THE NEUMANN PROBLEM FOR THE 2-D HELMHOLTZ

Download (0)

Full text

(1)

VOL. 21 NO. 2 (1998) 209-216

THE NEUMANN PROBLEM FOR THE 2-D HELMHOLTZ

EQUATION

IN A DOMAIN, BOUNDED BY CLOSED AND OPEN CURVES

P.A. KRUTITSKII

Dept.

ofMathematics, FacultyofPhysics,

Moscow State

University,

Moscow 119899,

Russia.

(Received September 23, 1996 and in revised form April 17, 1997)

ABSTRACT.

The

Neumann

problem for the dissipative Helmholtz equation in a connected planeregion boundedbyclosed andopencurves isstudied. Theexistenceof classical solutionis proved bypotentialtheory. The problemisreduced to theFredholmequation of the secondkind, which isuniquely solvable.

Our

approachholds for both internal and external domains.

KEY WORDS AND PHRASES:

Helmholtzequation,

Neumann

problem, boundaryintegral equation method.

1991

AMS SUBJECT CLASSIFICATION CODES:

35J05, 35J25, 31A25, 78A45.

1.

INTRODUCTION

The boundaryvalueproblemsindomains boundedbyclosed and opencurves werenot treated in the theory of 2-D

PDEs

before.

Even

in the case of Laplace and Helmholtz equations the problemsin domainsboundedbyclosedcurves

[1-2], [5-8]

and problemsin theexteriorof open arcs

[5], [9-11]

were treated separately, because different methods wereused in their analysis.

Previously the

Neumann

problemintheexteriorofanopenarc wasreducedtothehypersingular integral equation

[9-10]

ortotheinfinitealgebraic systemof equations

[11],

while the

Neumann

problemindomainsboundedbyclosedcurves wasreduced to the Fredholm equation of the second kind

[1], [6-8].

The combination of these methodsin caseof domains boundedbyclosed and open curves leadsto the integralequation, whichis algebraicorhypersingular onopen curvesand t is an equation of the second kind with compact integral operators on the closed curves. The integral equation on the whole boundary is too complicated and thegeneral theory ofsimilar equationsarenotconstructedcurrently. The approach suggestedinthepresentpaper enables to reduce the

Neumann problem

in domains bounded by closed and opencurves tothe Fredholm integral equationonthe wholeboundarywiththehelp ofthe nonclassicalangular potential. Since the boundaryintegral equationis

Fredholm,

the solvability theorem follows from the uniqueness theorem, which is ensured for the

Neumann

problem in the case ofthe dissipative Helmholtz equation This approachisbasedon

[3-4],

where theproblemsintheexteriorof opencurves were reducedtotheFredholm integral equations using theangularpotential.

2.

FORMULATION OF THE PROBLEM

By

a simple open curve we mean a non-closed smooth arc of finite length without self- intersections

[5]

In

theplanex

(xl,x2)

6

R

weconsider themultiply connecteddomainboundedbysimple

(2)

210 P. A. KRUTITSKII

opencurves

r,..., r, c-, , (0, ], n

simple closedcurves

r,...,r, c -,0,

othat the

curvesdonot have pointsin common.

We

will considerboththecaseofanexternal domain and thecaseofaninternal domain,when thecurve

F

encloses all other.

We

put

N N

r= Urn, r= Urn, r=rur .

The connected domain bondedby

F

will be calledD.

We

assumethat eachcurve

F

is para-

metricizedby thearc

lenh

s

r (. () ((),()), e [,b]}, 1,...,N,

k 1,2,

sothat

a < b} < < a, < b < a < b < < a < b

and the domn

D

is tothe right when theparameters increases on

F.

Thereforepoints x

F

and valuesoftheparameters are in

one-to-onecorrespondence except

a, b,

wchcorrespondtothesamepoint xforn 1,

N.

