TORNADO SOLUTIONS FOR SEMILINEAR ELLIPTIC EQUATIONS IN R 2 : APPLICATIONS

TORNADO SOLUTIONS FOR SEMILINEAR ELLIPTIC EQUATIONS IN R 2 : APPLICATIONS FRANCES HAMMOCK, PETER LUTHY, ALEXANDER M. MEADOWS, AND PHILLIP WHITMAN 1. Introduction In this pper we study positive solutions to equtions u = f(u) on domins of R n where f(u) is like u α ner zero. For the min result, when n = 2 nd 0 < α < 1, we show the nonexistence of torndo sequences of solutions. By results of the first pper [7], it follows tht uniformly bounded solutions of this eqution re equicontinuous. Existence of singulr solutions cn be proved then using degree rgument. The bulk of the proof lies in understnding the rdilly symmetric singulr solutions to u = u α. We estblish the existence of continuous fmily of solutions tht cn be used s brriers for the mximum principle. Importnt remining open questions re wht hppens when α 1 nd when n 3. Some work in this direction ws done in [6]. We recll the definition of torndo sequences. Definition 1. A torndo sequence of solutions to u = f(u) is given by number ɛ > 0, sequence r j > 0, j = 1, 2,..., with r j 0, nd positive solutions u j defined on blls B rj of rdius r j such tht u j > ɛ on B rj nd min Brj u j 0. We will ssume tht is Lipschitz nd tht f(u) = g(u)u α, where g stsifies the following: (1) 0 < α < 1, g(u) is Hölder continuous on (δ, ), δ > 0 0 C 1 < g(u) < C 2 <. Note tht ny solution u is subsolution to u = C 1 u α nd supersolution to u = C 2 u α. We my now stte the min results. Theorem 2. If 0 < α < 1 nd f stisfies the bove ssumptions, there does not exist torndo sequence of solutions to u = f(u). 1

Corollry 3. If 0 < α < 1, nd f stisfies the sme ssumptions, then sequence of positive solutions u k to u = f(u) on domin R 2 with u k M is equicontinuous on ny compct subdomin. We cll nonnegtive function u 0 singulr solution to u = f(u) if u 0 is limit of sequence of positive smooth solutions nd min u 0 = 0. Theorem 4. There exists lrge vriety of nonnegtive singulr solutions to u = f(u) on the disc B R 2. In [8] nd [9], Phillips studied the free boundry problem for u = (1 α)u α for 0 < α < 1. In this cse, the free boundry solution with boundry dt u = ψ on is the function u 0 which minimizes the integrl Du 2 + u 1 α. This solution is llowed to be identiclly zero (nd thus not stisfy the differentil eqution) on positive mesure subset 1 α 1,. The minimizer is loclly C 1+α, nd the free boundry hs loclly finite n 1-dimensionl Husdorff mesure. We note tht our results do not pply to the free boundry cse, but do pply to solutions to the PDE which re not minimizers of vritionl integrl, nor even stble with respect to vritionl integrl. In our cse, the vritionl integrl would be 1 u I(u) = 2 Du 2 + F (u), F (u) = f(z) dz. Solutions of the PDE u = f(u) re sttionry, nd they re stble if the second vrition of I is nonnegtive. The eqution u = u α lso rises in reltion to chemicl ctlyst kinetics (See [1] nd [4]). Similr equtions were studied in [2] nd [3]. Singulr nd stble solutions of the eqution u = 1 were studied in [6], where the work ws inspired u by similr study of the singulr miniml surfce eqution, s in [10]. The mjor results, which lso pply to u = u α, rely hevily on the ssumption tht solutions re stble with respect to the corresponding vritionl integrl. 1 2. Bsic Fcts We re prticulrly interested in singulr solutions of u = f(u), i.e. those tht re not differentible to second order. We construct such solutions s nonnegtive functions which re limits of positive smooth solutions. In the following, we show tht nonnegtive wek solutions which chieve the vlue zero re indeed singulr. 2

