Physics 299: Computational Physics II Project II

Transcription

1 Physics 99: Computational Physics II Project II Due: Feb 01 Handed out: 6 Jan 01 This project begins with a description of the Runge-Kutta numerical integration method, and then describes a project to implement it in our simple harmonic oscillator class. Introduction and Background We ve already discussed the Euler and the Euler-Cromer methods; although they might be your first recourse for numerically integrating Newton s laws (or some other differential equation) we already know that you have to be careful Euler s method, for instance, does not conserve energy in a simple harmonic oscillator, whereas the Euler- Cromer method does (at least over the span of one complete oscillation). The Euler and Euler-Cromer methods are first order methods meaning that the truncation error is proportional to the time step, t. 1 We can do better than this by applying higher order methods 1 Described by Carl Runge and Martin Kutta around the workhorse method is the 4th order Runge-Kutta method. Rather than jump into the description of this method, we ll get there gently, by discussing the Euler method, then the nd order Runge-Kutta method, and finally, the full-blown 4th order technique. Basic Idea Consider a first order differential equation of the form = f (x, t) so that = f (x, t), and therefore the Euler method would tell us that x n+1 = x n + f (x n, t n ) t. The picture that goes along with this method is shown in Figure 1. Notice that in the arbitrary case chosen that the estimated shift x is greater than the actual increase in x(t). Of course, in some cases, the shift might be less, but the basic idea is that if t is sufficiently small, we can use the Euler method to estimate the change in x(t) to very high accuracy. But, for any finite interval of time T, the total error will be proportional to the size of the time step, t. So, how can we improve our estimate? Notice that the Euler method relies solely on the information from the beginning of the interval. Clearly, if we evaluate the slope at other points in the interval, we can get a better estimate. The Runge-Kutta methods do just that. Figure 1: The Euler method estimates x n+1 by using the position and the slope at the beginning of the interval. Notice that in this example, the Euler method overestimates the value of the function a the end of the interval because of the high slope at the beginning of the interval.

2 physics 99: computational physics ii project ii Second Order Runge-Kutta The second order Runge-Kutta method uses the Euler method to extrapolate to the estimated position at the middle of the time interval (i.e. at t = t n + t/) and then calculates the slope at this estimated point. This slope is then used to make a refined estimate of the total change in x. Here s the scheme in algebraic terms: and then k 1 f (x n, t n ) t ( Euler method ), k f (x n + k 1, t n + t ) t ( x via midpoint method ), finally, we estimate x n+1 via x n+1 = x n + k + O[( t) 3 ] The picture illustrating this method may be seen in Figure. In this figure, we see that Euler method is used to estimate the midpoint value of the function; it does so by using the slope at the beginning of the interval. Then we use the slope at this estimated midpoint to then form a better estimate of the change of the funtion over the time step t. Clearly, as Figure illustrates, the second order Runge- Kutta method is a much better approximation to the final value of the function at the end of the interval. In the final analysis, this method approximates the ending value of the function by performing two slope evaluations; one at the beginning of the interval, and one at the estimated midpoint. Figure : The nd order Runge-Kutta method estimates x n+1 by using the slope at the midpoint (as estimated by the Euler method) of the interval rather than at the beginning of the interval. Notice that k = f (x n + k 1 /, t n + t/) t, and the second order estimate is clearly much closer to the actual final value of the function at the end of the interval. A numerical example A numerical example will help illustrate the improvement the nd order Runge-Kutta method gives over the Euler method. Suppose we have the differential equation = x; where x(t = 0) = 1.0. It is a simple matter so show that the exact solution to this differential equation is x(t) = e t, and we ll calculate in detail how to estimate x(t = 0.4), both by the Euler method, and the nd order Runge-Kutta method. Euler method The Euler method tells us that x n+1 = x n + f (x n, t n ) t,

