Chain-rule issue on Laplacian equation

[NinjaPeanut]

Homework Statement

"The flow of a fluid past a wedge is described by the potential

ψ(r,θ) = -cr^αsin(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."

Homework Equations

Δ is the Laplacian operator

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) = -αcr^αsinαθ 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.

 

[Dick]

Homework Statement

"The flow of a fluid past a wedge is described by the potential

ψ(r,θ) = -cr^αsin(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."

Homework Equations

Δ is the Laplacian operator

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) = -αcr^αsinαθ 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.

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.

 

[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^-2(δ²ψ/δθ²⁾
Plugging in, we have:
r^-1δ/δr[-r*cαr^(α-1)sin(αθ)] + r^-2[cα²r^αsin(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