Definition 5. By wek solution to u = f(u), we men function u L 2 loc (), Rn, such tht f(u) L 2 loc () nd u ζ = f(u)ζ, for ll ζ Cc (). Assuming f stisfies the ssumptions (1), wek solutions which hve zero on the interior of domin hve limited regulrity. Lemm 6. Let u C 1,β (B δ (0)) be wek solution of u = f(u) such tht u 0 nd u(0) = 0. Then β 1 α 1 + α. Proof: Let 0 < ρ < δ be rbitrry. Since u 0 nd u C 1, u(0) = 0, nd for ny x B δ (0), u(x) C x β, nd thus u(x) C x β+1. We now let ζ be rdilly symmetric function in Cc () with ζ = 1 on B ρ/2 (0), ζ C 1 ρ, nd ζ C 2. Then ρ 2 (2) B ρ u ζ C C 2 ρ 2 while by the wek eqution, (3) B ρ x β+1 C 2 ρ β+n 1, ( u ζ = f(u)ζ C ) x β+1 α C1 ρ n αβ α. B ρ B ρ B ρ/2 Thus, ρ n αβ α Cρ β+n 1 for rbitrrily smll ρ, from which necessrily β 1 α 1 + α. [ Note: Setting C α,n = (1+α) 2 2n(1+α) 4α ] 1 1+α, the function U α,n (x) = C α,n x 2 1+α is wek solution of u = u α chieving the mximum llowed regulrity. The sme rgument s bove cn be pplied to show lck of boundry regulrity. Lemm 7. Suppose R n is domin with n interior cone condition, nd suppose 0. Let u C 1,β ( ) be wek solution of u = f(u) such tht u 0 nd u(0) = 0. Then β 1 α 1 + α. Proof: The bove integrls (2) nd (3), tken insted over B ρ, cn be estimted in exctly the sme wy. There re some dditionl positivity results tht will be useful in the next section. 3

Lemm 8. Suppose u is positive nd smooth subsolution to u = C 1 u α on domin which includes bll B 2ρ of rdius 2ρ. Then (1) B 2ρ \B ρ u 1+α C 1 αω n ρ n+2 (2) sup u ( ) 1 C 1 α 2 n 1+α ρ 2 1+α 1 Here ω n is the volume of the unit bll in R n. Note tht the second property sttes tht positive solutions on with very smll boundry vlues do not exist, nd tht solutions defined on ll of R n must be unbounded. Proof: We use the wek inequlity Du Dζ C 1 ζu α for ll ζ Cc 1 () with ζ = u α ϕ 2 nd ϕ Lipschitz function which is equl to zero outside B 2ρ. Then αu α 1 ϕ 2 Du 2 + 2 u α ϕdu Dϕ C 1 ϕ 2. Using the Cuchy Schwrtz inequlity, ϕ ( ) 2 αu α 1 Du 2 + C 1 2 u α ϕ Du Dϕ ( ) ( ) 2 u α 1 2 ϕ Du u α+1 2 Dϕ α u α 1 ϕ 2 Du 2 + 1 u α+1 Dϕ 2. α Thus, (4) C 1 α ϕ 2 u α+1 Dϕ 2. We choose ϕ to be rdilly symmetric on B 2ρ so tht, using the rdil vrible r, ϕ(r) = 1 for 0 r ρ, nd ϕ(r) = 2 r/ρ for ρ r 2ρ. Then Dϕ = 1/ρ on B 2ρ \ B ρ. So, (4) becomes C 1 αω n ρ n C 1 α ϕ 2 1 u ρ B 1+α, 2 2ρ \B ρ which implies the first result. Now suppose u M on, so by the mximum principle, u M on B 2ρ. The first result implies which completes the proof. M α+1 ω n ((2ρ) n ρ n ) C 1 αω n ρ n+2, 4

3. Rdilly Symmetric Solutions The nlysis of solutions which re rdilly symmetric will be very importnt to us. Definition 9. A rdil solution to u = f(u) is solution u(x) which only depends on x. i.e. u(x) = u(r), r = x. Rdil solutions stisfy the ordinry differentil eqution u rr + n 1 r u r = f(u). We present the existence nd symptotics of rdil solutions to u = u α with vrious intitil-boundry conditions. Some nlysis of this ODE ws done by Bruner Nicolenko in [1]. We lso investigte the existence of rdil solutions to u = f(u). Lemm 10. For ny ɛ > 0, there is rdil solution to u = f(u) defined on R n with u(0) = ɛ. In cse f is loclly Lipschitz ner u = ɛ, the solution is unique. Proof: If we ssume f is Lipschitz on [ ɛ, 2ɛ], this cn be done by the 2 Contrction Mpping Theorem. Setting v = u ɛ, we seek solution v to the eqution ( ) (5) r n 1 v r = r rn 1 f(ɛ + v). We consider the mp T defined on {v C 0 [0, δ] : v < δ}, δ < ɛ/2, by Then nd T (v)(r) = r 0 1 t n 1 t 0 s n 1 f(ɛ + v(s)) ds dt. sup T (v) δ2 f(ɛ δ) 2n sup T (v 1 ) T (v 2 ) δ2 L 2n sup v 1 v 2. where L is the Lipschitz constnt. So, for smll enough δ, T is contrction mpping. Now T (v) C 2 [0, δ] with T (v) (0) = 0 nd the unique fixed point stisfies (5). Since f > 0, the solution v cnnot hve locl mximum, thus v 0, nd then by integrtion, v(r) Cr 2. By the ODE existence nd uniqueness theorem, this solution cn be continued to ll of [0, ). Thus, u = ɛ + v is rdil solution on R n. In the cse tht f is not loclly lipschitz, locl existence follows from the Schuder fixed point theorem pplied s in the proof of the next lemm. The following lemm, which we stte in generlity, will be pplied in the somewht simpler cse tht g(u) = C 1 = C 2 is single constnt. 5

