commit f27b84bbdd58e8f7df79cec5200b4b9d4b6f82e7
parent 03a7e444df478d56a19046bb7bf60731014541f2
Author: Ivan Gankevich <igankevich@ya.ru>
Date: Sun, 9 Apr 2017 14:30:40 +0300
Update three-dimensional velocity potential formulae.
Diffstat:
4 files changed, 49 insertions(+), 33 deletions(-)
diff --git a/Makefile b/Makefile
@@ -11,10 +11,10 @@ export TEXINPUTS=$(PWD)//:
all: build/$(PHD_RU).pdf build/$(PHD_EN).pdf
-build/$(PHD_RU).pdf: $(PHD_RU).tex bib/*
+build/$(PHD_RU).pdf: $(PHD_RU).tex preamble.tex bib/*
latexmk $(FLAGS) -f $(PHD_RU).tex
-build/$(PHD_EN).pdf: $(PHD_EN).tex bib/*
+build/$(PHD_EN).pdf: $(PHD_EN).tex preamble.tex bib/*
latexmk $(FLAGS) -f $(PHD_EN).tex
clean:
diff --git a/arma-thesis-ru.org b/arma-thesis-ru.org
@@ -1273,13 +1273,14 @@ eqref:eq-solution-2d-full до
\FourierY{\phi(x,y,z)}{u,v,w} = 0,
\end{equation*}
откуда имеем \(w=\pm{i}\sqrt{u^2+v^2}\). Решение уравнения будем искать в виде
-обратного преобразования Фурье \(\phi(x,y,z)=\InverseFourierY{E(u,v,w)}{x,y,z}\).
-Применяя полученное равенство, получаем
+обратного преобразования Фурье
+\(\phi(x,y,z)=\InverseFourierY{E(u,v,w)}{x,y,z}\). Подставляя
+\(w=i\sqrt{u^2+v^2}=i\Kveclen\) в исходную формулу, получаем
\begin{equation*}
\phi(x,y,z) = \InverseFourierY{
\left(
- C_1 e^{2\pi \sqrt{u^2+v^2} z}
- -C_2 e^{-2\pi \sqrt{u^2+v^2} z}
+ C_1 e^{2\pi \Kveclen z}
+ -C_2 e^{-2\pi \Kveclen z}
\right)
E(u,v)
}{x,y}.
@@ -1289,16 +1290,16 @@ eqref:eq-solution-2d-full до
\begin{equation}
\label{eq-guessed-sol-3d}
\phi(x,y,z) = \InverseFourierY{
- \Sinh{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)
+ \Sinh{2\pi \Kveclen (z+h)} E(u,v)
}{x,y}.
\end{equation}
Подставляя выражение для \(\phi\) в граничное условие, получим
\begin{equation*}
\arraycolsep=1.4pt
\begin{array}{rl}
- \zeta_t = & i f_1(x,y) \InverseFourierY{2 \pi u \Sinh{2\pi \sqrt{u^2+v^2} (z+h)}E(u,v)}{x,y} \\
- + & i f_2(x,y) \InverseFourierY{2 \pi v \Sinh{2\pi \sqrt{u^2+v^2} (z+h)}E(u,v)}{x,y} \\
- - & \InverseFourierY{2 \pi \sqrt{u^2+v^2} \Sinh{2\pi \sqrt{u^2+v^2} (z+h)}E(u,v)}{x,y}
+ \zeta_t = & i f_1(x,y) \InverseFourierY{2 \pi u \Sinh{2\pi \Kveclen (z+h)}E(u,v)}{x,y} \\
+ + & i f_2(x,y) \InverseFourierY{2 \pi v \Sinh{2\pi \Kveclen (z+h)}E(u,v)}{x,y} \\
+ - & \InverseFourierY{2 \pi \sqrt{u^2+v^2} \Sinh{2\pi \Kveclen (z+h)}E(u,v)}{x,y}
\end{array}
\end{equation*}
где \(f_1(x,y)={\zeta_x}/{\SqrtZeta{1+\zeta_x^2+\zeta_y^2}}-\zeta_x\) и
@@ -1308,9 +1309,9 @@ eqref:eq-solution-2d-full до
\arraycolsep=1.4pt
\begin{array}{rl}
\FourierY{\zeta_t}{u,v} = &
- \FourierY{i f_1(x,y) \InverseFourierY{2 \pi u \Sinh{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)}{x,y}}{u,v} \\
- + & \FourierY{i f_2(x,y) \InverseFourierY{2 \pi v \Sinh{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)}{x,y}}{u,v} \\
- - & 2 \pi \sqrt{u^2+v^2} \Sinh{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)
+ \FourierY{i f_1(x,y) \InverseFourierY{2 \pi u \Sinh{2\pi \Kveclen (z+h)} E(u,v)}{x,y}}{u,v} \\
+ + & \FourierY{i f_2(x,y) \InverseFourierY{2 \pi v \Sinh{2\pi \Kveclen (z+h)} E(u,v)}{x,y}}{u,v} \\
+ - & 2 \pi \Kveclen \Sinh{2\pi \Kveclen (z+h)} E(u,v)
\end{array}
\end{equation*}
Окончательное решение получается при подстановке выражения для \(E(u,v)\)
diff --git a/arma-thesis.org b/arma-thesis.org
@@ -1504,12 +1504,12 @@ of Laplace equation yields
\end{equation*}
hence \(w=\pm{i}\sqrt{u^2+v^2}\). We seek solution in the form of inverse Fourier
transform \(\phi(x,y,z)=\InverseFourierY{E(u,v,w)}{x,y,z}\). Plugging
-\(w=i\sqrt{u^2+v^2}\) into the formula yields
+\(w=i\sqrt{u^2+v^2}=i\Kveclen\) into the formula yields
\begin{equation*}
\phi(x,y,z) = \InverseFourierY{
\left(
- C_1 e^{2\pi \sqrt{u^2+v^2} z}
- -C_2 e^{-2\pi \sqrt{u^2+v^2} z}
+ C_1 e^{2\pi \Kveclen z}
+ -C_2 e^{-2\pi \Kveclen z}
\right)
E(u,v)
}{x,y}.
