-Δ u = λ u. u(x,y) = u 1. (x) u 2. (y) u(r,θ) = R(r) Θ(θ) Δu = 2 u + 2 u. r = x 2 + y 2. tan(θ) = y/x. r cos(θ) = cos(θ) r.

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1 The Laplace opeato in pola coodinates We now conside the Laplace opeato with Diichlet bounday conditions on a cicula egion Ω {(x,y) x + y A }. Ou goal is to compute eigenvalues and eigenfunctions of the Laplace opeato on this egion. -Δ u λ u As befoe, we seek a solution by the method of sepaation of vaiables. Howeve, given the geomety of the egion it does not make sense to seek solutions of the fom u(x,y) u 1 (x) u (y) Instead, we will change to pola coodinates and seek solutions of the fom u(,θ) R() Θ(θ) Given the geomety of the poblem at hand, this is a moe natual coodinate system to use, and as we will see, the poblem does sepaate cleanly into ODEs in and θ. Ou fist task it to ewite the Laplace opeato in pola coodinates. Initially we have that Δu u + u x y To change coodinates fom (x,y) to (,θ) we have to intoduce change of vaiable fomulas and use the chain ule to ewite x and y deivatives as and θ deivatives. Hee ae some of the details. x + y tan(θ) y/x To stat the pocess we develop some basic patial deivative facts. x x x + y cos(θ) cos(θ) y y x + y sin(θ) sin(θ) θ cos (θ) - x ( sec (θ) θ y - x x sin(θ) cos (θ) sec (θ) θ 1 y x ) - sin(θ) 1

2 θ y cos (θ) We ae now in position to stat applying the chain ule: 1 cos(θ) cos(θ) u u x x + u θ u cos(θ) + u θ x θ ( - sin(θ) ) u u y y + u θ u sin(θ) + u cos(θ) θ y θ ( ) The highe ode deivatives poceed similaly. I efe you to the text fo futhe details. At the end of the pocess we lean that u x + u u + y 1 u + θ 1 u Sepaation of Vaiables We ae now in position to implement the sepaation of vaiables. Once again, we want to compute eigenvalues and eigenfunctions of the Laplace opeato: To apply sepaation of vaiables we assume that - Δ u λ u u(,θ) R() Θ(θ) and substitute into the pola fom of the Laplace equation: - Δ u - u - 1 u - Θ(θ) d R() d - θ 1 u - R() Θ(θ) - 1 R() d Θ(θ) - 1 R() Θ(θ) - θ 1 R() Θ(θ) 1 Θ(θ) dr() λ u λ R() Θ(θ) d Dividing both sides by a facto of R() Θ(θ) multiplying both sides by gives R() d R() - 1 d Θ(θ) - Θ(θ) d R() dr() d λ If we move all of the tems that depend on and R() to one side and all of the tems that depend on θ and Θ(θ) to the othe we get - R() d R() d + R() dr() d + λ - 1 d Θ(θ) Θ(θ) The only way fo these two expessions to equal fo all possible values of and θ is to have them both equal a constant, γ.

3 - 1 d Θ(θ) Θ(θ) γ - R() d R() d + R() dr() d + λ γ Fo convenience, these two equations ae typically ewitten as d Θ(θ) + γ Θ(θ) 0 d R() d + 1 dr() d + ( λ - γ ) R() 0 The poblem is now fully sepaated. The fist of these two poblems is the easie to wok with. The appopiate bounday conditions to apply to this poblem state that the function Θ(θ) and its fist deivative with espect to θ ae peiodic in θ: The ODE Θ(-π) Θ(π) dθ(-π) dθ(π) d Θ(θ) + γ Θ(θ) 0 has eigenvalues γ 0 with associated eigenfunction and γ n with associated eigenfunctions Θ(-π) Θ(π) dθ(-π) dθ(π) Θ(θ) 1 Θ(θ) sin(n θ) Θ(θ) cos(n θ) As a side effect of solving the Θ(θ) poblem we have now detemined all of the legal values of γ: We can now focus ou attention on the R() equation. γ n fo n 0, 1,, 3

4 d R() d + 1 dr() d + λ - n ( ) R() 0 The standad technique fo solving this equation stats by multiplying both sides of the equation by a facto of. d R() d + dr() d + (λ - n ) R() 0 Next, and fo easons that will become clea below, we intoduce a simple change of vaiables. s λ With this change of vaiables we will ewite the equation in tems of a function S(s) R() R( s ) λ The equation uses vaious deivatives of R(), so we will have to detemine what happens to those deivatives once we make the change of vaiables. We now have d R() d dr() d d dr() d dr() d d ds(s) d S(s) ( λ ) d λ S(s) λ ( s λ ) λ d S(s) + s λ ds(s) λ + (s - n ) S(s) 0 s d S(s) + ds(s) s + (s - n ) S(s) 0 This equation is a well-known ODE, the Bessel equation of ode n. Solving the Bessel Equation The Bessel equation can be solved by assuming that the solution takes the fom Substituting this into the equation gives s (k+α)(k+α-1)s k+α- + s S(s) s α s k (k+α) s k+α-1 + s s k+α - n s k+α 0 To facilitate consolidating these sums into a single sum we shift the indices of summation in the thid tem. 4

