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The solution of Neumann problem, ∇²u= 0 if r < R , Uₙ (R,θ) = f(θ) is u(r,θ) = a'₀+ rⁿ(a'ₙ cosnθ + b'ₙ sinnθ) with boundary conditions uᵣ (r,θ) = ∑n R⁽ⁿ⁻¹⁾(Aₙ cosnθ + Bₙ sinnθ) = f(θ) and
Aₙ=∫(f(θ)cosnθ /π nR⁽ⁿ⁻¹⁾)dθ, where θ∈[-π, π]
Bₙ=∫(f(θ)sinnθ /π nR⁽ⁿ⁻¹⁾)dθ, where θ∈[-π, π]
Given that
The solution of Numann problem
∇²u= 0 if r < R , Uₙ (R,θ) = f(θ)
Use polar co-ordinates (r,θ)
uᵣᵣ + 1/r uᵣ+ 1/uᵣ (uθθ) = 0 ,0 < r< R,
0 <θ <2π and ∂u/∂r(R,θ) = f(θ) is directional derivative
r²d²u/dr² + rdu/dr + d²u/dθ² = 0
Let , r = ε⁻ᵗ , u(r(t),θ)
uₜ = uᵣ(rₜ) = - e⁻ᵗ uᵣ
uₜₜ =( - e⁻ᵗ uᵣ )ₜ = ε⁻ᵗuᵣ + e⁻²ᵗ uᵣᵣ
= r uᵣ+ r²uᵣᵣ
Thus we have, uₜₜ + uθθ = 0
Let u(t,θ) = X(t)Y(θ)
Which gives X''(t)Y(θ) + X(t)Y"(θ) = 0
X"(t)/X(t) = - Y"(θ)/Y(θ) = λ
From Y"(θ) + λ Y(θ) = 0
We get, Yₙ(θ) = aₙ cosnθ + bₙ sinnθ
λ= n² , n =0, 1, ...
With these values of λn we solve,
X"(t) - n² X(t) = 0
If n = 0 , X₀(t) = c₀t + d₀
X₀(r) = -c₀log (r) + d₀
If n not equal to 0 then
Xₙ (t) = cₙeⁿᵗ + dₙ e⁻ⁿᵗ
Xₙ(r) = cₙ(r)⁻ⁿ + dₙ (r)ⁿ
We have u₀(r, θ) = X₀(r)Y₀(θ)
= a₀ ( - c₀(log r) + d₀)
uₙ(r,θ) = Xₙ(r) Yₙ(θ)
= (cₙ r⁻ⁿ+ dₙrⁿ)(aₙ cosnθ + bₙ sinnθ)
But u must be positive at t =0
So, cₙ = 0 ; n = 0,1,2....
u₀ (r,θ) = a₀ d₀
uₙ(r,θ) = dₙ rⁿ( aₙ connθ + bₙ sinnθ)
By superposition , we can write as
u(r,θ) = a'₀+ rⁿ(a'ₙ cosnθ + b'ₙ sinnθ)
Boundary conditions gives
uᵣ (r,θ)=∑n R⁽ⁿ⁻¹⁾(Aₙ cosnθ + Bₙ sinnθ) = f(θ)
the coefficients aₙ , bₙ for n ≥ 1 are determined are Fourier series for f(θ)
but a₀ is not determined from f(θ) therefore , it may take arbitrary value. By using Fourier series,
Aₙ= Integration of(f(θ)cosnθ dθ/π n R⁽ⁿ⁻¹⁾) where θ∈[-π, π]
Bₙ= Integration of (f(θ) sinnθ dθ/π nR⁽ⁿ⁻¹⁾) where θ∈[-π, π]
To learn more about Directional derivative, refer:
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