1. Statement of the problem.

Transcription

1 218 Нелинейные граничные задачи 18, (2008 c M. A. Borodin THE STEFAN PROBLEM The Stefan problem in its classical statement is a mathematical model of the process of propagation of heat in a medium with different phase states, e.g., in a medium with liquid and solid phases.the process of propagation of heat in each phase is described by the parabolic equations. In this work we prove the existence of the global classical solution in a two-phase multidimensional Stefan problem. We apply a method which consists of the following. First, we construct approximating problems, then we prove some uniform estimates and pass to the limit. Keywords and phrases: free boundary problem, global classical solution, the Stefan problem MSC (2000: 35R35 1. Statement of the problem. The Stefan problem in its classical statement is a mathematical model of the process of propagation of heat in a medium with different phase states, e.g., in a medium with liquid and solid phases. As a result of melting or crystallization, the domains occupied by the liquid and solid phases undergo certain changes. This unknown interface is called a free boundary. The process of propagation of heat in each phase is described by the heat equation. Let D R 3 - the bounded domain, which boundary consists of two C 2+α surfaces D 1 and D 2 such, that D 1 contains inside D 2 and limits domain D, D T = D (0, T. The problem is to find a function u(x, t and domains Ω T, G T, satisfying to following conditions: u a(u u = 0 in Ω T G T, (1.1 Ω T ={(x, t D T : 0 < u(x, t < 1}, G T ={(x, t D T : u(x, t > 1}, where a(u - step function a 1 > 0 в Ω T and a 2 > 0 in G T ; on the known boundary D T u(x, t = 0 on D 1 (0, T, u(x, t = ϕ(x, t > 1 on D 2 (0, T ; (1.2

2 The Stefan problem 219 on the unknown (free boundary γ T = D T Ω T = D T G T u (x, t = u + (x, t = 1, 3 i=1 ( u x i u+ x i cos(n, x i +λ cos(n, t = 0; (1.3 where λ is a positive constant, n is the normal to the surface γ T directed to the side of increase of the function u(x, t; u + (x, t, u (x, t are the boundary values on the surface γ T taken from the domains G T, Ω T respectively. The function u(x, t is interpreted as the temperature of the medium, γ t is the interface between the liquid and solid phases, and u(x, t = 1 is the temperature of melting. The initial conditions are u(x, 0 = ψ(x > 0 in D, ψ(x = 0, D1 ψ(x = ϕ(x, 0 > 1, D2 (1.4 Ω 0 = {x D : 0 < ψ(x < 1}, G 0 = {x D : ψ(x > 1}, γ 0 = D Ω 0. 2.Construction of the approximating problem. Assume that problem has a classical solution. Multiply the equation (1.1 by a smooth function η(x, t which vanishes on D T and integrate by parts: D T ( u(x, t η(x, t + a(u u +λ χ(ψη(x, 0dx = 0, D η η(x, t + λχ(u dxdt+ (2.1 where χ(u = { 1, if u(x, t < 1, 0, if u(x, t > 1. For ε > 0 we introduce a function χ ε ( C (R 1 : χ ε ( = 1 1 ε, χ ε ( = 0 1, χ ε( 0, χ (n ε (x c ε, n Let a ε (x = a 1 + χ ε (x(a 2 a 1. Define the function {u ε (x, t} as solutions of the following problem:

3 220 M. A. Borodin u ε (x, t a ε (u ε (x, t uε (x, t = λ χ ε(u ε (x, t in D T (2.2 u ε (x, t = 0 on D 1 [0, T, u ε (x, t = ϕ(x, t on D 2 [0, T ; (2.3 u ε (x, 0 = ψ(x in D, ψ(x = 0 on D 1, ψ(x = ϕ(x, 0 > 1 on D 2. (2.4 Then, as is well known [1], takes place the statement Theorem 2.1. Let ψ(x C l+α l+α l+α, (D, ϕ(x, t H 2 (DT, 0 < α < 1, and assume that the corresponding consistency conditions hold at t = 0, x D. Then this problem is solvable and u ε (x, t c H l+α, l+α 2 (D T M(ε, (2.5 where positive constant c do not depend on ε,m(ε 0, if ε 0. The equation (2.2 we shall transform to a form u ε (x, t u ε (x,t [a ε ( λχ ε(]d = 0. 0 We divide the cylinder D T by the planes t = k, k = 1, 2,...N, N = T, integrate equation (2.2 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, where u ε k (x uε (x, k. After simple transformations we obtain где u ε k(x βk(x ε uε k (x = f ε k(x, (2.6 u ε k (0 = 0, uε k (l = ϕ(k = ϕ k, u ε 0 (x = ψ(x. (2.7 u ε k (x = uε k (x uε k 1 (x,

