# Couette flow with infinite depth

Consider a fluid below $$x$$ axis or $$xy$$ plane. Its top layer starts to move with velocity $$v$$ at and after time $$t_0$$. If the flow is not fully developed and evolving from standing water, I receive $$v_x=0$$ from continuity equation due to zero velocity in $$y$$ direction (and $$z$$ in 3D case). Which corresponds to fully developed flow since the acceleration is zero.

$$\nabla \cdot v = 0 \\v_t + (v \cdot \nabla) v = -(1/\rho) \nabla p + \nu \Delta v$$

Task itself was to calculate velocity at certain level $$h$$ below the surface after given time $$t_1$$. (Pressure at the surface is atmospheric).

I ended up reducing system to 1D heat equation whose solution is implicit.

Could one advise on what I’m doing wrong?

• This might be better suited for Physics or Engineering Stack Exchange. Also, without providing any equations it's hard to get where you are stuck. Feb 1, 2022 at 8:03
• Updated with system, idk PDE thought suitable here @Dan Doe Feb 1, 2022 at 8:43

Let the axes be oriented so that the fluid on the $$xy$$-plane moves impulsively at $$t = 0+$$ with velocity $$U$$ in the positive $$x$$-direction and the positive $$y$$-axis points into the body of fluid.

An analytical solution can be obtained assuming unsteady, unidirectional flow with one non-vanishing velocity component $$u$$ in the $$x$$-direction. The continuity equation requires that $$\frac{\partial u}{\partial x} = 0$$. Assuming the fluid is bounded by an infinite planar surface, we have that $$u(y,t)$$ is a function only of $$y$$ and $$t$$, and the Navier-Stokes equations and boundary conditions reduce to

$$\tag{*}\frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2}, \\ u(0,t) = U \,\,\text{ for }\,\, t \geqslant 0,\\ u(y,t) \to 0 \text{ as }y \to \infty,$$

where $$\nu = \mu/\rho$$ is the kinematic viscosity.

We can solve with a similarity transformation where

$$u(y,t) = f(\eta), \quad \eta = \frac{y}{\sqrt{\nu t}}$$

Then we have

$$\frac{\partial u }{\partial t}= - f'(\eta)\frac{y}{2 \sqrt{\nu}t^{3/2}}= - \frac{1}{2t}\eta f'(\eta),\\\frac{\partial u}{\partial y} = \frac{1}{\sqrt{\nu t}}f'(\eta),\\\frac{\partial^2 u}{\partial y^2}=\frac{1}{\sqrt{\nu t}}f''(\eta)\frac{1}{\sqrt{\nu t}} = \frac{1}{\nu t}f''(\eta)$$

Substituting for $$u$$ and its partial derivatives in (*) and canceling the $$1/t$$ factors , we get the second-order linear ordinary differential equation and conditions

$$f''(\eta) + \frac{1}{2}\eta f'(\eta) = 0,\\f(0) = U ,\\ f(\eta) \to 0 \,\, \text{ as } \, \, \eta \to \infty$$

It is not difficult to solve this by separating variables and integrating twice, yielding the solution

$$u = U \text{ erfc} \left(\frac{\eta}{2}\right) = U \text{ erfc} \left(\frac{y}{\sqrt{4\nu t}}\right),$$

where the complementary error function is given by

$$\text{erfc}(z) = \frac{2}{\sqrt{\pi}}\int_z^\infty e^{-t^2} \, dt$$

• Thanks, I also got ans in terms of erf but the problem how to calculate velocity at distance from surface after certain time, solution seems implicit Feb 2, 2022 at 1:30
• @2Napasa: What is implicit here? The velocity at depth $h$ and time $t_1$ is $u(h,t_1) = U \text{erfc}(\frac{h}{\sqrt{4\nu t_1}})$. You just need to be able to evaluate the complementary error function.
– RRL
Feb 2, 2022 at 1:43
• My bad, apologies Feb 2, 2022 at 2:00
• @2Napasa: No problem. You're welcome.
– RRL
Feb 2, 2022 at 2:03
• Yes — this is a special function related to the Gaussian distribution that cannot be evaluated in terms of elementary functions. You should be able to get numerical values directly with Wolfram alpha.
– RRL
Feb 2, 2022 at 4:18