Find equation for self similar solution to the thin shear layer equations

[fahraynk]

Homework Statement

An alternative to the usual no-slip wall boundary condition $u = 0, v = 0$, is the porous-wall condition $u = 0$, $v = v_w(x)$. A porous wall with negative $v_w$ (suction) is known to delay transition to turbulence, for example.
1a) Assuming a uniform $u_e(x)$, determine what differential equation needs to be satisfied by a self-similar boundary layer solution for a nonzero vw. Assume that $v_w≪u_e$ so that the TSL equations remain valid.

1b) Determine the restrictions that must be placed on $v_w(x)$ so that the similar boundary layer solution you derived in Part a) can occur.
1c) Now assuming a non-uniform $u_e(x)$ determine what differential equation needs to be satisfied by a self-similar boundary layer solution for a nonzero $v_w$
1d) Determine the restrictions that must be placed on $v_w(x)$ and $u_e(x)$ so that the similar boundary
layer solution you derived in Part c) can occur.

Homework Equations

density, pressure, velocity in x, velocity in y, velocity external, viscosity = $\rho$,$P$,$u$,$v$,$u_e$,$\vartheta$
$X_y=\frac{\partial X}{\partial Y}$
stream function : $-\phi_x=v$, $\phi_y=u$
stream function form of thin shear equations :
$$\phi_y * \phi_{xy} - \phi_x*\phi_{yy}=U_e*\frac{\partial U_e}{\partial x}+\vartheta * \phi_{yyy}$$

The Attempt at a Solution

So the answer for "non self similar solution" would be the regular thin shear layer equations with a boundary condition at the wall of $v(y=0)=v_w$
$$u_x+u_y=0\\\\
u_t+u*u_x+v*u_y=-\frac{1}{\rho}P_x+\vartheta*u{yy}$$
with a solution $u=u_e[1-e^\frac{v_wy}{\vartheta}]$

But the question wants to know the equation that needs to be solved by a self similar solution.

The professor gave a derivation of a self similar solution with no velocity at the wall, but I think there is a mistake in it. Please take a look :
Self similar solution condition :
$$BU(x,y)=U(x+dx,\frac{y}{A})\\\\
B=(1+\epsilon')\\\\
\frac{y}{A}=y+\epsilon y$$
through a taylor approximation, he gets a similar solution condition :
$$\frac{\partial U}{\partial x}+\frac{\epsilon}{dx}Y\frac{\partial U}{\partial y} = \frac{\epsilon'}{dx}U$$
with $\frac{\epsilon}{dx}=\theta$ and $\frac{\epsilon'}{dx}=\alpha$, put in stream function form :
$$\phi_{xy}+\theta Y \phi_{yy}= \alpha \phi_y$$
now take the integral of both sides in order to get a function for $\phi_x$ that can be used to make the stream function as a function of y alone. :
$$\phi_x=v_w-\theta y \phi_y + \phi(\alpha + \theta)-\alpha\phi(y=0)$$
So here is my first question/point of confusion. The professor explains that $\phi(y=0)=0$. He derived this for a situation with no suction, I am trying to modify it for suction. I think $\phi(y=0)=-v_w*x$ because $-\phi_x=v$. Is this correct?

Then, we plug I plug it into the stream equation I get :
$$(\alpha+\theta)(\phi_y)^2-[(\alpha+\theta) \phi+v_w-\alpha \phi (y=0)] \phi_{yy}=u_e \frac{\partial u_e}{\partial x} + \vartheta \phi_{yyy}$$
The professor leaves off a $\theta$ from the first term, with $\phi=0$ and $v_w=0$ he gets :
$$\alpha(\phi_y)^2-(\alpha+\theta) \phi \phi_{yy}=u_e \frac{\partial u_e}{\partial x} + \vartheta \phi_{yyy}$$

So... I think he just substituted wrong which is why the first term with $(\phi_y)^2$ does not equal $(\alpha + \theta)$

Is my equation the correct answer? Also should $\phi=-v_wx$?

 

[BigJDubs]

For part b), I think the restrictions on $v_w$ is that it should be much smaller than $u_e$ in order for the thin shear layer equations to remain valid. Is that correct? For parts c) and d), I am not sure how to even start. Any help would be appreciated.