4 The Stefan problem 221 β ε k(x = 1 {k ε [u ε k 1 + (u ε k u ε k 1] λχ ε[u ε k 1 + (u ε k u ε k 1]}d, 0 f ε k (x = 1 k (k 1 [ u ε (x, k u ε (x, t]dt. 3. The fundamental solutions and its properties. For studying of the problem (2.4, (2.5 we need in the integral representation of the solution. Let K R (x 0 be ball with center at the point x 0 and the radius R and Γ n k+1 ( x x 0 = i (sinh zr 1 sinh z(r x x 0 dz 2πa n 4π x x 0 (1 z a n (1 where L L = {z = ξ + iη : Rez > π2, z < ϱ}, R2 (, 0 a2,, 0,... ( a1 Property 1. Let x x 0 0, then Γ n k+1 a k Γ n k+1 Γ n k z a n 1 (1 z a k, ( an, 0 L, a i > 0, i = 1, 2,...n. = 0, k = 1...(n 1, (3.1 Γ 1 Γ 1 a n = 0, Γ 1( x x 0 = sinh an (R x x 0 4π x x 0 sinh a n. (3.2 Property 2. There is the estimation K R (x 0 K R (x 0 Γ n ( x x 0 dx K R (x 0 Γ n 1 ( x x 0 dx... Γ 1 ( x x 0 dx 1 ( an 1 R a n sinh a n 1. (3.3 R Property 3. Let K δ (x 0 denote the ball with its center at the point x 0 and radius δ. Then { Γ m k+1 1, if k = m, lim ds = δ 0 n 0, if k m, K δ (x 0

5 222 M. A. Borodin where n is the inner normal. All reduced above property can be received by immediate calculations. Property 4. Let {v k (x C 2 (D} and they satisfy the following equations: v k a kv k a k 1 v k 1 = f k f k 1, then there is an integral representation v m (x 0 = K R (x 0 (a 0 v 0 f 0 Γ m ( x x 0 + m k=1 K R (x 0 dx m k=1 K R (x 0 Γ m k+1 v k ds+ n Γ m k+1 Γ m k f k dx. (3.4 This integral representation follows from the previous properties of the fundamental solutions and Green s for elliptic equations. Property 5. The functions {Γ m k+1 ( x x 0 Γ m k ( x x 0 } change the sign on the interval 0 < x x 0 < R no more than once and Γ k 1 (r r Γ k (r r=r r. (3.5 r=r This property follows from property1 and from principle of the maximum. Property 6. We will denote by r k,k 1 the points, where the functions Γ k (r Γ k 1 (r, k = 1, 2...n, are equal to zero. Then we have the inequality r k,k 1 r k+1,k, and r 2,1 = ln a n 1 ln a n an 1 a n + o( 0, if a n 1 a n, (3.6 r 2,1 = 2 a n, if a n 1 = a n. Proof. The functions Γ k (r Γ k 1 (r satisfy the equation (Γ k+1 ( x x 0 Γ k ( x x 0 a m k Γ k+1 ( x x 0 Γ k ( x x 0 Γ k ( x x 0 Γ k 1 ( x x 0 = a m k+1. (3.7 =

6 The Stefan problem 223 We construct an integral representation at the center of the sphere K rk,k+1 (x 0 Γ k+1 (0 Γ k (0 = = K rk,k+1 (x 0 = K rk,k+1 (x 0 a m k+1 Γ k ( x x 0 Γ k 1 ( x x 0 Γ 1 ( x x 0 dx = Γ 1 ( x x 0 = (Γ k+1 ( x x 0 Γ k ( x x 0 Γ 1 n ds, am k sinh (r k+1,k x x 0. am k 4π x x 0 sinh r k+1,k Taking into account that the function Γ k+1 (r Γ k (r is equal to zero at the points r = 0, r k+1,k = 0, we obtain Γ k ( x x 0 Γ k 1 ( x x 0 0 = a m k+1 Γ 1 ( x x 0 dx. K rk,k+1 (x 0 If r k,k 1 > r k+1,k, then this equality is impossible. Formula (3.1 implies 4π x x 0 Γ 1 ( x x 0 = e x x am 0 + O (e R am, 4π x x 0 Γ 2 ( x x 0 = ( +O a ( m 1 e x x am 0 e x x 0 a m 1 a m e R a min, if a m 1 a m, am 1 4π x x 0 Γ 2 ( x x 0 = x x 0 2 am a m e x x 0 + O (e R am if a m 1 = a m. From here we obtain r 2,1 = ln a m 1 ln a m am 1 a m + o( 0. In particular, if a m 1 = a m, then r 2,1 = 2 am. +

