Lecture 26: Circular domains

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1 Introductory lecture notes on Partial Differential Equations - c Anthony Peirce. Not to be copied, used, or revised without eplicit written permission from the copyright owner. 1 Lecture 6: Circular domains (Compiled 15 May 18 In this lecture we consider the solution of Laplace s equations on domains that have a boundary that has at least one boundary segment that comprises a circular arc. Key Concepts: Laplace s equation; Circular domains; Pizza Slice-shaped regions, Dirichlet and Mied BC. Reference Section: Boyce and Di Prima Section General Analysis of Laplace s Equation on Circular Domains: 6.1 Laplacian in Polar Coordinates For domains whose boundary comprises part of a circle, it is convenient to transform to polar coordinates. For this purpose the Laplacian is transformed from cartesian coordinates (, y to polar coordinates (r, θ as follows: Differentiating with respect to and y we obtain: r = + y ( θ = tan 1 y rr = r = r y = y r r θ = ( y ( 1 + ( y = y + y = y r θ y = r u(, y = U(r, θ ( u = U r r + U θ θ = U r + Uθ ( y ( r r y ( u y = U r r y + U θ θ y = U r + U θ r r u = (U r r + U r r + (U θ θ + U θ θ = U rr r + U rθ θ r + U r r + U rθ r θ + U θθ θ + U θ θ r r r = r = y r 3 θ = y r 4 ( u = U rr r + U rθ y ( y y y r + U θθ r r 4 + U r r 3 + U θ r 4 u yy = U rr ry + U rθ r y θ y + U r r yy + U θr r y θ y + U θθ θy + U θ θ yy y ( ( y ( = U rr r + U rθ r + U θθ r r 4 + U r r 3 + U θ y r 4 u + u yy = U rr + 1 r U r + 1 r U θθ

2 Recall the Laplacian in polar coordinates: 6. Introductory remarks about circular domains Let = u = u + u yy = u rr + 1 r u r + 1 r u θθ r = ( + y 1/ θ = tan 1 (y/. (6.1 u(r, θ = R(rΘ(θ (6. r R + rr R(r which leads to r R + rr λ R = and Θ + λ Θ =. The R Equation: r R + rr λ R = : λ = : r R + rr =, R = r γ γ(γ 1 + γ = γ = R(r = C + D ln r = Θ Θ(θ = λ (6.3 λ : r R + rr λ =, R = r γ γ(γ 1 + γ λ = γ λ = R(r = Cr λ + Dr λ. The Θ Equation Θ + λ Θ = : Θ + λ Θ =, Θ = A cos λθ + B sin λθ, Θ = Aλ sin λθ + Bλ cos λθ Different Boundary Conditions and corresponding eigenfunctions: (I Θ( = = Θ( λ n = nπ/, n = 1,,..., Θ n (θ = sin λ n θ (II Θ ( = = Θ (, λ n = nπ/ n =, 1,,..., Θ n (θ {1, cos λ n θ} (III Θ( = = Θ (, λ n = (n 1π/ n = 1,,..., Θ n (θ = sin λ n θ (IV Θ ( = = Θ(, λ n = (n 1π/, n = 1,,..., Θ n (θ = cos λ n θ } Θ( π = Θ(π (V Θ ( π = Θ λ n = n, n =, 1,,..., Θ n (θ {1, cos λ n θ, sin λ n θ}. (π The most general solution is thus of the form Observations: u(r, θ = {A + ln r} { An r λ n + n r λ } n cos λ n θ (6.4 { Bn r λ n + β n r λ } n sin λ n θ. (6.5 For problems that include the origin, the condition u < as r dictates that =, n = and β n =. For problems that involve infinite domains the condition u < as r dictates that A n = and B n =. The values of λ n and the corresponding eigenfunctions depend on the boundary conditions (I (V that apply. 6.3 Wedge Problems Eample 6.1 Wedge with homogeneous BC on θ =, θ = < π u rr + 1 r u r + 1 r u θθ = < r < a, < θ < (6.6 u(r, = u(r, =, u(r, θ bounded as r, u(a, θ = f(θ (6.7