Lemm 11. For ny 0, there is rdil solution to u = f(u) defined on R n \ B (0) with u() = 0 nd u r () = 0. Proof: For δ > 0, let K be the closed convex subset of X = C 0 [, + δ] defined by { } K = v C 0 [, + δ] : A 1 (r ) 2 1+α v(r) A2 (r ) 2 1+α. Here the A i nd δ re constnts to be chosen. We define T : K X by r 1 t T (v)(r) = s n 1 f(v(s)) ds dt. t n 1 We will show for smll enough δ tht T : K K nd tht T is continuous nd compct in order to pply the Schuder Fixed Point Theorem. Now if v K, T (v)(r) = r 1 t n 1 r t C 2 r t t s n 1 g(v(s))v(s) α ds dt g(v(s))v(s) α ds dt A α 1 (s ) 2α 1+α ds dt C 2A α 1 (1 + α) 2 (r ) 2 1+α. 2(1 α) For the lower bound, we use lrge integer m, depending on nd α. We do this so tht δ is independent of, nd llows = 0. T (v)(r) = r r 1 t n 1 (m 1)+r m t m 1 n C 1 A α 2 s n 1 g(v(s))v(s) α ds dt t 1 n t (m 1)+r m r (m 1)+r m = m1 n C 1 A α 2 (1 + α) 1 α We choose m depending on α so tht s n 1 g(v(s))v(s) α ds dt t 1 n t (m 1)+r m (1 + α)m 2/(1+α) 2m + 1 α 2m 2/(1+α) 6 (s ) 2α 1+α ds dt ( (1 + α)m 2/(1+α) 2m + 1 α 2m 2/(1+α) 1 + α, 4 ) (r ) 2 1+α.

nd so T (v)(r) m1 n α C 1 A α 2 (1 + α) 2 (r ) 2 1+α. 4(1 α) For T to mp K to itself, we need the inequlities A 1 m1 n C 1 A α 2 (1 + α) 2 4(1 α) C 2A α 1 (1 + α) 2 2(1 α) A 2, which re stisfied if ssume (without loss of generlity) tht C 1 C 2 1 nd set the constnts ( ) m 1 n 1 C ( ) 1 α 1 2 (1 + α) 2 1 1+α A 1 =, 2C2 α 2(1 α) ( ) α ) 1 A 2 = (C ( ) 2 m (n 1)α 1 α 2C1 2 (1 + α) 2 1 1+α. 2(1 α) To see tht T is continuous, note tht T (v 1 )(r) T (v 2 )(r) nd we estimte r t t 1 n s n 1 f(v 1 (s)) f(v 2 (s)) ds dt f(v 1 (s)) f(v 2 (s)) v 1 (s) α g(v 1 (s)) g(v 2 (s)) +g(v 2 (s)) v1 (s) α v 2 (s) α nd v1 (s) α v 2 (s) α α δ min{v 1(s), v 2 (s)} α δ v 1 (s) δ v 2 (s) δ. Since T mps K into C 1, it is compct, nd by the Schuder Fixed Point Theorem, there exists solution in K. Remrk: By similr method, there exist solutions with u() = 0 nd u r () = A for ny A > 0. Remrk: In contrst, for α 1, there exists no rdil solution u to u = f(u) on 1 r < 1 + ɛ with u > 0 on (1, 1 + ɛ) nd u(1) = 0. For exmple in cse n = 2, we chnge to the vrible t = log r so tht u tt = e 2t f(u) on n intervl [0, ɛ ) with u(0) = 0. Since u t > 0, we hve u t u t (ɛ ) = C, so u Ct. But then u tt Ct α, nd so u t C+ C t s α ds. So, for α 1, u ɛ t is unbounded below on pproch to t = 0, contrdiction. A similr rgument holds for n 3. We now turn to the symptotics of rdil solutions to u = u α. We will show tht solutions re symptotic to the solution U α,n (x) = C α,n x 2 1+α, where s before, C 1 α α,n = 2 2α 2(n 1) + (1 + α) 2 1 + α. 7

