commit 78f90f20a696741f7c3be5d57d92bc93f1cfb66e
parent 33ce29db834efa19b28dde2f3bfc92e957cc6b93
Author: miksa <milutin@popovic.xyz>
Date: Sun, 19 Jun 2022 11:36:48 +0200
fix and add abstract
Diffstat:
6 files changed, 162 insertions(+), 143 deletions(-)
diff --git a/app_pde/abstract.tex b/app_pde/abstract.tex
@@ -0,0 +1,17 @@
+\begin{abstract}
+ The aim of this project is to give a master student a general idea of
+ fluid dynamic and further combining this knowledge to model a problem
+ focusing on the tsunami, generated by an earthquake in the Andaman Basin
+ in 2004. In this regard focusing on inviscid water flow, where the mass
+ density of water is taken to be constant. It is qualitatively show how to
+ derive Euler's Equations of Motion using basic multivariable analysis.
+ The whole problem of modeling water waves comes around to determining the
+ wave profile at the surface $z = h(x,y,t)$, where the need
+ to introduce boundary conditions on the governing equations comes into
+ play. To derive model hierarchies for different regimes (e.g. shallow
+ water, long-wave or small amplitude) dimensional analysis and scaling of
+ the parameters together with asymptotic expansion becomes essential.
+ Asymptotic analysis gives a model for the regime of the solitary wave and
+ the KdV equation which is the region $\varepsilon=O\left(\delta \right)$,
+ a key to modeling the tsunami wave before approaching the shore.
+\end{abstract}
diff --git a/app_pde/appendix.tex b/app_pde/appendix.tex
@@ -13,8 +13,8 @@ $b(t)$ respectively, by
\begin{align}
\frac{d\mathcal{I}}{dt} =
\frac{\partial \mathcal{I}}{\partial t}+
- \frac{\partial \mathcal{I}}{\partial a}\frac{\partial a}{\partial t}+
- \frac{\partial \mathcal{I}}{\partial b}\frac{\partial b}{\partial t}.
+ \frac{\partial \mathcal{I}}{\partial b}\frac{\partial b}{\partial t}+
+ \frac{\partial \mathcal{I}}{\partial a}\frac{\partial a}{\partial t}.
\end{align}
Which in integral representation reads
\begin{align}
@@ -30,11 +30,11 @@ We start off with the standard material derivative
\frac{D\mathbf{u}}{Dt} = \frac{\partial \mathbf{u}}{\partial t}
+(\mathbf{u}\nabla)\mathbf{u}.
\end{align}
-We will use Einstein's Summation Convention, where we sum over indices that
-both appear at as the bottom as the top index, to rewrite the second part of
-the material derivative $(\mathbf{u}\nabla)\mathbf{u}$ into
+We will use Einstein's Summation Convention, where we sum over indices
+appearing at the bottom and the top. To rewrite the second part of the
+material derivative $(\mathbf{u}\nabla)\mathbf{u}$ into
\begin{align}
- (\mathbf{u}\times (\nabla \times \mathbf{u}))_k
+ (\mathbf{u}\times (\nabla \times \mathbf{u}))_i
&= \varepsilon^{ijk}u_j(\nabla \times \mathbf{u})_k \\
&= \varepsilon^{ijk}u_j\varepsilon_{klm}\partial^l u^m\\
&=(\delta^i_l\delta^j_m-\delta^i_m\delta^j_l)u_j\partial^l u^m\\
@@ -62,24 +62,24 @@ In our case the implicit function for fixed time reads
\end{align}
The parametric representation is
\begin{align}
- \mathbf{\sigma} = \begin{pmatrix} x_1 \\ x_2 \\ h \end{pmatrix} .
+ \vec{\sigma} = \begin{pmatrix} x_1 \\ x_2 \\ h \end{pmatrix} .
