From 55137faa7847453d99a6170fc938ef55f0d1bc14 Mon Sep 17 00:00:00 2001 From: Mandar Chitre Date: Sat, 11 Jan 2025 23:45:38 +0800 Subject: [PATCH] doc(pm): working document on math for modal model --- docs/misc/modes.md | 113 +++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 113 insertions(+) create mode 100644 docs/misc/modes.md diff --git a/docs/misc/modes.md b/docs/misc/modes.md new file mode 100644 index 0000000..0400613 --- /dev/null +++ b/docs/misc/modes.md @@ -0,0 +1,113 @@ +# Mode propagation in a Pekeris waveguide + +This math is largely based on `kraken.pdf` by Michael B. Porter. + +## Basic theory + +The acoustic wave equation is: +$$ +\nabla^2p = \frac{1}{c^2}\frac{\partial^2p}{\partial t^2}. +$$ +In a Pekeris waveguide with a pressure-release surface and a pressure-release bottom (good approximation for a long range propagation problem, as it turns out), the boundary conditions in cylindrical coordinates are: +$$ +\begin{align*} +p(r,0) &= 0 \quad \forall \; r, \\ +p(r,-D) &= 0. +\end{align*} +$$ +For a monochromatic source of frequency $f$ Hz located at $(0,-z_s)$ boundary condition is: +$$ +p(0,-z_s) = -A\exp(-i\omega t), +$$ +where $A$ is the source amplitude, and $\omega = 2\pi f$. Due to symmetry of the Pekeris waveguide, we have dropped the azimuthal coordinate. + +The wave equation with these boundary conditions admits an approximate modal solution of the form: +$$ +p(r,z,t) = \frac{i}{4}\sum_m\psi_m(z_s)\psi_m(z)H_0^{(1)}(k_mr)\exp(-i\omega t), +$$ +where $\psi_m(z)$ is the mode of order $m$ and $H_0^{(1)}$ is the Hankel function of the first kind and order 0, $k_m$ is the horizontal wavenumber associated with mode. The mode $\psi_m$ is given by: +$$ +\psi_m(z) = \sqrt{\frac{2}{D}}\sin(\gamma_m z), +$$ +where $\gamma_m$ is the vertical wavenumber associated with mode $m$: +$$ +\gamma_m = \sqrt{k_0^2 - k_m^2}, +$$ +for wavenumber $k_0 = \omega/c$. + +$\gamma_m$ is chosen to ensure boundary conditions are satisfied: +$$ +\psi_m(0) = \psi_m(D) = 0, +$$ +i.e., +$$ +\gamma_m = m\frac{\pi}{D}. +$$ +If the bottom boundary condition is chosen for a rigid bottom: +$$ +\begin{align*} +& \left.\frac{d}{dz}\psi_m(z)\right|_{z=D} = 0, \\ +& \therefore \; \gamma_m = \left(m-\frac{1}{2}\right)\frac{\pi}{D}. +\end{align*} +$$ + +Substituting $\psi_m$ back, we have: +$$ +p(r,z,t) = \frac{i}{2D}\sum_m\sin(\gamma_m z_s)\sin(\gamma_m z)H_0^{(1)}(k_mr)\exp(-i\omega t), +$$ +The transmission loss is then given by: +$$ +\text{TL}(r,z) = -20\log\left| \frac{p(r,z)}{p^0_{r=1}} \right|, +$$ +where $p^0_{r=1}$ is the pressure due to the source at 1 m from the acoustic center of the source: +$$ +p^0_{r=1} = \frac{\exp(ik_0r)}{4\pi r} = \frac{\exp(ik_0)}{4\pi}. +$$ + +## Acousto-elastic boundary condition + +If we use an acousto-elastic boundary condition for the seabed with sound speed $c_b$ and density $\rho_b$, we have a boundary condition: +$$ +\left.