# Chain-rule issue on Laplacian equation

1. Jan 30, 2012

### NinjaPeanut

1. The problem statement, all variables and given/known data
"The flow of a fluid past a wedge is described by the potential

ψ(r,θ) = -crαsin(αθ),

where c and α are constants, and (r,θ) are the cylindrical coordinates of a point in the fluid (the potential is independent of z). Verify that this function satisfies Laplace's equation, Δψ = 0."

2. Relevant equations
Δ is the Laplacian operator

3. The attempt at a solution
I have proved the statement, but I feel like I had to fudge the Chain Rule in order for it to work out. I have that δ/δr(ψ) = -αcrαsinαθ but I think that since α is some constant I ought to have δ/δr(ψ) = -αcr(α-1)sinαθ, and the second partial derivative would yield -α(α-1)cr(α-2)sinαθ. If done this way, the terms will not cancel and I cannot conclude the Laplacian is equal to zero.

Why doesn't the power of α change if it's a constant? I thought that d/dx(xn) = nx(n-1), so why does that not seem to apply in this case?

Thanks for any insight.

2. Jan 30, 2012

### Dick

I don't think you are using the correct form for the laplacian in polar coordinates. Can you show what form you are using and your work in full? There shouldn't be any need to 'cheat' the chain rule.

Last edited: Jan 30, 2012
3. Jan 30, 2012

### NinjaPeanut

Thanks for the advice, it was spot on! The correct formula for the Laplacian in polar coordinates is
Δ2ψ = r-1δ/δr(r*δψ/δr) + r-22ψ/δθ2)
Plugging in, we have:
r-1δ/δr[-r*cαr(α-1)sin(αθ)] + r-2[cα2rαsin(αθ)]
Differentiating the first term and simplifying, we have
-cαr(α-2)sin(αθ) - cα2r(α-2)sin(αθ) + cαr(α-2)sin(αθ) + cα2r(α-2)sin(αθ) = 0