Higher dimensiona PEs and mutidimensiona eigenvaue probems 1 Probems with three independent variabes Consider the prototypica equations u t = u (iffusion) u tt = u (W ave) u zz = u (Lapace) where u = u xx + u yy (reca there are no t derivatives in the Lapacian). The domain for u(x, y, ) wi be a generaized cyinder where (x, y) and t > 0 or z (a, b). The panar region wi be bounded and have a smooth boundary. We hande a three cases together because one variabe can be separated out, eading to exacty the same eigenvaue probem in the x, y-variabes. Whie a variety of homogeneous boundary conditions on are possibe, we wi ony consider the two most common, u(x, y, ) = 0, (x, y) (irichet) (1) u(x, y, ) ˆn = 0, (x, y) (Neumann) (2) where ˆn is the norma vector to the boundary. Let s ook for soutions of the form u = T (t)v(x, y) (for the diffusion or wave equation) or u = Z(z)v(x, y) (for Lapace s equation). After separating variabes, we get T T T T Z Z = v v = λ (iffusion) = v v = λ (W ave) = v v = λ (Lapace) Therefore the resuting eigenvaue probem for v is the same in each case, v + λv = 0, pus boundary conditions (1) or (2). (3) For the time being, suppose that we aready know the eigenfunctions v n (x, y) and corresponding eigenvaues λ n, n = 1, 2, 3,... (in practice, it wi be easier to enumerate these using two indices). We wi show that with either of the above boundary conditions, is a sef adjoint operator. As a consequence, the eigenvaues are rea and the eigenfunctions are orthogona with respect to the inner product u, v = u(x)v(x) dx. (4) We wi aso show that the eigenvaues happen to be non-negative. The OEs impied by separation of variabes for T and Z and their soutions are the same as in the simper, two variabe case: T = λt, T = exp( λt) (iffusion) T = λt, T = sin( λt), cos( λt) (W ave) Z = λz, Z = exp(± λt) (Lapace). 1
Taking a superposition of a possibe separated soutions, the most genera soutions are u(x, y, t) = u(x, y, t) = u(x, y, z) = A n exp( λ n t)v n (x, y) n=1 (iffusion) [A n cos( λ n t) + B n sin( λ n t)]v n (x, y) (W ave) n=1 [A n exp( λ n z) + B n exp( λ n z)]v n (x, y) (Lapace) n=1 The main issue, therefore, is to sove the eigenvaue probem. This is in genera impossibe by hand uness the domain of interest has some symmetry. Our exampes incude three cases, where is a rectange, a disk, and the surface of a sphere. 2 Sef adjointness and properties of the Lapacian The Lapacian turns out to be sef-adjoint regardess of the domain we are working with, provided the boundary conditions are suitabe. To find the adjoint, we first need the equivaent of integration by parts in higher dimensions. 2.1 Green s identity Let u, v : R be smooth functions. Appying the divergence theorem to the vector fied u v, (u v) dx = u v ˆn dx. (5) Using the mutidimensiona product rue (u v) = u v + u v, in (5), one obtains u v dx = u v dx + u v ˆn dx. (6) This is known as Green s identity, or sometimes just Green s theorem. Note this resut is just ike integration by parts for definite integras, where derivatives are moved around inside an integra, whie paying the price of a boundary term. 2.2 Sef adjointness We can use (6) to compute the adjoint of, which is regarded as an operator acting on smooth functions (for exampe C ()) which have either type of boundary condition (1) or (2). The inner product used here is the standard L 2 one (4). It foows by using Green s identity twice that u, v = v u dx = v u dx + v u ˆn dx = (7) = u v dx u v ˆn dx = u, v. The integras on the boundary vanish because of the boundary conditions. The Lapacian is therefore sef-adjoint under either irichet or Neumann boundary conditions, just ike its one dimensiona counterpart xx. 2
