Keywords. Problem with free boundary, global solution, Stefan problem.

Transcription

1 Journal of Mathematical Sciences, Vol. 178, No. 1, October, 2011 Stefan problem Mikhail A. Borodin Presented by A. E. Shishkov Abstract. We prove the existence of a global classical solution of the multidimensional two-phase Stefan problem. The problem is reduced to a quasilinear parabolic equation with discontinuous coefficients in a fixed domain. With the help of a small parameter ε, we smooth coefficients and investigate the resulting approximate solution. An analytical method that enables one to obtain the uniform estimates of an approximate solution in the cross-sections t = const is developed. Given the uniform estimates, we make the limiting transition as ε 0. The limit of the approximate solution is a classical solution of the Stefan problem, and the free boundary is a surface of the class H 2+α,1+α/2. Keywords. Problem with free boundary, global solution, Stefan problem. 1. Statement of the problem The Stefan problem in the classical statement is a mathematical model of the propagation of heat in a medium, being in different phase states, e.g., liquid and solid ones ue to the melting or crystallization, the domains occupied by the liquid and solid phases will vary. Therefore, the surface separating these phases will vary as well. This unknown surface is called a free boundary. The process of propagation of heat in each of the phases is described by the heat equation. Let be a bounded domain in R 3 whose boundary consists of two C 2+α surfaces 1 and 2 such that 1 lies inside 2. Let a medium, being in two phase states, occupy the domain, and let ux, t be the medium temperature. The crystallization temperature, i.e., the temperature of transition from the liquid state to the solid one, is equal to 1. It is required to find a function ux, t в the domain T = 0,T, where T>0, that satisfies the following conditions: u au u =0 in Ω T G T, 1.1 Ω T = x, t T :0<ux, t < 1, G T = x, t T : ux, t > 1, where au is a piecewise constant function equal to a 1 > 0 in Ω T and to a 2 > 0 in G T. On the known boundary of the domain T, we set the conditions ux, t =0 on 1 [0,T, ux, t =ϕx, t > 1 on 2 [0,T. 1.2 On the free unknown surface γ T = T Ω T = T G T, the laws of conservation of the mass and the energy give two conditions: the equality of the temperature ux, t to the melting temperature and Translated from Ukrains kiĭ Matematychnyĭ Visnyk, Vol. 8, No. 1, pp , January February, Original article submitted November 5, /11/ c 2011 Springer Science+Business Media, Inc. 13

2 the Stefan condition which involves the heat release due to the latent heat of melting, u x, t =u + x, t =1, 3 i=1 u u+ cosn, x i +λ cosn, t = Here, u x, t and u + x, t mean the boundary values of the function ux, t on γ T which are taken, respectively, from the sides of the domains Ω T and G T, n isthenormaltothesurfaceω T or G T which is directed to the side where the functionux, t increases, and λ is the coefficient of latent melting heat. The initial conditions are as follows: ux, 0 = ψx > 0 in, ψx =0, ψx = ϕx, 0 > 1, 1 2 Ω 0 = x :0<ψx < 1, G 0 = x : ψx > 1, γ 0 = Ω 0. The multidimensional Stefan problem was studied by many researchers. Its investigation was started by works by O. A. Oleinik [22] and by S. A. Kamenomostkaya [15]. The conception of a generalized solution was first used there, which allowed one to prove the theorem of existence and uniqueness of a weak solution. The class of functions in which a solution of the Stefan problem is unique is wider than the class of functions that includes the generalized solution. Since every classical solution is also a generalized one, the question about the uniqueness of a solution of the Stefan problem was completely solved. But the question about the existence of the classical solution remained open. In 1973, G. uvaut [11] reduced the Stefan problem to some variational inequality, for which he proved the existence of a weak solution. In 1975, A. Friedman and. Kinderlehrer [12], by using the uvaut transformation, reduced the one-phase multidimensional Stefan problem to a variational inequality and then proved the Lipschitz property of the free boundary. However, this smoothness was not sufficient to prove that the solution is classical. In the same years, L. A. Caffarelli [7 9] obtained a number of fundamental results on the smoothness of free boundaries. Using the results of L. A. Caffarelli,. Kinderlehrer and L. Nirenberg [16] were succesful to prove the existence of the classical solution of the one-phase problem in In 1979, A. M. Meirmanov [20] proved the existence of the classical solution of the two-phase Stefan problem in small time intervals. In 1982, M. A. Borodin [4] proved the existence of a global solution of the two-phase problem and established the Lipschitz property of the free boundary. Then in the 1980s, the classical solvability in small time intervals was proved within various methods in works by E.-I. Hanzawa [14], B. V. Bazalii [3], and E. V. Radkevich [24]. The Lipschitz continuity of the free boundary in the two-phase problem was proved by R. H. Nochetto in [21]. In 1996, J. Athanasopoulos, L. Caffarelli, and S. Salsa showed in [1] that the viscous solution with a Lipschitz free boundary under the so-called condition of nondegeneracy is classical. In this case, the free boundary is a surface of the class C 1 in space and in time. We note else that the Stefan problem with regard for the free boundary curvature and the convection was studied, respectively, in [2, 6, 23] and [25]. The more detailed survey of the problem under consideration can be found in [10, 13]. In 1999, the existence of a global classical solution of the multidimensional Stefan problem was proved by the author in [5], and the free boundary was considered as a surface of the class H 2+α,1+α/2. Together with the ordinary requirement of smoothness, the sufficient condition was the requirement for the quantity ψx to have a fixed sign, where ψx means the initial distribution. This requirement leads to the monotonicity of the process described by a solution of the Stefan problem. In other words,

3 either the crystallization or the melting occurs. It is obvious that this requirement is sufficiently burdensome. In the present work, such a restriction is removed and is replaced by a more natural one, namely, ψx 0. As was shown in [20], the violation of this condition causes the nonexistence of the classical solution. Our main result consists in the following theorem: Theorem 1.1. Let be a bounded domain in R 3 whose boundary consists of two C 2+α surfaces 1 and 2 such that 1 lies inside of 2, T = 0,T, ψx C 2+α, ϕx, t H 2+α,2+α/2 T, min ψx > 0, min ϕx, t > 1, 2 [0,T ] and the corresponding conditions of matching at t =0,x are satisfied. Then problem is solvable and has the unique solution ux, t H 2+α,1+α/2 Ω T \ γ 0 G T \ γ 0 C T, and the free boundary is a surface of the class H 2+α,1+α/2. 2. Approximation of the problem We assume that problem has the classical solution. We now multiply equality 1.1 by the function ηx, t C 2,1 T which is zero on 1 0,T, 2 0,T, att = T, and will integrate by parts, where T ux, t ηx, t+au u χu = ηx, t+λχu η dx dt + λ 1, if ux, t < 1, 0, if ux, t > 1. χψηx, 0 dx =0, 2.1 For any ε>0, we define a function χ ε x C R 1 as follows: χ ε x =1 x 1, χ ε x =0 x 1+ε, χ εx 0, χ n ε x c ε n, where the positive constant c is independent of ε. Let a ε x =a 2 + χ ε xa 1 a 2. With the help of the functions a ε x,χ ε x, we smooth the coefficients in 2.1 and denote, by u ε x, t, the function that satisfies the integral identity T u ε x, t ηx, t+a ε u ε uε x, t ηx, t+λχ ε u ε ηx, t dx dt + λ χ ε ψηx, 0 dx =

