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Existence of weak backward solutions to a generalized Hele Shaw flow moving boundary problem
.V~th~enr .A~&w. Theory. Prmted m Great Bnram
.Mrrhods
&
Applrcnnom.Vol. 9.
.Vo 2. pp. ?O%?lS.
1985
0362-%X 85 $3 00 * .M) @ 1985 Pergamon Press Ltd.
EXISTENCE OF WEAK BACKWARD SOLUTIONS TO A GENERALIZED HELE SHAW FLOW MOVING BOUNDARY PROBLElM Department
of Mathematics, (Received
BJORN GUSTAFSSON Royal Institute of Technology, S-100 44 Stockholm 70, Sweden
3 January 1984, received for publication 19 April 1984)
Key worak and phrases: Hele Shaw flow, moving boundary problem,
variational inequality.
1. INTRODUCTION IN THE
present paper we show the existence of backward solutions in a weak sense for a moving boundary problem originally formulated (in [7]) for Hele Shaw flows in two dimensions. The moving boundary problem is, in a classical formulation (here stated rather vaguely), as follows. Let ,u be a positive measure with compact support in R” and let, for each bounded domain function in D defined by
D in 2” containing supp ,u, PO be the superharmonic
in D
-APD=P PD
=
0
(1)
on dD.
Then the problem is: given D as above, find a family of domains {Dt}for t in some time interval containing the origin such that Do = D and such that as f increases the boundary dD,moves with the velocity -gradpD, at each point on LID, and for every t (dD,has to have some degree of smoothness in order for the above condition to make sense). In the original Hele Shaw problem n = 2 and ,U= 6, the Dirac measure at the origin. A couple of, essentially equivalent, weak formulations of this problem have been given in [4, 8, 91 (and in [3] f or a variant of the above problem) in the case n = 2. It has been proved in [4, 81 that a classical solution (this notion been made precise in two different ways in [4, 81) also is a weak solution, and in [4, 8, 91 (and [3]) that a weak solution exists and is unique for the time interval [0, ~0) for arbitrary given initial domains Do. It has also been proved [4, 81 that, at least under some extra hypotheses which probably are fulfilled for almost all f > 0 in nonpathological cases, aD, consists of analytic curves when t > 0 if {D,: t2 0)is a weak solution. All the above results except the last one generalize without much trouble to arbitrary n B 2 (as noted also in [9, Section 31). Some variant of the last result probably also holds for n > 2, with “analytic curve” replaced by “real analytic hypersurface”. No results seem to have been published on the existence or uniqueness of weak solutions for time intervals of the kind [-T,0](T> 0)with Do = D given, i.e. backward solutions. The above mentioned results on the analyticity of dD,show (or indicate) that a backward solution 203
El. GLSTAFSSOS
204
cannot be expected to exist unless dD consists of real analytic hypersurfaces (certain kinds of singularities might be allowed). The aim of this paper is to prove such an existence result (theorem 1 below). 2.
STATEiMENTOFTHERESULT
We will state our results in terms of the concept of weak solution given in [4]. This is as follows. Let p be a positive measure in R” with compact support and of finite energy, i.e. such that ,UE H-‘([W”) (see list of notations at the end of the paper). Let T > 0 and let B be a large open ball in KY’with supp ,u C B. Then a family {D,: t E [0, T]} of open sets D, with supp ,u C D, CcB is a weak solution of the moving boundary problem described in the introduction if for each t E [0, T] the function UsE Hi(B) defined by XD, -
x00
= Au!
+
&
(2)
satisfies u,ZO (1 -
and XD,, k>
(3) (4)
= 0.
(In the definition of a weak solution in [4] there also occurs an auxiliary open set w which we here have abandoned.) The connection with the classical formulation at (1) is that the moving boundary condition there can be written a/atxD, =Apn, + ,U if the po, are extended to all R” by setting them equal to zero outside D,, and (2) is this equation integrated with respect to t, and uI = $hpn, d r. It is thus known [4, 8, 91 that given an open set D containing supp ,Uand given T > 0 there exists a unique weak solution {D,: t E [O, T]} in the above sense with Do = D if merely B is chosen large enough. (The solution does not depend on B then.) By “unique” above we mean that each D, is unique up to a set of n-dimensional Lebesgue measure zero. Such sets will be called null-sets in the rest of the paper. Actually the above result will be proved “in passing” also in this paper (h) of proposition 1.) It can be shown (in fact, it follows from (f) of proposition 1 below) that if {D,: t E [0, T]} and {0; : t E [0, T’]} are weak solution and Do = 0: for some tE [0, T’] then D, = D:-, (except for null-sets) for all t E [0, min{T, T’ - t}]. Therefore it is natural in this case to put De, = D’,, r for t E (0, r] and to call D-,: t E [0, ~1 a backward solution with Do = 0: given. It is for backward solutions in this sense that we prove existence. Observe that in order to construct {D_,: t E [0, t]} it is enough to construct D-, since the remaining D_, then are obtained as the solution of the forward problem with initial domain D_,. Therefore the following result, which is our main result, asserts the existence of backward solutions. 1. Let ,Uand B be as in the beginning of this section and let D be a domain with supp p C D C B and such that dD is a finite disjoint union of real analytic hypersurfaces. Then, if t > 0 is small enough, there exists a domain D_, with supp ,UC D_, C B such that the solution u E H;(B) of
THEOREM
XD -
XD-,
= Au
+
&
(5)
Existence
of weak backward
solutions
to a generalized
Hele Shaw flow
205
satisfies UZO
(6)
(l--xD,u)=o.
(7)
and
Remarks. (1) The theorem
still holds if the word “domain” (connected open set) is changed to “open set” both times it occurs, but unless one then adds to the conclusion that there shall be only one component of D_, in each component of D the theorem will become trivial, at least for certain interesting choices of ,u, If, for example p = 6 and D is arbitrary, then a trivial solution, consisting of open set, of the backward problem is {D-,: t E [0, T]} with D-, = D\At, where A, = {x E W:r(t) S 1x15 r(O)} with r(O) chosen so that B(0; r(0)) C D, T = lB(O; r(O)>1 (I. . .I d eno t es n-volume) and r(t) for 0 < t 5 T defined by IA[I = t. (2) The question of uniqueness for backward solutions is very interesting and perhaps hard, but I shall not deal with it in this paper. I want, however, to give an example which shows that uniqueness may fail to hold for certain ,u. (The same example has been used by Sakai [8, Example 12, p. 6f] in a related context.) Let S = {x E II??“:1x1 = 1) and let y be the surface measure on S (regarded as a distribution in KY). Then ,DE H-‘(R”). Let Do = {x E 54”: 1x1< r(O)} and 06 = {x E KY’:ri(0) < 1x1< Q(O)}, where r(O), Q(O), ~(0) are numbers chosen so that 0 c rl(0) < 1 c r(0) < r2(0) and 1DoI = / 06 I. Let {D,: t Z 0}, { 0; : t L 0) be the weak (forward) solutions with initial domains Do and 06 respectively. Then it can be shown that, up to null-sets, D, = {x E IW’:lxl
0; = {x E TZ”:rl(t) < 1x1< rt(t)},
where r(t), rl(t), r2(t) are certain functions with the properties that r(t) and rz(t) are increasing, r,(t) is decreasing and equals zero from a certain t = to and onwards, I D,I = I DoI f ct and ID; I =I D,$l + ct, where c denotes the total mass of ,u. It follows that D, # Dl (with I D, A 0; I > 0) for 0 5 t c to and, since I D,I = ID; /, that D, = 0: except for a null-set ((0)) for t 2 to. Therefore the backward problem with D, as initial domain has at least two different solutions. The proof of theorem 1 will be given in Section 4. We shall first (in Section 3) introduce and investigate a certain operator which will be used in the proof of theorem 1. 3. A USEFUL
It will turn out convenient operator
to reformulate
OPERATOR
the problem in terms of a certain nonlinear
F:H-‘(B)-+K’(B). (B is still an open bail in Rn.) F is simply defined to be the orthogonal projection onto the closed and convex set {YE H-‘(B): V5 1) in H-‘(B). Thus F(v) s 1 and F(v) minimizes IIF(v) - VII with this property. (See list of notations for I(.I1etc.) Since A :HA( B) + H-‘(B) is an isometric isomorphism the definition of F can also be written F(v) = Y+ Au
(V&H-‘(B)),
(8)
B. GLSTAFSSOS
206
where 11E HA(B) is the solution of the minimum norm problem UEK /lL~/ls Ii 4
for all u 5 K. 1
with K = {u E H&B) : Au f v I 1). The solution of (9) may also be characterized inequality uEK
(9)
as being the unique solution of the variational
and for all u E K, 1
(u, u - u) 2 0 or as the unique sohrtion of the compIementarity
(10)
problem
Au+vSl
(11)
uzo
(12)
(Au + Y- 1, u) = 0.
