AN EXISTENCE AND UNIQUENESS RESULT 21

AN EXISTENCE AND UNIQUENESS RESULT 21 research 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 solutions of elliptic partial dierential 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 equation to 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 dierential equations of elliptic type, pp. 69{94 in: Contributions to the Theory of Partial Dierential Equations, L. Bers, ed., Annals of Mathematics Studies No. 33, Princeton University Press, 1954. [4] Bers, L., John, F., and Schechter, M., Partial Dierential Equations, Interscience Publishers, Wiley, New York, 1964. [5] Berestycki, H., Nirenberg, L., and Varadhan, S. R. S., The principal eigenvalue and maximum 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 semiconductor or transonic gas dynamic ow: Existence theorem for potential ow, Comm. Pure Appl. Math., to appear. [7] Gilbarg, D., and Trudinger, N. S., Elliptic Partial Dierential Equtions of Second Order, Springer-Verlag, New York, 1983. [8] Grisvard, P., Elliptic Problems in Non-Smooth Domains, Monographs and Studies in Mathematics No. 24, Pitman, Boston, 1985. [9] Lady^zenskaja, O. A., and Ural'ceva, N. N., Equations aux Derivees Partielles de Type Elliptique, Dunod, Paris, 1968. Received February 1994.

20 IRENE M. GAMBA R = [C 1 ; C 2 ] [0; 1], such that F (@ i ) = @ i R with @ = 4 [ i=1 @ i K, each @ i K a side 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 @ i such that each @ i are (piecewise) C 2; -curves which joints the preceding one at the point! i making a 2 angle; that is, if i1, i are the parametrizations of @ i1 and @ i, respectively, i1t (! i ) it (! i ), where the subindex T denotes tangential derivative. Under these conditions, the proof is given by the classical result of conformal mapping that corresponds to an incompressible transformation. Take `2(~x) the real harmonic function that solves 8>< >: `2 = 0 in `2 = 0; `2 @4 = 1 @2 ; = 0 @1[@ 3 @' 2 @ `2 is a C 2; ( ) harmonic function as the boundary condition is compatible on the 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 the orthogonal eld to the level curves of `2, that is, `1 `1(x; y) = (x;y)(t) (x 0;y 0) (`2y `2x ) 0 (t)dt + `1(x 0 ; y 0 ) where (t) is a path joining (x 0 ; y 0 ) to (x; y) with (x 0 ; y 0 ) a xed point in @. Then `1 2 C 2; () satises @`1 @T = 0 on @ 1 [ @ 3, then `1 = C 1 = `1(x 0 ; y 0 ) if (x 0 ; y 0 ) 2 @ 1 R and `2 = C 2 where C 2 is xed by the path integral and by C 1. Moreover, the function `2;y satises `2;y = 0, `2;y > 0 on @ 4 [ @ 2 and r`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 satises F () = [C 1 ; C 2 ] [0; 1]. Acknowledgments. I would like to thank Robert Kohn, Cathleen S. Morawetz, and Louis Nirenberg for many helpful discussions and suggestions. This

AN EXISTENCE AND UNIQUENESS RESULT 19 where ( A(x) if x 2 R ~A(x) = (A 1 (x ); A 2 (x )) if x 2 R ~ n R and x 2 R is the reection of x 2 ~ R n R with respect to 1. In particular, for any test function ' 2 H 1 0( ~ R) (3.15) ~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 H 1 0 ( ~ R) for any given bounded function in ~ R as a right-hand side. Indeed, by estimate (3.8) ~! = ~ b 1 ~ b 2 is C ( ~ R). We solve for in H 1 0( ~ R) the adjoint problem (3.15) L = + ~ A(x)r = ~! in ~ R = 0 on @ ~ R : Since 2 H0 1 is an admissible test function, it satises (3.15). Then (3.16) 0 = r (r~! A(x)~!) ~ = ( + A(x)r ~ )~! = so that ~R 0 = ~R ~R ~! 2 : In particular, ~! = 0 in ~ R. The existence and uniqueness of problem (3.16) in ~ R can be found in a recent paper 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 at corner points. Moreover, it is easy to see from their proof that r 2 L 2 ( ~ R). In particular, 2 H 1 0( ~ R) so that (3.16) holds, and consequently, ~! = 0. Appendix Lemma (The Conformal Transformation). There exist two constants C 1 and C 2, and a unique conformal transformation F (z):! R, with