4 If the function u ε x, t satisfies identity 2.2 and is sufficiently smooth, then it is a solution of the following problem: u ε x, t a ε u ε x, t λχ εu ε x, t uε x, t =0in T, 2.3 u ε x, t =0on 1 [0,T, u ε x, t =ϕx, t on 2 [0,T, 2.4 u ε x, 0 = ψx in. 2.5 The function u ε x, t, being a solution of problem , is called an approximate solution of problem Our main goal is to prove the existence of an approximate solution and to show that ux, t = lim u ε x, t ε 0 is a solution of problem Since 0 <a 0 =mina 1,a 2 b ε u ε x, t = a ε u ε x, t λχ εu ε x, t c ε, where c>0 and is independent of ε, we have, as known [18], the following proposition. Theorem 2.1. Let ψx H r+α, ϕx, t H r+α,r+α/2 T, min 1 [0T ] ϕx, t > 1, 0 <α<1, r 2, and let the corresponding conditions of matching at t =0, x be satisfied. Then problem is solvable and has the unique solution u ε x, t H r+α,r+α/2 T. Moreover, the estimate u ε x, t H r+α,r+α/2 T c Mε holds, where the positive constant c is independent of ε, andmε 0, ifε 0. Let us represent Eq. 2.3 in the form 2.6 u ε x, t u ε x,t 0 [a ε λχ ε] d =0. We now cut the cylinder T by the planes t = k, where N = T, k =1, 2,...,N, and integrate with respect to the variable t from k 1 to k: k k 1 u ε x, d u ε k x u ε k 1 x [a ε λχ ε] d =0, 2.7 where u ε k x uε x, k. After the obvious transformations, we obtain u ε k x βε k x uε k x = fk ε x, 2.8 u ε k x =0on 1, u ε k x =ϕx, k > 1 on 2, u ε 0x =ψx in, 2.9 where 16 u ε k x = uε k x uε k 1 x,

5 1 βk ε x = a ε [u ε k 1 + uε k uε k 1 ] λχ ε[u ε k 1 + uε k uε k 1 ] d 0 = 1 0 b ε [u ε k 1 x+uε k x uε k 1 x] d, 2.10 f ε k x = 1 k [ u ε x, k u ε x, t] dt k 1 Our nearest purpose is to obtain the uniform estimates for solutions of problem , i.e., the estimates for u ε x, t on the cross-sections t = const. To this end, we will construct and study the integral representations which will be play the same role as the integral representations of the type of potentials in the theory of linear boundary-value problems for elliptic equations. 3. Integral representation of an approximate solution and properties of the fundamental solutions To study the properties of an approximate solution, we need an integral representation. Let K R x 0 be a ball of radius R centered at the point x 0 that belongs to the domain. We introduce the functions Γ m k+1 x x 0 = sinh[ zr x x 0 ] dz a m 2πi 4π x x 0 sinh zr1 z a m 1 z a m 1 1 z a k, 3.1 L where k =1, 2,...,m, L is the boundary of the domain, L = z : ϱ = z < 1 + q max 1 k N a k, Re z = b 0 > π2 2R 2, b 0 <ϱ, q>0. Since the numerator and the denominator of the integrand in 3.1 have the same branching point z =0, we can separate the univalent branch of the analytic function inside of the contour L. We denote ω m k+1 z = z a m Then relation 3.1 yields Γ m k+1 x x 0 = i 2πa m L z a m 1 z a k. sinh[ zr x x 0 ]ω m k+1 0 dz 4π x x 0 sinh. 3.2 zrω m k+1 z If a 1 = a 2 = = a m = a, then integral 3.1 takes the form Γ m k+1 x x 0 = 1 m k a m k 1 sinh[ zr x x 0 ] dz 2πi 4π x x 0 sinh zrz a. m k+1 The last integral is the Cauchy integral. Therefore, we have Γ m k+1 x x 0 = 1 m k a m k 1 m k sinh[ zr x x0 ] m k! z m k 4π x x 0 sinh zr L z= a. 17

6 If a 1,a 2,...,a m are different, then integral 3.1 is a separated difference of the corresponding order [19] and can be calculated by the formula Γ m k+1 x x 0 = 1 m k a m 1 a m 2 a k m k 1 m k sinh[ zr x x0 ] m k! z m k 4π x x 0 sinh zr where min 1 k m a k <a < max 1 k m a k. Below, we indicate some properties of the functions Γ m k+1 x x 0. z= a, 3.3 Property 3.1. Let x x 0 0. Then the functions Γ m k+1 x x 0 satisfy the following system of differential equations: Γ m k+1 Γ m k Γ m k+1 a k =0, k =1,...,m 1, Γ 1 Γ 1 a m =0, Γ 1 x x 0 = sinh am R x x π x x 0 sinh a m R. Proof. Let r = x x 0 be the spherical radius, r 0. Relation 3.1 yields d 2 [rγ m k+1 r] dr 2 = 1 a m 2πi = a m 1 If k = m, then relation 3.2 yields Γ 1 x x 0 = 1 a m 2πi z sinh zr r dz 4π sinh zr1 z a m 1 z a m 1 1 z a k L a k z a k sinh zr r dz 2πi 4π sinh zr1 z a m 1 z a m 1 1 z a k L L = a k x x 0 [Γ m k+1 x x 0 Γ m k x x 0 ]. sinh zr r dz 4π x x 0 sinh zr1 z a m = sinh an R x x 0 4π x x 0 sinh a m R. Property 3.2. Γ m x x 0 dx K R x 0 K R x 0 Γ m 1 x x 0 K R x 0 dx Γ 1 x x 0 dx = 1 a m 1 R a m sinh a m R. 3.5 Proof. If x x 0, then Γ m k+1 a k Γ m k+1 Γ m k h =0, k =1,...,m 1. 18