(13)
The equivalence between (9) and (10) is a theorem in elementary Hilbert space theory and the equivalence between (10) and (ll)-(13) is also an instance of a general and rather weilknown principle (cf. [Z, theorem 4]) and the simple proof is found, for example, in [4, theorem 31. Actually, the characterization (Il)-(13) of u is the one we most often will use. The operator Fcan also be defined in terms of the solution of another variational inequality. Given v E H-‘(B) let 1~E H;(B) be defined by A~/J= v-
1
(14)
and put o=u+tJ. Then u satisfies (ll)-(13)
(15)
iff u satisfies AudO
(16)
VZ$)
(17)
(Au, u - y) = 0.
(18)
Further, u satisfies (16)-(18) iff u solves the variational inequality VE L
and
( u, w - 0) B 0
for all 1~E L, 1
(19)
where L = {w E H,j(B): w 2 I#}. The first statement above is obvious and the second one is an instance of [2, theorem 41. (14) and (15) above show that Au + Y= Au + 1. It follows that the definition of F can be written F(Aly+
1) = Au f 1
(@H:(B)),
(20)
where u E H,$(B) is the solution of (16)-(18) or (19). Now F can be used to handle the concept of weak (forward and backward) solutions very
Existence of weak backuard solutions to a generalized Hels Sha\s flow
Xl:
conveniently. Let D be a given open set with supp ,u C D C B. Then {D,: r E [O. T]} (open sets uith supp !i C D, E B) is a weak forward solution with Do = D (except for a null-set) if and only if F(t;i -!- x0) = XD,
fort E [O. T].
(21)
In fact, if {D,} is a weak solution with Do = D then (2)-(J) show that ~1~E H:( B) defined bv (2) SatiSfieS (11)-(13) with v = t,u + XD. Hence F(~,u + ~0) = tu + XD + AllI = %D.by (2). ‘Conversely, if F(r;c + XD) = x0, (t E [0, T]) then, by the definition of F, x0, = r,u 1. x0 -L Art, where U, E Hd( B) satisfies (ll)-(13) with v = t,u + xo. For f = 0. 11= 0 solves (1 l)-( 13) so that ~0 = 0 and Do = D except for a null-set. Now it follows immediately that u, satisfies (2)-(4) and hence that {D,} is a weak solution. In a similar manner the concept of a backward solution can be expressed in terms of F. Thus given D { D_,:t E [0, T]} ( w h ere T > 0 and supp rc C D_, C B) is a backward solution with D as initial domain if and only if Do = D and F(fp - ;co_,) =%Dn_r (except for null-sets) for t E [O, T]. Therefore the conclusion of theorem 1 can be formulated as follovvs: if t > 0 is small enough there exists a domain D-, with supp ;l C D_, C B and F(~,u + x0_,) = %D.
1. The operator F: H-‘( B) += H- ‘(B) defined by (S)-(13) has the following properties (with V, vl, ~1, u E H-‘(B)). F(v) 5 1. If v 5 1 then F(v) = v. F(F(v)) = F(v). IIF - F(e)11 5 11vl - VII/. In particular F is continuous. If v1 5 r+ then F( zq) S F( v,). If vz Z 0 then F(F(u,) + e) = F(q + vz). ,u 5 F(v) whenever ;d 5 v and .I( I 1. In particular min{v, 1) 5 F(V) if Y is a measure (in H-‘(B)). Suppose I, can be written v = ,U + f, where ,u Z 0 and has compact support in B, f E L”(B) and, for some open set D with supp ,UC D G B,fZ 1 in D andfs 1 outside D. Then F(v) = ,yn + fxa*, where SJ = D U {x E B : u(x) > 0) and 11E H&B) is the function occurring in the definition (8) of F. If D is connected then so is S2. As a particular case, taking f = x0. we have F(,u + x0) = xn.
PROPOSITION
Remarks. (1) (g) of the proposition can be thought of as a generalization of known regularity results for solutions of the variational inequality (19) (or (10)) in the sense that it generalizes the inequality (23) below. (2) (h) of the proposition is a minor generalization of results in [4, 8, 91. It contains the existence and uniqueness theorem mentioned earlier for weak forward solutions. (3) The operator F is closely related to the “operator” Q(. , SL’) used by Sakai [8, 91. In fact, suppose u is a positive measure with compact support in ira’ and of finite energy. and let R be a bounded open set in W”. Choose B so that !G?G B and supp v C B. Then R E Q( V, SLi) implies F(v) = xn. Conversely, if F(v) = xn then there is an open set D with /DAR / = 0 such that D E Q(v, SL’) ([9, propositions 1 and 41 essentially). Before entering the proof of proposition 1 let us recall some known facts that will be used.
208
B. GCSTAFSSOS
1. Regularity of the solution of (19) [l, theoreme
1.31, [j], [6, theorem 2.3, p. 1081.
Given Y E Zfd( B), let u E Ho’(B) be the solution of (19). Then, if y E !f’,P( B) for some p with n
(22)
Suppose that v E LP( B) (n
(23)
u E Hd( B) and p E L*(B) C H-‘(B), then up E L’(B) and (p, u) = _faupdx. This easily from the definition of (* , a). (See list of notations.) u E H*+“(B) for some p 2 1 and u = 0 on a set EC B, then Au = 0 a.e. on E ([6. A.4, p. 531 applied twice).
Proof of proposition 1. (a)-(c) follow directly from the definition of F. and (d) is an elementary property shared by all projection operators onto closed convex sets in Hilbert spaces. (e) Due to the continuity (d) of F and the fact that the cone P = {ve H-‘(B): VZ 0) is closed in H-‘(B) it is sufficient to prove (e) for vI and Q in some subspace which is dense in H-‘(B) and whose intersection with P is dense in P. It is well-known that, for example. Cr( B), and hence any subspace of H-‘(B) containing C”(B), has this property. Thus we may assume that vl, y E L^( B) (say). Let ul, u2 EH& B) be the functions in (S) for Y= vl, y. Then, by (22) ul, u2 E H’+‘(B) for all p C 2. In particular u1 and u2 are (or. strictly speaking, have representatives in form of) continuous functions. Novv. let us first see that ui 5 u2 (assuming u1B y). Put W = ui - u*, Vi = {XE B: ui(x) > 0}, i = 1, 2. Then Ui are well-defined open sets. We want to prove that w d 0 in B. This obviously holds in B\U1. In Ii* Aui = 1 - ~1 by (13). Hence Aw = 1 - v1 - Au2 h 1 - y - Au2 2 0 there and since w = 0 on dU1 fl B and w E Hh( B) the maximum principle for subharmonic functions shows that w 5 0 also in UI. This proves that u1 5 u2 in B. Now in U2 Au2 + y = 1 by (13). Hence F(q) = Au, + v1 5 1 = Au2 + uz = F(ti) in r/z. In B\U2 u1 = u2 = 0 by the inequality u1 s ~2. Hence Aui = Au2 = 0 a.e. in B\U? by statement 3 before the proof. Therefore F(q) = Au1 + v1 = v1 Z y = Au2 + y = F(y) a.e. also in B\U:. Hence F(q)6 F(y) a.e. in all of B, from which (e) follows. (f) Assume y 2 0 and let ul, u2 E Hi(B) be defined by F(Q) = vl + Au1 F( F( q) + v2) = F(q)
respectively.
