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The Schwarz Potential of EvenDimensional Tori and Quadrature forHarmonic FunctionsDawit Aberra aa Department of Mathematics , University of Arkansas , Pine Bluff,P.O. Box 4989, Pine Bluff, AR, 71602Published online: 15 Sep 2010.

To cite this article: Dawit Aberra (2002) The Schwarz Potential of Even Dimensional Tori andQuadrature for Harmonic Functions, Complex Variables, Theory and Application: An InternationalJournal: An International Journal, 47:1, 1-15

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Complex Variables, 2002, Vol. 47, No. 1, pp. 1–15

The Schwarz Potential of Even Dimensional Toriand Quadrature for Harmonic Functions*

DAWITABERRA

Department of Mathematics,University of Arkansas at Pine Bluff,P.O. Box 4989, Pine Bluff, AR-71602

Communicated by R.P.Gilbert

(Received 4 June 2000; Revised12 October 2000)

The modified Schwarz potential of an axially symmetric solid torus � generated by a disc Dða,RÞ of center‘‘a’’ and radius R satisfies the equation

�ð�,wÞuþn� 2

�u� ¼ �D � T0,

where �D represents the characteristic function of the disc Dða,RÞ and T0 is a distribution supported on theðn� 2Þ-dimensional sphere traced by the center of the disc. For even dimensions, a method to explicitly cal-culate the modified Schwarz potential, hence the corresponding distribution T0, is suggested, by showing thata suitable transformation of the modified Schwarz potential satisfies polyharmonic equation of orderðn� 2Þ=2. Quadrature formula for functions harmonic and integrable over such tori is also established andexamples are provided.

Keywords: The Schwarz Potential; Quadrature; Harmonic functions; Polyharmonic functions

1991 Mathematics Subject Classification: 31A05; 31A30

1. INTRODUCTION AND PRELIMINARIES

Let �0 be a domain in Rn and ’ :¼ ’ðx1, x2, . . . , xnÞ 2 C1

ð�0Þ: Introduce polar coordi-nates for the first n� 1 variables;

x1 ¼ � sin �n�2 sin �n�3 � � � sin �2 cos �1

x2 ¼ � sin �n�2 sin �n�3 � � � sin �2 sin �1

x3 ¼ � sin �n�2 sin �n�3 � � � cos �2

*This work constitutes a part of the author’s Ph.D. dissertation at the University of Arkansas, Fayetteville.

ISSN 0278-1077 print: ISSN 0000-000X online/01/010000-00 � 2002 Taylor & Francis Ltd

DOI: 10.1080/02781070290002895

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xn�2 ¼ � sin �n�2 cos �n�3

xn�1 ¼ � cos �n�2,

0 �1 2�, 0 �i �, 2 i n� 2, � 0,

and let w :¼ xn. With this transformation of variables, the Laplacian of ’ becomes(see page 292, [6]):

�Rn’ðx1, . . . , xnÞ ¼ �

Rn�1’þ

@2’

@w2

¼ �ð�,wÞ’þn� 2

�

@’

@�þ

1

�2A’, ð1:1Þ

where,

A’ ¼1

h

Xn�2

i¼1

@

@�i

h

hi

@’

@�i

� �, ð1:2Þ

h ¼ sinn�3 �n�2 sinn�4 �n�3 � � � sin �2,

hn�2 ¼ 1, hn�3 ¼ sin2 �n�2 � � � ,

h1 ¼ sin2 �n � 2 sin2 �n � 3 � � � sin2 �2:

The Jacobian of this transformation is,

J ¼@ðx1, . . . , xn�1Þ

@ð�, �1, . . . , �n�2Þ

¼ h�n�2: ð1:3Þ

If ’ is a function of � and w only, hence independent of �i, 1 i n� 2, then A’ ¼ 0.In this case, ’ ¼ ’ð�,wÞ is called axially symmetric (with respect to the w-axis).

