Journal of Mathematical Sciences, Vol. 109, No. 5, 2002
L p ESTIMATES FOR THE SOLUTION OF THE MODEL VERIGIN PROBLEM E. V. Frolova
UDC 517.9
In this paper, we consider a model free-boundary problem related to the Verigin problem. Lp estimates of solutions are obtained with the help of results on Fourier multipliers. These estimates can be used to prove the solvability of the Verigin problem in Sobolev functional spaces. Bibliography: 14 titles.
The Verigin problem that describes the process of forcing a liquid in a porous medium is a free boundary problem. The unknowns are the pressure of the forcing liquid, the pressure of the displaced liquid, and the surface that separates these liquids. The behavior of the free boundary is subject to the Stefan condition. Also, we assume that the conormal derivative of pressure has zero jump. The classical solvability of the Verigin problem is proved in [1–3] with the help of a coordinate transform procedure suggested by E. I. Hanzawa [4]. By this transform, the free boundary problem is reduced to the initial boundary-value problem for a nonlinear parabolic equation in a fixed boundary domain. To prove the solvability of a linearized problem, a method based on constructing a regularizer [5] is used. This method requires an analysis of model “half-space” problems. The model problem related to the Verigin problem was analyzed in H¨ older function spaces by E. V. Radkevich [1], B. V. Bazaliy and S. P. Degtyarev [3], and in weighted H¨ older function spaces by G. I. Bizhanova and V. A. Solonnikov [6]. To prove that the Verigin problem is uniquely solvable in anisotropic Sobolev spaces, it is convenient to use the modification of the Hanzawa transform proposed in [7]. In this case, a model problem arises, which only slightly differs from the one considered in [1–3, 6]. This paper is concerned with Lp -estimates for the solution of this model problem. The proof is based on a theorem on Fourier multipliers [8]. This approach was used by L. R. Volevich [9] in the analysis of elliptic problems. Later, it was applied to the nonstationary Navier–Stokes system (see Mogilevskii [10]). Lp -estimates for the solution of a model Stefan problem were obtained in [11–13]. Let us introduce the following notation: R1 = Rn− = {x ∈ Rn | xn < 0}, Γ = {x ∈ Rn | xn = 0}, x = (x0 , xn ),
R2 = Rn+ = {x ∈ Rn | xn > 0},
RjT = Rj × [0, T ), j = 1, 2, ΓT = Γ × [0, T ), where x0 = (x1 , . . . , xn−1 ) ∈ Rn−1 .
The problem is to find two pairs of functions u (1) and Φ(1) and u(2) and Φ(2) defined on the half-spaces R1 and R2 . These functions satisfy the relations (k)
ut
−
n X
(k)
aij u(k) xi xj = 0,
x ∈ Rk , t > 0,
i,j=1
Translated from Zapiski Nauchnykh Seminarov POMI, Vol. 259, 1999, pp. 280–295. Original article submitted April 19, 1999. 2018
c 2002 Plenum Publishing Corporation 1072-3374/02/1095-2018 $27.00
(k) Φt
−
n X
(k)
x ∈ Rk , t > 0, k = 1, 2,
aij Φ(k) xi xj = 0,
i,j=1
u(1) xn =0 = u(2) xn =0 , µΦt + d1
Φ(1) xn =0 = Φ(2) xn =0 ,
∂u(1) ~ + b1 · ∇Φ(1) xn =0 = ψ(x0 , t), ∂xn
(1)
∂u(1) ∂u(2) ~ − d2 + b1 · ∇Φ(1) − ~b2 · ∇Φ(2) xn =0 = ϕ(x0 , t), ∂xn ∂xn u(1) t=0 = 0, u(2) t=0 = 0, Φ(1) t=0 = 0, Φ(2) t=0 = 0,
d1
(k)
where (aij ) i=1,... ,n are real symmetric matrices j=1,... ,n n X
