AN EXISTENCE AND UNIQUENESS RESULT 21research was supported by the National Science Foundation under Grant No.DMS 9206244. Bibliography[1] Agmon, S., Douglis, A., and Nirenberg, L., Estimates near the boundary for solutionsof elliptic partial di�erential equations satisfying general boundary conditions II, Comm.Pure Appl. Math. 17, 1964, pp. 35{92.[2] Backus, G. E., Application of a nonlinear boundary value problem for Laplace's equationto gravity and geomagnetic intensity surveys, Quart. J. Mech. Appl. Math. 21, 1968,pp. 195{221.[3] Bers, L., Function theoretical properties of solutions of partial di�erential equations ofelliptic type, pp. 69{94 in: Contributions to the Theory of Partial Di�erential Equations,L. Bers, ed., Annals of Mathematics Studies No. 33, Princeton University Press, 1954.[4] Bers, L., John, F., and Schechter, M., Partial Di�erential Equations, Interscience Pub-lishers, Wiley, New York, 1964.[5] Berestycki, H., Nirenberg, L., and Varadhan, S. R. S., The principal eigenvalue and maxi-mum principle for second-order elliptic operators in general domains, Comm. Pure Appl.Math. 47, 1994, pp. 47{92.[6] Gamba, I. M., and Morawetz, C. S., A viscous approximation for a 2-D steady semicon-ductor or transonic gas dynamic ow: Existence theorem for potential ow, Comm. PureAppl. Math., to appear.[7] Gilbarg, D., and Trudinger, N. S., Elliptic Partial Di�erential Equtions of Second Order,Springer-Verlag, New York, 1983.[8] Grisvard, P., Elliptic Problems in Non-Smooth Domains, Monographs and Studies inMathematics No. 24, Pitman, Boston, 1985.[9] Ladyzenskaja, O. A., and Ural'ceva, N. N., Equations aux D�eriv�ees Partielles de TypeElliptique, Dunod, Paris, 1968.Received February 1994.

20 IRENE M. GAMBAR = [C1; C2] � [0; 1], such that F (@i) = @iR with @ = 4[i=1 @iK, each @iK aside of K; where is the domain characterized in the previous section.Proof: It is essential for the regularity of the conformal map that@ = 4[i=1 @isuch that each @i are (piecewise) C2;�-curves which joints the preceding oneat the point !i making a �2 angle; that is, if i�1, i are the parametrizations of@i�1 and @i, respectively, i�1T (!i) ? iT (!i), where the subindex T denotestangential derivative.Under these conditions, the proof is given by the classical result of conformalmapping that corresponds to an incompressible transformation. Take `2(~x) thereal harmonic function that solves8>>><>>>: �`2 = 0 in `2���@4 = 0; `2���@2 = 1@'2@� ���@1[@3 = 0 ;`2 is a C2;�(�) harmonic function as the boundary condition is compatible onthe points !i, i = 1; . . . ; 4 because of the assumption of orthogonality on @ at!i.Moreover, by the Hopf maximum principle, 0 < `2 < 1 in and `2y > 0 on@4 [ @2.Now, let `1 be the conjugate harmonic function obtained by integrating theorthogonal �eld to the level curves of `2, that is, `1`1(x; y) = (x;y)(t)Z(x0;y0) (`2y � `2x) � 0(t)dt+ `1(x0; y0)where (t) is a path joining (x0; y0) to (x; y) with (x0; y0) a �xed point in @.Then `1 2 C2;�(�) satis�es @`1@T = 0 on @1 [ @3, then `1 = C1 = `1(x0; y0) if(x0; y0) 2 @1R and `2 = C2 where C2 is �xed by the path integral and by C1.Moreover, the function `2;y satis�es �`2;y = 0, `2;y > 0 on @4 [ @2 andr`2;y � n = 0 on @1 [ @3, then, by Hopf maximum principle, inf `2y ��inf@ `2y > 0. Thus, F (z) = `1 + i`2 is analytic and dF=dz = `1x + `2y + i(`1y �`2x) = 2(`2y� i`2x) has positive real part. Then F (z) is one to one. Hence, F (z)is a conformal transformation that satis�es F (�) = [C1; C2]� [0; 1].Acknowledgments. I would like to thank Robert Kohn, Cathleen S. Mor-awetz, and Louis Nirenberg for many helpful discussions and suggestions. This

AN EXISTENCE AND UNIQUENESS RESULT 19where ~A(x) = ( A(x) if x 2 �R(�A1(x�); A2(x�)) if x 2 �~R n �Rand x� 2 R is the re ection of x 2 �~R n �R with respect to �1.In particular, for any test function ' 2 H10( ~R)(3.15) Z~R r'(r~! � ~A(x)~!) = 0 :We prove that the boundary value problem (3.14) has a unique solution ~! = 0.In fact, it is enough to see that the adjoint operator to L is solvable in H10( ~R)for any given bounded function in ~R as a right-hand side. Indeed, by estimate(3.8) ~! = ~b1 � ~b2 is C�( ~R).We solve for in H10( �~R) the adjoint problem(3.15) L� = � + ~A(x)r = ~! in ~R = 0 on @ ~R :Since 2 H10 is an admissible test function, it satis�es (3.15). Then(3.16) 0 = Z~R r (r~! � ~A(x)~!) = � Z~R (� + ~A(x)r )~! =so that 0 = Z~R ~!2 :In particular, ~! = 0 in ~R.The existence and uniqueness of problem (3.16) in ~R can be found in a recentpaper by Berestycki, Nirenberg, and Varadhan; see [5]. They show that for given~! bounded in �~R, there exists a unique solution of the boundary value problem(3.16), where is W 2;p near all smooth boundary points and continuous atcorner points. Moreover, it is easy to see from their proof that r 2 L2( �~R). Inparticular, 2 H10 ( ~R) so that (3.16) holds, and consequently, ~! = 0.AppendixLemma (The Conformal Transformation). There exist two con-stants C1 and C2, and a unique conformal transformation F (z): ! R, with