18 IRENE M. GAMBA div (A(x)(v 1 v 2 )), with A(x) = @f @v (x; (v 1+(1)v 2 )(x)). Since A(x) 2 C ( R) using estimates (3.9) and (3.10) for this equation the following estimates hold: kb 1 b 2 k C 1;( R) C kak C ( R) kv 1 v 2 k 1; R + kak 1; R kv 1 v 2 k C ( R) CkAk C ( R) kv 1 v 2 k C ( R) : Thus, T is a continuous map. Therefore T has a xed point b which is the solution of the boundary value problem (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 b 1 and b 2 be two solutions in W 1;p ( R); p > 2, of the boundary value problem (1:3), then b 1 = b 2 on R. Proof: We prove uniqueness in the reected domain for the corresponding Dirichlet boundary value problem in a similar way as we worked out Lemma 3.1. Let! = b 1 b 2 on R. Then! satises where (3.12) A(x) =! = div (f(x; b 1 ) f(x; b 2 )) = div (A(x)!) 1 0 f b (x; t b 1 + (1 t) b 2 ) dt : Since f is Lipschitz in b and C in x in R, then A(x) is L 1 ( R). Then,! is a C 1; ( R) solution of linear elliptic operator in divergence form (3.13) 8 >< >: L! =! div (A(x)!) = 0 on R! = 0 on :! x = A 1 (x)! on 1 Now if ~ R dened as above is the rectangle that contains R, its reection with respect to 1 and 1, then, taking ~ b1 and ~ b2 the even reection of b 1 and b 2 respectively with respect to 1. Their dierence ~! is an even function that solves (3.14) L~! = ~! div ( ~ A(x)~!) = 0 in ~ R! = 0 on @ ~ R

AN EXISTENCE AND UNIQUENESS RESULT 17 In fact, if B = (B 1 ; B 2 ) solves uniquely B i = f i (x) in R, B i = 0 on @R, with f i 2 C ( R 1 ) \ C ( R 2 ), and e 1 = R 1 \ R 2 for i = 1; 2. By [1], each B i 2 W 2;p ( R 0 ) for 1 < p < 1, and, in particular, in C 1; (R 0 ) for 0 < < 1, for any regular subdomain R 0 of R. Hence, [B i ] = 0 = [(B i ) x ] on e 1. This implies that B i is the unique solution of the elliptic diraction problem dened above. By [9], each B i is C 2; ( k ); k = 1; 2, where k = (R 0 \ R k ) [ S for S any part of e 1 satisfying 1 S e 1. Moreover, the following estimate holds kb i k C 2;( k) Ckf ik C ( R) Ckf ik C ( R) ; i = 1; 2 k = 1; 2 where C depends only on R. Therefore, since h = div B i b solves h = 0 in R 0 with h = div B i on @R 0, then h is harmonic in R 0. Hence, h is C 1 in the interior of R 0, so that b is as good as div B i, and estimating b in R R 0 Therefore, kbk C 1;( R) Ckdiv B i k C 1;( R) CkB i k C 2;( R) Ckfk C ( R) : (3.9) kbk C 1;( R) C kf 1 ; f 2 k C ( R) where (3.10) kf 1 ; f 2 k C ( R) = kfk C ( R);Lip;v C kfk C ;x; R kfk 1; R + kfk Lip;v kvk C ( R) where C depends on R. Therefore (3.2) holds. The proof of Lemma (3.1) is now complete. In order to complete the proof of the existence part of Lemma 1.2, we make use 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 be estimated by (3.2) and (3.3). Indeed, estimate (3.3) provides the needed a priori estimate to show the operator T from the Banah space B = C ( R) into itself is bounded. Moreover, T (C (R)) C 1; ( R) is compactly embbeded in C ( R), then T is a compact operator, and, for any b in B = C 1; ( R) such that b = T b, for some 0 < < 1, from (3.3) we obtain the a priori estimate (3.11) kbk C 1; ( R) C kfk C ;x; R kfk 1; R + kfk Lip;b kfk 1; R M where 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 Lipschitz continuous. Indeed, T (v 1 ) T (v 2 ) = b 1 b 2 satises the equation (b 1 b 2 ) =

