WKB Solution to Approximate Slow Varying Wave
Long before, we discussed the simple harmonic equation (see this tutorial) with a general form of
\begin{equation}
\frac{d^2y}{dx^2} + m^2y = 0 \tag{1}
\end{equation}
where \(m\) is a constant and the sine/cosine solution basis may be compactly expressed by complex exponentials \( \exp(\pm imx) \). In some physical applications, though, \(m\) may not be a constant but a slowly varying function in \(x\), so now \( m = m(x) \). For example, \(m\) can be the reflective index in a non-homogeneous medium for an electromagnetic wave, or the buoyancy in a changing atmosphere. A naive way to obtain the new solution is to simply replace \( mx \) by \( \int m(x)dx \) which will be the same if \( m \) is indeed a constant, so for the canonical second-order linear ODE
\begin{equation}
\frac{d^2y}{dx^2} + m(x)^2y = 0 \tag{2}
\end{equation}
it may look like
\begin{equation}
y = \exp(\pm i\int m(x) dx) \tag{3}
\end{equation}
where \(\int m(x) dx\) is called the phase memory. (3) has captured the essence but still requires a tweak to be an appropriate approximation. Below, we shall heuristically derive the desired WKB solution to (2). The derivation is taken from the textbook Atmospheric and Oceanic Fluid Dynamics by Vallis.
We now assume a more flexible form of test solution as
\begin{equation}
y = A(x) \exp(i\theta(x)) \tag{4}
\end{equation}
For now, we may also take \(A(x)\) and \(\theta(x)\) to be real for convenience. Substituting this into (2) yields
\begin{align}
\begin{aligned} (\frac{d^2A}{dx^2} \exp(i\theta) + 2i\frac{dA}{dx}\frac{d\theta}{dx} \exp(i\theta) \\
+ iA\frac{d^2\theta}{dx^2}\exp(i\theta) -A(\frac{d\theta}{dx})^2\exp(i\theta)) + m(x)^2 A \exp(i\theta) \end{aligned} &= 0 \\
i(2\frac{dA}{dx}\frac{d\theta}{dx} + A\frac{d^2\theta}{dx^2}) -(A(\frac{d\theta}{dx})^2 -\frac{d^2A}{dx^2} -m^2A) &= 0 \tag{5}
\end{align}
By equating the real and complex parts, the terms in the two brackets must be zero respectively. The key to the WKB approximation then lies in assuming that
\begin{equation}
\left| \frac{1}{A} \frac{d^2A}{dx^2} \right| \ll m^2 \tag{6}
\end{equation}
This is the slowness condition for it to be valid. So, the \( \frac{d^2A}{dx^2} \) term in (5) can be neglected. Therefore, we have
\begin{align}
(\frac{d\theta}{dx})^2 &= m^2 \tag{7} \\
2\frac{dA}{dx}\frac{d\theta}{dx} + A\frac{d^2\theta}{dx^2} &= 0 \tag{8}
\end{align}
The solution to (7) is clearly the phase memory, \( \theta = \pm \int mdx \). Plugging this into (8) leads to
\begin{align}
2\frac{dA}{dx}m + A\frac{dm}{dx} &= 0 \\
\frac{dA}{dm} + \frac{A}{2m} &= 0 \\
\sqrt{m}\frac{dA}{dm} + \frac{A}{2\sqrt{m}} &= 0 \\
\frac{d}{dm}(\sqrt{m}A) &= 0 \\
\sqrt{m}A &= A_0 \\
A &= A_0m^{-\frac{1}{2}} \tag{9}
\end{align}
Hence the correct WKB solution will be
\begin{equation}
y = A_0m^{-\frac{1}{2}} \exp(\pm i\int m(x) dx) \tag{10}
\end{equation}
Compared to the naive guess (3), an extra addition of the \(m^{-\frac{1}{2}}\) factor is needed.
