commit b4929875a3d3b4db1a035c1827ba571de44c27eb
parent b7d967928bdb0edcc37ce77c4a6f16f8e9a63411
Author: Ivan Gankevich <igankevich@ya.ru>
Date: Mon, 12 Mar 2018 14:23:55 +0300
Merge github.com:igankevich/arma-thesis
Diffstat:
5 files changed, 450 insertions(+), 34 deletions(-)
diff --git a/LICENSE b/LICENSE
@@ -0,0 +1,396 @@
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diff --git a/README.org b/README.org
@@ -16,3 +16,9 @@ All the *code* is contained in another set of repositories:
In order to *build* PDF you need to execute every code block in ~setup.org~,
export the main file to LaTeX and compile it using ~make~.
+
+ARMA thesis is *licensed* under a Creative Commons Attribution 4.0 International
+License.
+
+You should have received a copy of the license along with this work. If not, see
+<http://creativecommons.org/licenses/by/4.0/>.
diff --git a/arma-thesis-ru.org b/arma-thesis-ru.org
@@ -1270,7 +1270,10 @@ arma.plot_nonlinear(file.path("build", "nit-standing"), args)
\begin{align}
\label{eq-problem-2d}
& \phi_{xx}+\phi_{zz}=0,\\
- & \zeta_t + \zeta_x\phi_x = \frac{\zeta_x}{\sqrt{1 + \zeta_x^2}} \phi_x - \phi_z, & \text{на }z=\zeta(x,t).\nonumber
+ & \zeta_t + \zeta_x\phi_x
+ = \FracSqrtZetaX{\zeta_x} \phi_x
+ - \FracSqrtZetaX{1} \phi_z,
+ & \text{на }z=\zeta(x,t).\nonumber
\end{align}
Для ее решения воспользуемся методом Фурье. Возьмем преобразование Фурье от
обоих частей уравнений Лапласа и получим
@@ -1313,7 +1316,7 @@ arma.plot_nonlinear(file.path("build", "nit-standing"), args)
\begin{equation*}
\zeta_t
=
- \left( i f(x) - 1 \right)
+ \left( i f(x) - 1/\SqrtZetaX \right)
\left[
\Fun{z}
\ast
@@ -1326,7 +1329,7 @@ arma.plot_nonlinear(file.path("build", "nit-standing"), args)
E(u) =
\frac{1}{2\pi u}
\frac{
- \FourierY{\zeta_t / \left(i f(x) - 1\right)}{u}
+ \FourierY{\zeta_t / \left(i f(x) - 1/\SqrtZetaX\right)}{u}
}{
\FourierY{\Fun{z}}{u}
}
@@ -1342,7 +1345,7 @@ arma.plot_nonlinear(file.path("build", "nit-standing"), args)
\InverseFourierY{
\frac{e^{2\pi u z}}{2\pi u}
\frac{
- \FourierY{ \zeta_t / \left(i f(x) - 1\right) }{u}
+ \FourierY{ \zeta_t / \left(i f(x) - 1/\SqrtZetaX\right) }{u}
}{
\FourierY{ \Fun{\zeta(x,t)} }{u}
}
@@ -1387,10 +1390,11 @@ arma.plot_nonlinear(file.path("build", "nit-standing"), args)
\phi(x,z) = \InverseFourierY{ \Sinh{2\pi u (z+h)} E(u) }{x}.
\end{equation*}
Подставляя \(\phi\) в граничное условие на свободной поверхности, получаем
-\begin{equation*}
- \zeta_t = f(x) \InverseFourierY{ 2\pi i u \Sinh{2\pi u (z+h)} E(u) }{x}
- - \InverseFourierY{ 2\pi u \SinhX{2\pi u (z+h)} E(u) }{x}.
-\end{equation*}
+\begin{align*}
+ \zeta_t & = f(x) \InverseFourierY{ 2\pi i u \Sinh{2\pi u (z+h)} E(u) }{x} \\
+ & - \frac{1}{\SqrtZetaX}
+ \InverseFourierY{ 2\pi u \SinhX{2\pi u (z+h)} E(u) }{x}.