Belowthesetsofthe intervalsonthe

Os as

N N Nk

n=l n=l k=l n=l

willbe denotedby

F, F2

d

F

also.

w ut (r)= {y() y()e [,], y()= y() }

d

(r) c(r).

n=l

The tangent vector to

r

at the point

z(s)

we denote by

COS

(8) X(8) sinG(8) X(8). Let , (siG(8)- cosG(8))

beaormalvector to 8t

x().

The rection of

n

ischosen such thatit will coincide with the rection of T if

n

is rotated

anticlockwisethrough anangleof

r/2.

We

say, that the nction

w(x) belongs

to the smoothns class

K

if

) e (r ) c(r),

2) vw e c(rrx),

here

X

is apoint-set, consistingofthend-pointsof

r

N

n=l

3)

intheneighborhood ofanypoint

x(d) X

forsomeconstantsC

>

0, e

>

-1 theinequality holds

v c (d) (.)

wherexx(d) andd=a, ord=b, n=l,...N,

4)

thereexistsaiformforall

z(s) e r

lit of

(n,, V())

as

alongthe normal

n.

REMARK. In

the definition of the class

K

weconsider

F

Accorng

to thisdefinition,

w(z)

and

Vw(z)

may haveajumpacross

rx.

Let

usformulatethe

Neumann

problem forthe ssipativeHelmholtzequation in the domn

PROBLEM U. To

findafction

w(x)

ofthe class

K

which satisfies theHelmholtzequation

,, () + ,,(z) + Z(z)

0,

e r , Z

cot,

Z >

0,

(2.2)

and theboundarycondition

(3)

If

D

is anexternal domain, thenweadd thefollowingcondition at infinity

(2.2c)

Allconditionsof theproblem

U

must be satisfiedinthe classicalsense.

By Ow/On

on

F

we meanthe limitensuredin the point

4)

of the definition of the smoothness class

K.

The normal derivative

0w/0nz

has tobe continuous across

FI\X

and has to take given valueson

FI\X. At

thesametime

w(x)

may haveajumpacross

FI\X.

On

the basisoftheenergy equalitiesand thetechniqueofequidistantcurves

[6],

wecaneasily prove thefollowingassertion.

THEOREM

1.

If F

6

C ’, A

6

(0, 1], F

6

C 2’0,

then theproblem U has at most one solution.

The theorem holdsforboth internal and externaldomain7).

3.

INTEGRAL EQUATIONS AT THE BOUNDARY

Below we assume that

f(s)

from

(2.2b)

is an arbitrary function from the Banach space

c,(r) c(r), e (0, }.

If

B1 (F ), B2(F 2)

areBanach spaces of functions givenon

F

and

F 2,

thenfor functions given on

F

weintroducetheBanach space

B(F )

gl

B2(F )

withthenorm

[[’l[l(r)nB(r2)= [l’l[(r)+

BY/...drwemeanaf,...da.

rk n=lNk

We

considertheangularpotential from

[3], [4]

forthe equation

(2.2a)

on

i/

w[,u](x) (r)V(x,a)da. (3.1)

F

Thekernel

V(x, r)

isdefinedonthe eachcurve

F,

n 1,

N

bytheformula

v(, ) / o,( (Z I y()l) d, = e [, b]

where

T/(0)(z)

isthe Hankel functionofthe first kind

(01)(z) v/exp(iz-ir/4)

? (

it

rv

0

exp(-t)t

-/ 1

+ -z

dr,

() ((),()), I ()1 v/(-()) + ( ()).

