updated fluid mechanics

Mathematics/2nd/Functions_of_several_variables/Functions_of_several_variables.tex

@@ -289,7 +289,7 @@
       \item If $m=1$ and $g(a)\ne0$, then $\displaystyle\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$ is differentiable at $a$ and: $$D\left(\frac{f}{g}\right)(a)=\frac{g(a)Df(a)-f(a)Dg(a)}{{g(a)}^2}$$
     \end{enumerate}
   \end{theorem}
-  \begin{theorem}[Chain rule]
+  \begin{theorem}[Chain rule]\label{FSV:chainrule}
     Let $U\subseteq\RR^n$ and $V\subseteq\RR^m$ be open sets. Let $\vf{f}:U\rightarrow\RR^m$ and $\vf{g}:V\rightarrow\RR^p$. Suppose that $\vf{f}(U)\subset V$, $\vf{f}$ is differentiable at $a\in U$ and $\vf{g}$ is differentiable at $\vf{f}(a)$. Then $\vf{g}\circ \vf{f}$ is differentiable at $a$ and: $$\vf{D}(\vf{g}\circ \vf{f})(a)=\vf{Dg}(\vf{f}(a))\circ \vf{Df}(a)$$
   \end{theorem}
   \begin{definition}

Physics/Advanced/Fluid_mechanics/Fluid_mechanics.tex

@@ -5,7 +5,7 @@
 \begin{multicols}{2}[\section{Fluid mechanics}]
   \subsection{Equations of motion}
   \subsubsection{Euler's equations}
-  In this section we will describe the motion of a fluid with a set of equation that result from the conservation of mass, momentum and energy. From what follows, let $D\subseteq \RR^3$ be a region filled with a fluid. For each time $t$ and $\vf{x}\in D$ we assume that the fluid has a well-defined mass density $\rho(\vf{x},t)$\footnote{The assumption that $\rho$ exists is a continuum assumption. Clearly, it does not hold if the molecular structure of matter is taken into account. For most macroscopic phenomena occurring in nature, it is believed that this assumption is extremely accurate.}. Finally, we denot by $\vf{u}(\vf{x},t)$ the velocity of the fluid at time $t$ and position $\vf{x}$. For the moment, we will also assume that $\rho$ and $\vf{u}$ are smooth functions.
+  In this section we will describe the motion of a fluid with a set of equation that result from the conservation of mass, momentum and energy. From what follows, let $D\subseteq \RR^3$ be a region filled with a fluid. For each time $t$ and $\vf{x}\in D$ we assume that the fluid has a well-defined mass density $\rho(\vf{x},t)$\footnote{The assumption that $\rho$ exists is a continuum assumption. Clearly, it does not hold if the molecular structure of matter is taken into account. For most macroscopic phenomena occurring in nature, it is believed that this assumption is extremely accurate.}. Finally, we denote by $\vf{u}(\vf{x},t)$ the velocity of the fluid at time $t$ and position $\vf{x}$. For the moment, we will also assume that $\rho$ and $\vf{u}$ are smooth functions.
   \begin{proposition}[Conservation of mass]\label{FLM:conservationofmass}
     Let $W\subseteq D$ be a fixed subregion of $D$. Then:
     $$
@@ -39,15 +39,15 @@
     Compute the time derivative of $\vf{u}(t,\vf{x}(t))$ using the Chain rule.
   \end{sproof}
   For any continuum, forces acting on a piece of material are of two types. First, there are forces of stress, whereby the piece of material is acted on by forces across its surface by the rest of the continuum. Second, there are external or body, forces such as gravity or a magnetic field, which exert a force per unit volume on the continuum.
-  \begin{definition}[Ideal fluid]
-    An \emph{ideal fluid} has the following property: for any motion of the fluid there is a function $p(\vf{x},t)$ called the \emph{pressure} such that if $S$ is a surface in the fluid with a chosen unit normal $\vf{n}$, the force of stress exerted across the surface $S$ per unit area at $\vf{x}\in S$ at time $t$ is $p(\vf{x},t)\vf{n}$. Thus, the total force of stress exerted inside a region $W\subseteq D$ is given by:
+  \begin{definition}[Ideal fluid]\label{FLM:idealfluid}
+    An \emph{ideal fluid} has the following property: for any motion of the fluid there is a function $p(\vf{x},t)$ called the \emph{pressure} such that if $S$ is a surface in the fluid with a chosen unit normal $\vf{n}$, the force of stress exerted across the surface $S$ per unit of area at $\vf{x}\in S$ at time $t$ is $p(\vf{x},t)\vf{n}$. Thus, the total force of stress exerted inside a region $W\subseteq D$ is given by:
     $$
       \vf{A}_{\partial W}:=\text{Force on $W$}=-\int_{\Fr{W}}p\vf{n}\dd{S}
     $$
     where the minus sign is because $\vf{n}$ points outwards.
