Existence of solution to parabolic equations with mixed boundary condition on non-cylindrical domain
arXiv:1611.06675v1 [math.AP] 21 Nov 2016
Tujin Kima,1,∗, Daomin Caob,2 a Institute b Institute
of Mathematics, Academy of Sciences, Pyongyang, Democratic People’s Republic of Korea of Applied Mathematics, AMSS, Chinese Academy of Sciences, Beijing 100190, P. R. China
Abstract In this paper we are concerned with the initial boundary value problems of linear and semilinear parabolic equations with mixed boundary conditions on non-cylindrical domains in spatialtemporal space. We obtain the existence of a weak solution to the problem. In the case of the linear equation the parts for every type of boundary condition are any open subsets of the boundary being nonempty the part for Dirichlet condition at any time. Due to this it is difficult to reduce the problem to one on a cylindrical domain by diffeomorphism of the domain. By a transformation of unknown function and penalty method we connect the problem to a monotone operator equation for functions defined on the non-cylindrical domain. In this way a semilinear problem is considered when the part of boundary for Dirichlet condition is cylindrical. Keywords: Parabolic equation, Non-cylindrical domain, Mixed boundary condition, Existence MSC: 35K20, 35K55, 35A15
1. Introduction There are vast literature for parabolic differential equations on non-cylindrical domain and various methods have been used to study them. In [6] the energy inequality for a linear equation with homogeneous Dirichlet boundary condition is proved, thus unique existence of solution is studied. For domains expanded along time existence and uniqueness of solution to initial boundary value problem of the linear(cf. [9], [14] and [15]), semilinear(cf. [13]) and nonlinear(cf. [14]) equations with homogeneous Dirichlet boundary condition are studied. For such domains and boundary conditions [13] also deals with attractor; and [2] considers unique existence of solution to a linear Schr¨odinger-type equation. In [3] dealing with the Dirichlet problem, they assume only H¨ older continuity on time-regularity of the boundary. In [25] semigroup theory is improved and the obtained result is applied to the initial boundary value problem of a linear parabolic equation with inhomogeneous Dirichlet condition on non-cylindrical domain. There are some literatures for unique existence of initial boundary value problems of linear equations relying on the method of potentials (see [8] and references therein). Domains in [20] and [21], where existence, uniqueness and regularity are studied, are more general, that is, ”initial” condition is given on a hypersurface in spatial-temporal space instead of the plane t = 0. In [17] optimal regularity of solution to a special kind of 1-dimensional problem is considered. Neumann problem of heat equations (cf. [11]), parabolic equation with Robin type boundary condition (cf. [12]) in non-cylindrical domains and behavior of solutions to the initial-boundary value problems of nonlinear equations (cf. [16] and [29]) are studied. In [23] and [26] optimal control and controllability of parabolic equation with homogenous Dirichlet condition on non cylindrical domain, respectively are studied. ∗ Corresponding
author. Email addresses:
[email protected] (Tujin Kim),
[email protected] (Daomin Cao ) 1 The author’s research was supported by TWAS, UNESCO and AMSS in Chinese Academy of Sciences. 2 Supported by Chinese Science Fund for Creative Research Group(10721101)
Preprint submitted to Elsevier
November 22, 2016
Also, there are many literatures for the initial boundary value problems with mixed boundary conditions. Under certain assumptions the non-cylindrical domains are transformed to a cylindrical one. The initial boundary value problems of linear parabolic equations with mixed time dependent lateral boundary condition on cylindrical domains are studied (cf. [4], [5], [28]). The boundary conditions on the lateral surfaces in [4] may be either two of the following classical ones: Dirichlet, Neumann and Robin, but one part for a kind of boundary condition is a connected and relatively open subset of the lateral surface and the boundary of the part is tangent to the plane t = 0. In [5] the first part is concerned with a classical problem on cylindrical domains and the result is applied to the problem with zero initial condition and lateral mixed boundary conditions on a cylindrical domain, where the non-cylindrical surface for boundary condition is transformed to a cylindrical one by a diffeomorphism. The lateral boundary surface of the cylindrical domain in [5] is also divided into two parts, and one part Γ1 for a kind of boundary condition is connected and relatively open subset of the boundary surface and at each point is transverse to the hyperplane t = const. Developing a method in abstract evolution equations, [28] is concerned with linear parabolic problems on cylindrical domains with mixed variable inhomogeneous Dirichlet and Neumann conditions. But, here change along time of distance of the sections of part of boundary for Dirichlet condition must be dominated by a Lipschitz continuous