In numerical analysis, the shooting method is a method for solving a boundary value problem by reducing it to an initial value problem. It involves finding solutions to the initial value problem for different initial conditions until one finds the solution that also satisfies the boundary conditions of the boundary value problem. In layman's terms, one "shoots" out trajectories in different directions from one boundary until one finds the trajectory that "hits" the other boundary condition.

Mathematical description

Suppose one wants to solve the boundary-value problem

$${\displaystyle y''(t)=f(t,y(t),y'(t)),\quad y(t_{0})=y_{0},\quad y(t_{1})=y_{1}.}$$

${\displaystyle y(t;a)}$

$${\displaystyle y''(t)=f(t,y(t),y'(t)),\quad y(t_{0})=y_{0},\quad y'(t_{0})=a.}$$

${\displaystyle y(t_{1};a)=y_{1))$

${\displaystyle y(t;a)}$

The shooting method is the process of solving the initial value problem for many different values of ${\displaystyle a}$ until one finds the solution ${\displaystyle y(t;a)}$ that satisfies the desired boundary conditions. Typically, one does so numerically. The solution(s) correspond to root(s) of

$${\displaystyle F(a)=y(t_{1};a)-y_{1}.}$$

${\displaystyle a}$

Roots of ${\displaystyle F}$ and solutions to the boundary value problem are equivalent. If ${\displaystyle a}$ is a root of ${\displaystyle F}$, then ${\displaystyle y(t;a)}$ is a solution of the boundary value problem. Conversely, if the boundary value problem has a solution ${\displaystyle y(t)}$, it is also the unique solution ${\displaystyle y(t;a)}$ of the initial value problem where ${\displaystyle a=y'(t_{0})}$, so ${\displaystyle a}$ is a root of ${\displaystyle F}$.

Etymology and intuition

The term "shooting method" has its origin in artillery. An analogy for the shooting method is to

• place a cannon at the position ${\displaystyle y(t_{0})=y_{0))$, then
• vary the angle ${\displaystyle a=y'(t_{0})}$ of the cannon, then
• fire the cannon until it hits the boundary value ${\displaystyle y(t_{1})=y_{1))$.

Between each shot, the direction of the cannon is adjusted based on the previous shot, so every shot hits closer than the previous one. The trajectory that "hits" the desired boundary value is the solution to the boundary value problem — hence the name "shooting method".

Linear shooting method

The boundary value problem is linear if f has the form

$${\displaystyle f(t,y(t),y'(t))=p(t)y'(t)+q(t)y(t)+r(t).}$$

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

${\displaystyle y_{(1)}(t)}$

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

${\displaystyle y_{(2)}(t)}$

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

Examples

Standard boundary value problem

A boundary value problem is given as follows by Stoer and Bulirsch^[2] (Section 7.3.1).

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

The initial value problem

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

Stoer and Bulirsch^[2] 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).

Eigenvalue problem

The shooting method can also be used to solve eigenvalue problems. Consider the time-independent Schrödinger equation for the quantum harmonic oscillator

$${\displaystyle -{\frac {1}{2))\psi _{n}''(x)+{\frac {1}{2))x^{2}\psi _{n}(x)=E_{n}\psi _{n}(x).}$$

${\displaystyle \psi _{n}(x)}$

$${\displaystyle \psi _{n}(x\rightarrow +\infty )=\psi _{n}(x\rightarrow -\infty )=0.}$$

${\displaystyle E_{n}=n+1/2}$

${\displaystyle n=0,1,2,\dots }$

• If ${\displaystyle \psi _{n}(x)}$ is an eigenfunction, so is ${\displaystyle C\psi _{n}(x)}$ for any nonzero constant ${\displaystyle C}$.
• The ${\displaystyle n}$-th excited state ${\displaystyle \psi _{n}(x)}$ has ${\displaystyle n}$ roots where ${\displaystyle \psi _{n}(x)=0}$.
• For even ${\displaystyle n}$, the ${\displaystyle n}$-th excited state ${\displaystyle \psi _{n}(x)=\psi _{n}(-x)}$ is symmetric and nonzero at the origin.
• For odd ${\displaystyle n}$, the ${\displaystyle n}$-th excited state ${\displaystyle \psi _{n}(x)=-\psi _{n}(-x)}$ is antisymmetric and thus zero at the origin.

To find the ${\displaystyle n}$-th excited state ${\displaystyle \psi _{n}(x)}$ and its energy ${\displaystyle E_{n))$, the shooting method is then to:

1. Guess some energy ${\displaystyle E_{n))$.
2. Integrate the Schrödinger equation. For example, use the central finite difference
   $${\displaystyle -{\frac {1}{2)){\frac {\psi _{n}^{i+1}-2\psi _{n}^{i}+\psi _{n}^{i-1))((\Delta x}^{2))}+{\frac {1}{2))(x^{i})^{2}\psi _{n}^{i}=E_{n}\psi _{n}^{i}.}$$

   
   • If ${\displaystyle n}$ is even, set ${\displaystyle \psi _{0))$ to some arbitrary number (say ${\displaystyle \psi _{n}^{0}=1}$ — the wavefunction can be normalized after integration anyway) and use the symmetric property to find all remaining ${\displaystyle \psi _{n}^{i))$.
   • If ${\displaystyle n}$ is odd, set ${\displaystyle \psi _{n}^{0}=0}$ and ${\displaystyle \psi _{n}^{1))$ to some arbitrary number (say ${\displaystyle \psi _{n}^{1}=1}$ — the wavefunction can be normalized after integration anyway) and find all remaining ${\displaystyle \psi _{n}^{i))$.
3. Count the roots of ${\displaystyle \psi _{n))$ and refine the guess for the energy ${\displaystyle E_{n))$.
   • If there are ${\displaystyle n}$ or less roots, the guessed energy is too low, so increase it and repeat the process.
   • If there are more than ${\displaystyle n}$ roots, the guessed energy is too high, so decrease it and repeat the process.

The energy-guessing can be done with the bisection method, and the process can be terminated when the energy difference is sufficiently small. Then one can take any energy in the interval to be the correct energy.