Linearized Shallow Water System
In this post, we will explore how Hermite polynomials can be relevant in expressing the structure of equatorial waves as a physical/atmospheric sciences application. We start by establishing the governing equations of the linearized shallow water system near the Equator:
\begin{align}
\left\{
\begin{aligned}
\frac{\partial u}{\partial t} -\beta y v &= -\frac{\partial \Phi}{\partial x} \\
\frac{\partial v}{\partial t} + \beta y u &= -\frac{\partial \Phi}{\partial y} \\
\frac{\partial \Phi}{\partial t} &= -gH(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y})
\end{aligned}\right. \tag{1}
\end{align}
The first two equations are the momentum equations, and the last one is the continuity equation. \( u, v, \Phi \) are the zonal, meridional winds, and the geopotential. The constants include \( \beta \), which denotes the beta effect (the change of the Coriolis parameter \(f\) over latitudes), and \(g,H\) as the gravitational acceleration and equivalent depth.
Assuming a traveling wave solution in the zonal (\(x\)-) direction:
\begin{align}
\left\{
\begin{aligned}
u &= \hat{u}(y)\exp(i(kx-\omega t)) \\
v &= \hat{v}(y)\exp(i(kx-\omega t)) \\
\Phi &= \hat{\Phi}(y)\exp(i(kx-\omega t)) \\
\end{aligned}\right. \tag{2}
\end{align}
so that \( \hat{u}(y), \hat{v}(y), \hat{\Phi}(y) \) are now functions of \(y\) only, representing the meridional structure. We now substitute (2) into (1) and work to derive the wave solution. First, we obtain
\begin{align}
-i\omega \hat{u}(y) e^{i(kx-\omega t)} -\beta y \hat{v}(y)e^{i(kx-\omega t)} &= -i k \hat{\Phi}(y)e^{i(kx-\omega t)} \\
-i\omega \hat{u}(y) -\beta y \hat{v}(y) &= -i k \hat{\Phi}(y) \tag{3}
\end{align}
\begin{align}
-i\omega \hat{v}(y)e^{i(kx-\omega t)} + \beta y \hat{u}(y)e^{i(kx-\omega t)} &= -\frac{d\hat{\Phi}(y)}{dy}e^{i(kx-\omega t)} \\
-i\omega \hat{v}(y) + \beta y \hat{u}(y) &= -\frac{d\hat{\Phi}(y)}{dy} \tag{4}
\end{align}
and
\begin{align}
-i\omega \hat{\Phi}(y)e^{i(kx-\omega t)} &= -gH\left(i k\hat{u}(y)e^{i(kx-\omega t)} + \frac{d\hat{v}(y)}{dy}e^{i(kx-\omega t)}\right) \\
-i\omega \hat{\Phi}(y) &= -gH \left(ik\hat{u}(y) + \frac{d\hat{v}(y)}{dy}\right) \tag{5}
\end{align}
Differentiating (5) gives
\begin{align}
-i\omega \frac{d\hat{\Phi}(y)}{dy} = -gH \left(i k\frac{d\hat{u}(y)}{dy} + \frac{d^2\hat{v}(y)}{dy^2}\right) \tag{6}
\end{align}
Substituting (4) into (6) leads to
\begin{align}
i\omega (-i\omega \hat{v}(y) + \beta y \hat{u}(y)) &= -gH\left(i k\frac{d\hat{u}(y)}{dy} + \frac{d^2\hat{v}(y)}{dy^2}\right) \\
\omega^2 \hat{v}(y) + i \omega\beta y \hat{u}(y) &= -gH\left(i k\frac{d\hat{u}(y)}{dy} + \frac{d^2\hat{v}(y)}{dy^2}\right) \tag{7}
\end{align}
Using (3) in (7) gives
\begin{align}
\omega^2 \hat{v}(y) + i \omega\beta y \hat{u}(y) &= -gH\left(-\frac{k}{\omega} \frac{d}{dy} (-i k \hat{\Phi}(y) + \beta y\hat{v}(y)) + \frac{d^2\hat{v}(y)}{dy^2}\right) \\