First, by Lemm 8, ll solutions re eventully bounded below by B 1 r 2 1+α nd thus bounded bove by B2 r 2 1+α for r > r0. Here the constnts B 1, B 2, nd r 0 my depend on ɛ for the solutions of Lemm 10 nd for the solutions of Lemm 11. We consider the function v where u = C α,n r 2 1+α (1+v). The eqution for v is v rr + 4 + n 1 1+α r v r + (1 + α)c 1 α α,n r 2 v = C 1 α α,n [ (1 + v) α 1 + αv ]. r 2 In order to mke the eqution utonomous, we chnge to the vrible t = log r nd we hve the eqution [ (6) v tt + βv t + γv = Cα,n 1 α (1 + v) α 1 + αv ], where β = n + 2 + (n 2)α 1 + α nd γ = 2n + 2(n 2)α. 1 + α Now we need to show tht s t, ll solutions v converge to the solution v = 0. Note tht solutions v now stisfy 0 < B 1 < 1 + v < B 2. The eqution (6) my be written s n utonomous system for v = (v, v t ): [ v vt ] [ ] [ ] 0 1 v = γ β vt t [ + Cα,n 1 α 0 (1 + v) α 1 + αv which we will write briefly s v t = A v + h( v). Note tht (0, 0) is the only criticl point of this system mong v with 1 + v > 0. For smll v, the nonliner prt h( v) is on the order of v 2. The liner prt with mtrix A hs eigenvlues λ = β 2 ± β 2 4 γ. For n = 2, β2 γ = 4 4 < 0, nd the eigenvlues re complex. 4 (1+α) 2 1+α For n 7, the eigenvlues re rel, nd for intermedite n, the sitution depends on α. From now on we will ssume tht n = 2. In this cse the rel prt of the eigenvlues is 2. 1+α First we will show tht for ny ɛ > 0, eventully v(t 0 ) < ɛ for some t 0. For nottionl convenience, we will substitute x = v, y = v t nd write the utonomous system s x t = P (x, y) = y y t = Q(x, y) = γx βy + Cα,2 1 α ((1 + x) α 1 + αx) 8 ],

First, there re no closed cycles of this system in the hlf-plne 1+x > 0 since by Green s Theorem, if Γ were cycle of period T enclosing region, then T 0 = x t y t dt y t x t dt = x t dy y t dx = P dy Q dx 0 Γ Γ = (P x + Q y ) da = β da. Now consider the region D consisting of rectngle with smll disk bout zero removed: D = {B 1 1 < x < B 2 1, R < y < R} \ { x 2 + y 2 < ɛ 2}. Since there re no criticl points or cycles in D, by the Poincré Bendixson Theorem, ny solution trjectory in D must leve in finite time. By our estimtes, no trjectory my leve through either side x = B 1 1 or x = B 2 1. On the other hnd, on the side y = R, we hve y t = γx βr + Cα,2 1 α ((1 + x) α 1 + αx) < 0 for R lrge enough, nd similrly y t > 0 long the side y = R for R lrge enough. Thus, the trjectory must leve through the circle x 2 + y 2 = ɛ 2 t some time t 0. Now the solution stisfies nd so v = e A(t t 0) v(t 0 ) + v e β 2 (t t 0) v(t 0 ) + t t 0 e A(t s) h( v(s)) ds t e β t 0 2 (t s) h( v(s)) ds Suppose t some time t 1 fter t 0 the trjectory grows to where v(t 1 ) = 2ɛ. Then we hve 2ɛ e β 2 (t 1 t 0 ) ɛ + t1 t 0 e β 2 (t1 s) C 0 ɛ 2 ds < ɛ + Cɛ 2, contrdiction for smll ɛ. Thus, replcing ɛ by ɛ/2, we hve e β 2 t v e β 2 t 0 v(t 0 ) + ɛ nd so by Gronwll s Inequlity, t v < Ce (ɛ β/2)t for t > t 0. Now, the homogeneous eqution w tt + βw t + γw = 0 9 t 0 e β 2 s v(s) ds,