2.3 Non-negativity of the eigenvaues It is often possibe to assess quantitative properties of eigenvaue probems without actuay soving them. One way to do this is with the Rayeigh quotient, which is formed by taking the inner product of the eigenvaue equation Lv + λv = 0 with v, resuting in a formua Lv, v λ = v, v. (8) The expression on the right depends on the unknown eigenfunction v, so it s hard to imagine how this might be usefu. If we speciaize to our situation, however, we have v v dx λ = v2 dx = v 2 dx v2 dx, where the Green s identity was used. The expression on the right is certainy non-negative, regardess of what v might be. Notice that if λ = 0, then it must be that v = 0, so that the eigenfunction is a constant. This can ony happen in the case of the Neumann boundary condition (2); for the irichet boundary condition (1), eigenvaues are stricty positive. There are other uses of Rayeigh quotient (8). For exampe, the minimum vaue this expression can take for any function v 0 (not just eigenfunctions) is equa to the smaest eigenvaue. Minimization of this expression is the basis for the Rayeigh-Ritz agorithm for computation of eigenvaue probems. 3 Eigenfunctions on the rectange We now speciaize to the region = {0 < x < π, 0 < y < π}. The natura idea to sove the eigenvaue probem is (of course!) separation of variabes. Letting v(v, y) = X(x)Y (y), pugging into (3) and separating gives X X + Y = λ (9) Y It foows that each term on the eft must be a constant, since the right hand side does not depend on either x or y. If those constants are λ x, λ y, respectivey, one gets X + λ x X = 0, Y + λ y Y = 0, λ = λ x + λ y. (10) Suppose that irichet boundary conditions on were imposed. This eads to the boundary conditions X(0) = 0 = X(π), Y (0) = 0 = Y (π). The eigenvaue probems for X and Y are famiiar, eading to eigenfunctions X = sin(nx) and Y = sin(my) for any n, m = 1, 2, 3,.... It foows that soutions to the mutidimensiona eigenvaue probem can be enumerated using a doube index v nm (x, y) = sin(nx) sin(my), λ nm = n 2 + m 2, n, m = 1, 2, 3,.... A superposition of these functions by themseves give one version of a mutidimensiona Fourier series. Since the underying operator is sef-adjoint, orthogonaity hods: v nm v n m dxdy = 0 uness n = n and m = m. 3
Figure 1: Noda sets (eve curves where v = 0) for the first severa eigenfunctions on a square domain with irichet boundary conditions. It is not too hard to extend this exampe to other boundary conditions and higher dimensions. For exampe, with the Neumann boundary condition v ˆn = 0, the boundary conditions for X, Y become X (0) = 0 = X (π), Y (0) = 0 = Y (π). We have soved the corresponding eigenvaue probems previousy; the soutions are λ x = n 2 for n = 0, 1, 2, 3,... and X = cos(nx), with simiar resuts for Y. Thus the two-dimensiona eigenfunctions are v nm (x, y) = cos(nx) cos(my), λ nm = n 2 + m 2, n, m = 0, 1, 2, 3,.... Notice that for this boundary condition, we get a constant eigenfunction corresponding to λ = 0. To provide some intuition about eigenfunctions, it is often usefu to visuaize them in various ways. A common way is to graph the nodes, which are the eve curves where v(x, y) = 0. The nodes for the first severa eigenfunctions on the square domain are given in figure (1). 4 Eigenfunctions on the disk Now et = {0 < r < a, 0 < θ < 2π}. In poar coordinates, (3) is v rr + 1 r v r + 1 r 2 v θθ = λv. 4
Looking for soutions of the form v = R(r)Θ(θ), separating variabes gives r 2 R + rr + λr 2 R R + Θ Θ = 0. (11) Again each term must be a constant; in particuar we can set Θ /Θ = λ θ. The boundary conditions for Θ are periodic, so the soutions are immediatey Θ = cos(nθ), sin(nθ), λ θ = n 2, n = 0, 1, 2,... The equation in R gives something new. Setting Θ /Θ = λ θ = n 2 in (11), we get r 2 R + rr + (λr 2 n 2 )R = 0 (12) In the case λ = 0, we get Euer s equation (reca the genera soution is R(r) = Ar n + Br n or R(r) = A + B n n if n = 0). For the irichet boundary condition, none of these soutions are admissibe, whereas for the Neumann boundary condition, we can have constant-vaued soutions. If λ > 0, we can simpify the probem a itte by a change of variabes ρ = λr. Thus R (r) = λr (ρ) and R (r) = λr (ρ), and R(ρ) soves ρ 2 R + ρr + (ρ 2 n 2 )R = 0. (13) This is known as Besse s equation of order n. Notice that