Series Solutions to Airy Equation
The Airy Equation, introduced in the last tutorial
\begin{equation}
\frac{d^2y}{dx^2} -xy = 0 \tag{1}
\end{equation}
is easily seen to have an ordinary point at \( x = 0 \) and hence we can find its series solutions there. Now substitute (1) and (3) of this tutorial into (1) above, leading to
\begin{align}
\sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} -x \sum_{n=0}^{\infty} a_n x^n &= 0 \\
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^{n} -\sum_{n=0}^{\infty} a_n x^{n+1} &= 0 \\
\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^{n} -\sum_{n=1}^{\infty} a_{n-1} x^{n} &= 0 \tag{2}
\end{align}
where we have adjusted the indices in the first/second summation by \(2\) and \(-1\). By comparing the coefficients, the recurrence relation is then
\begin{align}
(n+2)(n+1) a_{n+2} -a_{n-1} &= 0 \\
a_{n+2} &= \frac{a_{n-1}}{(n+2)(n+1)} \tag{3}
\end{align}
for \( n \geq 1 \), with a stride of \( 3 \). When \( n = 0 \), the second summation does not appear (By convention, \(a_{-1} = 0\)), and we must have \( a_2 = 0 \), hence \( a_2 = a_5 = a_8 = \ldots = a_{3m+2} = 0 \) for all \(m\). For \( a_{3m} \), we have
\begin{align}
\begin{aligned}
a_3 &= \frac{a_0}{(3)(2)} \\
a_6 &= \frac{a_3}{(6)(5)} = \frac{a_0}{(6)(5)(3)(2)} \\
& \vdots
\end{aligned} \tag{4}
\end{align}
and similarly, for \( a_{3m+1} \):
\begin{align}
\begin{aligned}
a_4 &= \frac{a_1}{(4)(3)} \\
a_7 &= \frac{a_4}{(7)(6)} = \frac{a_1}{(7)(6)(4)(3)} \\
& \vdots
\end{aligned} \tag{5}
\end{align}
The patterns keep going on. Therefore, two possible series solutions will then be
\begin{align}
y &= a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5 + a_6 x^6 + a_7 x^7 + \cdots \\
&= a_0 + a_1 x + (0) + \frac{a_0}{(3)(2)} x^3 + \frac{a_1}{(4)(3)} x^4 + (0) \\
&\quad + \frac{a_0}{(6)(5)(3)(2)} x^6 + \frac{a_1}{(7)(6)(4)(3)}x^7 + \cdots \\
&= a_0 (1 + \frac{x^3}{(3)(2)} + \frac{x^6}{(6)(5)(3)(2)} + \cdots) \\
&\quad + a_1 (x + \frac{x^4}{(4)(3)} + \frac{x^7}{(7)(6)(4)(3)} + \cdots) \tag{6}
\end{align}
These two series are analytic and converge for all \( x \). To see this, we apply the Ratio test. For the first series:
\begin{align}
&\quad \lim_{m \to \infty}\left| \frac{x^{3m+3}/(3m+3)(3m+2)(3m)(3m-1)\cdots}{x^{3m}/(3m)(3m-1)\cdots}\right| \\
&= \lim_{m \to \infty} \left| \frac{x^3}{(3m+3)(3m+2)}\right| = 0 \tag{7}
\end{align}
and it is similar for the second series. Therefore, any solution to the Airy equation can be expressed as the superposition of these two (clearly linearly independent) series.
Formulating the Airy Functions
Now we will show how the standard Airy Functions, the canonical forms of the two solutions to the Airy Equation, are constructed. These are built from some specific combinations of the two series solutions in (6) so as to abide by a certain convention.
Limits as \( x \to \infty \)
First, we need the observation that any solution of the Airy Equation \( y \) will either tend to \( 0 \) or \( \pm \infty \) as \( x \to \infty \). Assume the contrary that it tends to some non-zero finite constant (and without loss of generality, positive) value \( K \). By the definition of a limit, it implies that we can choose some sufficiently large \( x_l \) so that \( K/2 < y(x) < 3K/2 \) (and \( -K/2 < y'(x) < K/2 \) since \( y'(x) \to 0 \) when \( x \to \infty\)) as long as \( x \geq x_l \). Particularly, we take \( x_l > 1 \). So by (1), \( d^2y/dx^2 = xy > K/2 \) for \( x \geq x_l \). Integrating over a distance of \( \Delta x= 4 \) to the right of \( x_l \) leads to:
\begin{align}
[\frac{dy}{dx}]_{x_l}^x &> \frac{K}{2}(x-x_l) \\
\frac{dy(x)}{dx} &> \frac{K}{2}(x-x_l) -\frac{K}{2} \\
[y]_{x_l}^{x_l+\Delta x} &> \int_{x_l}^{x_l+\Delta x} (\frac{K}{2}(x-x_l) -\frac{K}{2}) dx \\
y(x_l+\Delta x) -\frac{K}{2} &> [\frac{K}{2}\frac{(x-x_l)^2}{2} -\frac{K}{2}(x-x_l)] _{x_l}^{x_l+\Delta x} \\
y(x_l+\Delta x) &> \frac{K}{2} + \frac{K}{2} \frac{(\Delta x)^2}{2} -\frac{K}{2} \Delta x \\
&= \frac{K}{2} + \frac{K}{2}\frac{4^2}{2} -\frac{K}{2}(4) = \frac{5}{2}K
\end{align}
which contradicts the assumption that \( y(x) < 2K \) and we are done.