\omega^2 \hat{v}(y) + i \omega\beta y \hat{u}(y) &= \begin{aligned} & \frac{-i gHk^2}{\omega}\frac{d\hat{\Phi}(y)}{dy} + \frac{gHk}{\omega}\left(\beta\hat{v}(y) + \beta y\frac{d\hat{v}(y)}{dy}\right) \\
&-gH \frac{d^2\hat{v}(y)}{dy^2} \end{aligned} \tag{8}
\end{align}
Applying both (4) and (5) in (8) yields, after some rearrangement
\begin{align}
\omega^2 \hat{v}(y) + i \omega\beta y \hat{u}(y) &= \begin{aligned} &\frac{i gHk^2}{\omega}(-i\omega \hat{v}(y) + \beta y \hat{u}(y)) \\
&+ \frac{gHk}{\omega}\left(\beta\hat{v}(y) + \beta y \left(\frac{i \omega}{gH} \hat{\Phi}(y) -i k\hat{u}(y)\right)\right) \\
& -gH \frac{d^2\hat{v}(y)}{dy^2} \end{aligned} \\
&= \begin{aligned} &gHk^2 \hat{v}(y) \color{red}{+ \frac{i gHk^2}{\omega} \beta y \hat{u}(y)} + \frac{gHk}{\omega}\beta\hat{v}(y) + i k \beta y \hat{\Phi}(y) \\
& \color{red}{-\frac{i gHk^2}{\omega} \beta y \hat{u}(y)} -gH \frac{d^2\hat{v}(y)}{dy^2} \end{aligned} \\
&= gHk^2 \hat{v}(y) + \frac{gHk}{\omega}\beta\hat{v}(y) + i k \beta y \hat{\Phi}(y) -gH \frac{d^2\hat{v}(y)}{dy^2} \tag{9}
\end{align}
Finally, adding the above by (3) times \(\beta y\), we have
\begin{align}
\omega^2 \hat{v}(y) -\beta^2 y^2 \hat{v}(y) = gHk^2 \hat{v}(y) + \frac{gHk}{\omega}\beta\hat{v}(y) -gH \frac{d^2\hat{v}(y)}{dy^2} \tag{10}
\end{align}
Cleaning this up, we obtain
\begin{align}
\frac{d^2\hat{v}(y)}{dy^2} + \left[\left(\frac{\omega^2}{gH} -k^2 -\beta\frac{k}{\omega}\right) -\frac{\beta^2}{gH}y^2\right]\hat{v}(y) = 0 \tag{11}
\end{align}
Equatorial Wave Solution
Finally, with a change of variable \(\tilde{y} = y/y_0, y_0 = (\sqrt{gH}/\beta)^{1/2} \) (non-dimensionalization via Rossby deformation radius), it becomes
\begin{align}
\frac{\beta}{\sqrt{gH}} \frac{d^2\hat{v}(\tilde{y})}{d\tilde{y}^2} + \left[\left(\frac{\omega^2}{gH} -k^2 -\beta\frac{k}{\omega}\right) -\frac{\beta^2}{gH}\left(\frac{\sqrt{gH}}{\beta}\tilde{y}^2\right)\right]\hat{v}(\tilde{y}) &= 0 \\
\frac{d^2\hat{v}(\tilde{y})}{d\tilde{y}^2} + \left[\frac{\sqrt{gH}}{\beta}\left(\frac{\omega^2}{gH} -k^2 -\beta\frac{k}{\omega}\right) -\tilde{y}^2\right]\hat{v}(\tilde{y}) &= 0 \tag{12}
\end{align}
Letting
\begin{align}
\mu = \frac{\sqrt{gH}}{\beta}\left(\frac{\omega^2}{gH} -k^2 -\beta\frac{k}{\omega}\right) \tag{13}
\end{align}
simplifies the ODE to
\begin{align}
\hat{v}^{\prime\prime} + (\mu -\tilde{y}^2)\hat{v} = 0 \tag{14}
\end{align}
This is known as the parabolic cylinder equation, and can be shown to be related to the Hermite equation as below. Making a change of variable \( \hat{v} = \hat{w} \exp(-\tilde{y}^2/2) \), then
\begin{align}
\hat{v}’ &= \hat{w}’ \exp(-\frac{\tilde{y}^2}{2}) -\tilde{y}\hat{w} \exp(-\frac{\tilde{y}^2}{2}) \tag{15} \\
\hat{v}^{\prime\prime} &= \hat{w}^{\prime\prime} \exp(-\frac{\tilde{y}^2}{2}) -\tilde{y}\hat{w}’ \exp(-\frac{\tilde{y}^2}{2}) -\hat{w} \exp(-\frac{\tilde{y}^2}{2}) \\
&\quad -\tilde{y}\hat{w}’ \exp(-\frac{\tilde{y}^2}{2}) + \tilde{y}^2\hat{w} \exp(-\frac{\tilde{y}^2}{2}) \\
&= (\hat{w}^{\prime\prime} -2\tilde{y}\hat{w}’ + (\tilde{y}^2-1)\hat{w})\exp(-\frac{\tilde{y}^2}{2}) \tag{16}
\end{align}