function in time t. In [28] as application of the result, unique existence of solution to a linear parabolic problem with homogeneous Dirichlet boundary condition on non-cylindrical domains is considered. Section 4 in [22] deals with the linear parabolic problem on non-cylindrical domain with Robin and Dirichlet boundary conditions, where the surface for Dirichlet condition is cylindrical type. [18] and [19] study existence, uniqueness and regularity of solutions to the initial boundary value problems of linear and semilinear parabolic equations on non-cylindrical domains, which is related to the the combustion phenomena. The domains in [18] and [19] are bordered with a part of cylindrical type surface where homogeneous Neumann condition is given, non-cylindrical hypersurfaces where Dirichlet boundary one is given and planes t = 0, t = T . Thus, by change of spatial independent variable they transform the problems to classical problems on cylindrical domains where Dirichlet and Neumann conditions, respectively, are given on cylindrical surfaces. In [24] some differential inclusions are studied and the result is applied to the following problem ∂ u ∈ β(u) on γ, ∂n u = 0 on Γ − γ, u(x, 0) = ξ in Ω0 ,
ut − ∆u ∈ F (u) in Ω, −
where Ω is a non-cylindrical domain in spatial-temporal space, Γ is its lateral surface, γ is a part of a cylindrical surface, β = ∂j and j is a proper lower semicontinuous convex function from R to [0, +∞] with j(0) = 0. On the other hand, in [27] a time-dependent Navier-Stokes problem on a non-cylindrical domain with a mixed boundary condition is considered. In [27] the part of boundary where homogeneous Dirichlet condition is given is cylindrical type and the boundary condition on the other part of boundary is such a special one that guarantee existence of solution to an elliptic operator equation obtained by penalty method. In this paper we are concerned with linear and semilinear parabolic equations on non-cylindrical domains with mixed boundary conditions which may include inhomogeneous Dirichlet, Neumann and Robin conditions together. In the case of linear equation the parts for every type of boundary condition are any open subsets of the boundary being nonempty the part for Dirichlet condition at any time. This rises difficulty in reducing the problem to one on cylindrical domains in [4], [5] and [28] or one in [18], [19], [22] and [24]. Our idea is to use a transformation of unknown function by which the problem is connected to a monotone operator equation for functions defined on the non-cylindrical domain. In this way we can also consider semilinear equation when the part of boundary for Dirichlet condition is cylindrical. This paper is composed of 5 sections. In Section 2 notation, the problem, the definition of weak solution and the main result are stated. In Section 3 by a change of unknown function an 2
equivalent problem is derived. Section 4 is devoted to an auxiliary penalized problem. In Section 5 the proof of the main result is completed. 2. Problem and main result S Let Ω(t) be bounded connected domains of RN , Q = t∈(0,T ) Ω(t) × {t}, 0 < T < ∞, Σ = S ¯ ¯ t∈(0,T ) ∂Ω(t) × {t} and Σ0 , Σ1 be open subsets of Σ such that Σ0 ∪ Σ1 = Σ and Σ0 (t) ≡ ¯ Σ0 ∩ Ω(t) 6= ∅ ∀t ∈ (0, T ). Let ν(x, t) be outward normal unit vector on the boundary Σ and n(x, t) be outward unit vector on R∂Ω(t) for fixed t. R normal Let kyk2Ω(t) = |∇y|2 dx and |y|2Ω(t) = |y|2 dx. Let H 1 (Q) = W21 (Q). Ω(t)
Ω(t)
For function y defined on Q define β(y) by β(y) =
Z
0
T
kyk2Ω(t) dt
12
whenever the integral make sense. Let ¯ ϕ|Σ0 = 0}, D(Q) = {ϕ : ϕ ∈ C ∞ (Q), V (Q) = {the completion of D(Q) under the norm β(y)}, W (Q) = {the completion of D(Q) in the space H 1 (Q)} ¯ and h , i be duality product between W (Q) and W (Q)∗ . By the condition Σ0 (t) ≡ Σ0 ∩ Ω(t) 6= ∅ ∀t ∈ (0, T ), β(·) is a norm in D(Q). We use the following Assumption 2.1. The hypersurface Σ belongs to C 2 for x and to C 1 for t and for any t ∈ [0, T ] there exist a diffeomorphisms X(t) on RN in the class C 2 which maps Ω(0) onto Ω(t), X(0) = I and is in C 1 for t, where I is the unit operator. Remark 2.1. Let ∂Ω(0) ∈ C 2 and Φi (x1 , · · · , xN , t) ∈ C 2,1 (RN × [0, T ]), i = 1, · · · , N be any DΦ functions such that Φi (x1 , · · · , xS N , 0) = xi and Jacobian Dx > 0, where Φ = {Φ1 , · · · , ΦN } and x = {x1 , · · · , xN }. Then Σ = t∈(0,T ) Σ(t) × {t}, where Σ(t) = {Φ(x, t)|x ∈ ∂Ω(0)} satisfies Assumption 2.1. We are concerned with the following initial boundary value problem N N X ∂y ∂ ∂y X ∂y bi (x, t) aij (x, t) + + c(x, t, y) = g(x, t), − ∂t i,j=1 ∂xi ∂xj ∂xi i=1 N X ∂y aij (x, t)ni y|Σ0 = y¯|Σ0 , k(x, t)y + = f (x, t), ∂xj Σ1 i,j=1
y(x, 0) = y0 (x) ∈ L2 (Ω(0)),
where aij (x, t), bi (x, t) and c(x, t, r) are functions satisfying the following conditions N P 1 aij (x, t)ξi ξj ≥ ρ|ξ|2 , ∃ρ > 0, ∀ ξ ∈ RN , i = 1, · · · , N, (A) aij (x, t) ∈ W∞ (Q), aij = aji ,
(2.1)
(2.2) (2.3)
i,j=1
(B) bi (x, t) ∈ L∞ (Q), i = 1, · · · , N, (C) c(x, t, r) is Lipschitz continuous with respect to r uniformly for (x, t) and measurable with respect to (x, t) for fixed r and c(x, t, 0) ∈ L2 (Q) and (D) y¯ ∈ H 1 (Q), k(x, t) ∈ L∞ (Σ1 ), g(x, t) ∈ L2 (Q), f (x, t) ∈ L2 (Σ1 ).
