Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem

A free boundary problem describing a filtration process in a porous medium is considered. An unknown interface divides the filtration domain into elliptic and parabolic regions. In the parabolic region the governing equation is degenerate. The existence of a smooth solution in the weighted Holder sp...

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Дата:	2011
Автори:	Bazaliy, B.V., Degtyarev, S.P.
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Мова:	English
Опубліковано:	Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України 2011
Назва видання:	Журнал математической физики, анализа, геометрии
Онлайн доступ:	http://dspace.nbuv.gov.ua/handle/123456789/106687
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Цитувати:	Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem / B.V. Bazaliy, S.P. Degtyarev // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 4. — С. 295-332. — Бібліогр.: 13 назв. — англ.

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spelling	irk-123456789-1066872016-10-03T03:02:19Z Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem Bazaliy, B.V. Degtyarev, S.P. A free boundary problem describing a filtration process in a porous medium is considered. An unknown interface divides the filtration domain into elliptic and parabolic regions. In the parabolic region the governing equation is degenerate. The existence of a smooth solution in the weighted Holder space is proved. 2011 Article Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem / B.V. Bazaliy, S.P. Degtyarev // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 4. — С. 295-332. — Бібліогр.: 13 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106687 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	A free boundary problem describing a filtration process in a porous medium is considered. An unknown interface divides the filtration domain into elliptic and parabolic regions. In the parabolic region the governing equation is degenerate. The existence of a smooth solution in the weighted Holder space is proved.
format	Article
author	Bazaliy, B.V. Degtyarev, S.P.
spellingShingle	Bazaliy, B.V. Degtyarev, S.P. Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem Журнал математической физики, анализа, геометрии
author_facet	Bazaliy, B.V. Degtyarev, S.P.
author_sort	Bazaliy, B.V.
title	Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem
title_short	Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem
title_full	Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem
title_fullStr	Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem
title_full_unstemmed	Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem
title_sort	classical solution of a degenerate elliptic-parabolic free boundary problem
publisher	Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
publishDate	2011
url	http://dspace.nbuv.gov.ua/handle/123456789/106687
citation_txt	Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem / B.V. Bazaliy, S.P. Degtyarev // Журнал математической физики, анализа, геометрии. — 2011. — Т. 7, № 4. — С. 295-332. — Бібліогр.: 13 назв. — англ.
series	Журнал математической физики, анализа, геометрии
work_keys_str_mv	AT bazaliybv classicalsolutionofadegenerateellipticparabolicfreeboundaryproblem AT degtyarevsp classicalsolutionofadegenerateellipticparabolicfreeboundaryproblem
first_indexed	2025-07-07T18:52:03Z
last_indexed	2025-07-07T18:52:03Z
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fulltext	Journal of Mathematical Physics, Analysis, Geometry 2011, vol. 7, No. 4, pp. 295–332 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem B.V. Bazaliy and S.P. Degtyarev Institute of Applied Mathematics and Mechanics National Academy of Sciences of Ukraine 74 R. Luxemburg Str., Donetsk 83114, Ukraine E-mail: bazaliy@iamm.ac.donetsk.ua Received March 11, 2011 A free boundary problem describing a filtration process in a porous medium is considered. An unknown interface divides the filtration domain into elliptic and parabolic regions. In the parabolic region the governing equation is degenerate. The existence of a smooth solution in the weighted Hölder space is proved. Key words: elliptic and parabolic equations, free boundary problems, classical solutions, a priori estimates. Mathematics Subject Classification 2000: 35K65, 35R35. To the memory of our teacher I.I. Danyljuk 1. Statement of the Problem In this paper we study the following problem. Let Ω = (0, l) ⊂ R1, ΩT = Ω× (0, T ), ∂c(u) ∂t − ∂2u ∂y2 = 0 in ΩT , (1.1) u(0, t) = g0(t) on [0, T ], (1.2) u(l, t) = g(t) on [0, T ], (1.3) c(u(y, 0)) = c(u0(y)) on [0, l]. (1.4) Here c(η) is a continuous function that is strictly increasing for η > 0 and c(η) = 0 for η ≤ 0, g0(t), g(t), u0(y) are given functions, g0(t) < 0 and g(t) > 0 ∀t ∈ [0, T ]. c© B.V. Bazaliy and S.P. Degtyarev, 2011 B.V. Bazaliy and S.P. Degtyarev Problem (1.1)–(1.4) arises in the theory of fluid flow through a partially sat- urated porous media (see [4–8, 10, 11, 13] and the references therein). The level set {u = 0} splits the domain ΩT in two regions in which equation (1.1) is re- spectively parabolic and elliptic. It was shown in [11, 7] that under appropriate conditions on the data there exists a function y = s(t) for which u(y, t) ≤ 0, c(u(y, t)) = 0 for 0 ≤ y ≤ s(t), and u(y, t) > 0, c(u(y, t)) > 0 for s(t) < y ≤ l. In this problem the function s(t) is the free (unknown) boundary, and our main concern is to describe the qualitative properties of s(t). It was shown in [4] that under the conditions c(η) ∈ C(R) ∩ C2+β(R+), c′(+0) = 0, and s(t) < l the function s(t) is continuously differentiable. Contrary to [4], we will study the case when c(η) = { η1−α, η ≥ 0, α ∈ (0, 1), 0, η ≤ 0. (1.5) This situation leads to the additional singularity in the free boundary problem. In the region where the medium is saturated, y ∈ (0, s(t)), we have u(y, t) = −g0(t) s(t) y + g0(t). (1.6) Note that the value s(0) is defined by the initial function u0(y) and it is assumed that there is the only point s(0) ∈ (0, l) that separates the saturated and unsaturated regions. In the unsaturated region, y ∈ (s(t), l), the function u(y, t) satisfies the equation ∂u ∂t − uα 1− α ∂2u ∂y2 = 0, (1.7) and since at y = s(t) we have u(s(t), t) = 0, equation (1.7) is a degenerate parabolic equation. On the free boundary we have the following conditions: u(s(t)− 0, t) = u(s(t) + 0, t) = 0, (1.8) ∂u ∂y (s(t)− 0, t) = ∂u ∂y (s(t) + 0, t). (1.9) 296 