Belowwesuppose

that/(a)

belongs to the Sanach space

C(F1),

w 6

(0,1],

q

e [0,1)

and

satisfies thefollowingadditionalconditions

/ l(a)

da

O,

n l,

N. (3.2)

We

say,

that/(s) e C(F )

if

gl

C ’(r

(4)

212 P.A. KRUTITSKII

where

C’(F1)

isaHolder spacewiththe index and

r=l IIcO,w(F1)

As

shown in

[3], [4]

forsuch

#(a)

theangularpotential

wl[#](x)

belongs to the class

K. In

particular, the inequality

(2.1)

holdswith e -q, ifq

(0, 1). Moreover,

integrating

wl[](x)

by partsand using

(3.2)

weexpress theangularpotentialinterms ofadoublelayerpotential

f p(a) ---) (Z lz y(a)l)

da,

(3.3)

[]() -

withthe density

p(c) ] #(f)d’,

a

e [al, b],

n 1,

N1. (3.4)

Consequently,

wl[#](x)

satisfiesboth equation

(2.2a)

outside

F

and the conditions at infinity

(2.2c).

Let

usconstructasolutionof theproblem U Thissolution canbeobtained withthehelpof potentialtheory fortheHelmholtzequation

(2.2a). We

seekasolutionoftheproblemin theform ofthe

anlar

pontialon

F

and thesingle-layer potentialon

F

w](x) w[](x) + w[](x) (3.5)

where

w[](x)is ven

by

(3.1), (3.3)

and

]()

F2

f (a)

)

(Z I y(a)l)d.

We

willsk

(s)

fromthe Bach space

c(r)(r=), e (0, ], e [0, )

withthenorm

]}’]lc(r)oc0(r=) l]’llv(r) + H’lle0(r=)

Besides,

(s)

mustsatisfyconditions

(3.2).

Itfollows from the properti of potentials

[1], [3-4], [6],

thatfor such

p(s)

the ction

(3.5)

belongstothecls

K

andsatisfiesall contionsoftheproblem

U

except theboundarycontion

(2.2b). In

theceof the

extern

domain the fction

(3.5)

satisfiesthe contion at inity

(.).

To

satisfythebodarycontionwe

pu (3.5)

in

(2.2b),

usethe litform for the

anlar

potential from

[3]

d aive the

ine

equation for the density

(s)

r

i

0((), u()) r

0

l u() I()- ()1 + _l u() (()’ ) e()u()+

2

rx

r

( I()- U()l) =/(), V, (.6)

+ f u()

where

6(s)

0 if and

g(s)

ifs

,

(z, ) h (x (()) d(, [a, b],

n

, 2,

By 0(z,)

we denote theanglebetween thevector and theNrectionof the normal

n.

The angle

0(, )

is takentobepositive ifit is meured anticlockwise from

n

and negative ifit is measured clockwise

om n.

Besides,

0(z, )

is continuous inz,

F

if z

.

(5)

Thus, if/(s)

is asolutionofequations

(3.2), (3.6)

from the space

C(F 1)

N

C(F2),

w

e (0,1],

q

e [0,1),

then the potential

(3.5)

satisfies all conditions of the problem

U.

The followingtheorem holds.

THEOREM

2.

/fr e c 2,, r e c ,, f(s) e c,(r)N co(r), A e (0,1],

equation

(S.6)

hasasolution

#(s) from

the Banach space

C(F )

f

C(F-),

w 6

(0, 1],

q6

[0, 1)

and condztions

(3.2)

hold, then the

function (3.5)

is asolution

of

theproblem

U

Belowwelook

for/z(s)

intheBanach space

C(F )

q

C(F).

Ifs 6

F -,

then

(3.6)

is anequation ofthe second kind withcompact integral operators. If s6

F ,

then

(3.6)

isasingular integral equation

[5].

Our

further treatmentwillbe aimed to the

proof

of the solvability of the system

(3.2), (3.6)

inthe Sanach space

C[(F 1)

q

C(F). Moreover,

wereduce thesystem

(3.2), (3.6)

toaFredholm equation of the secondkind,whichcanbe easilycomputed byclassicalmethods.

Equation

(3.6)

on

F

2we rewrite intheform

+ / I(a)A_(s, a)da -2f(s),

s 6

F , (3.7)

F

where

__o v ((), )+

A2(s,a) (1-6(a)) On

2

w’0n ( I()

and

V(x, q)

isthe kernelofthe

anlar

potential

(3.1).