7 In addition, we have lim 4π x x 0 Γ m k+1 x x 0 = x x 0 0 1, if k = m, 0, if k m, The principle of maximum yields Γ m k+1 R =0. Hence, k<m Γ m k+1 x x 0 > 0inK R x 0, K R x 0 a k Γ m k+1 Γ m k h = K R x 0 Γ m k+1 0on K R x 0. Γ m k+1 x x 0 dx 0. Property 3.3. Let K δ x 0 mean a ball of radius δ centered at the point x 0. Then 1, if k = m, lim δ 0 K δ x 0 Γ m k+1 ds = 0, if k m, 3.6 where n is the inner normal. Proof. Indeed, if k<m, then lim δ 0 i 2πa m L K δ x 0 sinh[ zr δ] 4πδ 2 sinh zr + z cosh[ zr δ] 4πδ sinh ds zr ω m k+10 dz = lim ω m k+1 z δ 0 i 2πa m L K δ x 0 ω m k+1 0 dz =0 ω m k+1 z since the sum of residues at all singular points including the infinitely remote one is equal to zero. If k = m, we obtain Γ 1 lim δ 0 ds = i dz 2πa m 1 z =1. a v K δ x 0 L Property 3.4. Let u k x C 2, and let u k a ku k a k 1 u k 1 = F k x, 3.7 h where F k C are the given functions. Then the following integral representation holds: u m x 0 = K R x 0 a 0 u 0 Γ m x x 0 dx K R x 0 u k Γ m k+1 ds + K R x 0 F k Γ m k+1 dx

8 Proof. The pair of functions ux,vx C 2 satisfies the Green identity [ u v ] v u u v dx = v u dx. Let us apply this identity to the pair of functions u k x, Γ m k+1 x x 0. Since the function Γ 1 x x 0 is not differentiable at the point x 0, we use the Green identity on the set K R x 0 \ K δ x 0 and then pass to the limit as δ 0. K R x 0 \K δ x 0 = [ Γm k+1 x x 0 u k x u k x Γ m k+1 x x 0 ] dx K R x 0 + Γ m k+1 x x 0 u kx K δ x 0 Γ m k+1 x x 0 u kx u k x Γ m k+1 x x 0 dx u k x Γ m k+1 x x 0 dx. 3.9 We now transform the left-hand side of equality 3.9, by using Property 3.1 and Eq. 3.7 and then pass to the limit as δ 0: lim δ 0 K R x 0 \K δ x 0 = lim lim [ uk xγ m k+1 x x 0 u k Γ m k+1 x x 0 ] dx δ 0 K R x 0 \K δ x 0 δ 0 K R x 0 \K δ x 0 + K R x 0 = = ak u k K R x 0 ak u k x a k 1 u k 1 x F k x Γ m k+1 dx u k x a k Γ m k+1 x x 0 Γ m k x x 0 K R x 0 F k Γ m k+1 x x 0 dx Γ m k x x 0 a k 1u k 1 Γ m k+1 x x 0 F k xγ m k+1 x x 0 dx + K R x 0 dx dx a 0 u 0 x Γ m x x 0 dx. Let us pass to the limit on the left-hand side of this identity with regard for 3.6. We obtain lim δ 0 K δ x 0 Γ m k+1 x x 0 u kx As a result, we obtain the integral representation 3.8. u k x Γ m k+1 x x 0 dx = u m x 0. 20

9 Property 3.5. The functions Γ k r Γ k 1 r, k =2,...,N, change the sign at most one time on the interval 0 <r<r. Proof. By r k,k 1, we denote the points at which the functions Γ k r Γ k 1 r,, 2...,n, vanish. Consider the function Γ 2 x x 0 Γ 1 x x 0. This function satisfies the following conditions: Γ 2 Γ 1 Γ 1 Γ 2 Γ 1 a n 1 = a n x :0< x x 0 <R, lim 0 Γ 2 Γ 1 = 1, x x 0 0 Γ 2 R Γ 1 R = Near the point x 0, the function Γ 2 x x 0 Γ 1 x x 0 is negative. We assume that this function changes the sign from minus to plus at the passage through the point r = r 2,1. Then, at the ends of the interval r 2,1,R, it vanishes. Inside this interval, it satisfies Eq whose right-hand side is negative. This implies that, inside the interval r 2,1,R, the function Γ 2 x x 0 Γ 1 x x 0 cannot take negative values. Let us consider the function Γ 3 x x 0 Γ 2 x x 0. It is a solution of the boundary-value problem Γ 3 Γ 2 Γ 2 Γ 1 Γ 3 Γ 2 a n 2 = a n 1 x :0< x x 0 <R, 3.11 Γ 3 0 Γ 2 0 = 0, Γ 3 R Γ 2 R =0. We will prove that, in a sufficiently small vicinity of the point r =0, the function Γ 3 r Γ 2 r is negative. We assume that there exists a point r 0 <r 2,1 at which the indicated function vanishes. Then relation 3.11 implies that it cannot have a positive maximum and, therefore, is negative on the interval 0,r 0. But if the function Γ 3 r Γ 2 r is positive in some vicinity of the point r = 0, then it cannot vanish on the interval 0,r 2,1, what was indicated above. On the interval r 2,1,R, the function cannot be zero as well, since it cannot have minimum. Hence, the function Γ 3 r Γ 2 r is positive on the whole interval 0,R. But this contradicts Property 3.2. We now assume that there exists a point r 3,2 such that, at its passage, the function Γ 3 r Γ 2 r changes the sign from minus to plus. Let now r 3,2 <r 2,1. On the interval r 3,2,r 2,1, the indicated function cannot vanish, since it would attain a positive maximum value on this interval, which contradicts the positiveness of the right-hand side of Eq By the analogous reasons, there are no changes of the sign on the interval r 2,1,R as well. But if r 3,2 >r 2,1, then the function Γ 3 r Γ 2 r > 0 on the interval r 3,2,R, because the right-hand side of Eq is negative on this interval. To complete the proof, it remains to apply the method of mathematical induction. Property 3.6. Let R = σ, 0 <σ< 1 2, N = T. Then the following inequality holds: Γ k R Γ k 1 R 0, k =2, 3,...,N Proof. We now show that it is sufficient to prove inequalities 3.12 at k = N. At k = m N, let the inequality Γ mr Γ m 1R 0 be valid. At k = m 1, we have Γ m 1R Γ m 2R >

10 Then, in some left half-neighborhood of the point r = R, the function Γ m 1 r Γ m 2 r < 0, since Γ m 1 R Γ m 2 R =0. On the other hand, by virtue of the previous property, this function is negative in the right halfneighborhood of the point r = 0 and can vanish not more than at one point. Therefore, Γ m 1 r Γ m 2 r < 0 r :0<r<R. On the indicated interval, the function Γ m 1 x x 0 Γ m 2 x x 0 satisfies the equation Γ m x x 0 Γ m 1 x x 0 [Γ m x x 0 Γ m 1 x x 0 ] a n m+1 Γ m 1 x x 0 Γ m 2 x x 0 = a n m whose right-hand side is positive. By virtue of the principle of maximum, this yields Γ m 1 x x 0 Γ m 2 x x 0 < 0 everywhere on the interval 0 < x x 0 <R. This contradicts inequality Then relation 3.2 yields Γ NR Γ N 1R = 2 1 a N a 1 2πi where Let We now calculate 22 L z 3 2 4πR sinhr z = 1 N a N 1a N 2 a 2 d N 1 z 3 z 2 N 1! N 2 dz N 1 4πR sinhr dz 1 z a N 1 z a N 1 1 z a 1 z= ξ a min <ξ<a max, c N = a N 1a N 2 a 2 4πRN 1! N 2, ϕ z= z 3 2 sinhr z =2z 3 exp R z 2 1 exp 2R z. = 1 N c N dn 1 ϕ z dz N 1 z= ξ F Rx = exp Rx 1 exp 2Rx = e 2k+1Rx = e Rx + e 3Rx e 2k 1Rx k=0 d N 1 N 1 ϕx dx N 1 =2 CN 1x k 3 k dn 1 k F x dx N 1 k k=0 =2x 3 dn 1 F x dx N 1 On the other hand, +6N 1x 2 dn 2 F x dx N 2 +6N 1N 2x N 3 F x x N 3 +2N 1N 2N 3 dn 4 F x dx N 4. d n F x dx n = 1 n R n e Rx + 1 n 3R n Oe 3Rx as 0, 3.15,

