Rahul Sharma (Editor)

Shooting method

Updated on
Edit
Like
Comment
Share on FacebookTweet on TwitterShare on LinkedInShare on Reddit

In numerical analysis, the shooting method is a method for solving a boundary value problem by reducing it to the solution of an initial value problem. Roughly speaking, we 'shoot' out trajectories in different directions until we find a trajectory that has the desired boundary value. The following exposition may be clarified by this illustration of the shooting method.

Contents

For a boundary value problem of a second-order ordinary differential equation, the method is stated as follows. Let

y ( t ) = f ( t , y ( t ) , y ( t ) ) , y ( t 0 ) = y 0 , y ( t 1 ) = y 1

be the boundary value problem. Let y(t; a) denote the solution of the initial value problem

y ( t ) = f ( t , y ( t ) , y ( t ) ) , y ( t 0 ) = y 0 , y ( t 0 ) = a

Define the function F(a) as the difference between y(t1; a) and the specified boundary value y1.

F ( a ) = y ( t 1 ; a ) y 1

If F has a root a then the solution y(t; a) of the corresponding initial value problem is also a solution of the boundary value problem. Conversely, if the boundary value problem has a solution y(t), then y(t) is also the unique solution y(t; a) of the initial value problem where a = y'(t0), thus a is a root of F.

The usual methods for finding roots may be employed here, such as the bisection method or Newton's method.

Linear shooting method

The boundary value problem is linear if f has the form

f ( t , y ( t ) , y ( t ) ) = p ( t ) y ( t ) + q ( t ) y ( t ) + r ( t ) .

In this case, the solution to the boundary value problem is usually given by:

y ( t ) = y ( 1 ) ( t ) + y 1 y ( 1 ) ( t 1 ) y ( 2 ) ( t 1 ) y ( 2 ) ( t )

where y ( 1 ) ( t ) is the solution to the initial value problem:

y ( 1 ) ( t ) = p ( t ) y ( 1 ) ( t ) + q ( t ) y ( 1 ) ( t ) + r ( t ) , y ( 1 ) ( t 0 ) = y 0 , y ( 1 ) ( t 0 ) = 0 ,

and y ( 2 ) ( t ) is the solution to the initial value problem:

y ( 2 ) ( t ) = p ( t ) y ( 2 ) ( t ) + q ( t ) y ( 2 ) ( t ) , y ( 2 ) ( t 0 ) = 0 , y ( 2 ) ( t 0 ) = 1.

See the proof for the precise condition under which this result holds.

Example

A boundary value problem is given as follows by Stoer and Burlisch (Section 7.3.1).

w ( t ) = 3 2 w 2 , w ( 0 ) = 4 , w ( 1 ) = 1

The initial value problem

w ( t ) = 3 2 w 2 , w ( 0 ) = 4 , w ( 0 ) = s

was solved for s = −1, −2, −3, ..., −100, and F(s) = w(1;s) − 1 plotted in the first figure. Inspecting the plot of F, we see that there are roots near −8 and −36. Some trajectories of w(t;s) are shown in the second figure.

Stoer and Burlisch state that there are two solutions, which can be found by algebraic methods. These correspond to the initial conditions w′(0) = −8 and w′(0) = −35.9 (approximately).

References

Shooting method Wikipedia


Similar Topics