# Shooting method

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. The following exposition may be clarified by this illustration of the shooting method.

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

$y''(t) = f(t, y(t), y'(t)), \quad y(t_0) = y_0, \quad y(t_1) = y_1$

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

$y''(t) = f(t, y(t), y'(t)), \quad y(t_0) = y_0, \quad y'(t_0) = a$

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 obviously the solution y(t; a) of the corresponding initial value problem is also a solution of the boundary value problem. On the other hand, 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:

$y(t) = y_{(1)}(t)+\frac{y_1-y_{(1)}(t_1)}{y_{(2)}(t_1)}y_{(2)}(t)$

where $y_{(1)}(t)$ is the solution to the initial value problem:

$y_{(1)}''(t) = p(t)y_{(1)}'(t)+q(t)y_{(1)}(t)+r(t),\quad y_{(1)}(t_0) = y_0, \quad y_{(1)}'(t_0) = 0,$

and $y_{(2)}(t)$ is the solution to the initial value problem:

$y_{(2)}''(t) = p(t)y_{(2)}'(t)+q(t)y_{(2)}(t),\quad y_{(2)}(t_0) = 0, \quad y_{(2)}'(t_0) = 1.$

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).

$w''(t) = \frac{3}{2} w^2, \quad w(0) = 4, \quad w(1) = 1$
$w''(t) = \frac{3}{2} w^2, \quad w(0) = 4, \quad w'(0) = s$

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.

Solutions of the initial value problem were computed by using the LSODE algorithm, as implemented in the mathematics package GNU Octave.

Stoer and Bulirsch 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).

The function F(s) = w(1;s) − 1.
Trajectories w(t;s) for s = w'(0) equal to −7, −8, −10, −36, and −40 (red, green, blue, cyan, and magenta, respectively). The point (1,1) is marked with a red diamond.