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem We drop the unessential factor 1/(1−α) in equation (1.7) and get the following free boundary problem for unknown functions u(y, t) and s(t): ∂u ∂t − uαuyy = 0 in y ∈ (s(t), l), t ∈ (0, T ), (1.10) u(y, 0) = u0(y), y ∈ (s(0), l), (1.11) u(l, t) = g(t), t ∈ [0, T ], (1.12) u(s(t), t) = 0, t ∈ [0, T ], (1.13) ∂u ∂y (s(t), t) = −g0(t) s(t) , t ∈ [0, T ], (1.14) s(0) = s0 ∈ (0, l). (1.15) The structure of the paper is as follows. In Sec. 2, we reduce the free bound- ary problem to a problem in a fixed domain and formulate our main result, Theorem 2.1. In Sec. 3, we reformulate the nonlinear free boundary problem as a nonlinear equation in Banach spaces by using Theorem 3.1. In Sec. 4, we formulate the results relating to the principal model problem for the degenerate parabolic equation. In Sec. 5, we finish the proof of Theorem 2.1. In Sec. 6, we study the properties of the model problem and derive the corresponding estimates by using an integral representation of its solution. R e m a r k 1.1. The similar approach can be used for a free boundary problem in the case of the constitutive function of the form c(η) = { η1+α, η ≥ 0, α ∈ (0, 1), 0, η ≤ 0. 2. Reduction of the Free Boundary Problem (1.10)–(1.15) to a Problem in a Fixed Domain and the Main Result Let s(t) = s0 + ρ(t) and introduce a spatial variable x = y − s0 − ρ(t) l − s0 − ρ(t) = y − s(t) l − s(t) (2.1) such that the segment [s(t), l] is mapped onto [0, 1] and the free boundary y = s(t) is mapped at x = 0. Denote v(x, t) = u(y(x, t), t). In the new variables we get the following problem in the fixed domain GT = G× (0, T ), G = [0, 1]: ∂v ∂t + x− 1 l − s0 − ρ(t) ∂v ∂x dρ dt − vα vxx (l − s0 − ρ(t))2 = 0 in GT , (2.2) v(x, 0) = v0(x), x ∈ [0, 1], (2.3) Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 297 B.V. Bazaliy and S.P. Degtyarev v(1, t) = g(t), t ∈ [0, T ], (2.4) v(0, t) = 0, t ∈ [0, T ], (2.5) ∂v ∂x \|x=0 = −g0(t) l − s0 − ρ(t) s0 + ρ(t) , t ∈ [0, T ], (2.6) ρ(0) = 0. (2.7) We will use the anisotropic Hölder spaces Cβ,γ(GT ) of smooth functions u(x, t) with the norm \|u\|(β,γ) GT = \|u\|(0) GT + 〈u〉(β) x,GT + 〈u〉(γ) t,GT , β, γ ∈ (0, 1), where \|u\|(0) GT = max GT \|u(x, t)\|, 〈u〉(β) x,GT = sup (x1,t),(x2,t)∈GT \|u(x1, t)− u(x2, t)\| \|x1 − x2\|β , 〈u〉(γ) t,GT = sup (x,t1),(x,t2)∈GT \|u(x, t1)− u(x, t2)\| \|t1 − t2\|γ , and the Hölder space C1+γ([0, T ]) with the norm \|u(t)\|(1+γ) [0,T ] = \|u\|(0) [0,T ] + \|ut\|(γ) [0,T ] with \|ut\|(γ) [0,T ] = \|ut\|(0) [0,T ] + 〈ut〉(γ) t,[0,T ]. We will also use the standard Hölder spaces Cγ([0, T ]), 0 < γ < 1, with the norm \|u\|(γ) [0,T ] = \|u\|(0) [0,T ] + 〈u〉(γ) t,[0,T ]. For our purposes the weighted Hölder space C 2+β,β/q α (GT ) is appropriate, where β ∈ (0, 1), q = 2− α, and ‖u‖ C 2+β,β/q α (GT ) ≡ ‖u‖(2+β) α,GT = \|u\|(0) GT + \|ux\|(0) GT + 〈ux〉 ( β+1−α q ) t,GT + \|xαuxx\|(β, β/q) GT + \|ut\|(β,β/q) GT . We define the space C2+γ α (G) in the similar way. We denote by Ċ 2+β,β/q α (GT ) the subspace of C 2+β,β/q α (GT ) such that u(x, t) ∈ Ċ 2+β,β/q α (GT ) if u(x, 0) = ut(x, 0) = 0 and similarly define the spaces Ċβ,γ(GT ) for the functions such that u(x, 0) = 0 and the space Ċγ([0, T ]). In problem (2.2)–(2.7), we assume that xαv0xx(x) ∈ C 2γ q ([0, 1]), v0(x) ∈ C1([0, 1]), ∂v0 ∂x (0) ≥ ν = const > 0, (2.8) g0(t), g(t) ∈ C1+γ/q([0, T1]), T1 > 0, γ > 2β/q 298 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem and that the consistency conditions of the first order are fulfilled. It means that v0(0) = 0, v0(1) = g(0), 1 l − s0 ∂v0 ∂x (0) ( dρ dt \|t=0 ) + 1 (l − s0)2 (vα 0 (x)v0xx(x))\|x=0 = 0, (2.9) ∂g ∂t (0)− 1 (l − s0)2 vα 0 (1)v0xx(1) = 0. Theorem 2.1. Let the conditions (2.8), (2.9) be fulfilled with q = 2 − α, 0 < 2β/q < γ, α + 2γ/q < 1. Then there exists a unique solution of prob- lem (2.2)–(2.7) for some 0 < T ≤ T1 such that v(x, t) ∈ C 2+β,β/q α (GT ), ρ(t) ∈ C1+β/q([0, T ]). This theorem asserts the existence and uniqueness of the classical solution to the elliptic-parabolic degenerate problem locally in time. The similar result is valid if we change the boundary conditions (1.2), (1.3) for the Neumann or mixed boundary conditions. 3. The Nonlinear Functional Equation Introduce the notations ρ(1) ≡ dρ dt (0), v(1)(x) ≡ ∂v ∂t (x, 0). Then from (2.9) ρ(1) = −(vα 0 (x)v0xx(x))\|x=0 (l − s0)v0x(0) , (3.1) and from equation (2.2) v(1)(x) = 1− x l − s0 v0x(x)ρ(1) + vα 0 (x)v0xx(x) (l − s0)2 . (3.2) Now we construct a function w(x, t) such that w(x, t) ∈ C 2+γ,γ/q α (GT ), γ > 2β/q, w(x, 0) = v0(x), wt(x, 0) = v(1)(x), w(0, t) ≡ 0. Consider a model problem ∂ũ ∂t − xα ∂2ũ ∂x2 = ṽ(1)(x)− xαṽ0xx(x), x > 0, t > 0, ũx\|x=0 = v0x(0), ũ(x, 0) = ṽ0(x), Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 299 B.V. Bazaliy and S.P. Degtyarev where ṽ0(x) and ṽ(1)(x) are the finite extensions of v0(x) and v(1)(x) on x ≥ 0. It follows from Theorem 4.1 below that there exists a unique solution of this problem and the estimate of the form (4.20) is valid so that in particular ũ(x, t) ∈ C 2+γ,γ/q α (GT ). Moreover, ũ(x, 0) = v0(x) and ũt(x, 0) = v(1)(x) on [0, 1] by the construction. Now we set w(x, t) = ũ(x, t)− ũ(0, t). Since ũ(0, 0) = v0(0) = 0 and ũt(0, 0) = v(1)(0) = 0 (due to consistency condi- tions), the function w(x, t) has the desired properties. We denote also σ(t) = ρ(1) · t so that σ(0) = 0, σt(0) = ρ(1) = ρt(0). To reduce problem (2.2)–(2.7) to a problem in the spaces Ċ 2+β,β/q α (GT ) and Ċ1+β/q([0, T ]) with zero initial data, we introduce the new unknown functions u(x, t) = v(x, t)− w(x, t), δ(t) = ρ(t)− σ(t) (3.3) such that δ(0) = 0, δt(0) = 0, u(x, 0) ≡ 0, ut(x, 0) ≡ 0. We rewrite equation (2.2) in the form L(ρ, v) = ∂v ∂t − a11(ρ, v) ∂2v ∂x2 + a1(ρ) dρ dt ∂v ∂x = 0, (3.4) where a11(ρ, v) = vα (l − s0 − ρ(t))2 , a1(ρ) = x− 1 l − s0 − ρ(t) . (3.5) For the new unknown functions δ(t) and u(x, t) we obtain the equation L(δ(t) + σ(t), u(x, t) + w(x, t)) = 0. (3.6) Next we single out the main linear part from L(δ+σ, u+w) with respect to (δ, u) so that equation (3.6) takes the form ∂u ∂t − a11(σ,w) ∂2u ∂x2 + a1(σ)δt ∂w ∂x = −L(σ,w)− [a11(σ,w)− a11(δ + σ, u + w)] ∂2u ∂x2 −[a11(σ,w)− a11(δ + σ, u + w)] ∂2w ∂x2 −[a1(δ + σ)(δt + σt)− a1(σ)σt] ∂u ∂x − a1(σ)σt ∂u ∂x 300 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem −[a1(δ + σ)(δt + σt)− a1(σ)σt − a1(σ)δt] ∂w ∂x = −L(σ,w) + F (δ, u) = F0(δ, u). One can check that the function F (δ, u) contains either “quadratic” terms with respect to (δ, u) or minor terms in the sense of smoothness. For instance, the