invLaplacian.pdf

8/11/2019 invLaplacian.pdf

1/5

INVARIANCE OF THE LAPLACE OPERATOR.

The goal of this handout is to give a coordinate-free proof of the invari-ance of the Laplace operator under orthogonal transformations ofRn (andto explain what this means).

Theorem. Let D Rn be an open set, f C2(D), and let U 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 transformationU L(Rn) is orthogonal if UU = U U = In (equivalently, U

1 = U.)Recall that, given A L(Rn), A L(Rn) is defined by:

Av w= v Aw, v, w Rn.

Orthogonal linear transformations are isometriesof Rn; they preserve theinner product:

Uv U w= v w, v, w Rn, ifU O(n),

and therefore preserve the lengths of vectors and the angle between two vec-tors. Orthogonal linear transformations form a group(inverse of orthogonalis orthogonal, composition of orthogonal is orthogonal), denoted by O(n).The matrix of an orthogonal transformation with respect to an orthonormalbasis ofRn is an orthogonal matrix: UtU = In, so that the columns ofU(or the rows ofU) give an orthonormal basis ofRn.

det(U) =1 ifU O(n). The orthogonal transformations with deter-minant one can be thought of as rotations ofRn. Ifn = 2, they are exactlythe rotation matricesR:

R = cos sin

sin cos .

If n = 3, one is always an eigenvalue, with a one-dimensional eigenspace(the axisof rotation); in the (two-dimensional) orthogonal complement ofthis eigenspace, Uacts as rotation by .

1


2/5

Change of variables in multiple integrals. LetA L(Rn) be an invertible

linear transformation. Then if D Rn

and f : A(D) R is integrable,then so is f A and:

A(D)

f dV =

D

(f A)|detA|dV.

(This is a special case of the change of variables theorem.)

Multivariable integration by parts. Let D Rn be a bounded open setwith C1 boundary, f C2(D),g C10 (D). Then:

D

(f)gdV =

D

f gdV.

Here C10(D) denotes the set ofC1 functions in D which are zero near the

boundary of D. This follows directly from Greens first identity, which inturn is a direct consequence of the divergence theorem.

Note that given D Rn open and P D , it is easy to find a function C10(D) so that 0 everywhere and(P) = 1. For example, one couldlet:

P(x) = max{0, 1 ||P x||2/2}.

This works for > 0 small enough, since P is positive only in a ball ofradius centered at P (which does not touch the boundary of D if issmall.)

This can be used to observe that ifh is continuous in D, then h 0 inD if, and only if, for any C10 (D) we have:

D

hdV = 0.

Indeed if h(P) = 0 for some P D we have (say) h(P) > 0, hence h >0 in a sufficiently small ball B centered at P (by continuity of h). Butthen hP 0 everywhere in D and yet (since P C

10 (D)) we must have

DhP= 0, so that hP 0 in D, contradicting the fact that it is positive

on B.

Proof of Theorem. By the observation just made, it is enough to showthat, for all C10 (D):

D

(f U)dV =

D

[(f) U]dV.

2


3/5

By multivariable integration by parts, for the left-hand side:

D

(f U)dV =

D

(f U) dV =

D

[(f)U] dV,

where we also used the chain rule to assert that (f U) = (f)U. On theother hand, for the right-hand side:

D

(f) UdV =

D

[(f)( U1)] U dV =

D

(f)( U1)dV,

using the change of variables theorem and the facts thatD is invariant underU and | det U| = 1.

Again from integration by parts and the chain rule we have:D

(f)( U1)dV =

D

f ( U1)dV =

D

f [()Ut]dV,

where we also use the fact thatU1 =Ut.Thus weve reduced the proof to showing that:

D

[(f)U] =

D

f [()Ut]dV.

But this follows from the (easily verified) fact that for any n n matrix Awe have:

(vA) w= v (wAt), v, w Rn.

Application: Laplacian of radial functions. A function f :B R(where B = {x Rn; |x| R} is a ball of some radius) is radial if it isinvariant under the orthogonal group: f = f U, for all U O(n). Thisimplies f is a function of distance to the origin only, that is (abusing thenotation):

f=f(r), ifx= r with r = |x|, 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)(r) =g(r), for some function g(r) depending on f, which we nowcompute.

3


4/5

From the divergence theorem on a ball BR of radius R centered at 0:BR

f dV =

SR

f

ndS,

or: R0

S

g(r)rn1ddr=

S

fr(R)Rn1d,

and since both integrals in just give the (n-1)-dimensional area ofS:

R0

g(r)rn1dr= fr(R)Rn1.

Differentiating in R we find:

g(R)Rn1 =frr(R)Rn1 + (n 1)fr(R)R

n2

(we assume n 2), so finally

g(r) = f(r) =frr+n 1

r fr.

Remark. For general (not necessarily radial) functions we have in polarcoordinates x= r :

f=frr+

n 1

r fr+

1

r2 Sf,

where Sis the spherical Laplacian:

Sf=f (n= 2), Sf=f+cos

sin f+

1

sin2 f (n= 3),

in polar coordinates (r, ) (resp. spherical coordinates (r,,), [0, ], [0, 2]). Note the analogy between Sfor n=3 and the Laplacian for n = 2:

f=frr+1

rfr+

1

r2Sf (n= 2).

Making the substitutions:

r sin ; 1

r= (log r)r

cos

sin = (log sin ), Sf f ,

we obtain Sffor n=3.

4


5/5

More generally, the spherical Laplacian Sn on the unite sphere Sn

R

n+1

can be described inductively in terms of the spherical Laplacian on theunit sphere Sn1 Rn. To do this, write Sn in spherical coordinates:as a vector in Rn+1,

= (sin , (cos)) R Rn, Sn1, [0, ].

(This representsSn1 as the equator = 0 in Sn.) In these coordinates, wehave, for a function f=f(, ) defined on Sn:

Snf=f+ (n 1)cos

sin f+

1

sin2 Sn1f.

Here the operator Sn1

f involves only differentiation in the variables Sn1.

5

invLaplacian.pdf

Documents

Transcript of invLaplacian.pdf