Integral Forms and Properties of Ai(x) and Bi(x)
While the two series solutions in (6) are valid, it is desired that, motivated by the argument above, we can find a specific form whose limit takes the value of zero at the point of infinity. Then, this shall be assigned as the Airy Function \( \text{Ai}(x) \) (of the first kind). It is known to have the integral form of
\begin{equation}
\text{Ai}(x) = \frac{1}{\pi} \int_0^{\infty} \cos (\frac{t^3}{3} + xt) dt \tag{8}
\end{equation}
To develop this expression requires advanced knowledge from Complex Analysis which is out of our scope. However, we can substitute (8) into (1) to see that it really satisfies the equation:
\begin{align}
\frac{d^2y}{dx^2} &= \frac{d^2}{dx^2}(\text{Ai}(x))\\
&= \frac{d^2}{dx^2} \left(\frac{1}{\pi}\int_0^{\infty} \cos (\frac{t^3}{3} + xt) dt\right) \\
&= \frac{1}{\pi} \int_0^{\infty} \frac{\partial^2}{\partial x^2}\cos (\frac{t^3}{3} + xt) dt \\
&= \frac{1}{\pi} \int_0^{\infty} -t^2\cos (\frac{t^3}{3} + xt) dt \\
&= \frac{1}{\pi} \int_0^{\infty} -\frac{t^2}{t^2 + x}\cos (\frac{t^3}{3} + xt) d(\frac{t^3}{3} + xt) \\
&= \frac{1}{\pi} \int_0^{\infty} (\frac{x}{t^2 + x}-1)\cos (\frac{t^3}{3} + xt) d(\frac{t^3}{3} + xt) \\
&= \frac{1}{\pi} \int_0^{\infty} \frac{x}{t^2 + x}\cos (\frac{t^3}{3} + xt) d(\frac{t^3}{3} + xt) \\
&\quad -\frac{1}{\pi} \int_0^{\infty} \cos (\frac{t^3}{3} + xt) d(\frac{t^3}{3} + xt) \\
&= \frac{x}{\pi} \int_0^{\infty} \frac{1}{t^2 + x}\cos (\frac{t^3}{3} + xt) d(\frac{t^3}{3} + xt) + 0 \\
&= \frac{x}{\pi} \int_0^{\infty} \cos (\frac{t^3}{3} + xt) dt = x\text{Ai}(x) = xy \tag{9}
\end{align}
There is a caveat to argue that the second integral has to converge to zero (the most reasonable value), but we will not go too deep into the issue here. However, notice that the integral expression of \(\text{Ai}(x)\) in (8) converges and so it is well-defined everywhere, as shown by Dirichlet’s test:
\begin{align}
\text{Ai}(x) &= \frac{1}{\pi} \int_0^{\infty} \cos (\frac{t^3}{3} + xt) dt \\
&= \frac{1}{\pi} \int_0^{\infty} \frac{1}{t^2 + x} \cos (\frac{t^3}{3} + xt) d(\frac{t^3}{3} + xt) \\
&= \frac{1}{\pi} \int_0^{\infty} \frac{1}{t^2 + x} (\cos u) du \tag{10}
\end{align}
where \( 1/(t^2 + x) \) will clearly decay to \( 0 \) as \( u = \frac{t^3}{3} + xt \) goes to infinity and \( \cos u \) is clearly bounded. This also partly suggests that what we have done in deriving (9) makes some sense. Moreover, as \(\text{Ai}(x)\) is shown to converge, and by the previous observation that an Airy function must either tend to \( 0 \) or \( \pm \infty \), we immediately know that \(\text{Ai}(x) \to 0 \) as \( x \to \infty \).
