Given the velocity field $\ (u,v) = UL( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}$), how can we find the Legrangian coordinates?

Question

I'm currently taking Graduate Fluid Dynamics and I'm using Stephen Childress' book, An intro to theoretical fluid mechanics. I've been stuck on one problem for quite some time and, so I'd like to see if someone else can assist me in finding the solution to this problem:

The "point vortex" flow in two dimensions has the velocity field

$\ (u,v) = UL( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}$)

where U, L are reference values of speed and length.

(a) Show that the Lagrangian coordinates for this flow may be written

$\ x(a, b, t) = R_0 cos(\omega t + \theta_0)$, $\ y(a, b, t) = R_0 sin(\omega t + \theta_0)$

where $\ R_0 = a^2 + b^2$, $\theta_0 = arctan (\frac{b}{a})$, $\omega = \frac{UL}{R_0^2}$

There's a secondary question to this problem, but I'd really like to understand (at the very least) this part because I don't even know where to start. When I tried solving it myself, I found a system of messy ODEs that I'm unable to solve, either due to the complexity of it or due to me forgetting something. Can anyone push me in the right direction?

Answer 1

The right direction is to use the polar coordinates from the scratch, which would turn the problem in a very easy one, but let's try a more physical observation:

Since your flow field is stationary, the Lagrangian coordinates (i.e. the trajectory of a parcel) coincide with the streamlines (and with the material lines). Since you know that the streamlines for a point vortex are circles centered on the origin, and that the modulus of the velocity at a certain distance from the center is $\frac{UL}{r}$, we can conclude that any parcel will have a circular motion around the center at an angular velocity $\omega = \frac{V}{r} = \frac{UL}{r^2}$, and hence you get exactly what you wrote in your question.

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In a more formal way, let's write the system of ODE you need to solve:

$$ \frac{dx}{dt} = u = UL \frac{-y}{x^2+y^2} \\ \frac{dy}{dt} = v = UL \frac{x}{x^2+y^2} $$

Hinted by the $x^2+y^2$, we would like to use polar coordinates, so we write:

$$ x = r \cos{\theta} \\ y = r \sin{\theta} $$

And we compute the derivatives of $x$ and $y$ as:

$$ \frac{dx}{dt} = \frac{dr}{dt} \cos{\theta} - r \sin{\theta} \frac{d\theta}{dt} = u = -UL \frac{\sin{\theta}}{r} \\ \frac{dy}{dt} = \frac{dr}{dt} \sin{\theta} + r \cos{\theta} \frac{d\theta}{dt} = v = UL \frac{\cos{\theta}}{r} $$

We can multiply the first by $\cos{\theta}$, the second by $\sin{\theta}$ and sum, thus obtaining:

$$ \frac{dr}{dt} \cos^2{\theta} + \frac{dr}{dt} \sin^2{\theta} = \frac{dr}{dt} = 0 $$

That means $r(t) = R_0 = \sqrt{a^2 + b^2}$, every parcel is always at the same distance from the center of the vortex.

Now we can multiply the first eq. by $\sin{\theta}$, the second by $\cos{\theta}$ and subtract the first from the second:

$$ r \frac{d \theta}{dt} = \frac{UL}{r} $$

Since we already know that $r$ is constant, we can trivially integrate this one:

$$ \theta(t) = \frac{UL}{{R_0}^2} t + \theta_0 $$

And going back to cartesian coordinates we obtain:

$$ x = r \cos{\theta} = R_0 \cos \left( \frac{UL}{{R_0}^2} t + \theta_0 \right) \\ y = r \sin{\theta} = R_0 \sin \left( \frac{UL}{{R_0}^2} t + \theta_0 \right) $$

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Easiest way, work from the scratch in polar coordinates. We know that the velocity $\vec{v}$ can be written with respect to the derivatives $\dot{r}$ and $\dot{\theta}$ as:

$$ \vec{v} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta} $$

$\hat{r}$ and $\hat{\theta}$ being respectively the radial and tangential versors (unit versors explained here).

Similary the velocity field induced by a point vortex can be written in polar coordinates:

$$ \vec{v} = \frac{UL}{r} \hat{\theta} $$

That is we get:

$$ \dot{r} = 0 \\ r \dot{\theta} = \frac{UL}{r} $$

Answer 2

While it's not nearly as conceptually helpful, the method of conformal mappings for 2D potential flows permits a very efficient calculation of the Lagrangian coordinates. Observe that the complex velocity of this vortex is given by $u-i\,v = -UL\left(\dfrac{y+i\,x}{x^2+y^2}\right)=\dfrac{UL}{i z}$ where $z=x+i\,y$. The differential equation defining the Lagrangian coordinates is then given by $u-i\,v=\dfrac{dx}{dt}-i \dfrac{dy}{dt}=\dfrac{d\overline{z}}{dt}$. This implies

$$UL = i z \dfrac{d\overline{z}}{dt}=i (r e^{i\theta})\dfrac{d}{dt}(r e^{-i\theta})=r^2 \dfrac{d\theta}{dt}+ir\dfrac{dr}{dt}$$

Note that the LHS is real, so the imaginary part of the RHS must vanish and so $r=R_0$ is constant. Therefore $\dfrac{d\theta}{dt}=\dfrac{UL}{R_0^2}=\omega\implies \theta(t)=\omega t+\theta_0$ and so $$z(t)=R_0 e^{i \phi(t)}=R_0 \cos(\omega t+\theta_0)+i R_0 \sin(\omega t+\theta_0)$$ Identifying real and imaginary parts with $(x,y)$ yields the desired results.