\end{align}
-The middle curvature of the surface parametrized by $\mathbf{\sigma}$ is
+The middle curvature of the surface parametrized by $\vec{\sigma}$ is
\begin{align}
\frac{1}{R} = \text{Tr}(G^{-1}B),
\end{align}
where $G$ and $B$ are given by
\begin{align}
- G_{ij} = \frac{\partial \mathbf{\sigma}}{\partial x_i} \frac{\partial
- \mathbf{\sigma}}{\partial x_j}, \\
- B_{ij} = -\mathbf{N} \frac{\partial^2 \mathbf{\sigma}}{\partial
+ G_{ij} = \frac{\partial \vec{\sigma}}{\partial x_i} \frac{\partial
+ \vec{\sigma}}{\partial x_j}, \\
+ B_{ij} = -\mathbf{N} \frac{\partial^2 \vec{\sigma}}{\partial
x_i\partial x_j},
\end{align}
where $i, j = 1, 2$ and $\mathbf{N}$ is the normal, normalized surface vector given by
\begin{align}
- \mathbf{N} &= \frac{\frac{\partial \mathbf{\sigma}}{\partial x_1}\times
- \frac{\partial \mathbf{\sigma}}{\partial x_2}}{\|\frac{\partial \mathbf{\sigma}}{\partial x_1}\times
- \frac{\partial \mathbf{\sigma}}{\partial x_2}\|} \\
+ \mathbf{N} &= \frac{\frac{\partial \vec{\sigma}}{\partial x_1}\times
+ \frac{\partial \vec{\sigma}}{\partial x_2}}{\|\frac{\partial \vec{\sigma}}{\partial x_1}\times
+ \frac{\partial \vec{\sigma}}{\partial x_2}\|} \\
&= \frac{1}{\sqrt{h_x^2 + h_y^2 +1}} \begin{pmatrix}
-h_x\\-h_y\\1 \end{pmatrix}.
\end{align}
diff --git a/app_pde/build/main.pdf b/app_pde/build/main.pdf
Binary files differ.
diff --git a/app_pde/chap1.tex b/app_pde/chap1.tex
@@ -426,131 +426,131 @@ As a consequence the \textbf{Integrated Mass Condition} is given by
\nabla_\perp \int_b^h \mathbf{u}_\perp\ dz + \underbrace{h_t -
b_t}_{=d_t} = 0.
\end{align}
-\subsection{Energy Equation}
-To derive the energy equation we start off with Euler's Equation of Motion
-\begin{align}
- \mathbf{u} _t + \nabla
- (\frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}+\Omega) = \mathbf{u}\times
- \mathbf{w},
-\end{align}
-multiplying the equation with $\mathbf{u}$ we get
-\begin{align}
- &\mathbf{u}\mathbf{u} _t \label{eq:energy1} \\
- &+(\mathbf{u}\nabla)(\frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}+\Omega)\label{eq:energy2}\\
- &= \mathbf{u}(\mathbf{u}\times
- \mathbf{w})\label{eq:energy3}.
-\end{align}
-The first equation given in \ref{eq:energy1} can we rewritten using inverse
-product rule of differentiation
-\begin{align}
- \mathbf{u}\frac{\partial \mathbf{u}}{\partial t}
- &= \frac{\partial
- }{\partial t} (\mathbf{u}\mathbf{u}) - \frac{\partial \mathbf{u}}{\partial t}
- \mathbf{u} \\
- &= \frac{\partial
- }{\partial t} (\mathbf{u}\mathbf{u}) - \mathbf{u}\frac{\partial
- \mathbf{u}}{\partial t}\\
- \Rightarrow\quad & \mathbf{u} \frac{\partial \mathbf{u}}{\partial t} =
- \frac{1}{2}\frac{\partial }{\partial t} (\mathbf{u}\mathbf{u}).
-\end{align}
-Then we may add
-\begin{align}
- \left(\frac{1}{2} \mathbf{u}\mathbf{u}+\frac{P}{\rho} +\Omega \right)
- \underbrace{(\nabla u)}_{=0} = 0,
-\end{align}
-to above not changing anything. Thereby getting
-\begin{align}
- \frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}\right)
- +(\mathbf{u}\nabla \mathbf{u})\left(
- \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho} \right)
- +\left( \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho} + \Omega \right)
- (\nabla \mathbf{u}) = 0.