Remark 2.2. On a part of Σ1 where k(x, t) = 0 we have Neumann condition. 3
¯ u ∈ D(Q), in view of (2.2) we have When y ∈ C 2 (Q), Z ∂y u dxdt = (y(x, T ), u(x, T ))Ω(T ) − (y0 , u(x, 0))Ω(0) Q ∂t Z Z ∂u y yu cos(ν,ˆ t) dσ − + dxdt, Q ∂t Σ1 −
N Z N X X ∂ ∂y ∂u ∂y aij (x, t) aij (x, t) u dxdt = dxdt ∂xj ∂xj ∂xi Q i,j=1 ∂xi i,j=1 Q Z Z f (x, t)u dσ, k(x, t)yu dσ − +
Z
(2.4)
(2.5)
Σ1
Σ1
where (ν,ˆ t) is the angle between ν and the positive direction of t-axis. Also, if y ∈ L2 (Q), then c(x, t, y) ∈ L2 (Q) (cf. Lemma 2.2, ch. 2 in [10]). In view of (2.4) and (2.5), we introduce the following Definition 2.1. A function y is called a solution to (2.1)-(2.3) if y satisfies the following y − y¯ ∈ V (Q), Z N Z N Z X X ∂v ∂y ∂y ∂v y − bi (x, t) aij (x, t) dxdt + v dxdt dxdt + ∂xj ∂xi ∂xi Q ∂t i=1 Q i,j=1 Q Z Z [yv cos(ν,ˆ t) + k(x, t)yv] dσ + c(x, t, y)v dxdt + Q Σ1 Z Z f (x, t)v dσ g(x, t)v dxdt + = (y0 , v(x, 0))Ω(0) + Σ1
Q
∀v ∈ W (Q) with v(x, T ) = 0.
Our main result of this paper is the following Theorem 2.1. Suppose that conditions (A), (B), (C) hold and that either c(x, t, y) is linear with respect to y or Σ0 = Γ0 × (0, T ), Γ0 ⊂ ∂Ω(0) and Γ0 is invariant under the diffeomorphism in Assumption 2.1. Then there exists a solution to problem (2.1)-(2.3) provided that (D) is valid. 3. Transformation of unknown function For the sake of simplicity, we will use the same constants in the estimates unless confusion will be caused. It is known that there exists a function ψ ∈ C 2 (Ω(0)) such that ψ(x) > 0 ∀x ∈ Ω(0), ψ|∂Ω(0) = 0 and |∇ψ| > 0 ∀x ∈ Ω(0)\ω0 , where ω¯0 ⊂ Ω(0) (cf. Lemma 1.1 in [7]). ¯ such that Lemma 3.1. There exists a function ϕ(x, t) ∈ C 2,1 (Q) ϕ(x, t) > 0 on Q, ϕ(x, t) = 0 on Σ and −
∂ϕ(x, t) > η > 0 on Σ, ∂n
¯ is the space of functions which are twice continuously differentiable with respect to where C 2,1 (Q) ¯ x and continuously differentiable with respect to t on Q.
4
Proof Take ϕ(x, t) = ψ(X −1 (t)x), where X −1 (t) is the diffeomorphism from Ω(t) onto Ω(0), which is the inverse of the one given in Assumption 2.1. Then the conclusion follows from the properties of function ψ and Assumption 2.1. Let us make a change by u = ek1 t+k2 ϕ(x,t) y,
(3.1)
¯ and (2.2) is satisfied. Then, where k1 and k2 are constants to be determined later. Let y ∈ C 2 (Q) ∂y ∂ϕ ∂u = − (k1 + k2 )u, ∂t ∂t ∂t N X ∂y ∂ aij (x, t) = − ek1 t+k2 ϕ(x,t) ∂xi ∂xj i,j=1
ek1 t+k2 ϕ(x,t)
N X X ∂ ∂ϕ ∂u ∂u aij k2 aij +2 ∂xi ∂xj ∂xj ∂xi i,j=1 ij
−
+ k2 u
ij
ek1 t+k2 ϕ(x,t)
N X
bi (x, t)
i=1
X X ∂2ϕ i ∂ϕ ∂ϕ ∂ϕ aij aij , − k2 + ∂xi ∂xj ∂xi ∂xj ∂xi ∂xj ij ij
h X ∂a
ij
N
Xh ∂y ∂u ∂ϕ i bi (x, t) = − bi (x, t)k2 u . ∂xi ∂xi ∂xi i=1
Also, we have that N X
aij (x, t)ni
i,j=1
=
N X
∂u = ∂xj Σ1
aij (x, t)ni
i,j=1
=
N X
aij (x, t)ni
i,j=1
N ∂y k1 t+k2 ϕ(x,t) X ∂ϕ(x, t) e + aij (x, t)ni k2 u ∂xj ∂xj Σ1 i,j=1
N ∂y k1 t+k2 ϕ(x,t) ∂ϕ(x, t) X aij (x, t)ni nj u e + k2 ∂xj ∂n i,j=1 Σ1
= f (x, t)ek1 t − k(x, t)u + k2
N ∂ϕ(x, t) X aij (x, t)ni nj u ∂n i,j=1 Σ1
where the fact that ϕ(x, t) = 0 on Σ and its corollary Taking into account these facts, we have
∂ϕ(x,t) ∂xj |Σ1
=
∂ϕ(x,t) ∂n nj |Σ1
have been used.