+\end{align*}
Здесь \(\sinh\) и \(\cosh\) дают схожие результаты вблизи свободной поверхности, и,
поскольку эта область является наиболее интересной с точки зрения практического
применения, положим \(\Sinh{2\pi{u}(z+h)}\approx\SinhX{2\pi{u}(z+h)}\). Выполняя
@@ -1403,7 +1407,7 @@ arma.plot_nonlinear(file.path("build", "nit-standing"), args)
\InverseFourierY{
\frac{\Sinh{2\pi u (z+h)}}{2\pi u}
\frac{
- \FourierY{ \zeta_t / \left(i f(x) - 1\right) }{u}
+ \FourierY{ \zeta_t / \left(i f(x) - 1/\SqrtZetaX\right) }{u}
}{
\FourierY{ \FunSecond{\zeta(x,t)} }{u}
}
@@ -1492,10 +1496,10 @@ Mathematica\nbsp{}cite:mathematica10. В линейной теории широ
\label{eq-problem-3d}
& \phi_{xx} + \phi_{yy} + \phi_{zz} = 0,\\
& \zeta_t + \zeta_x\phi_x + \zeta_y\phi_y
- =
- \frac{\zeta_x}{\SqrtZeta{1 + \zeta_x^2 + \zeta_y^2}} \phi_x
- +\frac{\zeta_y}{\SqrtZeta{1 + \zeta_x^2 + \zeta_y^2}} \phi_y
- - \phi_z, & \text{на }z=\zeta(x,y,t).\nonumber
+ = \FracSqrtZetaY{\zeta_x} \phi_x
+ + \FracSqrtZetaY{\zeta_y} \phi_y
+ - \FracSqrtZetaY{1} \phi_z, \nonumber\\
+ & \text{на }z=\zeta(x,y,t).\nonumber
\end{align}
Для ее решения также воспользуемся методом Фурье. Возьмем преобразование Фурье
от обоих частей уравнений Лапласа и получим
@@ -1530,11 +1534,11 @@ Mathematica\nbsp{}cite:mathematica10. В линейной теории широ
\begin{array}{rl}
\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}
+ - & f_3(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
-\(f_2(x,y)={\zeta_y}/{\SqrtZeta{1+\zeta_x^2+\zeta_y^2}}-\zeta_y\).
+где \(f_1(x,y)=\zeta_x/{\SqrtZetaY}-\zeta_x\),
+\(f_2(x,y)=\zeta_y/{\SqrtZetaY}-\zeta_y\) и \(f_3(x,y)=1/\SqrtZetaY\).
Также как и в разделе\nbsp{}[[#sec-pressure-2d]] мы предполагаем, что
\(\Sinh{2\pi{u}(z+h)}\approx\SinhX{2\pi{u}(z+h)}\) вблизи свободной поверхности,
@@ -1556,7 +1560,7 @@ Mathematica\nbsp{}cite:mathematica10. В линейной теории широ
\label{eq-phi-3d}
\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} }
+ \frac{ \FourierY{ \zeta_t / \left( i f_1(x,y) + i f_2(x,y) - f_3(x,y) \right)}{u,v} }
{ \FourierY{\mathcal{D}_3\left( x,y,\zeta\left(x,y\right) \right)}{u,v} }
}{x,y},
\end{equation*}
diff --git a/arma-thesis.org b/arma-thesis.org
@@ -1245,7 +1245,10 @@ Two-dimensional Laplace equation with Robin boundary condition is written as
\begin{align}
\label{eq-problem-2d}
& \phi_{xx}+\phi_{zz}=0,\\
- & \zeta_t + \zeta_x\phi_x = \frac{\zeta_x}{\sqrt{1 + \zeta_x^2}} \phi_x - \phi_z, & \text{at }z=\zeta(x,t).\nonumber
+ & \zeta_t + \zeta_x\phi_x
+ = \FracSqrtZetaX{\zeta_x}\phi_x
+ - \FracSqrtZetaX{1}\phi_z,
+ & \text{at }z=\zeta(x,t).\nonumber
\end{align}
Use Fourier method to solve this problem. Applying Fourier transform to both
sides of the equation yields
@@ -1286,7 +1289,7 @@ boundary condition yields
\begin{equation*}
\zeta_t
=
- \left( i f(x) - 1 \right)
+ \left( i f(x) - 1/\SqrtZetaX \right)
\left[
\Fun{z}
\ast
@@ -1299,7 +1302,7 @@ to both sides of this equation yields formula for coefficients \(E\):
E(u) =
\frac{1}{2\pi u}
\frac{
- \FourierY{\zeta_t / \left(i f(x) - 1\right)}{u}
+ \FourierY{\zeta_t / \left(i f(x) - 1/\SqrtZetaX\right)}{u}
}{
\FourierY{\Fun{z}}{u}
}
@@ -1314,7 +1317,7 @@ into\nbsp{}eqref:eq-guessed-sol-2d yields formula for \(\phi(x,z)\):
\InverseFourierY{
\frac{e^{2\pi u z}}{2\pi u}
\frac{
- \FourierY{ \zeta_t / \left(i f(x) - 1\right) }{u}
+ \FourierY{ \zeta_t / \left(i f(x) - 1/\SqrtZetaX\right) }{u}
}{
\FourierY{ \Fun{\zeta(x,t)} }{u}
}
@@ -1358,10 +1361,11 @@ into\nbsp{}eqref:eq-guessed-sol-2d-full yields
\phi(x,z) = \InverseFourierY{ \Sinh{2\pi u (z+h)} E(u) }{x}.