-\end{align}
-Applying the product rule we can simplify
-\begin{align}
- \frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}\right)
- +\nabla \left(\mathbf{u}\left(\mathbf{u}(
- \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}\right) \right) = 0,
-\end{align}
-additionally adding $\frac{\partial \Omega}{\partial t} =0$ leads us to
-\begin{align}
- \underbrace{\frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}
- +\Omega\right)}_{\text{change of total energy density}}
- +\underbrace{\nabla \left(\mathbf{u}\left(\mathbf{u}(
- \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}\right)
-\right)}_{\text{energy flow of the velocity field}} = 0.\label{eq:energy}
-\end{align}
-This is called the \textbf{energy equation} and is a general result for a
-inviscid and incompressible fluids, which we can apply to study water waves.
-We start off with replacing $\nabla = \nabla_\perp + \frac{\partial }{\partial
-z} $ and $\Omega = g z$ and multiplying by $\rho$, then our energy equation
-in \ref{eq:energy} becomes
-\begin{align}
- \frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
- g z\right) + \nabla_\perp\left( \mathbf{u}_\perp\left(
- \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right) \right)
- \frac{\partial}{\partial z} \left( w\left(
- \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho g z \right) \right) = 0.
-\end{align}
-Integrating from bottom to top, i.e. from bed to free surface gets us to
-\begin{align}
- &\int_b^h\frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
- g z\right)\ dz \label{eq:e-int1}\\
- &+ \int_b^h \nabla_\perp\left( \mathbf{u}_\perp\left(
- \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right) \right)\
- dz\label{eq:e-int2}\\
- &+ \left(\frac{\partial}{\partial z} \left( w\left(
- \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho g z \right)
-\right)\right)\Bigg|_b^h \label{eq:e-int3}
- = 0.
-\end{align}
-For equation \ref{eq:e-int1} we use Leibniz Rule of Integration, leaving us
-with
-\begin{align}
- \int_b^h\frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
- g z\right)\ dz
- &= \frac{\partial }{\partial t} \int_b^h
- \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho gz \ dz\\
- &+ \left( \frac{1}{2}\rho \mathbf{u}_s \mathbf{u}_s + \rho g h \right)
- h_t\\
- &- \left( \frac{1}{2}\rho \mathbf{u}_b \mathbf{u}_b + \rho g b \right)
- b_t
-\end{align}
-For equation \ref{eq:e-int2} we again take note of the Leibniz Rule of
-Integration, getting
-\begin{align}
- \int_b^h \nabla_\perp\left( \mathbf{u}_\perp\left(
- \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right) \right)\
- dz
- &= \nabla_\perp \int_b^h \mathbf{u}_\perp\left(
- \frac{1}{2}\rho\mathbf{u}\mathbf{u} + P + \rho g z \right) \ dz\\
- &- \left( \frac{1}{2}\rho \mathbf{u}_s\mathbf{u}_s + P + \rho g h \right)
- \left( \mathbf{u}_{\perp s} \nabla_\perp \right) h\\
- &+\left( \frac{1}{2}\rho \mathbf{u}_b\mathbf{u}_b + P + \rho g b \right)
- \left( \mathbf{u}_{\perp b} \nabla_\perp \right) b
-\end{align}
-Thereby transforming our equation into
-\begin{align}
- \frac{\partial }{\partial t} \underbrace{\int_b^h \frac{1}{2}\rho
- \mathbf{u}\mathbf{u}+\rho g z\ dz}_{=:\mathcal{E}}
- + \nabla_\perp&\underbrace{\int_b^h
- \mathbf{u}_\perp\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho g z
-\right)\ dz}_{:=\mathcal{F}}
-+ \underbrace{P_s h_t - P_b b_t}_{:=\mathcal{P}} = 0\\
-\nonumber\\
- &\frac{\partial \mathcal{E}}{\partial t}
- + \nabla_\perp \mathcal{F} + \mathcal{P} = 0,
-\end{align}
-where $\mathcal{E}$ represents the energy in the flow per unit horizontal
-area, since we are integrating from bed to free surface. Where $\mathcal{F}$
-is the horizontal energy flux vector and lastly $\mathcal{P} = P_s h_t -
-P_b b_t$ is the net energy input due to the pressure forces doing work on the
-upper and lower boundaries, i.e. bottom and free surface of the fluid.