the parameter λ has been eiminated; it wi re-appear when boundary conditions are invoked. Unfortunatey, (13) does not possess cosed form soutions. To get an idea of what soutions ook ike, et s approximate near ρ = 0. We can t eiminate the first two terms in (13) since we have no idea if the smaness of ρ is counteracted by the possibiity that R and R are arge. On the other hand, in the third term ρ is sma compared to n, except if n = 0. In this case, it turns out that ρ 2 R is sti smaer than the other terms, aowing us to ignore it. The approximate equation is then just ρ 2 R + ρr n 2 R = 0, which is just the Euer equation. Therefore for each n we expect two ineary independent soutions we ca J n (ρ) and Y n (ρ) having the behavior J n (ρ) { ρ n n > 0 1 n = 0, Y n (ρ) { ρ n n > 0 n ρ n = 0, as ρ 0 (note n does not index eigenvaues, but is simpy the order of the Besse equation). These soutions are caed Besse functions. The eigenvaues, of course, depend on boundary conditions. Take for exampe the irichet condition v(a, θ) = 0, which means R(r = a) = 0. Since the Besse functions Y n are singuar at the origin, they can be ignored for now. We need to know where the Besse functions J n have zeros. There is no simpe formua for these; we dea with this issue by simpy inventing notation β nm = mth positive zero of J n (ρ). (note that Haberman s text uses z mn instead). It turns out that each Besse function J n has an infinite number of zeros, so m = 1, 2, 3,.... (in fact, it is hepfu to think of Besse functions as distorted versions of cos(ρ)). It foows that the eigenvaues λ are determined by R(r = a) = J n ( λa) = 0. This means that for each n, λa must be a zero of J n, and the eigenvaues are λ nm = ( βnm a ) 2, 0 n <, 1 m <. 5
The corresponding eigenfunctions in origina variabes are { J 0 (β 0m r/a) n = 0 v(x, y) = J n (β nm r/a) cos(nθ), J n (β nm r/a) sin(mθ), n > 0. Other eigenvaue probems have simiar resuts. For exampe, in the case of Neumann boundary conditions, one needs the zeros of dj/dρ, not J n itsef. 4.1 No reay, what are the Besse functions? Unfortunatey, many soutions to differentia equations simpy cannot be written in terms of eementary functions. An aternative is to use power series; in our case we ook for soutions of Besse s equation of the form R = ρ α a k ρ k, a 0 0, k=0 where α and a n are to be determined. Pugging into equation (13) and re-indexing the sum on the ast term gives { } ρ α [(α + k)(α + k 1) + (α + k) n 2 ]a k ρ k 2 + a k 2 ρ k 2 = 0. k=0 Each coefficient of ρ k 2 must equa zero for this to be true. Thus for k = 0, 1, k=2 [α 2 n 2 ]a 0 = 0, [(α + 1) 2 n 2 ]a 1 = 0. The first equation impies that α is either n or n, so as caimed R ρ n or R ρ n when ρ is sma. The second equation impies a 1 = 0. In genera, the coefficients can be found recursivey by soving a k 2 Putting this a together gives for α = +n gives a k = (α + k) 2, k = 2, 3, 4,... n2 J n (ρ) = ( 1) k (ρ/2)n+2k k!(n + k)! k=0 where the conventiona choice a 0 = 2 n /n! was made. By the eementary ratio test, the series converges for a ρ. This is sti not a very satisfactory situation, since the power series is a mess. The best we can do is simpy utiize the notation J n, Y n which encapsuates a this compexity. In practice, there are severa other ways to quantify Besse functions, incuding integra representations, approximations, and computationa packages ike MATLAB. There are aso a miion identities concerning Besse functions and their derivatives (see, for exampe, Wikipedia!). 6