N N X ∂ ∂u ∂u X ∂u Bi (x, t) + C(x, t, u) = G(x, t), − aij (x, t) + ∂t i,j=1 ∂xi ∂xj ∂x i i=1
u|Σ0
N X ∂u aij (x, t)ni = u¯|Σ0 , K(x, t)u + = F (x, t), ∂xj Σ1 i,j=1
u(x, 0) = u0 ≡ y0 (x)ek2 ϕ(x,0) ∈ L2 (Ω(0)),
5
(3.2)
(3.3) (3.4)
where Bi (x, t) = bi (x, t) + 2
X
aij k2
j
∂ϕ , ∂xj
∂ϕ C(x, t, u) = ek1 t+k2 ϕ(x,t) c(x, t, e−k1 t−k2 ϕ(x,t) u) − (k1 + k2 )u ∂t h X ∂a ∂ϕ X X ∂ϕ i X ∂2ϕ ∂ϕ ∂ϕ ij aij bi aij , − k2 + − + k2 u ∂xi ∂xj ∂xi ∂xj ∂xi ∂xj ∂xi ij i ij ij
(3.5)
G(x, t) = ek1 t+k2 ϕ(x,t) g, F (x, t) = ek1 t f, u ¯ = ek1 t+k2 ϕ(x,t) y¯,
and K(x, t) = k(x, t) − k2
∂ϕ X aij ni nj . ∂n i,j
(3.6)
Now, we take
1 , (3.7) 2 which is possible by Lemma 3.1 and (A). Functions Bi (x, t), C(x, t, u), G(x, t), F (x, t), u ¯ and K(x, t), respectively, satisfy the conditions for bi (x, t), c(x, t, r), g(x, t), f (x, t), y¯ and k(x, t) in (B), (C) and (D). k2 > 0 as K(x, t) ≥
Lemma 3.2. In the sense of Definition 2.1 existence of solution to problems (2.1)-(2.3) and (3.2)(3.4) are equivalent Proof First, let us prove that if y is a solution in the sense of Definition 2.1 to problem (2.1)-(2.3), then u is a solution in the sense of Definition 2.1 to problem (3.2)-(3.4). For v ∈ W (Q), put v¯ = e−k1 t−k2 ϕ v. Then ∂ek1 t+k2 ϕ v¯ dxdt ∂t Q Z Z ∂ϕ ∂¯ v dxdt − (k1 + k2 )uv dxdt, =− u ∂t Q Q ∂t
∂v y dxdt = − − Q ∂t Z
N Z X
i,j=1
aij (x, t) Q
=
N Z X
i,j=1
ue−k1 t−k2 ϕ
∂y ∂v dxdt = ∂xj ∂xi aij (x, t)
Q
N Z X
Z
∂(e−k1 t−k2 ϕ u) ∂(ek1 t+k2 ϕ v¯) dxdt ∂xj ∂xi
h ∂u ∂ϕ i × e−k1 t−k2 ϕ − uk2 e−k1 −k2 ϕ ∂xj ∂xj i,j=1 Q h ∂ϕ i ∂¯ v + ek1 t+k2 ϕ k2 v¯ dxdt × ek1 t+k2 ϕ ∂xi ∂xi N Z h ∂u ∂¯ X ∂ϕ ∂u ∂ϕ ∂¯ v ∂ϕ ∂ϕ vi dxdt ≡ I. aij = + k2 v¯ − k22 u¯ v − k2 u ∂xj ∂xi ∂xi ∂xj ∂xj ∂xi ∂xj ∂xi i,j=1 Q
=
aij
6
Integrating by parts in the last term above, we have I=
N Z X
aij (x, t)
Q
i,j=1
+
N Z X
i,j=1
+ Also,
2k2 aij
Q
N Z X
i,j=1
N Z X ∂y ∂v ∂u ∂¯ v aij dxdt = dxdt ∂xj ∂xi ∂xj ∂xi i,j=1 Q
k2
Q
∂ϕ ∂u v¯ dxdt − ∂xi ∂xj
Z
aij k2
Σ1
∂ϕ ni nj u¯ v dσ ∂n
i ∂ϕ ∂ϕ ∂2ϕ ∂ϕ u¯ v − k2 aij u¯ v + aij u¯ v dxdt. ∂xi ∂xj ∂xj ∂xi ∂xi ∂xj
h ∂a
ij
XZ XZ ∂u ∂ϕ ∂y v dxdt = v¯ dxdt − u¯ v dxdt, bi bi k2 ∂x ∂x ∂x i i i Q Q Q i i i Z Z c(x, t, y)v dxdt = c(x, t, e−k1 −k2 ϕ u)ek1 t+k2 ϕ v¯ dxdt. XZ
bi
Q
Q
k1 t+k2 ϕ
The facts v ∈ W (Q) and v¯ ≡ e v ∈ W (Q) are equivalent, and so from above we can see that u is a solution to (3.2)-(3.4) in the sense of Definition 2.1. In the same way we can see that if u is a solution to problem (3.2)-(3.4) in the sense of Definition 2.1, then y is a solution in the sense of Definition 2.1 to (2.1)-(2.3). Therefore, in what follows we will consider the existence of a solution to problem (3.2)-(3.4). To this end, in the next section we will consider an auxiliary problem. 4. An auxiliary problem The main purpose in this section is to find a function um ∈ H 1 (Q) satisfying the following um − u¯ ∈ W (Q), Z Z Z X N 1 ∂um ∂v ∂v ∂um ∂v dxdt dxdt − um dxdt + aij (x, t) ∂t ∂xj ∂xi Q m ∂t ∂t Q Q i,j=1 +
Z X N
Bi (x, t)
Q i=1
∂um v dxdt + ∂xi
Z
C(x, t, um )v dxdt
Q
(4.1)
Z
[um v cos(ν,ˆ t) + K(x, t)um v] dσ + (um (x, T ), v(x, T ))Ω(T ) Z Z F (x, t)v dσ = (u0 , v(x, 0))Ω(0) + G(x, t)v dxdt + +
Σ1
Q
Σ1
∀v ∈ W (Q),
where m are positive integers. We have the following result. Theorem 4.1. Let k2 in (3.1) be as (3.7). Then, for some k1 in (3.1), which is taken before (4.7), there exists a unique solution to problem (4.1). Proof Set u = w + u ¯, define an operator Am ∈ (W (Q) 7→ W (Q)∗ ) and an element L ∈ W (Q)∗ ,
7
respectively, by ∀w, v ∈ W (Q); Z Z N Z X ∂v ∂u ∂v 1 ∂u ∂v u aij (x, t) dxdt dxdt − dxdt + hAm w, vi = m ∂t ∂t ∂t ∂x j ∂xi Q Q i,j=1 Q +
N Z X
∂u Bi (x, t) v dxdt + ∂xi Q
i=1
+ and
Z
Σ1
Z
C(x, t, u)v dxdt
Q
uv cos(ν,ˆ t) + K(x, t)uv dσ + (u(x, T ), v(x, T ))Ω(T )
hL, vi = (u0 , v(x, 0))Ω(0) +
Z
G(x, t)v dxdt + Q
Now, let us consider problem of finding w such that
Z
F (x, t)v dσ. Σ1
Am w = L.
