(6.2) -The Shooting Method for Nonlinear Problems

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1 (6.2) -The Shooting Method for Nonlinear Problems Consider the boundary value problems (BVPs) for the second order differential equation of the form (*) y f x,y,y, a x b, y a and y b. When f x,y,y is linear in y and y, the Shooting Method introduced in Section 6.1 solves a system of two initial value problems and the solution y x of the boundary value problem (*) is of the form y x y 1 x y 1 b y 2 x y 2 b where y 1 x and y 2 x are solutions of two initial value problems, respectively. Now we consider the cases where f x,y,y is not linear in y and y. Assume a given boundary value problem has a unique solution y x. We will approximate the solution y x by solving a sequence of initial value problems (**) y f x,y,y, a x b, y a, y a s k where s k are real numbers. Let y x,s k be solution of the initial value problem (**). We want to have a sequence s k so that lim y x,s k y x. k One of choices for s 0 is s 0 y y b y a a b a b a. How to choose s k for k 1? Consider choose s such that (***) y b,s 0, that is s is a solution of the equation. Observe that the equation y b,s 0 is a nonlinear equation in one variable. In Chapter 2, we have studied numerical methods: Bisection Method, Newton Method, Secant Method, Fixed-Point Method, for nonlinear equations of the form g s 0. For given s 0 and s 1, the Secant Method computes s k for k 2 as follows s k s k 1 g s k 1 s k 1 s k g s k 1 g s k 2. For given s 0, the Newton Method computes s k for k 1 as follows s k s k 1 g s k 1 g s k 1. These methods can be used here to solve the nonlinear equation (***). 1. Use the Secant Method to approximate the solution of y b,s 0. y b, s k 1 s k 1 s k 2 s k s k 1 y b, s k 1 y b, s k 2 Note that y b, s k 1 is the last element in the array y. Note also that we will need two initial choices s 0 and s 1 to compute s 2 in order to continue the iterations. 2. Use Newton s Method to approximate the solution of y b,s 0. s k s k 1 y b, s k 1 dy b, s ds k 1 Again, y b, s k 1 is the last element in the array y. Since we do not know y x explicitly, how can we determine dy b, s k 1? Let y x,s be the solution of the initial value problem (**). Then from ds (**), we have 1

2 y x,s f x,y x,s,y x,s, a x b, y a,s, y a,s s. By differentiating both sides of the differential equation with respect to s, wehave y x,s f x,y x,s,y x,s y x,s y f x x s f y f x,s y. Since x and s are independent, x s 0. Hence, y (****) x,s for a x b. The initial conditions are: y a,s Define z x,s y x,s we denote.since 3 y x,s x 2 f y y x,s d ds 0, y a,s z x,s f y y x,s ds d s 1. 2 y x,s x 2 y x,s, y x,s. Then the initial value problem (****) becomes the initial value problem (*****) z x,s f y z x,s f y z x,s, a x b, z a,s 0, z a,s 1. We can update s k using the information from z x,s as follows: s k s k 1 y b, s k 1 z b, s k 1 Shooting Method for nonlinear boundary value problems: y f x,y,y, a x b, y a and y b. Compute s 0.Fork 1, b a (i) Solve the system of two initial value problems: y f x,y,y, a x b, y a, y a s k 1 z f y z f y z, a x b, z a,s k 1 0, z a,s k 1 1 for y x and z x. (ii) Update s k s k 1 y b, s k 1. z b, s k 1 For a given accuracy requirement, the algorithm is terminated if y b, s k 1. Example Solve the BVP: y x3 yy, 1 x 3, y 1 17, y f x,y,y x3 yy. f y y, f y y Solve a system of 2 second-order initial value problems: 2

