Initial Value Problems

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1 Numerical Analysis, lecture 13: Initial Value Problems (textbook sections , 10.7) differential equations standard form existence & uniqueness y 0 y 2 solution methods x 0 x 1 h h x 2 y1 Euler, Heun, and Runge-Kutta adaptive step size

2 Differential equations describe change over time or space (p ) M z + Q z + Kz = 0 t T = i(κ T ) R 2 C R U b R R 4 U e R 3 U = ΔU i + C i Δ U i + I i (U) i R i i i y 1 = α 1 y 1 β 1 y 1 y 2 y 2 = α 2 y 2 + β 2 y 1 y 2 Numerical Analysis, lecture 13, slide 2

3 A high-order ODE can be written as a system of 1st-order ODEs z ( p) = F(x,z, z, z,,z ( p 1) ) a p-th order ODE y =z 1 y = z 2 y p =z ( p 1) y = f (x, y) p-vectors example (p. 311) M z + Q z + Kz = 0 y 1 = z, y 2 = z y 1 = y 2, y 2 = M 1 (Qy 2 + Ky 1 ) y 1 y 2 = y 2 M 1 (Qy 2 + Ky 1 ) f (x,y) exercise (p. 355) 3 z + 4x z + sin z = q(x) Numerical Analysis, lecture 13, slide 3

4 Initial value problem = differential equation & initial condition (p. 310) a differential equation specifies a direction field y = f (x, y) y integral curves x initial value problem Find the integral curve that passes through (x 0, y 0 ). initial condition Numerical Analysis, lecture 13, slide 4

5 Some initial value problems do not have unique solutions no solution ( y ) 2 + y 2 = 0, y(0) = 1 1 solution not unique y = y 1 3, y(0) = 0 y(x) = 0 if x α ( 2 (x α) ) 3 2 otherwise 3 y x blowup solution exists only locally y y = 1+ y 2, y(0) = 1 y(x) = tan(x + π 4) if 3π 4 < x < π x Numerical Analysis, lecture 13, slide 5

6 Lipschitz continuity of f ensures local existence and uniqueness (p ) Theorem (p. 312) If f (x, y) f (x, y) L y y for all x [a,b] and y, y Ω (closed and bounded), then for any interior point (x 0, y 0 ) of [a,b] Ω the IVP y = f (x, y), y(x 0 ) = y 0 has a unique solution in a neighbourhood of (x 0, y 0 ) that can be uniquely continued to the boundary of [a,b] Ω. helpful fact If f is differentiable then L = max [a,b] Ω f y examples f y = xy, y(0) = 1 : = x is bounded in nbhd of (0,1) y ( y ) 2 + y 2 = 0, y(0) = 1 : Not standard form y = y 1 3 f, y(0) = 0 : y = 1 3 y 2/3 is unbounded at y = 0 y = 1+ y 2, y(0) = 1 : f y = 2y is bounded in nbhd of (0,1) Numerical Analysis, lecture 13, slide 6

7 Euler s method can be derived in many ways y = f (x, y), y(x 0 ) = y 0 y n+1 = y n + hf (x n, y n ) finite difference (p ) geometry (p ) f (x n, y n ) = y (x ) y(x n + h) y(x n ) n h Taylor y 0 y 2 y(x n + h) = y(x n ) + h y (x n ) h2 y (ξ) y1 x 0 x 1 h h x 2 numerical integration y(x n + h) = y(x n ) + x n +h x n f (t, y(t))dt hf (x n, y(x n )) Numerical Analysis, lecture 13, slide 7

8 IVP solved by Euler s method y n+1 = y n + hf (x n, y n ) example (p. 314) y = xy, y(0) = 1 f(x,y)=xy with h = 0.2 : y 0 = 1 y 1 = = 1 y 2 = = 1.04 y 2 y(0.4) = exact solution y(x)=exp(x 2 /2) with h = 0.1: y 0 = 1 y 1 = = 1 y 2 = = 1.01 y 3 = = y 4 = = y 4 y(0.4) = Numerical Analysis, lecture 13, slide 8

9 Euler s method in Matlab (p ) function y = eulers(f,x,y0) N = length(x); h = x(2:end)-x(1:end-1); y = zeros(n,length(y0)); y(1,:) = y0(:).'; for n = 1:N-1 y(n+1,:) = y(n,:) + h(n)*f(x(n),y(n,:).').'; end >> f x*y; >> x = 0:.1:0.4; >> y = eulers(f,x,1) y = Numerical Analysis, lecture 13, slide 9

10 Euler s global truncation error is O(h) (p ) local truncation error proof is the difference between y n+1 and the value at x n+1 of the integral curve that passes through (x n, y n ). The local truncation error of Euler's method is O(h 2 ). global truncation error is the difference between y n+1 and the value at x n+1 of the integral curve that passes through (x 0, y 0 ). y 0 lte y 2 y1 The global truncation error of Euler's method is O(h). x 0 x 1 h h x 2 Numerical Analysis, lecture 13, slide 10