18 IRENE M. GAMBAdiv (A(x)(v1�v2)), with A(x) = @f@v (x; (�v1+(1��)v2)(x)). Since A(x) 2 C�( �R)using estimates (3.9) and (3.10) for this equation the following estimates hold:kb1 � b2kC1;�( �R) �� C �kAkC�( �R)kv1 � v2k1; �R + kAk1; �Rkv1 � v2kC�( �R)��� CkAkC�( �R)kv1 � v2kC�( �R) :Thus, T is a continuous map.Therefore T has a �xed point b which is the solution of the boundary valueproblem (1.3), and, by the estimate (3.11), (1.5) holds. Similarly from (3.2),(1.6) holds.(b) Uniqueness of the Solution of Problem (1.3)Theorem 3.2. Let b1 and b2 be two solutions in W 1;p( �R); p > 2, of theboundary value problem (1:3), then b1 = b2 on �R.Proof: We prove uniqueness in the re ected domain for the correspondingDirichlet boundary value problem in a similar way as we worked out Lemma 3.1.Let ! = b1 � b2 on R. Then ! satis�es�! = div (f(x; b1)� f(x; b2)) = div (A(x)!)where(3.12) A(x) = 1Z0 fb(x; t b1 + (1� t) b2) dt :Since f is Lipschitz in b and C� in x in �R, then A(x) is L1( �R).Then, ! is a C1;�( �R) solution of linear elliptic operator in divergence form(3.13) 8><>: L! = �! � div (A(x)!) = 0 on R! = 0 on �!x = A1(x)! on �1 :Now if ~R de�ned as above is the rectangle that contains R, its re ectionwith respect to �1 and �1, then, taking ~b1 and ~b2 the even re ection of b1 andb2 respectively with respect to �1. Their di�erence ~! is an even function thatsolves(3.14) L~! = �~! � div ( ~A(x)~!) = 0 in ~R! = 0 on @ ~R

AN EXISTENCE AND UNIQUENESS RESULT 17In fact, if B = (B1; B2) solves uniquely �Bi = fi(x) in R, Bi = 0 on @R,with fi 2 C�( �R1) \C�( �R2), and e�1 = �R1 \ �R2 for i = 1; 2.By [1], each Bi 2 W 2;p( �R0) for 1 < p < 1, and, in particular, in C1;�(R0)for 0 < � < 1, for any regular subdomain R0 of R.Hence, [Bi] = 0 = [(Bi)x] on e�1. This implies that Bi is the unique solutionof the elliptic di�raction problem de�ned above. By [9], each Bi is C2;�(�k); k =1; 2, where �k = (R0 \Rk) [ S for S any part of e�1 satisfying �1 �� S �� e�1.Moreover, the following estimate holdskBikC2;�(�k) �� CkfikC�( �R) �� CkfikC�( �R) ; i = 1; 2 k = 1; 2where C depends only on R.Therefore, since h = div Bi�b solves �h = 0 in R0 with h = div Bi on @R0,then h is harmonic in R0. Hence, h is C1 in the interior of R0, so that b is asgood as div Bi, and estimating b in �R � R0kbkC1;�( �R) �� Ckdiv BikC1;�( �R) �� CkBikC2;�( �R) �� CkfkC�( �R) :Therefore,(3.9) kbkC1;�( �R) �� C kf1; f2kC�( �R)where(3.10) kf1; f2kC�( �R) = kfkC�( �R);Lip;v�� C �kfkC�;x; �Rkfk1; �R + kfkLip;vkvkC�( �R)�where C depends on R. Therefore (3.2) holds. The proof of Lemma (3.1) is nowcomplete.In order to complete the proof of the existence part of Lemma 1.2, we makeuse of the Leray-Schauder �xed-point theorem; see [7], Chapter 11.Let the map T (v) = b be given by the solution of problem (3.1) which can beestimated by (3.2) and (3.3).Indeed, estimate (3.3) provides the needed a priori estimate to show theoperator T from the Banah space B = C�( �R) into itself is bounded. Moreover,T (C�(R)) � C1;�( �R) is compactly embbeded in C�( �R), then T is a compactoperator, and, for any b in B = C1;�( �R) such that b = �Tb, for some 0 < � < 1,from (3.3) we obtain the a priori estimate(3.11) kbkC1;�( �R) �� �C �kfkC�;x; �Rkfk1; �R + kfkLip;bkfk1; �R ��Mwhere C is independent of b and �, and so M is independent of � and b.Finally, since f is bounded and Lipschitz in v, then the map T is Lipschitzcontinuous. Indeed, T (v1) � T (v2) = b1 � b2 satis�es the equation �(b1 � b2) =