-Assuming stationary rigid bottom condition and constant surface pressure, we
-can set $P_s=0$, such that $\mathcal{P} =0$ leaving us with the equation
-\begin{align}
- \frac{\partial \mathcal{E}}{\partial t}
- + \nabla_\perp \mathcal{F} = 0.
-\end{align}
-We note that the assumption $P_s=0$ is only possible if the coefficient of
-surface tension is set to 0, which usually is not the case.
+%\subsection{Energy Equation}
+%To derive the energy equation we start off with Euler's Equation of Motion
+%\begin{align}
+% \mathbf{u} _t + \nabla
+% (\frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}+\Omega) = \mathbf{u}\times
+% \mathbf{w},
+%\end{align}
+%multiplying the equation with $\mathbf{u}$ we get
+%\begin{align}
+% &\mathbf{u}\mathbf{u} _t \label{eq:energy1} \\
+% &+(\mathbf{u}\nabla)(\frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}+\Omega)\label{eq:energy2}\\
+% &= \mathbf{u}(\mathbf{u}\times
+% \mathbf{w})\label{eq:energy3}.
+%\end{align}
+%The first equation given in \ref{eq:energy1} can we rewritten using inverse
+%product rule of differentiation
+%\begin{align}
+% \mathbf{u}\frac{\partial \mathbf{u}}{\partial t}
+% &= \frac{\partial
+% }{\partial t} (\mathbf{u}\mathbf{u}) - \frac{\partial \mathbf{u}}{\partial t}
+% \mathbf{u} \\
+% &= \frac{\partial
+% }{\partial t} (\mathbf{u}\mathbf{u}) - \mathbf{u}\frac{\partial
+% \mathbf{u}}{\partial t}\\
+% \Rightarrow\quad & \mathbf{u} \frac{\partial \mathbf{u}}{\partial t} =
+% \frac{1}{2}\frac{\partial }{\partial t} (\mathbf{u}\mathbf{u}).
+%\end{align}
+%Then we may add
+%\begin{align}
+% \left(\frac{1}{2} \mathbf{u}\mathbf{u}+\frac{P}{\rho} +\Omega \right)
+% \underbrace{(\nabla u)}_{=0} = 0,
+%\end{align}
+%to above not changing anything. Thereby getting
+%\begin{align}
+% \frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}\right)
+% +(\mathbf{u}\nabla \mathbf{u})\left(
+% \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho} \right)
+% +\left( \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho} + \Omega \right)
+% (\nabla \mathbf{u}) = 0.
+%\end{align}
+%Applying the product rule we can simplify
+%\begin{align}
+% \frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}\right)
+% +\nabla \left(\mathbf{u}\left(\mathbf{u}(
+% \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}\right) \right) = 0,
+%\end{align}
+%additionally adding $\frac{\partial \Omega}{\partial t} =0$ leads us to
+%\begin{align}
+% \underbrace{\frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}
+% +\Omega\right)}_{\text{change of total energy density}}
+% +\underbrace{\nabla \left(\mathbf{u}\left(\mathbf{u}(
+% \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}\right)
+%\right)}_{\text{energy flow of the velocity field}} = 0.\label{eq:energy}
+%\end{align}
+%This is called the \textbf{energy equation} and is a general result for a
+%inviscid and incompressible fluids, which we can apply to study water waves.
+%We start off with replacing $\nabla = \nabla_\perp + \frac{\partial }{\partial
+%z} $ and $\Omega = g z$ and multiplying by $\rho$, then our energy equation
+%in \ref{eq:energy} becomes
+%\begin{align}
+% \frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
+% g z\right) + \nabla_\perp\left( \mathbf{u}_\perp\left(
+% \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right) \right)
+% \frac{\partial}{\partial z} \left( w\left(
+% \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho g z \right) \right) = 0.
