From c2f1302501cef06c68655117e4f57693be60fda8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Mon, 9 Oct 2023 12:58:17 +0200 Subject: [PATCH] updated fluid mechanics + some stochastic calculus --- .../Stochastic_calculus.tex | 34 +++++++++++++ .../Fluid_mechanics/Fluid_mechanics.tex | 51 +++++++++++++++++-- preamble_formulas.sty | 3 ++ 3 files changed, 85 insertions(+), 3 deletions(-) diff --git a/Mathematics/5th/Stochastic_calculus/Stochastic_calculus.tex b/Mathematics/5th/Stochastic_calculus/Stochastic_calculus.tex index b53e07b..c5fa341 100644 --- a/Mathematics/5th/Stochastic_calculus/Stochastic_calculus.tex +++ b/Mathematics/5th/Stochastic_calculus/Stochastic_calculus.tex @@ -604,6 +604,13 @@ $$ Letting $t\to \infty$ we get that $\phi$, $\tilde{\phi}$ are indistinguishable. Finally, from the Lebesgue integral, we have that $\psi$, $\tilde{\psi}$ are indistinguishable. \end{proof} + \begin{proposition}\label{SC:ito_process_martingale} + Let $X={(X_t)}_{t\geq 0}$ be an Itô process such that $\dd{X_t}=\phi_t\dd{B_t}+\psi_t\dd{t}$. Then: + \begin{itemize} + \item $X$ is a local martingale if and only if $X_0\in L^1$ and $\psi=0$. + \item $X$ is a square-integrable martingale if and only if $X_0\in L^2$, $\phi\in\MM^2$ and $\psi=0$. + \end{itemize} + \end{proposition} \begin{definition} Let $X={(X_t)}_{t\geq 0}$ be an Itô process such that $\dd{X_t}=\phi_t\dd{B_t}+\psi_t\dd{t}$, and $Y={(Y_t)}_{t\geq 0}$ be a continuous adapted process. Then, $Y\phi\in \MM^2_{\text{loc}}$ and $Y\psi\in \MM^1_{\text{loc}}$ and we define: $$ @@ -695,5 +702,32 @@ where $\vf{X}:=(X^1,\ldots,X^d)$. \end{theorem} \subsubsection{Exponential martingales} + \begin{lemma}[Doléans-Dade exponential] + For any $\phi\in\MM^2_{\text{loc}}$, the process $Z^\phi={(Z_t^\phi)}_{t\geq 0}$ defined as + $$ + Z_t^\phi:=\exp{\int_0^t \phi_u\dd{B_u}-\frac{1}{2}\int_0^t{\phi_u}^2\dd{u}} + $$ + is a continuous local martingale. + \end{lemma} + \begin{proof} + Applying Itô's formula to $f(x)=\exp{x}$ and $X_t=\int_0^t \phi_u\dd{B_u}-\frac{1}{2}\int_0^t{\phi_u}^2\dd{u}$ we get: + \begin{equation*} + \dd{Z^\phi_t} = \exp{X_t}\left( \phi_t\dd{B_t}-\frac{1}{2}{\phi_t}^2\dd{t}\right)+\frac{1}{2}\exp{X_t}{\phi_t}^2\dd{t}= \exp{X_t}\phi_t\dd{B_t} + \end{equation*} + Since, $Z^\phi_0=1$, we obtain that $\forall t \geq 0$: + \begin{equation*} + Z^\phi_t=1+\int_0^t Z^\phi_u\phi_u\dd{B_u} + \end{equation*} + and the result follows from \mcref{SC:ito_process_martingale}. + \end{proof} + \begin{lemma} + If $M$ is a non-negative local martingale, then $M$ is a super-martingale. Moreover, for $T\in \RR_{\geq 0}$, ${(M_t)}_{t\in[0,T]}$ is a martingale if and only if $\Exp(M_T)\geq \Exp(M_0)$. + \end{lemma} + \begin{theorem}[Novikov's condition] + For ${(Z^\phi_s)}_{s\in[0,t]}$ to be a martingale, it suffices that: + $$ + \Exp\left(\exp{\frac{1}{2}\int_0^t{\phi_u}^2\dd{u}}\right)<\infty + $$ + \end{theorem} \end{multicols} \end{document} \ No newline at end of file diff --git a/Physics/Advanced/Fluid_mechanics/Fluid_mechanics.tex b/Physics/Advanced/Fluid_mechanics/Fluid_mechanics.tex index ea1e389..779c643 100644 --- a/Physics/Advanced/Fluid_mechanics/Fluid_mechanics.tex +++ b/Physics/Advanced/Fluid_mechanics/Fluid_mechanics.tex @@ -548,7 +548,7 @@ \displaystyle\div\vf{u}'=0 \end{cases} $$ - where $p'=p/(\rho_0 U^2)$. + where $p'=p/(\rho_0 U^2)$. In this equation, the term $(\vf{u}'\cdot \grad')\vf{u}'$ is called the \emph{convective term} (or \emph{inertia term}) and the term $\nu\laplacian'\vf{u}'$ is called the \emph{diffusive term} (or \emph{dissipation term}). \end{proposition} \begin{sproof} Use the \mnameref{FSV:chainrule}. @@ -556,7 +556,7 @@ \begin{definition} For a given problem, let $L$ be the unit of length (\emph{characteristic length}) and $U$ be the unit of velocity (\emph{characteristic velocity}). We define the \emph{Reynolds number} as: $$ - \mathrm{Re}=\frac{LU}{\nu} + \Rey=\frac{LU}{\nu} $$ \end{definition} \begin{remark} @@ -565,7 +565,7 @@ \begin{definition} Two flows with the same geometry and the same Reynolds number are said to be \emph{similar}. More precisely, let $\vf{u}_1$ and $\vf{u}_2$ be two flows on regions $D_1$ and $D_2$ that are related by a scale factor $\lambda$ so that $L_1 = \lambda L_2$. Let choices - of $U_1$ and $U_2$ be made for each flow, and let the viscosities be $\nu_1$ and $\nu_2$, respectively. If $\mathrm{Re}_1=\mathrm{Re}_2$, then the dimensionless velocity fields $\vf{u}_1$ and $\vf{u}_2$ satisfy exactly the same equation on the same region. + of $U_1$ and $U_2$ be made for each flow, and let the viscosities be $\nu_1$ and $\nu_2$, respectively. If $\Rey_1=\Rey_2$, then the dimensionless velocity fields $\vf{u}_1$ and $\vf{u}_2$ satisfy exactly the same equation on the same region. \end{definition} \begin{remark} This idea of the similarity of flows is used in the design of experimental models. For example, suppose we are contemplating a new design for an aircraft wing, and we wish to know the behavior of a fluid flow around it. Rather than build the wing itself, it may be faster and more economical to perform the initial tests on a scaled-down version. We design our model so that it has the same geometry as the full-scale wing, and we choose values for the undisturbed velocity, coefficient of viscosity, and so on, such that the Reynolds number for the flow in our experiment matches that of the actual flow. We can then expect the results of our experiment to be relevant to the actual flow over the full-scale wing. @@ -606,5 +606,50 @@ \vf{P}(\vf{w}):=\vf{u} $$ \end{definition} + \begin{remark} + The Navier-Stokes equations become thus: + $$ + \partial_t u=\vf{P}\left( -(\vf{u} \cdot \grad) \vf{u}+\frac{1}{\Rey} \laplacian \vf{u} \right) + $$ + \end{remark} + \begin{remark} + If $\Rey\gg 1$, then the Navier-Stokes equations for incompressible flow can be approximated by the \emph{Stokes' equations}: + \begin{important} + $$ + \begin{cases} + \displaystyle\partial_t\vf{u}=-\grad p + \frac{1}{\Rey}\laplacian\vf{u} \\ + \displaystyle\div\vf{u}=0 + \end{cases} + $$ + \end{important} + \end{remark} + \begin{proposition} + Consider an incompressible viscous flow confined in a region $D\subset\RR^3$ with smooth boundary. Then: + $$ + \dv{}{t}E_\text{kinetic}=-\mu \int_D \norm{\grad \vf{u}}^2\dd{V} + $$ + \end{proposition} + \begin{proof} + We have that: + \begin{align*} + \dv{}{t}E_\text{kinetic} & =\dv{}{t}\frac{1}{2}\int_D \rho \norm{\vf{u}}^2\dd{V}=\int_D \rho \vf{u}\cdot \matdv{\vf{u}}{t}\dd{V} \\ + & =\int_D \rho \vf{u}\cdot \left(-\grad p + \mu\laplacian\vf{u}\right)\dd{V} \\ + & =\mu\int_D \rho \vf{u}\cdot \laplacian\vf{u}\dd{V} + \end{align*} + because $\div \vf{u}=0$ and the orthogonality of $\vf{u}$ and $\grad p$. Now using that $\div(\vf{u} \cdot \grad \vf{u})=\vf{u}\cdot \laplacian\vf{u}+\norm{\grad \vf{u}}^2$, the \mnameref{RFA:divergencethm} and the boundary condition $\vf{u}=0$, we get the result. + \end{proof} + \begin{proposition} + Consider a three-dimensional viscous incompressible flow. Then, the vorticity equation becomes: + $$ + \matdv{\vf{\xi}}{t}-(\vf\xi\cdot\grad)\vf{u}=\frac{1}{\Rey}\laplacian\vf{\xi} + $$ + \end{proposition} + \begin{definition} + Assume we have an \emph{equation of state} $p=p(\rho)$, with $p'(\rho)>0$. Define $c=\sqrt{p'(\rho)}$, and called it the \emph{speed of sound}. Then, the \emph{Mach number} is defined as: + $$ + \Ma=\frac{u}{c} + $$ + where $u=\norm{\vf{u}}$. + \end{definition} \end{multicols} \end{document} \ No newline at end of file diff --git a/preamble_formulas.sty b/preamble_formulas.sty index 14411ec..342821e 100644 --- a/preamble_formulas.sty +++ b/preamble_formulas.sty @@ -414,6 +414,9 @@ %%% PHYSICS \newcommand{\const}{\text{const.}} +\newcommand{\Rey}{\mathrm{Re}} % Reynolds number +\newcommand{\Ma}{\mathrm{Ma}} % Mach number + %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%% for long-vector arrows %%%%%