(4.2)
By the conditions (A), (B) and (C), operator Am is Lipschitz continuous. For any w1 , w2 ∈ W (Q), letting w = w1 − w2 , we have that hAm w1 − Am w2 , wi = Z Z N Z X ∂w ∂w ∂w 1 ∂w ∂w w aij (x, t) dxdt dxdt − dxdt + m ∂t ∂t ∂t ∂x j ∂xi Q Q i,j=1 Q + +
N Z X
Zi=1
Σ1
∂w w dxdt + ∂xi
Z
w2 cos(ν,ˆ t) + K(x, t)w
2
Bi (x, t)
Q
Q
(4.3)
C(x, t, w1 + u¯) − C(x, t, w2 + u¯) w dxdt
dσ + |w(x, T )|2Ω(T ) .
On the other hand, by integrating by parts we get Z Z i ∂w 1h w − w2 cos(ν,ˆ t) dσ . dxdt = |w(0)|2Ω(0) − |w(T )|2Ω(T ) − ∂t 2 Q Σ1
(4.4)
From (4.3) and (4.4) we conclude that for any w1 , w2 ∈ W (Q) Z Z X N ∂w ∂w 1 ∂w ∂w hAm w1 − Am w2 , wi = aij (x, t) dxdt dxdt + ∂xj ∂xi Q m ∂t ∂t Q i,j=1 +
Z X N Q i=1
+
Z
Σ1
Bi (x, t)
∂w w dxdt + ∂xi
Z
Q
C(x, t, w1 + u¯) − C(x, t, w2 + u¯) w dxdt
(4.5)
i 1 1 w2 cos(ν,ˆ t) + K(x, t)w2 dσ + |w(0)|2Ω(0) + |w(x, T )|2Ω(T ) . 2 2 2
h1
It follows from (3.6) and the choice of k2 mentioned above that Z h i 1 2 w cos(ν,ˆ t) + K(x, t)w2 dσ ≥ 0. Σ1 2
(4.6)
Note that Bi (x, t) in (3.5) and K(x, t) in (3.6) are independent of k1 . Therefore, taking k1 in the expression of C(x, t, u) in (3.5) a negative number small enough independently of m, we have Z X Z X N N ∂w ∂w ∂w Bi (x, t) dxdt + w dxdt aij (x, t) ∂x ∂x ∂x j i i Q i=1 Q i,j=1 (4.7) Z Z ρ T 2 kwkΩ(t) dt C(x, t, w1 + u ¯) − C(x, t, w2 + u ¯) w dxdt ≥ + 2 0 Q 8
By (4.5)-(4.7) we have hAm w1 − Am w2 , w1 − w2 i ≥ αkwk2H 1 (Q) , ∃ α > 0, ∀ w1 , w2 ∈ W (Q) (Note that α depends on m.) Now, by the theory of monotone operator, there exists a unique solution wm to problem (4.2) (cf. Theorem 2.2, ch. 3 in [10]). Thus, um = wm + u ¯ is the solution asserted in the theorem. 5. Proof of Theorem 2.1 Let k1 be as in the proof of Theorem 4.1, and k2 as in (3.7). When um = wm + u ¯ is the solution to (4.1) asserted in Theorem 4.1, putting w1 = wm , w2 = 0, by (4.2), (4.3), (4.5)-(4.7) we have that Z Z X m 2 1 ∂wm 2 ∂w dxdt + |wm (x, 0)|2Ω(0) + |wm (x, T )|2Ω(T ) dxdt + ∂t ∂xi Q m Q i (5.1) ≤ c [hL, wm i + hAm u ¯, wm i] .
Applying Young inequality to the right hand side of (5.1) and taking into account um = wm + u¯, we have Z Z X m 2 1 ∂um 2 ∂u (5.2) dxdt + dxdt + |um (x, 0)|2Ω(0) + |um (x, T )|2Ω(T ) ≤ c, m ∂t ∂x i Q Q i
where c is independent of m.
