An incompressible viscous fluid of constant densite and kinematic viscosity occupies the space between porous walls at $y=0$ and $y=d$. The steady two dimensional flow is subject to a constant pressure gradient $$\frac{dp}{dx}=-G$$ Fluid enters through the wall at $y=d$ and leaves the wall at $y=0$ at a consant normal speed $V$.
Assume $\textbf u = (u(x,y_, v(x,y),0)$ and $p=p(x)$. Determine the pdes of $u$ and $v$.
$$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho} \frac{\partial p}{\partial x} + ν \bigg( \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} \bigg)$$
$$u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=ν \bigg( \frac{\partial^2 v}{\partial x^2} +\frac{\partial^2 v}{\partial y^2} \bigg)$$ That is what I got for the $x$ and $y$ components.
Are the conditions $u=0$ when $y=0$ or $y=d$?
The next part says, show that $v(x,y)=-V$ is a solution to the $y$ component. Do you just sub it in and then get zero on both sides since the derivatives of a constant are zero?
The next part is to find a second order pde for $u$. So we have: $$\frac{\partial^2 u}{\partial y^2} +\frac{V}{ν} \frac{\partial u}{\partial y} +\frac{G}{\rho ν} =0$$ Is this correct? I used the continuity equation which $\partial u / \partial x = \partial V / \partial y = 0$
And to solve this, can we just turn it into a first order by letting $\eta ' = u''$ and $\eta = u'$?
The differential equation you have seems to be correct, but you are missing terms, why did you set ∂V/∂y=0? It is correct that ∂u/∂x=∂V/∂y from the continuity equation but not that they are equal to 0.