   \end{definition}
-  \begin{proposition}[Conservation of momentum]
-    The balace of momentum for an ideal fluid is given by:
+  \begin{proposition}[Conservation of momentum]\label{FLM:conservationofmomentum}
+    The balance of momentum for an ideal fluid is given by:
     $$
       \rho\matdv{\vf{u}}{t}=-\grad p+\rho\vf{f}
     $$
@@ -140,7 +140,7 @@
     For any smooth enough function $f(\vf{x},t)$ we have:
     \begin{align*}
       \dv{}{t}\int_{W_t}\rho f\dd{V} & =\int_{W_t}\rho\matdv{f}{t}\dd{V}                     \\
-      \dv{}{t}\int_{W_t} f\dd{V}     & =\int_{W_t}\left[\dv{f}{t}+\div(f\vf{u})\right]\dd{V} \\
+      \dv{}{t}\int_{W_t} f\dd{V}     & =\int_{W_t}\left[\dv{f}{t}+\div(f\vf{u})\right]\dd{V}
     \end{align*}
   \end{corollary}
   \begin{definition}
@@ -325,7 +325,7 @@
     Let $\vf{u}=(u,v,w)$ be the velocity field of a fluid. The \emph{vorticity} is the vector field $\vf{\xi}:=\rotp\vf{u}$.
   \end{definition}
   \begin{proposition}
-    Let $\vf{x}\in \RR^3$ and let $\vf{u}(\vf{x})$ be a smooth vector field. Then, in a small neighbourhood of $\vf{x}$, $\vf{u}$ is the sum of a translation, a deformation and a rotation (with rotation vector $\omega/2$):
+    Let $\vf{x}\in \RR^3$ and let $\vf{u}(\vf{x})$ be a smooth vector field. Then, in a small neighbourhood of $\vf{x}$, $\vf{u}$ is the sum of a translation, a deformation and a rotation (with rotation vector $\vf\xi/2$):
     \begin{equation}\label{FLM:decomposition}
       \vf{u}(\vf{y})=\vf{u}(\vf{x})+\vf{D}(\vf{x})\cdot \vf{h}+\frac{1}{2} \vf{\xi}(\vf{x})\times \vf{h} +\O{\norm{\vf{h}}^2}
     \end{equation}
@@ -334,13 +334,13 @@
   \begin{proof}
     From \mnameref{FSV:taylor} we have:
     $$
-      \vf{u}(\vf{y})=\vf{u}(\vf{x})+\vf{Du}(\vf{x})\cdot \vf{h}+\O{\norm{\vf{h}}^2}
+      \vf{u}(\vf{y})=\vf{u}(\vf{x})+\grad\vf{u}(\vf{x})\cdot \vf{h}+\O{\norm{\vf{h}}^2}
     $$
     Now let:
     $$
-      \vf{D}=\frac{1}{2}\left[\vf{Du}(\vf{x})+\transpose{\vf{Du}(\vf{x})}\right]\quad \vf{S}=\frac{1}{2}\left[\vf{Du}(\vf{x})-\transpose{\vf{Du}(\vf{x})}\right]
+      \vf{D}=\frac{1}{2}\left[\grad\vf{u}(\vf{x})+\transpose{\grad\vf{u}(\vf{x})}\right]\quad \vf{S}=\frac{1}{2}\left[\grad\vf{u}(\vf{x})-\transpose{\grad\vf{u}(\vf{x})}\right]
     $$
-    Thus, $\vf{Du} = \vf{D}+\vf{S}$ and $\vf{S}\cdot\vf{h} = \frac{1}{2} \vf{\xi}(\vf{x})\times \vf{h}$.
