[ ] is a vector of size p.

Transcription

1 Lecture 11 Copyright by Hongyun Wang, UCSC Recap: General form of explicit Runger-Kutta methods for solving y = F( y, t) k i = hfy n + i1 j =1 c ij k j, t n + d i h, i = 1,, p + p j =1 b j k j A Runge-Kutta method is completely specified by 4 items (p, c, d and b). p: the number of stages (a scalar) c 11 c 1 p c = c p1 c pp d = d 1 d p is a p p matrix. [ ] is a vector of size p. [ ] is a vector of size p. b = b 1 b p Numerical error estimation in solving ODEs y N ( h): numerical approximation of y( Nh) obtained with time step h ( h) y N ( h) y( Nh) Global error: E N The global error is estimated as E N y N ( h) h ( h) y 2N p y N h ( h) y N p is NOT good! - 1 -

2 High order ODEs and first order ODE systems Example: A damped pendulum (Draw a pendulum with, L and mg) Newton s second law: () ml t MassAcceleration () = cl t mgsin t Drag sin ( ( t )) ( ()) Gravity ==> ( t) = c m ( t) g L This is a second order non-linear ODE. To solve it numerically, we need to convert it to a first order ODE system. Question: How to convert a second order ODE into a first order ODE system? Answer: We treat both the function and the derivative as unknowns. We introduce two unknown functions: () t and v()= t () t We have t = vt = ( t) = c m v t ( t) g L sin ( ( t )) = c m vt g L sin ( ( t ) ==> d dt () t vt () = vt () c m vt () g L sin () t This is a first order ODE system. It is in the form of W ( t) = F W ( t), t where W t = ( t) vt F W ( t), t = vt c m vt g L sin ( t ) - 2 -

3 Example: van der Pol equation y + μ y 2 1 y + y = 0 This is a second order non-linear ODE. (Go through the physical background and limit cycle of van der Pol equation) We compare van der Pol equation with damped pendulum. + c m Dissipation + g L sin ( )= 0 In van der Pol equation, the dissipation coefficient is μ( y 2 1). When y is large, the dissipation is positive (oscillation is damped). When y is small, the dissipation is negative (energy is injected to magnify oscillation). Thus, the oscillation converges to the steady state amplitude that balances the positive dissipation and the negative dissipation. This is called a limit cycle. We will see detailed behavior of limit cycle when we solve van der Pol equation in numerical demonstration. (Then skip the rest of the example in lecture) To solve it numerically, we need to convert it to a first order ODE system. We introduce two unknown functions: yt () and v()= t y() t We have y ( t) = vt v ( t) = y ( t) = μ y 2 ( t)1 d yt () ==> dt vt () = vt () μ( y 2 () t 1)vt This is a first order ODE system. We can write it in the form. W ( t) = F W ( t), t where W t = yt vt F W ( t), t = vt μ ( y 2 ( t)1)vt yt y ( t) yt = μ ( y 2 ( t)1)vt yt ()yt () - 3 -

4 Next, we look at how to implement Runge-Kutta methods to solve W = F ( W, t). sample code calc_srk4.m in directory ODE_sys_RK clear m=2; w0=[1, 0]; h=0.1; nstep=40/h; w=zeros(nstep+1,m); t=zeros(nstep+1,1); t(1)=0; w(1,1:m)=w0; p=4; d=[0, 1/2, 1/2, 1]; c=[0, 0, 0, 0; 1/2, 0, 0, 0; 0, 1/2, 0, 0; 0, 0, 1, 0]; b=[1/6, 1/3, 1/3, 1/6]; k=zeros(p,m); for j=1:nstep, for i=1:p, k(i,1:m)=h*f_sys(w(j,1:m)+c(i,1:i-1)*k(1:i-1,1:m), t(j)+d(i)*h); end w(j+1,1:m)=w(j,1:m)+b*k; t(j+1)=t(j)+h; end - 4 -

5 save data_srk4 sample code f_sys.m in directory ODE_sys_RK (for damped pendulum) function [z]=f_sys(w,t) mass=1.0; c=0.1; g=10; L=10; z=zeros(1,2); theta=w(1); v=w(2); z(1)=v; z(2)= -c/mass*v - g/l*sin(theta); Sample code f_sys.m for van der Pol oscillator. (Skip it in lecture) function [z]=f_sys(w,t) mu=0.1; z=zeros(1,2); y=w(1); v=w(2); z(1)=v; z(2)=-mu*(y^2-1)*v - y; - 5 -

