- #1
Qyzren
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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.
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.