(25) + u2 + Au2
(26)
This means, by definition, that Aui + ul Z 1
(27)
ui z 0
(28)
(Au1 + ul - 1, ul) = 0
(29)
Existence
of weak backward
solutions
to a generalized
109
Hele Shaw tlon
and Au?+F(v,)+
e%l
(30)
uzz 0
(31)
( Au1 + F( or) + yz - 1, u2) = 0.
(32)
To show that F( F(vJ + 5) = F(vi + ~2) is, by (25)-(26), equivalent F(Y~ + I+) = VI+ y + A(ul + UZ), i.e. that u1 + uz E H;(B) satisfies
to showing
that
A(ui + uz) + v1 + y 5 1
(33)
111 +
(34)
l.lz 2
0
(A(u, + uz) + vI + vz - 1, u1 + ~1~)= 0.
(35)
Now (33) follows immediately from (25) and (30), and (34) follows from (28) and (31). Since I+ Z 0 we have, by (c) and (e) of the proposition, that F(v,) = F( F( u,)) s F( F(Q) + y). Hence Au, + v1 ~5A(ul + UZ) + vl + Q. Therefore, using (33), (34). (25), (32), (28) and (29), 0 2 (A(u, + uz) + it+
fi-l,ul+uz)=(A(u,+~~l)+
v,+ vZ-l,rtl)
2 (Au1 + vl - 1, u,) = 0. This shows that also (35) holds and so proves (f). (g) follows by combining (b) and (e). (h) Let F(V) = v + Au, K E Hd( B). Since Y= ,u + f is a Radon measure 2 - I/f]‘_ (a) and (g) of the proposition show that - llf/lI 5 min{l, V}z5 F(v) s 1. In particular F(v) E L”(B).
Subtracting
(36)
v from the above inequality yields
- ,u - 211fllr I min{l - V,0) P Au Z 1 - v.
(37)
Since 1 - v5 1 + llfllx (37) sh ows that Au is bounded from above. Hence u can be written as the sum of a continuously differentiable function and a superharmonic function. In particular u has a unique representative in form of a lower semicontinuous function. In terms of this representative we define u = {x E B:u(x)
> O},
(38)
which is an open set. sow it follows from (ll)-( 13) that F(v) = 1 in II. In D we have v h 1, so F(V) = 1 also in D, by (36). Hence F(v) = 1 in R = D U U. Finally, in B\supp ,u Au is bounded both from above and below by (37) so u E H’.P in Bkupp p (for all p < =J). Since u = 0 in B\U this shows that Au = 0 a.e. in (B\(I) fl (Bkupp ,D) = B\( U U supp y). In particular this holds a.e. in B\( U U D) = B\Q so that F(v) = ki +f + Au =f there. Putting together, we get F(p + f) = XQ+ fxB,nas required. If D were connected and R disconnected then U would have a component w, disjoint from D. Then u > 0 in w, u = 0 on ao and Au = F(v) - v=l-v=l-fZOinw.This,however,
B. GL-ST.AFSSOY
210
contradicts
the maximum
principle.
Hence
R is connected
if D is. completing
the proof
of
(h). 4. PROOF
OF THE
THEOREM
The rest of the paper is devoted to the proof of theorem 1. The proof is based on two lemmas stated below. The first of these shows that the measure ~1 in the theorem can be assumed to be smooth and the second one contains the most technical parts of the proof. LEMMA
1. Let k1,fand D be as in (h) of proposition 1 and let h E LX(F) be a radial function (i.e. such that h(x) is a function of 1x1 only) with supp h C B(0; E), 112 0 and Jh dx = 1. where 0 < 2~ < dist(supp ;l, B\D). Then F(rc + f) = F(rc * h + f).
Remnrk. Lemma lemma 61).
1 is similar
to (actually
a slight generalization
of) [A, proposition
21 and [9.
LEbiht~ 2. Suppose JLE L”(B), ;lZ 0 and .I( has compact support in B and suppose D is a domain with supp !LC D C B and such that dD is a finite disjoint union of real analytic hypersurfaces. Then, if t > 0 is small enough, there exists a domain U with supp ~1C U @ D and a function u E Hd( B) such that
Lvhere p E H-‘(B).
11 =
0
in BLD
(39)
11 2
0
in B
(10)
An = xD,,C - tu - p in B.
(11)
p 2 0 in B. p = 0 outside
U and p E L” in a neighbourhood
of B\U.
of theorem 1. Let u, B and D be as in the theorem, i.e. B = {,KE 2” :;.r/ < R} say. Radon measure with compact support in B and belonging to H-‘(B). and D a domain with supp p C D C B and with dD consisting of a finite number of disjoint real analytic hypersurfaces. Consider Proof
,ua nonnegative
F(t;! + v) = %D
(Q)
as an equation to be solved for v E H-‘(B). According to the statement immediately before proposition 1, the assertion of the theorem is that if t > 0 is small enough (12) has a solution of the kind v = XQfor some domain 8 (to be D-,) with supp ~1C R. (Q G B is then automaticall> satisfied since Q C D by (h) of proposition 1.) It follows from lemma 1 that we can replace .I( by J( * h with h as in that lemma. This means that we may assume that ~1is arbitrarily smooth. Let us assume that $1E L”(B) (this will be enough). Now the idea of the proof is to first show that (42) has a solution of the form v = XC + p. where U is a domain with supp u C IiE D and p is a function in H-‘(B) with p Z 0 in U and p = 0 outside U. Such a solution can rather easily be constructed (this is lemma 2). Then we apply (h) of proposition 1 to get rid of the excessive function p (;cn = f’(~c + PI>. Thus, apply first lemma 2 to the situation in theorem 1. This gives us a domain U and functions u and p (if t > 0 is sufficiently small). Put v = XC + p. Then tu + v - Au = XD 5 1
Existence
of weak backward
solutions
to a generalized
Hele Shaw flow
211
by (41), u Z 0 by (40) and (f,u + Y + Au - 1, u) = (~0, u) = 0 by (39). This shows that F(tu -k v) = t,u + Y + Au = x0, by the definition of F. Next apply (h) of proposition 1 to v = xU + p above. This gives us a domain R with I/ c Q (hence supp .u C Q) and F(v) = xQ. (f) of proposition 1 then shows that F(t,u T xn) = F(t/l + F(v)) = F(t,u + v) = XD. Thus Y = ;cn is a solution of (42) of the required type and so the theorem is proven (except for the proofs of the lemmas). Proof of lemma 1. We first note that we may assume that f = 1 on D. For if we have proven the lemma for such f then, using (f) of proposition 1, we get for general f F(u + f) = F(CI + =
F(F(u
=
F(
F(
XD
+
+
XD
u * h
fxB\,D
+
+
+
fxB\D)
XD
+
(f
-
+
F(p*h
+
(f -
l)xD)
+
-
fxB?D>
=F(!~*h+XD+fXfhD+
=
l)xD)
(f
(f-
l>xD)
l)xD)
f).
Thus we assume that f = 1 on D. Let F(p + f) = y + f + Au, u E Hi( B). Then (h) of proposition 1 shows that p + f + Au = XQ + fxe,n where Q > D. Thus AIA = -u in D, in particular u is harmonic in D\supp ~1. Define u = u + xo(u * h - u),
(43)
where G = {x E B: dist(x, supp cl) < E}. Then G C D and dist( G, B\D) > E. The mean value property for harmonic functions shows that u * h = u in a neighbourhood of aG. Therefore the discontinuity of xo does not matter in (43), u E Hi(B) and taking the Laplacian yields Au = Au + ~o((Au) = ((Au) = We now aim at showing that u satisfies
that
F(u
-U
*
* h - Au)
* h)xc + Au *~a\o h + Au ’ XB\G.
* h + f) = u * h + f + Au. This
(44) amounts
to showing
!l*h+f+Avll
(45)
US.0
(46)
(u*h+f+Au-l,u)=O.
(47)
(44) shows that u*h+f+
Av=f+Au~~B\G=,u+f+Au-(Au.~G+p)=p+f+Au.