Suppose a > R > 0 and Dða,RÞ denotes the disc with center at a and radius R in theplane x2 ¼ x3 ¼ � � � ¼ xn�1 ¼ 0. By rotating this disk about the w-axis, we obtain anaxially symmetric solid torus � � R

n defined by

� ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix21 þ � � � þ x2n�1

q� a

�2þ w2 < R2,

or, letting � ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix21 þ � � � þ x2n�1

q,

ð�� aÞ2 þ w2 < R2:

In this process, each point of the disc Dða,RÞ traces a sphere.We will denote the space of distributions with support in a domain �0 by D0ð�0Þ:If � 2 D0ðDða,RÞÞ is supported at a, then the axially symmetric distribution obtained

by ‘‘rotating’’ � about the w axis and supported on the sphere swept by the center of the

2 D. ABERRA

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disk will be denoted by � �. Note that � :¼ �ð�,wÞ is actually the Lebesgue measureon the sphere centered at the origin and radius a.

The modified Schwarz potential of a torus � � Rnðn 2Þ is the solution of the

following Cauchy problem [2]:

�Rnu ¼ 1 near �,

u � 0 on �,

�ð1:4Þ

where � denotes the boundary of � and the notation u � 0 on � means u and its partialderivatives of order < 2 vanish on �: Hence, since the modified Schwarz potential ofan axially symmetric torus � is axially symmetric (a proof is outlined in the followingsection), it satisfies the axially symmetric equation:

�ð�,wÞuþn� 2

�u� ¼ 1 near @Dða,RÞ,

u � 0 on @Dða,RÞ:

(ð1:5Þ

In an effort to prove the Schwarz potential conjecture (see [2,3,5]) and to study quad-rature for harmonic functions, the Schwarz potential of the first few nontrivial surfaces(spheres, cylinders, cones and ellipsoids) has been considered by Khavinson andShapiro [3]. For detailed background and references on quadrature identities, werefer the readers to Shapiro’s book [5].

In this paper, we will first establish some properties of the solutions of axially sym-metric equations (see the following section). Some of the results obtained in thissection has already been known to Weinstein and his school (see [7] and referencestherein).

The symbols T and T0 will denote the distributions defined by the equations:

�Rnu ¼ �� � T, ð1:6Þ

and

�ð�,wÞuþn� 2

�u� ¼ �D � T0 , ð1:7Þ

where, in the first equation above, u is the solution of the Eq. (1.4) and in the second,u is the solution of the Eq. (1.5). In Section 3, we will establish the relationship betweenT and T0 and derive quadrature formula for functions harmonic in the torus andintegrable over the torus. The linear space of all functions harmonic in the torus andintegrable over the torus � will be denoted by HL1ð�Þ:

Finally, we will close our note with the examples.

2. AXIALLY SYMMETRIC FUNCTIONS

The results in this section will be used in the sequel. For more properties of axially sym-metric functions, we refer the readers to Weinstein’s papers (see [7] and referencestherein).

SCHWARZ POTENTIAL 3

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THEOREM 2.1 Let ’ð�,wÞ be a function in C1ðDÞ satisfying the equation

�’þk

�’� ¼ 0:

Then,

(1) The function 1 :¼ ð1=�Þ ’� satisfies the equation

� þkþ 2

� � ¼ 0:

(2) The function 2 :¼ �k�1’ satisfies the equation

� �k� 2

� � ¼ 0: ð2:1Þ

Moreover, if k is even, 2 satisfies the equation

�k=2 ¼ 0:

(3) The function 3 :¼ �k’ ð¼ � 2Þ satisfies the equation

� � k1

�

� ��

¼ 0:

Moreover, if k is even, 3 satisfies the equation

�ðkþ2Þ=2 ¼ 0:

Proof (1) follows from the identity

Lkþ21

�’�

� �¼

1

�ðLk’Þ�, where, Lk :¼ �þ

k

�

@

@�: ð2:2Þ

To prove (2), first note that if ¼ �k�1’, then

�k�1 �’þk

�’�

� �¼ � �

k� 2

� �: ð2:3Þ

Hence, the first part of (2) and the second part for k ¼ 2 follow. Assume k > 2.By (1), the function v2 ¼ ð1=�Þ � satisfies the equation

