- #1
skrat
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Homework Statement
I know that the following problem is really long but bear with me. The questions I have are quite simple.
The problem: Calculate the speed profile of an inviscid fluid around a perturbated cylinder. The radius of the cylinder is given as $$r=R(1-\varepsilon \sin ^2 \Theta)$$ for ##\varepsilon \ll 1##.Velocity far away from the cylinder is constant.
Homework Equations
The Attempt at a Solution
Ok, using continuity equation and assuming the field is irrotational (##\nabla \times \vec v=0##) than as expected one has to solve Laplace equation $$\nabla ^2 v(r,\Theta)=0.$$ The solution to this equation is $$\Phi (r,\Theta)=a ln(r)+b+\sum_{m=1}[A_mr^m+\frac{B_m}{r^m}][c_m\cos(m\Theta)+D_m\sin(m\Theta)]$$ Now let's assume a completely general (orientation of the) velocity far away from the cylinder $$v_\infty=v_0(\cos \varphi,-\sin \varphi)$$ and knowing ##v=-\nabla \Phi## brings me to a boundary condition far away from the cylinder $$\Phi _\infty=-v_0\cos(\Theta +\varphi).$$ Other boundary condition (I seriously doubt this is ok) is that we have a symmetry in $\Theta$ direction, meaning since ##\sin## is an odd function only ##\cos \Theta## can stay in my equation for scalar field. And now my question: What is the next boundary condition? I have a strong feeling that I should say that normal component to the cylinder is 0, written in equation, this should be $$\nabla \Phi=0$$ of course evaluated on the cylinder.
Since ##\nabla ## is in polar coordinates, my guestion here is probably a stupid one: Would it be enough to say that ##\frac{\partial }{\partial r}\Phi=0## or not?
I hope nobody gets angry with me if I stop here for a second and continue after "the boundary conditions" part is clear.