We claim that for any v ∈ W (Q) Z 1 ∂um ∂v dxdt → 0 as m → ∞. Q m ∂t ∂t
(5.3)
Indeed, by H¨ older inequality Z Z 1 ∂um ∂v i 21 h Z ∂v 2 i 12 1 h 1 ∂um 2 dxdt ≤ √ √ dxdt , dxdt · m Q m ∂t Q m ∂t ∂t Q ∂t
R 1 ∂um 2 which shows (5.3) since by (5.2) one has that Q m ∂t dxdt ≤ c. By (5.2) we can also choose a subsequence, which is still denoted by {um }, such that wm ≡ um − u¯ ⇀ w, weakly in V (Q), um (T ) ⇀ r weakly in L2 (Ω(T )).
(5.4)
First, let c(x, t, y) be linear with respect to y, that is, c(x, t, y) ≡ c(x, t)y. Then, C(x, t, u) is also linear with respect to u, that is C(x, t, u) ≡ C(x, t)u. Now put u ≡ w + u¯. Then, using (5.3) and (5.4) and passing to the limit in (4.1), we have w ≡ u − u¯ ∈ V (Q), Z Z X Z X N N ∂u ∂u ∂v ∂v Bi (x, t) aij (x, t) u − dxdt + v dxdt dxdt + ∂t ∂x ∂x ∂x j i i Q Q i=1 Q i,j=1 Z Z [uv cos(ν,ˆ t) + K(x, t)uv] dσ C(x, t)uv dxdt + + Σ1 Q Z Z F (x, t)v dσ G(x, t)v dxdt + = (u0 , v(x, 0))Ω(0) + Σ1
Q
∀v ∈ D(Q) with v(T ) = 0,
which shows that u is a solution to problem (3.2)-(3.4). 9
(5.5)
Next, let Σ0 = Γ0 × (0, T ), Γ0 ⊂ ∂Ω(0) and Γ0 is invariant under the diffeomorphism in Assumption 2.1. Following the method in [27], we will prove that the set {um } of solutions to problem (4.1) is relatively compact in L2 (Q). First, let us prove that the following two norms in V (Q) Z T Z T kw(t)kΩ(t) dt and kw(t)kH 1 (Ω(t)) dt are equivalent, (5.6) 0
1
where H (Ω(t)) ≡ t such that
0
W21 (Ω(t)).
It is enough to show that there exists a constant C independent of
kw(t)kH 1 (Ω(t)) ≤ Ckw(t)kΩ(t) ∀t ∈ [0, T ] ′
′
′
(5.7) ′
Let w(x) be a function defined on Ω(t) and x ∈ Ω(0). Then, w (x ) ≡ w(X(t)x ) is a function defined on Ω(0). By Friedrichs inequality Z Z ′ ′ 2 ′ |∇w′ (x′ )|2 dx′ . |w (x )| dx ≤ C(Ω(0), Γ0 ) Ω(0)
Ω(0)
′
Denoting Jacobian of transformation x′ = X −1 (t)x by J = Dx Dx , we have Z Z Z |w(x)|2 |J| dx = |w′ (x′ )|2 dx′ ≤ C(Ω(0), Γ0 ) |∇w′ (x′ )|2 dx′ Ω(t) Ω(0) Ω(0) Z |∇w(x)|2 |J|−1 dx, ≤ C(Ω(0), Γ0 ) Ω(t)
where it was considered that in ∇w′ (x′ ) and ∇w(x) operators ∇ are, respectively, with respect to x′ and x. From this we get Z Z |w(x)|2 dx ≤ C(t) |∇w(x)|2 dx, Ω(t)
Ω(t)
where C(t) is continuous in t ∈ [0, T ]. This implies (5.7). On the other hand, by (5.2) Z T kwm k2Ω(t) dt ≤ c.
(5.8)
0
Let Ω ⊂ RN such that Ω(t) ⊂ Ω ∀t ∈ [0, T ]. For any m let us make w ¯ m (x, t) ∈ H 1 (Ω × (0, T )), an extension of wm (x, t) ∈ W (Q) as follows. ′m ′ m ′ Let w (x ) ≡ w (X(t)x ) on Ω(0) and denote bounded extensions in H 1 (RN ) by the same(cf. Lemma 1.29, ch. 2 in [10]). Then, Z Z ′ ′m ′ 2 ′m ′ 2 |w′m (x′ )|2 + |∇w′m (x′ )|2 dx′ |w (x )| + |∇w (x )| dx ≤ c N Ω(0) R Z |wm (x, t)|2 |J(t)| + |∇wm (x, t)|2 |J(t)|−1 dx ≤c Ω(t) Z |wm (x, t)|2 + |∇wm (x, t)|2 dx. ≤c Ω(t)
We take the restriction on Ω of a function defined by w ¯ m (x, t) = w′m (X −1 (t)x) on RN . Then, we have Z Z m 2 m 2 |w′m (x′ )|2 + |∇w′m (x′ )|2 dx′ . |w ¯ (x, t)| + |∇w ¯ (x, t)| dx ≤ c RN
Ω
By (5.6), (5.8) and two inequalities above, we get Z T kw ¯m k2H 1 (Ω) dt ≤ c. 0
10
(5.9)
Also, by (5.2) we get Z
Ω×(0,T )
1 ∂ w ¯ m 2 dxdt ≤ c, m ∂t |w ¯m (x, 0)|Ω ≤ c,
Z
Ω×(0,T )
˜ 2 1 ∂ u dxdt ≤ c, m ∂t
|w ¯m (x, T )|Ω ≤ c.