16 IRENE M. GAMBA 0 = = ~R ~R = 2 R r ~' r ~ b f ~ dx ~' x ~ bx + ~' y ~ by ( ~' x ~ f1 + ~' x ~ f2 ) dxdy r' r ~ b f dx ; since ~' x ; ~ b x, and ~ f 1 are odd functions and ~' y ; ~ b y, and ~ f 2 are even ones. Hence (3.1.w) holds for b = ~ bj R, making this b the unique weak solution of the boundary value problem (3.1). In addition, by denition of ~ f, Then by estimate (3.7) k ~ fk 1; ~ R = kfk 1; R : (3.8) kbk C ( ~ R) Ckfk L 1 ( R) ; where C depends only on R, so that estimate (3.2) holds. In order to obtain the C 1; regularity and estimate (3.3), note that ~ fj R 2 C ( R) and ~ f ~RnR 2 C ( ~ R n R) ( ~ f is not C across 1 ) and, by the boundary conditions given on problem (3.1), ~ f 1 = 0 on @ 2 ~ R [ @4 ~ R. R 1.. ~ 1 @ 2 ~ R ~R. R 1 R 0. @ 4 ~ R S. R 2 Figure 3. The reected domain R. The C 1; estimate for b in R is obtained rst for b by dierentiating C 2; solutions of elliptic diraction problems dened in R where e 1 is the diraction boundary and the given right-hand sides are C on each side of e 1 ; see Ladyzenskaja and Ural'ceva, [9].

AN EXISTENCE AND UNIQUENESS RESULT 15 and 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 @ 1 R, respectively. See Figure 3. Recalling Agmon-Douglis-Nirenberg (see [1]) interior regularity results for elliptic problems, it follows that the solution b of problem (3.4.1) in the domain R for the subdomain ~ R R satises the following estimate (3.6) kbk W 1;p( ~ R) Ckfk Lp ( R) Ckfk L 1 ( R) ; where C depends only on R, since the denition of f only involves even or odd reections of the components of f. Since ~ b = bj ~R, the Embedding Theorem yields ~ b 2 C ( ~ R) along with the estimate (3.7) k ~ bk C ( ~ R) Ckfk L 1( 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 of problem (3.1) in the sense (3.1.w). Due to the uniqueness of the boundary value problem (3.4.1){(3.4.2), it is enough to show that ~ b(x ), where x is the reection of x with respect to 1, is also 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 and r x = @ x + @ y as dierential operators, hence and, by denition of ~ f, x ~ b(x ) = x ~ b(x ) r x ~ f(x; ~v(x)) = r x ~ f(x; ~v(x )) = r x ~ f(x ; ~v(x )) : Therefore if ~ b is the solution of (3.4.1){(3.4.2) then so that x ~ b(x ) = r x ~ f(x ; ~v(x )) ; x ~ b(x (x)) = r x ~ f(x; ~v(x)) : In particular, since ~ b(x ) = 0 on @ ~ R, then ~ b(x ) = ~ b(x), so that ~ b is an even function in ~ R. Now, let ' be a test function in H 1 0(R 2 n ). Let ~' be the even reection of ' to ~ R, then ~' 2 H 1 0( ~ R), and, for the solution ~ b of problem (3.4)

14 IRENE M. GAMBA problem, and so keeps the interior regularity up to the boundary. Methods discussed 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 reection with respect to 1, and 1 ; and let ~v dene on ~ R be the even reection 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 (f 1 (x ; ~v(x )); f 2 (x ; ~v(x )) ) if x 2 ~ R n R where x 2 R is the reection of x 2 ~ R n R with respect to 1, that is, f 1 is reected oddly and f 2 is reected evenly with respect to 1, respectively. Now, since v 2 C ( R) and ~v is the even reection of v with respect to 1, then ~v 2 C ( ~ R), so that ~ f is Lipchitz in ~v. However, since f1 does not vanish on 1, then ~ f is discontinuous across 1 but ~ f remains bounded and measurable in ~ R. The existence of an H 1 0 solution of problem (3.4) follows from standard results. In fact, a weak solution of problem (3.4.1) must satisfy (3.5) r'r ~ b = r' f(x; v) for all ' 2 H 1 0( ~ R) ; ~R ~R Since f 2 L 1 ( ~ R), then `(') = ~R r' f(v) is a continuous linear form in H 1 0 satisfying k`(')k H 1 0 ( ~ R) Ck'k H 1 0 kfk 1; ~ R where C depends on the domain. By the Riesz representation theorem, there exists a unique solution ~ b in H 1 0 of problem (3.4.1) that satises k'k H 1 0 ( ~ R) Ckfk 1; ~ R : Next, the solution ~ b can be reected oddly across all boundary sections @ ~ Ri, and it denes a b that solves a similar problem for the enlarged domain R in the space H 1 0(R) with a right-hand side f. Now this right-hand side f is measurable