(k)
aij ξi ξj ≥ γ|ξ|2 ,
∀ξ ∈ Rn , γ > 0
i,j=1 j (without loss of generality one may assume that a(1) ij = α1 δi (α1 > 0)). Formally applying the Fourier–Laplace transform Z +∞ Z 0 1 F u(ξ, xn , s) = u ˜(ξ, xn , s) = √ dt u(x0 , xn , t)e−ts+ix ·ξ dx0 , ( 2π)n−1 0 Rn−1
(2)
we convert problem (1) to a system of ordinary differential equations: (k)
se u
−
d2 u e(k) a(k) nn dx2n
n−1 X
n−1 X (k) de u(k) +2 ajn ξj i+ aij ξj ξj u e(k) = 0, dx n j=1 i,j=1
n−1 n−1 2 e (k) X X (k) de u(k) (k) (k) d Φ e + 2 a ξ i + aij ξi ξj u e(k) = 0, sΦ − ann jn j dx2n dx n j=1 i,j=1
where xn < 0 for k = 1, xn > 0 for k = 2, u e(1) xn =0 = u e(2) xn =0 , e (1) + d1 µsΦ d1
(3)
e (1) e (2) Φ =Φ , xn =0 xn =0
e (1) de u(1) dΦ e (1) e s), + b1n − i~b 01 · ξ Φ = ψ(ξ, xn =0 dxn dxn
e (1) e (2) de u(1) dΦ u(2) dΦ e (1) − d2 de e (2) + b1n − i~b 01 · ξ Φ − b2n + i~b 02 · ξ Φ = ϕ(ξ, ˜ s). xn =0 dxn dxn dxn dxn
Solutions of system (3) that vanish as |xn | → +∞ have the form u e(1) = C1 er1 xn ,
u e(2) = C3 e−r2 xn ,
e (1) = C2 er1 xn , Φ
e (2) = C4 e−r2 xn , Φ
(4)
2019
where r1 and −r2 are the roots of the quadratic equations n−1 n−1 X (k) X (k) 2 a(k) ajn ξj λ − s + aij ξi ξj = 0, k = 1, 2, nn λ − 2i j=1
i,j=1
and Re(r1 ) > 0 and Re(r2 ) > 0. Therefore, Pn−1 r1 =
(1) j=1 ajn ξj i (1) ann
q +
D (1) 4
,
r2 =
−
where (k)
D 4
n−1 X
= a(k) nn s + ann
Pn−1
(2) j=1 ajn ξj i (2) ann
aij ξi ξj − (k)
i,j=1
n−1 X
q +
D (2) 4
,
(5)
2 ajn ξj . (k)
j=1
From the continuity conditions, it follows that C1 = C3 and C2 = C4 . Substituting the functions determined by relations (4) into the boundary conditions of problem (3), we arrive at the following system of linear equations for the coefficients C1 and C2 : ( e s) d1 r1 C1 + (µs + b1n r1 − i~b 01 · ξ)C2 = ψ(ξ, ˜ s). (d1 r1 + d2 r2 )C1 + (b1n r1 + b2n r2 − i~b 0 · ξ + i~b 0 · ξ)C2 = ϕ(ξ, 1
2
From these equations we find C1 =
µs + b1n r1 − i~b 01 · ξ b1n r1 + b2n r2 + i(~b 02 − ~b 01 ) · ξ e ϕ(ξ, e s) − ψ(ξ, s), µ(d1 r1 + d2 r2 )(s + D) µ(d1 r1 + d2 r2 )(s + D) C2 = −
where
(6)
d1 r1 1 e s), ϕ(ξ, e s) + ψ(ξ, µ(d1 r1 + d2 r2 )(s + D) µ(s + D)
r1 r2 d1 r1 1 0 0 0 ~ ~ ~ (d2 b1n − d1 b2n ) − ib 1 · ξ + i(b 1 − b 2 ) · ξ D= µ d1 r1 + d2 r2 d1 r1 + d2 r2
(7)
2 2 1 d1 |r1 |2 r2 + d2 |r2 |2 r1 d |r | + d d r r ¯ 1 1 2 1 2 1 0 0 0 = (d2 b1n − d1 b2n ) − i~b 1 · ξ + i(~b 1 − ~b 2 ) · ξ . µ |d1 r1 + d2 r2 |2 |d1 r1 + d2 r2 |2 Thus, we have explicit formulas for the solution of problem (3). Applying some results on Fourier multipliers [8, 9] and using these formulas, we obtain L p -estimates for the solution of problem (1). m,m/2 As usual, we denote by Wp (Rk,T ) (m/2 ∈ N) the space of functions u(x, t) that are defined in Rk,T , possess the generalized derivatives Dta Dxα u(x, t) (where α = (α1 , . . . , αn ), a, αi ∈ N ∪ {0}, |α| + 2a ≤ m), and are such that the norm kukp,Rk,T = (m)
X |α|+2a≤m
2020
Z
T 0
Z Rk
1/p |Dxα Dta u(x, t)|p dx dt
m,m/2
is finite (for the definition of the norm in the space Wp for m/2 6∈ N, see, e.g., [5]). m−1/p,m/2−1/2p m,m/2 Thus, Wp (ΓT ) is the space of traces of functions from the space Wp (Rk,T ) m,m/2 m−1/p,m/2−1/2p (m ≥ 1) on ΓT . We denote by W p (Rk,T )(W p (ΓT )) the subspace of the m,m/2 (Rk,T ) Wp
space conditions.