3 Laplace s Equation 3 Figure 1. Homogeneous Dirichlet Boundary conditions on a wedge shaped domain (6.7 Let u(r, θ = R(r Θ(θ. r (R + 1 r R R = Θ (θ Θ(θ = λ r R + rr λ R = Euler Eq. Θ + λ Θ = u(r, = R(rΘ( = Θ( = ; u(r, = R(rΘ( = Θ( = Eigenvalue Problem { Θ + λ Θ = Θ( = = Θ( Θ = A cos λθ + B sin(λθ Θ( = A = Θ( = B sin(λ = (6.8 λ n = (nπ/ n = 1,,... Θ n = sin nπθ. (6.9 To solve the Euler Eq. let R(r = r γ, R = γr γ 1, R = γ(γ 1r γ. γ(γ 1 + γ λ = γ λ = γ = ±λ. (6.1 R(r = c 1 r λ + c r λ. (6.11 Now since u(r, θ < as r we require c =. u(r, θ = nπθ sin (6.1 { u(a, θ = f(θ = c n a ( nπ } nπθ sin. (6.13 This is just a Fourier Sine Series for f(θ: c n a ( nπ = c n = nπ a ( nπθ f(θ sin dθ (6.14 f(θ sin nπθ dθ. (6.15

4 4 u(, θ = nπθ sin. (6.16 Eample 6. A wedge with Inhomogeneous BC Figure. Inhomogeneous Dirichlet Boundary conditions on a wedge shaped domain (6.18 u rr + 1 r u r + 1 r u θθ = < r < a, < θ < (6.17 u(r, = u, u(r, = u 1, u(r, θ < as r, u(a, θ = f(θ (6.18 Let us look for the simplest function of θ only that satisfies the inhomogeneous BC of the from: w(θ = (u 1 u θ +u. Note that w θθ = and that w( = u and w( = u 1. Then let u(r, θ = w(θ + v(r, θ. u rr + 1 r u r + 1 r u θθ = v rr + 1 r v r + 1 r v θθ = The solution is where v(r, = v(r, = v(a, θ = f(θ w(θ u(r, θ = (u 1 u θ + u + c n = nπ a ( Essentially the problem solved in Eample 6.1 nπθ sin [ f(θ w(θ ] sin ( nπθ Eample 6.3 A wedge with insulating BC on θ = and θ = < π. (6.19 (6. dθ. (6.1 u rr + 1 r u r + 1 r u θθ = u θ (r, =, u θ (r, =, u(a, θ = f(θ (6. Let u(r, θ = R(rΘ(θ r ( R + 1 r R /R(r = Θ /Θ = λ (6.3

5 Laplace s Equation 5 Figure 3. Mied Boundary conditions on a wedge shaped domain (6. Θ equation Θ + λ Θ = Θ ( = = Θ ( R equation r R n + rr n λ nr n =. } Θ(θ = A cos λθ + B sin(λθ Θ ( = Bλ = λ = or B = ; n = : rr + R = (rr = rr = d R (r = c + d ln r. n 1: r R n + rr n λ R n = R n = c n r λ n + D n r λ n. (6.4 Θ (θ = Aλ sin(λθ + Bλ cos(λθ Θ ( = Aλ sin(λ = λ n = nπ ; n =, 1,... (6.5 Since u(r, θ < (i.e. must be bounded as r we require d = = D n. u(r, θ = c + nπθ cos f(θ = u(a, θ = c + c n a ( nπ nπθ cos c = f(θdθ c n = nπ a ( (6.6 (6.7 nπθ f(θ cos dθ (6.8 u(r, θ = c + nπθ cos. (6.9 Eample 6.4 Mied BC - a crack like problem. u = u rr + 1 r u r + 1 r u θθ = (6.3 subject to u u(r, = (r, π = (6.31 θ u(a, θ = f(θ.

6 6 Figure 4. Mied crack-like boundary conditions on a circular domain as prescribed in (6.31 Let u(r, θ = R(rΘ(θ. Θ equation Θ + λ Θ = Θ( = Θ (π = r ( R + 1 r R R = Θ (θ Θ(θ = λ (6.3 Θ = A cos λθ + B sin λθ Θ = Aλ sin λθ + Bλ cos λθ Θ( = A = Θ (π = Bλ cos(λπ = πλ 1 = π, 3π,... (6.33 or λ n = (n n =, 1,... λ as this would be trivial. R equation r R + rr λ R = R(r = r γ γ λ = γ = ±λ. u n (r, θ = ( c n r λ n + d n r λ n sin λ n θ. (6.34 Since u should be bounded as r we conclude that d n =. The general solution is thus (n + 1 u(r, θ = c n r (n+1/ sin θ (6.35 n= ( n + 1 f(θ = u(a, θ = c n a (n+1/ sin θ. (6.36 Check orthogonality π ( m + 1 sin θ sin n= c n = a (n+ 1 π u(r, θ = n= (( n + 1 π { m n θ dθ = π/ m = n. (6.37 (( f(θ sin n + 1 θ dθ (6.38 (( c n r (n+ 1 sin n + 1 θ (6.39