INVARIANCE OF THE LAPLACE OPERATOR.

Transcription

1 INVARIANCE OF THE LAPLACE OPERATOR. The goal of this handout is to give a coordinate-free proof of the invariance of the Laplace operator under orthogonal transformations of R n (and to explain what this means). Theorem. Let R n be an open set, f C 2 (), and let U O(n) be an orthogonal linear transformation leaving invariant (U() = ). Then: (f U) = ( f) U. Before giving the proof we recall some basic mathematical facts. Orthogonal linear transformations. An invertible linear transformation U L(R n ) is orthogonal if U U = UU = I n (equivalently, U 1 = U.) Recall that, given A L(R n ), A L(R n ) is defined by: A v w = v Aw, v, w R n. Orthogonal linear transformations are isometries of R n ; they preserve the inner product: Uv Uw = v w, v, w R n, if U O(n), and therefore preserve the lengths of vectors and the angle between two vectors. Orthogonal linear transformations form a group (inverse of orthogonal is orthogonal, composition of orthogonal is orthogonal), denoted by O(n). The matrix of an orthogonal transformation with respect to an orthonormal basis of R n is an orthogonal matrix: U t U = I n, so that the columns of U (or the rows of U) give an orthonormal basis of R n. det(u) = ±1 if U O(n). The orthogonal transformations with determinant one can be thought of as rotations of R n. If n = 2, they are exactly the rotation matrices R θ : [ ] cos θ sin θ R θ =. sin θ cos θ If n = 3, one is always an eigenvalue, with a one-dimensional eigenspace (the axis of rotation); in the (two-dimensional) orthogonal complement of this eigenspace, U acts as rotation by θ. 1

2 Change of variables in multiple integrals. Let A L(R n ) be an invertible linear transformation. Then if R n and f : A() R is integrable, then so is f A and: fdv = (f A) deta dv. A() (This is a special case of the change of variables theorem.) Multivariable integration by parts. Let R n be a bounded open set with C 1 boundary, f C 2 (), g C0 1 (). Then: ( f)gdv = f gdv. Here C 1 0 () denotes the set of C1 functions in which are zero near the boundary of. This follows directly from Green s first identity, which in turn is a direct consequence of the divergence theorem. Note that given R n open and P, it is easy to find a function ϕ C0 1 () so that ϕ 0 everywhere and ϕ(p ) = 1. For example, one could let: ϕ P (x) = max{0, 1 P x 2 /ɛ 2 }. This works for ɛ > 0 small enough, since ϕ P is positive only in a ball of radius ɛ centered at P (which does not touch the boundary of if ɛ is small.) This can be used to observe that if h is continuous in, then h 0 in if, and only if, for any ϕ C0 1 () we have: hϕdv = 0. Indeed if h(p ) 0 for some P we have (say) h(p ) > 0, hence h > 0 in a sufficiently small ball B ɛ centered at P (by continuity of h). But then hϕ P 0 everywhere in and yet (since ϕ P C0 1 ()) we must have hϕ P = 0, so that hϕ P 0 in, contradicting the fact that it is positive on B ɛ. Proof of Theorem. By the observation just made, it is enough to show that, for all ϕ C0 1(): (f U)ϕdV = [( f) U]ϕdV. 2

3 By multivariable integration by parts, for the left-hand side: (f U)ϕdV = (f U) ϕdv = [( f)u] ϕdv, where we also used the chain rule to assert that (f U) = ( f)u. On the other hand, for the right-hand side: ( f) UϕdV = [( f)(ϕ U 1 )] UdV = ( f)(ϕ U 1 )dv, using the change of variables theorem and the facts that is invariant under U and det U = 1. Again from integration by parts and the chain rule we have: ( f)(ϕ U 1 )dv = f (ϕ U 1 )dv = f [( ϕ)u t ]dv, where we also use the fact that U 1 = U t. Thus we ve reduced the proof to showing that: [( f)u] ϕ = f [( ϕ)u t ]dv. But this follows from the (easily verified) fact that for any n n matrix A we have: (va) w = v (wa t ), v, w R n. Application: Laplacian of radial functions. A function f : B R (where B = {x R n ; x R} is a ball of some radius) is radial if it is invariant under the orthogonal group: f = f U, for all U O(n). This implies f is a function of distance to the origin only, that is (abusing the notation): f = f(r), if x = rω with r = x, ω S (the unit sphere). The theorem implies the Laplacian of a radial function is also radial: f = f U U f = (f U) = ( f) U U, or ( f)(rω) = g(r), for some function g(r) depending on f, which we now compute. 3

4 or: From the divergence theorem on a ball B R of radius R centered at 0: f fdv = B R S R n ds, R 0 S g(r)r n 1 dωdr = S f r (R)R n 1 dω, and since both integrals in ω just give the (n-1)-dimensional area of S: R ifferentiating in R we find: (we assume n 2), so finally 0 g(r)r n 1 dr = f r (R)R n 1. g(r)r n 1 = f rr (R)R n 1 + (n 1)f r (R)R n 2 g(r) = f(r) = f rr + n 1 f r. r Remark. For general (not necessarily radial) functions we have in polar coordinates x = rω: where S is the spherical Laplacian: f = f rr + n 1 f r + 1 r r 2 Sf, S f = f θθ (n = 2), S f = f φφ + cos φ sin φ f φ + 1 sin 2 φ f θθ (n = 3), in polar coordinates (r, θ) (resp. spherical coordinates (r, φ, θ), φ [0, π], θ [0, 2π]). Note the analogy between S for n=3 and the Laplacian for n = 2: Making the substitutions: f = f rr + 1 r f r + 1 r 2 Sf (n = 2). r sin φ; 1 r = (log r) r cos φ sin φ = (log sin φ) φ, S f f θθ, we obtain S f for n=3. 4

5 More generally, the spherical Laplacian S n on the unite sphere S n R n+1 can be described inductively in terms of the spherical Laplacian on the unit sphere S n 1 R n. To do this, write ω S n in spherical coordinates: as a vector in R n+1, ω = (sin φ, (cosφ)θ) R R n, θ S n 1, φ [0, π]. (This represents S n 1 as the equator φ = 0 in S n.) In these coordinates, we have, for a function f = f(φ, θ) defined on S n : S nf = f φφ + (n 1) cos φ sin φ f φ + 1 sin 2 φ S n 1f. Here the operator S n 1f involves only differentiation in the variables θ S n 1. 5