0 m−1/p,m/2−1/2p (Wp (ΓT ))
0
that contains the functions satisfying zero initial
Theorem 1. Let the coefficients in problem (1) satisfy the conditions d2 b1n − d1 b2n > 0,
µ > 0,
d1 > 0,
d2 > 0,
(8)
d2 b1n − d1 b2n k~b 02 − ~b 01 kp ≤ , 2C ∗
(9)
where the constant C ∗ is chosen so that estimates (26) take place. m+1−1/p,m/2+1/2−1/2p Then, for any ϕ, ψ ∈ W p (ΓT ), there exists a unique solution of problem 0
(1) u(k) ∈ W m+2,m/2+1 (Rk,T ), p 0
Φ(k) ∈ W m+2,m/2+1 (Rk,T ) p 0
and for this solution the estimate X (m+2) (m+2) (m+1−1/p) (m+1−1/p) ku(k) kp,Rk,T + kΦ(k) kp,Rk,T ≤ c kϕkp,ΓT + kψkp,ΓT
(10)
k=1,2
holds. Proof. We extend the known functions in problem (1) to the set Γ∞ . Let ϕ1 , ψ1 ∈ W m+1−1/p,m/2+1/2−1/2p (Γ∞ ) p 0
satisfy the conditions
ϕ1 t≤T = ϕ,
ϕ1 t≥T +ε = 0, and
(m+1−1/p)
kψ1 kp,Γ∞
ψ1 t≤T = ψ,
ψ1 t≥T +ε = 0,
(m+1−1/p)
≤ ckψkΓT
,
ε > 0,
(m+1−1/p)
kϕ1 kp,Γ∞
(m+1−1/p)
≤ ckϕkΓT
.
Let us extend all the functions to t < 0 by assigning them zero values. Now, we can use the Fourier–Laplace transform (2) and pass to problem (3). For further analysis, it is convenient to introduce the following quantities: A1 =
1 , d1 r1 + d2 r2
A2 =
b1n r1 − i~b 01 · ξ , µ(d1 r1 + d2 r2 ) (11)
A3 = −
b1n r1 + b2n r2 + i(~b 02 − ~b 01 ) · ξ , µ(d1 r1 + d2 r2 )
A4 = −
d1 r1 . µ(d1 r1 + d2 r2 ) 2021
˜ (k) of problem (3) can be rewritten in the form According to (4), (6), the solution u ˜(k) , Φ A1 s + A2 A3 e ϕ e1 (ξ, s)e−rk |xn| + ψ1 (ξ, s)e−rk |xn| , s+D s+D A4 1 = ϕ e1 (ξ, s)e−rk |xn | + ψe1 (ξ, s)e−rk |xn | . s+D µ(s + D)
u e(k) = e (k) Φ
(12)
The solution u(k) , Φ(k) , k = 1, 2, of problem (1) can be found with the help of the inverse Fourier–Laplace transform. The uniqueness of the solution of problem (1) can be proved by arguments similar to those presented in [14] (see Theorem 11.1) for parabolic problems. Now, we verify that this solution satisfies estimate (10). The proof is based on the P. I. Lizorkin theorem on Fourier multipliers [8], which generalizes the results of S. G. Mikhlin. Also, our proof uses the results of Volevich [9]. For the sake of further application, we formulate these theorems