7 224 M. A. Borodin The function Γ 2 (r Γ 1 (r changes the sign once. Therefore, as follows from the equations (3.7, the functions Γ k (r Γ k 1 (r change the sign once too. It means that the inequalities (3.6 hold. Property 7. We have following estimate { Γ N 1 n M 1 q N R exp{ M R 2 } + exp { T }} π2, (3.8 R 2 a max where q 2, and positive constants M 1, M 2 do not depend on N,, R. Proof. Let us estimate integral Γ N = i z n 2πa N 2πR sinh( zr dz (1 z a 1 (1 z a 2...(1 z a N, KR (x 0 where L is the boundary of the domain L L = {z : ϱ = z < (1+q max a k 1 k N, Rez = b 0 > π2 R 2, b 0 < 0, ϱ > max 1 k N a k }. (3.9 Let us represent the integral (3.9 as a sum of two terms: I 1 and I 2, where I 1 denotes the integral along the part of the curve L which is an arch of a circle, and I 2 denotes the integral along the part of the contour which lies inside the straight line Rez = b 0. Let us estimate the integral I 1. The estimates z (1 (1 z...(1 z N a 1 a 2 a N z 1 a max q N, sinh( zr sinh[ z R cos(argz/2] = sinh[ z R cos ϕ], where ϕ π, if 0, imply 4 1 I 1 c 1 q N R exp{ c R 2 }, where the constants c 1 and c 2 do not depend on. Let us now estimate the integral I 2. As Rez = b 0, we obtain z (1 (1 z...(1 z ( a 1 a 2 a N 1 + b 0 a max T.

8 Assume b 0 = π2 2R 2. Then The Stefan problem 225 z R sinh zr c 4. Thus we obtain I 2 c 5 exp { T } π2. R 2 a max The constants c 3, c 4, c 5 do not depend on. 4. Uniform estimates. Passage to the limit. Theorem 4.1. Let ψ(x C l+α l+α l+α, (D, ϕ(x, t H 2 (DT, 0 < α < 1, l 3, and assume that the corresponding consistency conditions hold at t = 0, x D. Then exists the constant c, which not depend ε, that the following estimate holds: max x D u ε (x, t + max D T \{ω ε (x,t ω ε (x,0} u ε (x, t x c, (4.1 where ω ε (x, t = {(x, t D T : 1 < u ε (x, t < 1 + ε}. Let x 0 be an arbitrary point in D. Denote by K R (x 0 the sphere with the center at the point x 0 of radius R = σ, K R (x 0 D. Let s transform the equation (2.6 to the following form ( u ε k (x = [f ε k (x f ε k 1 (x] + + βε k (x βε k (x 0 βε k (x 0 u ε k (x + βε k 1 (x 0 u ε k 1 (x = u ε k (x βε k 1 (x βε k 1 (x 0 u ε k 1 (x. In order to obtain the estimate for the functions uε k (x the integral representation (Property 4. It will give we will use u ε n(x 0 = K R (x 0 u ε 0 (xγ n( x x 0 dx

9 226 M. A. Borodin n k=1 K R (x 0 u ε k Γ n k+1 ds n n k=1k R (x 0 n k=1k R (x 0 (f ε k (x f ε k 1 (x { β ε k (x β ε k (x 0 Γ n k+1 ( x x 0 dx u ε k (x βε k 1 (x βε k 1 (x 0 Γ n k+1 ( x x 0 dx = I 1 + I 2 + I 3 + I 4. u ε k 1 (x } Let s estimate every term. From (3.4 follows 1 I 1 uε 0 (x Γ n( x x 0 dx max u ε 0 (x 1. x D a n K R (x 0 Let assume 0 < σ < 1. Then from (3.8 follows 2 I 2 T c { 1 1 M(ε q N R exp{ c R 2 } + exp { T }} π2 R 2 a max T c { 1 1 M(ε q N exp{ c σ σ 2 } + exp { T }} π2 2σ a max where c(σ do not depend ε. From (2.5, (3.4 follows + I 3 n k=1k R (x 0 K R \K r2,1 (x 0 K r2,1 (x 0 f ε k (x Γ n k+1 ( x x 0 Γ n k ( x x 0 max x D,1 k N fk ε(x Γ 1( x x 0 dx+ ( max fk ε(x 1 dx c x x x D,1 k N 0 where constant c do not depend ε,. From (3.5, (3.6 follows I 4 n βk ε(x βε k (x 0 k=1k R (x 0 uε k (x Γ n k+1( x x 0 Γ n k ( x x 0 c(σ M(ε, dx + 2σ M(ε M(ε dx+,

