# Laplace equation (PDE)

• Jun 15th 2012, 01:25 PM
arbolis
Laplace equation (PDE)
Hi all, I'm stuck on the following problem:
We consider the flux of an imcompressible and irrotational fluid around a clynder of radius a, so long that we consider it as infinitely long in the z direction. We assume a stationary regime such that the velocity does not depend on the axial component (I guess they mean the z component), its axial component is null and far from the cylinder the velocity $\vec u$ is uniform and constant.
Mathematically, $\vec u$ tends asymptotically to $(v,0)$ when $r=\sqrt {x^2+y^2}$ tends to infinity.
From the imcompressibility and the mass conservation, one has that $\nabla \cdot \vec u= \frac{\partial u_1 }{\partial x }+ \frac{\partial u_2 }{\partial y}=0$. (1)
Introducing the function "potential current" $\psi$ such that $\frac{\partial \psi }{\partial x }=-u_2$ and $\frac{\partial \psi }{\partial y}=u_1$ so that (1) is satisfied and using the irrotationability of the fluid, one reaches Laplace equation for the potential current: $\triangle \psi =0$ for $r>a$. Furthermore, $\psi =0$ for $r=a$.
Using the method of separation of variables in appropriate coordinates, determine $\psi$ and $\vec u$.
------------------------------------------------------------------------
My attempt:
First, I made a sketch of the situation. I realize that $\vec u = (u_1, u_2)$. My plan to solve the problem is first to determine $\psi (r, \theta )$ (in polar coordinates), then translate it to $\psi (x,y)$ and then by integration I could get $u_1(x,y)$ and $u_2(x,y)$ and therefore $\vec u$.
So I assumed that $\psi (r, \theta )=R(r)\Theta (\theta )$. I then took the Laplacian in polar coordinates and equated to 0.
It simplified the problem into 2 ODE's, namely:
1) $\frac{rR'}{R}+r^2\frac{R''}{R}=m^2$
2) $-\frac{\Theta '' }{\Theta}=m^2$
The solution to 1) is of the form $R(r)=Ar^m+Br^{-m}$ but since it's a physical problem and psi must remain finite inside the disk of radius a, $B=0$. Thus $R(r)=Ar^m$.
The solution to 2) is $\Theta (\theta ) =C \cos (\theta m )+ B \sin (\theta m )$.
Therefore $\psi _m (r, \theta )=r^m [E \cos ( \theta m) + F \sin ( \theta m )]$ and thus $\psi (r, \theta ) = \sum _{m=0} ^{\infty } r^m [E_m \cos (\theta m )+ F_m \sin (\theta m) ]$.
Now this is where I'm stuck. I'm looking to apply the boundary condition, $\psi (a, \theta )=0$. I have reached $\sum _{m=0}^{\infty } a^m [E_m \cos (\theta m ) + F_m \sin (\theta m )]=0$. I don't know how to get the constants $E_m$ and $F_m$ from here and also I'm skeptical on the solution. Because I have an infinite series in which each terms are linearly independent of each other and the whole series is worth 0, hence it makes me think that all coefficients (therefore $E_m$'s and $F_m$'s) must equal 0... And I would get the trival solution.
Is there something wrong in what I've done so far? How could I proceed further? Thanks in advance.