11 ξ since x = a min, R = σ, 0 <σ< 1 2, and therefore, Rx +, if 0. Thus, d N 1 ϕx dx N 1 = 1 N 1 2 RN 3 e Rx 1+e 2Rx [x3 R 3 3R 2 x 2 N 1 +3N 1N 2xR N 1N 2N 3] + 1 N 2 3R N 2 Oe 3Rx [x 3 3R 3 33R 2 x 2 N 1 +3N 1N 2x3R N 1N 2N 3]. The signs of the whole relation and the square bracket are determined by the first and last terms, respectively. Therefore, [ d N 1 ] ϕx sign =sign[ 1 N ] Relation 3.15 yields dx N 1 Γ NR Γ N 1R = 1 N 1 c N dn 1 ϕ z = 1 N 1 c N dz n 1 [ ϕ N 1 2 ξ + 1 N 1 c N ξ N 1 [ z= ξ N 1N 2 1! NN 1N 2N 3 2! ϕ N 2 ξ ] N ξ 2 ϕ N 3 ξ ] N+1 < 0. 2 ξ Property 3.7. Let R = σ, 0 <σ< 1 2, N = T,T > 0, 0 <a 0 a k a max c ε,wherec and a 0 are positive constants that are independent of ε. Then the following estimate holds: Γ N 1 M 1 q N R exp amax R Tπ 2 M 2 + exp M 3 2R a max KR x 0 Here, q 2, and the positive constants M 1,M 2, and M 3 are independent of N,,R,ε. Proof. Let us estimate the integral Γ N = i 2πa N KR x 0 L z 4πR sinh zr dz 1 z a 1 1 z a 2 1 z a N, 3.18 where L is the boundary of the domain: L = z : ϱ = z < 1 + q max 1 k N a k, Re z = b 0 > π2 R 2, b 0 < 0, ϱ > max 1 k N a k. We present the closed curve L as two ones, L 1 and L 2, where the curve L 1 is an arc of the circle whose equation is z = ρe iϕ, ϕ 0 ϕ ϕ 0, and π 2 ϕ 0, lim ϕ 0 = π

12 The curve L 2 is a segment of the straight line whose equation is Re z = b 0 = π2 2R 2, ρ sin ϕ 0 Im z ρ sin ϕ 0. By I 1 and I 2, we denote, respectively, integral 3.18 over the curve L 1 and the integral over the curve L 2. Let us estimate the integral I 1. Since 1 z 1 z 1 z z 1 N q N, a 1 a 2 a N a max sinh zr sinh[ z R cosarg z/2] = sinh[ z R cosϕ 0 /2], we have 1 I 1 c 1 q N R exp amax R c 2, where the constants c 1 and c 2 are independent of,ε. We now estimate the integral I 2. Since Re z = b 0 = π2,wehave 2R 2 1 z 1 z 1 z 1+ b T 0, a 1 a 2 a N a max z R sinh c 3. zr As a result, we obtain I 2 c 3 R 2 exp Tπ2 2R 2 c 4 exp c 5Tπ 2 a max 2R 2, a max where the constants c 3,c 4,c 5 are independent of,ε. 4. Uniform estimates of an approximate solution Let the conditions of Theorem 2.1 be satisfied. It follows from Eq. 2.3, u ε x, t a ε u ε x, t λχ εu ε x, t uε x, t =0 in T, and the principle of maximum that the solution u ε x, t takes the maximum and minimum values on the parabolic boundary, i.e., either on i [0,T, i=1, 2, oratt =0. Therefore, we have max u ε x, t max T max ψx, max ϕx, t T Let x 0 be any point of the domain. We transform Eq. 2.7 to the form u ε k x βk ε x 0 uε k x = [βk ε x βε k x 0] uε k x where k =2, 3,...,N. If k =1, then the equation takes the form u ε 1 x βε 1 x 0 u ε 1 x + 1 uε 0x = βε 1 x βε 1 x fε k x, u ε 1 x.

13 Theorem 4.1. Let ψx H r+α, ϕx, t H r+α,r+α/2 T, 0 <α<1, r 3, min ϕx, t > 1, 2 [0,T ] σ 1/3 ε 2 M 2 ε, 1/3 <σ<1/2, 4.2 and let the corresponding conditions of matching be satisfied at t =0, x. Then there exists a positive constant c independent of ε and such that the estimate u ε x, t ϕx, t max c max max ψx, max 4.3 x T x 2 [0,T ] is true. Proof. Let K R x 0 be a ball of radius R = σ centered at the point x 0, and m = T 1 T. We will use the integral representation 3.8, where we take β ε k x 0asa k, u ε mx 0 = + K R x 0 u ε 0x Γ m x x 0 K R x 0 K R x 0 u ε k K R x 0 dx Γ m k+1 ds K R x 0 fk εx Γ m k+1 x x 0 dx [βk ε x βε k x 0] uε k x Γ m k+1 x x 0 dx [βk 1 ε x βε k 1 x 0] uε k 1 x Γ m k+1 x x 0 dx = I 1 + I 2 + I 3 + I 4 + I 5. Let us estimate each of the integrals. It follows from 3.5 that Γ 1 x x 0 I 1 max ψ dx max ψ βmx ε max ψ 0 mina 1,a 2 max ψ. a 0 K R x 0 Let us estimate I 2. In view of the well-known inequality x µ e x µ µ e µ, x 0, +, µ>0, Property 3.7, estimate 4.2, and the estimate a max c ε, we obtain I 2 T 1c 1 1 Mε q N R exp R T c 2 1 π 2 + exp M 3 R 2 a max = T 1c 1 1 Mε q N σ exp σ εt c 2 1 π 2 + exp c 3 2σ where the constant c is independent of and ε. Relations 2.6 and 2.11 yield fε k x c Mε. c Mε, 25