\rho_b\frac{d\psi_m}{dz}\right|_{z=D} = -\rho\sqrt{k_m^2 - k_b^2} \; \psi_m(r,D). +$$ +for $k_b = \omega/c_b$. Substituting $\psi_m(r,z)$: +$$ +\rho_b\gamma_m\cos(\gamma_m D) + \rho\sqrt{k_m^2 - k_b^2} \; \sin(\gamma_m D) = 0. +$$ + +## Group velocity + +The group velocity of mode $m$ is $d\omega/dk_m$. Given: +$$ +\begin{align*} +k_m^2 &= k_0^2 - \gamma_m^2 \\ +\therefore \; 2k_m\frac{dk_m}{d\omega} &= \frac{2k_0}{c} - 2\gamma_m\frac{d\gamma_m}{d\omega} \\ +\therefore \; \frac{dk_m}{d\omega} &= \frac{k_0}{c k_m} - \frac{\gamma_m}{k_m}\frac{d\gamma_m}{d\omega} +\end{align*} +$$ +Also: +$$ +\begin{align*} +&\rho_b\gamma_m\cos(\gamma_m D) + \rho\sqrt{k_m^2 - k_b^2} \; \sin(\gamma_m D) = 0 \\ +\therefore \quad &\rho_b\gamma_m\cos(\gamma_m D) + \rho\sqrt{k_0^2 - \gamma_m^2 - k_b^2} \; \sin(\gamma_m D) = 0 \\ +\therefore \quad &\rho_b\frac{d\gamma_m}{d\omega}\cos(\gamma_m D) - \rho_b\gamma_m\sin(\gamma_m D)D\frac{d\gamma_m}{d\omega} + \rho\zeta\cos(\gamma_m D)D\frac{d\gamma_m}{d\omega} \\ +&- \frac{\rho}{2\zeta}\sin(\gamma_m D)\left(\frac{2k_0}{c} - 2\gamma_m\frac{d\gamma_m}{d\omega} - \frac{2k_b}{c}\right) = 0 \\ +\therefore \quad &\left[\rho_b\cos(\gamma_m D) - \rho_bD\gamma_m\sin(\gamma_m D) + \rho D\zeta\cos(\gamma_m D) - \frac{\rho\gamma_m}{\zeta}\sin(\gamma_m D)\right]\frac{d\gamma_m}{d\omega} \\ +&= -\frac{\rho k_0}{c\zeta}\sin(\gamma_m D) - \frac{\rho k_b}{c\zeta} \; \sin(\gamma_m D) \\ +\therefore \quad &\frac{d\gamma_m}{d\omega} = -\frac{\rho\sin(\gamma_m D)(k_0 + k_b)}{c\zeta[\rho_b\cos(\gamma_m D) - \rho_bD\gamma_m\sin(\gamma_m D) + \rho D\zeta\cos(\gamma_m D) - \frac{\rho\gamma_m}{\zeta}\sin(\gamma_m D)]} \\ +&= -\frac{\rho\sin (\gamma_m D)\omega\left(c^2-c_b^2\right)}{c^2 c_b^2 \gamma_m \sin (\gamma_m D) \left(D\text{$\rho $b} \zeta+\rho \right)+\cos(\gamma_m D) \left(c^2 \left(D\rho\omega ^2-c_b^2 \text{$\rho $b} \zeta\right)-c_b^2 D\rho\omega^2\right)+c^2 c_b^2 D\rho\gamma_m^2\cos(\gamma_m D)}. +\end{align*} +$$ +where $\zeta = \sqrt{k_0^2 - \gamma_m^2 - k_b^2}$. + + +Mathematica code to do the simplification: +``` +Derivative[1][\[Gamma]][\[Omega]] /. + Solve[Collect[ + D[\[Rho]b \[Gamma][\[Omega]] Cos[\[Gamma][\[Omega]] h] + \[Rho] \ +Sqrt[(\[Omega]/c)^2 - \[Gamma][\[Omega]]^2 - (\[Omega]/ + cb)^2] Sin[\[Gamma][\[Omega]] h], \[Omega]], + Derivative[1][\[Gamma]][\[Omega]]] == 0, + Derivative[1][\[Gamma]][\[Omega]]][[1]] // Simplify +```