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 di erent 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...

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Дата:2008
Автор: Borodin, M.A.
Формат: Стаття
Мова:Англійська
Опубліковано: Інститут прикладної математики і механіки НАН України 2008
Онлайн доступ:https://nasplib.isofts.kiev.ua/handle/123456789/124263
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Назва журналу:Digital Library of Periodicals of National Academy of Sciences of Ukraine
Цитувати:The Stefan problem / M.A. Borodin // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 218-229. — Бібліогр.: 2 назв. — англ.

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Digital Library of Periodicals of National Academy of Sciences of Ukraine
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author Borodin, M.A.
author_facet Borodin, M.A.
citation_txt The Stefan problem / M.A. Borodin // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 218-229. — Бібліогр.: 2 назв. — англ.
collection DSpace DC
description The Stefan problem in its classical statement is a mathematical model of the process of propagation of heat in a medium with di erent 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.
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fulltext 218 Нелинейные граничные задачи 18, 218-229 (2008) c©2008. 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 C2+α surfaces ∂D1 and ∂D2 such, that ∂D1 contains inside ∂D2 and limits domain D, DT = D × (0, T ). The problem is to find a function u(x, t) and domains ΩT , GT , satisfying to following conditions: ∆u− a(u) ∂u ∂t = 0 in ΩT ∪GT , (1.1) ΩT ={(x, t) ∈ DT : 0 < u(x, t) < 1}, GT ={(x, t) ∈ DT : u(x, t) > 1}, where a(u) - step function a1 > 0 в ΩT and a2 > 0 in GT ; on the known boundary ∂DT u(x, t) = 0 on ∂D1 × (0, T ), u(x, t) = ϕ(x, t) > 1 on ∂D2 × (0, T ); (1.2) The Stefan problem 219 on the unknown (free) boundary γT = DT ∩ ∂ΩT = DT ∩ ∂GT u−(x, t) = u+(x, t) = 1, 3 ∑ i=1 (∂u− ∂xi − ∂u+ ∂xi ) cos(n, xi)+λ cos(n, t) = 0; (1.3) where λ is a positive constant, n is the normal to the surface γT 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 γT taken from the domains GT , ΩT respectively. The function u(x, t) is interpreted as the tempe- rature of the medium, γt 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) ∣ ∣ ∣ ∂D1 = 0, ψ(x) ∣ ∣ ∣ ∂D2 = ϕ(x, 0) > 1, Ω0 = {x ∈ D : 0 < ψ(x) < 1}, G0 = {x ∈ D : ψ(x) > 1}, γ0 = D ∩ ∂Ω0. (1.4) 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 ∂DT and integrate by parts: ∫ DT ( ∇u(x, t)∇η(x, t) + a(u)∂u ∂t η(x, t) + λχ(u)∂η ∂t ) dxdt+ +λ ∫ D χ(ψ)η(x, 0)dx = 0, (2.1) where χ(u) = { 1, if u(x, t) < 1, 0, if u(x, t) > 1 . For ∀ε > 0 we introduce a function χε(τ) ∈ C∞(R1): χε(τ) = 1 ∀τ ≤ 1 − ε, χε(τ) = 0 ∀τ ≥ 1, χ′ ε(τ) ≤ 0, χ(n) ε (x) ≤ c εn , Let aε(x) = a1 + χε(x)(a2 − a1). Define the function {uε(x, t)} as solutions of the following problem: 220 M. A. Borodin ∆uε(x, t) − aε(u ε(x, t)) ∂uε(x, t) ∂t = −λ∂χε(u ε(x, t)) ∂t in DT (2.2) uε(x, t) = 0 on ∂D1 × [0, T ), uε(x, t) = ϕ(x, t) on ∂D2 × [0, T ); (2.3) uε(x, 0) = ψ(x) in D, ψ(x) = 0 on ∂D1, ψ(x) = ϕ(x, 0) > 1 on ∂D2. (2.4) Then, as is well known [1], takes place the statement Theorem 2.1. Let ψ(x) ∈ C l+α(D), ϕ(x, t) ∈ H l+α, l+α 2 (DT ), 0 < α < 1, and assume that the corresponding consistency conditions hold at t = 0, x ∈ ∂D. Then this problem is solvable and ∥ ∥ ∥ uε(x, t) ∥ ∥ ∥ Hl+α, l+α 2 (DT ) ≤ c 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 ∆uε(x, t) − ∂ ∂t uε(x,t) ∫ 0 [aε(τ) − λχ′ ε(τ)]dτ = 0. We divide the cylinder DT by the planes t = kτ, k = 1, 2, ...N,Nτ = T, integrate equation (2.2) with respect to the variable t, from (k−1)τ to kτ . kτ ∫ (k−1)τ ∆uε(x, τ)dτ − uε k (x) ∫ uε k−1 (x) [aε(τ) − λχ′ ε(τ)]dτ = 0, where uεk(x) ≡ uε(x, kτ). After simple transformations we obtain 4uεk(x) − βεk(x) ∂uεk(x) ∂t = f εk(x), (2.6) uεk(0) = 0, uεk(l) = ϕ(kτ) = ϕk, uε0(x) = ψ(x). (2.7) где ∂uεk(x) ∂t = uεk(x) − uεk−1(x) τ , The Stefan problem 221 βεk(x) = 1 ∫ 0 {kε[uεk−1 + τ(uεk − uεk−1)] − λχ′ ε[u ε k−1 + τ(uεk − uεk−1)]}dτ, f εk(x) = 1 τ kτ ∫ (k−1)τ [∆uε(x, kτ) − ∆uε(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 KR(x0) be ball with center at the point x0 and the radius R and Γn−k+1(|x− x0|) = iτ 2πan ∮ ∂L (sinh √ zR)−1 sinh √ z(R− |x− x0|)dz 4π|x− x0|(1 − zτ an )(1 − zτ an−1 )(1 − zτ ak ) , (3.1) where L = {z = ξ + iη : Rez > − π2 R2 , |z| < %}, (a1 τ , 0 ) , (a2 τ , 0 ) , ... (an τ , 0 ) ∈ L, ai > 0, i = 1, 2, ...n. Property 1. Let |x− x0| 6= 0, then ∆Γn−k+1 − ak Γn−k+1 − Γn−k τ = 0, ∀k = 1...(n− 1), ∆Γ1 − an Γ1 τ = 0, Γ1(|x− x0|) = sinh √ an τ (R − |x− x0|) 4π|x− x0| sinh √ an τ . (3.2) Property 2. There is the estimation ∫ KR(x0) Γn(|x− x0|) τ dx ≤ ∫ KR(x0) Γn−1(|x− x0|) τ dx ≤ ... ≤ ∫ KR(x0) Γ1(|x− x0|) τ dx ≤ 1 an ( 1 − √ an τ R sinh √ an τ R ) ≤ 1. (3.3) Property 3. Let Kδ(x0) denote the ball with its center at the point x0 and radius δ. Then lim δ→0 ∮ ∂Kδ(x0 ∂Γm−k+1 ∂n ds = { 1, if k = m, 0, if k 6= m, 222 M. A. Borodin where n is the inner normal. All reduced above property can be received by immediate calculations. Property 4. Let {vk(x) ∈ C2(D)} and they satisfy the following equations: ∆vk − akvk − ak−1vk−1 τ = −fk − fk−1 τ , then there is an integral representation vm(x0) = ∫ KR(x0) (a0v0 − f0)Γm(|x− x0| τ dx− m ∑ k=1 ∫ ∂KR(x0) vk ∂Γm−k+1 ∂n ds+ + m ∑ k=1 ∫ KR(x0) fk Γm−k+1 − Γm−k τ 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 {Γm−k+1(|x − x0|) − Γm−k(|x − x0|)} change the sign on the interval 0 < |x− x0| < R no more than once and ∣ ∣ ∣ ∣ ∂Γk−1(r) ∂r ∣ ∣ ∣ ∣ r=R ≤ ∣ ∣ ∣ ∣ ∂Γk(r) ∂r ∣ ∣ ∣ ∣ r=R . (3.5) This property follows from property1 and from principle of the maximum. Property 6. We will denote by rk,k−1 the points, where the functions Γk(r) − Γk−1(r), k = 1, 2...n, are equal to zero. Then we have the inequality rk,k−1 ≤ rk+1,k, and r2,1 = √ τ ln an−1 − ln