Hi, friends:

I'd like to ask a simple question on bounday layer simulation. The problem is : given free stream velocity U(t)=A*cos(wt), where w is angular frequency , how to solve a linear laminar boundary layer numerically (my problem is rough wall turbulence, but I simplify it to laminar case here for simplicity). Seems that it's a quite simple homework problem. We have governing equation:

Equation(1): d(u)/dt = nu* d^(u)/dz^2

look for solution u(z,t), where z is distance from wall and nu is kinematic viscosity. Boundary conditions are :

Equation(2): a)u=0 on z =0;

Equation(3): b)u=U on z=h

where h is is the thickness of the boundary or

Equation(4): c)tau=nu*d(u)/dz =0 on z=h.

Normally h(t) is not a constat, if we choose h large enough, we should be able to set h=constant. My question is really: how can we satisfy both (3) and (4)?? Seems that if we choose to use (3), then we won't be able to satisfy (4), or if we want to satisfy (4), we won't be able to satisfy (3). Because this is a 2 point BVP, only two boundary conditions are needed, and (2) can't be neglected. My problem is that : on the top of the bounday layer, if we want shear stress tau(h)=0, then we won't have solution u(h) =U; if we want u(h)=U, then tau(h)\=0. This is puzzling me for sometime, and I want to know your opinion. I also know that there is analytical solution to this problem (Stokes 1847, Lamb 1932) which says tau=0 at infinity, u=U at infinity. But I want to use finite difference for a far more complicated application, so I have to choose a finite domain 0<= z <=h. Now, I'm kind of biased to using (3) instead of (4), because it's important to keep u(h)=U(t), otherwise after some time of time integration, the solution u(t,z) will be out of whack of the free stream velocity U(t), but I don't know whether this is a standard practice for oscillatory boundary layer modeling.

Best,

Wen

You don't try to satisfy both; e.g. If you take the full nonlinear problem

u_t + uu_x + vu_y = -P_x + u_yy,

u_x + v_y = 0, P_y = 0,

(You can linearize these if you want it won't change what I'm saying)

with the boundary condition u=v=0 on y=0. For the upper boundary condition, as y-> infinity, you have two equivalent options:

(1) set u->U(x,t) => -P_x = U_t + UU_x (Bernoulli),

(2) set P_x using Bernoulli and U_y -> 0.

Basically you choose one of these and, provided the lid is highing enough, the other condition is automatically satisfied.

Tom:

Thank you for the tip.

Wen