in terms of the present paper (ξn ←→ s). It is worth noting that the presence of the weight multiplier e−s1 t in our estimates is stipulated by the use of the Laplace transform with respect to the variable t instead of the Fourier transform. Definition. A function g(ξ, s), ξ ∈ Rn−1 , s = s1 + is2 , s2 ∈ R, s1 ≥ 0, is called a Fourier multiplier of the class (p, p) if for any v ∈ C0∞ the estimate ke−ts1 F −1 g(ξ, s)F vkp ≤ cke−ts1 vkp ,
1 ≤ p < +∞,
holds, where F −1 denotes the integral transform inverse to the Fourier–Laplace transform defined by (2). Let us introduce the notation Mpp (g) = sup
v∈C0∞
ke−ts1 F −1 g(ξ, s)F vkp . ke−ts1 vkp
Theorem 2 ([9]). Let η(x, t) ∈ Lp (Rn+,∞ ), let N (ξ, xn , s) be a Fourier multiplier of the class (p, p), and let the estimate c Mpp (N (·, xn , ·) ≤ (13) xn hold. Then, for the function Z +∞ Hη = F −1 [N (ξ, xn + y, s)F η(ξ, y, s)]dy 0
the estimate
ke−s1 t Hηkp,Rn+,∞ ≤ cke−s1 t ηkp,Rn+,∞
is true. Remark. Obviously, Theorem 2 can be reformulated for the function Z 0 Hη = F −1 [N (ξ, xn + y, s)F η(ξ, y, s)]dy. −∞
2022
(14)
Theorem 3 ([8]). Assume that a function g(ξ, s) (ξ ∈ Rn−1 , s = s1 + is2 , s1 ≥ 0, s2 ∈ R) and the mixed derivatives ∂ l g(ξ, s) ∂ l g(ξ, s) , , ∂ξk1 . . . ∂ξkl ∂ξk1 , . . . , ∂ξkl−1 ∂s where l ≥ 1, ki 6= kj for i 6= j, are continuous for |s| + ξ 2 > 0 and satisfy the estimates ξk1 . . . ξkl
∂ l g(ξ, s) ≤ M, ∂ξk1 . . . ∂ξkl
l g(ξ, s) ∂ ξk1 . . . ξkl−1 s ≤ M, ∂ξk1 . . . ∂ξkl−1 ∂s
(15.1)
M > 0.
(15.2)
Then, the function g(ξ, s) is a Fourier multiplier of the class (p, p) and Mpp(g) ≤ CM. Based on Theorem 3, we can prove the following lemma. Lemma. Let w(ξ, s), r(ξ, s), and B(ξ, s) (|s| + ξ2 > 0, Re s > 0) be smooth functions that satisfy the following homogeneity conditions: B(λξ, λ2 s) = λB(ξ, s),
r(λξ, λ2 s) = λr(ξ, s),
λ>0
(i.e., they are homogeneous functions of order 1), and w(λξ, λ2 s) = λ2 w(ξ, s), λ > 0 (i.e., w is a homogeneous function of order 2). Assume that they are subject to the following inequalities:
Then, the function R = the estimate
Re(r(ξ, s)) ≥ c|r(ξ, s)| ≥ c(|s| + ξ2 )1/2 ,
(16)
Re(B(ξ, s)) ≥ c|B(ξ, s)| ≥ c(|s| + ξ2 )1/2 .