AN EXISTENCE AND UNIQUENESS RESULT 13 3. Proof of Lemma 1.2 Without loss of generality, we assume that 1 is contained in the axis x = 0 and that its lower end is the origin. (a) Existence We rst show the existence of a weak solution to the boundary value problem (1.3) by constructing a precompact continuous map T from an appropriate Banach 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 dene the map T by letting b = T v be the unique weak solution in C 1; ( R) of the linear boundary value problem (3.1) b = div f(x; v(x)) on R b = 0 on and b x = f 1 ((0; y); v(0; y)) on 1 f 1 (x; 0) = 0 for any x 2 \ (@ 2 R [ @ 4 R) : By weak solution we mean that (3.1.w) Remark. Note that @R for all ' 2 H 1 0(R 2 n ). Lemma 3.1. r' rb = r' f(x; v) for all ' 2 H0(R 1 2 n ) : R R '( rb f(x; v) ) nds = 1 ' b x f 1 ((0; y); v(0; y)) dy = 0 The problem (3:1) has a unique weak solution b in C 1; ( R) in the sense of (1:4) or (3.1.w) satisfying the following estimates: (3.2) (3.3) kbk C ( R) Ckfk 1; R kbk C 1; ( R) C kfk C ;x; R + kfk Lip;v kfk1; R where C depends only on R. Proof: The proof of this lemma is an application of existence and regularity theorems for elliptic equations in general smooth domain. Although our domain is a rectangle, the nature of the boundary data permits reecting the boundary appropriately such that we deal with a Dirichlet boundary value transmission

12 IRENE M. GAMBA where C depends on kgk C 1;(@ 3) and on jf z j L 1, i.e., C depends on the data g(x) and on the domain. Finally, by (2.4), we take (2.35) ' = Re 8 < : z z 0 wdz 9 = ; ; ' = 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 C 1; () functions, then ' 2 C 2; () and by estimates (2.32) and (2.34) (2.36) k'k C 1;() CeCkhk 1; 1 + khk1; k'k C 2; () CeCkhk 1; and 1 + khk C ()khk 1; + khk1; 2 1 + khk1; where C depends on the C 1; -norm of g(x) and on the domain, so that estimates in (1.2) hold. In addition, by (2.33) (2.37) 0 < k e CsupH C inf < jr'j < Ke CsupH C sup where H is the upper bound of jhj, C sup and C inf bounds depending on g(x) and, and k, K depends only on. We stress once more that C inf > 0, because of the 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(w 1 ) < 0, implies r' n < 0 on @ 1. Therefore ' increases along the paths orthogonal to the level surfaces that start at the boundary @ 1, and, in particular, (2.38) 0 ' e K near @1 : In general, (2.39) e K ' e K in ; where e K, both in (2.38) and (2.39), depends on the diameter of and on the upper 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 boundary value problem (2.21), (2.23).

AN EXISTENCE AND UNIQUENESS RESULT 11 transformation of the domain onto R = [C 1 ; C 2 ] [0; 1]. Recalling w the complex 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) = (h 1 + ih 2 )(z) : Now, by (2.5), (2.6), and (2.7), (2.31) w = u df dz = uf z where u is the unique solution of the ow equation in the rectangle F () = R of the boundary value problem given by Lemma 3.1, where 1 0 df = (h 1 + ih 2 ) : dz Hence, by Lemma 2.1, w = f(z)e s(z) F z ; jf(z)j 6= 0 : Since F is an analytic function in, which is C 1 () (due to the special geometrical properties of ) and F z 6= 0 in and jf z j is uniformly bounded away from zero. See Lemma in the Appendix. Then <w and =w are C 1; ( ) and by (2.10) the following estimates hold, kwk C () Ckfk C () (inf jfj) 2 e Ck0 fk 1; 1 + k 0 k 1; kfk 1; ; and (2.32) kwk C 1;() Ckfk C 1; () (inf jfj) 2 e Ck0 fk 1; 1+ k 0 k C ()k 0 k 1; + k 0 k 2 1; and the estimate for the absolute value of w (2.33) 0 < C inf kfk 2 1; 1 + k 0 k 1; kfk 1; ; n jfj jf z je inf <s(z)o n jwj C sup jfj jf z je sup <s(z)o : The constant C and jf z j depend only on, the inf jfj is positive and depends on g(x) on @ 3 R, 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 z 1 j, and from ~ h = (h 1 ; h 2 ) the given coecients 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 estimated by Ckfk C () (inf jfj) 2 e Ck0 k 1; 1 + Ck 0 k 1; ; and (2.34) Ckfk C 1; () (inf jfj) 2 e Ck0 k 1; 1 + Ck 0 k C ()k 0 k 1; + Ck 0 k1; 2 1 + Ck 0 k 1; ;