(48)
Hence (45) is satisfied since ,u + f + Au S 1. Moreover (46) holds, for in B\G, u = u 2 0 and in G Av = -,u * h 5 0 by (44), so (46) follows from the minimum principle for superharmonic functions. Finally, since u is superharmonic in D, u * h 5 u in G, showing that u 5 u in B, and (17) follows from O=(p+f+Au-1, u)=(,u*hff+Au-1, u)l(,u*h+f+Au-1,
212
B. GLSTAFSSON
u) S 0. (In the first inequality we used u S u and (45) and in the second (45) and (46).) This proves that F( ,u * h + f) = ,u * h + f + Au. Now the lemma is proven for the assertion F( u * h + f) = F( y + f) is nothing else than (48).
Proof of lemma 2. By the Cauchy-Kovalevska theorem there is an open neighbourhood of dD and a real analytic function u defined in V such that
u =Oandgradv
=OonaD
V
(49)
ho = 1 in V.
(50)
We may assume that D\v is connected and that ri fl supp ,U= 0. For LY> 0 put V, = the union of those components of {XE V: v(x) < a} which intersect aD. Then V, is open and aD C V, C V. Now we want to choose (Y> 0 so that: (a) V, is relatively compact in V, (b) u L 0 in V,. (c) grad v # 0 on av,. Let us see that this is possible. That aD is a finite disjoint union of real analytic hypersurfaces means that there are open sets Gj (i belonging to some index set) and real analytic functions vi: G,+ iR such that {Gj} covers dD, aD II Gj = {X E G,: qj(X) = 0) and grad qj # 0 on dD. Since dD is compact the collection {Gj} can be taken to be finite. We may also assume that grad qj is bounded away from zero on aD n Gj. The Cauchy data (49) for u imply that there are real analytic functions a,: G, + 2 such that U(X)
= a,(X)Qji(X)’
on Gj.
(51)
Taking the Laplacian of (51) gives (in view of (50)) that 1 = Au = 2aJgrad FjCij’ on aD f? Gj. Hence aj(x)Zc,>O where cj are constants. Moreover,
onaDnGj,
(52)
taking the gradient of (51) gives
grad v = 2aj grad Cpi+ qj grad aj V
in Gj\a D.
Now (51) and (52) (together with grad qj f 0 on dD) show that given any neighbourhood W of aD (e.g. W = lJjG, or W = V) V, C W for a> 0 sufficiently small. This immediately shows (a) and (b) for cy> 0 sufficiently small. Moreover, (c) follows (with cu> 0 sufficiently small) since the right member of (53) is continuous in Gj and nonzero on JD. From now on let cu> 0 be fixed such that (a)-(c) hold. From (a) and the definition of V, follow (d)
v =
a
on av,
and from (b), (d) and the definition of V, (e) grad v is directed out from V, on aV,.
Existence
Moreover (f) D\v,
of weak backward
solutions
to a generalized
Hele Shaw flow
213
we have is connected.
In fact, D\ri C D\ri, and D\v was assumed connected
so if D\vi, were not connected it would have a component, o, disjoint from D\ri. Then o C V (since o C D) so that u is defined in o. Since V, is a neighbourhood of aD dw II SD = 0. Hence aw C aV,. Therefore u = (Yon aw by (d) and now (50) and (e) immediately yields a contradiction. Hence (f) holds Now we define U = D\ri,.
Then I/ is open and connected, supp ,u C II E D and dU = D n dV,. It follows from (a), (c) and (d) and the fact that u is a real analytic that dU is a finite disjoint union of real analytic hypersurfaces. Therefore, and by the Cauchy-Kovalevska theorem again, there is an open neighbourhood W of dU and real analytic function w defined in W such that w = u = crand
(54)
gradw=gradu Aw=O
ondU
inW.
(55) (56)
In a similar manner as for u it follows from (c), (e) and the definition of w that there exists a p > (Yand a neighbourhood Wp of aU, Wp C W such that Wp is relatively compact in W.
w = /30naWPn
U.
We assume that W is chosen so small so that W rl supp ,u = 0. Now choose E > 0 such that cy+ ER’ < /3 (where B = {x E W”: jx j < R}) and define a function u in B by u=o
in B\D
(57)
u=u
onDnv,=D\U
(58)
u(x) = min{w(x), (Y+ E(R* - Ix 1’) forx E U n Wg u(x) = a+ E(R* - Ix/*)
forx E u\Wp
(59) (60)
We claim that the set U and the function u have the properties stated in the lemma (with p defined by (41)). As already remarked U is a domain with supp ,u C U G D. Since CY < LY+ E(R* - /xl’) < p for all x E B (by the definition of E), w = CY on au = a(un W,) n (D\U) and w = p on JWb n U = a( U fl W,) fl (U\Wp) we have, in U n W,, u = w in a neighbourhood of aU and u(x) = (Y+ E(R’ - jx12) in a neighbourhood of aWp n U. From this it follows that u is continuous in D (and in all of B). In view of (49), (54) and (55) it also follows that u is continuously differentiable in a neighbourhood of B\U. Moreover, (50) and (56) show that Au = xD\u in a neighbourhood of
B. GLWAFSSOS
211
B\U. (There will be no contributions to Au from the surfaces of discontinuity. JD and dl’ since these are real analytic and the zeroth and first order derivatives of u are continuous over them.) Next observe that CYY+ E( R’ - ;xl’) is superharmonic: A(@+
E(R’-
ixi’)) = -2m<
0.
(61)
Since w is (super-)harmonic and the minimum of two superharmonic functions is superharmanic, it follows from (59)-(60) that u is superharmonic in I/. Moreover, uiL. E H’(U) because both w and CY+ E(R’ - 1x1’) belong to H’ where they are defined and the maximum of two functions in H’ is a function in H’ [IO. proposition 28.21. By (61) Au = -2nE in a neighbourhood of supp ,u. Therefore, and since J! E L”(B), Au 5 -@ in U for t > 0 small enough. This shows that (41) holds with p defined by p = -Au - t,u in U, p = 0 in B\U and t > 0 so small so that p Z 0. Since u belongs to H’ in U p belongs to H-’ in U. Moreover, p E L” in a neighbourhood of B\U. It follows that p E H-‘(B) and that the statements about p in the lemma hold. Finally, it follows immediately from the definition of u that (39) and (40) hold. Since 11E HJ( B) follows from (39), (41) and that p E H-‘(B) the lemma is now proven.
APPENDIX-LIST
OF
NOTATIONS
(u,b)={.rER:a
denotes
AU = c$, 2
H”.p( B), H”(B)
(A also denotes
symmetric
difference
between
sets.)
the closure
of D in ?-.
)
= kfm.:( B). fij( B):
Sobole\
spaces as defined in [lo]. Hd( B) is a Hilbert space with inner product
(B always
bounded
here) and norm
H-‘(B) is also a Hilbert space, equipped with that inner product and norm (also denoted ) xvhich makes A : Hd(B) -, H-‘(B) an isometric isomorphism [ 10, theorem 23. I]. u E H.]( B)) when H-‘(B) is regard as the dual space (u. o) the pairing between H-‘(B) and Hd(B) (u E H-‘(B), of I&j(B) in the usual way. Then (u, u) = -(Au, u) (u, v E f&‘(B)) [lo, section 231. All functions, elements in Sobolev spaces and measures in this paper are regarded as, and treated formally as. distributions. Thus ~5 1 (in an open set) e.g. means that 1 - v is a positive distribution (in that open set). B) “measure” we always understand Radon measure, i.e. Bore1 measure which is finite on compact sets.
REFERENCES 1. BREZISH., Problkmes unilatkraux. J. murk. pures uppl. 51, l-158 (1972). 2. CRYER C. W. & DEMPSTER M. A. H., Equivalence of linear complementarity problems and linear programs in vector lattice Hilbert spaces, SMM /. Conrrol Oprim. 18, 76-90 (1980). 3. ELLIOTT C. M. & JANOVSKY V., A variational inequality approach to Hele Shaw flow with a moving boundary. Proc. R. Sot. Edinb. 88A, 93-107 (1981).
