Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a 2D flow velocity field $\bar{V} = y^2\hat{i} + 2\hat{j}$. I'd like to find the equations for the streamlines and pathlines. Since $\bar{V}$ has 0 time derivative, the flow is steady, and so the equations for the streamlines should be identica, right?

Streamlines: $$\left. \frac{dy}{dx} \right)_{streamline} = \frac{v}{u} = \frac{2}{y^2}$$ $$ \therefore y^2 dy = 2 dx \Rightarrow \int_{y_0}^y y^2 \, dy = \int_{x_0}^x 2 \, dx $$ $$ \therefore \tfrac{1}{3}y^3 - \tfrac{1}{3}y_0^3 = 2(x - x_0) \Rightarrow y^3 - 6x = y_0^3 - 6x_0$$

Pathlines ($x_p$ and $y_p$ the particle coordinates):

$$ \frac{dx_p}{dt} = y^2 \Rightarrow x_p = y^2(t-t_0) + x_{p,0} $$ $$ \frac{dy_p}{dt} = 2 \Rightarrow y_p = 2(t - t_0) + y_{p,0} $$ Now eliminating $(t-t_0)$: $$x_p = y^2 \left( \frac{y_p - y_{p,0}}{2} \right) + x_{p,0}$$ Without expanding this final equation for the pathline, it is obviously different from the equation of the streamlines; but it shouldn't be, as the flow is steady? Normally this might suggest I've been fast and loose with the math, but I just can't see where I'm wrong here...

share|cite|improve this question
y is changing in time, you integrated the x equation as if it were constant. – Ron Maimon Aug 25 '11 at 23:49
@Ron: you might want to post that as an answer (and maybe expand on it a bit). – David Z Sep 4 '11 at 0:18
up vote 0 down vote accepted

y is changing in time, but in your second solution you integrated the x equation as if it were constant, which is illegitimate.

$${dy_p\over dt} = 2 \implies y=2t+y_0$$ $${dx_p\over dt} = y^2 = (2t+y_0)^2 \implies x_p = {4\over 3}t^3 + 2y_0 t^2 + y_0^2 t + x_0$$

Which is consistent with the first solution.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.