Solving Linear Boundary Value Problems with Shooting Method

Ordinary Differential Equations Boundary Value Problems Dr. Rajib Mandal Department of Mathematics Raiganj University 1

14.1 Shooting Method for Solving Linear BVPs We investigate the second-order, two-point boundary-value problem of the form ) , , ( y y x f y = a , x b with Dirichlet boundary conditions , ) ( = a y = (b ) y Or Neuman boundary conditions ) ( = a y ) (b = y , Or mixed boundary condition ) ( 1 + c a y + = = ( ) ( ) y b c y b ( ) , y a 2 2

14.1 Shooting Method for Solving Linear BVPs Simple Boundary Conditions = = = + + ( ) , ( ) y a y y b y ( ) ( ) ( ), , y p x y q x y r x a x b a b The approach is to solve the two IVPs = = ( ) , 0 u a = + + ( ) , u a y ( ) ( ) ( ), u p x u q x u r x a . 1 = = ( ) v a = + ( ) , 0 v a ( ) ( ) , v p x v q x u If the solution of the original two-point BVP is given by y(x) = u(x) + Av(x) is found from the requirement that y(b) = u(b) + Av(b) = yb ( ) , 0 v b b ( ) y u b b ( ) y u b = A = + ( ) ( ) ( ). y x u x v x ( ) v b ( ) v b 3

14.3 Shooting Method for Solving Linear BVPs EX14.1 2y x = = ) 1 ( y 10 , ) 2 ( y , 0 = , y convert to the pair of initial-value problems = , 2 ) 1 ( = = ) 1 ( u 10 , , 0 u u u x 2 = v v ) 1 ( = = ) 1 ( v , 0 , 0 v x 2 2 w = w = = , w = , , w w x w , w w 1 2 3 4 2 4 4 x x 0 ( ) w b 1 b = + ( ) ( ). y x w x w x 1 3 ( ) w 3 4

14.1 Shooting Method for Solving Linear BVPs General Boundary Condition at x = b = + + ( ) ( ) ( ) y p x y q x y r x The condition at x=b involves a linear combination of y(b) and y (b) = + = ( ) , ( ) ( ) y a y y b cy b y a b The approach is to solve the two IVPs ) ( q v x p v + = = = ( ) , 0 u a = + + ( ) , u a y ( ) ( ), u p x u q x u r x a . 1 = = ( ) v a ( ) , 0 v a ( ) ( ) , x u + ( ) ( ) 0 If there is a unique solution, given by ( ) ( ) ( b v v b cv b ) ( ) y u b + cu b b = + ( ). y x u x v x ( ) ( ) cv b 5

14.3 Shooting Method for Solving Linear BVPs General Boundary Condition at Both Ends of the Interval a = a + ) + ( ) y ( ) ( ) y y p + x c y ( q = x y , r x y + = ( ) ( ) ( ) . a y b c y b b y 1 2 the approach is to solve two IVPs ) ( v x p v + = = = = + + ( ) , u a y ( ) , 0 u a ( ) ( ), u p x u q x u r x a = 1 c = ( ) . v a ( ) , 1 v a ( ) ( ) , q x u If , there is a unique solution, given by 0 ) ( ) ( 2 + b v c b v ( ) ( ) y u b + c u b b 2 = + ( ) ( ) ( ). y x u x v x ( ) ( ) v b c v b 2 6

14.3 Shooting Method for Solving Linear BVPs Ex 14.4 2 2 2 x ) 1 ( = + = y = + , 1 + ) 0 ( y , 1 ) 1 ( y . 0 y 2 y y x + + 2 1 1 x x = / ) 6 + 4 2 ( 3 6 y x x x Ex 14.5 2 2 2 x y = + , 1 + 2 y y x ) 0 ( ) 1 ( + = = ) 0 ( y , 0 ) 1 ( y . 3 y y + + 2 1 1 x x = + + 4 2 / 6 3 / 2 1 y x x x 7

14.2 Shooting Method for Solving NonLinear BVPs Nonlinear Shooting Based on the Secant Method use an iterative process based on the secant method presented in Chapter 2 b ), = = = ( ) , y a ay ( ( ), ( )) , 0 h y b y ( , , y f x y y the initial slope t , begin with u (a) = t(1) = 0, error is m(1) Unless the absolute value of m(1) is less than the tolerance, we continue by solving eq. ) 1 i ( ( ) 2 t i t i = ) 1 ( ) ( ( 1 ). t i t i m i ) 1 ( ( ) 2 m i m 8

