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…
Best Answer
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.