We

note

A(s,a) (F

2x

F),

because

F: C ’.

It

canbe easily provedthat

sin 0

(x(s), y(a))

e C’(r

x

r )

() ()l

(see [3], [4]

for

details).

Thereforewe canrewrite

(3.6)

on

F

intheform

/ ():

ds

+ / .()Y(. )d -2/(). e r

F F

where

1--6(cr)) [ ( sinO(x(s)’y(a))lx(s

s

Vo ((s), a)

()i o()on,

0

(Z I() ()l) } e c.o(r r).

P0=Aif0<A<l andp0=l-e0foranye06(0,1)

ifA=l.

(3.8)

4.

THE FREDHOLM INTEGRAL EQUATION AND THE SOLUTION OF THE PROBLEM

Invertingthe singular integraloperatorin

(3.8)

wearriveat thefollowingintegral equation of thesecond kind

[5]:

1 N,-1

#(a)Ao(s, a)da + F1

Q (s) G"s (I)0(s),

s6

(4.1)

"() + O()

r =o

Q()

where

Go, GNI-1

arearbitraryconstantsand

2Q(a)f(a)da

Ao(s, a) _lr /

F

Y ( a) Q (I)0(s) lr

F

/

a s

(6)

214 P.A. KRUTITSKII

To

derive equationsfor

Go, GN-I

wesubstitute

#(s)

from

(4.1)

inthe conditions

(3.2),

then

weobtain

N

/tt(a)l,(a)da + Sn,G, Hn,

n 1,...,Nx

(4.)

F m--O

where

r r

B,, -/Q(s)s’ds.

r

(4.3)

By B

we denote the

N1

x

N1

matrixwith the elements

B,,

from

(4.3). As

shown in

[4],

the

matrix

B

isinvertible. Theelementsof theinverse matrixwill becalled

(B-1),.

Invertingthe matrix

B

in

(4.2)

weexpress the constants

Go, GN-

interms of

tt(s)

G. (B-I).. H. (a)l(a)da

We

substitute

G,

in

(4.1)

and obtain the integral equationfor

#(s)

on

F

1/

(s) + tt(a)A (s, a)da qli’s) O1 (s),

s

e F , (4.4)

where

N1 N1

A (s, a) Ao(s, a) _,

s

n--0 rn=l

N N1

Ol(s) Po(S)

s

(B-I),,H,

n=O

It

canbe shown using the properties of singularintegrals

[2], [5],

that

0(s), A0(s, a)

areHolder functions ifsE

I ’1,

o E1". Therefore,

O(s), A(s, a)

arealso Holder functions ifs

F x,

a

F.

Consequently, any solutionof

(4.4)

belongsto

C’/(r 1)

andbelowwe look for

(s)

on ’1 inthis

space.

We

put

Q(s) (1 (s)) (s) + 5(s),

s

e F.

Instead of

(s) 6’1/(r )c6(r)

weintroduce thenewunknown function

.(s) (s)(s) C’(I"1)

C

C(1’)

andrewrite

(:3.7), (4.4)

intheformofoneequation

#,(s) + f #,(r)Q-x(a)A(s,a)da (s),

s

e F, (4.5)

where

A(s,r) (1 -6(s))A(s,a)+8(s)A(s,a), (s)= (1- 8(s)) @x(s) 26(s)y(s).

Thus, the systemofequations

(3.2), (3.6)

for

tz(s)

has been reduced to the equation

(4 5)

for the function

#,(s). It

isclearfromourconsideration that any solution of

(4.5)

givesasolution of system

(3.2), (3.6).

As

notedabove,

l(s)

and

A(s,a)

areHolder functions ifs6

F ,

cr6F.

More

precisely

(see

[4], [5]), (s) e C’(F1),

p

min{1/2, A}

and

Al(S,a)

belongsto

C’(F l)

in suniformly with respecttoaE

F. We

arriveatthe followingassertion.

