Finding pdes of velocity component and pressure An incompressible viscous ﬂuid of constant density $ρ$
and kinematic viscosity $ν$ occupies the space above a solid boundary at $y = 0$ in two-dimensional Cartesian coordinates $(x,y)$. For time $t < 0$, the liquid and solid wall are stationary. At time $t = 0$, the solid wall is suddenly jerked into motion with constant velocity $(U,0,0)$. For time $t > 0$,
the wall continues to move with the same speed. In the absence of gravity, the continuity and Navier-Stokes equations are $$∇·\textbf u = 0$$ $$ \frac{∂ \textbf u}{ ∂ t}
+(\textbf u·∇)\textbf u = −
\frac{1}ρ ∇p+
ν ∇^2 \textbf u$$
where $p$ is the pressure.
Assume $\textbf u = (u(y,t),0,0)$ and $p = p(y,t)$. Determine partial differential equations for $u(y,t)$ and $p(y,t)$.

Can someone guide me on how to do this please. Do we just rearrange the navier equation to find pressure. And for $u(y,t)$, isn't this just some function of $y$ and $z$? Because if the continuity equation holds, and the second and third of the velocity components are zero, then their partial derivatives would also be zero so you are just left with $\partial u / \partial x=0$ which gives $u=f(y,z)$
 A: Derivation of the basic PDE requires a combination of mathematical and physical reasoning to eliminate variables, yet still render an acceptable solution to the Navier-Stokes equations.
Without an external body force component and a surface force component at the plate in the $z-$direction, there will be no velocity component in that direction, $w = 0,$ and no gradient $\frac{\partial u}{\partial z} = 0.$ Similarly, without mass injection or suction there will be no velocity component in the $y-$direction, $v = 0$. At high-speeds, instabilities could grow, inducing turbulence and non-zero velocity components / gradients in those directions, but the intent here is most likely to solve Stokes' first problem for the basic laminar flow.
The equation of continuity for incompressible flow, expressing conservation of mass, is
$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}= 0.$$
With $v = w = 0$ we have
$$\frac{\partial u}{\partial x} = 0,$$
and since $\frac{\partial u}{\partial z} = 0$ as argued above, it follows that  $ u = u(y,t)$.
The Navier-Stokes equations reduce to
$$\frac{\partial u}{\partial t } = -\frac{1}{\rho}\frac{\partial p}{\partial x } + \nu \frac{\partial^2 u}{\partial y^2 }, \\ \frac{\partial p}{\partial y } = 0,\\ \frac{\partial p}{\partial z } = 0.$$
Hence, $p$ cannot depend on $y$ or $z$, and since $u$ depends only on $y$ and $t$ the $u-$momentum equation implies $\frac{\partial p}{\partial x} = C(t).$
If the pressure is uniform far from the plate, then in the absence of body forces (with gravity neglected) we can assume $C(t) = 0.$ 
The governing equation is 
$$\frac{\partial u}{\partial t } = \nu \frac{\partial^2 u}{\partial y^2 }, $$
and the boundary / initial conditions are
$$u(y,t) \to 0 \,\,\, \text{as} \,\,\, y \to \infty,\\ u(0,t) = U \,\,\, \text{for} \,\,\, t \geqslant 0+, \\u(y,0) = 0.$$
