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15. Second-order_BVP

# 15. Second-order_BVP - 3.5.1 Example Shooting Method nd 2...

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1/4 3.5.1 Example: Shooting Method 2 nd order ODE example x e y dx dy x dx y d = + + 2 2 Two boundary conditions ( ) 3 0 = y , ( ) 2 1 = y Find ( ) x y for 1 0 x (range defined by BCs) Solution: Use methods developed in 3.2 for solving IVPs. To do this, write the 2 nd order ODE as a set of two 1 st order ODEs F dx dy = where v F = G dx dv = y xv e G x + + = ( ) 3 0 = y , ( ) 2 1 = y Set-up a grid Choose say 2 . 0 = Δ x - B.C.s give the two y values 3 0 = y 2 5 = y - unknowns are 4 3 2 1 , , , y y y y To solve the set of two 1 st order ODEs using methods developed in 3.2 (e.g., explicit Euler method) need y 0 and v 0 to start the solution. However, with BCs, know only y 0 . Method: guess v 0 and solve for y(x). If y(x=1)=2, i.e., matches the BC, we have the solution; if not, make another guess for v 0 and repeat. Can use any method from 3.2 here. Illustrate with the (explicit) Euler method. First “shot” Second “shot” x i y i v i F i G i 0 3 0 0.000 -2.000 0.2 3.000 -0.400 -0.400 -1.699 0.4 2.920 -0.740 -0.740 -1.132 0.6 2.772 -0.966 -0.966 -0.370 0.8 2.579 -1.040 -1.040 0.479 1 2.371 Third “shot” x i y i v i F i G i 0 3 1 1.000 -2.000 0.2 3.200 0.600 0.600 -2.099 0.4 3.320 0.180 0.180 -1.900 0.6 3.356 -0.200 -0.200 -1.414 0.8 3.316 -0.483 -0.483 -0.704 1 3.220 x i y i v i F i G i 0 3 -1 -1.000 -2.000 0.2 2.800 -1.400 -1.400 -1.299 0.4 2.520 -1.660 -1.660 -0.364 0.6 2.188 -1.733 -1.733 0.674 0.8 1.842 -1.598 -1.598 1.662 1 1.522 After a few “shots”, can start “homing in” on the solution by extrapolating/interpolating between

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