14 Therefore, we have I 3 K R x 0 fε k x Γ n k+1 x x 0 dx c Mε, where the constant c is independent of ε and. Relations 2.6 and 2.10 yield Therefore, [β ε k x βε k x 0] uε k x I 4 + I 5 K R x 0 [β ε k 1 x βε k 1 x 0] uε k 1 x c ε 2 M 2 ε x x 0 Γ m k+1 x x 0 dx c x x 0 ε 2 M 2 ε. c σ ε 2 M 2 ε <c1/3. Thus, with regard for 4.2, we obtain uε mx 0 c 1 max u ε 0x, 4.4 x where the constant c 1 is independent of and ε. We now show that ω ε m = x :1<u ε mx < 1+ε R = x :distx, >R. Let x 1 R, x 2 2 and distx 1,x 2 =R = σ. Relation 4.2 yields u ε mx 2 u ε mx 1 max u ε mx R c Mε σ cε 2 Mε cε. Since min 2 [0,T ] ϕx, t > 1, we can consider, without any loss of generality, that, at sufficiently small ε, min 2 [0,T ] ϕx, t 1+ε + cε. Whence, we obtain u ε k x 1 ϕx 2,k cε 1+ε. Analogously, if x 1 R, x 2 1 and distx 1,x 2 =R, we have u ε mx 1 max u ε mx R c Mε σ cε < 1. Thus, in the domain \ R = x :distx, <R, Eq. 2.3 is a differential equation with constant coefficients. Therefore, the principle of maximum yields max uε mx max uε mx ϕx, t, max, max u 0 x [0,T ] x \ R, 1 m N max x, 1 m N max x R, 1 m N As a result, relations 4.4 and 4.5 yield uε mx c max 26 max u 0 x, max 2 [0,T ] ϕx, t, 4.6

15 where the constant c is independent of and ε. Let x, t be any point from T, t [m 1,m] at some m, where 1 m N. Then u ε x, t uε mx = 1 m m 1 u ε x, t This yields uε x, t uε mx + 2 max 2 u ε x, t x,t T 2 ϕx, t c max max u 0 x, max 2 [0,T ] c 1 max Estimate 4.3 is proved. We denote uε x, η dη. η + c Mε max u 0 x, max 2 [0,T ] ϕx, t. ω ε x, t =x, t T :1<u ε x, t < 1+ε, R T = R 0,T, ω x, ε t =x, t T : u ε x, t < 1, ω+x, ε t =x, t T : u ε x, t > 1+ε. Theorem 4.2. Let the conditions of Theorem 4.1 be satisfied. Then x, t, y, t T ω ε ± x, t, and there exists a constant c independent of ε, t, x, y and such that the estimate uε x, t uε y, t c x y α u 0 x C 2+α, ϕx, t H 2+α,2+α/2 T, 4.7 where 0 <α<1, is satisfied. Proof. Equation 2.3 in T \ ω ε x, t =ω x, ε t ω+x, ε t takes the form u ε x, t =a ε u ε x, t uε x, t. Let x 0,y 0 be any points from ω ε x, m R or ω+ ε x, m R. In each of the indicated sets, a ε u ε x, t = const. It follows from the above assumption that βmx ε 0 =βmy ε 0. For definiteness, we assume that K R x 0 ω+x, ε m, K R y 0 ω+x, ε m. Using 3.8, we construct the integral representation for uε mx 0 uε my 0 and estimate the differences of one-type terms. Among these differences of terms, the first difference is principal, since the rest ones tend to zero as 0, as follows from the proof of Theorem 4.1. Thus, I 1 = u ε 0x Γ m x x 0 dx u ε 0x Γ m x y 0 dx. K R x 0 K R x 0 K R y 0 In the second term, we change the integration variable, x = ω + y 0 x 0, and obtain I 1 = u ε 0x Γ m x x 0 dx u ε 0ω + y 0 x 0 Γ m ω x 0 dω. K R x 0 27

16 This yields I 1 max u 0 x u 0 x + y 0 x 0 x K R x 0 Γ 1 x x 0 c max u 0 x u 0 x + y 0 x 0 c x 0 y 0 α u 0 x C x 2+α. We estimate the remaining terms, as in the proof of Theorem 4.1. As a result, we obtain uε mx 0 uε my 0 dx c x 0 y 0 α u 0 x C 2+α + c σ 1 x 0 y 0 ε 2 M 2 ε c x 0 y 0 α u 0 x C 2+α + c 2 1/3 x 0 y Let now x 0,tandy 0,t be any points from ω ε x, t R T or ωε + x, t R T, and t [m 1,m] at some m. Using estimate 4.8, we obtain u ε x 0,t uε y 0,t u ε m x 0 = 1 m u ε x 0,t This yields uε x 0,t m 1 uε y 0,t uε my 0 uε y 0,t u ε x 0,η η = 1 m m 1 dη uε y 0,η dη η x 0 y 0 2 u ε ξ,t ξ 2 u ε ξ,η dξ. η ξ uε mx 0 uε my u ε x, t max 2 x,t T x 2 x 0 y 0 t c 1 x 0 y 0 α u 0 x C 2+α + x c 2 0 y 0 Mε c 3 x 0 y 0 α u 0 x C 2+α, 4.9 where the constants c 1,c 2,c 3 are independent of and ε. Equation 2.3 in ω ε x, t ω ε +x, t is a linear equation with constant coefficients. Since estimate 4.3 is proved, the boundary T C 2+α, and the function ϕx, t H r+α,r+α/2 T, estimate 4.8 is satisfied up to the boundary of the domain T. Theorem 4.3. Let the conditions of Theorem 4.1 be satisfied, and let σ 1/3 <ε 2 M 3 1 ε, 3 <σ<1 2, min ψx = c 0 > Then there exist positive constants c 1 and c 2 independent of ε and such that the estimates max u ε x, t c 1 max max u ε 0x, max ϕx, t, 4.11 T \ω ε x,0 ω ε x,t x T min u ε x, t c 2 > T R\ωε x,t 28

17 Proof. are valid. Let us differentiate both sides of Eq. 2.7 with respect to one of the variables x i. We obtain u ε k x b εu ε uε k k x x b ε u ε k 1 x uε k 1 x = fε k x. Let a point x 0, and let K R x 0. We denote wk εx = uε k x. The equation can be transformed to the following form: w ε k x b εu ε k x 0w ε k x b εu ε k 1 x 0w ε k 1 x + b εu ε k x b εu ε k x 0 = fε k x wk ε x b εu ε k x b εu ε k 1 x 0 wk 1 ε x. Using Property 3.4 of the fundamental solutions, we construct the integral representation w ε mx at the point x 0 : w ε mx 0 = K Rx0 + b ε u 0 x 0 u 0x Γ m x x 0 dx K R x 0 K Rx0 K Rx0 w ε k x Γ m k+1 ds K R x 0 fk εx Γ m k+1 dx b ε u ε k x b εu ε k x 0 wk ε xγ m k+1 x x 0 dx b ε u ε k 1 x b εu ε k 1 x 0 wk 1 ε xγ m k+1 x x 0 dx = I 1 + I 2 + I 3 + I 4 + I Let us estimate I 1. Property 3.2 yields I 1 b ε u 0 x 0 max u 0x K Rx0 Γ 1 x x 0 dx cb ε u 0 x 0 max u 0 x. We now estimate I 2 as the analogous integral in the proof of Theorem 4.1: I 2 Tc 1 1 Mε q N R exp R c 2 + exp Tπ2 R 2 a max Tc 1 1 Mε q N σ exp σ c 2 + exp where the constant c is independent of and ε. We also have I 3 n K Rx0 max fε k x Γ 1 x x 0 dx x Tπ2 2σ a max c Mεa 0, c Mε, 29