6 Implicit Runge-Kutta methods We describe briefly the general form of implicit Runge-Kutta methods. k i = hf( y n + c i1 k 1 + c i 2 k 2 ++ c ip k p, t n + d i h), i = 1,, p Or in the summation notation k i = hfy n + p j =1 c ij k j, t n + d i h, + b 1 k 1 + b 2 k b p k p Or in the summation notation + b j k j p j =1 i = 1,, p Below we write the two implicit methods we learned so far (backward Euler method and trapezoidal method) in the form of implicit Runger-Kutta methods. Backward Euler method: + hf y n +1, t n +1 Let k 1 = hf(, t n+1 ). We have k 1 = hf( y n + k 1, t n + h) + k 1 In terms of p, c, d, b, backward Euler method is specified by p = 1, c = (), 1 d = (), 1 b = () 1 Trapezoidal method: + h 2 Fy n, t n ( + Fy ( n +1, t n +1 )) Let k 1 = hf( y n, t n ) and k 2 = hf(, t n+1 ). We have k 1 = hf( y n, t n ) k 2 = hf y n k k 2, t n + h - 6 -

7 + 1 2 k k 2 In terms of p, c, d, b, trapezoidal method is specified by 0 0 p = 2, c = 1 1, d = ( 0, 1), b = , 1 2 Stiff problems and advantages of implicit methods Although implicit methods are difficult to implement numerically, in some situations it is necessary to use implicit methods. Question: What happens if we use Euler method to solve y = ( y cos() t ), = 10 8? y( 0)= 0 To answer this question, let us first look at a simpler model problem. A model problem: y = y, > 0 and is large Exact solution: yt = y( 0)e t 0 as t The numerical solution should follow the general trend of the exact solution. We require that y n be bounded as n. Behavior of Euler method: + hf y n,t n ==> hy n ==> = ( 1 h)y n ==> y n = ( 1 h) n y 0 y n is bounded if and only if 1 h 1 iff 1 h 1 1 iff 0 h 2-7 -

8 iff h 2 For large, time step h has to be very small. = 10 8 ==> h 2 = ==> To reach t final = 10, we need N = t final h = steps. That is, we need 500 million steps! Summary of Euler method applied to y = y, > 0 If the exponential decay is resolved (when time step h is small enough), then the numerical solution is good. If the exponential decay is not resolve (when time step h is not small enough), then the numerical solution blows up. In other words, the numerical solution is well behaved only if the exponential decay is resolved. Behavior of backward Euler method: + hf,t n +1 ==> hy n +1 ==> + h ==> = h y n 1 ==> y n = 1 + h n y 0 ==> y n 0 as n for any value of h. Summary of backward Euler method applied to y = y, > 0 Even if the exponential decay is not resolved (when time step h is not small enough), the numerical solution retains the property that y n 0 as n. In other words, the numerical solution is always well behaved even if the exponential decay is not resolved. We are ready to study the original problem - 8 -

9 y = ( y cos() t ), = 10 8 y( 0)= 0 Exact solution: yt = cos( t) sin t e t (See Appendix for the derivation of the exact solution) The system has two very different time scales: fast decay of e t and slow evolution of cos(). t ==> It is a stiff problem (see the definition below). Definition of stiff problem: A problem is called stiff if it has (at least) two very different time scales. Behavior of Euler method for solving y = ( y cos() t ): + hf ( y n,t n ) ==> h ( y n cos( t n )) y n is bounded if and only if h 2. When we use Euler method or other explicit methods to solve this problem, we are constrained by two requirements: we need to use h 2 to keep y n bounded (due to the fast decay of e t ) we need to run the simulation to t final > 2 to see the evolution of cos t (due to the slow evolution of cos()) t Together these two requirements imply that N = t final h > 2 2 = > That is, we need 300 million steps! In general, when we use an explicit method to solve a stiff problem, ()

10 we need to use time step h small enough to resolve the fast time evolution we need to run simulation to a final time large enough to capture the slow time evolution ==> It is very expensive to solve a stiff problem using an explicit method. Behavior of backward Euler method for solving y = ( y cos() t ): + hf(,t n+1 ) ==> h ( y n +1 cos( t n +1 )) ==> + h + h cos t n +1 1 ==> = 1 + h y + h cos t n ( n+1) y n is bounded for any value of h. In general, when we use an implicit method to solve a stiff problem, the numerical solution is always well behaved even if the fast time evolution is not resolved. Appendix: y = y cos( t) Exact solution of y( 0) = 0 First, moving y to the left, we have y + y = cos( t) Multiplying by e t, we get y e t + ye t = cos( t)e t ==> ( ye t ) = cos( t)e t Integrating from t = 0 to t = s yields ys e s y 0 s = cos t 0 e t dt

11 ==> ys = e s cos t s 0 e t dt Here we have used the initial condition y( 0) = 0. Using the integration formula e ax cos( bx)dx = eax ( acos( bx) + bsin( bx) ) + C a 2 + b 2 we have s cos ( t )e t dt = et cos( t) + sin( t) = e s [ ] [ cos( s) + sin( s) ] Substituting this into the expression of ys, we obtain ys = cos( s) sin s e s t =s t =0-11 -