reoplaced ODE with space

Mathematics/3rd/Differential_geometry/Differential_geometry.tex

@@ -885,7 +885,7 @@
     Let $S\subseteq\RR^3$ be a surface and $\vf{X}$, $\vf{Y}$ be vector fields tangent to $S$ along a curve $\vf\alpha:I\rightarrow S$ of class $\mathcal{C}^\infty$ such that they are parallel. Then, $t\mapsto\langle \vf{X}(t),\vf{Y}(t)\rangle$ is constant. In particular, the norms $\|\vf{X}(t)\|$, $\|\vf{Y}(t)\|$ as well as the angle between $\vf{X}(t)$ and $\vf{Y}(t)$ are constant.
   \end{proposition}
   \begin{proposition}
-    Let $S\subseteq\RR^3$ be a surface, $(V,\vf\varphi(u,v))$ is a parametrization of $S$ and $\vf\alpha: I\rightarrow S$ be a parametrized curve of class $\mathcal{C}^\infty$ such that $\vf\alpha=\vf\varphi(u(t),v(t))$. Then, given $t_0\in I$ and $\vf{w}\in T_{\vf\alpha(t_0)}S$ there exists a unique parallel vector field $\vf{X}=a\vf\varphi_u+b\vf\varphi_v$ along $\vf\alpha$ such that $\vf{X}(t_0)=\vf{w}$. This vector field is called \emph{parallel transport} of the vector $\vf{w}$ along $\vf{\alpha}$, and it is defined on the entire interval $I$. It can be found by solving this system ofODEs:
+    Let $S\subseteq\RR^3$ be a surface, $(V,\vf\varphi(u,v))$ is a parametrization of $S$ and $\vf\alpha: I\rightarrow S$ be a parametrized curve of class $\mathcal{C}^\infty$ such that $\vf\alpha=\vf\varphi(u(t),v(t))$. Then, given $t_0\in I$ and $\vf{w}\in T_{\vf\alpha(t_0)}S$ there exists a unique parallel vector field $\vf{X}=a\vf\varphi_u+b\vf\varphi_v$ along $\vf\alpha$ such that $\vf{X}(t_0)=\vf{w}$. This vector field is called \emph{parallel transport} of the vector $\vf{w}$ along $\vf{\alpha}$, and it is defined on the entire interval $I$. It can be found by solving this system of ODEs:
     $$\left\{
       \begin{aligned}
         a'+\Gamma_{11}^1au'+\Gamma_{12}^1av'+\Gamma_{21}^1bu'+\Gamma_{22}^1bv' & =0 \\
@@ -991,7 +991,7 @@
   \end{lemma}
   \begin{definition}
     Let $U\subseteq\RR^n$ be an open set and $\vf{X}=\sum X^i\pdv{}{x^i}\in\mathcal{X}(U)$. We say that a parametrized curve $\vf{\gamma}:I\rightarrow\RR^n$ is an \emph{integral curve} of $\vf{X}$ if: $$\vf\gamma'(t)=\vf{X}(\vf\gamma(t))\qquad \forall t\in I$$
-    That is, the integral curve $\vf\gamma(t)=(x^1(t),\ldots,x^n(t))$ of $\vf{X}$ satisfies the following system ofODEs:
+    That is, the integral curve $\vf\gamma(t)=(x^1(t),\ldots,x^n(t))$ of $\vf{X}$ satisfies the following system of ODEs:
     $$
       \left\{
       \begin{aligned}

