trying to solve some integral curve of vector field I am really confused about the notation used in many textbook.

I can see problem of this sort :

Find the integral curves of the following vector field : \begin{align} X(x,y) = x^2\frac{\partial}{\partial x} + xy\frac{\partial}{\partial y} \end{align}

So to set up the notation in this setting, a curve should be $\gamma : \mathbb{R} \rightarrow \mathbb{R}^2$ and any vector field of the curve can be written as : \begin{align} X_{{\gamma},p} = \frac{\partial }{\partial x^i}.(x^i\circ\gamma)'(0) \end{align}

Where $\gamma(0) = p$. So if we want to find the integral curve of the vector field we have to make sure at any point of the curve (any $t$) the tangent vector of the curve is the same as the vector field evaluated at $\gamma(t)$.

The problem is when equating the two equation we end up with something like $x^2 = (x^i\circ\gamma)'(0)$ which do not make sense.

Are we abusing notation in this sort of problems where $x^2$ is actually the composition of the coordinate function $x$ with the curve at the specific point of interest ?

If this is the case, how inaccurate is the drawing of the vector field in 2D ? ( I can see in my textbook the drawing of the vector field where at each $(x,y)$ an arrow is drawn with direction $(x^2, xy)$.

I can see some other textbook which define the curve to be $\gamma(t) = (x(t), y(t))$ which again do not make sense as it is conflicting with the coordinate function definition $x : \mathbb{R^2} \rightarrow \mathbb{R}$.

I am asking this question because the books go ahead solving an ODE where they have no problem solving stuff like : $x' = x^2$ and $y' = xy$.

So is this abusing the notation ?


$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$If $x$ and $y$ denote the Cartesian coordinate functions on $\Reals^{2}$ and if $\gamma$ is a curve in $\Reals^{2}$, then $$ \gamma(t) = ((x \circ \gamma)(t), (y \circ \gamma)(t)), $$ so that $$ \gamma'(t) = ((x \circ \gamma)'(t), (y \circ \gamma)'(t)). $$ This notation is cumbersome in practice, so one usually writes $$ \gamma(t) = (x(t), y(t)), \tag{1} $$ allowing $x$ to denote both a Cartesian coordinate in the plane and the first component of $\gamma$.

You're right: Technically (1) is abuse of notation. It would be more honest to introduce new symbols, writing $$ \gamma(t) = (u(t), v(t)), \tag{2} $$ so that $(x \circ \gamma)(t) = u(t)$, $(y \circ \gamma)(t) = v(t)$, and (for your specific vector field $X$) \begin{align*} \gamma'(t) = (u'(t), v'(t)) &= u'(t)\, \frac{\dd}{\dd x} + v'(t)\, \frac{\dd}{\dd y}, \\ (X \circ \gamma)(t) &= u(t)^{2}\, \frac{\dd}{\dd x} + u(t) v(t)\, \frac{\dd}{\dd y}. \end{align*} The flow equation then becomes the ODE system $$ u'(t) = u(t)^{2},\qquad v'(t) = u(t) v(t). $$ The thing is, this same reasoning holds if we write (1) instead of (2), with the conclusion $$ x'(t) = x(t)^{2},\qquad y'(t) = x(t) y(t), $$ i.e., $x' = x^{2}$, $y' = xy$.

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    $\begingroup$ Thank you for this clarification. Exactly what I wanted $\endgroup$ – user149705 Jan 18 '17 at 7:00
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    $\begingroup$ Thanks to this (+1) answer, I finally understand more than one half of the question :-( But still a notation like $\;a \circ b\;$ is pure mathemagics to me. Can someone explain what it means, and how it is pronounced? $\endgroup$ – Han de Bruijn Jan 18 '17 at 11:01
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    $\begingroup$ @HandeBruijn: The notation refers to composition of mappings (with $(a \circ b)(x) = a(b(x))$ for all $x$), and is often read "$a$ of $b$" or "$a$ composed with $b$". $\endgroup$ – Andrew D. Hwang Jan 18 '17 at 13:01
  • $\begingroup$ @HandeBruijn it is also often pronounced "a after b" for obvious raisons :) $\endgroup$ – user149705 Jan 18 '17 at 15:27

Relevant to the above is the following reference at Mathematics Stack Exchange:

The OP's specific problem is found there at : $\;b)\; X = x^2\,\partial_x+xy\,\partial_y\;$ .

  • $\begingroup$ I am not sure how this reference helps me understand if the problem posted is an abuse of notation. Could you please try to direct the reasoning toward my goal ? $\endgroup$ – user149705 Jan 17 '17 at 16:58
  • $\begingroup$ @user149705: I see that you already have an excellent accepted answer. But in case someone else is still interested, the solution is (I guess you know it): $$ \gamma(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} \Large \frac{x(0)}{1-t\,x(0)} \\ \Large \frac{y(0)}{1-t\,x(0)} \end{bmatrix} $$ $\endgroup$ – Han de Bruijn Jan 18 '17 at 11:09
  • $\begingroup$ Thank you for your comment anyways. I was getting lost in the reference you linked but I'm sure it will be great help for other ! $\endgroup$ – user149705 Jan 18 '17 at 15:22

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