@@ -1519,32 +1519,44 @@ two-dimensional case) yields
\begin{equation}
\label{eq-guessed-sol-3d}
\phi(x,y,z) = \InverseFourierY{
- \Sinh{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)
+ \Sinh{2\pi \Kveclen (z+h)} E(u,v)
}{x,y}.
\end{equation}
Plugging \(\phi\) into the boundary condition on the free surface yields
\begin{equation*}
\arraycolsep=1.4pt
\begin{array}{rl}
- \zeta_t = & i f_1(x,y) \InverseFourierY{2 \pi u \Sinh{2\pi \sqrt{u^2+v^2} (z+h)}E(u,v)}{x,y} \\
- + & i f_2(x,y) \InverseFourierY{2 \pi v \Sinh{2\pi \sqrt{u^2+v^2} (z+h)}E(u,v)}{x,y} \\
- - & \InverseFourierY{2 \pi \sqrt{u^2+v^2} \SinhX{2\pi \sqrt{u^2+v^2} (z+h)}E(u,v)}{x,y}
+ \zeta_t = & i f_1(x,y) \InverseFourierY{2 \pi u \Sinh{2\pi \Kveclen (z+h)}E(u,v)}{x,y} \\
+ + & i f_2(x,y) \InverseFourierY{2 \pi v \Sinh{2\pi \Kveclen (z+h)}E(u,v)}{x,y} \\
+ - & \InverseFourierY{2 \pi \Kveclen \SinhX{2\pi \Kveclen (z+h)}E(u,v)}{x,y}
\end{array}
\end{equation*}
where \(f_1(x,y)={\zeta_x}/{\SqrtZeta{1+\zeta_x^2+\zeta_y^2}}-\zeta_x\) and
-\(f_2(x,y)={\zeta_y}/{\SqrtZeta{1+\zeta_x^2+\zeta_y^2}}-\zeta_y\). Applying
-Fourier transform to both sides of the equation yields formula for coefficients
-\(E\):
+\(f_2(x,y)={\zeta_y}/{\SqrtZeta{1+\zeta_x^2+\zeta_y^2}}-\zeta_y\).
+
+Like in Section\nbsp{}[[#sec:pressure-2d]] we assume that
+\(\Sinh{2\pi{u}(z+h)}\approx\SinhX{2\pi{u}(z+h)}\) near free surface, but in
+three-dimensional case this is not enough to solve the problem. In order to get
+analytic formula for coefficients \(E\) we need to assume, that all Fourier
+transforms in the equation have radially symmetric kernels, i.e. replace \(u\)
+and \(v\) with \(\Kveclen\). There are two points supporting this assumption.
+First, in numerical implementation integration is done over positive wave
+numbers, so the sign of \(u\) and \(v\) does not affect the result. Second, the
+growth of \(\cosh\) term of the integral kernel is much higher than the growth
+of \(u\) or \(\Kveclen\), so the substitution has small effect on the magnitude
+of the solution. Despite these two points, a use of more mathematically rigorous
+approach would be preferable.
+
+Then applying Fourier transform to both sides of the equation and plugging the
+result into eqref:eq-guessed-sol-3d yields formula for \(\phi\):
\begin{equation*}
- \arraycolsep=1.4pt
- \begin{array}{rl}
- \FourierY{\zeta_t}{u,v} = &
- \FourierY{i f_1(x,y) \InverseFourierY{2 \pi u \Sinh{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)}{x,y}}{u,v} \\
- + & \FourierY{i f_2(x,y) \InverseFourierY{2 \pi v \Sinh{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)}{x,y}}{u,v} \\
- - & 2 \pi \sqrt{u^2+v^2} \SinhX{2\pi \sqrt{u^2+v^2} (z+h)} E(u,v)
- \end{array}
+ \phi(x,y,z,t) = \InverseFourierY{
+ \frac{ \Sinh{\smash{2\pi \Kveclen (z+h)}} }{ 2\pi\Kveclen }
+ \frac{ \FourierY{ \zeta_t / \left( i f_1(x,y) + i f_2(x,y) - 1 \right)}{u,v} }
+ { \FourierY{\mathcal{D}_3\left( x,y,\zeta\left(x,y\right) \right)}{u,v} }
+ }{x,y},
\end{equation*}
-Final solution is obtained after plugging \(E(u,v)\) into eqref:eq-guessed-sol-3d.
+where \(\FourierY{\mathcal{D}_3\left(x,y,z\right)}{u,v}=\Sinh{\smash{2\pi\Kveclen{}z}}\).
* Numerical methods and experimental results
** The shape of ACF for different types of waves
diff --git a/preamble.tex b/preamble.tex
@@ -57,4 +57,8 @@
\newcommand{\InverseFourierX}[3]{\mathcal{F}^{-1}_{#2}\!\left\{#1\right\}\!\left(#3\right)}
% properly aligned version of sqrt for \zeta_y^2
-\newcommand{\SqrtZeta}[1]{\sqrt{\vphantom{\zeta_x^2}\smash[b]{#1}}}-
\ No newline at end of file
+\newcommand{\SqrtZeta}[1]{\sqrt{\vphantom{\zeta_x^2}\smash[b]{#1}}}
+
+% wave vector
+\newcommand{\Kvec}{\vec{k}}
+\newcommand{\Kveclen}{\lvert\smash[b]{\Kvec}\rvert}