Mathematics/4th/Dynamical_systems/Dynamical_systems.tex

@@ -36,7 +36,7 @@
     \end{enumerate}
   \end{theorem}
   \begin{definition}
-    Let $f,g:\RR^2\rightarrow\RR$ be two functions and consider the system ofODEs:
+    Let $f,g:\RR^2\rightarrow\RR$ be two functions and consider the system of ODEs:
     \begin{equation}\label{DS:plane}
       \left\{
       \begin{aligned}
@@ -645,14 +645,14 @@
     A semiestable limit cycle $\Gamma_\mu$ of a family of rotated vector fields splits into two simple limit cycles, one stable and one unstable, as the parameter $\mu$ is varied in one sense and it disappears as $\mu$ is varied in the opposite sense.
   \end{theorem}
   \begin{theorem}[Melnikov's method]
-    Let $\vf{f}\in\mathcal{C}^1(\RR^2)$, $\vf{g}\in\mathcal{C}^1(\RR^2\times\RR^m)$ and $\varepsilon\simeq 0$. Consider the followingODE:
+    Let $\vf{f}\in\mathcal{C}^1(\RR^2)$, $\vf{g}\in\mathcal{C}^1(\RR^2\times\RR^m)$ and $\varepsilon\simeq 0$. Consider the following ODE:
     \begin{equation}\label{DS:melnikov}
       \vf{x}'=\vf{f}(\vf{x})+\varepsilon\vf{g}(\vf{x},\vf{\mu})
     \end{equation}
     Suppose that for $\varepsilon =0$ the system has a one-parameter family of periodic orbits $\vf\gamma_h(t)$ of period $T_h$. Then for any simple zero $(\vf\mu_0,h_0)$ of the function $$M(\vf\mu, h)=\int_{0}^{T_h}\vf{f}(\vf\gamma_h(t))\times \vf{g}(\vf\gamma_h(t))\dd{t}$$ there exists a unique limit cycle $\vf\Gamma_\varepsilon$ for $\varepsilon\simeq 0$ such that $\displaystyle\lim_{\varepsilon\to 0}\vf\Gamma_\varepsilon=\vf\gamma_{h_0}$. On the other hand, if $M(\vf\mu_0,h_0)\ne 0$, for sufficiently small $\varepsilon$, the system of \mcref{DS:melnikov} with $\vf\mu=\vf\mu_0$ has no limit cycle in any sufficiently small neighborhood of $\vf\gamma_{h_0}$.
   \end{theorem}
   \begin{corollary}[Melnikov's method]
-    Let $H\in\mathcal{C}^2(\RR^2)$, $P,Q\in\mathcal{C}^1(\RR^2\times\RR^m)$ and $\varepsilon\simeq 0$. Consider the following system ofODEs:
+    Let $H\in\mathcal{C}^2(\RR^2)$, $P,Q\in\mathcal{C}^1(\RR^2\times\RR^m)$ and $\varepsilon\simeq 0$. Consider the following system of ODEs:
     \begin{equation*}
       \left\{
       \begin{aligned}

Mathematics/4th/Harmonic_analysis/Harmonic_analysis.tex

@@ -166,7 +166,7 @@
     Let $f(x)=\exp{-a x^2}$. Then, $\F f(\xi)=\sqrt{\frac{\pi}{a}}\exp{-\frac{{(\pi \xi)}^2}{a}}$ and moreover $\F^2f=f$. In particular if $a=\pi$, then $\F f=f$.
   \end{lemma}
   \begin{sproof}
-    $f$ satisfies theODE $y'=-2a x y$. Taking $\ \widehat{}\ $ on this expression and using \mcref{HA:diffFourierXf,HA:diffFourierTransf} we obtain that $\widehat{f}$ must satisfy the followingODE:
+    $f$ satisfies the ODE $y'=-2a x y$. Taking $\ \widehat{}\ $ on this expression and using \mcref{HA:diffFourierXf,HA:diffFourierTransf} we obtain that $\widehat{f}$ must satisfy the following ODE:
     $$y'=-\frac{2\pi^2\xi}{a} y$$
     with initial condition $y(0)=\int_{-\infty}^{+\infty}\exp{-a x^2}\dd{x}=\sqrt{\frac{\pi}{a}}$.
   \end{sproof}
@@ -1266,7 +1266,7 @@
     $$
       \partial_t \widehat{E}+4\pi^2a^2\norm{\vf\xi}^2\widehat{E}=\delta_t
     $$
-    because $\delta=\delta_{\vf{x}}\delta_t$. It can be seen that a solution of thisODE is:
+    because $\delta=\delta_{\vf{x}}\delta_t$. It can be seen that a solution of this ODE is:
     $$
       \widehat{E}(t,\xi)=\indi{[0,\infty)}(t)\exp{-4\pi^2a^2\norm{\vf\xi}^2t}
     $$

Mathematics/4th/Numerical_calculus/Numerical_calculus.tex

@@ -84,7 +84,7 @@
     Moreover, we say that the algorithm has \emph{order of convergence} $p$ if $\norm{\vf{e}_n}=\O{h^p}$.
   \end{definition}
   \begin{remark}
-    Note that in a consistent method the difference equation for the method approaches theODE as the step size goes to zero, whereas in a convergent method is the solution to the difference equation that approaches the solution to theODE as the step size goes to zero.
+    Note that in a consistent method the difference equation for the method approaches the ODE as the step size goes to zero, whereas in a convergent method is the solution to the difference equation that approaches the solution to the ODE as the step size goes to zero.
   \end{remark}
   \begin{theorem}\label{NC:errorLipschitz}
     Consider a consistent one-step explicit method such that its incremental function $\vf\phi$ is Lipschitz continuous (with constant $L$) with respect to $\vf{x}$. Then:

Mathematics/4th/Partial_differential_equations/Partial_differential_equations.tex