an√ an−1 − √ an + o( √ τ ) τ → 0, if an−1 6= an, (3.6) r2,1 = √ τ 2 an , if an−1 = an. Proof. The functions Γk(r) − Γk−1(r) satisfy the equation ∆(Γk+1(|x−x0|)−Γk(|x−x0|))−am−k Γk+1(|x− x0|) − Γk(|x− x0|) τ = = −am−k+1 Γk(|x− x0|) − Γk−1(|x− x0|) τ . (3.7) The Stefan problem 223 We construct an integral representation at the center of the sphere Krk,k+1 (x0) Γk+1(0) − Γk(0) = = ∫ Krk,k+1 (x0) am−k+1 Γk(|x−x0|)−Γk−1(|x−x0|) τ Γ1(|x− x0|)dx = = − ∫ ∂Krk,k+1 (x0) (Γk+1(|x− x0|) − Γk(|x− x0|) ∂Γ1 ∂n ds, Γ1(|x− x0|) = sinh √ am−k τ (rk+1,k − |x− x0|) 4π|x− x0| sinh √ am−k τ rk+1,k . Taking into account that the function Γk+1(r)−Γk(r) is equal to zero at the points r = 0, rk+1,k = 0, we obtain 0 = ∫ Krk,k+1 (x0) am−k+1 Γk(|x− x0|) − Γk−1(|x− x0|) τ Γ1(|x− x0|)dx. If rk,k−1 > rk+1,k, then this equality is impossible. Formula (3.1) implies 4π|x− x0|Γ1(|x− x0|) = e−|x−x0| √ am τ +O ( e−R √ am τ ) , 4π|x−x0|Γ2(|x−x0|) = am−1 am−1 − am ( e−|x−x0| √ am τ − e−|x−x0| √ am−1 τ ) + +O ( e−R √ amin τ ) , if am−1 6= am, 4π|x− x0|Γ2(|x− x0|) = |x− x0| 2 √ amτ e−|x−x0| √ am τ +O ( e−R √ am τ ) if am−1 = am. From here we obtain r2,1 = √ τ ln am−1 − ln am√ am−1 − √ am + o(τ) τ → 0. In particular, if am−1 = am, then r2,1 = √ τ 2√ am . 224 M. A. Borodin The function Γ2(r)−Γ1(r) changes the sign once. Therefore, as follows from the equations (3.7), the functions Γk(r) − Γk−1(r) change the sign once too. It means that the inequalities (3.6) hold. Property 7. We have following estimate ∣ ∣ ∣ ∣ ∂ΓN ∂n ∣ ∣ ∣ ∣ ≤M1 { 1 qNR exp{−M2 R√ τ } + exp { − Tπ2 R2amax }} , (3.8) where q ≥ 2, and positive constants M1,M2 do not depend onN, τ, R. Proof. Let us estimate integral ∂ΓN ∂n ∣ ∣ ∣ ∣ ∣ ∂KR(x0) = −iτ 2πaN ∮ L √ z 2πR sinh( √ zR) · dz (1 − zτ a1 )(1 − zτ a2 )...(1 − zτ aN ) , (3.9) where ∂L is the boundary of the domain L = {z : % = |z| < (1+q) max 1≤k≤N ak τ , Rez = b0 > − π2 R2 , b0 < 0, % > max 1≤k≤N ak τ }. Let us represent the integral (3.9) as a sum of two terms: I1 and I2, where I1 denotes the integral along the part of the curve ∂L which is an arch of a circle, and I2 denotes the integral along the part of the contour which lies inside the straight line Rez = b0. Let us estimate the integral I1. The estimates ∣ ∣ ∣ ∣ (1 − zτ a1 )(1 − zτ a2 )...(1 − zτ aN ) ∣ ∣ ∣ ∣ ≥ ∣ ∣ ∣ ∣ |z|τ amax − 1 ∣ ∣ ∣ ∣ N ≥ qN , ∣ ∣sinh( √ zR) ∣ ∣ ≥ sinh[ √ |z|R cos(argz/2)] = sinh[ √ |z|R cosϕ], where ϕ→ π 4 , if τ → 0, imply |I1| ≤ c1 1 qNR exp{−c2 R√ τ }, where the constants c1 and c2 do not depend on τ . Let us now esti- mate the integral I2. As Rez = b0, we obtain ∣ ∣ ∣ ∣ (1 − zτ a1 )(1 − zτ a2 )...(1 − zτ aN ) ∣ ∣ ∣ ∣ ≥ ( 1 + τ |b0| amax ) T τ . The Stefan problem 225 Assume b0 = − π2 2R2 . Then ∣ ∣ ∣ ∣ √ |z|R sinh √ zR ∣ ∣ ∣ ∣ ≤ c4. Thus we obtain |I2| ≤ c5 exp { − Tπ2 R2amax } . The constants c3, c4, c5 do not depend on τ . 