� �k� 4

� � ¼ 0: ð2:4Þ

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Taking the Laplacian of both sides of (2.1) and using the above result, we obtain,

�2 ¼ ðk� 2Þ�1

� �

� �

¼ðk� 2Þðk� 4Þ

�ðv2Þ�,

where v2 satisfies (2.4).We claim that for any integer m,

�m ¼ðk� 2Þðk� 4Þ � � � ðk� 2mÞ

�ðvmÞ�, ð2:5Þ

where vm is a function satisfying the equation

�vþ2m� k

�v� ¼ 0: ð2:6Þ

Assume (2.5) is true for some m > 2, where vm is a function satisfying (2.6). Taking theLaplacian of both sides of (2.5), and using the result of (1) for (2.6), the claim followsfor mþ 1 replacing m.

To prove (3), suppose ¼ �k’, i.e., ð1=�Þ ¼ �k�1’.From (2.3), we find that

�k�1 �’þk

�’�

� �¼ �

1

�

� ��k� 2

�

1

�

� ��

: ð2:7Þ

But, writing ¼ �ð1=�Þ , we also have

� ¼ ��1

�

� �þ 2

1

�

� ��

: ð2:8Þ

Using this in Eq. (2.7) above, we obtain

�k�1 �’þk

�’�

� �¼

1

�� � k

1

�

� ��

!,

and hence,

�k �’þk

�’�

� �¼ � � k

1

�

� ��

: ð2:9Þ

But, from the hypothesis, we know that �’þ ðk=�Þ’� ¼ 0: Therefore, the first part of(3) follows from (2.9).

SCHWARZ POTENTIAL 5

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Also, applying the result in (2) to the function ð1=�Þ ð¼ �k�1’Þ, we obtain

�k=2 1

�

� �4

� �¼ 0:

Using this and (2.9), together with the hypothesis �’þ ðk=�Þ’� ¼0, we obtain

�ðkþ2Þ=2 ¼ �k=2ð� Þ

¼ k�k=2 1

�

� ��

¼ 0: œ

THEOREM 2.2 The modified Schwarz potential of an axially symmetric torus is axiallysymmetric.

Proof If u� is the solution of (1.4) and A is a rotation (in Rn) about the w-axis, then

Dxm ðu� � AÞ ¼Xnj¼1

tjmðDxjuÞ � A, ð2:10Þ

and

4ðu� � AÞ ¼ ð4u�Þ � A, ð2:11Þ

where Dxm denotes the partial derivative with respect to the m-th variable and ½tjk� is thematrix of A with respect to the standard basis of R

n. Using (2.10) and (2.11), it is easy tosee that u� � A solves the Cauchy problem (1.4), hence must coincide with u�. œ

3. QUADRATURE FOR HARMONIC FUNCTIONS

For polyharmonic functions defined on a disk D � R2, the following quadrature

formula was already known (see, for example, [1]):

1

�R2

ZD

’ dA ¼Xp�1

n¼0

R

2

� �2n�n’ðP0Þ

ðn!Þ2ðnþ 1Þ, ð3:1Þ

where ’ 2 C2pðDÞ is a solution of �p’ ¼ 0, P0 is the center and R is the radius of thedisc.

Hence, (3.1), together with Theorem 2.1(3), can be used to obtain quadratureformula for axially symmetric harmonic functions in even dimensional spaces.

Examples 3.1 Let ’ ¼ ’ðx1, x2, x3, x4Þ ¼ ’ð�,wÞ be axially symmetric harmonic func-tion in the axially symmetric torus � � R

4ð� :¼ ðx21 þ x22 þ x23Þ

1=2, w :¼ x4Þ: Usingspherical coordinates in the first three variables, we obtain:

Z�

’ d� ¼ 4�

Zð��aÞ2þw2<R2

�2’ð�,wÞ d� dw

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and, by Theorem 2.1(3), �2’ð�,wÞ is bi-harmonic in the disc ð�� aÞ2þ w2 < R2: Hence,using (3.1), we obtainZ