(5.10) (5.11)
where u ˜ is a bounded extension of u ¯ and c is independent of m. Put w ¯m (x, t) = 0 for −T < t < 0, T < t < 2T . Let Z 1 t m wh (x, t) = w ¯m (x, s) ds for |h| < T. h t−h Then,
1 m ∂whm (x, t) = w ¯ (x, t) − w ¯ m (x, t − h) , whm (x, t)|Σ0 = 0, ∂t h which means whm |Q ∈ W (Q). Replacing v by whm |Q in (4.1), we have Z 1 ∂um (x, t) 1 m w ¯ (x, t) − w ¯m (x, t − h) dxdt ∂t h Q m Z m 1 wm (x, t) w ¯ (x, t) − w ¯m (x, t − h) dxdt − h Q Z Z X N ∂wm (x, t) ∂um ∂whm u ¯(x, t) h − aij (x, t) dxdt dxdt + ∂t ∂xj ∂xi Q Q i,j=1 Z X N
(5.12)
Z ∂um m wh dxdt + C(x, t, um )whm dxdt Bi (x, t) + ∂xi Q Q i=1 Z m m u wh cos(ν,ˆ t) + K(x, t)um whm dσ + (um (x, T ), whm (x, T ))Ω(T ) + Σ1 Z Z m m F (x, t)whm dσ. G(x, t)wh dxdt + = (u0 , wh (x, 0))Ω(0) + Σ1
Q
¯ × [0, T ]), let us estimate Assuming w(x, ¯ t) ∈ C (Ω Z 1 2 I1 ≡ ¯ t) − w(x, ¯ t − h) dxdt. w(x, h 1
Ω×(0,T )
First, let h > 0. Then Z I1 ≤
t
2 ∂ w(x, ¯ s) ds dxdt t−h ∂s Ω×(h,T ) Z t Z i2 ∂ 1h w(x, ¯ 0) + w(x, ¯ s) ds dxdt + 2 0 ∂s Ω×(0,h) h Z Z 1 t ∂ w(x, 2 ¯ s) 2 ≤ ¯ 0)|2Ω ds dxdt + |w(x, h ∂s h Ω×(h,T ) t−h Z Z t 2 2 ∂ + 2 ¯ s) ds · h dxdt w(x, h Ω×(0,h) 0 ∂s Z ∂ 2 1n (T − h) ¯ t) dxdt + 2|w(x, ¯ 0)|2Ω ≤ w(x, h Ω×(0,T ) ∂t Z 2 ∂ o ¯ t) dxdt + 2h w(x, Ω×(0,T ) ∂t Z ∂ 2 o 1n ≤ (T + h) ¯ t) dxdt + 2|w(x, ¯ 0)|2Ω . w(x, h Ω×(0,T ) ∂t 1 h2
Z
11
(5.13)
If h < 0, using I1 ≤
Z
Ω×(0,T −|h|)
+
Z
1 h2
Ω×(T −|h|,T )
t
2 ∂ w(x, s) ds dxdt ∂s t−h Z t i2 1h ∂ w(x, ¯ T ) + w(x, s) ds dxdt, h2 T ∂s
Z
in the same way above we get Z ∂ 2 o 1 n (T + |h|) ¯ t) dxdt + 2|w(x, ¯ T )|2Ω . I1 ≤ w(x, |h| Ω×(0,T ) ∂t
(5.14)
By (5.2) and (5.15), we get Z 1 ∂um (x, t) 1 h i w ¯m (x, t) − w ¯m (x, t − h) dxdt · m ∂t h Q 21 Z 1 ∂um 2 i 2 12 1 Z 1 h ≤ ¯m (x, t) − w ¯m (x, t − h) dxdt dxdt √ w ∂t m Q h Q m c ≤ p . |h|
(5.16)
¯ × [0, T ]) is dense in H 1 (Ω × (0, T )), by (5.10), (5.11), (5.13), (5.14) for any w Since C 1 (Ω ¯ m we have Z 1h i 2 o 12 1 n c √ ≤ p . ¯m (x, t) − w ¯m (x, t − h) dxdt w (5.15) m h |h| Ω×(0,T )
We have Z T Z N Z X 1 t ∂um ∂whm m dxdt ≤ c ku kΩ(t) k w ¯m (x, s) dskΩ(t) dt aij (x, t) ∂xj ∂xi h t−h 0 Q i,j=1 Z Z T 12 p 1 t kw ¯m (x, s)k2Ω ds dt ≤ c/ |h|, kum kΩ(t) p ≤c |h| t−h 0 Z N X ∂um m Bi (x, t) wh dxdt ∂x i Q i=1 Z T Z 12 p 1 t m ≤c ku kΩ(t) p |w ¯m (x, s)|2Ω ds dt ≤ c/ |h|, |h| t−h 0 Z h Z i p C(x, t, um )whm dxdt ≤ c |um | + |C(x, t, 0)| |whm | dxdt ≤ c/ |h|. Q
(5.17)
Q
By Assumption 2.1, sin(1ν,t) ˆ ≥ δ > 0 on Σ1 . Taking this and the trace theorem into account, we get Z um whm cos(ν,ˆ t) + K(x, t)um whm dσ Σ1
Z =
≤c
1 dxdt ˆ sin(ν, t) 12 p kw ¯ m (x, s)k2Ω ds dt ≤ c/ |h|,
[um whm cos(ν,ˆ t) + K(x, t)um whm ]
Σ1 Z T 0
1 kum kΩ(t) p |h|
Z
t
t−h
where (5.6) and (5.8) were used. Also Z 12 p 1 T m m p |w ¯m (x, s)|2Ω ds dt ≤ c/ |h|, (u (x, T ), wh (x, T ))Ω(T ) ≤ c |h| T −h p m (u0 , wh (x, 0))Ω(0) ≤ c/ |h|. 12
(5.18)
(5.19)
Similarly to (5.17), (5.19), let us estimate −
R
Q
u¯(x, t)
m ∂wh (x,t) ∂t
dxdt.