(7)

LEMMA. /fF C 2’a, A (0,1], F C 2,, (s) co(r=),

p=

min(A, 1/2},

and

l.(s) from C(F) satisfies

the equation

(,.5)

then

#.(s) C’(F 1)

N

C(r2).

Thecondition

(s) C(r x)

N

C(F2)

holds if

f(s) C,(F 1)

N

C(F=).

Hence

belowwewill

seek/.(s)

from

C(F).

Since

A(s, a) C(F

x

F),

theintegral operatorfrom

(4.5):

Ap. f tz.(a)Q-(a)A(s,a)da

F

is a compact operatormapping

C(F)

intoitself.

Therefore, (4.5)

is aFredholmequation of the second kind in the Banach space

C(F).

Let

usshow thathomogeneousequation

(4.5)

hasonlyatrivial solution.

Then,

according to Fredholm’stheorems,the inhomogeneous equation

(4.5)

hasaunique solutionforanyright-hand side.

We

will prove thisbyacontradiction.

Let/z.(s) C(F)

beanon-trivialsolutionoftheho- mogeneousequation

(4.5).

Accordingtothelemma

.(s) C’(F 1)

N

C(F),

p

min{A, 1/2}.

Thereforethefunction

#(s) #.(s)Q-(s) C/=(F )

N

C(F2)

converts thehomogeneousequa- tions

(3.7), (4.4)

intoidentities. Usingthehomogeneousidentity

(4.4)

wecheck,

that/(s)

satisfies

conditions

(3.2).

Besides,actingonthehomogeneous identity

(4.4)

withasingular operatorwith thekernel

(s- t)

-x wefind that

#(s)

satisfies thehomogeneous equation

(3.8).

Consequently,

#(s)

satisfiesthehomogeneousequation

(3.6). On

the basisof theorem

2, w[#](x)

isasolution

ofthe homogeneous problem

U.

According totheorem 1

w[/l(x) 0,

x

D\F x.

Using the limitformulas for tangentderivativesofanangularpotential

[3],

weobtain

0 0 0

lira

wu ](x)-

lim

w[/](x) #(s)

--0, s

F

-x(s)e(rl)+ x-,(s)e(rl)-

OT

By (F)

+ wedenote the sideof

F

whichis ontheleftas aparametersincreasesandby

(F)

we

denotetheother side.

Hence, w[#](x) w[#](x) 0,

x

,

and

/z(s)

satisfies the following homogeneous equation

" if

o, 0 .,O)

( lx(s y(a)l

da 0,

sF . (4.6)

-_ u() +

F

u o--:no

TheFredholm equation

(4.6)

iswell-knowninclassical mathematical physics

[1], [6]. We

arrive

at

(4.6)

when solvingthe

Neumann

problemfor theHelmholtz

equation. (2.2a)

inthe domain

:D

by the singlelayerpotential.

It

iswell-known

[1],

that the equation

(4.6)

hasonlythe trivial solution

/(s)

0in

C(F).

Thisistrueforboth internal and external domain

:D.

Consequently, if s

F,

then

/z(s)

_=

0, /z.(s) I(s)Q-(s) =--

0 and we arrive at the

contradiction to theassumption that

p.(s)

is anon-trivialsolutionofthehomogeneousequation

(4.5). Thus,

thehomogeneous Fredholmequation

(4.5)

hasonlyatrivial solutionin

C(r). We

haveprovedthefollowingtheorem.

THEOREM

3.

If F C 2’, F C ’, A (0,1],

then

(.5)

is aredholmequation

of

the

second kindinthespace

C(F). Moreover,

equation

(,.5)

hasaunique solutwn

#.(s) C(F) for () c(r).

As

aconsequence ofthetheorem 3 and the lemmaweobtainthe corollary.