term [a11(σ,w)− a11(δ + σ, u + w)] ∂2u ∂x2 = [ wα (l − s0 − σ(t))2 − (u + w)α (l − s0 − σ(t)− δ(t))2 ] ∂2u ∂x2 = [ wα (l − s0 − σ(t))2 − wα (l − s0 − σ(t)− δ(t))2 ] ∂2u ∂x2 + wα − (u + w)α (l − s0 − σ(t)− δ(t))2 ∂2u ∂x2 (3.7) consists of the terms ”quadratic” in (δ, u), and in the expression [a1(δ + σ)(δt + σt)− a1(σ)σt − a1(σ)δt] ∂w ∂x = dδ dt ( 1 l − s0 − σ(t)− δ(t) − 1 l − s0 − σ(t) ) ∂w ∂x + dσ dt ( 1 l − s0 − σ(t)− δ(t) − 1 l − s0 − σ(t) ) ∂w ∂x the first term is “quadratic” and the second one is minor. Note also that by the construction of σ(t), w(x, t) and (3.2) L(σ,w)\|t=0 = 0. Thus problem (2.2)–(2.7) is transformed to the nonlinear problem for the func- tions u(x, t) and δ(t) ∂u ∂t − a11(σ,w) ∂2u ∂x2 + a1(σ)δt ∂w ∂x = F0(δ, u) in GT , (3.8) u(x, 0) = 0, δ(0) = 0, x ∈ [0, 1], (3.9) u(0, t) = 0, u(1, t) = g(t)− w(1, t), t ∈ [0, T ], (3.10) ∂u ∂x (0, t) = F1(δ(t)), t ∈ [0, T ], (3.11) where F1(δ(t)) = −g0(t) l − s0 − σ(t)− δ(t) s0 + σ(t) + δ(t) − ∂w ∂x (0, t), (3.12) Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 301 B.V. Bazaliy and S.P. Degtyarev and moreover, ut(x, 0) = 0, δt(0) = 0. (3.13) First we study the linear problem ∂u ∂t − a11(σ,w) ∂2u ∂x2 + a1(σ)δt ∂w ∂x = f0(x, t) in GT , (3.14) u(x, 0) = 0, δ(0) = 0, x ∈ [0, 1], (3.15) u(0, t) = 0, u(1, t) = ϕ(t), t ∈ [0, T ], (3.16) ∂u ∂x (0, t) = f1(t), t ∈ [0, T ], (3.17) where ϕ(t) = g(t)− w(1, t), f0(x, t) and f1(t) correspond to the right hand sides of (3.8) and (3.11) for some fixed u(x, t) and δ(t) from classes Ċ 2+β,β/q α (GT ) and Ċ1+β/q([0, T ]), respectively. So we assume f0(x, t) ∈ Ċβ,β/q(GT ), f1(t) ∈ Ċ β+1−α q ([0, T ]), ϕ(t) ∈ Ċ1+β/q([0, T ]). (3.18) From (3.5) we have a11(σ,w) = wα (l − s0 − σ(t))2 = 1 (l − s0 − σ(t))2 ( w(x, t) x )α xα ≡ a(x, t)xα, (3.19) where µ ≤ a(x, t) ≤ µ−1, µ = const > 0, for small t since σ(0) = 0 and w(x, t) is the smooth function with w(0, t) = 0, wx(0, t) ≥ ν > 0. Similar arguments give a1(σ)δt ∂w ∂x (x, t) = x− 1 l − s0 − σ(t) ∂w ∂x (x, t) dδ(t) dt ≡ −b(x, t) dδ(t) dt (3.20) with b(x, t) ≥ µ > 0 for x ∈ [0, 1/2], t ∈ [0, T ]. Theorem 3.1. Under conditions (3.18) there exists a unique solution of prob- lem (3.14)–(3.17) u(x, t) ∈ Ċ 2+β,β/q α (GT ), δ(t) ∈ Ċ1+β/q([0, T ]) for some T > 0 and ‖u‖(2+β) α,GT + \|δ\|(1+β/q) [0,T ] ≤ C ( \|f0\|(β,β/q) GT + \|f1\| (β+1−α q ) [0,T ] + \|ϕ\|(1+β/q) [0,T ] ) . (3.21) The proof of Theorem 3.1 uses the well-known procedure (see [12], Ch. 4): i) partition of unity on G, ii) investigation of model problems in R+ T or RT , iii) construction of a regularizator. 302 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem In the next section we describe the main model problem related to problem (3.14)–(3.17). We will not discuss problems i) and iii). For discussion of these problems see, for example, [1–3]. Problem (3.8)–(3.12) has the form A(u, δ) = F(u, δ), (3.22) where A(u, δ) is the linear bounded operator, and F(u, δ) is the nonlinear operator F(u, δ) = (F0(δ, u), F1(δ), g(t)− w(1, t)). (3.23) Theorem 3.1 means that the operator A has the bounded inverse operator A−1. Hence, equation (3.22) can be written as (u, δ) = A−1F(u, δ), and in Sec. 5 we will show that the operator A−1F is contractive. 4. The Model Problem Consider the differential operator defined by the left hand side in (3.14) and freeze the coefficients a(x, t) from (3.19) and b(x, t) from (3.20) at the point (x, t) = (0, 0). Let a(0, 0) = a0, b(0, 0) = b0, R+ = {x ≥ 0}, R+ T = R+ × [0, T ]. We are looking for a solution (u(x, t), δ(t)) of the problem ∂u ∂t − a0x α ∂2u ∂x2 − b0 dδ dt = f(x, t) in R+ T , (4.1) u(x, 0) = 0, δ(0) = 0, x ∈ R+, (4.2) u(0, t) = 0, ∂u ∂x (0, t) = f1(t), t ∈ [0, T ] (4.3) with f(x, t) ∈ Cβ,β/q(R+ T ), f1(t) ∈ C β+1−α q ([0, T ]), (4.4) where function f(x, t) has a finite support. Note that by all norms in Hölder spaces over unbounded domains R+ and R+ T we mean supremum in M > 0 of the corresponding norms over sets R+ ∩ {\|x\| ≤ M} and R+ T ∩ {\|x\| ≤ M}. Define the function θ(x, t) = u(x, t)− b0δ(t). (4.5) Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 303 B.V. Bazaliy and S.P. Degtyarev The function θ(x, t) can be found from the equations ∂θ ∂t − a0x α ∂2θ ∂x2 = f(x, t) in R+ T , (4.6) θ(x, 0) = 0, x ∈ R+, (4.7) ∂θ ∂x (0, t) = f1(t), t ∈ [0, T ]. (4.8) To find the function δ(t) we have the relation θ(0, t) + b0δ(t) = 0. (4.9) Keeping in mind also the problem of constructing of the function w(x, t) from Sec. 3, we consider the next model problem for the unknown function u(x, t) ∂u ∂t − xα ∂2u ∂x2 = f(x, t) in R+ T , (4.10) ∂u ∂x (0, t) = f1(t), t ∈ [0, T ], (4.11) u(x, 0) = u0(x), x ∈ R+, (4.12) where the functions f(x, t) and u0(x) have finite supports. In this case the fol- lowing conditions are required in addition to (4.4): u0(x) ∈ C2+2β/q α (R+), f(x, 0) ∈ C2β/q(R+). (4.13) To find a general solution of the equation ∂u ∂t − xα ∂2u ∂x2 = 0, we apply the Laplace transform in t such that pũ− xαũxx = 0, (4.14) where ũ(x, p) is the Laplace image of u(x, t). The general solution of (4.14) is (see [9], 8.491(7)) ũ(x, p) = c1x 1/2I−1/q( 2 q √ pxq/2) + c2x 1/2K−1/q( 2 q √ pxq/2), where q = 2 − α, Iµ(z), Kµ(z) are the modified Bessel functions, and c1, c2 are arbitrary constants. The Green function to the Neumann problem for equation (4.14) is G̃(x, ξ, p) = {2 q I−1/q(2 qp1/2xq/2)K−1/q(2 qp1/2ξq/2)x1/2ξ1/2ξ−α, x < ξ, 2 qK−1/q(2 qp1/2xq/2)I−1/q(2 qp1/2ξq/2)x1/2ξ1/2ξ−α, x > ξ, 304 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem and the inverse Laplace transform gives (see [9], 6.653) the Green function for problem (4.10)–(4.12) G(x, ξ, t) = c(q)t−1+1/q ( xq/2ξq/2 q2t )1/q I−1/q ( 2 (xξ)q/2 q2t ) e −xq+ξq q2t ξ−α. (4.15) Denote u = ξq/2 qt1/2 , v = xq/2 qt1/2 and rewrite the Green function as G(x, ξ, t) = c(q)t−1/q(uv)1/qI−1/q(2uv)e−(u2+v2)u−2α/q. (4.16) We define the constant c(q) in (4.16) by the condition ∞∫ 0 G(x, ξ, t)dξ = 1. (4.17) By direct calculations one can show that the function G(x, ξ, t − τ) satisfies the equations ∂G ∂t − xα ∂2G ∂x2 = 0 and ∂G ∂τ + ∂2 ∂ξ2 (ξαG) = 0, τ < t. (4.18) We use equations (4.18) and the Green formula to get the integral representation of the solution to problem (4.10)–(4.12) u(x, t) = t∫ 0 dτ ∞∫ 0 G(x, ξ, t− τ)f(ξ, τ)dξ + ∞∫ 0 G(x, ξ, t)u0(ξ)dξ − t∫ 0 (ξαG(x, ξ, t− τ))\|ξ=0f1(τ)dτ. (4.19) Theorem 4.1. Let in (4.10)–(4.13) and (4.4) the consistency condition f1(0)= u0x(0) be fulfilled. Then there exists a unique solution u(x, t) ∈ C 2+β,β/q α (R+ T ), α + β < 1, q = 2− α, such that 〈xαuxx〉(β,β/q) BR,T + 〈ut〉(β,β/q) BR,T + 〈ux〉((β+1−α)/q) t,BR,T ≤ C(R, T )(〈f〉(β,β/q) R+ T + 〈f1〉 (β+1−α q ) [0,T ] + \|f(x, 0)\|(2β/q) R+ + \|xαu0xx\|(2β/q) R+ ), (4.20) where BR,T = {0 ≤ x ≤ R} × [0, T ], R > 0, and the constant C(R, T ) is bounded for bounded R and T . Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 305 B.V. Bazaliy and S.P. Degtyarev The proof of (4.20) is based on the estimates of the potentials on the right hand side of (4.19) and will be done in Sec. 6. 5. Proof of Theorem 2.1 Now we return to problem (3.22) and the equation (u, δ) = A−1F(u, δ), (5.1) where F(u, δ) is defined in (3.23). We introduce the space M = Ċ2+β,β/q α (GT )× Ċ1+β/q([0, T ]) (5.2) with the elements z = (u(x, t), δ(t)) and the norm ‖z‖M = ‖u‖(2+β) α,GT + \|δ\|(1+β/q) [0,T ] , (5.3) and the space W = Ċβ,β/q(GT )× Ċ β+1−α q ([0, T ])× Ċ1+β/q([0, T ]) (5.4) with the elements f = (f(x, t), f1(t), ϕ(t)) and the norm ‖f‖W = \|f \|(β,β/q) GT + \|f1\| (β+1−α q ) [0,T ] + \|ϕ\|(1+β/q) [0,T ] . (5.5) It is straightforward to check that analogously to the inequality for the stan- dard Hölder norms \|u\|(l)GT from [12] \|u\|(l′)GT ≤ CT l−l′ 2 \|u\|(l)GT , (5.6) which is valid for the functions u ∈ Ċ l,l/2(GT ), l′ < l, l′ integer, (see [12], Ch. 4), in our case we have ‖u‖(2+β) α,GT ≤ CT γ−β q ‖u‖(2+γ) α,GT , \|f \|(β,β/q) GT ≤ CT γ−β q \|f \|(γ,γ/q) GT , β < γ < 1 (5.7) for the functions u ∈ Ċ 2+γ,γ/q α (GT ) and f ∈ Ċγ,γ/q(GT ), and analogous inequality is valid for the functions defined on [0, T ]. We will also use a known inequality \|fg\|(β,β/q) GT ≤ CT β/q\|f \|(β,β/q) GT \|g\|(β,β/q) GT (5.8) for the functions f, g ∈ Ċβ,β/q(GT ). We give here the short outline of the proof of inequalities (5.7). 306 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem Consider, for example, a weighted Hölder constant 〈xαuxx〉(β) x,GT in the defini- tion of the norm ‖u‖(2+β) α,GT in the space C 2+β,β/q α (GT ). Let function u(x, t) be in fact more smooth, namely, u ∈ C 2+γ,γ/q α (GT ) with 0 < β < γ < 1. According to the definition, 〈xαuxx〉(β) x,GT = sup (x,t),(x,t)∈GT \|v(x, t)− v(x, t)\| \|x− x\|β , where v(x, t) = xαuxx. Consider the ratio \|v(x,t)−v(x,t)\| \|x−x\|β . Consider two cases. If \|x− x\| < T 1/q, then \|v(x, t)− v(x, t)\| \|x− x\|β = \|v(x, t)− v(x, t)\| \|x− x\|γ \|x− x\|γ−β ≤ 〈v〉γ x,GT T γ−β q . If now \|x− x\| ≥ T 1/q, then, as v(x, 0) ≡ 0, \|v(x, t)− v(x, t)\| \|x− x\|β ≤ \|v(x, t)\|+ \|v(x, t)\| T 1/q ≤ 2〈v(x, t)〉γ/q t,GT tγ/q T β/q ≤ 2〈v(x, t)〉γ/q t,GT T γ−β q . Consequently, in both cases, 〈xαuxx〉(β) x,GT ≤ CT γ−β q ‖u‖(2+γ) α,GT . Other terms in the definition of the norm ‖u‖(2+β) α,GT can be treated in a quite similar way which leads to (5.7). Due to our assumptions on the smoothness of the initial data u0(x), the boundary functions g0(t) and g(t), and consistency conditions (2.9) with (5.7), we obtain ‖F(0, 0)‖W ≤ PT γ−β q , (5.9) where P is a constant independent of T . We set Md = {z ∈ M : ‖z‖M ≤ d}, where d > 0 is sufficiently small and will be given below. Lemma 5.1. There holds an inequality ‖F(z2)−F(z1)‖W ≤ C(d + Tκ)‖z2 − z1‖M , (5.10) where κ > 0 is a certain positive number, and the constant C is bounded for bounded d and T . Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 307 B.V. Bazaliy and S.P. Degtyarev It follows from this lemma and from (5.9) that if T and d are sufficiently small, then the mapping z → A−1F(z) maps Md into itself and is a contraction in Md. Hence this mapping has a unique fixed point. This completes the proof of Theorem 2.1. P r o o f of Lemma 5.1. The origin of the estimate (5.10) is that the expression for F(z) = F(u, δ) contains, as it was mentioned, either “quadratic” terms or minor terms. Thus in the factor (d+Tκ) in (5.10) the value d appears as we estimate “quadratic” terms and Tκ appears for minor terms in view of (5.7). As a typical example of calculations in the proof of Lemma 5.1, we show the estimate of the term E(z) ≡ wα − (u + w)α (l − s0 − σ(t)− δ(t))2 ∂2u ∂x2 in (3.7). By the mean value theorem one can write wα − (u + w)α = −αu 1∫ 0 [εw + (1− ε)(u + w)]α−1dε and hence E(z) = (xα ∂2u ∂x2 ) (u x ) α (l − s0 − σ(t)− δ(t))2 × 1∫ 0 [ εw + (1− ε)(u + w) x ]α−1 dε. (5.11) Since α− 1 < 0, some additional arguments are required to show the smoothness of E(z) in z. Taking into account that u(0, t) = w(0, t) = 0, we use the following represen- tation with some sufficiently small a > 0: εw(x, t) + (1− ε)(u(x, t) + w(x, t)) x = {∫ 1 0 [εwx(ωx, t) + (1− ε)(ux(ωx, t) + wx(ωx, t))]dω, 0 ≤ x ≤ a, εw(x,t)+(1−ε)(u(x,t)+w(x,t)) x , x > a, (5.12) where the mean value theorem is applied. By the properties of the initial function u0(x) and w(x, t), wx(0, 0) ≥ ν > 0, u0x(0) ≥ ν > 0, and because u(x, t) ∈ Md, for sufficiently small T and d we get wx(x, t) ≥ ν/2, (wx + ux)(x, t) ≥ ν/2 for 0 ≤ x ≤ a, (5.13) 308 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem and for x ≥ a εw(x, t) + (1− ε)(u(x, t) + w(x, t)) x ≥ ν1, ν1 = const > 0 (5.14) as far as w(x, t) ≥ ν2 for x ≥ a. Moreover, the integrand in (5.12) is bounded from above and the same is true for the left hand side of (5.14). Thus from (5.12), inequalities (5.13), (5.14) and (u, δ) ∈ Md it follows ν2 ≤ εw(x, t) + (1− ε)(u(x, t) + w(x, t)) x ≤ ν−1 2 , ν2 = const > 0. (5.15) We observe also that for small d and T and t ≤ T ν3 ≤ l − s0 − σ(t)− δ(t) ≤ ν−1 3 , ν3 = const > 0. (5.16) So we obtain the representation E(z) = ( xα ∂2u ∂x2 ) (u x ) Φ(z), (5.17) Φ(z) = α (l − s0 − σ(t)− δ(t))2 1∫ 0 [ εw + (1− ε)(u + w) x ]α−1 dε, and from (5.15), (5.16) it follows that Φ(z) is the smooth function in z for z ∈ Md and small t > 0. The difference E(z2)− E(z1), z1, z2 ∈ Md, is evaluated as follows: \|E(z2)−E(z1)\|(β,β/q) GT ≤ ∣∣∣∣ ( xα ∂2u2 ∂x2 − xα ∂2u1 ∂x2 )(u2 x ) Φ(z2) ∣∣∣∣ (β,β/q) GT + ∣∣∣∣ ( xα ∂2u1 ∂x2 )(u2 x − u1 x ) Φ(z2) ∣∣∣∣ (β,β/q) GT + ∣∣∣∣ ( xα ∂2u1 ∂x2 ) (u1 x ) (Φ(z2)− Φ(z1)) ∣∣∣∣ (β,β/q) GT . To continue this estimate we note that u x = 1∫ 0 ux(εx)dε, \|Φ(z)\|(β,β/q) GT ≤ const, z ∈ Md, Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 309 B.V. Bazaliy and S.P. Degtyarev \|Φ(z2)− Φ(z1)\|(β,β/q) GT ≤ const · ‖z2 − z1‖M , z1, z2 ∈ Md. This leads to \|E(z2)−E(z1)\|(β,β/q) GT ≤ const · (‖z2‖M + ‖z1‖M )‖z2 − z1‖M ≤ const · d ‖z2 − z1‖M , z1, z2 ∈ Md. Similar calculations of the rest of the terms in the difference F(z2) − F(z1) prove Lemma 5.1. 6. Proof of Theorem 4.1 In this section we use representation (4.19) to obtain the estimates from The- orem 4.1. First we deduce some properties of the Green function G (x, ξ, t) from (4.16). The Bessel function Iµ (z) has the series expansion Iµ (z) = ∞∑ k=0 1 k!