Standard boundary value problem

Figure 1. Trajectories w(t;s) for s = w'(0) equal to −7, −8, −10, −36, and −40. The point (1,1) is marked with a circle.

Figure 2. The function F(s) = w(1;s) − 1.

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

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

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^[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

When searching for the ground state of the harmonic oscillator with energy ${\displaystyle E_{0}=1/2}$, 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 ${\displaystyle 0.495}$ and ${\displaystyle 0.500}$ (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

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

In quantum mechanics, one seeks normalizable wavefunctions ${\displaystyle \psi _{n}(x)}$ and their corresponding energies subject to the boundary conditions

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

The problem can be solved analytically to find the energies ${\displaystyle E_{n}=n+1/2}$ for ${\displaystyle n=0,1,2,\dots }$, but also serves as an excellent illustration of the shooting method. To apply it, first note some general properties of the Schrödinger equation:

• 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.