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If I have a three dimensional ODE for $x_1,x_2,x_3$ and for the first equation I have

$\dot{x}_1=x_2$

what does this mean?

That the change of $x_1$ is in the direction of $x_2$?

Background:

In this paper it is said on page 436 (Case I.) that since $u'=v$, we have that $S_{\epsilon}\cap M_{\theta}$ has only one common point. Why does this hold?

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    $\begingroup$ Are $x_1$, $x_2$ and $x_3$ functions? Then $\dot{x}_i$ is sometimes used to denote the derivative of $x_i$. $\endgroup$ – Matias Heikkilä Nov 10 '15 at 11:28
  • $\begingroup$ Yes, functions. But I do not know how to interprate that the derivative of $x_1$ is $x_2$. $\endgroup$ – M. Meyer Nov 10 '15 at 11:28
  • $\begingroup$ $x_2$ describes the rate of change of $x_1$. What kind of interpretation would you be looking for? $\endgroup$ – Alvin Lepik Nov 10 '15 at 11:31
  • $\begingroup$ It is just another way of saying $f'(t)=g(t)$. $\endgroup$ – Joe Johnson 126 Nov 10 '15 at 11:35
  • $\begingroup$ In an example I have an intersection of a three-dim- function with some plane. And then one says that because $x_1'=x_2$ there is only one common point. $\endgroup$ – M. Meyer Nov 10 '15 at 11:36
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(I'm using primes instead of dots, since MathJax renders dots so lightly in my browser.) Suppose generally that your system has the form \begin{align*} x'_{1} &= F_{1}(x_{1}, x_{2}, x_{3}), \\ x'_{2} &= F_{2}(x_{1}, x_{2}, x_{3}), \\ x'_{3} &= F_{3}(x_{1}, x_{2}, x_{3}). \end{align*} Geometrically, this system defines a vector field, an association of the vector $$ \mathbf{F}(x_{1}, x_{2}, x_{3}) = \bigl(F_{1}(x_{1}, x_{2}, x_{3}), F_{2}(x_{1}, x_{2}, x_{3}), F_{3}(x_{1}, x_{2}, x_{3})\bigr) $$ to each point of space (or to each point in the region where the ODE is defined).

A soluton of the ODE system, a.k.a., a flow line of the vector field $\mathbf{F}$, is a path $\mathbf{x}(t)$ (often interpreted as the position of a point particle at time $t$) whose velocity at each instant is the value of the vector field at the particle's location: $$ \mathbf{x}'(t) = \mathbf{F}\bigl(\mathbf{x}(t)\bigr). $$

The equation $x'_{1} = x_{2}$ says the $x_{1}$-component of the velocity of an arbitrary solution is equal to the $x_{2}$-position of the particle at each instant. (If $x_{2} \neq 0$, the particle therefore crosses each plane $x_{1} = \text{const}$ at most once; haven't read your linked article carefully enough to know if that's the gist of the argument, however.)

In my experience, extracting any satisfying intuition from this piece of qualitative information is tricky: As the particle moves, its position changes, which causes the velocity to change, which affects the position...in an endless cycle of feedback. (Based on other questions you've asked, I hope it's not presumptuous to suspect you're trying to overcome a similar issue.)

Instead, plot the vector field $\mathbf{F}$ (or imagine the vector field $\mathbf{F}$ existing throughout space as a static piece of information), then think of a solution as the trajectory of a point particle that "follows the flow".

Here are a couple of (possibly CPU-intensive) interactive visualization links that run in a web browser:

  • A wind map of the earth, using (more or less) real-time data;

  • The Lorenz differential equation.

