Shooting method

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 ${\displaystyle a=y^{\prime }(t_0)}$ 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 $${\displaystyle f(t,y(t),y^{\prime }(t))=p(t)y^{\prime }(t)+q(t)y(t)+r(t).}$$ In this case, the solution to the boundary value problem is usually given by: $${\displaystyle y(t)=y_{(1)}(t)+\frac{y_1-y_{(1)}(t_1)}{y_{(2)}(t_1)}{y}_{(2)}(t)}$$ where ${\displaystyle y_{(1)}(t)}$ is the solution to the initial value problem: $${\displaystyle y_{(1)}^{″}(t)=p(t)y_{(1)}^{\prime }(t)+q(t)y_{(1)}(t)+r(t),y_{(1)}(t_0)=y_0,y_{(1)}^{\prime }(t_0)=0,}$$ and ${\displaystyle y_{(2)}(t)}$ is the solution to the initial value problem: $${\displaystyle y_{(2)}^{″}(t)=p(t)y_{(2)}^{\prime }(t)+q(t)y_{(2)}(t),y_{(2)}(t_0)=0,y_{(2)}^{\prime }(t_0)=1.}$$ See the proof for the precise condition under which this result holds.

Examples

Standard boundary value problem

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

$${\displaystyle w^{″}(t)=\frac{3}{2}{w}^2(t),w(0)=4,w(1)=1}$$

The initial value problem $${\displaystyle w^{″}(t)=\frac{3}{2}{w}^2(t),w(0)=4,w^{\prime }(0)=s}$$ 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 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 $${\displaystyle -\frac{1}{2}{\psi }_n^{″}(x)+\frac{1}{2}{x}^2{\psi }_n(x)=E_n{\psi }_n(x).}$$ In quantum mechanics, one seeks normalizable wave functions ${\displaystyle {\psi }_n(x)}$ and their corresponding energies subject to the boundary conditions $${\displaystyle {\psi }_n(x\to +\mathrm{\infty })={\psi }_n(x\to -\mathrm{\infty })=0.}$$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$${\displaystyle -\frac{1}{2}\frac{{\psi }_n^{i+1}-2{\psi }_n^i+{\psi }_n^{i-1}}{{\mathrm{\Delta }x}^2}+\frac{1}{2}(x^i{)}^2{\psi }_n^i=E_n{\psi }_n^i.}$$
   • If ${\displaystyle n}$ is even, set ${\displaystyle {\psi }_0}$ to some arbitrary number (say ${\displaystyle {\psi }_n^0=1}$ — the wave function 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 wave function 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.