hs solutions w 1 = e λ1t nd w 2 = e λ2t, where λ i re the eigenvlues of A given bove. The Wronskin W of these hs order e β. By the vrition of prmeters formul, we my now write the solution v s v = A 1 w 1 + A 2 w 2 + w 1 t 0 h(v) w 2 W + w 2 t 0 h(v) w 1 W, where h(v) = (1 + v) α 1 + αv. Since the integrls in this formul re convergent, we my rewrite s We hve shown the following v = Ã1w 1 + Ã2w 2 + O ( e (ɛ β)t). Lemm 12. Let u be rdil solution to u = u α in dimension 2. Then ( ) u C α,2 r 2 1+α = A cos( log r) + B sin( log r) + O r ɛ 2 1+α for ny ɛ nd suitble. The eqution u = u α hs nturl scling so tht if u is ny solution nd C > 0 ny constnt, then the function ũ C (x) = C 2 1+α u(cx) is lso solution. The rdilly symmetric singulr solution U α,2 is invrint under this scling. The bove symptotics show tht ny rdilly symmetric solution u stisfies u U α,2 < k for some constnt k. Thus, for ny constnt C, ũ C U α,2 < C 2 1+α k, nd so s C, the sclings of ny rdilly symmetric solution converge uniformly to U α,2. So, if we choose ll the sclings of rdil solution with u(1) = u (1) = 0, nd ll sclings of rdil solution with u(0) = 1, u (0) = 0, these form continuous fmily of solutions including U α,2. Corollry 13. There exists continuous fmily W s, s R of rdilly symmetric solutions to u = u α, where - W 0 = U α,2, - for s < 0, W s is defined on ( s, ) with W s ( s) = W s( s) = 0, nd - for s > 0, W s is complete with W s (0) = t, W s(0) = 0. Also, becuse of the oscilltory symptotics in lemm 12, we hve Corollry 14. The Dirichlet problem { u = u α on B 1 R 2 u = C α,2 on B 1 10

hs infinitely mny distinct positive solutions. The free boundry problem u = u α on B 1 R 2 u = C α,2 on B 1 u = u = 0 on B 1 lso hs infinitely mny positive solutions. Similr sttements hold for suitble α in dimensions 3 through 6. 4. Torndos We re now redy to prove Theorem 2. Proof of Theorem 2: Suppose {u j } is torndo sequence of solutions to u = f(u) s in Definition 1, where f(u) stisfies the hypotheses in (1). Consider the continuous fmily of functions W s (x) = W s ( C 2 x), which stisfy W α s = C 2 W s, nd re thus supersolutions to the eqution. For ech j, we consider s j := sup{s : Ws < u j }. Then < s j < nd W sj u j with W sj (x j ) = u j (x j ) for some x j B rj. Now for very lrge j, Winf uj < ɛ on B rj. Thus, Wsj < ɛ on B rj nd x j B rj. Also, 0 < W sj u j on = {x B rj : Wsj (x) > 0}. But ( W sj u j ) > C 2 ( W α s j u α j ) > 0. Since W sj u j hs n interior zero mximum t x j, by the Hopf Mximum Principle, Wsj = u j, contrdiction. Corollry 3 now follows from the result of [7]. Remrk: The sme proof holds showing there is no torndo sequence of clss µ with µ(δ) = log δ m for ny 0 < m < 1 s in [7]. We cn lso modify the proof of the torndo theorem to show boundry continuity on the disk. Lemm 15. A sequence of positive solutions u j to u = f(u) on disk B R 2 with u j C 1 ( B) M nd u j B > ɛ is equicontinuous on B. Proof: Suppose not. Then there is fixed δ > 0 nd sequence of solutions u j nd points x j, y j B with u j (x j ) u j (y j ) > δ nd x j y j 0. Using Corollry 3 nd pssing to subsequence, then x j, y j x 0 B. By stndrd elliptic regulrity (for instnce Theorem 9.14 in [5]), then there re points z j with z j z 0 B nd u j (z j ) 0. Now we use the functions W s s in the bove proof to get contrdiction. 5. Singulr Solutions The construction of singulr solutions to u = f(u) follows the Lery Schuder degree scheme used in [6] nd [10]. We re ble here 11

