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