(17)
w −rx , s+B e
where x > 0, is a Fourier multiplier of the class (p, p) and Mpp (R) ≤
c x
(18)
holds. Proof of the lemma. The continuity of the function R(ξ, s) and its mixed derivatives for |s| + ξ 2 > 0 directly follows from the assumptions of the lemma. Thus, it is sufficient to prove estimates (15) with M = Cx . In what follows, we denote any homogeneous function of order k by U{k} . By Dξα , where l
α = (α1 , ..., αn−1), αi ∈ {0, 1}, i = 1, . . . , n − 1, we denote the differential operator ∂ξk ∂...∂ξk , 1 l l = |α|. 1 The derivative Dξα R is a linear combination of the functions Dξβ1 w Dξβ2 s+B Dξβ3 e−rx , where |β1 | + |β2 | + |β3 | = l. Since w is assumed to be a homogeneous function of the second order, while B and r are assumed to be homogeneous functions of the first order, the above derivative is a linear combination of the functions U{2−|β1 |} U{m−|β2 |}
1 xj U{j−|β3 |} e−rx , m+1 (s + B) 2023
where 0 < m ≤ |β2 |, 0 < j ≤ |β3 | for |β2 |, |β3 | > 0, m = 0 for |β2 | = 0, and j = 0 for |β3 | = 0. Since the function B satisfies estimate (17), for Re s > 0 we have 1 1 1 . ≤ ≤c |s + B| Re(s + B) (|s| + ξ2 )1/2 Multiplying Dξα R by ξk1 . . . ξkl , we arrive at the estimate |ξk1 . . . ξkl Dξα R(ξ, s)|
≤c
l l−m X X m=0 j=0
≤c
l l−m X X
|ξ| U{2+m+j−l} l
|ξ|l−2−m−j |U{2+m+j−l} |
m=0 j=0
1 j −rx x e (s + B)m+1
|ξ|m+1 xj |ξ|j+1 e− Re rx 2 (m+1)/2 (|s| + ξ )
l X ≤c (|s| + ξ 2 )(j+1)/2 xj e− Re rx .
(19)
j=0
Now, we apply a method similar to what was used in [10] for deriving analogous estimates. We use the inequality c y j e−yz ≤ j , j ∈ N, y > 0, z > 0, z from which it follows that (Re r)j xj−1 e− Re rx ≤
c , x
j ∈ N.
(20)
In view of (16), we estimate the right-hand side of inequality (19) with the help of (20) and arrive at (15.1): c |ξk1 . . . ξkl Dξα R(ξ, s)| ≤ . x Let us now proceed to the proof of (15.2). We see that s
∂R ∂ =s ∂s ∂s
=
w −rx e = s+B
! ∂B ∂r sw 1 + s ∂w swx ∂s ∂s ∂s − − e−rx 2 s+B (s + B) s+B
U{2} −rx U{3} U{3} x −rx −rx e + e + e . s+B (s + B)2 s+B
The first term has the same structure as R and, therefore, for thisterm estimate (15.1) −rx U e {3} holds. To estimate the derivatives of the second term, we represent Dξα (s+B)2 as a linear combination of the functions 1 β1 β2 Dξ U{3} Dξ Dξβ3 e−rx , (s + B)2 2024
where |β1 | + |β2 | + |β3 | = l. By virtue of the homogeneity of the functions B and r, we arrive at the estimate −rx U e {3} α ξk1 . . . ξkl D ξ (s + B)2 ≤c
l l−m X X
|ξ|l |U{3−|β1 |} U{m−|β2 |}
m=0 j=0
≤c
l l−m X X
|ξ|
l−3−m−j
m=0 j=0
1 xj U{j−|β3 |} e−rx | (s + B)m+2
|ξ|m+2 |U{3+m+j−l} | xj |ξ|j+1 e− Re rx 2 m/2+1 (|s| + ξ )
l X c ≤c (|s| + ξ 2 )(j+1)/2 xj e− Re rx ≤ . x j=0
The derivatives of the third term are estimated in a similar way: ξk1 . . . ξkl Dξα U{3} x e−rx s+B ≤c
l l−m X X
|ξ|l |U{3+m+j−l}
m=0 j=0
≤c
1 (|s|
+ ξ 2 )(m+1)/2
xj+1 e− Re rx
l X c (|s| + ξ 2 )(j+2)/2 xj+1 e− Re rx ≤ . x j=0
Thus, estimates (15.1) are valid for the mixed derivatives of the function s ∂R ∂s . This means that C for the function R estimates (15.2) hold with M = x . By Theorem 3, the function R(ξ, s) is a Fourier multiplier of the class (p, p) and Mpp (R) ≤ Cx . Now, we continue the proof of Theorem 1. m+1,m/2+1/2 (Rk,∞ ) be extensions of the functions ϕ1 and ψ1 on the halfLet Gk , Hk ∈ W p 0
spaces Rk,∞ , k = 1, 2, and also (m+1−1/p) kGk k(m+1) , p,Rk ,∞ ≤ ckϕ1 kp,Γ∞ (m+1)
(m+1−1/p)
kHk kp,Rk,∞ ≤ ckψ1 kp,Γ∞
(21)
.