14.2 Shooting Method for Solving NonLinear BVPs EX 14.6 ) 1 ( + y = = = 2 y y ) 1 ( y . 0 25 . 0 , ) 0 ( y , 1 y y = 1/(x+1) 9

14.2 Shooting Method for Solving NonLinear BVPs Nonlinear Shooting Using Newton s Method = = = ( , , ), y f x y y ( ) , ( ) . y a ay y b by begin by solving the initial-value problem = = = ( ) , ( ) , u a y u a t ( , x , ), u f x u u a k ), = . 1 = = + ( ) , 0 ( ) v a ' ' v ( , , ) ( , , v a vf u u v f x u u u u Check for convergence: , ( t b u m = ) ; y k b if |m| < tol, stop; Otherwise, update t: = / ( , ); t t m v b t + 1 k k k 10

14.2 Shooting Method for Solving NonLinear BVPs EX 14.8 2 [ ] y = , y = = ) 1 ( y , 2 ) 0 ( y , 1 y = ) 0 ( = = 2 1 ) 0 ( u , 1 [ ] , u u u . u kt = ) 1 u 2 2 ( , , ) [ ] u ( , f x u u u u u = [ 2 ( , , ) ] . f x u u 1 u = ) 0 ( v v [ 2 ) 0 ( v , 0 . 1 = = 2 2 [ ] ] , v u u v u u 1 11

14.3 Finite Difference Method for Solving Linear BVPs Replace the derivatives in the differential equation by finite-difference approximations (discussed in Chapter 11). We now consider the general linear two-point boundary-value problem = + + ( ) ( ) ( ), , y p x y q x y r x a x b with boundary conditions = = ( ) , ( ) , y a y b To solve this problem using finite-differences, we divide the interval [a, b] into n subintervals, so that h=(b-a)/n. To approximate the function y(x) at the points we use the central difference formulas from Chapter 11: y y y x y i ) ( 2 = + = + ) 1 , ( , x a h x a n h 1 1 n + 2 y y + + 1 1 1 1 , ( ) i i i i i y x i 2 h h 12

Finite Difference Method for Solving Linear BVPs Substituting these expressions into the BVP and writing as ____ as and as gives ) ( , i i x q p , ) ( r y y p h 2 ( ) p ix ix y 2 iq ir + 2 y y = + + + + 1 1 1 1 i i i i i q y r i i i i h Further algebraic simplification leads to a tridiagonal system for the unknowns viz. , , , 1 1 y + + ny + h h = = , 1 2 2 1 2 ( ) 1 , , 1 , p y h q y p y h r i n + 1 1 i i i i i i i 2 2 = = where and . = = ) ( 0 a y y (b ) yn y 13

Finite Difference Method for Solving Linear BVPs Expanding this expression into the full system gives + h h 2 ( + + = 2 2 ) 1 1 , h q y p y h r p 1 1 1 2 1 1 2 2 + h h 2 ( + + = 2 2 1 ) 1 , p y h q y p y h r 1 1 2 2 2 3 2 2 2 + y i + h p i h 2 ( + = 2 2 1 ) 1 , p h q y y h r + 1 1 i i i i i 2 2 + h + y h 2 ( + = 2 2 1 ) 1 , p y h q y p h r 1 2 3 2 2 2 1 2 n n n n n n n 2 2 + h h 2 ( + = 2 2 1 ) 1 , p y h q y h r p 1 2 1 1 1 1 n n n n n n 2 2 14

Example 14.9 A Finite-Difference Problem Use the finite-difference method to solve the problem + = x y y ( ), 4 0 x 4, x with y(0)=y(4)=0 and n=4 subintervals. Using the central difference formula for the second derivative, we find that the differential equation becomes the system. 2 ) ( 2 = i h + y y y y + = + 1 1 ( ), 4 i 1,2,3. i i i x y x x i i i For this example, h = 1 ,i =1, y = 0, and i = 3, y = 0. Substituting this values, we obtain 0 2 1 2 = + y y + = 1 ( 1 + 4 ), y y y 1 y 2 ( 2 + 2 4 ), y y y 3 2 + 1 2 = 3 ( 3 + 0 2 4 ). y 3 2 3 15