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  • $\begingroup$ My problem in general is the following: I have a point in $\mathbb{R}^3$, namely $(x_1(t),x_2(t),x_3(t))$, and then I have its derivative, i.e. $(x_1'(t), x_2'(t),x_3'(t))$. Now, I know that $(x_1'(t), x_2'(t),x_3'(t))$ is nothing else but the vector that the vector field $F$ "gives" to the point $(x_1(t),x_2(t),x_3(t))$, right? This vector that F gives to this point has a special direction. But what are the components $x_1'(t)$, $x_2'(t)$ and $x_3'(t)$? How can I imagine them? $\endgroup$ – M. Meyer Nov 10 '15 at 12:43
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    $\begingroup$ To "plot" a vector field is to draw a bunch of arrows from $\mathbf{x}$ to $\mathbf{x} + \mathbf{F}(\mathbf{x})$. (In principle, this arrow is drawn for every $\mathbf{x}$. In any case, the domain of the vector field looks like it's covered with/filled with fur.) The components of the arrow based at $\mathbf{x}$ are $(x_{1}', x_{2}', x_{3}')$, in the sense that a particle finding itself at $\mathbf{x}$ necessarily finds its velocity to be $(x_{1}', x_{2}', x_{3}') = \mathbf{F}(\mathbf{x})$. $\endgroup$ – Andrew D. Hwang Nov 10 '15 at 12:52
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    $\begingroup$ If it happens that $x_{2} \neq 0$, i.e., the curve never hits the plane $x_{2} = 0$, then $x_{1}' = x_{2}$ is non-vanishing, which means the particle always has a non-zero component of motion in the $x_{1}$-direction, and therefore cannot cross any plane $x_{1} = \text{const}$ more than once. Loosely, to cross one of these planes twice, the particle would have to double back in the $x_{1}$-direction. :) (Again, I'm not positive that's what's going on in the paper you're reading, but that's a typical way of reading out qualitative information about motion.) $\endgroup$ – Andrew D. Hwang Nov 10 '15 at 13:00
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    $\begingroup$ Everything here sits in a Cartesian $3$-dimensional space with coordinates $(x_{1}, x_{2}, x_{3})$, and these equations have their conventional meanings as planes, e.g., $x_{2} = 0$ defines the $(x_{1}, x_{3})$-plane, and $x_{1} = \text{const}$ signifies an arbitrary plane parallel to the $(x_{2}, x_{3})$-plane. :) $\endgroup$ – Andrew D. Hwang Nov 10 '15 at 13:16
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    $\begingroup$ 1. Yes, I couldn't tell whether that was the case, either. It does appear the author is establishing a local result, and quite possibly he's considering a region of space away from $v = 0$. 2. The ratio $x_{3}'/x_{1}'$ can be interpreted as "the slope of the vector $\mathbf{F}(\mathbf{x})$ as you look down the $x_{2}$-axis", or alternatively as $$\frac{dx_{3}}{dx_{1}} = \frac{dx_{3}/dt}{dx_{1}/dt}.$$ $\endgroup$ – Andrew D. Hwang Nov 10 '15 at 13:29
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First off we have three function $x_1$, $x_2$ and $x_3$.

Since the dot notation is used it is safe to assume that these are functions of time, i.e. $x_i$ could be written more obviously as $x_i(t)$, for $i = 1,2,3$. Note that the meaning of $\dot{x}$ is normally $\frac{dx}{dt}$.

The first equation is then simply saying that the second function is the derivative of the first, or more explicitly $\frac{dx_1}{dt} = x_2$, this might make more sense if we turn it round, so $x_2(t) = \frac{dx_1(t)}{dt}$.

Consider the concrete example where $x_1(t) = 3t^2$, the the above is just saying that $\frac{dx_1}{dt} = 6t = x_2$, so $x_2 = 6t$.

Having seen the question you linked in the comments section the notation of the paper is actually $x_1 ' = x_2$. This notation indicates that we are dealing with a function of one variable and taking the derivative with respect to that variable. Essentially we don't necessarily have a function of time, but the above still all holds, it's just that $t$ would be a variable that doesn't necessarily have a specific meaning attached to it.

In the specific question you cited we do seem to have a function of time.

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  • $\begingroup$ Nonetheless, I do not understand why $u'=v$ ensure that there is only one point in $S_{\varepsilon}\cap M_{\theta}$. $\endgroup$ – M. Meyer Nov 10 '15 at 12:11
  • $\begingroup$ Have you read and understood the proof of this in the paper (Lemma 2.1)? Beacuase this shows why that happens. Note the proof contains more than just the fact that $u' =v $. $\endgroup$ – EHH Nov 10 '15 at 12:17
  • $\begingroup$ No, Lemma 2.1 refers to another $M_{\theta}$ which is used in Case II. $\endgroup$ – M. Meyer Nov 10 '15 at 12:19
  • $\begingroup$ For Case I it really seems to be enough that $u'=v$. $\endgroup$ – M. Meyer Nov 10 '15 at 12:22
  • $\begingroup$ Yes but it's not clear from the paper exactly how much work is involved in showing that that conditions leads to the result. I think you're better reposting your question with that being the topic. And tag it more specifically with wave equations etc.. $\endgroup$ – EHH Nov 10 '15 at 12:31
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If $x_1$ is a single variable function that depends on $t$, we have $$\dot{x_1}=\frac{d}{dt}x_1=x_2$$ If $x_1$ is a multivariable function with that depends on $t$, we have $$\dot{x_1}=\frac{\partial}{\partial t}x_1=x_2$$ Usually Newton's notation for differentiation is used to denote derivatives with respect to time. So if $x_1$ represents displacement, then $x_2$ is the instantaneous velocity at any time $t$. Until you and more context to your post, I cannot assist any further.

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  • $\begingroup$ I have also seen the the notation that if $x_1$ is a multivariable function then the partial derivative of $x_i$ with respect to $t$ is denoted as $f_{x_i}$. $\endgroup$ – Mr Pie Feb 19 '18 at 16:16

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