18 where the constant c is independent of ε,, and n. Since b εu ε k x b εu ε k x 0 wk ε x b εu ε k 1 x b εu ε k 1 x 0 wk 1 ε x x x 0 c ε 2 M 2 ε c σ σ ε 2 M 3 ε, we obtain, by using Property 3.2: I 4 + I 5 c σ ε 2 M 3 ε, where the constant c is independent of ε,, and m. In view of 4.10, we obtain This yields the following estimate: uε mx 0 where the constant c is independent of ε,, and x 0. If then the previous estimate yields uε mx 0 I 2 + I 3 + I 4 + I 5 c 1/ = w ε m x 0 b ε u 0 x 0 max x 0 x :1<u 0 x < 1+ε, = w ε m x 0 c 3 max u 0x 1 + c 1/3, a 0 u 0 x + c 1/3 c 4 max u 0 x. Let x, t T R, and let t [m 1,m] at some integer 1 m N. With regard for 4.10, we obtain uε x, t = u ε mx t + m 2 u ε x, η dη u ε mx + t m max 2 u ε x, η x,t T c 4 max where the constant c is independent of ε, x, and t. In the domain \ R = x :distx, <R, Eq. 2.3 u ε x, t =a ε u ε x, t uε x, t uε mx + c Mε u 0 x + c 2 1/3 c max u 0 x, can be considered at every t 0,T as a Poisson equation with bounded right-hand side. Since the boundary C 2+α and the function ϕx, t H r+α,r+α/2 T, estimate 4.11 is satisfied up to the boundary of the domain T. By the condition of the theorem, min ψx = c 0 > 0. Therefore, at any point of, at least one of the partial derivatives of the function ψx satisfies the inequality ψx = u 0x c 0 /3 > 0. For definiteness, we assume that u 0x 0 c 0 /3. Then 30

19 I 1 b ε u 0 x 0 u 0x 0 We note that Whence, we have c 0 3a 0 K Rx0 K Rx0 b ε u 0 x 0 a 0 c 0 1 3b ε u m x 0 Γ 1 x x 0 K Rx0 Γ 1 x x 0 dx u 0x 0 u 0x Γ 1 x x 0 dx dx b ε u 0 x 0 c 1 b R εu mx 0 sinh R b εu mx 0 K Rx0 x x 0 Γ 1 x x 0 σ b ε u 0 x 0 1 b ε u m x 0 dx b R εu mx 0 sinh R R = σ, 0 <a 0 b ε u m x 0 c x, x 0, + ε sinh x 3! x 2. b R εu mx 0 sinh R b εu mx 0 3! 1 2σ a 0 0 if 0. Therefore, relation 4.14 yields uε mx 0 c b ε u m x 0 min ψx I 2 + I 3 + I 4 + I 5 Hence, for any point x R \ ω ε m, the inequality c b ε u m x 0 min b εu mx 0. ψx c 1/ uε mx 0 c1 min ψx, where the constant c 1 is independent of ε, x 0,,and m, is satisfied. As in the first part of the theorem, this estimate, which is valid on the cross-section t = m, yields estimate 4.12 in T R\ωε x, t. Corollary 4.1. Let the conditions of Theorem 4.3 be satisfied. Then there exists a positive constant c independent of ε and such that the estimate u ε x, t W 1,1 2 T c, 4.16 where the constant c is independent of ε, is satisfied. Proof. Let us multiply Eq. 2.3 by u ε tx, t and integrate by parts in the domain t = 0,t. We obtain 0= b ε u ε x, tu ε tx, t ux, t ε x, t]u ε tx, t dx dt t = t b ε u ε x, t[u ε tx, t] 2 3 [u ε x x, tu ε tx, t] + 1 i 2 [uε x i x, t] 2 dx dt. i=1 31

20 This yields 1 2 u ε x, t 2 dx + t b ε u ε x, t[u ε tx, t] 2 dx dt = 1 2 t u ε xx, 0 2 dx + 0 u ε x, t u ε tx, t ds dt. It follows from 4.3 and 4.11 that all terms on the right-hand side are bounded, and b ε u ε x, t a 0. Thus, estimate 4.16 is proved. Theorem 4.4. Let the conditions of Theorem 4.1 be satisfied, and let σ 1/3 <ε 4 M 4 ε Then there exists a positive constant c independent of ε and such that the estimate 2 u ε x, t ψx c max C α ω± ε C x,t 2+α, ϕx, t H 2+α,2+α/2 T 4.18 holds. Proof. We differentiate both sides of Eq. 2.7 twice with respect to one of the variables x i. As a result, we obtain 2 u ε k x β ε u ε k x 2 u ε k x β ε u ε k 1 x 2 u ε k 1 x = 2 f ε k x + β εu ε uε k x k x 2 β ε u ε k 1 x u ε k 1 x 2. Let a point x 0, K R x 0. We transform the obtained equation to the following form: 2 u ε k x β ε u ε k x 0 2 u ε k x + 1 β ε u ε k 1 x 0 2 u ε k 1 x = 2 fk εx + β εu ε uε k x k x [β ε u ε k x β εu ε k x 0] 2 u ε k x 2 β ε u ε k 1 x u ε k 1 x 2 [β ε u ε k 1 x β εu ε k 1 x 0] 2 u ε k 1 x Let a point x 0, K R x 0. enotew m x = 2 u ε m x. Using Property 3.4 of the fundamental solutions, we construct the integral representation w m x at the point x 0 : 32 w ε mx 0 = K Rx0 b ε u 0 x 0 2 u 0 x Γ m x x 0 x 2 dx i.