to get better results thn those chieved for the eqution u = 1 in u [6]. We will consider the Bnch spce B = C 2,µ () of functions with Hölder continuous second derivtives on domin R 2 with norm 2,µ. First, by Schuder estimtes, for ny δ > 0, there is number M δ such tht ny solution u of u = f(u) with δ u 1 stisfies δ u 2,µ < M δ. We set U δ = {u B : u > δ, u 2,α < M δ }. Given boundry dt ϕ 0 with δ < ϕ 0 < 1, we consider the nonliner δ opertor T 0 : U δ B, defined by T 0 (u) = v, where v is the solution of { v = f(u) on v = ϕ 0 on Lemm 16. For ech R 2, there exists M() such tht if ϕ 0 > M(), then the Lery Schuder degree deg(i T 0, U δ, 0) = 1 for ll 0 < δ < 1/M(). Proof: We choose n origin 0. We gin use the fmily W s of solutions to W α s = C 2 W s. We choose s 0 such tht W s0 > 1, nd set M() = sup s s 0,x W s (x). Now consider T t : U δ B, for 0 t 1, defined by T t (u) = v, where v is the solution to { v = (1 t)f(u) on v = ϕ 0 on Since T 1 is constnt function with imge in U δ, deg(i T 1, U δ, 0) = 1. So, the result holds unless there is fixed point of T t on U δ for some 0 t < 1. But ny such fixed point u stisfies u 2,µ < M δ by the Schuder estimte, nd u is lso subsolution of u = C 2 u α nd so the continuous fmily of brrier functions W s force u to be greter thn W s0 nd thus greter thn δ on. So, there re no fixed points on U δ nd the result is proved. Conversely, it follows from Lemm 8 tht there is ɛ() > 0 such tht if ϕ 0 < ɛ, then deg(i T 0, U δ, 0) = 0. Now we suppose ϕ t is smooth fmily of boundry dt with ϕ 0 > M() nd ϕ 1 < ɛ(). We consider T t : U δ B defined by T t (u) = v, the solution of { v = f(u) on v = ϕ t 12 on

By the properties of the degree, there must be n intermedite t δ nd fixed point u δ of T tδ on U δ. Thus, u δ stisfies { uδ = f(u δ ) on u = ϕ tδ on with u δ δ nd min u δ = δ. We choose sequence δ j 0 nd construct corresponding u j = u δj. By Corollry 3, there is subsequence u j u 0 on ny compct subset of, where u 0 is singulr solution to u = f(u). If is disc B, there is subsequence u j u 0 on B with u 0 = ϕ t0 on B. The lrge vriety of possible fmilies ϕ t implies lrge vriety of singulr solutions to the eqution u = f(u). This completes the proof of Theorem 4. References [1] Bruner, C.M. nd Nicolenko, B., On nonliner eigenvlue problems which extend into free boundries problems, Bifurction nd nonliner eigenvlue problems (Proc., Session, Univ. Pris XIII, Villetneuse, 1978), pp. 61 100, Lecture Notes in Mth., 782, Springer, Berlin-New York, 1980. [2] Dávil, Jun, Globl regulrity for singulr eqution nd locl H 1 minimizers of nondifferentible functionl, Commun. Contemp. Mth., vol. 6, no. 1, pp. 165 193, 2004. [3] Dávil, Jun nd Montenegro, Mrcelo, Positive versus free boundry solutions to singulr elliptic eqution, J. Anl. Mth., vol. 90, pp. 303 335, 2003. [4] Diz, J.I. nd Morel, J.M. nd Oswld, L., An elliptic eqution with singulr nonlinerity, Comm. P.D.E., vol. 12, pp. 1333 1344, 1987. [5] Gilbrg, Dvid nd Trudinger, Neil S., Elliptic Prtil Differentil Equtions of Second Order, Springer, New York, revised 2nd editon, 2001. [6] Medows, Alexnder, Stble nd singulr solutions of the eqution u = 1 u, Indin Univ. Mth. J., 2005. [7] Medows, Alexnder, Torndo solutions for semiliner elliptic equtions in R 2 : regulrity theory, in preprtion, 2004. [8] Phillips, Dniel, A minimiztion problem nd the regulrity of solutions in the presence of free boundry, Indin Univ. Mth. J., vol. 32, no. 1, pp. 1 17, 1983. [9] Phillips, Dniel, Husdorff mesure estimtes of free boundry for minimum problem, Comm. P.D.E., vol. 8, pp. 1409 1454, 1983. [10] Simon, Leon, Some exmples of singulr miniml hypersurfces, in preprtion, 2001. University of Cliforni Berkeley, CA Connecticut College 13

Deprtment of Mthemtics Cornell University Ithc, NY 14853 emil: medows@mth.cornell.edu University of Texs t Austin 14