�

’ d� ¼ �R2ð4�Þ aR2’ða, 0Þ þR2

84ð�2’Þða, 0Þ

� �

¼ �R2ð4�Þ a2 þR2

4

� �’ða, 0Þ þ

R2a

4’�ða, 0Þ

� �¼ ð�R2Þðð4�a2 þ �R2Þ’ða, 0Þ þ ð�R2aÞ’�ða, 0ÞÞ: ð3:2Þ

In the following more general theorem, L ð¼ LkÞ denotes the operator definedby (2.2) and L� denotes the formal adjoint of L defined by

L�’ ¼ �’� k1

�’

� ��

:

THEOREM 3.2 Suppose T and T0 are the distributions defined by (1.6) and (1.7). Then,we have

(1)

hT0, ’i ¼

ZD

’ dA 8’, L�’ ¼ 0:

Hence, T0 ¼ T00 þ T00

0, where T00 and T00

0 are distributions supported at the point asuch that

hT00, ’i ¼

ZD

’ dA 8’, L�’ ¼ 0,

and,

hT000, ’i ¼ 0 8 ’, L�’ ¼ 0:

(2)

hT 000 �, hi ¼ 0 8h 2 HL1ð�Þ, and hence,

hT , hi ¼ hT0 �, hi

¼ hT 00 �, hi 8h 2 HL1ð�Þ:

The proof of the above theorem will be given after the following two lemmas.

LEMMA 3.3 Let (1)–(3) be the following statements.

(1) If � is a distribution supported at a and h�, ’i ¼ 0 8’, L�’ ¼ 0, then, there exists adistribution � supported at a such that L� ¼ �.

(2) If there exist constants c�,N such that

Xj�jN

c�D�’ja ¼ 0 8’ L�’ ¼ 0, ð3:3Þ

then there exist constants d�,M such that

Xj�jN

c�D�’ja ¼

Xj�jM

d�D� � L�’ja 8’ 2 C1ðDÞ:

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(3) If there exist constants fcmng, 0 m N � n, 0 n 1, such that,

X1n¼0

XN�n

m¼0

cmnDm�D

nw’ja ¼ 0 8’, L�’ ¼ 0, ð3:4Þ

then, cmn ¼ 0, 8m, n; 0 m N � n, 0 n 1:

Then, ð1Þ()ð2Þ and ð3Þ¼)ð2Þ. Moreover, for all even integers k, ð3Þ always holds.

Proof The first two statements in the lemma are equivalent by the Theorem ofSchwartz on the representation of point distributions (see [4], p. 165).

To see (3) implies (2), first note that everywhere ’ww appears in the sum

Xj�jN

c�D�’,

we can replace it by L�’� ðk=�2Þ’þ ðk=�Þ’� � ’�� and rewrite the restriction of thissum at the point a, after combining similar terms, as

Xj�jN

c�D�’ja ¼

X1n¼0

XN�n

m¼0

ccmnDm�D

nw’ja þ

Xj�jM

d�D� � L�’ja 8’ 2 C1ðDÞ,

for some constants ccmn, d� and M.Using this, the fact that statement (3) implies statement (2) is obvious.Finally, we will show that the last statement ð3Þ is true for all non-negative even inte-

gers k: We proceed by induction on k:For the case k ¼ 0, we observe that L ¼ L� ¼ 4. Define harmonic functions umn as

follows: um, 0ð�, wÞ :¼ <eððz� aÞmÞ and um, 1ð�, wÞ :¼ =mððz� aÞmÞ: Then, u0, 1 ¼ 0 andDðm�nÞ� �DðnÞ

w umnja ¼ m!: Using these and substituting umn for ’ in (3.4), we find thatcmn ¼ 0, 0 m N � n, 0 n 1:

Assume the result is true for all even positive integers up to k� 2. Let’ ¼ ð2� kÞvþ �v�. We have,

L�k’ ¼ ð2� kÞL�

k�2vþ �ðL�k�2vÞ�:

Hence, if v is any solution of the equation L�k�2v ¼ 0, then ’ is a solution of the equation

L�k’ ¼ 0. Also, observe that

Dm�D

nw’ ¼ ðmþ 2� kÞDm

�Dnwvþ �ðD

mþ1� Dn

wÞv

Substituting this in the sum that appear in the hypothesis of ð3Þ, we obtain:

X1n¼0

XN�n

m¼0

cmnDm�D

nw’ja ¼

X1n¼0

XN�n

m¼0

ðmþ 2� kÞcmnDm�D

nwvja

þ cmnð�ðDmþ1� Dn

wÞvÞja , 8v, L�k�2v ¼ 0:

Hence, rearranging coefficients and applying the induction assumption, we obtain,

ð2� kÞc00 ¼ acN, 0 ¼ 0, ðmþ 2� kÞcm, 0 þ acm�1, 0 ¼ 0, 1 m N,

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and,

ð2� kÞc01 ¼ acN�1, 1 ¼ 0, ðmþ 2� kÞcm, 1 þ acm�1, 1 ¼ 0,

1 m N � 1:

Hence, cmn ¼ 0 8m, n; 0 m N � n, 0 n 1: œ

LEMMA 3.4 If � is a distribution supported at the center of a disc Dða,RÞ and � is theLebesgue measure on the sphere traced by the center of the disc, then

ðL�Þ � ¼ �Rn ð� �Þ ðas distributionsÞ:

Proof Assume f is an axially symmetric smooth function in Rn. Then,

�Rn f ðx1, . . . , xnÞ ¼ ðLf Þð�,wÞ:

Hence, as distributions,

�Rn f ðx1, . . . , xnÞ ¼ ðLf Þð�,wÞ �:

Also, since f is axially symmetric, we can write (as distributions):

f ðx1, . . . , xnÞ ¼ f ð�,wÞ �,

or,

�Rn f ðx1, . . . , xnÞ ¼ �R

n ðf ð�,wÞ �Þ:

Hence, as distributions,

ðLf Þð�,wÞ � ¼ �Rn ðf ð�,wÞ �Þ: ð3:5Þ

Next, we observe that if � is any distribution in D0ðDða,RÞÞ, then for any test function’ 2 Dð�Þ,

h� �, ’i ¼ h�, ’i, where ð3:6Þ

’ ¼ ’ð�,wÞ ¼ �ðn�2Þ ð�,wÞ and,

ð�,wÞ ¼ ð�,wÞ½’� ¼

ZUð�̂�Þ

hð�̂�Þ’ð�,w, �̂�Þ d �̂�,

with

Uð�̂�Þ : ¼ fð�1, . . . , �n�2Þ : 0 �1 2�, 0 �i �, 2 i n� 2g,

�̂� ¼ ð�1, . . . , �n�2Þ and

d �̂� ¼ d�1d�2 . . . d�n�2, see notes in Section 1:

SCHWARZ POTENTIAL 9

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Using this, we see that

hðL�Þ �, ’i ¼ hL�, ’i

¼ h�, L�’i and, ð3:7Þ

h�Rn ð� �Þ, ’i ¼ h� �, �R

n’i

¼ h�, �Rn’i ð3:8Þ

Also, since � 2 D0ðDða,RÞÞ, there is a sequence of smooth functions ffig � C1ðR2Þ such

that

h�, ’i ¼ limih fi, ’i

¼ limi

ZDða,RÞ

fi ’ dA,

for all test functions ’ with support in Dða,RÞ (see [4], p.173). Hence, using this with(3.7), (3.8) and (3.5) we obtain

hðL�Þ �, ’i ¼ h�, L�’i

¼ limihfi, L

�’i

¼ limihðLfiÞ �, ’i

¼ limih�R

nðfi �Þ, ’i

¼ limihfi, �R

n’i

¼ h�, �Rn’i

¼ h�Rnð� �Þ, ’i 8’ 2 Dð�Þ,

giving us the required result. œ

Proof of Theorem 3.2: Let L� be the formal adjoint of L. Then, we have

hT0, ’i ¼ h�D � Lu, ’i

¼

ZD

’ dA�

ZD

uL�’ dA 8 ’ 2 C1ðDÞ:

Hence,

hT0, ’i ¼

ZD

’ dA 8’ 2 C1ðDÞ, L�’ ¼ 0:

To prove ð2Þ, let T0 ¼ T00 þ T00

0, where T00 and T00

0 are as in the Theorem.Using axial symmetry, we write

T ¼ T0 �

¼ T00 � þ T 00

0 �: ð3:9Þ

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Using Lemma 3.3 and Lemma 3.4 we obtain,

hT000 �, hi ¼ hðL�Þ �, hi

¼ h�Rn ð� �Þ, hi 8h 2 HL1ð�Þ:

Hence,

hT000 �, hi ¼ 0 8 h 2 HL1ð�Þ œ

We note that from (1.6), we obtain

hT, ’i ¼

Z�

’ d� 8’ 2 HL1ð�Þ:

Hence, from Theorem 3.2, we conclude that the integral of a harmonic function over anaxially symmetric even dimensional solid torus generated by a disc Dða, RÞ is equal toan integral with respect to a distribution sitting on the ðn� 2Þ-dimensional spheretraced by the center of the generating disc Dða, RÞ:

4. EXAMPLES

The equation of the Schwarz potential in variables �,w can be rewritten as

�ð�,wÞuþk

�u� ¼ 1 near @D,

uj@D� 0,

8<: ð4:1Þ

where k :¼ n� 2:If we substitute

v ¼ u��2

2ðkþ 1Þand V ¼ �k�1v, ð4:2Þ

then v satisfies the equation

�ð�,wÞvþk

�v� ¼ 0 near @D,

vj@D� �

�2

2ðkþ 1Þ,

8>><>>: ð4:3Þ

and, by Theorem 2.1, V satisfies the equation

�k=2ð�,wÞV ¼ 0 near @D,

Vj@D¼ �

�kþ1

2ðkþ 1Þ, V�j@D

¼ ��k

2, Vwj@D

¼ 0:

8><>: ð4:4Þ

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Note also that from (2.3), we have,

�k�1 �vþk

�v�

� �¼ �V �

k� 2

�V�, ð4:5Þ

which can also be written as, using complex variables z :¼ �þ iw, zz :¼ �� iw,

�k �vþk

�v�

� �¼ 2g½ððz� aÞ þ ð zz� aÞ þ 2aÞg�Vz zz þ ð2� kÞðVz þ V zzÞ: ð4:6Þ

For the case n ¼ 4 ð k ¼ 2 Þ, (4.4) becomes

�V ¼ 0 near @D,

Vj@D� �

�3

6:

(ð4:7Þ

Using that V is harmonic near @D and is real on @D, we seek for the solution in theform V ¼ f ðzÞ þ f ð zzÞ, where f is real analytic function near @D: Differentiating withrespect to z, restricting on @D, using the given boundary condition for Vz and analy-ticity of the function f , we find that Vz near @D is given by:

Vz ¼�1

24z� aþ

R2

z� aþ 2a

� �2

¼ f 0ðzÞ:

Integrating with respect to z and using the given boundary condition for V , we obtain

V ¼�1

3ð24Þððz� aÞ3 þ ð zz� aÞ3Þ þ 6aððz� aÞ2 þ ð zz� aÞ2Þ

�þ ð12a2 þ 6R2Þððz� aÞ þ ð zz� aÞÞ þ 12aR2ðlogðz� aÞ þ logð zz� aÞÞ

þ� 3R4 1

z� aþ

1

zz� a

� ��þ

�1

3ð24Þð12aR2 þ 8a3 � 12aR2logðR2ÞÞ:

Hence, the equation of u follows from (4.2) and, observing that

T0 ¼ � �vþk

�v�

� �, ð4:8Þ

and using (4.5), we obtain

T0 ¼ �1

��V

¼aR2

4��ðlogðz� aÞ þ logð zz� aÞÞ �

R4

16��

1

z� aþ

1

zz� a

� �:

