Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

Show that the solution of the Neumann problem vều = 0 if r < R, un(R, ) = f(0) (where UN du/aN is the directional derivative in the direction of the outer normal) is u(r, 6) = Ao + r™(An cos nd + Bn sin në) B) n=1 with arbitrary Ao and 1 An focos no do, TORN-1 -TT 1 Bn = fe sin no do. TTnR7 -1 -TT

Sagot :

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:

https://brainly.com/question/12873145

#SPJ4

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.