3 physics 99: computational physics ii project ii 3 so we start with x(0) = 1.0 and then calculate successive values of x by noting that f (x, t) = x, and with t = 0.1, we find x(0.1) = ( 1.0)0.1 = 0.9 x(0.) = ( 0.9)0.1 = 0.81 x(0.3) = ( 0.81)0.1 = 0.79 x(0.4) = ( 0.79)0.1 = and, since the exact value is e 0.4 = , we see that the Euler method (with time step 0.1) gives a.1% error. nd order Runge-Kutta Since the nd order Runge-Kutta method makes twice as many evaluations of the derivative in the interval, we will increase our time step by a factor of to make a more fair comparison; this way we perform about the same number of computational steps to traverse the interval, and we ll see if we still improve our estimate. So, we use t = 0. and then calculate carry through the nd order calcuations; recall that the second order Runga-Kutta steps are k 1 f (x n, t n ) t and k f (x n + k 1 t n + t ) t where in our case, f (x, t) = x, so that we have for the first time step: k 1 = f (x 0, t 0 ) 0. = 0. and for the second time step: k = f ( ) 0. = 0.18 x(0.) = x(0) + k = 0.8 Notice that in this example, the derivative, f (x, t), has no time dependence. Thus, in the calculation of k 1, we merely have f (x 0, t 0 ) = f (1, 0) = 1. k 1 = (0.8) 0. = k = f ( ) 0. = x(0.) = x(0.) + k = and, recalling that the exact answer is , we see that the nd order method (with time step 0.) gives a 0.3% error, almost 10 times smaller than the Euler method! 4th order Runge-Rutta This is the real deal; it s the workhorse numerical integration technique. It s a forth order method, so that when integrating over a finite

4 physics 99: computational physics ii project ii 4 time T, the truncation error is of order ( t) 4. The 4th order Runge- Kutta method evaluates the derivative at the beginning of the interval, at two different points in the middle of the interval, and once at the end of the interval. It weights the two middle estimates more heavily than the endpoints. Here s the method: and then, k 1 = f (x n, t n ) t (slope at beginning) k = f (x n + k 1, t n + t ) t (same as nd order) k 3 = f (x n + k, t n + t ) t (slope at middle) k 4 = f (x n + k 3, t n + t) t (slope at end) x n+1 = x n (k 1 + k + k 3 + k 4 ) + O[( t) 5 ] 4th Order Runge-Kutta & Newton s Law More often than not, we have to integrate a second order equation such as d x = 1 F(x, v, t) = a(x, v, t); m to accomplish this, we use the standard trick of breaking this up into two first order equations: = v(t) and dv = a(x, v, t). Now we have the following scheme for implementing 4th order Runge- Kutta: k 1v = a(x n, v n, t n ) t k 1x = v n t k v = a(x n + k 1x, v n + k 1v, t n + t ( ) ) t k x = v n + k 1v t k 3v = a(x n + k x, v n + k v, t n + t ( ) ) t k 3x = v n + k v t and finally: k 4v = a(x n + k 3x, v n + k 3v, t n + t) t k 4x = (v n + k 3v ) t v n+1 = v n (k 1v + k v + k 3v + k 4v ) x n+1 = x n (k 1x + k x + k 3x + k 4x )

5 physics 99: computational physics ii project ii 5 Project For this project, take your code from Project 1, and add the ability to numerically integrate using the second order Runge-Kutta method or the 4th order method. Then, I d like you to 1. Compare the accuracy of the Euler-Cromer method, nd order Runge- Kutta, and 4th order Runge-Kutta method; typeset a nice table in LATEX that shows numerically the solutions given by the methods. For instance, you might want to let the oscillator make 1,000,000 oscillations and see if it returns to it s starting point. How does the accuracy depend on the time step for the different methods? How much bigger can your time step be in 4th order Runge-Kutta be and achieve the same accuracy as the Euler-Cromer method?. Make a nice plot (or likely several) that illustrate the accuracy of the methods. 3. Write this all up in a nice LATEX document; I wrote this using the Tufte-LaTeX handout class (google it).