\end{equation*}
Plugging \(\phi\) into the boundary condition on the free surface yields
-\begin{equation*}
- \zeta_t = f(x) \InverseFourierY{ 2\pi i u \Sinh{2\pi u (z+h)} E(u) }{x}
- - \InverseFourierY{ 2\pi u \SinhX{2\pi u (z+h)} E(u) }{x}.
-\end{equation*}
+\begin{align*}
+ \zeta_t & = f(x) \InverseFourierY{ 2\pi i u \Sinh{2\pi u (z+h)} E(u) }{x} \\
+ & - \frac{1}{\SqrtZetaX}
+ \InverseFourierY{ 2\pi u \SinhX{2\pi u (z+h)} E(u) }{x}.
+\end{align*}
Here \(\sinh\) and \(\cosh\) give similar results near free surface, and since this
is the main area of interest in practical applications, we assume that
\(\Sinh{2\pi{u}(z+h)}\approx\SinhX{2\pi{u}(z+h)}\). Performing analogous to the
@@ -1373,7 +1377,7 @@ previous section transformations yields final formula for \(\phi(x,z)\):
\InverseFourierY{
\frac{\Sinh{2\pi u (z+h)}}{2\pi u}
\frac{
- \FourierY{ \zeta_t / \left(i f(x) - 1\right) }{u}
+ \FourierY{ \zeta_t / \left(i f(x) - 1/\SqrtZetaX\right) }{u}
}{
\FourierY{ \FunSecond{\zeta(x,t)} }{u}
}
@@ -1459,10 +1463,10 @@ Three-dimensional version of\nbsp{}eqref:eq-problem is written as
\label{eq-problem-3d}
& \phi_{xx} + \phi_{yy} + \phi_{zz} = 0,\\
& \zeta_t + \zeta_x\phi_x + \zeta_y\phi_y
- =
- \frac{\zeta_x}{\SqrtZeta{1 + \zeta_x^2 + \zeta_y^2}} \phi_x
- +\frac{\zeta_y}{\SqrtZeta{1 + \zeta_x^2 + \zeta_y^2}} \phi_y
- - \phi_z, & \text{at }z=\zeta(x,y,t).\nonumber
+ = \FracSqrtZetaY{\zeta_x} \phi_x
+ + \FracSqrtZetaY{\zeta_y} \phi_y
+ - \FracSqrtZetaY{1} \phi_z, \nonumber\\
+ & \text{at }z=\zeta(x,y,t).\nonumber
\end{align}
Again, use Fourier method to solve it. Applying Fourier transform to both sides
of Laplace equation yields
@@ -1496,11 +1500,11 @@ Plugging \(\phi\) into the boundary condition on the free surface yields
\begin{array}{rl}
\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}
+ - & f_3(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
-\(f_2(x,y)={\zeta_y}/{\SqrtZeta{1+\zeta_x^2+\zeta_y^2}}-\zeta_y\).
+where \(f_1(x,y)={\zeta_x}/{\SqrtZetaY}-\zeta_x\),
+\(f_2(x,y)={\zeta_y}/{\SqrtZetaY}-\zeta_y\) and \(f_3(x,y)=1/\SqrtZetaY\).
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
@@ -1522,7 +1526,7 @@ and plugging the result into\nbsp{}eqref:eq-guessed-sol-3d yields formula for
\label{eq-phi-3d}
\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} }
+ \frac{ \FourierY{ \zeta_t / \left( i f_1(x,y) + i f_2(x,y) - f_3(x,y) \right)}{u,v} }
{ \FourierY{\mathcal{D}_3\left( x,y,\zeta\left(x,y\right) \right)}{u,v} }
}{x,y},
\end{equation}
diff --git a/preamble.tex b/preamble.tex
@@ -63,6 +63,12 @@
% properly aligned version of sqrt for \zeta_y^2
\newcommand{\SqrtZeta}[1]{\sqrt{\vphantom{\zeta_x^2}\smash[b]{#1}}}
+% normalisation
+\newcommand{\SqrtZetaX}{\sqrt{1 + \zeta_x^2}}
+\newcommand{\SqrtZetaY}{\SqrtZeta{1 + \zeta_x^2 + \zeta_y^2}}
+\newcommand{\FracSqrtZetaX}[1]{\frac{#1}{\SqrtZetaX{}}}
+\newcommand{\FracSqrtZetaY}[1]{\frac{#1}{\SqrtZetaY{}}}
+
% wave vector
\newcommand{\Kvec}{\vec{k}}
\newcommand{\Kveclen}{\lvert\smash[b]{\Kvec}\rvert}