16 IRENE M. GAMBA0 = Z~R r ~'�r~b� ~f � dx= Z~R ~'x~bx + ~'y~by � ( ~'x ~f1 + ~'x ~f2) dxdy= 2 ZR r'�r~b� f� dx ;since ~'x;~bx, and ~f1 are odd functions and ~'y;~by, and ~f2 are even ones.Hence (3.1.w) holds for b = ~bjR, making this b the unique weak solution ofthe boundary value problem (3.1). In addition, by de�nition of ~f ,k ~fk1;�~R = kfk1; �R :Then by estimate (3.7)(3.8) kbkC�(�~R) �� CkfkL1( �R) ;where C depends only on R, so that estimate (3.2) holds.In order to obtain the C1;� regularity and estimate (3.3), note that ~f j �R 2C�( �R) and ~f �� ~RnR 2 C�( ~R nR) ( ~f is not C� across �1) and, by the boundaryconditions given on problem (3.1), ~f1 = 0 on @2 ~R [ @4 ~R...................................................................................... ..............................................................................................R ~R�1S R2R0 ~�1@2 ~R@4 ~RR1Figure 3. The re ected domain R.The C1;� estimate for b in �R is obtained �rst for b by di�erentiating C2;�solutions of elliptic di�raction problems de�ned in R where e�1 is the di�rac-tion boundary and the given right-hand sides are C� on each side of e�1; seeLady�zenskaja and Ural'ceva, [9].

AN EXISTENCE AND UNIQUENESS RESULT 15and bounded in R and C� on each side of R up to, but not across, e�1 and @1 ~R,the extension to R of �1 and @1R, respectively. See Figure 3.Recalling Agmon-Douglis-Nirenberg (see [1]) interior regularity results forelliptic problems, it follows that the solution b of problem (3.4.1) in the domainR for the subdomain �~R �� R satis�es the following estimate(3.6) kbkW1;p(�~R) �� CkfkLp( �R) �� CkfkL1( �R) ;where C depends only on R, since the de�nition of f only involves even or oddre ections of the components of f .Since ~b = bj ~R, the Embedding Theorem yields ~b 2 C�( �~R) along with theestimate(3.7) k~bkC�(�~R) �� CkfkL1( �R) ;where C depends only on R.Next, we show �rst that ~b is an even function in ~R and later that any even,and hence any solution of the problem (3.4.1){(3.4.2) yields a weak solution b ofproblem (3.1) in the sense (3.1.w).Due to the uniqueness of the boundary value problem (3.4.1){(3.4.2), it isenough to show that ~b(x�), where x� is the re ection of x with respect to �1, isalso a solution. Consequently, ~b(x�) = ~b(x), i.e., ~b is an even function.Note that if �1 � fx = 0g, then x� = (x; y)� = (�x; y). Thus �x = �x� andrx� � = �@x + @y as di�erential operators, hence�x~b(x�) = �x�~b(x�)and, by de�nition of ~f ,rx � ~f (x; ~v(x)) = rx � ~f(x; ~v(x�)) = rx� � ~f (x�; ~v(x�)) :Therefore if ~b is the solution of (3.4.1){(3.4.2) then�x�~b(x�) = rx� � ~f (x�; ~v(x�)) ;so that �x~b(x�(x)) = rx � ~f (x; ~v(x)) :In particular, since ~b(x�) = 0 on @ ~R, then ~b(x�) = ~b(x), so that ~b is an evenfunction in ~R.Now, let ' be a test function in H10(R2 n �). Let ~' be the even re ection of' to ~R, then ~' 2 H10 ( ~R), and, for the solution ~b of problem (3.4)

14 IRENE M. GAMBAproblem, and so keeps the interior regularity up to the boundary. Methods dis-cussed by Grisvard in [8] would also be adequate to obtained the same regularity.Let ~R be the rectangle given by the union of R, its re ection with respect to�1, and �1; and let ~v de�ne on ~R be the even re ection of v with respect to �1.We consider the following boundary value problem(3.4.1) �~b = div ~f (x; ~v(x)) in ~R~b = 0 on @ ~R ;where ~f (x; ~v(x)) is the extension of f(x; v(x)) to ~R given by(3.4.2) ( f(x; ~v(x)) if x 2 �R(�f1(x�; ~v(x�)); f2(x�; ~v(x�)) ) if x 2 �~R n �Rwhere x� 2 R is the re ection of x 2 �~R n �R with respect to �1, that is, f1 isre ected oddly and f2 is re ected evenly with respect to �1, respectively.Now, since v 2 C�( �R) and ~v is the even re ection of v with respect to �1,then ~v 2 C�( �~R), so that ~f is Lipchitz in ~v. However, since f1 does not vanishon �1, then ~f is discontinuous across �1 but ~f remains bounded and measurablein �~R.The existence of anH10 solution of problem (3.4) follows from standard results.In fact, a weak solution of problem (3.4.1) must satisfy(3.5) Z~R r'r~b = � Z~R r'f(x; v) for all ' 2 H10( ~R) ;Since f 2 L1( ~R), then `(') = Z~R r'f(v)is a continuous linear form inH10 satisfying k`(')kH10 ( ~R) �� Ck'kH10kfk1; ~R whereC depends on the domain.By the Riesz representation theorem, there exists a unique solution ~b in H10of problem (3.4.1) that satis�esk'kH10 ( ~R) �� Ckfk1; ~R :Next, the solution ~b can be re ected oddly across all boundary sections @ ~Ri,and it de�nes a b that solves a similar problem for the enlarged domainR in thespace H10 (R) with a right-hand side f . Now this right-hand side f is measurable