5 (k+α)(k+α-1)s k+α + (k+α) s k+α + - s k+α - n s k+α 0 If we conside only tems with the fom s α we see that o k a 0 (α (α-1) + α - n ) 0 α - n 0 This tells us that solutions must have the fom o S(s) s n s k S(s) s -n s k The latte fom does not poduce easonable solutions, since fo n 1 this function is unbounded at s 0. Thus we will continue with the assumpation that α n. With this assumption we have that (k+n)(k+n-1)s k+n + (k+n) s k+n + - s k+n - n s k+n 0 Consideing tems with factos of s n+1 we see that o o Fo tems beyond s n+ we have o k a 1 ((n+1 ) n + ( n+1 ) - n ) 0 a 1 ( n + 1) 0 a 1 0 ((k+n)(k+n-1) + (k+n) - n ) ((k+n) - n ) This lea to a ecuence elation that allows us to compute the coefficients in tems of -. 5

6 - - - (k+n) - n - k (k + n) Since a 1 0, the ecuence elation tells us that 0 fo all odd k. Fo the even k we have that a a - 0 ( + n) Moe geneally, fo k j we have that a a 4-4 (4 + n) a a (6 + n) a 0! 4 (n+1)(n+) a 0 3! 6 (n+3)(n+)(n+1) a j ( -1 ) j a 0 j! j (n+1)(n+) (n+j) Since we ae fee to pick a 0, we select fo ou convenience a 0 1 n n! so that a j (-1) j j+n j! (n+j)! This now gives S(s) (-1) j j+n s j 0 j!(n+j)! ( ) This function is known as the Bessel function of ode n, usually witten J n (s). Eigenvalues and Eigenfunctions of the Laplacian Following the easoning above we have that u(,θ) R() Θ(θ) whee Θ(θ) sin(n θ) o Θ(θ) cos(n θ) 6

7 and R() S(s) J n (s) If we ae going to equie that u(a,θ) 0 we must have that R(A) S( λ A) J n ( λ A) 0 o that λ A must be a zeo of J n (s). It tuns out that fo each value of n 0 the Bessel function J n (s) has an infinite sequence s n,m of zeoes. Coesponding to each of these zeoes is an eigenvalue of the Laplacian: λ n,m A s n,m (s λ n,m n,m ) A Coesponding to each of these eigenvalues ae two eigenfunctions φ (1) s n,m(,θ) J n,m n cos(n θ) ( A ) φ () s n,m(,θ) J n,m n sin(n θ) ( A ) In the n 0 case we have instead φ 0,m (,θ) J 0 ( A s 0,m ) Othogonality Popeties of the Eigenfunctions Since the opeato -Δ is a symmetic opeato, eigenfunctions coesponding to distinct eigenvalues will have to be othogonal. It is also possible to show that the eigenfunctions with the same eigenvalues ae also othogonal. ( φ(1) n,m(,θ), φ () n,m(,θ) ) 0 Solving PDEs on the disk We ae now in a position to solve PDEs containing the Laplace opeato on the disk. Conside fo example the Poisson equation with Diichlet bounday conditions on the disk. -Δ u f(,θ) Using the method of eigenfunction expansions, we stat by assuming that u(,θ) a 0,m φ 0,m (,θ) + m 1 n 1 m 1 ( a n,m φ (1) n,m(,θ) + b n,m φ () n,m(,θ) ) Since φ (1) n,m(,θ) and φ () n,m(,θ) ae both eigenfunctions of the negative Laplacian with eigenvalues of λ n,m we 7

8 have that -Δ u a 0,m λ 0,m φ 0,m (,θ) + m 1 n 1 m 1 λ n,m( a n,m φ (1) n,m(,θ) + b n,m φ () n,m(,θ) ) To solve fo the coefficients a n,m and b n,m we need to constuct the eigenfunction expansion fo the focing function f(,θ) c 0,m φ 0,m (,θ) + m 1 n 1 m 1 ( c n,m φ (1) n,m(,θ) + d n,m φ () n,m(,θ) ) The coefficients c n,m and d n,m ae computed in the usual way via the othogonality popeties of the eigenfunctions. c 0,m ( f,φ 0,m(,θ) ) ( φ 0,m(,θ), φ 0,m (,θ) ) c n,m ( f,φ(1) n,m(,θ) ) ( φ(1) n,m(,θ), φ (1) n,m(,θ) ) d n,m ( f,φ() n,m(,θ) ) ( φ() n,m(,θ), φ () n,m(,θ) ) Once we have computed these coefficients we can solve fo the coefficients a n,m and b n,m. a 0,m c 0,m A c 0,m λ 0,m ( s 0,m) a n,m c n,m A c n,m λ n,m ( s n,m ) d b n,m n,m A d n,m λ n,m ( s n,m ) 8