Γ (µ + k + 1) (z 2 )µ+2k , (6.1) where Γ (x) is the Gamma function, and the asymptotic expansion for large z Iµ (z) ∼ ez √ 2πz ( 1 + C z + . . . ) . (6.2) Here and below we will denote by C various positive constants. From these representations it follows that I−1/q ∼ C { z−1/q for z ≤ 1, ezz−1/2 for z > 1. (6.3) Using (6.1)–(6.3), we can get the following estimates: \|G (x, ξ, t)\| ≤ Ct−1/qu−2α/q { e−(u2+v2), uv ≤ 1, e−γ(u−v)2 (uv)1/q−1/2 , uv > 1, (6.4) \|Gt (x, ξ, t)\| ≤ Ct−1−1/qu−2α/q { e−(u2+v2), uv ≤ 1, e−γ(u−v)2 (uv)1/q−1/2 , uv > 1, (6.5) \|Gx (x, ξ, t)\| ≤ Ct−1/q u−2α/q x { e−(u2+v2) ( v2 + u2v2 ) , uv ≤ 1, e−γ(u−v)2 (1 + v) (uv)1/q−1/2 , uv > 1, (6.6) 310 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem \|Gxt (x, ξ, t)\| ≤ Ct−1−1/q u−2α/q x { e−(u2+v2) ( v2 + u2v2 ) , uv ≤ 1, e−γ(u−v)2 (1 + v) (uv)1/q−1/2 , uv > 1, (6.7) \|Gtt (x, ξ, t)\| ≤ Ct−2−1/qu−2α/q { e−(u2+v2), uv ≤ 1, e−γ(u−v)2 (uv)1/q−1/2 , uv > 1, (6.8) where γ is some positive constant. To check, for example, the estimate of Gx (x, ξ, t) for uv ≤ 1 in (6.6) we note that for small z z1/qI−1/q (z) ∼ C ( 1 + 2− α 1− α z2 4 + . . . ) , and vx = Cv/x, zx = Cz/x, where z = 2uv, u, v — from (4.16). Therefore Gx (x, ξ, t) ∼ −Ct−1/qu−2α/q2ve−(u2+v2) v x z1/qI−1/q (z) +Ct−1/qu−2α/qe−(u2+v2) z x d dz ( z1/qI−1/q (z) ) ∼ Ct−1/qu−2α/q 1 x e−(u2+v2) ( v2 + u2v2 ) . Denote J1 (v) = ∞∫ 0 uae−γ(u−v)2du, J2 (v) = ∞∫ 1/2v uae−γ(u−v)2du, J3 (v) = 1/2v∫ 0 uae−γ(u−v)2du. Lemma 6.1. Let a > −1, γ > 0, v ≥ 0. The next inequalities are valid: J1 (v) ≤ C { va for v ≥ 1, 1, for v < 1, (6.9) J2 (v) ≤ C { va for v ≥ 1, e−γ1/v2 , for v < 1, (6.10) J3 (v) ≤ C { e−γ1v2 , for v ≥ 1, 1, for v < 1, (6.11) where 0 < γ1 ≤ γ. Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 311 B.V. Bazaliy and S.P. Degtyarev P r o o f. To prove (6.9) we split the integral J1 (v) J1 (v) =   v/2∫ 0 + 3v/2∫ v/2 + ∞∫ 3v/2   ( uae−γ(u−v)2du ) ≡ A1 (v) + A2 (v) + A3 (v) . (6.12) We have \|u− v\|2 ≥ v2/4 for the integrand in A1 and then A1 (v) ≤ v/2∫ 0 uae−γv2/4du ≤ Cva+1e−γv2/4 ≤ Ce−γv2/8. Since u ∈ (v/2, 3v/2) in A2 (v) , A2 (v) ≤ Cva 3v/2∫ v/2 e−γ(u−v)2du ≤ Cva ∞∫ −∞ e−γ(u−v)2du ≤ Cva. For a ≥ 0 A3 (v) = ∞∫ v/2 (v + z)a e−γz2 dz ≤ C ∞∫ −∞ (va + za) e−γz2 dz ≤ C (va + 1) , and for a ∈ (−1, 0) A3 (v) = ∞∫ v/2 (v + z)a e−γz2 dz ≤ Cva ∞∫ −∞ e−γz2 dz ≤ Cva. The estimates of Ai (v) , i = 1, 2, 3, give (6.9). We first consider the integral J2 (v) for v ≤ 1/2. Then in J2 (v) u − v ≥ 1 2v − v = 1−2v2 2v ≥ 1/4v and J2 (v) = ∞∫ 1/2v uae−γ(u−v)2du ≤ ∞∫ 1/2v uae− γ 2 (u−v)2e− γ 32 1 v2 du ≤ e− γ 32v2 ∞∫ 0 uae− γ 2 (u−v)2du ≤ Ce− γ 32v2 va ≤ Ce−γ1/v2 , where we used estimate (6.9). The estimates for J2 (v) for v > 1 and 1/2 < v < 1 follow from (6.9). 312 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem One can check that for v > 1 and u > 1/2v there is v − u ≥ v/2, then J3 (v) ≤ 1/2v∫ 0 uae−γv2/4du ≤ e−γv2/4 1/2v∫ 0 uadu ≤ Ce−γv2/4. For v ≤ 1 we can estimate J3 (v) by (6.9). This completes the proof of (6.11) and Lemma 6.1. Lemma 6.2. Let x, ξ > 0, q = 2− α, α ∈ (0, 1) . i) For ξ > x ξq/2 − xq/2 ≥ C { (ξ − x) xq/2−1, ξ − x ≤ x, (ξ − x)q/2 , ξ − x > x. (6.13) ii) For ξ < x xq/2 − ξq/2 ≥ C (x− ξ) xq/2−1. (6.14) iii) For δ > 0, 2δ/q < 1, and f (x) ∈ C2δ/q (R+) \|f (x)− f (ξ)\| ≤ C (x) \|f \|(2δ/q) R+ ∣∣∣xq/2 − ξq/2 ∣∣∣ 2δ/q , (6.15) where C(x) = const ·max ( 1, xαδ/q ) . P r o o f. In the case i) let A1 = ξq/2 − xq/2. Successively using the change of variables η = x + z and y = z/ (ξ − x) , we get A1 = 2 q ξ∫ x η 2 q −1 dη = 2 q ξ−x∫ 0 (x + z) 2 q −1 dz = 2 q (ξ − x) 1∫ 0 dy (x + (ξ − x) y)1−q/2 . For ξ − x > x A1 = 2 q (ξ − x)q/2 1∫ 0 dy ( x ξ−x + y )α/2 ≥ 2 q (ξ − x)q/2 1∫ 0 dy (1 + y)α/2 ≥ C (ξ − x)q/2 . For ξ − x ≤ x A1 = 2 q (ξ − x) 1∫ 0 dy xα/2 ( 1 + ξ−x x y )α/2 ≥ C (ξ − x)x−α/2. These estimates prove (6.13). Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 313 B.V. Bazaliy and S.P. Degtyarev In the case ii) let B (x, ξ) = xq/2 − ξq/2. Similarly to the case i), we have B (x, ξ) = 2 q (x− ξ) 1∫ 0 dy xα/2 ( 1− x−ξ x y )α/2 ≥ 2 q (x− ξ) xq/2−1 1∫ 0 dy = C (x− ξ) xq/2−1 and, hence, (6.14). Let F (x, ξ) = \|f (x)− f (ξ)\|∣∣xq/2 − ξq/2 ∣∣2δ/q , ∆f = f (x)− f (ξ) . We prove that F (x, ξ) is bounded if x is bounded. For ξ > x and ξ − x ≤ x by (6.13) F (x, ξ) ≤ C \|∆f \| ( (ξ − x) x−α/2 )2δ/q ≤ Cxαδ/q 〈f〉(2δ/q) x . For ξ > x and ξ − x > x again by (6.13) F (x, ξ) ≤ C \|∆f \| (ξ − x)δ ≤ C 〈f〉(2δ/q) x \|x− ξ\|2δ/q \|x− ξ\|δ = C 〈f〉(2δ/q) x \|x− ξ\|δα/q , and we obtain (6.15) under \|x− ξ\| ≤ 1. For \|x− ξ\| > 1 F (x, ξ) ≤ C max \|f \| . Finally, for ξ < x F (x, ξ) ≤ C \|∆f \| ( (x− ξ) x−α/2 )2δ/q ≤ Cxαδ/q 〈f〉(2δ/q) x . Lemma 6.2 is proved. Lemma 6.3. Let f (x) ∈ C2δ/q (R+) , 2δ/q < 1. Then the following integral exists and lim t→0 ∞∫ 0 G (x, ξ, t) (f (x)− f (ξ)) dξ = 0. (6.16) P r o o f. In the integral K (x, t) = ∞∫ 0 G (x, ξ, t) (f (x)− f (ξ)) dξ 314 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem we change the variable according to ξ = ( qt1/2u )2/q = C (q) t1/qu2/q. Below we will frequently use this change ξ → u. Using (6.15) and (6.4), we get \|f (x)− f (ξ)\| ≤ C (x) \|f \|(2δ/q) R+ \|v − u\|2δ/q tδ/q and K (x, t) ≤ C (x) \|f \|(2δ/q) R+ tδ/q (K1 (x, t) + K2 (x, t)) , (6.17) K1 (x, t) = 1/v∫ 0 e−(u2+v2)u− 2α q + 2 q −1 \|v − u\|2δ/q du, K2 (x, t) = ∞∫ 1/v e−γ(u−v)2 (uv)1/q−1/2 u − 2α q + 2 q −1 \|v − u\|2δ/q du. To estimate K1 (x, t), we use the inequality \|v − u\|2δ/q ≤ C ( v2δ/q + u2δ/q ) and (6.9), then \|K1 (x, t)\| ≤ C    e−v2 1/v∫ 0 e−u2 u −α q + 2δ q du + e−v2 v2δ/q 1/v∫ 0 e−u2 u−α/qdu    ≤ const. In the case of K2 (x, t) , we use (6.10) to obtain \|K2 (x, t)\| ≤ C ∞∫ 1/v e−γ1(u−v)2 (uv)1/q−1/2 u − 2α q + 2 q −1 du = Cvα/2q ∞∫ 1/v e−γ1(u−v)2u−α/2qdu ≤ const. Now Lemma 6.3 follows from (6.17). Lemma 6.4. Let w (x, t) = ∞∫ 0 G (x, ξ, t) f (ξ) dξ, f (x) ∈ C2β/q ( R+ ) , 2β/q < 1, β ∈ (0, 1) . Then 〈w〉(β) x,BR,T + 〈w〉(β/q) t,BR,T ≤ C (R) \|f \|(2β/q) R+ . (6.18) Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 315 B.V. Bazaliy and S.P. Degtyarev P r o o f. Recall the notation: Ω̃T = ([0,∞) ∩BR)T . From (4.17) it follows w (x, t) = ∞∫ 0 G (x, ξ, t) [f (ξ)− f (x)] dξ + f (x) = z (x, t) + f (x) , so that we need to estimate the function z (x, t). First we evaluate the Hölder constant of z (x, t) in t. Let 0 < t < t. We have ∣∣z (x, t)− z ( x, t )∣∣ ≤ ∣∣∣∣∣∣∣ t∫ t zτ (x, τ) dτ ∣∣∣∣∣∣∣ (6.19) and from the representation of z (x, t) and Lemma 6.2 \|zt\| ≤ C (R) \|f \|(2β/q) R+ ∞∫ 0 \|Gt (x, ξ, t)\| ∣∣∣xq/2 − ξq/2 ∣∣∣ 2β q dξ. The change of variable ξ → u leads to \|zt\| ≤ C (R) \|f \|(2β/q) R+ t−1+β/q (K1 (x, t) + K2 (x, t)) ≤ C (R) \|f \|(2β/q) R+ t−1+β/q, where K1 (x, t) , K2 (x, t) were introduced in the proof of Lemma 6.3. Hence, ∣∣z (x, t)− z ( x, t )∣∣ ≤ C (R) \|f \|(2β/q) R+ t∫ t τ−1+β/qdτ ≤ C (R) \|f \|(2β/q) R+ ∣∣t− t ∣∣β/q . (6.20) To estimate the Hölder constant of z(x, t) with respect to x we consider two cases. Let ∆x = x − x > 0 and ∆x ≥ t1/q. By Lemma 6.3 z (x, 0) = 0 and by (6.20), we get \|z (x, t)− z (x, t)\| \|∆x\|β ≤ \|z (x, t)− z (x, 0)\| tβ/q + \|z (x, t)− z (x, 0)\| tβ/q ≤ C (R) \|f \|(2β/q) R+ , (6.21) i.e., we have the required estimate. In the case ∆x < t1/q we consider two possibilities. The first one is ∆x < x/2. Then z (x, t)− z (x, t) = ∞∫ 0 (G (x, ξ, t)−G (x, ξ, t)) (f (ξ)− f (x)) dξ 316 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem +(f (x)− f (x)) ∞∫ 0 G (x, ξ, t) dξ = i1 (x, x, t) + i2 (x, x, t) . Using (4.17), we can evaluate i2 (x, x, t) \|i2 (x, x, t)\| ≤ \|f \| (2β/q) R+ \|x− x\|2β/q = \|f \|(2β/q) R+ \|x− x\|β \|x− x\|αβ/q ≤ \|f \|(2β/q) R+ \|x− x\|β \|∆x\|αβ/q ≤ C (R) \|f \|(2β/q) R+ \|x− x\|β . (6.22) To estimate i1 (x, x, t) we apply the mean value theorem. Let θ ∈ [x, x] , then \|i1 (x, x, t)\| ≤ C (R) \|f \|(2β/q) R+ ∞∫ 0 Gθ (θ, ξ, t) \|∆x\| ∣∣∣xq/2 − ξq/2 ∣∣∣ 2β/q dξ. Due to ∆x < x/2 the values θ, x, x are equivalent. Therefore we change θ by x below. We change ξ to u in the integral and use estimate (6.6) that gives \|i1 (x, x, t)\| ≤ C (R) \|f \|(2β/q) R+ ∆x x tβ/q (i11 (x, x, t) + i12 (x, x, t)) , i11 (x, x, t) = 1/v∫ 0 e−(u2+v2)v2 ( 1 + u2 ) u −α q \|v − u\|2β/q du, i12 (x, x, t) = ∞∫ 1/v e−γ(u−v)2 (1 + v) (uv)1/q−1/2 u −α q \|v − u\|2β/q du. Since e−(u2+v2) \|v − u\|2β/q ≤ Ce−γ(u2+v2), 0 < γ < 1, we get \|i11 (x, x, t)\| ≤ Cv2e−γv2 1/v∫ 0 ( 1 + u2 ) u −α q e−γu2 du, and one can see that either the estimate \|i11 (x, x, t)\| ≤ Cv (6.23) or \|i11 (x, x, t)\| ≤ Cv2 (6.24) is valid. Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 317 B.V. Bazaliy and S.P. Degtyarev Similarly, with Lemma 6.1 \|i12 (x, x, t)\| ≤ C (1 + v) v1/q−1/2 ∞∫ 1/v e−γ(u−v)2u −α q du ≤ C (1 + v) vα/2q { v−α/2q, v ≥ 1, e−δ1/v2 , v < 1, and again \|i12 (x, x, t)\| ≤ Cv, (6.25) or \|i12 (x, x, t)\| ≤ Cv2. (6.26) Now we use estimates (6.23) and (6.25) and can write \|i1 (x, x, t)\| ≤ C (R) \|f \|(2β/q) R+ ∆x x tβ/qv. Note that in the case under consideration v = xq/2/qt1/2 and \|∆x\| \|∆x\|β xq/2 xt1/2−β/q = ∣∣∣∣ ∆x x ∣∣∣∣ 1−q/2 ( ∆x t1/q )q/2−β ≤ const. It means \|i1 (x, x, t)\| ≤ C (R) \|f \|(2β/q) R+ \|∆x\|β . (6.27) Thus for ∆x ≤ x/2 and ∆x ≤ t1/q \|z (x, t)− z (x, t)\| ≤ C (R) \|f \|(2β/q) R+ \|∆x\|β . (6.28) Finally, we consider the case ∆x ≥ x/2 and ∆x ≤ t1/q . We have \|z (x, t)− z (x, t)\| \|∆x\|β ≤ C ( \|z (x, t)− z (0, t)\| xβ + \|z (x, t)− z (0, t)\| xβ ) since 2∆x ≥ x and 3∆x ≥ x. We evaluate, as an example, the second term on the right hand side. To this end we use the inequality \|z (x, t)− z (0, t)\| ≤ ∞∫ 0 \|G (x, ξ, t)−G (0, ξ, t)\| \|f (ξ)− f (x)\| dξ + ∣∣∣∣∣∣ ∞∫ 0 G (0, ξ, t) (f (0)− f (x)) dξ ∣∣∣∣∣∣ = i3 (x, x, t) + i4 (x, x, t) . 318 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem The term i4 (x, x, t) is evaluated by (4.17) and 2∆x ≥ x as follows: \|i4 (x, x, t)\| ≤ \|f (0)− f (x)\| ≤ 〈f〉(2β/q) x,R+ x2β/q = 〈f〉(2β/q) x,R+ xβx2β/q−β ≤ C (R) 〈f〉(2β/q) x,R+ \|∆x\|β . (6.29) The term i3 (x, x, t) is similar to i1 (x, x, t) and to estimate it we use the mean value theorem. Let θ ∈ [0, x] and v = θq/2/qt1/2. In this case we apply inequalities (6.23) and (6.25) and get \|i3 (x, x, t)\| ≤ C (R) \|f \|(2β/q) R+ x θ tβ/qv2 ≤ C (R) \|f \|(2β/q) R+ xβ ≤ C (R) \|f \|(2β/q) R+ \|∆x\|β , (6.30) since x1−β θ tβ/q ( θq/2 t1/2 )2 = x1−βt−1+β/qθq−1 ≤ t 1−β q −1+β q + q−1 q = 1. Inequalities (6.20), (6.21), (6.28), (6.29), and (6.30) lead to inequality (6.18). Lemma 6.4 is proved. R e m a r k 6.1. The reader will see that in our subsequent evaluations in Lemmas 6.5–6.8 we will apply the approach which is analogous to the approach used in the proof of Lemma 6.4. We will use the change of variable ξ → u, split the integration domain according to the small and large values of z = 2uv, and apply the estimates of the Green function. Lemma 6.5. Let w (x, t) = ∞∫ 0 G (x, ξ, t) f (ξ, t) dξ, f (x, t) ∈ Cβ,β/q ( R+ T ) , f (x, 0) = 0. (6.31) Then 〈w〉(β) x,R+ T + 〈w〉(β/q) t,R+ T ≤ C 〈f〉(β/q) t,R+ T . (6.32) P r o o f. First we estimate the Hölder constant of w (x, t) in t. Let 0 < t < t, ∆t = t − t. In the case of ∆t > t/2 it follows that ∆t = t − t > t − 2∆t. Then ∆t > t/3 and hence ∣∣w ( x, t )− w (x, t) ∣∣ \|∆t\|β/q ≤ C (∣∣w ( x, t )∣∣ t β/q + \|w (x, t)\| tβ/q ) . Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 319 B.V. Bazaliy and S.P. Degtyarev Taking into account f (x, 0) = 0, we can estimate the terms on the right hand side. For example, by (4.17) \|w (x, t)\| tβ/q ≤ ∞∫ 0 G (x, ξ, t) \|f (ξ, t)\| tβ/q dξ ≤ 〈f〉(β/q) t,R+ T ∞∫ 0 G (x, ξ, t) dξ ≤ 〈f〉(β/q) t,R+ T . Therefore in this case 〈w〉(β/q) t,R+ T ≤ C 〈f〉(β/q) t,R+ T . (6.33) Let now ∆t < t/2 < t/2. We use the representation w ( x, t )− w (x, t) = ∞∫ 0 G ( x, ξ, t ) [ f ( ξ, t )− f (ξ, t) ] dξ + ∞∫ 0 [ G ( x, ξ, t )−G (x, ξ, t) ] f (ξ, t) dξ = w1 (x, t) + w2 (x, t) . For the function w1 (x, t) we obtain \|w1 (x, t)\| ≤ 〈f〉(β/q) t,R+ T \|∆t\|β/q ∞∫ 0 ∣∣G ( x, ξ, t ) ∣∣ dξ = 〈f〉(β/q) t,R+ T \|∆t\|β/q . (6.34) We apply the mean value theorem to estimate w2 (x, t) and note that estimate (6.5) means \|Gt (x, ξ, t)\| ≤ Ct−1 \|G (x, ξ, t) \| . Moreover, the inequality ∆t < t/2 implies that for any θ ∈ [ t, t ] the values θ, t, and t are equivalent, θ ∼ t ∼ t. Then \|w2 (x, t)\| ≤ C ∆t θ 〈f〉(β/q) t,R+ T tβ/q ∞∫ 0 \|G (x, ξ, θ) \| dξ ≤ C ∆t t 〈f〉(β/q) t,R+ T tβ/q = C 〈f〉(β/q) t,R+ T \|∆t\|β/q \|∆t\|1−β/q t1−β/q ≤ C 〈f〉(β/q) t,R+ T \|∆t\|β/q . (6.35) From (6.33)–(6.35) it follows that 〈w〉(β/q) t,R+ T ≤ C 〈f〉(β/q) t,R+ T (6.36) in all cases. 