Application to Airy Equation
We demonstrate how the WKB approximation will actually look by applying it to the Airy equation (see this tutorial, (1)):
\begin{equation}
\frac{d^2y}{dx^2} -xy = 0 \tag{11}
\end{equation}
Comparing to (2), we easily see that \( m(x) = (-x)^{\frac{1}{2}} = ix^{\frac{1}{2}}\), and by (10) the two corresponding WKB solutions are represented by
\begin{equation}
y_{1,2} = A_{\pm}|x|^{-\frac{1}{4}} \exp(\pm i \int ix^{\frac{1}{2}} dx) = A_{\pm}|x|^{-\frac{1}{4}} \exp(\frac{\mp 2x^{\frac{3}{2}}}{3}) \tag{12}
\end{equation}
where we have added an absolute value to deal with the fourth root. The problem is that in (12), the factor \( |x|^{-\frac{1}{4}} \) blows up at \(x = 0\). Like the actual Airy equation, the \( \exp(\frac{\mp 2x^{\frac{3}{2}}}{3}) \) part is exponential when \( x > 0 \) but wave-like when \( x < 0 \). To bridge the gap and make the full WKB solution continuous at \( x = 0 \), we need two different sets of integration constants \( A_{\pm} \) so as to “patch” the two regions together. We demand that (12) be consistent with the asymptotic expressions of \( \text{Ai}(x) \) as shown by (13) in this tutorial: For the exponential regime,
\begin{align}
y = Ay_1 &= \frac{e^{-\frac{2x^{3/2}}{3}}}{2\sqrt{\pi}x^{1/4}} \\
A &= \frac{1}{2\sqrt{\pi}} \tag{13}
\end{align}
Meanwhile, for the sinusoidal regime, by noting that \( x^{\frac{3}{2}} = -i|x|^{\frac{3}{2}} \) for negative \( x \):
\begin{align}
y = By_1 + Cy_2 &= \frac{1}{\sqrt{\pi}|x|^{1/4}} \sin(\frac{2|x|^{3/2}}{3} +\frac{\pi}{4}) \\
B |x|^{-\frac{1}{4}} \exp(i\frac{2|x|^{\frac{3}{2}}}{3}) + C|x|^{-\frac{1}{4}} \exp(-i\frac{2|x|^{\frac{3}{2}}}{3}) &= \frac{1}{2i\sqrt{\pi}|x|^{1/4}} (e^{i\frac{\pi}{4}} e^{i\frac{2|x|^{3/2}}{3}} – e^{-i\frac{\pi}{4}} e^{-i\frac{2|x|^{3/2}}{3}}) \\
\Rightarrow B &= \frac{e^{i\frac{\pi}{4}}}{2i\sqrt{\pi}}, C= \frac{e^{-i\frac{\pi}{4}}}{2i\sqrt{\pi}} \tag{14}
\end{align}
They are then known as the connection formulas for the Airy Equation. This is particularly useful in modeling quantum particles, which may have a varying energy potential \( V(x) \), i.e. the normalized Schrödinger Equation may read like
\begin{equation}
\frac{d^2\psi}{dx^2} + V(x) \psi = 0 \tag{15}
\end{equation}
The WKB approximation alone fails due to blowing up at the turning point again (\( m = \sqrt{V(x)} = 0\)), and we have to further linearize the energy potential near the turning point, \( V(x) \approx kx \). This turns into an Airy equation, and the connection formulas then reenable the validity of the WKB approximation.
Exercise
Internal gravity wave in the atmosphere (see Chapter 7 of Vallis) has the governing equation of
\begin{equation}
[\frac{\partial^2}{\partial t^2} (\nabla^2 + \frac{\partial^2}{\partial z^2}) + N^2 \nabla^2] w = 0 \tag{16}
\end{equation}
where \( w \) is the perturbation vertical velocity and here \(N^2\) is the Brunt-Väisälä frequency, \( \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \) is the two-dimensional horizontal Laplacian. By assuming the ansatz of
\begin{equation}
w = W(z) e^{i(kx+ly-\omega t)} \tag{17}
\end{equation}
derive a second-order ODE for solving the vertical structure \( W(z) \) and find its corresponding WKB solution.
Answer
Substituting (17) into (16) yields the relation
\begin{align}
(k^2+l^2-\frac{d^2W}{dz^2})\omega^2 -N^2(k^2+l^2)W &= 0 \\
\frac{d^2W}{dz^2} + \frac{(N^2-\omega^2)K^2}{\omega^2}W &= 0
\end{align}
where \( K^2 = k^2 + l^2 \) is the total horizontal wavenumber. The WKB solution then has the form of (10) with \( m = (\frac{(N^2 -\omega^2)K^2}{\omega^2})^{\frac{1}{2}} \). Locally, the flow behaves like a plane wave with a vertical wavenumber \( m \) and its amplitude scaled by \( m^{-\frac{1}{2}} \).
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