# Shooting method

In numerical analysis, the shooting method is a method for solving a boundary value problem by reducing it to the system 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.

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 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:

$y(t)=y_{(1)}(t)+{\frac {y_{1}-y_{(1)}(t_{1})}{y_{(2)}(t_{1})}}y_{(2)}(t)$ 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.

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).