+    Thus, $\grad\vf{u} = \vf{D}+\vf{S}$ and $\vf{S}\cdot\vf{h} = \frac{1}{2} \vf{\xi}(\vf{x})\times \vf{h}$.
   \end{proof}
   \begin{remark}
     The physical intuition behind $\vf{D}$ is the following. Because $\vf{D}$ is symmetric, for each $\vf{x}$ fixed, there is an orthonormal basis $\mathcal{B}$ such that:
@@ -351,7 +351,7 @@
         0   & 0   & d_3
       \end{pmatrix}
     $$
-    Now if we ignore all the terms in \eqref{FLM:decomposition} except $\vf{D}(x)\cdot \vf{h}$, we see that:
+    Now if we ignore all the terms in \eqref{FLM:decomposition} except $\vf{D}(\vf{x})\cdot \vf{h}$, we see that:
     $$
       \dv{\vf{h}}{t}=\vf{D}(\vf{x})\cdot \vf{h}
     $$
@@ -475,5 +475,97 @@
     One of the troubles with the 3-dimensional case is that given $\vf\xi$, the vector field $A$ is not uniquely determined (we cannot impose boundary condition such as $\vf{A} = 0$ on $\Fr{D}$ because $\vf{A}$ need not be constant on $\Fr{D}$ as was the case with $\psi$).
   \end{remark}
   \subsubsection{Navier-Stokes equations}
+  \begin{theorem} [Cauchy's stress theorem]
+    The force acting on a surface $S$ of a fluid is a linear function of the normal vector $\vf{n}$ to $S$.
+  \end{theorem}
+  In \mcref{FLM:idealfuid} we defined an ideal fluid as one in which forces across a surface were normal to that surface. We now consider more general fluids. We will assume now that the force exerted across a surface $S$ per unit of area is given by:
+  $$
+    \text{Force on $S$}=-p(\vf{x},t)\vf{n}+\vf\sigma(\vf{x},t)\cdot\vf{n}
+  $$
+  where the matrix $\vf\sigma$ is called \emph{stress tensor}\footnote{The new feature is that $\vf\sigma\cdot\vf{n}$ need not be parallel to $\vf{n}$. The separation of the forces into pressure and other forces in is somewhat ambiguous because $\vf\sigma\cdot\vf{n}$ may contain a component parallel to $\vf{n}$. This issue will be resolved later when we give a more definite functional form to $\vf\sigma$.}.
+  \begin{remark}
+    We will make the following assumptions to the stress tensor:
+    \begin{itemize}
+      \item $\vf\sigma$ is a linear function ($ax+b$) of the velocity gradients $\grad \vf{u}$.
+      \item $\vf\sigma$ is invariant under rigid body rotations. That is, if $\vf{U}$ is an orthogonal matrix:
+            $$
+              \vf\sigma(\vf{U}\cdot\grad \vf{u}\cdot\vf{U}^{-1})=\vf{U}\cdot\vf\sigma(\grad \vf{u})\cdot\vf{U}^{-1}
+            $$
+      \item $\vf\sigma$ is symmetric.
+    \end{itemize}
+  \end{remark}
+  \begin{proposition}
+    The stress tensor $\vf\sigma$ can be written as:
+    $$
+      \vf\sigma=2\mu\left[\vf{D} -\frac{1}{3}(\div\vf{u}) \vf{I}\right]+\zeta(\div\vf{u}) \vf{I}
+    $$
+    where $\vf{I}$ is the identity matrix, $\vf{D}$ is the deformation tensor, $\mu$ is the \emph{first viscosity coefficient} and $\zeta=\lambda+\frac{2}{3}\mu$ is the \emph{second viscosity coefficient}.
