Editing Shooting method (section)

== Examples ==

=== Standard boundary value problem ===
[[Image:Shooting method trajectories.svg|thumb|400px|Figure 1. Trajectories ''w''(''t'';''s'') for ''s'' = ''w''<nowiki>'</nowiki>(0) equal to −7, −8, −10, −36, and −40. The point (1,1) is marked with a circle.]][[Image:Shooting method error.svg|thumb|400px|Figure 2. The function ''F''(''s'') = ''w''(1;''s'') − 1.]]A [[boundary value problem]] is given as follows by Stoer and Bulirsch<ref name = "Stoer1980">Stoer, J. and Bulirsch, R. ''Introduction to Numerical Analysis''. New York: Springer-Verlag, 1980.</ref> (Section 7.3.1).
<math display="block"> w''(t) = \frac{3}{2} w^2(t), \quad w(0) = 4, \quad w(1) = 1 </math>

The [[initial value problem]]
<math display="block"> w''(t) = \frac{3}{2} w^2(t), \quad w(0) = 4, \quad w'(0) = s</math>
was solved for ''s'' = −1, −2, −3, ..., −100, and ''F''(''s'') = ''w''(1;''s'') − 1 plotted in the Figure 2.
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 Figure 1.

Stoer and Bulirsch<ref name = "Stoer1980"/> 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).{{clear}}

=== Eigenvalue problem ===
[[File:Shooting method.svg|alt=Illustration of the shooting method for finding the ground state of the quantum harmonic oscillator|thumb|400x400px|When searching for the ground state of the harmonic oscillator with energy <math>E_0 = 1/2</math>, the shooting method produces wavefunctions that diverge to infinity. Here, the correct wavefunction should have zero roots and go to zero at infinity, so it lies somewhere between the orange and green lines. The energy is therefore between <math>0.495</math> and <math>0.500</math> (with numerical accuracy).]]

The shooting method can also be used to solve eigenvalue problems. Consider the [[time-independent Schrödinger equation]] for the [[quantum harmonic oscillator]] <math display="block">-\frac{1}{2} \psi_n''(x) + \frac{1}{2} x^2 \psi_n(x) = E_n \psi_n(x).</math> In quantum mechanics, one seeks normalizable wavefunctions <math>\psi_n(x)</math> and their corresponding energies subject to the boundary conditions <math display="block">\psi_n(x \rightarrow +\infty) = \psi_n(x \rightarrow -\infty) = 0.</math>The problem can be solved analytically to find the energies <math>E_n = n + 1/2</math> for <math>n = 0, 1, 2, \dots</math>, but also serves as an excellent illustration of the shooting method. To apply it, first note some general properties of the Schrödinger equation:

* If <math>\psi_n(x)</math> is an eigenfunction, so is <math>C \psi_n(x)</math> for any nonzero constant <math>C</math>.
* The <math>n</math>-th excited state <math>\psi_n(x)</math> has <math>n</math> roots where <math>\psi_n(x) = 0</math>.
* For even <math>n</math>, the <math>n</math>-th excited state <math>\psi_n(x) = \psi_n(-x)</math> is symmetric and nonzero at the origin.
* For odd <math>n</math>, the <math>n</math>-th excited state <math>\psi_n(x) = -\psi_n(-x)</math> is antisymmetric and thus zero at the origin.

To find the <math>n</math>-th excited state <math>\psi_n(x)</math> and its energy <math>E_n</math>, the shooting method is then to:

# Guess some energy <math>E_n</math>.
# Integrate the Schrödinger equation. For example, use the central [[Finite difference method|finite difference]]<math display="block">-\frac{1}{2} \frac{\psi^{i+1}_n - 2 \psi^i_n + \psi^{i-1}_n}{{\Delta x}^2} + \frac{1}{2} (x^i)^2 \psi^i_n = E_n \psi^i_n.</math>
#* If <math>n</math> is even, set <math>\psi_0</math> to some arbitrary number (say <math>\psi^0_n = 1</math> — the wavefunction can be normalized after integration anyway) and use the symmetric property to find all remaining <math>\psi_n^i</math>.
#* If <math>n</math> is odd, set <math>\psi^0_n = 0</math> and <math>\psi^1_n</math> to some arbitrary number (say <math>\psi^1_n = 1</math> — the wavefunction can be normalized after integration anyway) and find all remaining <math>\psi_n^i</math>.
# Count the roots of <math>\psi_n</math> and refine the guess for the energy <math>E_n</math>.
#* If there are <math>n</math> or less roots, the guessed energy is too low, so increase it and repeat the process.
#* If there are more than <math>n</math> 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.