Existence +* GI_~T.C,FSSO%B., Applications +. 5. 6. 7. 8. 9. 10.
of weak backward
solutions
to a generalized
Hele Shaw flow
215
of variational ineoualities to a movine boundary . problem for Hele Shaw flows. . SIAM 1. marh. An&is (to appear). GUSTAFSSO?; B.. A simple proof of the regularity theorem for the variational inequality of the obstacle problem, Applicable Analysis (to appear). KWDERLEHRER D. & STA.MPACCHIA G., An Introduction IO Variational Inequalities and their Applications, Academic Press, New York (1980). RICHARDSON S., Hele Shaw flows with a free boundary produced by the injection of fluid into a narrow channel, J. Nuid lMechanics 56. 609-618 (1972). SAK~\I M., Quadrature domains. Lecture Notes in Mathematics 934, Springer. Berlin (1982). SAKAI M.. Applications of variational inequalities to the existence theorem on quadrature domains. Trans. Am. math. Sot. 276. 267-279 (1983). TREVES F., Basic Linear Partial Differential Equations. Academic Press, New York (1975).
.Mrrhods
&
Applrcnnom.Vol. 9.
.Vo 2. pp. ?O%?lS.
1985
0362-%X 85 $3 00 * .M) @ 1985 Pergamon Press Ltd.
EXISTENCE OF WEAK BACKWARD SOLUTIONS TO A GENERALIZED HELE SHAW FLOW MOVING BOUNDARY PROBLElM Department
of Mathematics, (Received
BJORN GUSTAFSSON Royal Institute of Technology, S-100 44 Stockholm 70, Sweden
3 January 1984, received for publication 19 April 1984)
Key worak and phrases: Hele Shaw flow, moving boundary problem,
variational inequality.
1. INTRODUCTION IN THE
present paper we show the existence of backward solutions in a weak sense for a moving boundary problem originally formulated (in [7]) for Hele Shaw flows in two dimensions. The moving boundary problem is, in a classical formulation (here stated rather vaguely), as follows. Let ,u be a positive measure with compact support in R” and let, for each bounded domain function in D defined by
D in 2” containing supp ,u, PO be the superharmonic
in D
-APD=P PD
=
0
(1)
on dD.
Then the problem is: given D as above, find a family of domains {Dt}for t in some time interval containing the origin such that Do = D and such that as f increases the boundary dD,moves with the velocity -gradpD, at each point on LID, and for every t (dD,has to have some degree of smoothness in order for the above condition to make sense). In the original Hele Shaw problem n = 2 and ,U= 6, the Dirac measure at the origin. A couple of, essentially equivalent, weak formulations of this problem have been given in [4, 8, 91 (and in [3] f or a variant of the above problem) in the case n = 2. It has been proved in [4, 81 that a classical solution (this notion been made precise in two different ways in [4, 81) also is a weak solution, and in [4, 8, 91 (and [3]) that a weak solution exists and is unique for the time interval [0, ~0) for arbitrary given initial domains Do. It has also been proved [4, 81 that, at least under some extra hypotheses which probably are fulfilled for almost all f > 0 in nonpathological cases, aD, consists of analytic curves when t > 0 if {D,: t2 0)is a weak solution. All the above results except the last one generalize without much trouble to arbitrary n B 2 (as noted also in [9, Section 31). Some variant of the last result probably also holds for n > 2, with “analytic curve” replaced by “real analytic hypersurface”. No results seem to have been published on the existence or uniqueness of weak solutions for time intervals of the kind [-T,0](T> 0)with Do = D given, i.e. backward solutions. The above mentioned results on the analyticity of dD,show (or indicate) that a backward solution 203
El. GLSTAFSSOS
204
cannot be expected to exist unless dD consists of real analytic hypersurfaces (certain kinds of singularities might be allowed). The aim of this paper is to prove such an existence result (theorem 1 below). 2.
STATEiMENTOFTHERESULT
We will state our results in terms of the concept of weak solution given in [4]. This is as follows. Let p be a positive measure in R” with compact support and of finite energy, i.e. such that ,UE H-‘([W”) (see list of notations at the end of the paper). Let T > 0 and let B be a large open ball in KY’with supp ,u C B. Then a family {D,: t E [0, T]} of open sets D, with supp ,u C D, CcB is a weak solution of the moving boundary problem described in the introduction if for each t E [0, T] the function UsE Hi(B) defined by XD, -
x00
= Au!
+
&
(2)
satisfies u,ZO (1 -
and XD,, k>
(3) (4)
= 0.
(In the definition of a weak solution in [4] there also occurs an auxiliary open set w which we here have abandoned.) The connection with the classical formulation at (1) is that the moving boundary condition there can be written a/atxD, =Apn, + ,U if the po, are extended to all R” by setting them equal to zero outside D,, and (2) is this equation integrated with respect to t, and uI = $hpn, d r. It is thus known [4, 8, 91 that given an open set D containing supp ,Uand given T > 0 there exists a unique weak solution {D,: t E [O, T]} in the above sense with Do = D if merely B is chosen large enough. (The solution does not depend on B then.) By “unique” above we mean that each D, is unique up to a set of n-dimensional Lebesgue measure zero. Such sets will be called null-sets in the rest of the paper. Actually the above result will be proved “in passing” also in this paper (h) of proposition 1.) It can be shown (in fact, it follows from (f) of proposition 1 below) that if {D,: t E [0, T]} and {0; : t E [0, T’]} are weak solution and Do = 0: for some tE [0, T’] then D, = D:-, (except for null-sets) for all t E [0, min{T, T’ - t}]. Therefore it is natural in this case to put De, = D’,, r for t E (0, r] and to call D-,: t E [0, ~1 a backward solution with Do = 0: given. It is for backward solutions in this sense that we prove existence. Observe that in order to construct {D_,: t E [0, t]} it is enough to construct D-, since the remaining D_, then are obtained as the solution of the forward problem with initial domain D_,. Therefore the following result, which is our main result, asserts the existence of backward solutions. 1. Let ,Uand B be as in the beginning of this section and let D be a domain with supp p C D C B and such that dD is a finite disjoint union of real analytic hypersurfaces. Then, if t > 0 is small enough, there exists a domain D_, with supp ,UC D_, C B such that the solution u E H;(B) of
THEOREM
XD -
XD-,
= Au
+
&
(5)
Existence
of weak backward
solutions
to a generalized
Hele Shaw flow
205
satisfies UZO
(6)
(l--xD,u)=o.
(7)
and
Remarks. (1) The theorem
still holds if the word “domain” (connected open set) is changed to “open set” both times it occurs, but unless one then adds to the conclusion that there shall be only one component of D_, in each component of D the theorem will become trivial, at least for certain interesting choices of ,u, If, for example p = 6 and D is arbitrary, then a trivial solution, consisting of open set, of the backward problem is {D-,: t E [0, T]} with D-, = D\At, where A, = {x E W:r(t) S 1x15 r(O)} with r(O) chosen so that B(0; r(0)) C D, T = lB(O; r(O)>1 (I. . .I d eno t es n-volume) and r(t) for 0 < t 5 T defined by IA[I = t. (2) The question of uniqueness for backward solutions is very interesting and perhaps hard, but I shall not deal with it in this paper. I want, however, to give an example which shows that uniqueness may fail to hold for certain ,u. (The same example has been used by Sakai [8, Example 12, p. 6f] in a related context.) Let S = {x E II??“:1x1 = 1) and let y be the surface measure on S (regarded as a distribution in KY). Then ,DE H-‘(R”). Let Do = {x E 54”: 1x1< r(O)} and 06 = {x E KY’:ri(0) < 1x1< Q(O)}, where r(O), Q(O), ~(0) are numbers chosen so that 0 c rl(0) < 1 c r(0) < r2(0) and 1DoI = / 06 I. Let {D,: t Z 0}, { 0; : t L 0) be the weak (forward) solutions with initial domains Do and 06 respectively. Then it can be shown that, up to null-sets, D, = {x E IW’:lxl
0; = {x E TZ”:rl(t) < 1x1< rt(t)},
where r(t), rl(t), r2(t) are certain functions with the properties that r(t) and rz(t) are increasing, r,(t) is decreasing and equals zero from a certain t = to and onwards, I D,I = I DoI f ct and ID; I =I D,$l + ct, where c denotes the total mass of ,u. It follows that D, # Dl (with I D, A 0; I > 0) for 0 5 t c to and, since I D,I = ID; /, that D, = 0: except for a null-set ((0)) for t 2 to. Therefore the backward problem with D, as initial domain has at least two different solutions. The proof of theorem 1 will be given in Section 4. We shall first (in Section 3) introduce and investigate a certain operator which will be used in the proof of theorem 1. 3. A USEFUL
It will turn out convenient operator
to reformulate
OPERATOR
the problem in terms of a certain nonlinear
F:H-‘(B)-+K’(B). (B is still an open bail in Rn.) F is simply defined to be the orthogonal projection onto the closed and convex set {YE H-‘(B): V5 1) in H-‘(B). Thus F(v) s 1 and F(v) minimizes IIF(v) - VII with this property. (See list of notations for I(.I1etc.) Since A :HA( B) + H-‘(B) is an isometric isomorphism the definition of F can also be written F(v) = Y+ Au
(V&H-‘(B)),
(8)
B. GLSTAFSSOS
206
where 11E HA(B) is the solution of the minimum norm problem UEK /lL~/ls Ii 4
for all u 5 K. 1
with K = {u E H&B) : Au f v I 1). The solution of (9) may also be characterized inequality uEK
(9)
as being the unique solution of the variational
and for all u E K, 1
(u, u - u) 2 0 or as the unique sohrtion of the compIementarity
(10)
problem
Au+vSl
(11)
uzo
(12)
(Au + Y- 1, u) = 0.