Example 14.9 A Finite-Difference Problem Combining like terms and simplifying gives + = , 3 3 + y y 1 2 = , 4 3 y y y 1 2 3 = , 3 3 y y 2 3 Solving, we find that y_1=13/7, y_2 = 18/7, and y_3 = 13/7. We note for comparison that the exact solution of this problem is = e e e 4 4 1 ( 2 ) 1 ( 2 ) e e + 2 x x 4 . 2 y e e x x 4 4 4 4 e 16

Example 14.9 A Finite-Difference Problem 17

Example 14.10 A Matlab Script for a Linear FDP. The Matlab script that follows solves the BVP. , 2 2 y y y = = = 3e ) 0 ( y , 1 . 0 ) 3 ( y 1 . 0 cos(3). function S_linear_FD aa = 0; bb = 3; n = 300; p = 2*ones(1, n-1); q = -2*ones(1, n-1); r = zeros(1, n-1); ya = 0.1; yb = 0.1*exp(3)*cos(3); h = (bb-aa)/n; h2 = h/2; hh= h*h; x = linspace(aa+h, bb, n); a = zeros(1, n-1); b = a; a(1:n-2) = 1 - p(1, 1:n-2)*h2; d = -(2 + hh*q); b(2:n-1) = 1 + p(1, 2:n-1)*h2; c(1) = hh*r(1) - (1+p(1)*h2)*ya; c(2:n-2) = hh*r(2:n-2); c(n-1) = hh*r(n-1) - (1 - p(n-1)*h2)*yb; y = Thomas(a, d, b, c) xx = [aa x]; yy = [ya y yb]; out = [xx' yy']; disp(out) plot(xx, yy), grid on, hold on plot(xx, 0.1*exp(xx).*cos(xx)) hold off 18

Example 14.10 A Matlab Script for a Linear FDP. 19

14.4 FDM for Solving Nonlinear BVPs We consider the nonlinear ODE-BVP of the form , 2 2 y y y = = = 3e * ) 0 ( y , 1 . 0 Q and ) 3 ( y 1 . 0 P ( x y 2 y i cos(3). * Assume that there are constants and such that * ) , , ( 0 y y x f Q y *Q , | ) * | h = , , Q f y y P * if * Use a finite-difference grid with spacing and let denote the result of evaluating at using for . The ODE then becomes the system 2 2 h ) 1 ( 2 + . = / ) , P /( + iy ( 2 ) y h f ix 1 1 i + y y y = + 1 1 . 0 i i i f i An explicit iteration scheme, analogous to the SOR method 1 i i y y = , + + 2 2 y y h f + 1 1 i i i where and The process will converge for = n y w 0y 2Q * . 2 / h 20

Example 14.12 Solving a Nonlinear BVP by Using FDM Consider again the nonlinear BVP y 2 y y = = = , ) 0 ( y , 1 ) 1 ( y . 2 We illustrate the use of the iterative procedure just outlined by taking a grid with h = . The general form of the difference equation is ) 1 ( 2 + , 1 = + + 2 2 y y y y h f + 1 1 i i i i i where 2 + 2 + i 2 i ( ) /( 2 ) 2 y y h y y y y = = + + 1 1 1 1 1 1 . i i y i i f i + 2 4 1 ( i ) h y i 21

Example 14.12 Solving a Nonlinear BVP by Using FDM Substituting the rightmost expression for f_i into the equation for y_i, we obtain 1 1 + + 2 + i 2 i 2 y y y y = + + + + 1 1 y 1 1 2 . i i y y y y + 1 i i i i 1 ( 2 ) 4 i The computed solution after 10 iterations agrees very closely with the exact solution. function S_nonlinear_FD ya = 1; yb = 2; a = 0; b = 1; max_it = 10; n = 4; w = 0.1; ww = 1/(2*(1+w)); h = (b-a)/n y(1:n-1) = 1 for k = 1:max_it y(1) = ww*(ya+2*w*y(1)+y(2)+(ya^2-2*ya*y(2)+y(2)^2)/(4*y(1))); y(2) = ww*(y(1)+2*w*y(2)+y(3)+(y(1)^2-2*y(1)*y(3)+y(3)^2)/(4*y(2))); y(3) = ww*(y(2)+2*w*y(3)+yb+(y(2)^2-2*y(2)*yb+yb^2)/(4*y(3))); end x = [ a a+h a+2*h a+3*h b ]; z = [ ya y yb ]; plot(x, z), hold on, zz = sqrt(3*x+1); plot(x, zz), hold off 22

Example 14.12 Solving a Nonlinear BVP by Using FDM 23