10 + n K r2,1 (x 0 k=1k R \K r2,1 (x 0 + c 1 x x 0 ε 2 M(ε The Stefan problem 227 β ε k (x βε k (x 0 uε k (x Γ n k+1( x x 0 Γ n k ( x x 0 K R \K r2,1 (x 0 Γ 1 ( x x 0 dx+ c 1 x x 0 ε 2 M(ε 1 dx x x 0 dx c ( 1/2 + 3σ 1. ε 2 M(ε In result we shall receive u ε n (x 0 c 1 max u ε 3σ 1 0(x + c 2 x D ε 2 M(ε, where constants c 1, c 2 do not depend from, ε. Let us assume that 3σ 1 < ε 2 M(ε, 1/3 < σ < 1/2. Then u ε (x 0, t uε k (x 0 1 k { } u ε (x, t + uε (x, η dη (k 1 c 1 max u ε 0 (x + c 3σ 1 2 x D ε 2 M(ε + c 3 M(ε c 4, where the constant c 4 do not depend from ε. Near of boundary of domain D the equation (2.6 become the linear equation with the constant coefficients. Therefore the appropriate estimate can be easily received. We differentiate the equation (2.6 with respect to one of the variables x i. ( 2 u ε k (x x 2 x b ε(u ε k (xuε k (x b ε(u ε k 1 (xuε k 1 (x We use the property 4. It gives n (x 0 = β0 εuε 0 (x Γn( x x 0 dx u ε { n k=1 n K R (x 0 K R (x 0 k=1k R (x 0 u ε k (x Γ n k+1 n ds h K R (x 0 f ε k (xγ n k+1dx (βk ε(x 0 βk ε(xuε k (x Γ m k+1( x x 0 Γ m k ( x x 0 dx h } = f ε k (x x. (4.3

11 228 M. A. Borodin From this integral representation, applying the same reasoning as above, we obtain the second part of the estimate (4.1. The estimation (4.2 can be proved similarly. Let s return to integral representation (4.3 and we will estimate from below the first term. We will assume, that ψ(x 0 everywhere x in D. For definiteness we will assume ψ(x > 0. Then x K R (x 0 β ε (ψ(x 0 u 0(x x a min min D (ψ (x = a min min D (ψ (x Γ n( x x 0 dx K R (x 0 { Γ n( x x 0 dx = n 1 K R (x 0 k=1 + K R (x 0 1 β ε k (x 0 Γ 1 ( x x 0 dx. Γ n k+1 ( x x 0 dx+ n } = I 1 + I 2. From (3.8 follows I 1 T c { 1 M(ε q N exp{ c σ σ 2 } + exp { T }} π2 2σ a max where c, c(σ do not depend ε. From (3.3 follows Γ 1 ( x x 0 I 2 = dx = 1 ( 1 βn ε(x 0 sinh K R (x 0 βn(x ε 0 σ c(σ M(ε, βn(x ε 0 σ Similarly to the previous theorem it is possible to prove that all other terms in the received integral representation (4.3 have limits equal to zero when σ 0. From here follows Theorem 4.2. Let ψ(x C 2+α 2+α 2+α, (D, ϕ(x, t H 2 (DT, min D ψ(x > 0, (0 < α < 1 and assume that the corresponding consistency conditions hold at t = 0, x D. Then exists the constant c, which not depend ε, that the following estimate holds: u ε (x, t c > 0 (x, t D T \ ω ε (x, t.

12 The Stefan problem 229 Let the function η(x, t C 2,1 (D T be equal to zero on ( D (0, T (D (t = T. We multiply (2.2 by η(x, t, integrate it over D T. After simple transformations we obtain ( u ε (x, t η(x, t + a(u uε η(x, t + λχ ε(u ε η dxdt+ D T +λ χ ε (ψη(x, 0dx = 0, D The possibility of passage to the limit follows from the statements proved above. As a result we obtain Theorem 4.3. Let the following conditions be satisfied: ψ(x C 2+α (D, ϕ(x, t H 2+α,1+α/2 (D T, min ψ(x > 0, D (0 < α < 1 and we will assume that corresponding consistency conditions hold at t = 0, x D. Then T > 0 there exists a solution of the problem (1.1-(1.4 and u(x, t C(D T ( H 2+α,1+α/2 (Ω T \ γ 0 H 2+α,1+α/2 (G T \ γ 0, the free boundary is given by the graph of a function from H 2+α,1+α/2 class. In this work existence of classical solution is proved at more natural limitations, than in work [2]. 1. Ladyshenskaja O. A., Solonnikov V. A. and Uraltceva N. N. Linear and Quasi-linear Parabolic Equations, Nauka, Moscow, Borodin M.A. Existence of the clobal classical solution for a two-phase Stefan problem // SIAM J. Math. Anal., vol. 30, No. 6(1999, pp Donetsk Nationality university, Dept. of math. physics, st. Universitetskaya 24, Donetsk 83055, Ukraine Received