+%\end{align}
+%Integrating from bottom to top, i.e. from bed to free surface gets us to
+%\begin{align}
+% &\int_b^h\frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
+% g z\right)\ dz \label{eq:e-int1}\\
+% &+ \int_b^h \nabla_\perp\left( \mathbf{u}_\perp\left(
+% \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right) \right)\
+% dz\label{eq:e-int2}\\
+% &+ \left(\frac{\partial}{\partial z} \left( w\left(
+% \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho g z \right)
+%\right)\right)\Bigg|_b^h \label{eq:e-int3}
+% = 0.
+%\end{align}
+%For equation \ref{eq:e-int1} we use Leibniz Rule of Integration, leaving us
+%with
+%\begin{align}
+% \int_b^h\frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
+% g z\right)\ dz
+% &= \frac{\partial }{\partial t} \int_b^h
+% \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho gz \ dz\\
+% &+ \left( \frac{1}{2}\rho \mathbf{u}_s \mathbf{u}_s + \rho g h \right)
+% h_t\\
+% &- \left( \frac{1}{2}\rho \mathbf{u}_b \mathbf{u}_b + \rho g b \right)
+% b_t
+%\end{align}
+%For equation \ref{eq:e-int2} we again take note of the Leibniz Rule of
+%Integration, getting
+%\begin{align}
+% \int_b^h \nabla_\perp\left( \mathbf{u}_\perp\left(
+% \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right) \right)\
+% dz
+% &= \nabla_\perp \int_b^h \mathbf{u}_\perp\left(
+% \frac{1}{2}\rho\mathbf{u}\mathbf{u} + P + \rho g z \right) \ dz\\
+% &- \left( \frac{1}{2}\rho \mathbf{u}_s\mathbf{u}_s + P + \rho g h \right)
+% \left( \mathbf{u}_{\perp s} \nabla_\perp \right) h\\
+% &+\left( \frac{1}{2}\rho \mathbf{u}_b\mathbf{u}_b + P + \rho g b \right)
+% \left( \mathbf{u}_{\perp b} \nabla_\perp \right) b
+%\end{align}
+%Thereby transforming our equation into
+%\begin{align}
+% \frac{\partial }{\partial t} \underbrace{\int_b^h \frac{1}{2}\rho
+% \mathbf{u}\mathbf{u}+\rho g z\ dz}_{=:\mathcal{E}}
+% + \nabla_\perp&\underbrace{\int_b^h
+% \mathbf{u}_\perp\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho g z
+%\right)\ dz}_{:=\mathcal{F}}
+%+ \underbrace{P_s h_t - P_b b_t}_{:=\mathcal{P}} = 0\\
+%\nonumber\\
+% &\frac{\partial \mathcal{E}}{\partial t}
+% + \nabla_\perp \mathcal{F} + \mathcal{P} = 0,
+%\end{align}
+%where $\mathcal{E}$ represents the energy in the flow per unit horizontal
+%area, since we are integrating from bed to free surface. Where $\mathcal{F}$
+%is the horizontal energy flux vector and lastly $\mathcal{P} = P_s h_t -
+%P_b b_t$ is the net energy input due to the pressure forces doing work on the
+%upper and lower boundaries, i.e. bottom and free surface of the fluid.
+%Assuming stationary rigid bottom condition and constant surface pressure, we
+%can set $P_s=0$, such that $\mathcal{P} =0$ leaving us with the equation
+%\begin{align}
+% \frac{\partial \mathcal{E}}{\partial t}
+% + \nabla_\perp \mathcal{F} = 0.
+%\end{align}
+%We note that the assumption $P_s=0$ is only possible if the coefficient of
+%surface tension is set to 0, which usually is not the case.
diff --git a/app_pde/main.tex b/app_pde/main.tex
@@ -6,6 +6,8 @@
\begin{document}
\maketitle
+
+\include{./abstract.tex}
\tableofcontents
\include{./chap1.tex}
diff --git a/app_pde/preamble.tex b/app_pde/preamble.tex
@@ -100,6 +100,6 @@
\title{University of Vienna\\
\vspace{1cm}Seminar:\\Applied PDE Seminar\\
\vspace{0.5cm}
-Mathematical Modeling of Some Water-Waves
+Mathematical Modeling of Water-Wave Problems
}
\author{Milutin Popovic}