320 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem Our next step is to evaluate the Hölder constant of w (x, t) with respect to x. Let 0 < ∆x = x− x. For ∆x > t1/q due to w (x, 0) = 0 we obtain \|w (x, t)− w (x, t)\| \|∆x\|β ≤ \|w (x, t)− w (x, 0)\| tβ/q + \|w (x, t)− w (x, 0)\| tβ/q ≤ 2 〈w〉(β/q) t,R+ T ≤ C 〈f〉(β/q) t,R+ T . (6.37) Next we suppose ∆x ≤ t1/q and ∆x ≤ x/2. Consider the difference w (x, t)− w (x, t) = ∞∫ 0 [G (x, ξ, t)−G (x, ξ, t)] f (ξ, t) dξ. (6.38) The mean value theorem gives G (x, ξ, t)−G (x, ξ, t) = Gx (θ, ξ, t) (x− x) , θ ∈ [x, x] , and since θ ∼ x ∼ x for ∆x ≤ x/2 G (x, ξ, t)−G (x, ξ, t) ≤ C \|Gx (x, ξ, t)\| \|∆x\| . We use this inequality in (6.38) and arrive at the situation which we already en- countered in the proof of Lemma 6.4 (see inequalities (6.23) and (6.25)). There- fore \|w (x, t)− w (x, t)\| ≤ C 〈f〉(β/q) t,R+ T ∆x x tβ/q xq/2 t1/2 = C 〈f〉(β/q) t,R+ T \|∆x\|β \|∆x\|1−β x xq/2tβ/q−1/2 = C 〈f〉(β/q) t,R+ T \|∆x\|β ( ∆x x )1−q/2 ( ∆x t1/q )q/2−β ≤ C 〈f〉(β/q) t,R+ T \|∆x\|β . (6.39) In the case ∆x > x/2 and ∆x ≤ t1/q we use the inequality \|w (x, t)− w (x, t)\| \|∆x\|β ≤ C { \|w (x, t)− w (0, t)\| xβ + \|w (x, t)− w (0, t)\| xβ } , and, for instance, the second term on the right hand side is estimated as follows (see the estimate of i3 in the proof of Lemma 6.4): \|w (x, t)− w (0, t)\| xβ ≤ C 〈f〉(β/q) t,R+ T x xβ tβ/q ∞∫ 0 \|Gx (θ, ξ, t)\| dξ ≤ C 〈f〉(β/q) t,R+ T x1−βtβ/q 1 θ ( θq/2 t1/2 )2 ≤ C 〈f〉(β/q) t,R+ T . Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 321 B.V. Bazaliy and S.P. Degtyarev In this case we obtain \|w (x, t)− w (x, t)\| ≤ C 〈f〉(β/q) t,R+ T \|∆x\|β . (6.40) Inequalities (6.36), (6.37), (6.39), and (6.40) give (6.32). Lemma 6.5 is proved. Corollary 6.1. Let f (x, t) ∈ Cβ,β/q ( R+ T ) , f (x, 0) ∈ C2β/q(R+), w (x, t) = ∞∫ 0 G (x, ξ, t) f (ξ, t) dξ. The inequality 〈w〉(β) x,BR,T + 〈w〉(β/q) t,BR,T ≤ C (R) ( 〈f〉(β/q) t,R+ T + \|f (x, 0)\|(2β/q) R+ ) (6.41) is valid. Inequality (6.41) is the consequence of (6.18) and (6.32). Lemma 6.6. (The potential of the initial data) Let w (x, t) = ∞∫ 0 G (x, ξ, t) u0 (ξ) dξ, xαu0xx ∈ C2β/q ( R+ ) . (6.42) Then 〈wt〉(β) x,Ω̃T + 〈wt〉(β/q) t,Ω̃T + 〈wx〉(β+1−α)/q t,Ω̃T ≤ C (R) \|xαu0xx\|2β/q R+ . (6.43) P r o o f. We can consider the case of u0 (0) = u0x (0) = 0 since otherwise we can introduce the function v (x, t) = u (x, t) − u0 (0) − u0x (0)x in problem (4.10)–(4.12). Using (4.18) and integrating by parts, we get wt (x, t) = ∞∫ 0 Gt (x, ξ, t) u0 (ξ) dξ = ∞∫ 0 (ξαG (x, ξ, t))ξξ u0 (ξ) dξ = ∞∫ 0 G (x, ξ, t) ξαu0ξξ (ξ) dξ. Lemma 6.4 gives the estimates of wt(x, t), and the estimate of wx(x, t) can be obtained similarly. 322 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem Lemma 6.7. (The volume potential) Let w (x, t) = t∫ 0 dτ ∞∫ 0 G (x, ξ, t− τ) f (ξ, τ) dξ, f (x, t) ∈ Cβ,β/q ( R+ T ) , (6.44) f (x, 0) ∈ C2β/q ( R+ ) . Then 〈wt〉(β) x,BR,T + 〈wt〉(β/q) t,BR,T + 〈wx〉(β+1−α)/q t,Ω̃T ≤ C (R) ( 〈f〉(β/q) t,R+ T + \|f (x, 0)\|(2β/q) R+ ) . (6.45) P r o o f. First we derive the representation for wt (x, t) . Let wh (x, t) = t−h∫ 0 dτ ∞∫ 0 G (x, ξ, t− τ) f (ξ, τ) dξ, then ∂wh ∂t (x, t) = t−h∫ 0 dτ ∞∫ 0 Gt (x, ξ, t− τ) f (ξ, τ) dξ + ∞∫ 0 G (x, ξ, h) f (ξ, t− h) dξ = t−h∫ 0 dτ ∞∫ 0 Gt (x, ξ, t− τ) [f (ξ, τ)− f (ξ, t)] dξ − ∞∫ 0 dξf (ξ, t) t−h∫ 0 dτGτ (x, ξ, t− τ) + ∞∫ 0 G (x, ξ, h) f (ξ, t− h) dξ = t−h∫ 0 dτ ∞∫ 0 Gt (x, ξ, t− τ) [f (ξ, τ)− f (ξ, t)] dξ − ∞∫ 0 G (x, ξ, h) f (ξ, t) dξ + ∞∫ 0 G (x, ξ, t) f (ξ, t) dξ + ∞∫ 0 G (x, ξ, h) f (ξ, t− h) dξ. Now we go to the limit as h → 0 and get ∂w ∂t (x, t) = t∫ 0 dτ ∞∫ 0 Gt (x, ξ, t− τ) [f (ξ, τ)− f (ξ, t)] dξ + ∞∫ 0 G (x, ξ, t) f (ξ, t) dξ = w1 (x, t) + w2 (x, t) . (6.46) Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 323 B.V. Bazaliy and S.P. Degtyarev The estimate of the function w2 (x, t) is given by (6.41). Since f (ξ, τ)− f (ξ, t) = (f (ξ, τ)− f (ξ, 0))− (f (ξ, t)− f (ξ, 0)) , we can assume that in the representation of w1 (x, t), f (x, t) has the property f (x, 0) = 0. Consider the smoothness of w1 (x, t) with respect to x. Let 0 < ∆x = x − x and ∆x ≤ x/2. We have w1 (x, t)− w1 (x, t) = t∫ t−(∆x)q dτ ∞∫ 0 Gt (x, ξ, t− τ) [f (ξ, τ)− f (ξ, t)] dξ − t∫ t−(∆x)q dτ ∞∫ 0 Gt (x, ξ, t− τ) [f (ξ, τ)− f (ξ, t)] dξ + t−(∆x)q∫ 0 dτ ∞∫ 0 [Gt (x, ξ, t− τ)−Gt (x, ξ, t− τ)] [f (ξ, τ)− f (ξ, t)] dξ = i1 (x, x, t) + i2 (x, x, t) + i3 (x, x, t) . (6.47) The comparison of (6.4) and (6.5) gives \|Gt\| ≤ Ct−1 \|G\| . With this remark and (4.17) we have \|i1 (x, x, t)\| ≤ C 〈f〉(β/q) t,R+ T t∫ t−(∆x)q dτ ∞∫ 0 (t− τ)−1+β/q \|G (x, ξ, t− τ)\| dξ ≤ C 〈f〉(β/q) t,R+ T t∫ t−(∆x)q dτ (t− τ)−1+β/q ≤ C 〈f〉(β/q) t \|∆x\|β . The integral i2 (x, x, t) is evaluated in a similar way. To estimate i3 (x, x, t) , we change the integration variable ξ → u, use inequality (6.7), and note that due to ∆x ≤ x/2 the values θ ∈ [x, x] , x, and x are equivalent. As the result, we obtain \|i3 (x, x, t)\| ≤ C 〈f〉(β/q) t,R+ T \|∆x\| t−(∆x)q∫ 0 (t− τ)β/q dτ ∞∫ 0 \|Gtx (θ, ξ, t− τ)\| dξ 324 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem ≤ C 〈f〉(β/q) t,R+ T \|∆x\| x t−(∆x)q∫ 0 (t− τ)−1+β/q dτe−v2 v2 1/v∫ 0 e−u2 ( 1 + u2 ) u−α/qdu + C 〈f〉(β/q) t,R+ T \|∆x\| x t−(∆x)q∫ 0 (t− τ)−1+β/q dτv1/q−1/2 × (1 + v) ∞∫ 1/v u1/q−1/2−α/qe−γ(v−u)2du. We drop the estimates of the internal integrals (see, for example, the proof of Lemma 6.4) and write down \|i3 (x, x, t)\| ≤ C 〈f〉(β/q) t,R+ T \|∆x\| x t−(∆x)q∫ 0 (t− τ)−1+β/q xq/2 (t− τ)1/2 dτ ≤ C 〈f〉(β/q) t,R+ T \|∆x\| x xq/2 ∞∫ \|∆x\|q y−1−1/2+β/qdy. The last integral exists since, under our assumptions, 1 2 − β q > 0 . Finally we get \|i3 (x, x, t)\| ≤ C 〈f〉(β/q) t,R+ T \|∆x\| x xq/2 \|∆x\|q(−1/2+β/q) = C 〈f〉(β/q) t,R+ T \|∆x\|β ( ∆x x )1−q/2 ≤ C 〈f〉(β/q) t \|∆x\|β . Now let ∆x ≥ x/2. In this case we use the equality w1 (x, t)− w1 (x, t) = [w1 (x, t)− w1 (0, t)] + [w1 (x, t)− w1 (0, t)] . To estimate the second term on the right hand side, we represent it analogously to the way it was done in (6.47) w1 (x, t)− w1 (0, t) = a1 (x, t) + a2 (x, t) + a3 (x, t) . The values of a1 (x, t) , a2 (x, t) are estimated similarly to i1 (x, x, t) , and a3 (x, t) is estimated similarly to i3 (x, x, t) , but in the case of a3 (x, t) we use the inequal- ities in (6.23) and (6.25). We arrive at \|w1 (x, t)− w1 (x, t)\| ≤ C 〈f〉(β/q) t,R+ T \|∆x\|β . Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 325 B.V. Bazaliy and S.P. Degtyarev So, we have proved that 〈w1 (x, t)〉(β) x,R+ T ≤ C 〈f〉(β/q) t,R+ T . (6.48) To evaluate the Hölder constant of the function w1 (x, t) with respect to t we use the representation (0 < ∆t = t− t) w1 ( x, t )− w1 (x, t) = t∫ t−2∆t dτ ∞∫ 0 Gt ( x, ξ, t− τ ) [ f (ξ, τ)− f ( ξ, t )] dξ − t∫ t−∆t dτ ∞∫ 0 Gt (x, ξ, t− τ) [f (ξ, τ)− f (ξ, t)] dξ + t−∆t∫ 0 dτ ∞∫ 0 [ Gt ( x, ξ, t− τ )−Gt (x, ξ, t− τ) ] [f (ξ, τ)− f (ξ, t)] dξ + t−∆t∫ 0 dτ ∞∫ 0 Gt ( x, ξ, t− τ ) [ f (ξ, τ)− f ( ξ, t )] dξ = A1 ( x, t, t ) + A2 ( x, t, t ) + A3 ( x, t, t ) + A4 ( x, t, t ) . (6.49) We can estimate A4 ( x, t, t ) as follows: ∣∣A4 ( x, t, t )∣∣ ≤ ∣∣∣∣∣∣ ∞∫ 0 [ f (ξ, t)− f ( ξ, t )] dξ t−∆t∫ 0 dτGτ ( x, ξ, t− τ ) ∣∣∣∣∣∣ ≤ C 〈f〉(β/q) t,R+ T ∣∣t− t ∣∣β/q ∞∫ 0 [∣∣G ( x, ξ, t )∣∣ + \|G (x, ξ, ∆t)\|] dξ ≤ C 〈f〉(β/q) t,R+ T \|∆t\|β/q . The integrals A1 ( x, t, t ) , A2 ( x, t, t ) are estimated similarly ∣∣A1 ( x, t, t )∣∣ ≤ C 〈f〉(β/q) t,R+ T t∫ t−2∆t ( t− t )−1+β/q dτ ∞∫ 0 ∣∣G ( x, ξ, t− τ )∣∣ dξ 326 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem ≤ C 〈f〉(β/q) t,R+ T t∫ t−2∆t ( t− t )−1+β/q dτ ≤ C 〈f〉(β/q) t,R+ T \|∆t\|β/q . In the integral A3 ( x, t, t ) it is natural to assume ∆t ≤ t. With this condition for any θ ∈ [ t, t ] the values θ, t, t are equivalent, so that if we apply the mean value theorem, we obtain Gt ( x, ξ, t− τ )−Gt (x, ξ, t− τ) = Gtt (x, ξ, θ − τ)∆t ∼ Gtt (x, ξ, t− τ) ∆t. We observe that from (6.8) and (6.4) it follows that \|Gtt (x, ξ, t− τ)\| ≤ C \|t− τ \|−2 \|G (x, ξ, t− τ)\| . Therefore ∣∣A3 ( x, t, t )∣∣ ≤ C 〈f〉(β/q) t,R+ T ∆t t−∆t∫ 0 dτ (t− τ)−2+β/q ≤ C 〈f〉(β/q) t,R+ T ∆t t−∆t∫ −∞ dτ (t− τ)−2+β/q ≤ C 〈f〉(β/q) t,R+ T (∆t)β/q . Thus it is proved that 〈w1 (x, t)〉(β/q) t,R+ T ≤ C 〈f〉(β/q) t,R+ T . (6.50) The inequalities (6.48), (6.50) together with the estimate of the function w2 (x, t) give (6.45) for wt(x, t). The estimate of wx(x, t) is obtained in a similar way. This completes the proof. R e m a r k 6.2. Since the function w (x, t) from (6.44) is a solution of the equation wt − xαwxx = f (x, t) , we get 〈xαwxx〉(β) x,BR,T + 〈xαwxx〉(β/q) t,BR,T ≤ C (R) ( 〈f〉(β/q) t,R+ T + \|f (x, 0)\|(2β/q) R+ ) (6.51) The kernel of the simple layer potential in (4.19) is g (x, t) = G (x, ξ, t) ξα \|ξ=0= Ct−1+1/qe − xq q2t . (6.52) Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 327 B.V. Bazaliy and S.P. Degtyarev Lemma 6.8. (The simple layer potential) Let w (x, t) = t∫ 0 g (x, t− τ) f1 (τ) dτ, f1 (t) ∈ C 1+β−α q ([0, T ]) , f1 (0) = 0. (6.53) Then 〈wt〉(β) x + 〈wt〉(β/q) t + 〈wx〉(β+1−α)/q t ≤ C 〈f1〉 ( 1+β−α q ) t . (6.54) P r o o f. To prove Theorem 4.1 it is sufficient to consider the case of f1 (0) = 0 since otherwise we can introduce the other unknown function as it was proposed in the proof of Lemma 6.7. We represent the derivative wt (x, t) in the form wt (x, t) = t∫ 0 gt (x, t− τ) f1 (τ) dτ = t∫ 0 gt (x, t− τ) [f1 (τ)− f1 (t)] dτ − f1 (t) t∫ 0 gτ (x, t− τ) dτ = t∫ 0 gt (x, t− τ) [f1 (τ)− f1 (t)] dτ − f1 (t) g (x, t) = w1 (x, t) + w2 (x, t) . (6.55) For the function w2 (x, t) the Hölder constant in t is obtained as follows. We represent w2 (x, t) as w2 (x, t) = g1 (t) g2 (x, t) , g1 (t) = f1 (t) /t(1−α)/q, g2 (x, t) = e − xq q2t and use the inequality 〈w2 (x, t)〉(β/q) t ≤ 〈g1 (t)〉(β/q) t max \|g2 (x, t)\|+ max \|g1 (t)\| 〈g2 (x, t)〉(β/q) t . (6.56) The value 〈g2 (x, t)〉(β/q) t can be evaluated as follows: 〈g2 (x, t)〉(β/q) t ≤ t1−β/q max \|g2t\| ≤ Ct1−β/qt−1 = Ct −β q . Since g1 (t) ≤ 〈f1〉 ( 1+β−α q ) t tβ/q, max \|g1 (t)\| 〈g2 (x, t)〉(β/q) t ≤ C 〈f1〉 ( 1+β−α q ) t . 328 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem The term 〈g1 (t)〉(β/q) t max \|g2 (x, t)\| is estimated similarly, thus 〈w2 (x, t)〉(β/q) t ≤ C 〈f1〉 ( 1+β−α q ) t . (6.57) For the function w1 (x, t) we consider the difference w1 ( x, t )− w1 (x, t) = t∫ t−2∆t gt ( x, t− τ ) [ f1 (τ)− f1 ( t )] dτ − t∫ t−∆t gt (x, t− τ) [f1 (τ)− f1 (t)] dτ + t−∆t∫ 0 [ gt ( x, t− τ )− gt (x, t− τ) ] [f1 (τ)− f1 (t)] dτ + [ f1 (t)− f1 ( t )] t−∆t∫ 0 gt ( x, t− τ ) dτ = 4∑ k=1 bk ( x, t, t ) . (6.58) The estimate of b1 ( x, t, t ) has the form ∣∣b1 ( x, t, t )∣∣ ≤ C 〈f1〉 ( 1+β−α q ) t t∫ t−2∆t ( t− τ )−2+ 1 q + 1+β−α q e − xq q2(t−τ) dτ ≤ C 〈f1〉 ( 1+β−α q ) t 2∆t∫ 0 y −2+ 2 q +β−α q dy ≤ C 〈f1〉 ( 1+β−α q ) t \|∆t\|β/q . The integral b2 ( x, t, t ) is estimated in the same way. It is natural to assume that ∆t ≤ t in b3 ( x, t, t ) , otherwise this integral is absent. The application of the mean value theorem leads to (γ = const > 0) ∣∣b3 ( x, t, t )∣∣ ≤ C 〈f1〉 ( 1+β−α q ) t \|∆t\| t−∆t∫ 0 (t− τ)−3+ 1 q + 1+β−α q e −γ xq q2(t−τ) dτ ≤ C 〈f1〉 ( 1+β−α q ) t \|∆t\| t∫ ∆t y −3+ 1 q + 1+β−α q dy ≤ C 〈f1〉 ( 1+β−α q ) t \|∆t\|β/q . Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 329 B.V. Bazaliy and S.P. Degtyarev Finally, ∣∣b4 ( x, t, t )∣∣ ≤ C 〈f1〉 ( 1+β−α q ) t \|∆t\| 1+β−α q t−∆t∫ 0 (t− τ)−2+1/q dτ ≤ C 〈f1〉 ( 1+β q ) t \|∆t\| 1+β−α q ∞∫ ∆t y −2+ 1 q dy ≤ C 〈f1〉 ( 1+β−α q ) t \|∆t\|β/q . As the result, we have shown that 〈wt〉(β/q) t ≤ C 〈f1〉 ( 1+β−α q ) t . (6.59) To estimate the Hölder constant for wt (x, t) in x we set 0 < ∆x = x−x and first consider the case of ∆x ≥ t1/q. Since wt (x, 0) = 0, we can write \|wt (x, t)− wt (x, t)\| \|∆x\|β ≤ \|wt (x, t)\| tβ/q + \|wt (x, t)\| tβ/q ≤ C 〈f1〉 ( 1+β−α q ) t by using (6.59). Now we consider the case of ∆x < t1/q, first for the function w2 (x, t) . By the mean value theorem g (x, t)− g (x, t) = Ct−1+1/q [ e − xq q2t − e − xq q2t ] = Ct−1+1/q∆x θq−1 t e − θq q2t , θ ∈ [x, x] , so that \|w2 (x, t)− w2 (x, t)\| ≤ C 〈f1〉 ( 1+β−α q ) t ∆xt −1+ 1+β−α q ≤ C 〈f1〉 ( 1+β−α q ) t \|∆x\|β t −1+ 1+β−α q + 1−β q ≤ C 〈f1〉 ( 1+β−α q ) t \|∆x\|β . For the function w1 (x, t) we study the difference w1 (x, t)− w1 (x, t) = t∫ t−(∆x)q gt (x, t− τ) (f1 (τ)− f1 (t)) dτ − t∫ t−(∆x)q gt (x, t− τ) (f1 (τ)− f1 (t)) dτ 330 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4 Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem + t−(∆x)q∫ 0 [gt (x, t− τ)− gt (x, t− τ)] (f1 (τ)− f1 (t)) dτ = 3∑ k=1 ak (x, x, t) . It is easy to estimate a1 (x, x, t) and a2 (x, x, t) , for example, taking into account ∆x < t1/q, \|a1 (x, x, t)\| ≤ 〈f1〉 ( 1+β−α q ) t t∫ t−(∆x)q (t− τ) 1+β−α q −2+ 1 q e − xq q2(t−τ) dτ ≤ C 〈f1〉 ( 1+β−α q ) t (∆x)q∫ 0 y 1+β−α q −2+ 1 q dy ≤ C 〈f1〉 ( 1+β−α q ) t \|∆x\|β . Finally we estimate a3 (x, x, t) by the mean value theorem \|a3 (x, x, t)\| ≤ C 〈f1〉 ( 1+β−α q ) t ∆x t−(∆x)q∫ 0 (t− τ)−2+ 1+β−α q e −γ θq q2(t−τ) dτ ≤ C 〈f1〉 ( 1+β−α q ) t ∆x ∞∫ (∆x)q y −2+ 1+β−α q dy ≤ C 〈f1〉 ( 1+β−α q ) t ∆x 1+q ( −1+ 1+β−α q ) ≤ C 〈f1〉 ( 1+β−α q ) t \|∆x\|β . Thus 〈wt〉(β) x ≤ C 〈f1〉 ( 1+β−α q ) t . (6.60) The estimates of the function wt(x, t) are proved by inequalities (6.59), (6.60). The estimate of wx(x, t) is obtained in a similar way. Theorem 4.1 is the consequence of Lemmas 6.6–6.8. References [1] B.V. Bazaliy and S.P. Degtyarev, Solvability of a Problem with an Unknown Bound- ary Between the Domains of a Parabolic and an Elliptic Equations. — Ukr. Math. J. 41 (1989), No. 10, 1155–1160. 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Mannucci and J.L. Vazquez, Viscosity Solutions for Elliptic-Parabolic Problems. — Nonlinear Differ. Equ. Appl. 14 (2007), Nos. 1–2, 75–90. 332 Journal of Mathematical Physics, Analysis, Geometry, 2011, vol. 7, No. 4

Classical Solution of a Degenerate Elliptic-Parabolic Free Boundary Problem

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