4. Uniform estimates. Passage to the limit. Theorem 4.1. Let ψ(x) ∈ C l+α(D), ϕ(x, t) ∈ H l+α, l+α 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 ∣ ∣ ∣ ∣ ∂uε(x, t) ∂t ∣ ∣ ∣ ∣ + max DT \{ωε(x,t)∩ωε(x,0)} ∣ ∣ ∣ ∣ ∂uε(x, t) ∂x ∣ ∣ ∣ ∣ ≤ c, (4.1) where ωε(x, t) = {(x, t) ∈ DT : 1 < uε(x, t) < 1 + ε}. Let x0 be an arbitrary point in D. Denote by KR(x0) the sphere with the center at the point x0 of radius R = τσ, KR(x0) ⊂ D. Let’s transform the equation (2.6) to the following form ∆ ( ∂uε k (x) ∂t ) − βε k (x0) τ ∂uε k (x) ∂t + βε k−1 (x0) τ ∂uε k−1 (x) ∂t = = ∂[fε k (x)−fε k−1 (x)] ∂t + + βε k (x)−βε k (x0) τ ∂uε k (x) ∂t − βε k−1 (x)−βε k−1 (x0) τ ∂uε k−1 (x) ∂t . In order to obtain the estimate for the functions ∂uε k (x) ∂t we will use the integral representation (Property 4). It will give ∂uεn(x0) ∂t = − ∫ KR(x0) ∆uε0(x) Γn(|x− x0|) τ dx− 226 M. A. Borodin n ∑ k=1 ∫ ∂KR(x0) ∂uε k ∂t ∂Γn−k+1 ∂n ds− − n ∑ k=1 ∫ KR(x0) ∂(fε k (x)−fε k−1 (x)) ∂t Γn−k+1(|x− x0|)dx− − n ∑ k=1 ∫ KR(x0) { βε k (x)−βε k (x0) τ ∂uε k (x) ∂t − βε k−1 (x)−βε k−1 (x0) τ ∂uε k−1 (x) ∂t } × ×Γn−k+1(|x− x0|)dx = I1 + I2 + I3 + I4. Let’s estimate every term. From (3.4) follows |I1| ≤ ∫ KR(x0) 1 τ |∆uε0(x)|Γn(|x− x0|)dx ≤ max x∈D |∆uε0(x)| 1 an . Let assume 0 < σ < 1 2 . Then from (3.8) follows |I2| ≤ Tc1 τM(ε) { 1 qNR exp{−c2 R√ τ } + exp { − Tπ2 R2amax }} ≤ ≤ Tc1 τM(ε) { 1 qNτσ exp{−c2 τσ√ τ } + exp { − Tπ2 τ 2σamax }} ≤ c(σ) M(ε) τ, where c(σ) do not depend ε. From (2.5), (3.4) follows |I3| ≤ n ∑ k=1 ∫ KR(x0) |f εk(x)| ∣ ∣ ∣ Γn−k+1(|x−x0|)−Γn−k(|x−x0|) τ ∣ ∣ ∣ dx ≤ ≤ ∫ Kr2,1 (x0) max x∈D,1≤k≤N |f εk(x)|Γ1(|x−x0|) τ dx+ + ∫ KR\Kr2,1 (x0) max x∈D,1≤k≤N |f εk(x)| 1 τ |x−x0|dx ≤ c ( τ M(ε) + τ2σ M(ε) ) , where constant c do not depend ε, τ. From (3.5), (3.6) follows |I4| ≤ n ∑ k=1 ∫ KR(x0) |βεk(x) − βεk(x0)|× ×|∂u ε k (x) ∂t | ∣ ∣ ∣ Γn−k+1(|x−x0|)−Γn−k(|x−x0|) τ ∣ ∣ ∣ dx+ The Stefan problem 227 + n ∑ k=1 ∫ KR\Kr2,1 (x0) |βεk(x) − βεk(x0)|× ×|∂u ε k (x) ∂t | ∣ ∣ ∣ Γn−k+1(|x−x0|)−Γn−k(|x−x0|) τ ∣ ∣ ∣ dx ≤ ≤ ∫ Kr2,1 (x0) c1|x−x0| ε2M(ε) Γ1(|x−x0|) τ dx+ + ∫ KR\Kr2,1 (x0) c1|x−x0| ε2M(ε) 1 τ |x−x0|dx ≤ c ε2M(ε) (τ 1/2 + τ 3σ−1). In result we shall receive ∣ ∣ ∣ ∣ ∂uεn(x0) ∂t ∣ ∣ ∣ ∣ ≤ c1 max x∈D |∆uε0(x)| + c2 τ 3σ−1 ε2M(ε) , where constants c1, c2 do not depend from τ, ε. Let us assume that τ 3σ−1 < ε2M(ε), 1/3 < σ < 1/2. Then ∣ ∣ ∣ ∣ ∂uε(x0, t) ∂t ∣ ∣ ∣ ∣ ≤ ∂uεk(x0) ∂t + ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 τ kτ ∫ (k−1)τ { ∂uε(x, t) ∂t − ∂uε(x, η) ∂t } dη ∣ ∣ ∣ ∣ ∣ ∣ ∣ ≤ ≤ c1 max x∈D |∆uε0(x)| + c2 τ 3σ−1 ε2M(ε) + c3 τ M(ε) ≤ c4, where the constant c4 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 appro- priate estimate can be easily received. We differentiate the equation (2.6) with respect to one of the variables xi. ∂2 ∂x2 ( ∂uεk(x) ∂x ) − bε(u ε k(x))u ε′ k (x) − bε(u ε k−1(x))u ε′ k−1(x) τ = ∂f εk(x) ∂x . We use the property 4. It gives uε′n (x0) = ∫ KR(x0) βε0u ε′ 0 (x)Γn(|x−x0|) h dx− − n ∑ k=1 { ∫ ∂KR(x0) uε′k (x)∂Γn−k+1 ∂n ds− ∫ KR(x0) f ε′k (x)Γn−k+1dx } − − n ∑ k=1 ∫ KR(x0) (βεk(x0) − βεk(x))u ε′ k (x)Γm−k+1(|x−x0|)−Γm−k(|x−x0|) h dx (4.3) 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) ∂x 6= 0 everywhere in D. For definiteness we will assume ∂ψ(x) ∂x > 0. Then ∫ KR(x0) βε(ψ(x0)) ∂u0(x) ∂x Γn(|x−x0|) τ dx ≥ ≥ amin minD(ψ′(x) ∫ KR(x0) Γn(|x−x0|) τ dx = = amin minD(ψ′(x) { ∫ ∂KR(x0) n−1 ∑ k=1 1 βε k (x0) ∂Γn−k+1(|x−x0|) ∂n dx+ + ∫ KR(x0 Γ1(|x−x0|) τ dx. } = I1 + I2. From (3.8) follows |I1| ≤ Tc τM(ε) { 1 qNτσ exp{−c2 τσ√ τ } + exp { − Tπ2 τ 2σamax }} ≤ c(σ) M(ε) τ, where c, c(σ) do not depend ε. From (3.3) follows |I2| = ∫ KR(x0) Γ1(|x− x0|) τ dx = 1 βεn(x0) ( 1 − √ βε n(x0) τ τσ sinh √ βε n(x0) τ τσ ) 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) ∈ C2+α(D), ϕ(x, t) ∈ H2+α, 2+α 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: |∇uε(x, t)| ≥ c > 0 ∀(x, t) ∈ DT \ ωε(x, t). The Stefan problem 229 Let the function η(x, t) ∈ C2,1(DT ) be equal to zero on (∂D × (0, T ))∪ (D× (t = T )). We multiply (2.2) by η(x, t), integrate it over DT . After simple transformations we obtain ∫ DT ( ∇uε(x, t)∇η(x, t) + a(u)∂u ε ∂t η(x, t) + λχε(u ε)∂η ∂t ) dxdt+ +λ ∫ D χε(ψ)η(x, 0)dx = 0, 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) ∈ C2+α(D), ϕ(x, t) ∈ H2+α,1+α/2(DT ), min D |∇ψ(x)| > 0, (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(DT ) ∩ ( H2+α,1+α/2(ΩT \ γ0) ×H2+α,1+α/2(GT \ γ0) ) , 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, 1967. 2. 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. 1264-1281. Donetsk Nationality university, Dept. of math. physics, st. Universitetskaya 24, Donetsk 83055, Ukraine Received 6.10.08
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institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
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language English
last_indexed 2025-12-07T18:35:56Z
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publisher Інститут прикладної математики і механіки НАН України
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spelling Borodin, M.A.
2017-09-23T09:45:02Z
2017-09-23T09:45:02Z
2008
The Stefan problem / M.A. Borodin // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 218-229. — Бібліогр.: 2 назв. — англ.
0236-0497
MSC (2000): 35R35
https://nasplib.isofts.kiev.ua/handle/123456789/124263
The Stefan problem in its classical statement is a mathematical model of the process of propagation of heat in a medium with di erent 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.
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Інститут прикладної математики і механіки НАН України
The Stefan problem
Article
published earlier
spellingShingle The Stefan problem
Borodin, M.A.
title The Stefan problem
title_full The Stefan problem
title_fullStr The Stefan problem
title_full_unstemmed The Stefan problem
title_short The Stefan problem
title_sort stefan problem
url https://nasplib.isofts.kiev.ua/handle/123456789/124263
work_keys_str_mv AT borodinma thestefanproblem
AT borodinma stefanproblem