+  \end{proposition}
+  \begin{proof}
+    Since $\vf\sigma$ is symmetric, it follows that only depends on the symmetric part of $\grad \vf{u}$, that is, on the deformation tensor $\vf{D}$. Thus, the eigenvalues of $\vf\sigma$ are linear functions of those of $\vf{D}$. By property 2, they must also be symmetric because we can choose $\vf{U}$ to permute two eigenvalues of $\vf{D}$ and this must permute the corresponding eigenvalues of $\vf{\sigma}$. Thus, the eigenvalues $\sigma_i$ of $\vf\sigma$ can be written as $\sigma_i=a d_i + b$, or equivalently:
+    $$
+      \sigma_i=\lambda \div \vf{u} + 2\mu d_i
+    $$
+    for some constants $\lambda$ and $\mu$ and $i=1,2,3$. Using again, property 2, we can reconstruct $\vf\sigma$ as:
+    $$
+      \vf\sigma=\lambda (\div \vf{u}) \vf{I} + 2\mu \vf{D}
+    $$
+    Or equivalently:
+    $$
+      \vf\sigma=2\mu\left[\vf{D} -\frac{1}{3}(\div\vf{u}) \vf{I}\right]+\zeta(\div\vf{u}) \vf{I}
+    $$
+  \end{proof}
+  \begin{corollary}
+    Introducing the stress tensor, the balance of momentum equation yields the \emph{Navier-Stokes equations}:
+    $$
+      \rho \matdv{\vf{u}}{t}=-\grad p + (\lambda+\mu)\grad(\div\vf{u}) + \mu\laplacian\vf{u}
+    $$
+  \end{corollary}
+  \begin{sproof}
+    Follow the proof of \mnameref{FLM:conservationofmomentum} adapted to the stress tensor.
+  \end{sproof}
+  \begin{remark}
+    This equation together with the continuity equation and energy equation completely describe the flow if a compressible viscous fluid. In the case of an incompressible homogeneous fluid with $\rho=\rho_0=\const$,  the
+    complete set of equations becomes the \emph{Navier-Stokes equations for incompressible flow}:
+    $$
+      \begin{cases}
+        \displaystyle\matdv{\vf{u}}{t}=-\grad p' + \nu\laplacian\vf{u} \\
+        \displaystyle\div\vf{u}=0
+      \end{cases}
+    $$
+    where $p'=p/\rho_0$ and $\nu=\mu/\rho_0$ is the \emph{kinematic viscosity}. To this we should add boundary condition, which for an ideal fluid we use $\vf{u}\cdot \vf{n}=0$ and if the solid wall that bounds the fluid is stationary, we use $\vf{u}=\vf{0}$ on the walls.
+  \end{remark}
+  \subsubsection{Reynolds number}
+  \begin{proposition}
+    For a given problem, let $L$ be the unit of length (\emph{characteristic length}) and $U$ be the unit of velocity (\emph{characteristic velocity}). This choice determines then the unit of time $T=L/U$. If we convert $\vf{u}$, $\vf{x}$ and $t$ to dimensionless quantities by
+    $$
+      \vf{u}':= \frac{\vf{u}}{U}\quad \vf{x}':=\frac{\vf{x}}{L}\quad t':=\frac{t}{T}
+    $$
+    then the Navier-Stokes equations for incompressible flow become:
+    $$
+      \begin{cases}
+        \displaystyle\pdv{\vf{u}'}{t'}+(\vf{u}'\cdot\grad')\vf{u}=-\grad p' + \frac{\nu}{LU}\laplacian'\vf{u}' \\
+        \displaystyle\div\vf{u}'=0
+      \end{cases}
+    $$
+    where $p'=p/(\rho_0 U^2)$.
+  \end{proposition}
+  \begin{sproof}
+    Use the \mnameref{FSV:chainrule}.
+  \end{sproof}
+  \begin{definition}
+    For a given problem, let $L$ be the unit of length (\emph{characteristic length}) and $U$ be the unit of velocity (\emph{characteristic velocity}). We define the \emph{Reynolds number} as:
+    $$
+      \mathrm{Re}=\frac{LU}{\nu}
+    $$
+  \end{definition}
+  \begin{remark}
+    Note that the dimensionless Navier-Stokes equations for incompressible flow only depend on the Reynolds number.
+  \end{remark}
+  \begin{definition}
+    Two flows with the same geometry and the same Reynolds number are said to be \emph{similar}. More precisely, let $\vf{u}_1$ and $\vf{u}_2$ be two flows on regions $D_1$
+    and $D_2$ that are related by a scale factor $\lambda$ so that $L_1 = \lambda L_2$. Let choices
+    of $U_1$ and $U_2$ be made for each flow, and let the viscosities be $\nu_1$ and $\nu_2$, respectively. If $\mathrm{Re}_1=\mathrm{Re}_2$, then the dimensionless velocity fields $\vf{u}_1$ and $\vf{u}_2$ satisfy exactly the same equation on the same region.
+  \end{definition}
 \end{multicols}
 \end{document}