Mathematics desk
< April 8	<< Mar \| April \| May >>	April 10 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

April 9

Calculating flow

There is a channel of some incompressible liquid that stretches away indefinitely from the origin (x = 0) in either direction. This is at a constant depth d. A flow is induced (I am not concerned about how) such that the speed of flow at coordinate x is given by v = kx for some positive constant k. Positive v means flow in the direction of increasing x; negative v means flow in the direction of decreasing x. When I do the calculations, I get that after time t the water is at constant depth d/exp(kt), but I can't figure out how the depth at x = 0 can have changed when there is no flow there. I find it quite hard to visualise what ought to happen around x = 0, actually. Am I doing this right? 86.179.5.189 (talk) 20:19, 9 April 2014 (UTC)[reply]

Let the depth be called y. Choose units of time and length such that k = dv/dx = 1 and y₀ = 1. Use letter d for differentiation. The rate of change of depths , dy/dt = −y dv/dx = −y , doesn't depend on v but only on dv/dx which is unity. The differential equation dy/y = −dt is integrated to log(y) = −t+const. Choose the zeropoint of time such that the constant is zero. Then the depth is y = e^−t , independently of x. Bo Jacoby (talk) 03:53, 10 April 2014 (UTC).[reply]

To help visualization, imagine a bucket of water, from which you suddenly remove the sides. The water flows equally in all directions, but the water at the centre has zero radial velocity, only a (depth-dependent) downwards velocity. Your scenario is qualitatively the same. —Quondum 04:22, 10 April 2014 (UTC)[reply]

Here reality differs from theory. In reality you would get all kinds of flow at the center. If you think about it as the molecules of water, there's no way the top molecules can go lower, and remain at the center, without the molecules underneath it moving out of the way. StuRat (talk) 04:34, 10 April 2014 (UTC)[reply]

I don't understand Bo Jacoby's approach. I want an answer that involves d and k. Is d/exp(kt) correct? 86.160.87.195 (talk) 11:05, 10 April 2014 (UTC)[reply]

I'm confused here as well. Are we sure that such a flow is possible? If not then that would explain the counterintuitive results. I'm thinking that the y-component of velocity must be constant, else the surface does not remain level. Also the y-component can't be 0 else the flow violates incompressibility, but it must be 0 at the y-axis, giving a contradiction. If this reasoning is incorrect then can someone give an example of the flow with the given properties? --RDBury (talk) 19:43, 10 April 2014 (UTC)[reply]

One thing that I didn't mention, but implicitly assumed, is that my v is intended to be the x-component of the velocity only. 86.171.43.111 (talk) 20:23, 10 April 2014 (UTC)[reply]

I was working on the assumption that v was length of velocity. If v is the x-coordinate then (kx, -ky) is an incompressible flow with the desired properties. An individual "molecule" of water follows a hyperbolic path of the form xy=c, following an axis if c=0. On the y-axis they are all moving downward toward the origin but since the volume of this set is 0 there is no conflict with incompressibility. --RDBury (talk) 22:06, 10 April 2014 (UTC)[reply]

"Is d/exp(kt) correct?". Yes, it is correct. The special case d=k=1 gives y = e^−t. The general case d = y₀ and k ≠ 1 is

y = y₀ e^−kt.

So a molecule which is initially (for t=0) in the position

(x,y) = (x₀,y₀)

will at time t be in the position

(x,y) = (x₀ e^kt, y₀ e^−kt).

The velocity vector is

(v_x,v_y) = (kx₀ e^kt, −ky₀ e^−kt) = (kx, −ky).

The divergence is

{\frac {\partial v_{x}}{\partial x}}+{\frac {\partial v_{y}}{\partial y}}=(k)+(-k)=0

so this flow is indeed incompressible.

Bo Jacoby (talk) 03:17, 11 April 2014 (UTC).[reply]