2)-(2/t^2)*x(1)+t*log(t) Enter the initial condition X(t0) = [x1 x2...xk] Enter value of t0: t0 = 1 Enter 2 initial values in the form [a b.. etc]: X(t0) = [1 0] Enter right end of solution interval: b = 2 The solution at t = b is W = f e e+00 f2 B&F 5.9 #3b Enter the stepsize: h = 0.1
15 From rk4th t u 1 (t) u 2 (t) From true soln t y(t) t absolute error e e e e e e e e e e e e e e e e e e e e e-06 True soln. Absolute error
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