21 + K R x 0 K Rx0 K Rx0 K Rx0 w ε k x Γ m k+1 ds K R x 0 2 fk εx x 2 Γ m k+1 dx i β ε u ε k x β εu ε k x 0 wk ε xγ m k+1 x x 0 dx β ε u ε k 1 x β εu ε k 1 x 0 wk 1 ε xγ m k+1 x x 0 dx b εu ε k x u ε k x 2 b εu ε k 1 x uε k 1 x 2 Γ m k+1 x x 0 dx = I 1 + I 2 + I 3 + I 4 + I 5 + I Let us estimate all integrals in the integral representation We note that the estimates of the integrals I k,, 2, 3, 4, 5, can be carried out as for the integral representation 4.14 see the proof of Theorem 4.3. Therefore, it is sufficient to estimate only I 6.Let u Fk ε x =b εu ε ε k x k x 2. We transform I 6 as follow: I 6 = K Rx0 [ F ε k x F ε k x 0 We note that relation 2.6 yields Fk εx F k εx 0 F k 1 ε x F k 1 ε x 0 ] Γ m k+1 x x 0 dx K Rx0 Fk εx 0 Fk 1 ε x 0 Γ m k+1 x x 0 dx = I6 1 + I 2 6. F ε k 1 x F ε k 1 x 0 c x x 0 1 ε 4 M 4 ε, where the constant c 1 is independent of k,, and ε. We will estimate each term separately. We have I 1 6 c 1 σ ε 3 M 2 ε K Rx0 The second integral should be firstly transformed: I 2 6 = K Rx0 Fk ε x 0 m 1 + Γ m k+1 x x 0 dx Fk εx 0 Fk 1 ε x 0 Γ m k+1 x x 0 dx = F ε mx 0 K Rx0 Γ m k+1 x x 0 Γ m k x x 0 c 1 σ a 0 ε 4 M 4 ε. K Rx0 dx F ε 0 x 0 Γ 1 x x 0 K Rx0 dx Γ m x x 0 dx 33

22 + F ε mx 0 Fk ε x 0 m 1 = K Rx0 = m 1 K Rx0 Γ 1 x x 0 F ε k x 0 β ε k x 0 + F ε mx 0 Let us now apply Property 3.2. We obtain K Rx0 K Rx0 Γ m k+1 x x 0 β ε k x 0 dx F ε 0 x 0 K Rx0 Γ m k+1 x x 0 Γ 1 x x 0 dx Γ m x x 0 ds dx F ε 0 x 0 I6 2 c Tc 1 1 ε 2 M 2 ε q N σ exp σ c 2 + exp Tπ2 2σ a max u + c 2 b εu ε ε k x 2 mx + b u x ε u ε ε 0 x 2 0x i c 3 u ε 2 M 2 ε + c 2 b εu ε ε k x mx where the constants c 2 and c 3 are independent of m,, and ε. As in the proof of Theorem 4.3, we obtain firstly the estimate dx K Rx0 2 + b ε u ε 0x Γ m x x 0 dx. u ε 0 x 2, 2 u ε mx 0 cb ε u 0 x 0 max 2 u 0 x 0 u + c 2 b εu ε ε k x 2 mx + b u x ε u ε ε 0 x 2 0x. i If x, t ω ε x, t, then the obtained estimate takes the form 2 u ε mx 0 cb ε u 0 x 0 max 2 u 0 x 0 u + c 2 b εu ε ε 0 x 2, 0x where the constants c and c 2 are independent of ε, x 0,m, and. u ε mx =1andu ε mx =1+ε, which are the boundary of the set We note that, on the surfaces x :1<u ε mx < 1+ε, this estimate takes the form 2 u ε mx 0 c max 2 u 0 x 0 + In view of relations 2.6 and 4.18, this estimate yields u ε 0 x 2. 2 u ε x, t cb ε u 0 x max 2 u 0 x u + b εu ε ε 0 x 2. 0x As in the proof of Theorem 4.1, we obtain the uniform estimate of the Hölder constant for the function 2 u ε x,t. 34

23 Equation 2.3 in the domains ω ε ±x, t at every fixed t 0,T, u ε x, t =a ε u ε x, t uε x, t, is the Poisson equation with the right-hand side which belongs to the space C α T \ω ε x, t. Moreover, the boundary T \ ω ε x, t H 2+α,2+α/2 T, and the function ϕx, t H r+α,r+α/2 T. Therefore, it follows from the well-known results of the theory of elliptic boundary-value problems that u ε x, t C 2+αω ψx ± ε x,t c max C 2+α, ϕx, t H 2+α,2+α/2 T, where the constant c is independent of ε and t. Theorem 4.5. Let be a bounded domain in R 3 whose boundary consists of two C 2+α surfaces 1 and 2 such that 1 lies inside of 2, T = 0,T, ψx C 2+α, ϕx, t H 2+α,2+α/2 T, min ψx > 0, min ϕx, t > 1, 2 [0,T ] σ 1/3 <ε 4 M 4 ε, and let the corresponding conditions of matching at t =0, x be satisfied. Then the function u ε x, t, which is a solution of problem , satisfies estimates 4.3, 4.7, 4.11, 4.12, 4.16, and4.18. Proof. The uniform estimates 4.3, 4.7, 4.11, 4.12, 4.16, and 4.19 were obtained under the assumption that ψx C r+2, ϕx, t H r+α,r+α/2 T, where r 3. However, the righthand sides of these estimates depend only on the norms ψx C 2+α, ϕx, t H 2+α,2+α/2 T. Therefore, we can get rid of the superfluous requirements of smoothness for the functions ψx and ϕx, t in all theorems of this section. To this end, we will approximate the functions ψx C α+2 and ϕx, t H 2+α,2+α/2 T by sequences of the functions ψ k x C r+2 andϕ k x, t H r+α,r+α/2 T, prove the indicated uniform estimates for the sequence of the functions u ε k x, t, and then make the limiting transition in these estimates. 5. Limiting transition Let the conditions of Theorem 4.5 be satisfied. It follows from estimates 4.3 and 4.11 that u ε x,t is uniformly bounded everywhere in T,and u ε x, t is bounded everywhere in T except for the set ω ε x, 0 ω ε x, t whose measure tends to zero, if ε 0. Therefore, from the family of functions u ε x, t, we can separate a subsequence which converges almost everywhere in T.Let ux, t = lim ε 0 u ε x, t. In any closed subdomain of the domain \ γ 0, the families of functions uε x,t, u ε x, t are uniformly bounded, and the sequence u ε x, t converges uniformly. The limiting function ux, t is uniformly continuous in the domain \ γ 0. Therefore, the function ux, t can be continued to γ 0 and in the unique way so that the obtained function is continuous. Let Ω T = x, t T :0<ux, t < 1, G T = x, t T : ux, t > 1, 35