Hence, in terms of the point evaluation � :¼ �fag, T0 can be rewritten as

T0 ¼ �R2 ��R4

4a2

� ���

�R4

4a��:

12 D. ABERRA

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If ’ðx1, . . . , x4Þ 2 HL1ð�Þ ð� � R4Þ, using polar coordinates we write

’ðx1, . . . , x4Þ ¼ ’ð� cos �1 cos �2, � cos �1 sin �2, � cos �2,wÞ

¼: �ð�1, �2, �,wÞ

Using these,

Z�

’ d� ¼ hT , ’i

¼ hT0 �, ’i ¼ hT0, ’i,

where,

’ ¼ ’ð�,wÞ ¼ �2Z 2�

0

Z �

0

�ð�1, �2, �,wÞ sin �2 d�2 d�1:

Using ð�2�Þ� ¼ 2��þ �2��, we obtain,

ð’Þ� ¼2

�’’þ ’�’�

Hence, using this and simplifying, we finally obtain

Z�

’ d� ¼ �R2a2 þ�R4

4

� �Z 2�

0

Z �

0

�ð�1, �2, a, 0Þ sin �2 d�2 d�1

þ�R4a

4

Z 2�

0

Z �

0

��ð�1, �2, a, 0Þ sin �2 d�2 d�1: ð4:9Þ

In particular, if ’ is axially symmetric, the first integral above reduces to 4�’ða, 0Þand the second to 4�’�ða, 0Þ, hence, in this case, (4.9) reduces to (3.2).

For the case n ¼ 6 ðk ¼ 4Þ, (4.4) is the bi-harmonic equation given by:

�2V ¼ 0 near @D,

Vj@D¼ �

�5

10, V�j@D

¼ ��4

2, Vwj@D

¼ 0:

8<: ð4:10Þ

By switching to polar coordinates, we find that second order boundary conditions of Vare given by (on @D):

V�� ¼ �2�3 þ�3ð�� aÞ2

R2

Vww ¼ �3 ��3ð�� aÞ2

R2

V�w ¼�3ð�� aÞw

R2:

8>>>>>>><>>>>>>>:

ð4:11Þ

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From (4.3) and (4.6), we see that ðzþ zzÞVz zz ¼ Vz þ V zz (near @DÞ, and hence,

Vzz zz ¼1

zþ zzVzz ðnear @DÞ:

Hence, using this, (4.11) and (4.10), and rewriting in terms of complex variables, weobtain (on @D):

V ¼�1

265ðz� aþ

R2

z� aþ 2aÞ5

Vz ¼ V zz ¼�1

26z� aþ

R2

z� aþ 2a

� �4

Vz zz ¼�1

25z� aþ

R2

z� aþ 2a

� �3

Vzz zz ¼1

zþ zzVzz ¼

1

25z� aþ

R2

z� aþ 2a

� �2

�2þR2

ðz� aÞ2

� �:

8>>>>>>>>>>>>><>>>>>>>>>>>>>:

As in the previous example, using the fact that Vz zz is harmonic near @D and is real on@D, we let Vz zz ¼ f ðzÞ þ f ð zzÞ where f is (real) analytic function near @D: Hence,

f 0ðzÞ ¼ Vzz zz

¼1

25z� aþ

R2

z� aþ 2a

� �2

�2þR2

ðz� aÞ2

� �:

Solving for V , by integrating and using the boundary conditions, we finally obtain:

Vðz, zÞ :¼�1

4a3R2 þ �

1

4a3 �

5

32aR2

� �ðz� aÞ2 �

1

192ðz� aÞ4 ðz� aÞ

þð1=8Þa2R4 þ ð1=64ÞR6

z� aþ �

1

8a2 �

1

64R2

� �ðz� aÞ3

þ �1

4a3 �

5

32aR2

� �ðz� aÞ2 þ

ð1=8Þa2R4 þ ð1=64ÞR6

z� a

þ �1

8a2 �

1

64R2

� �ðz� aÞ3 �

1

192ðz� aÞ ðz� aÞ4

þ1

8a2R2 lnðR2Þ �

1

4a4 �

1

8a2R2 lnðz� aÞ �

1

8a2R2 lnðz� aÞ �

3

8a2R2

� �ðz� aÞ

þ1

8a2R2 lnðR2Þ �

1

4a4 �

1

8a2R2 lnðz� aÞ �

1

8a2R2 lnðz� aÞ �

3

8a2R2

� �ðz� aÞ

�1

32a ðz� aÞ4 �

1

16R4 aþ

1

8aR4 lnðR2Þ �

1

10a5 þ

1

4a3R2 lnðR2Þ þ

1

192

R6 ðz� aÞ

ðz� aÞ2

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þ1

192

R6 ðz� aÞ

ðz� aÞ2þ �

1

8aR2 lnðz� aÞ �

1

4a3 �

1

8aR2 lnðz� aÞ

�

�1

8aR2 þ

1

8aR2 lnðR2Þ

�ðz� aÞ ðz� aÞ �

1

24a ðz� aÞ3 ðz� aÞ

þ �1

8a2 �

3

64R2

� �ðz� aÞ ðz� aÞ2 �

1

32a ðz� aÞ4

�1

320ðz� aÞ5 þ

1

16

R4 a ðz� aÞ

z� aþ

1

16

R4 a ðz� aÞ

z� a�

1

320ðz� aÞ5

�1

8aR4 lnðz� aÞ �

1

4a3R2 lnðz� aÞ �

1

24a ðz� aÞ ðz� aÞ3

þ �1

8a2 �

3

64R2

� �ðz� aÞ2 ðz� aÞ �

1

8lnðz� aÞR4 a

�1

4lnðz� aÞa3R2 þ

1

96

aR6

ðz� aÞ2þ

1

96

R6 a

ðz� aÞ2

Hence, the equation for u follows from (4.2) and T0 can be calculated using (4.6) and(4.8) with k ¼ 4:

Finally, we note that the modified Schwarz potential and the corresponding quadra-ture distribution of an axially symmetric solid torus in even n-dimensional spaces can becalculated similarly as in the above two examples. In general, from the first few ex-amples or from the special case of quadrature for axially symmetric harmonic functionsusing (3.1), one should expect that the corresponding potential has logarithmic singu-larity and polar singularity of order ðn� 2Þ=2 at the center of the generating disk,giving the corresponding quadrature distribution, which has support the ðn� 2Þ-dimen-sional sphere traced by the center of the generating disk and has order ðn� 2Þ=2:However, explicit calculations are very tedious; even for the case n ¼ 6, we had touse the software Maple V, Release V:

Acknowledgements

I would like to thank Professor Dmitry Khavinson for suggesting the problem as well ashelping during this work. I also thank the National Science Foundation for financialsupport during the summers of 1998 and 1999 under the grant DMS-970395:

References

[1] R. Courant and D. Hilbert (1962). Methods of Mathematical Physics. Interscience, New York, 1962.[2] D. Khavinson (1995). Holomorphic Partial Differential Equations and Classical Potential Theory.

Universidad de La Laguna, Tenerife, Spain.[3] D. Khavinson and H.S. Shapiro (1989). The Schwarz Potential in R

n and Cauchy’s Problem for the LaplaceEquation. TRITA-MAT-1989-36. pp. 112. Royal Institute of Technology, Stockholm.

[4] W. Rudin (1991). Functional Analysis, 2nd Edn. Mc-Graw Hill, Inc., New York, 1991.[5] H.S. Shapiro. The Schwarz Function and its Generalization to Higher Dimensions. University of Arkansas

lecture notes in the mathematical sciences. John Wiley & Sons, Inc., New York, 1992.[6] I.N. Vekua (1967). New Methods for Solving Elliptic Equations. North-Holland series in Applied Math.,

Amsterdam, translated from Russian.[7] A. Weinstein (1953). Generalized axially symmetric potential theory. Bull. Amer. Math. Soc., 59, pp. 20–

38.

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