AN EXISTENCE AND UNIQUENESS RESULT 133. Proof of Lemma 1.2Without loss of generality, we assume that �1 is contained in the axis x = 0and that its lower end is the origin.(a) ExistenceWe �rst show the existence of a weak solution to the boundary value prob-lem (1.3) by constructing a precompact continuous map T from an appropriateBanach space B into itself. Then we use the Leray-Shauder �xed-point theorem(see Gilbarg-Trudinger, [7], Chapter 11), where B is C�( �R).Thus, we let v 2 C�( �R) such that v � 0 on �.We de�ne the map T by letting b = Tv be the unique weak solution inC1;�( �R) of the linear boundary value problem(3.1) �b = div f(x; v(x)) on Rb = 0 on � andbx = f1((0; y); v(0; y)) on �1f1(x; 0) = 0 for any x 2 � \ (@2R [ @4R) :By weak solution we mean that(3.1.w) ZR r' � rb = ZR r' � f(x; v) for all ' 2 H10 (R2 n �) :Remark. Note thatZ@R '(rb� f(x; v) ) � nds = Z�1 '�bx � f1((0; y); v(0; y))� dy = 0for all ' 2 H10 (R2 n �).Lemma 3.1. The problem (3:1) has a unique weak solution b in C1;�( �R) inthe sense of (1:4) or (3.1.w) satisfying the following estimates:kbkC�( �R) �� Ckfk1; �R(3.2) kbkC1;�( �R) �� C �kfkC�;x; �R + kfkLip;v� kfk1; �R(3.3)where C depends only on �R.Proof: The proof of this lemma is an application of existence and regularitytheorems for elliptic equations in general smooth domain. Although our domainis a rectangle, the nature of the boundary data permits re ecting the boundaryappropriately such that we deal with a Dirichlet boundary value transmission

12 IRENE M. GAMBAwhere C depends on kgkC1;�(@3) and on jFzjL1 , i.e., C depends on the datag(x) and on the domain .Finally, by (2.4), we take(2.35) ' = Re8<: zZz0 wdz9=; ; ' = 0 on @1 ;then ' is a real function that solves the boundary value problem (1.1) and,since w has real and imaginary parts in the class of C1;�(�) functions, then' 2 C2;�(�) and by estimates (2.32) and (2.34)(2.36) k'kC1;�(�) �� CeCkhk1;� �1 + khk1;�� andk'kC2;�(�) �� CeCkhk1;���1 + khkC�(�)khk1;� + khk21;�� �1 + khk1;��where C depends on the C1;�-norm of g(x) and on the domain , so that esti-mates in (1.2) hold.In addition, by (2.33)(2.37) 0 < k e�CsupHCinf < jr'j < KeCsupHCsupwhere H is the upper bound of jhj, Csup and Cinf bounds depending on g(x) and, and k, K depends only on . We stress once more that Cinf > 0, because ofthe assumption inf g(x) > 0 on @3; see estimate 2.16.Furthermore, since the given data ' = 0 on @1 then r' � � = 0 along@1. Since jr'j never vanishes in � then r' cannot change sign in @1. Thus,r' � nj@1(w1) < 0, implies r' � n < 0 on @1. Therefore ' increases along thepaths orthogonal to the level surfaces that start at the boundary @1, and, inparticular,(2.38) 0 �� ' �� eK near @1 :In general,(2.39) � eK �� ' �� eK in � ;where eK , both in (2.38) and (2.39), depends on the diameter of and on theupper bound of jr'j.Now the proof of Theorem 1 is complete.In the last section, we prove our \key lemma" (1.2) to solve the boundaryvalue problem (2.21), (2.23).