3 y x3 yy, 1 x 3, y 1 17, y 1 s k z f y z f y z 1 8 y z yz, 1 x 3, z 1 0, z 1 1 Let u 1 y, u 2 y, u 3 z, u 4 z. Solve a system of 4 first-order initial value problems: using MatLab function ode45.m. u 2 u 4 u 1 u x3 u 1 u 2 u 3 u u 2u 3 u 1 u 4, u u 2 1 s k u u y"=(32+2x 3 -yy)/8, [1,3], y(1)=17, y(3)=43/3 (Secant) 20 - solution y=x 2 +16/x y(x,s 0 ) y(x,s 2 ) 12 y(x,s 1 ) MatLab program: lect6_2_ex1.m Shooting Method for Nonlinear BVP clear clf alpha 17; beta 43/3; a 1; b 3; the first shoot: s(1) (beta-alpha)/(b-a); [X,Y] ode45( funsys1,[a b],[alpha;s(1);0;1]); plot(x,y(:,1), -. ) hold text(2.6,19.8, y(x,s_0) ) the second shoot: 3

4 s(2) s(1) (beta-y(n,1))/y(n,3); clear X; clear Y; [X,Y] ode45( funsys1,[a,b],[alpha;s(2);0;1]); plot(x,y(:,1), ) text(2.6,11.0, y(x,s_1) ) the third shoot: s(3) s(2) (beta-y(n,1))/y(n,3); clear X; clear Y; [X,Y] ode45( funsys1,[a,b],[alpha;s(3);0;1]); plot(x,y(:,1), : ) text(2.6,12.5, y(x,s_3) ) Comparisons title( y" (32 2x^3-yy)/8, [1,3], y(1) 17, y(3) 43/3 (Secant) ) ysol X.^2 16./X; plot(x,ysol) text(1.2,20, - solution y x^2 16/x ) hold off MatLab program: funsys1.m function yv funsys1(x,y); yv(1,1) y(2,1); yv(2,1) 1/8*(32 2*x^3-y(1,1)*y(2,1)); yv(3,1) y(4,1); yv(4,1) -y(2,1)/8*y(3,1)-y(1,1)/8*y(4,1); MatLab program: lect6_2_ex2.m lect6_2_ex2 clear clf alpha 17; beta 43/3; a 1; b 3; epsilon input( epsilon ); First shooting: s(1) (beta-alpha)/(b-a); 4

5 [X,Y] ode45( funsys1,[a b],[alpha;s(1);0;1]); plot(x,y(:,1), -. ) hold iterations: flag 0; diff abs(y(n,1)-beta); if diff epsilon flag 1; end k 1; while flag 0, s(k 1) s(k) (beta-y(n,1))/y(n,3); clear X; clear Y; [X,Y] ode45( funsys1,[a,b],[alpha;s(k 1);0;1]); plot(x,y(:,1), ) diff abs(y(n,1)-beta); if diff epsilon flag 1; end k k 1; end title( y" (32 2x^3-yy)/8, [1,3], y(1) 17, y(3) 43/3 (Secant) ) ysol X.^2 16./X; plot(x,ysol) text(1.2,20, - solution y x^2 16/x ) hold off Exercises: 1. Use the Shooting Method to approximate the solution to the boundary value problem y y 2 y lnx, 1 x 2, y 1 0, y 2 ln2 to within Compare your results to the actual solution y ln x by plotting the absolute values of the differences. 2. Use the Shooting Method to approximate the solution to each of the following boundary value problems to within Compare your results with given true solution by plotting the absolute values of the differences. a. y y 3 yy,1 x 2, y 1 1, y The true solution: y x x 1 1. b. y y 2 y lnx 3 1 x,2 x 3, y 2 1 ln 2, f 3 1 ln3 2 3 The true solution: y x x 1 ln x. c. y 2y 3 6y 2x 3,1 x 2, y 1 2. y The true solution: y x x 1 x. 5

6 3. Write an MatLab program to implement the Shooting Method for nonlinear boundary value problems using the Newton Method to compute s 1 and then use the Secant Method to update s k for k 2,3, The Van der Pol equation y y 2 1 y y 0, 0, governs the flow of current in a vacuum tube with three internal elements. Let 1, y 0 0, and 2 y 2 1. Approximate y x for 0 x 2 to within