10 IRENE M. GAMBA Once 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) = (b y + p 1 (x; b); b x + p 2 (x; b)) where a(x) = 0 on @ 3 R. In particular, a = div p(x; b) with a n = 0 on @ 2 R[@ 4 R, a n = p(x; 0) n on @ 1 R, and a = 0 on @ 3 R. Therefore, a similar result to lemma 3.1 yields that a(x) is a C 1; ( R) function, and by (2.24) and (2.25), satises the estimates (2.27) kak C 1;( R) C kpk C ;x; R + kpk Lip;b kpk1; R and (2.28) kak C ( R) Ckpk 1; R ; 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 be given, with a and b real C 1; ( R) functions which are the unique solutions of the boundary value problem (2.19), (2.21), and (2.23). Then s(z) solves the boundary value problem (2.13){(2.14) uniquely. Therefore, by (2.11), (2.17), and (2.13), u = f(z)e a+ib solves the boundary value problem (2.8), (2.9) in the rectangle R uniquely. Moreover, using estimates (2.24), (2.25), (2.27), and (2.28), along with the form of p given by (2.20), yields the estimates (2.29) kuk C ( R) Ckfk C ( R) e sup j<s(z)j R 1 + kskc ( R) Ckfk C ( R) (inf jfj) 2 e Ck0 fk 1; R 1 + k 0 fk1; R kuk C 1;( R) Ckfk C 1;( R) e sup R Ckfk C 1; ( R) (inf jfj) 2 e CkAi k k 1; R j<s(z)j 1 + kskc 1;( R) 1 + kskc ( R) and 1 + ka ik k C ( R) ka ik k 1; R + ka ik k 2 1; R 1 + kaik k 1; R Ckfk C 1; ( R) (inf jfj) 2 e Ck0 fk 1; R 1 + k 0 fk C ( R) k 0 fk 1; R + k 0 fk 2 1; 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 the resulting rectangle given in the Appendix; then F :! R is the unique conformal

AN EXISTENCE AND UNIQUENESS RESULT 9 where A + ib = 0 f, and f = f I + if R, then the terms p 1 and p 2 from (2.18) can be written as (2.19) p 1 (x; b) = 2jfj 2 [f R A(1 + cos 2b) + f I B(1 cos 2b) (f R B + f I A) sin 2b] p 2 (x; b) = 2jfj 2 [f I A(1 + cos 2b) + f R B(1 cos 2b) (f R A + f I B) sin 2b] : Clearly, (2.20) p i (x; b) = 3X k=1 A ik (x)`ik (b) ; i = 1; 2 where A ik (x) are real C functions in R and `ik (b) bounded and Lipschitz. Since p i does not depend on a = <s(z), system (2.19) yields an elliptic equation for b(x) alone, namely (2.21) b = p 2x p 1y : The boundary conditions for b are obtained from (2.14) and the relationship (2.19). Since <s = a = 0 on @ 3 R, then a y = 0 on @ 3 R, and by (2.19), b x = p 2 (x; b) on @ 3 R. Also, =s = b = 0 on @R n @ 3 R. In particular, since f is real at @ 3 R [ @ 4 R and at @ 3 R [ @ 2 R, we have the compatibility condition (2.22) p 2 (x; 0) = 0 if x 2 @ 3 R \ @ 4 R or x 2 @ 3 R [ @ 2 R : Therefore, we must show that there exists a unique solution of (2.21), along with the data b = 0 on @R n @3 R (2.23) : b x = p 2 ((C 2 ; y); b) on @ 3 R In the next section, we prove Lemma 1.2 which ensures the existence of a unique weak solution b(x) in the class C 1; ( R) of the nonlinear boundary value problem (2.21){(2.23) that satises the estimates (2.24) kbk C 1;( R) C kpk C ;x; R + kpk Lip;b kpk1; R ; and (2.25) kbk C ( R) Ckpk 1; R where C depends on R.