Figure 2: Spherica coordinates used here. 4.2 Orthogonaity Notice that Besse functions J n (β nm r/a) are themseves eigenfunctions of a differentia operator, namey L = d2 dr 2 + 1 d r dr n2 r 2. (14) Its not hard to show that this operator is sef-adjoint with respect to the inner product u, v = a 0 u(r)v(r)rdr. Notice the weight r in the integrand, which shoud come as no surprise, since the operator (14) arises from the use of poar coordinates. It foows that the scaed Besse functions are orthogona: J n (β nm r/a), J n (β nk r/a) = 0, m k. (note that n needs to be the same since different n impies different operators!). 5 Spherica harmonics In spherica coordinates (r, φ, θ) (see figure 2), Lapace s equation is u = u rr + 2 r u r + 1 [ uφφ r 2 sin 2 (θ) + 1 ] sin θ (sin θ u θ) θ = 0. (15) Here 0 φ < 2π is the ongitudina variabe, whereas 0 θ < π measures the atitude where θ = 0 is the north poe (note that in some texts, this notation is opposite). Suppose we want soutions to u = 0 of the form u = R(r)v(φ, θ). After separating out the r variabe, we are eft with an eigenvaue probem for v(φ, θ) s v + λv = 0, s v v φφ sin 2 (θ) + 1 sin θ (sin θ v θ) θ. The operator s is the spherica Lapacian (which is aso known as the Lapace-Betrami operator for the surface of a sphere). Its eigenfunctions wi be derived next. 5.1 Eigenfunctions on the surface of a sphere Letting v = p(θ)q(φ), separating variabes gives q ( (φ) sin θ(sin θ p q(φ) + (θ)) ) + λ sin 2 θ = 0. (16) p(θ) 7
The terms invoving φ and θ must therefore each equa constants λ q and λ p, which sum to 0. The q equation is q + λ q q = 0, equipped with periodic boundary conditions q(0) = q(2π) and q (0) = q (2π). We therefore immediatey know the soutions, q = cos(mφ), sin(mφ), λ q = m 2, m = 0, 1, 2,.... If u is compex vaued, the eigenfunctions are q = exp(imφ), m = 0, ±1, ±2,... instead. Then λ p = λ q = m 2, so the equation for p in (16) reads 1 sin θ (sin θ p ) + (λ m2 ) sin 2 p = 0. θ This appears very unfavorabe, but a ucky change of variabes s = cos θ heps: [(1 s 2 )p ] + [λ m 2 /(1 s 2 )]p = 0, (17) which is known as the associated Legendre s equation. The boundary conditions on p simpy come from the fact that the compete soution must be bounded at θ = 0, π. In the s variabe, this impies p(s = ±1) is bounded. First set m = 0. As with the Besse equation, we ook for a soution of the form Substituting into the equation, we have k=2 p(s) = a k s k. k=0 (k + 2)(k + 1)a k+2 s k (k 2 + k λ)a k s k = 0. k=0 Equating coefficients of s k eads to the recursion reation a k+2 = a k k(k + 1) λ (k + 2)(k + 1). Notice if λ = ( + 1) for some = 0, 1, 2, 3,... something remarkabe happens: the series soution has zero coefficients for k >. In other words, we get soutions which are Legendre poynomias, denoted P (s), and the eigenvaues are λ = ( + 1). The first three are P 0 =, P 1 = s and P 2 = 1 2 (3s2 1). What if λ ( + 1) for some positive integer? The ratio of successive coefficients a k+2 /a k in the power series goes to one, and therefore the power series has an interva of convergence ( 1, 1). This means that the series must diverge at s = ±1, which can t happen since the eigenfunctions must be bounded there. Now for m > 0. It turns out the soution of the associated Legendre equation (17) can be written (remarkaby!) in terms of the Legendre poynomias themseves by p(s) = P m (s) (1 s 2 m/2 dm ) ds m P (s). The functions P m (s) are caed the associated Legendre functions. Notice that if m >, the right hand side is zero since P is a poynomia. Thus we have a funny enumeration scheme for the 8
m P m Y m 0 0 1 1 1 0 s = cos θ cos θ = z 1 1 (1 s 2 ) 1/2 = sin θ e iφ sin θ = x+iy r 2 0 1 2 (3s2 1) = 1 2 (3 cos2 θ 1) 1 2 (3 cos2 θ 1) = 1 2 2z 2 x 2 y 2 r 2 2 1 3s(1 s 2 ) 1/2 = 3 cos θ sin θ 3e iφ cos θ sin θ = 3 (x+iy)z r 2 2 2 3(1 s 2 ) = 3 sin 2 θ 3e 2iφ sin 2 θ = 3 (x+iy)2 r 2 Tabe 1: Some of the associated Legendre functions P m (s), where s = cos θ, and corresponding spherica harmonics Y m (φ, θ). eigenfunctions, specificay = 0, 1, 2,... and m = 0, 1, 2,...,. The first severa are given in tabe 1. Now et s put it a together. The compete set of eigenfunctions is cos(mφ)p m (cos θ), sin(mφ)p m (cos θ), = 0, 1, 2,... m = 0, 1, 2,...,, with corresponding eigenvaues λ = ( + 1). Often, the eigenfunctions are written in compex form, which gives Y m (φ, θ) = e imφ P m (cos θ), = 0, 1, 2,... m = 0, ±1, ±2,..., ±. These are the famous spherica harmonics, the first severa of which are given in tabe 1. Notice that for each