Applying the Newton–Leibnitz formula to (12), we deduce the integral representations for u e(k) , e (k) . For example, Φ A1 s + A2 e A3 e k k −rk (|xn |+|y|) =− Gk (ξ, (−1) y, s) + Hk (ξ, (−1) y, s) e dy s+D s+D 0 Z +∞ A1 s + A2 e A3 e k k k = (−1) rk Gk (ξ, (−1) y, s) + Hk (ξ, (−1) y, s) e−rk (|xn|+|y|) dy s+D s+D 0 Z +∞ A1 s + A2 ∂ e A3 ∂ e k k − Gk (ξ, (−1) y, s) + Hk (ξ, (−1) y, s) e−rk (|xn|+|y|) dy. s + D ∂y s + D ∂y 0 Z
+∞
∂ ∂y
u e(k) (ξ, xn , s)
2025
Consequently, for j = 2, . . . , m + 2 we have F
−1
Z (rkj F u(k) )
+∞
=
F
−1
0
e A1 s + A2 2 −rk (|xn |+|y|) j−2 ∂ Gk k j−1 e (−1) rk Gk − rk rk e s+D ∂y
e A3 2 −rk (|xn|+|y|) j−2 ∂ Hk k j−1 e (−1) rk Hk − rk + r e dy. s+D k ∂y
(22)
Similarly, F
−1
Z (rkj F Φ(k) )
+∞
=
F
−1
0
e A4 2 −rk (|xn |+|y|) j−2 ∂ Gk k j−1 e (−1) rk Gk − rk r e s+D k ∂y
e 1 j−2 ∂ Hk 2 −rk (|xn|+|y|) k j−1 e + (−1) rk Hk − rk r e dy. µ(s + D) k ∂y
(23)
Let us check that the functions A1 s + A2 2 −rk |xn | A3 2 −rk |xn | rk e , r e , s+D s+D k A4 2 −rk |xn | 1 rk e r2 e−rk |xn | , s+D µ(s + D) k
(24)
are Fourier multipliers of the class (p, p) and that estimate (13) for Mpp (·) holds. For this purpose, we show that these functions satisfy the conditions of the above lemma. It is clear that the functions rk (ξ, s), k = 1, 2, and D(ξ, s) are homogeneous of order 1 and the functions A2 rk2 , A3 rk2 , and A4 rk2 are homogeneous of order 2 and smooth for |s| + ξ 2 > 0 (see (5), (7), (11)). P One can show that the positive definiteness of the quadratic form ni,j=1 aij ξi ξj implies that (k) is positive for Re s > 0 and, also, rk , k = 1, 2, satisfy estimate (16). Re D4 By formula (7), Re D = I1 + I2 , where I1 =
(d2 b1n − d1 b2n ) (d1 |r1 |2 Re r2 + d2 |r2 |2 Re r1 ) , · µ |d1 r1 + d2 r2 |2 I2 = −
d1 d2 ~ 0 ~ 0 Im(r1 r¯2 ) (b 1 − b 2 ) · ξ . µ |d1 r1 + d2 r2 |2
Condition (8) of Theorem 1 and inequality (16) for rk imply the estimate I1 ≥ c(Re r1 + Re r2 ) ≥ c(|s| + ξ 2 )1/2 .
(25)
The term I2 is not positive definite and can be estimated using the fact that the norm 0 ~ kb 1 − ~b 02 k is small. As is follows from (16) for rk , there exists a positive constant C ∗ such that d2 |ξ| ≤ C ∗ |r1 | 2026
d1 |ξ| ≤ C ∗ |r2 |.