COROLLARY. /f F C ’a, A (0, 1], F C ’

and

p(s) C’(F 1)

N

C(F),

where

p

min{A, 1/2},

then the unzque solution

of (.5)

in

C(F),

ensured by theorem

3,

belongs to

c0,(r ) c0(r).

We

recall that

(s)

belongs to the class ofsmoothness required in the corollary if

f(s)

C,(F )

N

C(F2). As

mentionedabove, if

#.(s) C’(F ) C(F2)

is asolution of

(4.5),

then

(8)

216 P. A. KRUTITSKII

,(s) ,.(s)Q-() e cz=(r)nc(r =)

is asolutionof system

(3.2), (3.6). We

obtainthe following statement.

THEOREM

4.

If F

E

C ’, F C -’, f(s) C’(F1)f C(F), A (0,1],

then the

system of

equations

(3.2], (3.6)

hasasolution

(s) e CT/=(F )

N

C(F),

p

min{1/2, A},

which s ezpressed bythe

formula t(s) I.(s)Q-l(s),

where

p.(s) e C’P(F 1)

N

C(F )

is the unique

solution

of

the Fredholmequation

(.5

in

c(r’).

On

the basis of the theorem2wearriveat the final result.

THEOREM

5.

If F e C ’, F e C 2’, f(s) e C’(F)NC(F), A e (0,1},

then the solution

of

theproblem

U

exists andisgivenby

(3.5),

where

#(s)

is asolution

of

equations

(3.), (3.6) from It C/2(F ) n C(F2),

p

min{1/2, A}

ensuredbythe theorem

.

can be checked directly that the solutionofthe problem

U

satisfies condition

(2.1)

with

-1/2.

Explicit expressionsfor singularities ofthe solutiongradient at the end-pointsofthe opencurves canbe easily obtained with thehelp of formulas presentedin

[4].

REFEPNCES

[1] COLTON, D.

and

KRESS R.,

Integral equation methodsinscatteringtheory, John Wiley

&

Sons, N.Y.,

1983.

[2] GAKHOV, F.D.,

Boundaryvalueproblems,

Pergamon Press, Oxford;

Addison-Wesley,Read- ing,

Mass.,

1966.

[3] KRUTITSKII, P.A.,

Dirichletproblem forthe Helmholtz equation outside cutsin aplane,

Comp.

Maths. Math.

Phys.

34

(1994),

1073-1090.

[4] KRUTITSKII, P.A., Neumann

problem fortheHelmholtzequation outsidecutsinaplane,

Comp.

Maths. Math. Phys.34

(1994),

1421-1431.

[5] MUSKHELISHVILI, N.I.,

Singularintegral equations,

Noordhoff,

Groningen, 1972.

[6] VLADIMIROV, V.S.,

Equations

of

mathematical physics, Marcel

Dekker, N.Y.,

1971.

[7] TORRES, R.H.

and

WELLAND, G.V.,

TheHelmholtz equation andtransmissionproblems withLipschitz interfaces, Indiana Univ. Math.

J.

42

(1993),

1457-1485.

[8] PETERSDORF, T.V.,

Boundaryintegral equationsformixedDirichlet,

Neumann

and trans- missionproblems, Math. Meth.

Appl.

Sci. 11

(1989),

185-213.

[9] DURAND, M., Layer

potentialsand boundaryvalueproblems for the Helmholtzequation in thecomplement ofathinobstacle, Math. Meth. Appl. Sci. 5

(1983),

389-421.

[10] PANASYUK, V.V., SAVRUK, M.P.

and

NAZARCHUK, Z.T.,

The method

of

singularzntegral

equationsintwo-dimensional

diffraction

problems, Naukova

Dumka,

Kiev, 1984.

(in Russian).

[11] TUCHKIN, Y. A.,

Scattering ofwaves by an unclosed cylindrical screen of an arbitrary profilewith

Neumann

boundarycondition.Dokl. Akad. Nauk

SSSR

293

(1987),

343-345.

(in

Russian).

Figure

Updating...

References

Related subjects :