AN EXISTENCE AND UNIQUENESS RESULT 11transformation of the domain onto R = [C1; C2] � [0; 1]. Recalling w thecomplex velocity representation in , we have by (2.1), (2.2), and (2.3) that(2.30) w�z = �(z)w + ��(z) �w ; w = 'x � i'y and�(z) = (h1 + ih2)(z) :Now, by (2.5), (2.6), and (2.7),(2.31) w = udFdz = uFzwhere u is the unique solution of the ow equation in the rectangle F () = Rof the boundary value problem given by Lemma 3.1, where�0 = (h1 + ih2)�dFdz ��1 :Hence, by Lemma 2.1,w = f(z)es(z) � Fz ; jf(z)j 6= 0 :Since F is an analytic function in , which is C1(�) (due to the specialgeometrical properties of ) and Fz 6= 0 in and jFzj is uniformly boundedaway from zero. See Lemma in the Appendix.Then <w and =w are C1;�(�) and by (2.10) the following estimates hold,kwkC�(�) ��CkfkC�(�) (inf jfj)�2 eCk�0fk1;� �1 + k�0k1;�kfk1;�� ; andkwkC1;�(�) �� CkfkC1;�(�) (inf jfj)�2 eCk�0fk1;�(2.32) �� 1+ �k�0kC�(�)k�0k1;� + k�0k21;�� kfk21;�� �1 + k�0k1;�kfk1;�� ;and the estimate for the absolute value of w(2.33) 0 < C inf njfj jFzjeinf <s(z)o �� jwj �� C supnjfj jFzjesup<s(z)o :The constant C and jFzj depend only on , the inf jfj is positive and dependson g(x) on @3R, and sup and inf of <s(z) = a(x) are bounded and, by (2.26),are bounded from above and below by the upper and lower bounds of jf(z)j from(2.16), from jF�1z j, and from ~h = (h1; h2) the given coe�cients of equation (1.1).Estimate (2.33) thus ensures that r' is positive throughout �.Furthermore, by estimate (2.16), the right-hand sides of (2.32) can be esti-mated by(2.34) CkfkC�(�) (inf jfj)�2 eCk�0k1;� �1 + Ck�0k1;�� ; andCkfkC1;�(�) (inf jfj)�2 eCk�0k1;���1 + Ck�0kC�(�)k�0k1;� +Ck�0k21;�� �1 + Ck�0k1;�� ;

10 IRENE M. GAMBAOnce we have the unique function b(x) = =s(z), then using equation (2.20)we �nd a unique a(x) = <s(z) by integrating the �eld(2.26) ra(x) = (by + p1(x; b);�bx + p2(x; b))where a(x) = 0 on @3R. In particular, �a = div p(x; b) with an = 0 on @2R[@4R,an = p(x; 0) � n on @1R, and a = 0 on @3R.Therefore, a similar result to lemma3.1 yields that a(x) is a C1;�( �R) function,and by (2.24) and (2.25), satis�es the estimateskakC1;�( �R) �� C �kpkC�;x; �R + kpkLip;b� kpk1; �R(2.27)and kakC�( �R) �� Ckpk1; �R ;(2.28)where C depends on R.We may now �nish the proof of Lemma 2.1.End of Proof of Lemma (2.1): Let the complex function s(z) = a + ib begiven, with a and b real C1;�( �R) functions which are the unique solutions of theboundary value problem (2.19), (2.21), and (2.23).Then s(z) solves the boundary value problem (2.13){(2.14) uniquely. There-fore, by (2.11), (2.17), and (2.13), u = f(z)ea+ib solves the boundary valueproblem (2.8), (2.9) in the rectangle R uniquely.Moreover, using estimates (2.24), (2.25), (2.27), and (2.28), along with theform of p given by (2.20), yields the estimateskukC�( �R) �� CkfkC�( �R) esup�R j<s(z)j �1 + kskC�( �R)��� CkfkC�( �R) (inf jfj)�2 eCk�0 fk1; �R �1 + k�0 fk1; �R� andkukC1;�( �R) �� CkfkC1;�( �R) esup�R j<s(z)j �1 + kskC1;�( �R)� �1 + kskC�( �R)��� CkfkC1;�( �R) (inf jfj)�2 eCkAikk1; �R(2.29) � �1 + kAikkC�( �R)kAikk1; �R + kAikk21; �R� �1 + kAikk1; �R��� CkfkC1;�( �R) (inf jfj)�2 eCk�0 fk1; �R� �1 + k�0 fkC�( �R)k�0 fk1; �R + k�0 fk21; �R� �1 + k�0 fk1; �R�where C depends only on R. Hence, (2.10) holds.Thus, the proof of Lemma 2.1 is now completed.Proof of Theorem 1: Let F be the conformal transformation and R theresulting rectangle given in the Appendix; then F : ! R is the unique conformal

AN EXISTENCE AND UNIQUENESS RESULT 9where A + iB = �0f, and f = fI + ifR, then the terms p1 and p2 from (2.18)can be written as(2.19) p1(x; b) = 2jfj�2[fRA(1 + cos 2b)+ fIB(1 � cos 2b)� (fRB + fIA) sin 2b]p2(x; b) = 2jfj�2[�fIA(1 + cos 2b)+ fRB(1� cos 2b)� (fRA+ fIB) sin 2b] :Clearly,(2.20) pi(x; b) = 3Xk=1Aik (x)`ik(b) ; i = 1; 2where Aik(x) are real C� functions in �R and `ik (b) bounded and Lipschitz.Since pi does not depend on a = <s(z), system (2.19) yields an ellipticequation for b(x) alone, namely(2.21) �b = p2x � p1y :The boundary conditions for b are obtained from (2.14) and the relationship(2.19).Since <s = a = 0 on @3R, then ay = 0 on @3R, and by (2.19), bx = p2(x; b)on @3R.Also, =s = b = 0 on @R n @3R. In particular, since f is real at @3R[ @4R andat @3R [ @2R, we have the compatibility condition(2.22) p2(x; 0) = 0 if x 2 @3R \ @4R or x 2 @3R [ @2R :Therefore, we must show that there exists a unique solution of (2.21), along withthe data(2.23) � b = 0 on @R n @3Rbx = p2((C2; y); b) on @3R :In the next section, we prove Lemma 1.2 which ensures the existence of aunique weak solution b(x) in the class C1;�( �R) of the nonlinear boundary valueproblem (2.21){(2.23) that satis�es the estimateskbkC1;�( �R) �� C �kpkC�;x; �R + kpkLip;b� kpk1; �R;(2.24)and kbkC�( �R) �� Ckpk1; �R(2.25)where C depends on R.