(13)
The equivalence between (9) and (10) is a theorem in elementary Hilbert space theory and the equivalence between (10) and (ll)-(13) is also an instance of a general and rather weilknown principle (cf. [Z, theorem 4]) and the simple proof is found, for example, in [4, theorem 31. Actually, the characterization (Il)-(13) of u is the one we most often will use. The operator Fcan also be defined in terms of the solution of another variational inequality. Given v E H-‘(B) let 1~E H;(B) be defined by A~/J= v-
1
(14)
and put o=u+tJ. Then u satisfies (ll)-(13)
(15)
iff u satisfies AudO
(16)
VZ$)
(17)
(Au, u - y) = 0.
(18)
Further, u satisfies (16)-(18) iff u solves the variational inequality VE L
and
( u, w - 0) B 0
for all 1~E L, 1
(19)
where L = {w E H,j(B): w 2 I#}. The first statement above is obvious and the second one is an instance of [2, theorem 41. (14) and (15) above show that Au + Y= Au + 1. It follows that the definition of F can be written F(Aly+
1) = Au f 1
(@H:(B)),
(20)
where u E H,$(B) is the solution of (16)-(18) or (19). Now F can be used to handle the concept of weak (forward and backward) solutions very
Existence of weak backuard solutions to a generalized Hels Sha\s flow
Xl:
conveniently. Let D be a given open set with supp ,u C D C B. Then {D,: r E [O. T]} (open sets uith supp !i C D, E B) is a weak forward solution with Do = D (except for a null-set) if and only if F(t;i -!- x0) = XD,
fort E [O. T].
(21)
In fact, if {D,} is a weak solution with Do = D then (2)-(J) show that ~1~E H:( B) defined bv (2) SatiSfieS (11)-(13) with v = t,u + XD. Hence F(~,u + ~0) = tu + XD + AllI = %D.by (2). ‘Conversely, if F(r;c + XD) = x0, (t E [0, T]) then, by the definition of F, x0, = r,u 1. x0 -L Art, where U, E Hd( B) satisfies (ll)-(13) with v = t,u + xo. For f = 0. 11= 0 solves (1 l)-( 13) so that ~0 = 0 and Do = D except for a null-set. Now it follows immediately that u, satisfies (2)-(4) and hence that {D,} is a weak solution. In a similar manner the concept of a backward solution can be expressed in terms of F. Thus given D { D_,:t E [0, T]} ( w h ere T > 0 and supp rc C D_, C B) is a backward solution with D as initial domain if and only if Do = D and F(fp - ;co_,) =%Dn_r (except for null-sets) for t E [O, T]. Therefore the conclusion of theorem 1 can be formulated as follovvs: if t > 0 is small enough there exists a domain D-, with supp ;l C D_, C B and F(~,u + x0_,) = %D.
1. The operator F: H-‘( B) += H- ‘(B) defined by (S)-(13) has the following properties (with V, vl, ~1, u E H-‘(B)). F(v) 5 1. If v 5 1 then F(v) = v. F(F(v)) = F(v). IIF - F(e)11 5 11vl - VII/. In particular F is continuous. If v1 5 r+ then F( zq) S F( v,). If vz Z 0 then F(F(u,) + e) = F(q + vz). ,u 5 F(v) whenever ;d 5 v and .I( I 1. In particular min{v, 1) 5 F(V) if Y is a measure (in H-‘(B)). Suppose I, can be written v = ,U + f, where ,u Z 0 and has compact support in B, f E L”(B) and, for some open set D with supp ,UC D G B,fZ 1 in D andfs 1 outside D. Then F(v) = ,yn + fxa*, where SJ = D U {x E B : u(x) > 0) and 11E H&B) is the function occurring in the definition (8) of F. If D is connected then so is S2. As a particular case, taking f = x0. we have F(,u + x0) = xn.
PROPOSITION
Remarks. (1) (g) of the proposition can be thought of as a generalization of known regularity results for solutions of the variational inequality (19) (or (10)) in the sense that it generalizes the inequality (23) below. (2) (h) of the proposition is a minor generalization of results in [4, 8, 91. It contains the existence and uniqueness theorem mentioned earlier for weak forward solutions. (3) The operator F is closely related to the “operator” Q(. , SL’) used by Sakai [8, 91. In fact, suppose u is a positive measure with compact support in ira’ and of finite energy. and let R be a bounded open set in W”. Choose B so that !G?G B and supp v C B. Then R E Q( V, SLi) implies F(v) = xn. Conversely, if F(v) = xn then there is an open set D with /DAR / = 0 such that D E Q(v, SL’) ([9, propositions 1 and 41 essentially). Before entering the proof of proposition 1 let us recall some known facts that will be used.
208
B. GCSTAFSSOS
1. Regularity of the solution of (19) [l, theoreme
1.31, [j], [6, theorem 2.3, p. 1081.
Given Y E Zfd( B), let u E Ho’(B) be the solution of (19). Then, if y E !f’,P( B) for some p with n
(22)
Suppose that v E LP( B) (n
(23)
u E Hd( B) and p E L*(B) C H-‘(B), then up E L’(B) and (p, u) = _faupdx. This easily from the definition of (* , a). (See list of notations.) u E H*+“(B) for some p 2 1 and u = 0 on a set EC B, then Au = 0 a.e. on E ([6. A.4, p. 531 applied twice).
Proof of proposition 1. (a)-(c) follow directly from the definition of F. and (d) is an elementary property shared by all projection operators onto closed convex sets in Hilbert spaces. (e) Due to the continuity (d) of F and the fact that the cone P = {ve H-‘(B): VZ 0) is closed in H-‘(B) it is sufficient to prove (e) for vI and Q in some subspace which is dense in H-‘(B) and whose intersection with P is dense in P. It is well-known that, for example. Cr( B), and hence any subspace of H-‘(B) containing C”(B), has this property. Thus we may assume that vl, y E L^( B) (say). Let ul, u2 EH& B) be the functions in (S) for Y= vl, y. Then, by (22) ul, u2 E H’+‘(B) for all p C 2. In particular u1 and u2 are (or. strictly speaking, have representatives in form of) continuous functions. Novv. let us first see that ui 5 u2 (assuming u1B y). Put W = ui - u*, Vi = {XE B: ui(x) > 0}, i = 1, 2. Then Ui are well-defined open sets. We want to prove that w d 0 in B. This obviously holds in B\U1. In Ii* Aui = 1 - ~1 by (13). Hence Aw = 1 - v1 - Au2 h 1 - y - Au2 2 0 there and since w = 0 on dU1 fl B and w E Hh( B) the maximum principle for subharmonic functions shows that w 5 0 also in UI. This proves that u1 5 u2 in B. Now in U2 Au2 + y = 1 by (13). Hence F(q) = Au, + v1 5 1 = Au2 + uz = F(ti) in r/z. In B\U2 u1 = u2 = 0 by the inequality u1 s ~2. Hence Aui = Au2 = 0 a.e. in B\U? by statement 3 before the proof. Therefore F(q) = Au1 + v1 = v1 Z y = Au2 + y = F(y) a.e. also in B\U:. Hence F(q)6 F(y) a.e. in all of B, from which (e) follows. (f) Assume y 2 0 and let ul, u2 E Hi(B) be defined by F(Q) = vl + Au1 F( F( q) + v2) = F(q)
respectively.