8 IRENE M. GAMBA reected rectangular domain ~ R whose sides @i ~ R @i R for i = 1; 3. Let f <h be the reected function of <h. Also, let G be the even reection of `n ~g across the endpoints of @ 3 R dened on @ 3 ~ R. Then f <h is a harmonic function in ~ R that satises @f<h @n = 0 on @ 1 ~ R[@ 2 ~ R[@4 ~ R and f <h = G on @3 ~ R. In addition, f <h is C 1 ( ~ R) [ C 1; ( ~ R [ T ) where T is any regular portion of the boundary of ~ R. Then <h = f <hj R is in C 1 (R) [ C 1; ( R). Therefore, since the harmonic conjugate of <h inherits its regularity, =h is C 1 (R) [ C 1; ( R) and the following estimates hold k<h ; =hk C 1; ( R) Ck~gk C 1; ( R) ; where C depends on R. Moreover, by the Hopf maximum principle, sup @ 3R ln ~g sup @R <h sup R <h inf <h R inf <h @R inf ln ~g : @ 3R Thus h is analytic in R with its real part bounded by a constant that depends only on the bounds for `n ~g on @ 3 R. In particular, we obtain f(z) = exp(h(z)) has its absolute value bounded below away from zero by a positive constant that depends on the lower bound of ~g(y) in @ 3 R. In addition, the solution of (2.12) satises (2.16) kfk C 1;( R) k~gk C 1; 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 write the corresponding equations and boundary conditions. We recall (2.17) s z = 1 2 (s x + is y ) = 1 2 f(a x b y ) + i(b x + a y )g : Hence, equations (2.17) and (2.13) yield the following equations (a x b y ) = < 0 + 0 ff e2ib = p 1 (x; b) ; (2.18) (b x a y ) = = 0 + 0 ff e2ib = p 2 (x; b) : On the other hand, the right hand-side of equation (2.13) can be written as f jfj 2 (0 f + 0 fe 2ib ) = jfj 2 (f R if I )((A + ib) + (A ib)(cos 2b i sin 2b) ;

AN EXISTENCE AND UNIQUENESS RESULT 7 so that (2.13) s z = 0 + 0 f f with the boundary conditions e2i =s(z) (2.14) =s = 0 on @R n @ 3 R and <s = 0 on @ 3 R : We shall show that the boundary value problem (2.12) has a unique nonvanishing solution. Moreover, the boundary value problem (2.13){(2.14) has a unique solution as well. In order to show uniqueness of a nonvanishing u solution of the boundary value problem (2.8){(2.9), let ~u be another nonvanishing solution of the same problem and set ~s = `n( f ~u ), 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 = 0 on @R n @ 3 R and <~s = 0 on @ 3 R. Dierentiating ~s with respect to z it is easy to see that ~s solves equation (2.13) since f z = 0. Indeed, ~u ~s z = `n f z = ~u z ~u = 0 + 0 ~u ~u = 0 + 0 f f e2i=~s : Therefore, by the uniqueness of problem (2.13){(2.14), ~s = s. Then, by the representation (2.11), ~u(z) = f(z) e s(z) = u(z) on R. Thus, the uniqueness of nonvanishing 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 @ 3 R ; <h = `n ~g on @ 3 R : 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. In particular, =h = 0, =h = 0 on @R n @ 3 R and (=h) n = ~g0 ~g on @ 3R, where ~g0 ~g is C (@ 3 R) and j~g0 j C, C depending on the lower and upper bounds of ~g(0; y) ~g and ~g 0 (0; y). Both, <h and =h are C 1 (R) functions. We use reection techniques in order to obtain the C 1; regularity up to the boundary of R. Reect the obtained harmonic function <h (which satises @<h @n = 0 on @ 1R[ @ 2 R [ @ 4 R and <h = `n ~g on @ 3 R) evenly across @R 2 and @ 4 R to a larger