8 IRENE M. GAMBAre ected rectangular domain ~R whose sides @i ~R �� @iR for i = 1; 3. Let f<h bethe re ected function of <h. Also, let G be the even re ection of `n ~g across theendpoints of @3R de�ned on @3 ~R.Then f<h is a harmonic function in ~R that satis�es @f<h@n = 0 on @1 ~R[@2 ~R[@4 ~Rand f<h = G on @3 ~R.In addition, f<h is C1( ~R) [ C1;�( ~R [ T ) where T is any regular portion ofthe boundary of ~R. Then <h = f<hj �R is in C1(R) [C1;�( �R).Therefore, since the harmonic conjugate of <h inherits its regularity, =h isC1(R) [C1;�( �R) and the following estimates holdk<h ; =hkC1;�( �R) �� Ck~gkC1;�( �R) ;where C depends on R.Moreover, by the Hopf maximum principle,sup@3R ln ~g �� sup@R <h �� sup�R <h �� inf�R <h �� inf@R <h �� inf@3R ln ~g :Thus h is analytic in R with its real part bounded by a constant that dependsonly on the bounds for `n ~g on @3R.In particular, we obtain f(z) = exp(h(z)) has its absolute value boundedbelow away from zero by a positive constant that depends on the lower boundof ~g(y) in @3R. In addition, the solution of (2.12) satis�es(2.16) kfkC1;�( �R) �� k~gkC1;� and 0 < k �� jf(z)j �� K ;where C depends on R, k = inf ~g and K = sup ~g, respectively.Next, we solve (2.13){(2.14). For this problem we set s = a+ ib, and we writethe corresponding equations and boundary conditions.We recall(2.17) s�z = 12 (sx + isy) = 12f(ax � by) + i(bx + ay)g :Hence, equations (2.17) and (2.13) yield the following equations(2.18) (ax � by) = <��0 + �0 ffe�2ib� = p1(x; b) ;(bx � ay) = =��0 + �0 ffe�2ib� = p2(x; b) :On the other hand, the right hand-side of equation (2.13) can be written asfjfj2 (�0f+ ��0fe�2ib) = jfj�2(fR � ifI)((A + iB) + (A� iB)(cos 2b� i sin 2b) ;

AN EXISTENCE AND UNIQUENESS RESULT 7so that(2.13) s�z = �0 + �0 ff e�2i=s(z)with the boundary conditions(2.14) =s = 0 on @R n @3R and <s = 0 on @3R :We shall show that the boundary value problem (2.12) has a unique non-vanishing solution. Moreover, the boundary value problem (2.13){(2.14) has aunique solution as well.In order to show uniqueness of a nonvanishing u solution of the boundaryvalue problem (2.8){(2.9), let ~u be another nonvanishing solution of the sameproblem and set ~s = `n( ~uf ), with f the unique nonvanishing solution of problem(2.12).Since u and ~u have the same data on @R, then it is easy to see that =~s = 0on @R n @3R and <~s = 0 on @3R. Di�erentiating ~s with respect to �z it is easy tosee that ~s solves equation (2.13) since f�z = 0. Indeed,~s�z = �`n� ~uf���z = ~u�z~u = �0 + ��0 �~u~u = �0 + ��0 �ffe�2i=~s :Therefore, by the uniqueness of problem (2.13){(2.14), ~s = s. Then, by therepresentation (2.11), ~u(z) = f(z) es(z) = u(z) on �R. Thus, the uniqueness ofnonvanishing solutions follows.We �rst solve problem (2.12). We set `n f(z) = `n jf(z)j+ i arg f(z). Thus,if jf(z)j 6= 0, then f�z = 0 if and only if (`n f)�z = 0.Thus, we set h = `n f(z) and solve(2.15) h�z = 0 ; =h = 0 on @R n @3R ;<h = `n ~g on @3R :Problem (2.15) is solved using standard methods. Indeed, take the Cauchy-Riemann equations and solve the corresponding harmonic equation for <h, with@<h@n = 0 on @R n @3R, and �nd =h as the conjugate harmonic function. Inparticular, �=h = 0, =h = 0 on @R n @3R and (=h)n = � ~g0~g on @3R, where ~g0~gis C�(@3R) and j~g0j~g �� C, C depending on the lower and upper bounds of ~g(0; y)and ~g0(0; y).Both, <h and =h are C1(R) functions.We use re ection techniques in order to obtain the C1;� regularity up to theboundary of R.Re ect the obtained harmonic function <h (which satis�es @<h@n = 0 on @1R[@2R [ @4R and <h = `n ~g on @3R) evenly across @R2 and @4R to a larger