(26)
Thus, if condition (9) of Theorem 1 holds, then we arrive at the estimate |I2 | ≤ k~b 01 − ~b 02 k
d1 d2 |ξ| (|r1 | Re r2 + |r2 | Re r1 ) µ|d1 r1 + d2 r2 |2
C∗ I1 2 2 (d |r | Re r + d |r | Re r ) ≤ . 1 1 2 2 2 1 µ|d1 r1 + d2 r2 |2 2
≤ k~b 01 − ~b 02 k
(27)
Therefore, we see that Re D ≥ I1 − |I2 | ≥
I1 ≥ c(|s| + ξ 2 )1/2 . 2
(28)
Taking into account (16), (25), and (27) we estimate the function Im D: | Im D| ≤
+
d2 b1n − d1 b2n 1 (d1 |r1 |2 | Im r2 | + d2 |r2 |2 | Im r1 |) + k~b 01 k|ξ| 2 µ|d1 r1 + d2 r2 | µ
k~b 01 − ~b 02 kt|ξ| 2 (d1 |r1 |2 + d1 d2 | Re(r1 r¯2 )|) ≤ c(I1 + k~b 01 k |ξ|) ≤ c Re D. 2 µ|d1r1 + d2 r2 |
(29)
Condition (17) for D(ξ, s) follows from (28) and (29). By the lemma, the functions Am 2 −rk |xn | r e , s+D k
rk2 e−rk |xn | , µ(s + D)
m = 2, 3, 4,
k = 1, 2,
are Fourier multipliers of the class (p, p) and estimate (18) holds. 2 sA1 rk Let us represent the function s+D e−rk |xn | in the form A1 rk2 e−rk |xn | −
A1 Drk2 −rk |xn | e . s+D
(30)
Since the function A1 rk2 is a homogeneous function of order 1, the first term in (30) can be analyzed by the results of Mogilevskii (see [10], Lemma 3.2). The function A1 Drk2 is homogeneous of order 2. Therefore, the second term in (30) can be analyzed by the above lemma. Thus, all the functions in (24) are Fourier multipliers of the class (p, p) and condition (13) of Theorem 2 holds for Mpp (·). In view of Theorem 2, integral representations (22) yield the estimates ke−s1 t F −1 (rkj F u(k) )kp,Rk,∞ ≤ c(ke−s1 t F −1 (rkj−1 F Gk )kp,Rk,∞
−s t −1 j−2 ∂
1
+ e F rk F Gk
∂xn
+ ke−s1 t F −1 (rkj−1 F Hk )kp,Rk,∞ p,Rk,∞
−s t −1 j−2 ∂
1
+ e F rk F Hk
∂xn
≡ F,
(31)
p,Rk,∞
ke−s1 t F −1 (rkj F Φ(k) kp,Rk,∞ ≤ cF. 2027
It is clear that (see (12)) ∂ j F −1 (F u(k) )
=F
∂xjn
−1
∂j ∂xjn
(k)
(F u
) = F −1 ((−1)j(k−1) rj F u(k) ),
∂ j F −1 (F Φ(k) ) = F −1 ((−1)j(k−1) rj F Φ(k) ), ∂xjn
(32)
j = 1, 2, . . . , m + 2.
