invlaplacian.pdf
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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 .
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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.
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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.
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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.
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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.
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