We note in passing that further results from Complex Analysis and the Gamma Function \(\Gamma\) show that
\begin{align}
& \text{Ai}(0) = \frac{1}{3^{2/3}\Gamma(\frac{2}{3})} & & \text{Ai}'(0) = -\frac{1}{3^{1/3}\Gamma(\frac{1}{3})} \tag{11}
\end{align}
and hence by the form of (6), the series expansion of \(\text{Ai}(x)\) will be just
\begin{align}
\text{Ai}(x) &= \text{Ai}(0) (1 + \frac{x^3}{(3)(2)} + \frac{x^6}{(6)(5)(3)(2)} + \cdots) \\
&\quad + \text{Ai}'(0) (x + \frac{x^4}{(4)(3)} + \frac{x^7}{(7)(6)(4)(3)} + \cdots) \tag{12}
\end{align}
\(\text{Ai}(0) = a_0\) and \(\text{Ai}'(0) = a_1\), so as to match the set of initial conditions in (11). Moreover, the asymptotic expressions of \(\text{Ai}(x)\) as \( x \) goes to a positive/negative value at large are (the saddle point method required)
\begin{align}
\text{Ai}(x) \sim
\left\{\begin{aligned}
& \frac{1}{\sqrt{\pi}|x|^{1/4}} \sin(\frac{2|x|^{3/2}}{3} +\frac{\pi}{4}) & &\text{as} \quad x \to -\infty \\
& \frac{e^{-\frac{2x^{3/2}}{3}}}{2\sqrt{\pi}x^{1/4}} & &\text{as} \quad x \to \infty
\end{aligned}\right. \tag{13}
\end{align}
corresponding to the oscillatory/exponential decaying regimes. Meanwhile, the Airy Function of the second kind \(\text{Bi}(x)\) is chosen in a way such that it oscillates \(\pi/2\) out of phase relative to \(\text{Ai}(x)\) in the negative region. Here we simply state its integral form:
\begin{equation}
\text{Bi}(x) = \frac{1}{\pi} \int_0^{\infty} [\exp(-\frac{t^3}{3} + xt) + \sin (\frac{t^3}{3} + xt)] dt \tag{14}
\end{equation}
initial conditions:
\begin{align}
& \text{Bi}(0) = \frac{1}{3^{1/6}\Gamma(\frac{2}{3})} & & \text{Bi}'(0) = \frac{3^{1/6}}{\Gamma(\frac{1}{3})} \tag{15}
\end{align}
and asymptotic formulae:
\begin{align}
\text{Bi}(x) \sim
\left\{\begin{aligned}
& \frac{1}{\sqrt{\pi}|x|^{1/4}} \cos(\frac{2|x|^{3/2}}{3} +\frac{\pi}{4}) & &\text{as} \quad x \to -\infty \\
& \frac{e^{\frac{2x^{3/2}}{3}}}{\sqrt{\pi}x^{1/4}} & &\text{as} \quad x \to \infty
\end{aligned}\right. \tag{16}
\end{align}
Note that \( \text{Bi}(x) \to \infty \) as \(x \to \infty\), in contrast to \( \text{Ai}(x) \to 0 \). So they must be linearly independent. Moreover, it implies that any other solution to the Airy Equation that also decays to zero at \( x = \infty \) must be linearly dependent on and hence a constant multiple of \(\text{Ai}(x)\) (otherwise, these two solutions alone cannot simulate unboundedness at infinity), showing the uniqueness of \(\text{Ai}(x)\) up to a constant factor. The diagram below sketches the graphs of \( \text{Ai}(x) \) and \( \text{Bi}(x) \). It can be observed that in the oscillatory regime, as \(x \to -\infty \) moves to the left, for both of them, the frequency slowly increases while the amplitude slowly decreases.
Exercise
For a quantum particle that is confined inside a constant force field with a potential of \( V(x) = kx \), the corresponding time-independent Schrödinger Equation is
\begin{align}
-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + V(x)\psi &= E\psi \\
-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} +kx\psi &= E\psi \tag{17}
\end{align}
where \( E \) is the energy level and \( \psi(x) \) is the wavefunction. By a suitable change of variable, show that it is actually an Airy Equation and find the physical solution to \( \psi \).
Answer
We rewrite (17) as
\begin{align}
\frac{d^2\psi}{dx^2} -\frac{2m}{\hbar^2}(kx-E)\psi &= 0
\end{align}
and by a variable transformation (non-dimensionalization) \( x = (\hbar^2/2mk)^{1/3}\tilde{x} \), it becomes
\begin{align}
(\frac{2mk}{\hbar^2})^{2/3}\frac{d^2\psi}{d\tilde{x}^2} -\frac{2m}{\hbar^2}(k(\frac{\hbar^2}{2mk})^{1/3}\tilde{x}-E)\psi &= 0 \\
\frac{d^2\psi}{d\tilde{x}^2} -(\tilde{x}-(\frac{2m}{\hbar^2k^2})^{1/3}E)\psi &= 0
\end{align}
Denote \( \tilde{E} = (2m/\hbar^2k^2)^{1/3}E \) as the non-dimensionalized energy level, then it will appear as an Airy equation with the independent variable \(\tilde{x}\) displaced:
\begin{align}
\frac{d^2\psi}{d\tilde{x}^2} -(\tilde{x}-\tilde{E})\psi &= 0
\end{align}
The solution will then be formed by \( \text{Ai}(\tilde{x}-\tilde{E}) \) and \( \text{Bi}(\tilde{x}-\tilde{E}) \). However, to have a physically bounded solution, we will just keep \( \text{Ai}(\tilde{x}-\tilde{E}) \) and discard \( \text{Bi}(\tilde{x}-\tilde{E}) \) which blows up at infinity. \( \tilde{E} \) will then be the turning point of the wavefunction.
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