6 IRENE M. GAMBATherefore, we want to prove the following lemma.Lemma 2.1. Let �0 be a C�( �R) function. The boundary value problem(2.8) u�z = �0u+ �0�uon a rectangle R = [C1; C2]� [0; 1] with boundary data(2.9) � juj = ~g(y) on @R3 = fC2 + iy; 0 �� y �� 1g=u = 0 @R n @R3 and u > 0 on @1R \ @2Rhas a unique nonvanishing solution that takes the formu = f(z)es(z) = f(z)e<s(z)+i=s(z)where <s(z) and =s(z) are C1;�( �R) functions and f(z) is analytic in R, nonva-nishing in �R and satis�es the same boundary data as u.In addition, the following estimates hold:(2.10) kukC�( �R) �� KkfkC�( �R) (inf jfj)�2 eKk�0 fk1; �R �1 + k�0 fk1; �R� andkukC1;�( �R) �� KkfkC1;�( �R) (inf jfj)�2 eKk�0 fk1; �R� �1 + k�0 fk1; �R� �1 + k�0 fkC�( �R)k�0 fk1; �R + k�0 fk21; �R�where K depends on R, and, kfkC1;�( �R) and inf jfj depend on C1;�-norm andthe lower bound for ~g respectively.Proof: We �rst reduce the problem for u to a problem for a = <s andb = =s. The solution u of equation (2.8) admits a representation; see Bers andNirenberg representation in [4], Chapter II, Section 6.(2.11) u(z) = f(z)es(z) :Here, f is an analytic function satisfying(2.12) f�z = 0 in �R ; along with=f = 0 on @R n @3R ;f > 0 on @1R \ @2R ; andjfj = ~g(x) on @3R :The function s(z) is complex valued and satis�esf(z)es(z)s�z = u�z = �0fes(z) + �0fes(z) ;

AN EXISTENCE AND UNIQUENESS RESULT 5and obtain the \complex ow equation"w�z = �(z)w + �(z) �w(2.2)with �(z) = (h1 + ih2)(z) ; �(z) = �(z) :(2.3)Thus, Lemma 3 of [3] states that if w is a solution of (2.2){(2.3) in a simplyconnected domain, then we have a solution ' of (1.1) with the form(2.4) '(z) = Re�Z zz0 wdz� :We �rst show that equation (2.2) preserves its form under any conformal trans-formation. Indeed, let z0 = F (z) be a conformal transformation from into 0,where w = 'x � i'y is the complex velocity in (z). In 0(z0) let(2.5) u = 'z0 = 'z dzdz0 + '�z d�zdz0 = 'z dzdz0 ;A short computation yields from (2.2)u�z0 = (�w + �� �w) d�zd�z0 dzdz0 = ��u d�zd�z 0 + ���u dzdz0�= � d�zd�z0u+ �� dzdz0 �uor(2.6) u�z0 = �0u+ �0�ude�ned in 0, with(2.7) �0 = �� dzdz0� (z0) and �0 = �0 :In particular, if F is the conformal transformation (see Appendix) that takes into 0 = R, the corresponding rectangle, we solve equations (2.6), (2.7) in R sothat, using (2.5), the function w = udz0dz = uFz solves (2.1) and the solution ' isthen given by (2.4). The boundary data from (1.1) is equivalent to =u = 0 on@1R [ @2R [ @4R and juj = jwj �� dzdz0 �� = g(x)jFzj = ~g(0; y) > 0 in @3R.In addition, the condition r' � nj@1 < 0 yields that w > 0 on @1 \ @2.Since Fz��@1\@2 > 0 (see Appendix), then u > 0 on @1R\ @2R. This conditionwill be su�cient to see that @1 is an in ow boundary.

4 IRENE M. GAMBAb=0b=0 rb�n=f�n@1R @4R�=@1R[@2R[@3Rb=0 R@2R @3R=�1Figure 2. Domain F and boundary conditions on b.Section 3 deals with the proof of Lemma 1.2.Remark 1. The nature of estimate (1.2) is due to the boundary conditionsfrom (1.1) and (1.3). Otherwise, the right-hand side of (1.2) might not havelinear growth in khkC�(�).Remark 2. In case speed is prescribed all over the boundary for the boundaryvalue problem (1.1), even for ~h � 0, uniqueness is lost for some special geometriesas shown by G. E. Backus in [2].2. Proof of Theorem 1The proof of this theorem makes use of three lemmas. The �rst one (seeAppendix) is a simple proof, included for completeness of our arguments, of theexistence of a unique conformal transformation of the domain under consid-eration into a rectangle R with one �xed given side. This transformation is asregular as the given regularity for each portion @i of the boundary @.Next, after showing that the equation (1.1) conserves the same form in thenew variables, Lemma 2.1 will show how to solve the new equation in a rect-angular domain with prescribed data as in (1.1). There, we shall see that thesolution has a representation (Bers-Nirenberg representation), whose two com-ponents solve a linear homogeneous elliptic problem and a nonlinear ellipticboundary value problem, (1.4), respectively, (Lemma 1.2).Thus, we reconstruct the solution ' by taking the inverse conformal trans-formation of the solution given by the Bers-Nirenberg representation (see [4]),which is constructed uniquely by solving the equations for their components inthe rectangle with the corresponding boundary data.From now on we use the complex representation of the ow equation. Fol-lowing Bers (see [3]), we set(2.1) w = 2'z = 'x � i'y ;