(25) + u2 + Au2
(26)
This means, by definition, that Aui + ul Z 1
(27)
ui z 0
(28)
(Au1 + ul - 1, ul) = 0
(29)
Existence
of weak backward
solutions
to a generalized
109
Hele Shaw tlon
and Au?+F(v,)+
e%l
(30)
uzz 0
(31)
( Au1 + F( or) + yz - 1, u2) = 0.
(32)
To show that F( F(vJ + 5) = F(vi + ~2) is, by (25)-(26), equivalent F(Y~ + I+) = VI+ y + A(ul + UZ), i.e. that u1 + uz E H;(B) satisfies
to showing
that
A(ui + uz) + v1 + y 5 1
(33)
111 +
(34)
l.lz 2
0
(A(u, + uz) + vI + vz - 1, u1 + ~1~)= 0.
(35)
Now (33) follows immediately from (25) and (30), and (34) follows from (28) and (31). Since I+ Z 0 we have, by (c) and (e) of the proposition, that F(v,) = F( F( u,)) s F( F(Q) + y). Hence Au, + v1 ~5A(ul + UZ) + vl + Q. Therefore, using (33), (34). (25), (32), (28) and (29), 0 2 (A(u, + uz) + it+
fi-l,ul+uz)=(A(u,+~~l)+
v,+ vZ-l,rtl)
2 (Au1 + vl - 1, u,) = 0. This shows that also (35) holds and so proves (f). (g) follows by combining (b) and (e). (h) Let F(V) = v + Au, K E Hd( B). Since Y= ,u + f is a Radon measure 2 - I/f]‘_ (a) and (g) of the proposition show that - llf/lI 5 min{l, V}z5 F(v) s 1. In particular F(v) E L”(B).
Subtracting
(36)
v from the above inequality yields
- ,u - 211fllr I min{l - V,0) P Au Z 1 - v.
(37)
Since 1 - v5 1 + llfllx (37) sh ows that Au is bounded from above. Hence u can be written as the sum of a continuously differentiable function and a superharmonic function. In particular u has a unique representative in form of a lower semicontinuous function. In terms of this representative we define u = {x E B:u(x)
> O},
(38)
which is an open set. sow it follows from (ll)-( 13) that F(v) = 1 in II. In D we have v h 1, so F(V) = 1 also in D, by (36). Hence F(v) = 1 in R = D U U. Finally, in B\supp ,u Au is bounded both from above and below by (37) so u E H’.P in Bkupp p (for all p < =J). Since u = 0 in B\U this shows that Au = 0 a.e. in (B\(I) fl (Bkupp ,D) = B\( U U supp y). In particular this holds a.e. in B\( U U D) = B\Q so that F(v) = ki +f + Au =f there. Putting together, we get F(p + f) = XQ+ fxB,nas required. If D were connected and R disconnected then U would have a component w, disjoint from D. Then u > 0 in w, u = 0 on ao and Au = F(v) - v=l-v=l-fZOinw.This,however,
B. GL-ST.AFSSOY
210
contradicts
the maximum
principle.
Hence
R is connected
if D is. completing
the proof
of
(h). 4. PROOF
OF THE
THEOREM
The rest of the paper is devoted to the proof of theorem 1. The proof is based on two lemmas stated below. The first of these shows that the measure ~1 in the theorem can be assumed to be smooth and the second one contains the most technical parts of the proof. LEMMA
1. Let k1,fand D be as in (h) of proposition 1 and let h E LX(F) be a radial function (i.e. such that h(x) is a function of 1x1 only) with supp h C B(0; E), 112 0 and Jh dx = 1. where 0 < 2~ < dist(supp ;l, B\D). Then F(rc + f) = F(rc * h + f).
Remnrk. Lemma lemma 61).
1 is similar
to (actually
a slight generalization
of) [A, proposition
21 and [9.
LEbiht~ 2. Suppose JLE L”(B), ;lZ 0 and .I( has compact support in B and suppose D is a domain with supp !LC D C B and such that dD is a finite disjoint union of real analytic hypersurfaces. Then, if t > 0 is small enough, there exists a domain U with supp ~1C U @ D and a function u E Hd( B) such that
Lvhere p E H-‘(B).
11 =
0
in BLD
(39)
11 2
0
in B
(10)
An = xD,,C - tu - p in B.
(11)
p 2 0 in B. p = 0 outside
U and p E L” in a neighbourhood
of B\U.
of theorem 1. Let u, B and D be as in the theorem, i.e. B = {,KE 2” :;.r/ < R} say. Radon measure with compact support in B and belonging to H-‘(B). and D a domain with supp p C D C B and with dD consisting of a finite number of disjoint real analytic hypersurfaces. Consider Proof
,ua nonnegative
F(t;! + v) = %D
(Q)
as an equation to be solved for v E H-‘(B). According to the statement immediately before proposition 1, the assertion of the theorem is that if t > 0 is small enough (12) has a solution of the kind v = XQfor some domain 8 (to be D-,) with supp ~1C R. (Q G B is then automaticall> satisfied since Q C D by (h) of proposition 1.) It follows from lemma 1 that we can replace .I( by J( * h with h as in that lemma. This means that we may assume that ~1is arbitrarily smooth. Let us assume that $1E L”(B) (this will be enough). Now the idea of the proof is to first show that (42) has a solution of the form v = XC + p. where U is a domain with supp u C IiE D and p is a function in H-‘(B) with p Z 0 in U and p = 0 outside U. Such a solution can rather easily be constructed (this is lemma 2). Then we apply (h) of proposition 1 to get rid of the excessive function p (;cn = f’(~c + PI>. Thus, apply first lemma 2 to the situation in theorem 1. This gives us a domain U and functions u and p (if t > 0 is sufficiently small). Put v = XC + p. Then tu + v - Au = XD 5 1
Existence
of weak backward
solutions
to a generalized
Hele Shaw flow
211
by (41), u Z 0 by (40) and (f,u + Y + Au - 1, u) = (~0, u) = 0 by (39). This shows that F(tu -k v) = t,u + Y + Au = x0, by the definition of F. Next apply (h) of proposition 1 to v = xU + p above. This gives us a domain R with I/ c Q (hence supp .u C Q) and F(v) = xQ. (f) of proposition 1 then shows that F(t,u T xn) = F(t/l + F(v)) = F(t,u + v) = XD. Thus Y = ;cn is a solution of (42) of the required type and so the theorem is proven (except for the proofs of the lemmas). Proof of lemma 1. We first note that we may assume that f = 1 on D. For if we have proven the lemma for such f then, using (f) of proposition 1, we get for general f F(u + f) = F(CI + =
F(F(u
=
F(
F(
XD
+
+
XD
u * h
fxB\,D
+
+
+
fxB\D)
XD
+
(f
-
+
F(p*h
+
(f -
l)xD)
+
-
fxB?D>
=F(!~*h+XD+fXfhD+
=
l)xD)
(f
(f-
l>xD)
l)xD)
f).
Thus we assume that f = 1 on D. Let F(p + f) = y + f + Au, u E Hi( B). Then (h) of proposition 1 shows that p + f + Au = XQ + fxe,n where Q > D. Thus AIA = -u in D, in particular u is harmonic in D\supp ~1. Define u = u + xo(u * h - u),
(43)
where G = {x E B: dist(x, supp cl) < E}. Then G C D and dist( G, B\D) > E. The mean value property for harmonic functions shows that u * h = u in a neighbourhood of aG. Therefore the discontinuity of xo does not matter in (43), u E Hi(B) and taking the Laplacian yields Au = Au + ~o((Au) = ((Au) = We now aim at showing that u satisfies
that
F(u
-U
*
* h - Au)
* h)xc + Au *~a\o h + Au ’ XB\G.