eigenvaue ( + 1), there are 2n + 1 ineary independent eigenfunctions. The set Y m, m forms one such set, but others are possibe. As with Besse functions, a ot of compexity can be hidden inside the notation Y m. Notice that we can work directy with this notation in some cases, for exampe s Y m = ( + 1)Y m simpy because the spherica harmonics are eigenfunctions of s. Additionay, the operator s is (unsurprisingy) sef-adjoint if one uses the inner product v 1, v 2 = π 2π 0 0 v 1 (φ, θ)v 2 (φ, θ) sin θdφdθ. (18) It foows that < Y m, Y m >= 0 uness = and m = m by orthogonaity. Figure 3 shows a visuaization of some of the spherica harmonics. Their noda ines are curves on the surface of a sphere, and intersect orthogonay much ike the eigenfunctions for the rectange. 5.2 Lapace s equation inside the sphere We return to equation (15), which is to be soved on the interior of a sphere of radius a, subject to a boundary condition u(a, φ, θ) = f(φ, θ). Reca that separation of variabes u = R(r)v(φ, θ) produced r 2 R 2rR R 9 = sv v = λ
Figure 3: Visuaization of some of the spherica harmonics thus s v + λv = 0. We immediatey have a soution of the eigenvaue probem: v = Y m (φ, θ), λ = ( + 1), = 0, 1, 2,... m = 0, ±1, ±2,..., ±. (even if f(φ, θ) is rea, it is more convenient to use the compex form for the eigenfunctions). For each eigenvaue λ = ( + 1), the radia equation is just r 2 R + 2rR ( + 1)R = 0. (19) This is an Euer equation with soutions of the form R = r α. Substituting into (19) gives α(α 1) + 2α ( + 1) = (α )(α + + 1) = 0, so that α = or α = 1. For domains which contain the origin, soutions with α < 0 must be rejected. A genera soution is given by a superposition of separated soutions u = =0 m= A m r Y m (φ, θ), where A m can be compex vaued. We now set r = a and use the boundary condition to give f(φ, θ) = =0 m= A m a Y m (φ, θ). This is just an orthogona expansion, and the coefficients are determined in the usua way by taking an inner products with each eigenfunction using (18). This eads to A m = 1 a f, Y m Y m, Y m, v 1, v 2 = π 2π 0 0 v 1 (φ, θ)v 2 (φ, θ) sin θdφdθ. 10
5.3 Spherica harmonics in retrospect It seems ike a ot of guesswork went into deriving spherica harmonics, and their characterization is sti somewhat hidden behind the mystery of the Legendre functions. Remarkaby, in terms of Cartesian coordinates, spherica harmonics are easy to understand! To see why this is the case, et P (x) = r Y m (φ, θ) be a separated soution to Lapace s equation, where x is a vector coordinate for (r, φ, θ). P (x) has a specia property it is homogeneous of degree, which means P (αx) = α P (x), α > 0. A we known theorem about homogeneous functions is a foows: If P (x) : R n R is continuous and homogeneous of degree, then P (x) is a poynomia. (the proof is easy: just differentiate P (αx) = α P (x) with respect to α exacty + 1 times). It foows that Y m = P (x) r, P (x) is a homogeneous poynomia of degree which soves Lapace s equation. For = 0, the ony possibiity is P = 1 and Y0 0 = 1. For = 1, there are three ineary independent choices for homogeneous poynomias of degree one: P = x, y, z. These ead to z/r, which is Y1 0, and x/r and y/r which are the rea and imaginary parts of Y1 0. The converse of the idea above is aso true: If P (x) is a homogeneous poynomia of degree which soves Lapace s equation, then P (x/r), where r = x, is an eigenfunction of s with eigenvaue ( + 1). Since P is homogeneous of degree, then P (x/r) = P (x)/r or P (x) = r P (x/r). u = r P (x/r) into (15) gives Pugging [( 1) + 2]r 2 P (x/r) + r 2 s P (x/r) = 0, (20) so that s P (x/r) + ( + 1)P (x/r) = 0 as caimed. Note that in genera, P (x/r) wi not be one of the standard eigenfunctions Y m, but rather a inear combination of them. For exampe, it is easy to see that P = xz is a homogeneous poynomia of degree 2 and P = 0. It foows that P (x/r) = xy/z 2 is an eigenfunction with eigenvaue 2(2 + 1) = 6. It happens to be a inear combination of the spherica harmonics xz r 2 = 1 1 (Y2 + Y2 1 ). 6 11