11 Get better accuracy by using several stages for each time step (p ) Heun s 2-stage method has gte = O(h 2 ) k 1 = f (x n, y n ) k 2 = f (x n + h, y n + hk 1 ) y n+1 = y n + h 2 k 1 + h 2 k 2 proof that lte = O(h 3 ): example (p. 320) y = xy, y(0) = 1 with h = 0.2 : k 1 = 0, k 2 = 0.2, y 1 = 1.02 k 1 = 0.204, k 2 = , y 2 = y 2 y(0.4) = y n+1 = y n hf(x n, y n ) h f(x n, y n ) + hf x (x n, y n ) + hk 1 f y (x n, y n ) + O(h 3 ) = y n + hf(x n, y n ) h2 f x (x n, y n ) + f(x n, y n )f y (x n, y n ) + O(h 3 ) ŷ(x n + h) = ŷ(x n ) + hŷ (x n ) h2 ŷ (x n ) + O(h 3 ) = y n + hf(x n,y n )+ 1 2 h2 d dx f(x n,y n )+O(h 3 ) = y n + hf(x n,y n )+ 1 2 h2 f x (x n,y n )+f(x n,y n )f y (x n,y n ) + O(h 3 ) Numerical Analysis, lecture 13, slide 11

12 The classic 4-stage Runge-Kutta method has been popular since 1905 (p ) This method has gte O(h 4 ) k 1 = f (x n, y n ) k 2 = f (x n h, y n hk 1 ) k 3 = f (x n h, y n hk 2 ) k 4 = f (x n + h, y n + hk 3 ) y n+1 = y n h ( k 1 + 2k 2 + 2k 3 + k ) 4 example (p. 322) y = xy, y(0) = 1 with h = 0.4 : y 1 y(0.4) = Numerical Analysis, lecture 13, slide 12

13 The Runge-Kutta method in Matlab (p ) function y = rungekutta(f,x,y0) N = length(x); h = x(2:end)-x(1:end-1); y = zeros(n,length(y0)); y(1,:) = y0(:).'; for n = 1:N-1 k1 = f(x(n), y(n,:).'); k2 = f(x(n)+h(n)/2, y(n,:).'+h(n)*k1/2); k3 = f(x(n)+h(n)/2, y(n,:).'+h(n)*k2/2); k4 = f(x(n)+h(n), y(n,:).'+h(n)*k3); y(n+1,:) = y(n,:) + h(n)/6*(k1+2*k2+2*k3+k4).'; end 10 5 Error with various h >> f x*y; >> N = 2.^(0:13); >> for i=1:length(n) >> y = rungekutta(f,0:0.4/n(i):0.4,1); >> err(i) = abs(y(end)-exp(0.4^2/2)); >> end >> loglog(n,err,'o') N Numerical Analysis, lecture 13, slide 13

14 The Runge-Kutta method can solve the rocket problem from lecture 1 (p. 3-4) function dy = rocket(t,y) v = y(2); m = max(180-10*t,0); M = 120+m; dy = [v (( *v)*(t<=18)-0.1*v*abs(v))/M ]; >> t = 0:.1:40; >> Y = rungekutta(@rocket,t,[0 0]); >> plot(t,y(:,1)) h t Numerical Analysis, lecture 13, slide 14

15 Adaptive solvers change the step size according to local truncation error (p. 334) strategy Compute y n+1 with local truncation error O(h p+1 ) and y n+1 with l.t.e. O(h p+2 ). Estimate l.t.e. of y n+1 as d = y n+1 y n+1. If d > τ, redo the step with smaller step size = max 0.8h τ d 1 ( p+1), h 5 ; Otherwise, accept y n+1 and do next step with larger step size = min 0.8h τ d 1 ( p+1), 5h. Numerical Analysis, lecture 14, slide 15

16 Matlab s ODE solvers are adaptive (p ) example (p. 336) y 1 = 1+ y 2 1 y 2 4y 1, y 1 (0) = 1.5 y 2 = 3y 1 y 2 1 y 2, y 2 (0) = 3 >> f =@(x,y) [1+y(1)*(y(1)*y(2)-4); y(1)*(3-y(1)*y(2))]; >> options = odeset('abstol', 1e-6, 'RelTol', 1e-3, 'Stats','on'); >> [x,y] = ode45(f,[0 20],[1.5;3],options); 46 successful steps 12 failed attempts 349 function evaluations a step is accepted if d AbsTol OR d/y RelTol 5 >> ii = 1:4:length(x); >> plot(x,y,'r-',... x(ii),y(ii,1),'bo',... x(ii),y(ii,2),'b*') y x Numerical Analysis, lecture 14, slide 16

17 what happened ODE initial value problems > standard form y = f(x,y), y(x 0 ) = y 0 > existence & uniqueness if f/ y is bounded step-by-step solution > Euler s method has local truncation error O(h 2 ), global t. e. O(h) > 2-stage Heun method has g.t.e. O(h 2 ) > 4-stage Runge-Kutta has g.t.e. O(h 4 ) modern codes use a pair of formulas to estimate local truncation error & thereby automatically adjust the step size Numerical Analysis, lecture 13, slide 17

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