INVARIANCE OF THE LAPLACE OPERATOR. The goal of this handout is to give a coordinate-free proof of the invari- n ance of the Laplace operator under orthogonal transformations of R (and to explain what this means). n 2 Theorem. Let D ⊂ R be an open set, f 2 C (D), and let U 2 O(n) be an orthogonal linear transformation leaving D invariant (U(D) = D). Then: ∆(f ◦ U) = (∆f) ◦ U: Before giving the proof we recall some basic mathematical facts. Orthogonal linear transformations. An invertible linear transformation n ∗ ∗ −1 ∗ U 2 L(R ) is orthogonal if U U = UU = In (equivalently, U = U .) n ∗ n Recall that, given A 2 L(R ), A 2 L(R ) is defined by: ∗ n A v · w = v · Aw; 8v; w 2 R : n Orthogonal linear transformations are isometries of R ; they preserve the inner product: n Uv · Uw = v · w; 8v; w 2 R ; if U 2 O(n); and therefore preserve the lengths of vectors and the angle between two vec- tors. 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 n t basis of R is an orthogonal matrix: U U = In, so that the columns of U n (or the rows of U) give an orthonormal basis of R . det(U) = ±1 if U 2 O(n). The orthogonal transformations with deter- n minant one can be thought of as rotations of R . If n = 2, they are exactly the rotation matrices Rθ: 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 n Change of variables in multiple integrals. Let A 2 L(R ) be an invertible n linear transformation. Then if D ⊂ R and f : A(D) ! R is integrable, then so is f ◦ A and: Z Z fdV = (f ◦ A)jdetAjdV: A(D) D (This is a special case of the change of variables theorem.) n Multivariable integration by parts. Let D ⊂ R be a bounded open set 1 2 1 with C boundary, f 2 C (D), g 2 C0 (D). Then: Z Z (∆f)gdV = − rf · rgdV: D D 1 1 Here C0 (D) denotes the set of C functions in D which are zero near the boundary of D. This follows directly from Green's first identity, which in turn is a direct consequence of the divergence theorem. n Note that given D ⊂ R open and P 2 D, it is easy to find a function 1 ' 2 C0 (D) so that ' ≥ 0 everywhere and '(P ) = 1. For example, one could let: 2 2 'P (x) = maxf0; 1 − jjP − xjj / g: This works for > 0 small enough, since 'P is positive only in a ball of radius centered at P (which does not touch the boundary of D if is small.) This can be used to observe that if h is continuous in D, then h ≡ 0 in 1 D if, and only if, for any ' 2 C0 (D) we have: Z h'dV = 0: D Indeed if h(P ) 6= 0 for some P 2 D we have (say) h(P ) > 0, hence h > 0 in a sufficiently small ball B centered at P (by continuity of h). But 1 then h'P ≥ 0 everywhere in D and yet (since 'P 2 C (D)) we must have R 0 D h'P = 0, so that h'P ≡ 0 in D, contradicting the fact that it is positive on B. Proof of Theorem. By the observation just made, it is enough to show 1 that, for all ' 2 C0 (D): Z Z ∆(f ◦ U)'dV = [(∆f) ◦ U]'dV: D D 2 By multivariable integration by parts, for the left-hand side: Z Z Z ∆(f ◦ U)'dV = − r(f ◦ U) · r'dV = − [(rf)U] · r'dV; D D D where we also used the chain rule to assert that r(f ◦ U) = (rf)U. On the other hand, for the right-hand side: Z Z Z (∆f) ◦ U'dV = [(∆f)(' ◦ U −1)] ◦ UdV = (∆f)(' ◦ U −1)dV; D D D using the change of variables theorem and the facts that D is invariant under U and j det Uj = 1. Again from integration by parts and the chain rule we have: Z Z Z (∆f)(' ◦ U −1)dV = − rf · r(' ◦ U −1)dV = − rf · [(r')U t]dV; D D D where we also use the fact that U −1 = U t. Thus we've reduced the proof to showing that: Z Z [(rf)U] · r' = rf · [(r')U t]dV: D D But this follows from the (easily verified) fact that for any n × n matrix A we have: t n (vA) · w = v · (wA ); 8v; w 2 R : Application: Laplacian of radial functions. A function f : B ! R n (where B = fx 2 R ; jxj ≤ Rg is a ball of some radius) is radial if it is invariant under the orthogonal group: f = f ◦ U, for all U 2 O(n). This implies f is a function of distance to the origin only, that is (abusing the notation): f = f(r); if x = r! with r = jxj;! 2 S (the unit sphere): The theorem implies the Laplacian of a radial function is also radial: f = f ◦ U 8U ) ∆f = ∆(f ◦ U) = (∆f) ◦ U 8U; or (∆f)(r!) = g(r), for some function g(r) depending on f, which we now compute. 3 From the divergence theorem on a ball BR of radius R centered at 0: Z Z @f ∆fdV = dS; BR SR @n or: Z R Z Z n−1 n−1 g(r)r d!dr = fr(R)R d!; 0 S S and since both integrals in ! just give the (n-1)-dimensional area of S: Z R n−1 n−1 g(r)r dr = fr(R)R : 0 Differentiating in R we find: n−1 n−1 n−2 g(R)R = frr(R)R + (n − 1)fr(R)R (we assume n ≥ 2), so finally n − 1 g(r) = ∆f(r) = f + f : rr r r Remark. For general (not necessarily radial) functions we have in polar coordinates x = r!: n − 1 1 ∆f = f + f + ∆ f; rr r r r2 S where ∆S is the spherical Laplacian: cos φ 1 ∆Sf = f (n = 2); ∆Sf = f + f + f (n = 3); θθ φφ sin φ φ sin2 φ θθ in polar coordinates (r; θ) (resp. spherical coordinates (r; φ, θ), φ 2 [0; π]; θ 2 [0; 2π]). Note the analogy between ∆S for n=3 and the Laplacian for n = 2: 1 1 ∆f = f + f + ∆ f (n = 2): rr r r r2 S Making the substitutions: 1 cos φ r ! sin φ; = (log r) ! = (log sin φ) ; ∆ f ! f ; r r sin φ φ S θθ we obtain ∆Sf for n=3. 4 n More generally, the spherical Laplacian ∆Sn on the unite sphere S ⊂ n+1 R can be described inductively in terms of the spherical Laplacian on the n−1 n n unit sphere S ⊂ R . To do this, write ! 2 S in spherical coordinates: as a vector in Rn+1, n n−1 ! = (sin φ, (cosφ)θ) 2 R × R ; θ 2 S ; φ 2 [0; π]: (This represents Sn−1 as the equator φ = 0 in Sn.) In these coordinates, we have, for a function f = f(φ, θ) defined on Sn: cos φ 1 ∆Sn f = f + (n − 1) f + ∆ n−1 f: φφ sin φ φ sin2 φ S Here the operator ∆Sn−1 f involves only differentiation in the variables θ 2 Sn−1. 5.