Solving the Laplace Equation on a Circular Wedge

[NT123]

Homework Statement

I need to solve the Laplace equation, u_rr + (1/r)u_r + (1/r^2)u_{theta}{theta} = 0,
on a circular wedge with radius R, angle {alpha}, where u(r,0) = 0, u(R,{theta}) = 0, and
u(r,{alpha}) = 50.

Homework Equations

The Attempt at a Solution

Separate variables - u(r,{theta}) = P(r)Q({theta}), and we have the equations

Q'' + (k^2)Q = 0, (r^2)P'' + rP' - (k^2)P = 0, where k is a constant. Q_n = Asin(n{theta})+Bcos(n{theta}), P_n = C(r^n).

Since u(r,0) = 0, B = 0. Now, since u(R,{theta}) = 0, this must mean R^n = 0, hence
R = 0, so u = 0 for all r, {theta}. Is this reasoning correct?

 

[LCKurtz]

Homework Statement

I need to solve the Laplace equation, u_rr + (1/r)u_r + (1/r^2)u_{theta}{theta} = 0,
on a circular wedge with radius R, angle {alpha}, where u(r,0) = 0, u(R,{theta}) = 0, and
u(r,{alpha}) = 50.

Homework Equations

The Attempt at a Solution

Separate variables - u(r,{theta}) = P(r)Q({theta}), and we have the equations

Q'' + (k^2)Q = 0, (r^2)P'' + rP' - (k^2)P = 0, where k is a constant. Q_n = Asin(n{theta})+Bcos(n{theta}), P_n = C(r^n).

Since u(r,0) = 0, B = 0. Now, since u(R,{theta}) = 0, this must mean R^n = 0, hence
R = 0, so u = 0 for all r, {theta}. Is this reasoning correct?

No, it isn't. You don't get just rⁿ but you also get r^-n. And I would suggest using something other than n since it isn't restricted to integer values with only one theta boundary condition. Perhaps μ would be better. So you have sin(μθ) and you have {r^μ,r^-μ}, or equivalently {e^μln(r),e^-μln(r)} or, even better, {cosh(μln(r)),sinh(μln(r)}. This would give a two parameter family Acosh(μln(r)) + Bsinh(μln(r)), which looks like an addition formula. Since you have a zero boundary condition when r = R, this suggests even another choice for a two parameter family:

P(r) = A sinh(μln(r) + B)

Then P(R) = A sinh(μln(R) + B) and this can be made to be 0 by picking B = -μln(R).

This gives P(r) = A sinh(μln(r)-μln(R)) = A sinh(μln(r/R))

So at this point you have:

Q_μ = sin(μθ) and P_μ=sinh(μln(r/R)).

These work in your DE and solve the homogeneous BC's. Hopefully, you can take it from there because it's been too long since I have looked at this stuff.

 

[NT123]

No, it isn't. You don't get just rⁿ but you also get r^-n

I thought the r^(-n) part was removed because then the solution would go to infinity at r = 0.

 

[LCKurtz]

I thought the r^(-n) part was removed because then the solution would go to infinity at r = 0.

You didn't give careful statements of your boundary conditions but I would expect they are something like this:

u(r,0) = 0, 0 < r < R
u(R,θ) = 0, 0 < θ < α
u(r,α) = 50, 0 < r < R

In particular you aren't giving u(0,0) and the first and third conditions would disagree if you tried. So I don't see where keeping the r^-μ is ruled out. And you need it to get your second boundary condition as I have shown. Your separated equation for P(r) has a singular point at r =0.

You can verify that u(r,θ) = Asin(μθ)sinh(μln(r/R)) satisfies the DE and the homogeneous BC's. But, like I said, you're on your own from here.