Substituting these into (14), it becomes
\begin{align}
(\hat{w}^{\prime\prime} -2\tilde{y}\hat{w}’ + (\tilde{y}^2-1)\hat{w})\exp(-\frac{\tilde{y}^2}{2}) + (\mu -\tilde{y}^2)\hat{w} \exp(-\frac{\tilde{y}^2}{2}) &= 0 \\
\hat{w}^{\prime\prime} -2\tilde{y}\hat{w}’ + (\mu-1)\hat{w} &= 0 \tag{17}
\end{align}
This shows that it is Hermite’s equation in disguise, where we have \(\mu = 2\nu + 1 \), \( \nu = 0, 1, 2, \ldots \), if the solutions have to be bounded. (see the last tutorial) Hence, the equatorial wave modes are given in terms of \( \hat{w} = H_n(\tilde{y}) \) and
\begin{align}
\hat{v}(\tilde{y}) &= \hat{v}_0H_n(\tilde{y})\exp(-\frac{\tilde{y}^2}{2}) \tag{18}
\end{align}
From this, \(\hat{u}(\tilde{y})\) and \(\hat{\Phi}(\tilde{y})\) can be found via any two of (3), (4), (5). Note that (13) embodies the dispersion relation for wavelength and frequency.
(Schematic from Kiladis et al., 2009)
Exercise
The linearized shallow water system (1) permits a special solution which is not included in (18): the meridional wind component is identically zero, \( v = 0 \), and only zonal wind persists. By setting \( \hat{v} = 0 \) in (3)-(5), derive this special solution.
Answer
We can first take the new (3) and (5), which are
\begin{align}
-i\omega \hat{u}(y) &= -i k \hat{\Phi}(y) \\
-i\omega \hat{\Phi}(y) &= -gH ik\hat{u}(y)
\end{align}
Multiplying them together leads to the dispersion relation of
\begin{align}
-\omega^2 \hat{u}(y)\hat{\Phi}(y) &= -gHk^2\hat{\Phi}(y)\hat{u}(y) \\
\frac{\omega}{k} &= \sqrt{gH}
\end{align}
Eliminating \( \hat{u} \), (4) becomes
\begin{align}
\beta y (\frac{k}{\omega}\hat{\Phi}(y)) &= -\frac{d\hat{\Phi}(y)}{dy} \\
\frac{d\hat{\Phi}(y)}{dy} + \frac{\beta}{\sqrt{gH}} y\hat{\Phi}(y) &= 0 \\
\end{align}
Again, making the change of variable \(\tilde{y} = y/y_0, y_0 = (\sqrt{gH}/\beta)^{1/2} \), it is reduced to
\begin{align}
\frac{1}{(\sqrt{gH}/\beta)^{1/2}}\frac{d\hat{\Phi}(\tilde{y})}{d\tilde{y}} + \frac{\beta}{\sqrt{gH}}(\sqrt{gH}/\beta)^{1/2} \tilde{y}\hat{\Phi}(\tilde{y}) &= 0 \\
\frac{d\hat{\Phi}(\tilde{y})}{d\tilde{y}} + \tilde{y}\hat{\Phi}(\tilde{y}) &= 0 \\
\end{align}
This is a simple first-order ODE that can be solved using an integrating factor:
\begin{align}
e^{\tilde{y}^2/2}(\frac{d\hat{\Phi}}{d\tilde{y}} + \tilde{y}\hat{\Phi}) &= 0 \\
\frac{d}{d\tilde{y}} (e^{\tilde{y}^2/2} \hat{\Phi}) &= 0 \\
e^{\tilde{y}^2/2} \hat{\Phi} &= A \\
\hat{\Phi} &= A e^{-\tilde{y}^2/2}
\end{align}
The solution for \( \hat{u} \) is similar. So in this special solution, the amplitude of the zonal wind and geopotential decay at a rate of \( \exp(-\frac{\tilde{y}^2}{2}) \) just like (18) without any further complications from the Hermite polynomials. This physically corresponds to the Kelvin wave (Subfigure (f) in the schematic above). We usually denote this solution by \( n = -1 \), because \( H_{-1} \) is conventionally taken to be \( 0 \), and so \( \hat{v} \) by (18) will also be zero.
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