* h + f) = u * h + f + Au. This
(44) amounts
to showing
!l*h+f+Avll
(45)
US.0
(46)
(u*h+f+Au-l,u)=O.
(47)
(44) shows that u*h+f+
Av=f+Au~~B\G=,u+f+Au-(Au.~G+p)=p+f+Au.
(48)
Hence (45) is satisfied since ,u + f + Au S 1. Moreover (46) holds, for in B\G, u = u 2 0 and in G Av = -,u * h 5 0 by (44), so (46) follows from the minimum principle for superharmonic functions. Finally, since u is superharmonic in D, u * h 5 u in G, showing that u 5 u in B, and (17) follows from O=(p+f+Au-1, u)=(,u*hff+Au-1, u)l(,u*h+f+Au-1,
212
B. GLSTAFSSON
u) S 0. (In the first inequality we used u S u and (45) and in the second (45) and (46).) This proves that F( ,u * h + f) = ,u * h + f + Au. Now the lemma is proven for the assertion F( u * h + f) = F( y + f) is nothing else than (48).
Proof of lemma 2. By the Cauchy-Kovalevska theorem there is an open neighbourhood of dD and a real analytic function u defined in V such that
u =Oandgradv
=OonaD
V
(49)
ho = 1 in V.
(50)
We may assume that D\v is connected and that ri fl supp ,U= 0. For LY> 0 put V, = the union of those components of {XE V: v(x) < a} which intersect aD. Then V, is open and aD C V, C V. Now we want to choose (Y> 0 so that: (a) V, is relatively compact in V, (b) u L 0 in V,. (c) grad v # 0 on av,. Let us see that this is possible. That aD is a finite disjoint union of real analytic hypersurfaces means that there are open sets Gj (i belonging to some index set) and real analytic functions vi: G,+ iR such that {Gj} covers dD, aD II Gj = {X E G,: qj(X) = 0) and grad qj # 0 on dD. Since dD is compact the collection {Gj} can be taken to be finite. We may also assume that grad qj is bounded away from zero on aD n Gj. The Cauchy data (49) for u imply that there are real analytic functions a,: G, + 2 such that U(X)
= a,(X)Qji(X)’
on Gj.
(51)
Taking the Laplacian of (51) gives (in view of (50)) that 1 = Au = 2aJgrad FjCij’ on aD f? Gj. Hence aj(x)Zc,>O where cj are constants. Moreover,
onaDnGj,
(52)
taking the gradient of (51) gives
grad v = 2aj grad Cpi+ qj grad aj V
in Gj\a D.
Now (51) and (52) (together with grad qj f 0 on dD) show that given any neighbourhood W of aD (e.g. W = lJjG, or W = V) V, C W for a> 0 sufficiently small. This immediately shows (a) and (b) for cy> 0 sufficiently small. Moreover, (c) follows (with cu> 0 sufficiently small) since the right member of (53) is continuous in Gj and nonzero on JD. From now on let cu> 0 be fixed such that (a)-(c) hold. From (a) and the definition of V, follow (d)
v =
a
on av,
and from (b), (d) and the definition of V, (e) grad v is directed out from V, on aV,.
Existence
Moreover (f) D\v,
of weak backward
solutions
to a generalized
Hele Shaw flow
213
we have is connected.
In fact, D\ri C D\ri, and D\v was assumed connected
so if D\vi, were not connected it would have a component, o, disjoint from D\ri. Then o C V (since o C D) so that u is defined in o. Since V, is a neighbourhood of aD dw II SD = 0. Hence aw C aV,. Therefore u = (Yon aw by (d) and now (50) and (e) immediately yields a contradiction. Hence (f) holds Now we define U = D\ri,.
Then I/ is open and connected, supp ,u C II E D and dU = D n dV,. It follows from (a), (c) and (d) and the fact that u is a real analytic that dU is a finite disjoint union of real analytic hypersurfaces. Therefore, and by the Cauchy-Kovalevska theorem again, there is an open neighbourhood W of dU and real analytic function w defined in W such that w = u = crand
(54)
gradw=gradu Aw=O
ondU
inW.
(55) (56)
In a similar manner as for u it follows from (c), (e) and the definition of w that there exists a p > (Yand a neighbourhood Wp of aU, Wp C W such that Wp is relatively compact in W.
w = /30naWPn
U.
We assume that W is chosen so small so that W rl supp ,u = 0. Now choose E > 0 such that cy+ ER’ < /3 (where B = {x E W”: jx j < R}) and define a function u in B by u=o
in B\D
(57)
u=u
onDnv,=D\U
(58)
u(x) = min{w(x), (Y+ E(R* - Ix 1’) forx E U n Wg u(x) = a+ E(R* - Ix/*)
forx E u\Wp
(59) (60)
We claim that the set U and the function u have the properties stated in the lemma (with p defined by (41)). As already remarked U is a domain with supp ,u C U G D. Since CY < LY+ E(R* - /xl’) < p for all x E B (by the definition of E), w = CY on au = a(un W,) n (D\U) and w = p on JWb n U = a( U fl W,) fl (U\Wp) we have, in U n W,, u = w in a neighbourhood of aU and u(x) = (Y+ E(R’ - jx12) in a neighbourhood of aWp n U. From this it follows that u is continuous in D (and in all of B). In view of (49), (54) and (55) it also follows that u is continuously differentiable in a neighbourhood of B\U. Moreover, (50) and (56) show that Au = xD\u in a neighbourhood of
B. GLWAFSSOS
211
B\U. (There will be no contributions to Au from the surfaces of discontinuity. JD and dl’ since these are real analytic and the zeroth and first order derivatives of u are continuous over them.) Next observe that CYY+ E( R’ - ;xl’) is superharmonic: A(@+
E(R’-
ixi’)) = -2m<
0.
(61)
Since w is (super-)harmonic and the minimum of two superharmonic functions is superharmanic, it follows from (59)-(60) that u is superharmonic in I/. Moreover, uiL. E H’(U) because both w and CY+ E(R’ - 1x1’) belong to H’ where they are defined and the maximum of two functions in H’ is a function in H’ [IO. proposition 28.21. By (61) Au = -2nE in a neighbourhood of supp ,u. Therefore, and since J! E L”(B), Au 5 -@ in U for t > 0 small enough. This shows that (41) holds with p defined by p = -Au - t,u in U, p = 0 in B\U and t > 0 so small so that p Z 0. Since u belongs to H’ in U p belongs to H-’ in U. Moreover, p E L” in a neighbourhood of B\U. It follows that p E H-‘(B) and that the statements about p in the lemma hold. Finally, it follows immediately from the definition of u that (39) and (40) hold. Since 11E HJ( B) follows from (39), (41) and that p E H-‘(B) the lemma is now proven.
APPENDIX-LIST
OF
NOTATIONS
(u,b)={.rER:a
denotes
AU = c$, 2
H”.p( B), H”(B)
(A also denotes
symmetric
difference
between
sets.)
the closure
of D in ?-.
)
= kfm.:( B). fij( B):
Sobole\
spaces as defined in [lo]. Hd( B) is a Hilbert space with inner product
(B always
bounded
here) and norm
H-‘(B) is also a Hilbert space, equipped with that inner product and norm (also denoted ) xvhich makes A : Hd(B) -, H-‘(B) an isometric isomorphism [ 10, theorem 23. I]. u E H.]( B)) when H-‘(B) is regard as the dual space (u. o) the pairing between H-‘(B) and Hd(B) (u E H-‘(B), of I&j(B) in the usual way. Then (u, u) = -(Au, u) (u, v E f&‘(B)) [lo, section 231. All functions, elements in Sobolev spaces and measures in this paper are regarded as, and treated formally as. distributions. Thus ~5 1 (in an open set) e.g. means that 1 - v is a positive distribution (in that open set). B) “measure” we always understand Radon measure, i.e. Bore1 measure which is finite on compact sets.
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to a generalized
Hele Shaw flow
215
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