@@ -135,7 +135,7 @@
   \end{definition}
   \begin{proposition}[Fermat's principle]
     \emph{Fermat's principle} states that the path taken by a ray between two given points $a$ and $b$ is the path that can be traveled in the least time. Mathematically, we want to minimize the functional: $$\mathcal{T}(\vf{x})=\int_a^b\frac{\abs{\dd{\vf{x}}}}{v(\vf{x})}$$
-    So we shall solve the equation $\delta \mathcal{T}=0$, which is equivalent to solve: $$\delta\int_a^bn(\vf{x})\dd{s}=0$$ where $s$ is the arc-length parameter. From the Euler-Lagrange equations, we get the followingODE: $$\dv{}{s}\left(n\dv{\vf{x}}{s}\right)=\grad{n}$$
+    So we shall solve the equation $\delta \mathcal{T}=0$, which is equivalent to solve: $$\delta\int_a^bn(\vf{x})\dd{s}=0$$ where $s$ is the arc-length parameter. From the Euler-Lagrange equations, we get the following ODE: $$\dv{}{s}\left(n\dv{\vf{x}}{s}\right)=\grad{n}$$
   \end{proposition}
   \begin{proposition}[Eikonal equation]
     The time $T(x)$ taken by the light to travel from a fixed point $x_0$ to $x$ in a medium of refractive index $n$ is given by: $${\norm{\grad{T}}}^2=n^2$$
@@ -635,7 +635,7 @@
     for certain constants $C_1, C_2\in\RR$.
   \end{proposition}
   \begin{sproof}
-    Observe that $u(x,t) = f(\frac{x}{\sqrt{t}})=:f(s)$ and the heat equation is transformed into $-f's=2\alpha f''$. The solution of thisODE is straightforward.
+    Observe that $u(x,t) = f(\frac{x}{\sqrt{t}})=:f(s)$ and the heat equation is transformed into $-f's=2\alpha f''$. The solution of this ODE is straightforward.
   \end{sproof}
   \subsubsection{Distributions}
   \begin{definition}

Mathematics/4th/Stochastic_processes/Stochastic_processes.tex

@@ -1165,7 +1165,7 @@
     Let ${(X_t)}_{t\geq 0}$ be a CTHMC. Then, ${(X_t)}_{t\geq 0}$ is said to be \emph{stable} if $\forall i\in I$, $q_i<\infty$, and is said to be \emph{conservative} if $\forall i\in I$, $q_i=\sum_{\substack{k\in I\\k\ne i}}q_{ik}$.
   \end{definition}
   \begin{theorem}
-    Let ${(X_t)}_{t\geq 0}$ be a CTHMC and a regular jump process. Then, the two KolmogorovODEs are satisfied.
+    Let ${(X_t)}_{t\geq 0}$ be a CTHMC and a regular jump process. Then, the two Kolmogorov ODEs are satisfied.
   \end{theorem}
   \subsubsection{Limit and stationary distributions}
   \begin{definition}

Mathematics/5th/Montecarlo_methods/Montecarlo_methods.tex

@@ -320,7 +320,7 @@
     Note that Euler scheme reduces to generating independent increments $\vf{B}_{t_{i+1}}-\vf{B}_{t_i}\sim \sqrt{t_{i+1}-t_i}N_d(0,\vf{I}_d)$.
   \end{remark}
   \begin{remark}
-    Trying to build an implicit Euler scheme for SDEs is much more complicated than for ODEs, as we need to ensure that the process is still adapted.
+    Trying to build an implicit Euler scheme for SDEs is much more complicated than for  ODEs, as we need to ensure that the process is still adapted.
   \end{remark}
   \begin{definition}
     Let $\vf{X}$ be the solution to \mcref{MM:SDE}. We define the \emph{continuous Euler scheme} as:

Physics/Basic/Classical_mechanics/Classical_mechanics.tex

@@ -300,7 +300,7 @@
         \label{CM_RLC-genS}
       \end{minipage}
     \end{center}
-    TheODE of the system for the charge $q(t)$ is: $$\ddot{q}+\frac{R}{L}\dot{q}+\frac{q}{LC}=\frac{V_\text{in}}{L}$$ Thus, in steady-state part we have: $$q(t)=\frac{\epsilon_0/\omega}{\sqrt{{\left(L\omega-\frac{1}{C\omega}\right)}^2+R^2}}\cos(\omega t-\delta)$$ where $\delta=-\arctan\left(\frac{R}{L\omega-\frac{1}{C\omega}}\right)$.
+    The ODE of the system for the charge $q(t)$ is: $$\ddot{q}+\frac{R}{L}\dot{q}+\frac{q}{LC}=\frac{V_\text{in}}{L}$$ Thus, in steady-state part we have: $$q(t)=\frac{\epsilon_0/\omega}{\sqrt{{\left(L\omega-\frac{1}{C\omega}\right)}^2+R^2}}\cos(\omega t-\delta)$$ where $\delta=-\arctan\left(\frac{R}{L\omega-\frac{1}{C\omega}}\right)$.
     And finally:
     \begin{equation}\label{CM_Vout}
       V_\text{out}=RI=-\frac{R}{\sqrt{{\left(L\omega-\frac{1}{C\omega}\right)}^2+R^2}}\epsilon_0\sin(\omega t-\delta)
@@ -321,7 +321,7 @@
   %       \label{CM_RLC-genP}
   %     \end{minipage}
   %   \end{center}
-  %   TheODE of the system for the charge $q(t)$ is: $$\ddot{q}+\frac{R}{L}\dot{q}+\frac{q}{LC}=\frac{V_\text{in}}{L}$$ Thus, in steady-state part we have: $$q(t)=\frac{\epsilon_0/\omega}{\sqrt{{\left(L\omega-\frac{1}{C\omega}\right)}^2+\frac{1}{R^2}}}\cos(\omega t-\delta)$$ where $\delta=-\arctan\left(\frac{1/R}{L\omega-\frac{1}{C\omega}}\right)$.
+  %   The ODE of the system for the charge $q(t)$ is: $$\ddot{q}+\frac{R}{L}\dot{q}+\frac{q}{LC}=\frac{V_\text{in}}{L}$$ Thus, in steady-state part we have: $$q(t)=\frac{\epsilon_0/\omega}{\sqrt{{\left(L\omega-\frac{1}{C\omega}\right)}^2+\frac{1}{R^2}}}\cos(\omega t-\delta)$$ where $\delta=-\arctan\left(\frac{1/R}{L\omega-\frac{1}{C\omega}}\right)$.
   %   And finally:
   %   \begin{equation}\label{CM_Vout}
   %     V_\text{out}=RI=-\frac{R}{\sqrt{{\left(L\omega-\frac{1}{C\omega}\right)}^2+R^2}}\epsilon_0\sin(\omega t-\delta)
@@ -341,11 +341,11 @@
         f_0 & \text{if }0\leq t\leq \Delta t \\
         0   & \text{if }t>\Delta t
       \end{cases}$$
-    That is, $f$ is a piecewise function\footnote{We shall suppose that the system was at equilibrium for $t<0$.}. Moreover, assuming that $x(0)=\dot{x}(0)=0$, the general solution to thisODE when $0<t<\Delta t$ is:
+    That is, $f$ is a piecewise function\footnote{We shall suppose that the system was at equilibrium for $t<0$.}. Moreover, assuming that $x(0)=\dot{x}(0)=0$, the general solution to this ODE when $0<t<\Delta t$ is:
     $$x(t)=\frac{f_0}{{\omega_0}^2}+\exp{-\beta t}\left(c_1\cos(\tilde{\omega}t)+c_2\sin(\tilde{\omega}t)\right)$$
     where $\tilde{\omega}=\sqrt{{\omega_0}^2-\beta^2}$ and $c_1=-\frac{f_0}{{\omega_0}^2}$, $c_2=-\frac{\beta f_0}{\omega_1{\omega_0}^2}$.
     Therefore: $$x(\Delta t)=\text{O}({\Delta t}^2)\quad\text{and}\quad\dot{x}(\Delta t)=f_0\Delta t+\text{O}({\Delta t}^2)$$
-    If $t>\Delta t$, then the general solution to theODE is:
+    If $t>\Delta t$, then the general solution to the ODE is:
     $$x(t)=\exp{-\beta t}\left(k_1\cos(\tilde{\omega}t)+k_2\sin(\tilde{\omega}t)\right)$$ where $k_1=\text{O}({\Delta t}^2)$, $k_2=\frac{f_0\Delta t}{\tilde{\omega}}+\text{O}({\Delta t}^2)$
     Therefore, $\forall t>\Delta t$: $$x(t)=f_0\Delta t\frac{\exp{-\beta t}}{\tilde{\omega}}\sin(\tilde{\omega}t)+\text{O}({\Delta t}^2)$$
   \end{proposition}
@@ -391,7 +391,7 @@
   \end{theorem}
   \subsubsection{Non linear oscillations}
   \begin{definition}
-    Consider a pendulum whose rod (of length $L$) is in a non-small angle $\theta_0$ at initial time. TheODE that satisfies $\theta(t)$ is:
+    Consider a pendulum whose rod (of length $L$) is in a non-small angle $\theta_0$ at initial time. The ODE that satisfies $\theta(t)$ is:
     $$\ddot{\theta}+\frac{g}{L}\sin\theta=0$$
     Then, the period of the pendulum does depend on $\theta_0$. Indeed:
     \begin{multline*}