# Laplace equation

1. Oct 19, 2007

### Qyzren

u(r, θ) satisfies Laplace's equation inside a 90º sector of a circular annulus with
a < r < b ; 0 < θ < π/2 . Use separation of variables to find the solution that
satisfies the boundary conditions
u(r, 0) = 0 u(r, π/2) = f(r) ; a < r < b
u(a, θ) = 0 u(b, θ) = 0 ; 0 < θ < π/2
Consider all possible cases (negative, zero, positive) for the separation constant. Give
integral expressions for the constants in the final form for u(r, θ).

so Laplace's equation.
∂²u/∂r² + 1/r ∂u/∂r + 1/r² ∂²u/∂θ² = 0,
use U(r, θ) = G(r)Φ(θ)

r²G''/G + rG'/G = -Φ''/Φ = -λ <--seperation constant.

i've shown when λ < 0 and λ = 0, there's only trival solutions.
for λ > 0, i get r²G'' + rG' + λG = 0. using Caucy Euler, i get the general sol:
G = A cos(√λ log r) + B sin(√λ log r)
subbing in my boundary conditions i get λ = [nπ/log (b/a)]² as my eigenvalues.
So i proceed to solve for Φ.
Φ'' - [nπ/log (b/a)]²Φ = 0
Φ = A sinh [nπθ/log (b/a)] + B cosh [nπθ/log (b/a)]
Boundary condition Φ(0) = 0 => B = 0.
so we're left with Φ = A sinh [nπθ/log (b/a)]

u(r,θ) = ∑{A sinh [nπθ/log (b/a)](cos(nπ log r / log (b/a)) + sin(nπ log r / log (b/a))}
so i split it up to u(r,θ) = ∑(A sinh [nπθ/log (b/a)]cos(nπ log r / log (b/a)) + ∑(A sinh [nπθ/log (b/a)] sin(nπ log r / log (b/a)))

Now the answer says: u(r,θ) = ∑{A sinh [nπθ/log (b/a)](sin(nπ log (r/a) / log (b/a))
where A sinh [nπ²/2log (b/a)] = 2/log(b/a) ∫ f(r) sin(nπ log (r/a) / log (b/a)) dr/r (the integral is from a to b)

which seems abit different to what i have, or are they equivalent? can someone show me how to get the answer? or tell me where i went wrong? thank you.

2. Oct 19, 2007

### Kreizhn

Hmm...it's getting a little bit hard to read this. Next time you should just clearly define what $$\lambda$$ is and just use it from then on. Now, you'll notice that there is a discrepency between your eigenfunctions and the solutions eigenfunctions, notably with

$$\cos{\sqrt{\lambda} \log r}$$ and the solution's $$\cos{\sqrt{\lambda} \log{\frac{r}{a}}}$$

Now I disagree with what you found as far as substituting your initial condition and solving for lambda, since the logarithm is contained within the argument of the trigonometric function, it cannot be as easily manipulated as you seem to have done. To remedy getting two rather disgusting equations which cannot easily be solved for, the solutions have instead used their result of $$\cos{\sqrt{\lambda} \log{\frac{r}{a}}}$$

to ensure that $$u(a,\theta) = A=0$$. This can be done since we can easily see that substituting this result into the PDE doesn't affect the answer. This is a general result that can usually simplify your solutions. Continuing on like this, you find B quickly, and derive the same conclusion as the solutions. I don't know how you claim to have gotten your $$\lambda$$, but unless you used $$\log{\frac{r}{a}}$$ you shouldn't have gotten anything remotely close to the lambda of the solutions.

You also seem to have completely forgotten to apply the initial conditions. You have an as yet, undetermined value for A, which can be solved by applying said initial conditions. The eigenfunction set is complete and orthonormal, so you can use it to model a function space much akin to what you doing using Fourier Series (though I'm not sure that this counts as a Fourier Series because of the composition of the basis functions). To get their result for A, use the orthonormality of the basis set, and integrate over the domain.

Don't forget that both your expressions for G and $$\Theta$$ should have constants outside of them. Normally this means you would need to solve for 4 constants, but because you have eliminated cosh from the solution, you now only need to solve for two. You can take the constant from sinh, and multiply it with the constants for sin/cos to define two new constants, but you should still have two constants in your solution before applying the initial conditions.

Last edited: Oct 19, 2007