6 IRENE M. GAMBA Therefore, 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 = 0 u + 0 u on a rectangle R = [C 1 ; C 2 ] [0; 1] with boundary data (2.9) juj = ~g(y) on @R3 = fc 2 + iy; 0 y 1g =u = 0 @R n @R 3 and u > 0 on @ 1 R \ @ 2 R has a unique nonvanishing solution that takes the form u = f(z)e s(z) = f(z)e <s(z)+i=s(z) where <s(z) and =s(z) are C 1; ( R) functions and f(z) is analytic in R, nonvanishing in R and satises the same boundary data as u. In addition, the following estimates hold: (2.10) kuk C ( R) Kkfk C ( R) (inf jfj) 2 e Kk0 fk 1; R 1 + k 0 fk1; R kuk C 1; ( R) Kkfk C 1; ( R) (inf jfj) 2 e Kk0 fk 1; R 1 + k 0 fk 1; R 1 + k 0 fk C ( R) k 0 fk 1; R + k 0 fk 2 1; R and where K depends on R, and, kfk C 1; ( R) and inf jfj depend on C 1; -norm and the lower bound for ~g respectively. Proof: We rst reduce the problem for u to a problem for a = <s and b = =s. The solution u of equation (2.8) admits a representation; see Bers and Nirenberg representation in [4], Chapter II, Section 6. (2.11) u(z) = f(z)e s(z) : Here, f is an analytic function satisfying (2.12) f z = 0 in R ; along with =f = 0 on @R n @ 3 R ; f > 0 on @ 1 R \ @ 2 R ; and jfj = ~g(x) on @ 3 R : The function s(z) is complex valued and satises f(z)e s(z) s z = u z = 0 fe s(z) + 0 fe s(z) ;

AN EXISTENCE AND UNIQUENESS RESULT 5 and obtain the \complex ow equation" (2.2) w z = (z)w + (z) w with (2.3) (z) = (h 1 + ih 2 )(z) ; (z) = (z) : Thus, Lemma 3 of [3] states that if w is a solution of (2.2){(2.3) in a simply connected domain, then we have a solution ' of (1.1) with the form (2.4) '(z) = Re z z 0 w dz We rst show that equation (2.2) preserves its form under any conformal transformation. Indeed, let z 0 = F (z) be a conformal transformation from into 0, where w = ' x i' y is the complex velocity in (z). In 0 (z 0 ) let dz (2.5) u = ' z 0 = ' z dz 0 + ' dz z dz 0 = ' dz z dz 0 ; or A short computation yields from (2.2) u z 0 = (w + w) dz dz 0 dz dz 0 = u dz dz 0 + u dz dz 0 = dz dz 0 u + dz dz 0 u (2.6) u z 0 = 0 u + 0 u dened in 0, with (2.7) 0 = dz dz 0 (z 0 ) 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 so that, using (2.5), the function w = u dz0 dz = uf z solves (2.1) and the solution ' is then given by (2.4). The boundary data from (1.1) is equivalent to =u = 0 on @ 1 R [ @ 2 R [ @ 4 R and juj = jwj dz dz 0 = g(x)jfz j = ~g(0; y) > 0 in @ 3 R. In addition, the condition r' nj @1 < 0 yields that w > 0 on @ 1 \ @ 2. Since F z @1\@ 2 > 0 (see Appendix), then u > 0 on @ 1R \ @ 2 R. This condition will be sucient to see that @ 1 is an inow boundary. :

4 IRENE M. GAMBA =@ 1R[@ 2R[@ 3R @ 2R b=0 @ 3R= 1 b=0 rbn=fn @ 1R b=0 R @ 4R Figure 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 conditions from (1.1) and (1.3). Otherwise, the right-hand side of (1.2) might not have linear growth in khk C (). Remark 2. In case speed is prescribed all over the boundary for the boundary value problem (1.1), even for ~ h 0, uniqueness is lost for some special geometries as shown by G. E. Backus in [2]. 2. Proof of Theorem 1 The proof of this theorem makes use of three lemmas. The rst one (see Appendix) is a simple proof, included for completeness of our arguments, of the existence of a unique conformal transformation of the domain under consideration into a rectangle R with one xed given side. This transformation is as regular as the given regularity for each portion @ i of the boundary @. Next, after showing that the equation (1.1) conserves the same form in the new variables, Lemma 2.1 will show how to solve the new equation in a rectangular domain with prescribed data as in (1.1). There, we shall see that the solution has a representation (Bers-Nirenberg representation), whose two components solve a linear homogeneous elliptic problem and a nonlinear elliptic boundary value problem, (1.4), respectively, (Lemma 1.2). Thus, we reconstruct the solution ' by taking the inverse conformal transformation of the solution given by the Bers-Nirenberg representation (see [4]), which is constructed uniquely by solving the equations for their components in the rectangle with the corresponding boundary data. From now on we use the complex representation of the ow equation. Following Bers (see [3]), we set (2.1) w = 2' z = ' x i' y ;