Using the equivalent normalization of the Sobolev spaces, related to the Laplace–Fourier trans(m+2) form, from (31), (32), (21) we deduce the following estimate for the norm ku(2) kp,R2,T : −s1 t (2) (m+2) ku(2)k(m+2) u kp,R2,∞ = cke−s1 t F −1 F u(2) k(m+2) p,R2,T ≤ cke p,R2,∞
−s t ∂ j+2 F −1 F u(2) 1
≤c ke + e
∂xj+2 n p,R2,∞ j≤m
X
−s t −1 j ∂
−s1 t −1 j+1 1 ≤c (ke F (r F G2 )kp,R2,∞ + F r F G2
e
∂xn p,R2,∞ j≤m X
−s1 t
+ke
−s1 t
F
−1
F −1 r2j+2 F u(2) kp,R2,∞
j+1
(r
(m+1)
−s t −1 j ∂
1
F H2 )kp,R2,∞ + e F r F G 2
∂xn p,R2,∞ (m+1)
(m+1−1/p)
≤ c(kG2 kp,R2,∞ + kH2 kp,R2,∞ ) ≤ c(kϕ1 kp,Γ∞ (m+1−1/p)
≤ c(kϕkp,ΓT
(m+1−1/p)
+kψkp,ΓT
(m+1−1/p
+kψ1 kp,Γ∞
)
). (m+2)
(m+2)
The same method can be used to estimate the norms ku(1) kp,R1,T , kΦ(k) kp,Rk,T , k = 1, 2. Hence, we see that the solution u(k) , Φ(k) , k = 1, 2, of problem (1) satisfies the required estimate (10), and, thus, the proof is complete. Remark. In proving the classical solvability of the Verigin problem [1–3], it is assumed that (1) (2) the pressures u 0 and u0 at the initial moment satisfy the condition (2)
(1)
∂u0 ∂u0 − > 0 on ∂n ∂n
Γ0 ,
(33)
where Γ0 is the initial position of the free boundary and ~n is the vector of the unit outward (1) normal defined for the domain Ω 0 . Assuming that condition (33) holds, we arrive at model problem (1). In this problem, the coefficients satisfy condition (8) of Theorem 1. Condition (9) of Theorem 1 is also not restrictive. This fact can be easily established if the free boundary problem is analyzed in accordance with the scheme proposed in [7]. The author wishes to express her gratitude to Professor V. A. Solonnikov for his interest in this paper and for several helpful suggestions. 2028
Translated by E. V. Frolova. REFERENCES 1. E. V. Radkevich, “On the solvability of general nonstationary free boundary problems,” in: Applications of Functional Analysis to Mathematical Physics, Novosibirsk, (1986), pp. 85– 111. 2. E. V. Radkevich and A. S. Melikulov, Free Boundary Problems [in Russian], FAN, Tashkent (1988). 3. B. V. Bazaliy and S. P. Degtyarev, “On the solvability of the many-dimensional filtration free-boundary problem,” Preprint JPMM Acad. Nauk Ukraine, Donetsk 89.10, 3–49 (1989). 4. E. I. Hanzawa, “Classical solution of the Stefan problem,” Tohoku Math. J., 33, 297–335 (1981). 5. O. A. Ladyzhenskaya, V. A. Solonnikov, and N. N. Ural’tseva, Linear and Quasilinear Equations of Parabolic Type [in Russian], Nauka, Moscow (1967). 6. G. I. Bizhanova and V. A. Solonnikov, “On model problems for second order parabolic equations with time derivative in the boundary conditions,” Algebra Analiz, 6, No. 6, 30–50 (1994). 7. V. A. Solonnikov and E. V. Frolova, “Lp-theory for the Stefan problem,” Zap. Nauchn. Semin. POMI, 243, 299–323 (1997). (r) 8. P. I. Lizorkin, “Generalized Liouville differentiation and functional spaces Lp (Rn ),” Math. Sb., 60, No. 3, 325–353 (1963). 9. L. R. Volevich, “Solvability of boundary-value problems for general elliptic systems,” Math. Sb., 68, No. 3, 373–416 (1965). 10. I. Sh. Mogilevskii, “Estimates for a solution of a general initial boundary value problem for the linear nonstationary Navier–Stokes system in a half-space,” Zap. Nauchn. Semin. LOMI, 84, 147–173 (1979). 11. E. V. Frolova, “Lp -estimates for solutions of the model problems with time derivative in the boundary condition or in the conjugation condition,” Izv. St. Petersburg Electrotechnical Univ., 501, 69–79 (1996). 12. E. V. Frolova, “Solvability in Sobolev spaces of a problem for a second order parabolic equation with time derivative in the boundary condition,” Portugaliae Mathematica, 56, No. 4, 419–441 (1999). 13. H. Koch, “Classical solutions to phase transition problems are smooth,” Preprint SFB 359 No. 96-46 (1996). 14. M. S. Agranovich and M. I. Vishik, “Elliptic problems with a parameter and parabolic problems of the general type,” Usp. Mat. Nauk, 19, No. 3, 53–161 (1964).
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