AN EXISTENCE AND UNIQUENESS RESULT 3Moreover, @1 is an in ow boundary sincer' � n < 0 on @1We show in the Appendix that � can be uniquely conformally transformedinto a �xed rectangle R such that the four boundary sections @i; i = 1; 4 aretransformed into the sides of R, in the same order.Thus, we prove Theorem 1 using the two-dimensional complex representationof solutions of the equation �' = r' � ~h (Bers-Nirenberg representation), andthe following key lemma in the transformed domain R:Lemma 1.2. There exists a unique weak solution b in C1;�( �R) for the prob-lem(1.3) 8><>: div (rb� f(x; b)) = 0 in Rb = 0 in �bn = f(x; b) � n in �1 :where n denotes the outer normal to @R, and, the function f(x; b) = (f1; f2)(x; b)is bounded, C�( �R) in x and and Lipschitz as function of b.The domain R is a rectangle where @R = � [ �1 with �1 one side of R andf(x; 0) � n = 0 in @2R [ @4R, where @2R and @4R are the sides of R adjacent to�1 (compatibility condition for regularity up to the boundary). See Figure 2.Note: b is a C1;�( �R) weak solution if(1.4) ZR r'(rb� f(x; b)) = 0 for all ' 2 H10 (R2 n �) ;where H10 (R2n�) is the closure of C10(R2n�). That means the boundary conditionon �1 is a natural boundary condition for weak solutions in the sense (1:4).In addition, if we denote kfkC�;x; �R the C�( �R)-norm of f with respect to the�rst variable and kfkLip;b denotes the Lipschitz norm of f with respect to thesecond variable, the following estimates hold:(1.5) kbkC1;�( �R) �� C �kfkC�;x; �R + kfkLip;b kfk1; �Rand(1.6) kbkC�( �R) �� Ckfk1; �R ;where C depends on R.

2 IRENE M. GAMBA.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... r'�n��@2=0@4@1'��@1=0 jr'j��@3=g(x)@3@2r'�n��@4=0Figure 1. Domain and boundary conditions on '.We consider the following elliptic problem: given ~h = (h1; h2) a C�(�) func-tion, 0 < � < 1,(1.1) 8>>>>>>><>>>>>>>: �' = r' �~h in '���@1 = 0 and r' � n���@1(w1) < 0 in ow conditions@'@n ���@2[@4 = 0 streamline wallsjr'j���@3 = g(x) prescribed speed :Here n denotes the exterior normal to @ and g(x) is a C1;�(@3) strictly positivefunction.We show the following theorem.Theorem 1. There exists a unique solution ' 2 C2;�(�) of the bound-ary value problem (1:1) with nonvanishing gradient. In addition, the solution 'satis�es(1.2) k'kC1;�(�) �� KeKkhk1;� �1 + khk1;��andk'kC2;�(�) �� KeKkhk1;� �1 + khkC�(�)khk1;� + khk21;���1 + khk1;��where K depends on and on the C1;�-norm of g; and the following estimateshold, 0 < ke�KH �� jr'j �� KeKH ; and � eK �� ' �� eK on � ;where k and K depend on and g(x), H depends on the upper bound for j~hj.eK depends on the diameter of and on the upper bound for jr'j.

An Existence and Uniqueness Resultof a Nonlinear Two-DimensionalEllipticBoundary Value ProblemIRENE M. GAMBACourant InstituteAbstractWe consider a boundary value problem for the generalized two-dimensional ow equation�' = r' �~h for a ~h a C� vector �eld, where the speed is prescribed on a part of the boundary.By using Bers theory combined with elliptic operator theory in nonsmooth domains, we showexistence and uniqueness of a C2;� solution with nonvanishing gradient, and we �nd positivelower and upper bounds for jr'j along with C2;� estimates of ', in terms of the C� and L1norms of ~h. c 1995 John Wiley & Sons, Inc.1. IntroductionWhen studying compressible potential ow, the following equation arises nat-urally:Given a strictly positive C1;� density � in a ow region , �nd the owpotential function ' with nonvanishing gradient that satis�es the equationdiv (�r') = 0with boundary conditions given by r' � n = 0, n the outer unit normal to theboundary, along the lateral walls, ' constant at the in ow boundary, and thespeed jr'j prescribed on the remaining boundary section.We are able to solve uniquely this problem in two dimensions by using theBers theory for the ow equation.In fact, for our application in compressible hydrodynamics (see [6]), we needto consider a general C�, 0 < � < 1, vector �eld ~h, instead of r`n�, and wemust obtain existence of a unique C2;� solution ' with positive lower bound forthe speed jr'j, along with a C2;� estimate for ' that has linear growth in theC� norm of h.Let be a domain in two dimensions with a C2;� boundary except for fourpoints called wi, i = 1; . . . ; 4.Let @i be the C2;� component of @ and let satisfy the condition that@i \ @i+1 meet at points wi forming a �=2 angle. Thus is a \smoothrectangular" domain. See Figure 1.Communications on Pure and Applied Mathematics, Vol. XLVIII, 1{21 (1995)c 1995 John Wiley & Sons, Inc. CCC 0010{3640/94/070001{21