24 γ T = x, t T : ux, t =1, γ + T = G T T, γ T =Ω T T. We now consider identity 2.2, u ε x, t ηx, t+a ε u ε uε x, t ηx, t+λχ ε u ε η dx dt + λ T χ ε ψηx, 0 dx =0. Relation 4.16 yields u ε x, t W 1,1 2 T c. Therefore, the sequence u ε x, t weakly converges in W 1,1 2 T and strongly converges in L 2 T. enote A k = x, t T : ux, t >k, A k = x, t T : ux, t =k. Byux, t, we construct the function u k x, t = maxux, t k, 0. As known [17], if u ε x, t converge in W 1,1 2 T to ux, t, then u ε 1,1 kx, t converge in W2 T to u k x, t. In addition, mes[a k \ A ε k A k] 0, mes[a ε k \ Aε k A k A k ] 0 as ε 0. With regard for the above consideration, we make the limiting transition in identity 2.2: ux, t ηx, t dx dt + a 1 a 2 T Ω T + T a 2 ux, t ηx, t dx dt ux, t ηx, t dx dt + λ + λ where 1 T = x, t T : ux, t =1. Let 0 <δ<1 be any positive number, and let Then, for sufficiently small ε, we have 1 T Ω T η dx dt Ω δ T = x, t T :0<ux, t < 1 δ. Ω δ T ωε x, t =x, t T :0<u ε x, t < 1. lim χ εu ε η dx dt + λ ηx, 0 dx =0, 5.1 ε 0 Ω 0 Therefore, in Ω δ T and, hence, in any closed subdomain of the domain Ω T, the family of functions u ε x, t satisfies all assertions of Theorem 4.5. The analogous result is true for the domain G T.This implies that the limiting function ux, t satisfies the estimates 2 u ψx c max C α Ω t C T Gt T 2+α, ϕx, t H 2+α,2+α/2 T, 5.2 max 0<t<T where Ω t T,Gt T stands for the cross-sections of the domains Ω T,G T by the plane t =const, max ux, t c max max ψx, max ϕx, t, T x T min T \ 1 T ux, t c 1 > 0,

25 ux, t ϕx, t max c max max ψx, max, 5.4 T x T ux, t max ψx 0<t<T c max C α T t C 2+α, ϕx, t H 2+α,2+α/2 T, 5.5 where c and c 1 are some positive constants. With regard for these estimates, the function ux, t and its derivatives can be uniquely continued on γ ± T so that the continued function and the corresponding derivatives are continuous in Ω T γ T and G T γ + T. Every point of the surfaces γ± T possesses a vicinity in which one of the partial derivatives ux,t is nonzero, e.g., ux,t x 1. Then, the equation of the surface can be presented in every vicinity in the form x 1 = sx 2,x 3,t. Moreover, this function is differentiable with respect to t, and, at every fixed t, we have sx 2,x 3,t C 2+α. The integral identity yields ux, t a 1 u =0in Ω T, λ 1 T ux, t a 2 u =0in G T, lim χ εu ε η dx dt =0. ε 0 Since ηx, t is an arbitrary function equal to zero on 1 2 0,T, the last equality yields mes x, t 1 T : lim χ εu ε > 0 =0. ε 0 Then identity 5.1 takes the form ux, t ηx, t dx dt + T Whence, we obtain + Ω T a 1 T \Ω T a 2 ux, t ηx, t dx dt ux, t ηx, t dx dt + λ Ω T η dx dt + λ ηx, 0 dx = Ω 0 ux, t a 1 u =0в Ω T, ux, t a 2 u =0в \ Ω T = G T, γ + = γ = Ω T T = G T T = γ T. On the free boundary γ T, the following conditions are satisfied: 3 i=1 u x, t =u + x, t =1, u u+ cosn, x i +λ cosn, t = Here, u x, t and u + x, t mean the boundary values of the function ux, t on γ T taken, respectively, from the sides of the domains Ω T and G T,andnisanormaltothesurfaceΩ T,G T which is directed to the side where the function ux, t increases. We now prove that the derivatives ux,t satisfy the Hölder condition in t with exponent 1/2. The domains Ω T and G T satisfy the cone condition. In other words, there exists a fixed globular cone of 37

26 height d such that if its vertex is positioned at any point of Ω T or G T, then the cone itself can be turned so that it will be completely, as the whole, in Ω T or G T. For arbitrary points x and x from Ω T, the following relation holds: I = x x [ ux, t ux,t ] t dl l l = ux,t ux,t ux,t +ux 1,t. 5.8 t Here, l means the differentiation along the segment [x,x ]. By virtue of 5.5, the right-hand side of 5.8 does not exceed c t t. Estimates 5.2 and 5.4 yield In addition, we have x x I c 1 t t, [ ux, t ux,t ] dl l l = x x [ ux, t l x x [ ux, t ux,t ] dl l l c 2 x 1 x ux,t ] dl + l x x [ ux,t ux,t ] dl l l = I 1 + I 2 I 2 I 1, 5.10 I 2 = x x ux,t ux,t l l We now obtain the upper bound for I 1. At the passage through the surface γ T in the cross-section t = t along the segment [x x ], the function ux,t l has a discontinuity of the jump type. Let us divide the integration interval into two ones: [x,c 0 ] and [c 0,x ], where c 0 is the discontinuity point of the function ux,t l. Therefore, in view of 5.2, we obtain c 0 [ ux, t I 2 ux,t ] x [ dl l l + ux, t ux,t ] dl l l c x x 2, 5.12 x where c is a positive constant depending on 2 u x 2 C α Ω t i T Gt T. Relations yield x x Whence, we obtain 38 ux,t l [ ux, t ux,t ] dl l l x x ux,t l ux,t l 1 x x c 0 x x ux,t l c x x 2. [ ux, t ux,t ] dl l l + c x x. 5.13

27 With regard for 5.8, we now estimate the integral on the right-hand side of inequality 5.13: x x [ ux, t ux,t ] dl l l x x In view of 5.12, we obtain [ ux, t ux, ] t dl l l + ux,t l ux,t l x x [ ux, t ux,t ] dl l l t t µ 1 x x + µ 2 x x. c 1 t t + c 2 x x 2. Taking x as the cone vertex, we minimize the right-hand side with respect to x along a segment of any ray l with length d which belongs to the cone. As a result, we obtain ux,t ux,t µ2 µ1 + t t 1/2 c t t α/2, l l µ 1 α 0, 1. Since the ray l can be chosen to be arbitrarily directed inside of the cone, we obtain the required assertion ux,t ux,t c t t α/ Let x 0,t 0 γ T, and let one of the partial derivatives ux,t be nonzero, e.g., ux0,t 0 x 3 0. Then, by virtue of the implicit function theorem, there exists a vicinity = x 1 x 0 1 <b 1, x 2 x 0 2 <b 1, x 3 x 0 3 <b 3, t t 0 <δ, that cuts a part of the surface described by the function µ 2 x 3 = sx 1,x 2,t, x 1,x 2,t = x 1 x 0 1 <b 1, x 2 x 0 2 <b 1, t t 0 <δ from γ T. Relations 5.7 and 5.14 yield sx 1,x 2,t H 2+α,1+α/2. Thus, Theorem 1.1 is proved. REFERENCES 1. I. Athanasopoulos, L. Caffarelli, and S. Salsa, Regularity of the free boundary in parabolic phase transition problems, Acta Math., 1762, I. Athanasopoulos, L. A. Caffarelli, S. Salsa, Stefan-like problems with curvature, J. Geom. Anal., 13, No. 1, B. V. Bazalii, Stefan problem, okl. AN UkrSSR. Ser. A, No. 1, M. A. Borodin, On the solvability of the two-phase nonstationary Stefan problem, okl. AN SSSR, 263, No. 5,

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