AN EXISTENCE AND UNIQUENESS RESULT 3 Moreover, @ 1 is an inow boundary since r' n < 0 on @ 1 We show in the Appendix that can be uniquely conformally transformed into a xed rectangle R such that the four boundary sections @ i ; i = 1; 4 are transformed into the sides of R, in the same order. Thus, we prove Theorem 1 using the two-dimensional complex representation of solutions of the equation ' = r' ~ h (Bers-Nirenberg representation), and the following key lemma in the transformed domain R: There exists a unique weak solution b in C 1; ( R) for the prob- Lemma 1.2. lem (1.3) 8 >< >: div (rb f(x; b)) = 0 in R b = 0 in : b n = f(x; b) n in 1 where n denotes the outer normal to @R, and, the function f(x; b) = (f 1 ; f 2 )(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 and f(x; 0) n = 0 in @ 2 R [ @ 4 R, where @ 2 R and @ 4 R are the sides of R adjacent to 1 (compatibility condition for regularity up to the boundary). See Figure 2. Note: b is a C 1; ( R) weak solution if (1.4) R r'(rb f(x; b)) = 0 for all ' 2 H 1 0(R 2 n ) ; where H 1 0(R 2 n) is the closure of C 1 0(R 2 n). That means the boundary condition on 1 is a natural boundary condition for weak solutions in the sense (1:4). In addition, if we denote kfk C ;x; R the C ( R)-norm of f with respect to the rst variable and kfk Lip;b denotes the Lipschitz norm of f with respect to the second variable, the following estimates hold: (1.5) kbk C 1; ( R) C kfk C ;x; R + kfk Lip;b kfk 1; R and (1.6) kbk C ( R) Ckfk 1; R ; where C depends on R.

2 IRENE M. GAMBA ' @1 =0 @ 2 r'n.... @ 2 =0 @ 1 @ 3 r'n =0 @ 4 @ 4 jr'j @3 =g(x) Figure 1. Domain and boundary conditions on '. We consider the following elliptic problem: given ~ h = (h 1 ; h 2 ) a C ( ) function, 0 < < 1, (1.1) 8 >< >: ' = r' ~ h ' = 0 and r' n @1 (w 1) < 0 @1 @2[@ = 0 4 @' @n jr'j = g(x) @3 in inow conditions streamline walls prescribed speed Here n denotes the exterior normal to @ and g(x) is a C 1; (@ 3 ) strictly positive function. We show the following theorem. Theorem 1. There exists a unique solution ' 2 C 2; ( ) of the boundary value problem (1:1) with nonvanishing gradient. In addition, the solution ' satises (1.2) k'k C 2; () KeKkhk 1; k'k C 1;() KeKkhk 1; and 1 + khk1; 1 + khk C ()khk 1; + khk1; 2 1 + khk1; : where K depends on and on the C 1; -norm of g; and the following estimates hold, 0 < ke KH jr'j Ke KH ; and e K ' e K 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 Result of a Nonlinear Two-Dimensional Elliptic Boundary Value Problem IRENE M. GAMBA Courant Institute Abstract We 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 show existence and uniqueness of a C 2; solution with nonvanishing gradient, and we nd positive lower and upper bounds for jr'j along with C 2; estimates of ', in terms of the C and L1 norms of ~ h. c 1995 John Wiley & Sons, Inc. 1. Introduction When studying compressible potential ow, the following equation arises naturally: Given a strictly positive C 1; density in a ow region, nd the ow potential function ' with nonvanishing gradient that satises the equation div (r') = 0 with boundary conditions given by r' n = 0, n the outer unit normal to the boundary, along the lateral walls, ' constant at the inow boundary, and the speed jr'j prescribed on the remaining boundary section. We are able to solve uniquely this problem in two dimensions by using the Bers theory for the ow equation. In fact, for our application in compressible hydrodynamics (see [6]), we need to consider a general C, 0 < < 1, vector eld ~ h, instead of r`n, and we must obtain existence of a unique C 2; solution ' with positive lower bound for the speed jr'j, along with a C 2; estimate for ' that has linear growth in the C norm of h. Let be a domain in two dimensions with a C 2; boundary except for four points called w i, i = 1;...; 4. Let @ i be the C 2; component of @ and let satisfy the condition that @ i \ @ i+1 meet at points w i forming a =2 angle. Thus is a \smooth rectangular" 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