Z ∂wm (x, t) dxdt u ¯(x, t) h − ∂t Q Z Z ∂u ¯ (x, t) u ¯(x, T )whm (x, T ) dx whm (x, t) dxdt − = ∂t Ω(T ) Q Z Z u¯(x, t)whm (x, t) cos(ν,ˆ t) dσ + u¯(x, 0)whm (x, 0) dx − Ω(0)
Also, we get
(5.20)
Σ1
p ≤ c/ |h|.
Z Z m G(x, t)wh dxdt ≤ c Q
0
Z
Z m F (x, t)wh dσ ≤ c
0
Σ1
Let us estimate
I≡−
1 h
Z
Q
T
T
Z 21 p 1 t p |w ¯m (x, s)|2Ω ds dt ≤ c/ |h|, |h| t−h Z 12 p 1 t p kw ¯ m (x, s)k2Ω ds dt ≤ c/ |h|. |h| t−h
(5.21)
m wm w ¯ (x, t) − w ¯m (x, t − h) dxdt.
Putting Ω(t) = Ω(T ) for t > T , Ω(t) = Ω(0) for t < 0 and using −ab = 21 [(a − b)2 − a2 − b2 )], we have the following estimate. 1 I =− h
Z
T
0
Z T m 2 1 w dt + ¯ (x, t − h) Ω(t) dt Ω(t) 2h 0 0 Z T m 2 1 w − ¯ (x, t) − w ¯m (x, t − h) Ω(t) dt 2h 0 Z T Z T m 1 1 wm (x, t) 2 w (x, t) 2 =− dt − dt Ω(t)∩Ω(t+h) Ω(t)\Ω(t+h) 2h 0 2h 0 Z T Z T m 2 m 2 1 1 w w ¯ (x, t − h) Ω(t)∩Ω(t−h) dt + ¯ (x, t − h) Ω(t)\Ω(t−h) dt + 2h 0 2h 0 Z T m 2 1 w − ¯ (x, t) − w ¯m (x, t − h) Ω(t) dt 2h 0 Z T Z T m m 1 1 w (x, t) 2 w (x, t) 2 =− dt − dt Ω(t)∩Ω(t+h) Ω(t)\Ω(t+h) 2h 0 2h 0 Z T −h Z T m 2 m 2 1 1 w w + ¯ (x, t) Ω(t)∩Ω(t+h) dt + ¯ (x, t − h) Ω(t)\Ω(t−h) dt 2h −h 2h 0 Z T m 2 1 w ¯ (x, t) − w ¯m (x, t − h) Ω(t) dt. − 2h 0 1 =− 2h
Z
T
w ¯m (x, t), w ¯ m (x, t) − w ¯ m (x, t − h) Ω(t) dt m w (x, t) 2
Now, put p = 4 for N = 1, · · · , 4, p =
2N N −2
for N ≥ 5 and let
13
2 p
+
1 q
(5.22)
= 1. Applying H¨ older
inequality, (5.9), Assumption 2.1 and the fact that H 1 (Ω) ֒→ Lp (Ω), we have 1 Z T |w ¯m (x, t − h)|2Ω(t)\Ω(t−h) dt 2h 0 Z T hZ i p2 1 1 ≤ |w ¯m (x, t)|p dx · mes Ω(t + h) \ Ω(t) q dt 2|h| 0 Ω(t+h)\Ω(t) Z T 1 c c ≤ . kw ¯ m (x, t)k2H 1 (Ω) dt · |h| q ≤ 2|h| 0 |h|(1−1/q) Substituting (5.23) in the right hand side of (5.22), we have that Z m 1 − wm (x, t) w ¯ (x, t) − w ¯m (x, t − h) dxdt ≤ h Q Z T m 2 c 1 w ¯ (x, t) − w ¯ m (x, t − h) Ω(t) dt if h > 0, − (1−1/q) 2h 0 h Z m 1 m − w (x, t) w ¯ (x, t) − w ¯m (x, t − h) dxdt ≥ h Q Z T m 2 1 c w ¯ (x, t) − w ¯m (x, t − h) Ω(t) dt if h < 0. − (1−1/q) + 2|h| 0 |h|
(5.23)
(5.24)
Formulas (5.8), (5.12), (5.16)-(5.21) and (5.24) imply Z
T
0
and
h
Z
m 2 w ¯ (x, t + h) − w ¯m (x, t) Ω(t) dt ≤ c|h|1/q
m 2 w ¯ (x, t)
dt + Ω(t)
0
Next, let
m
Z
T
T −h
w ˜ (x, t) =
(
m 2 w ¯ (x, t)
Ω(t)
for h ∈ R
dt ≤ ch1/q
for h > 0.
(5.25)
(5.26)
wm (x, t) if (x, t) ∈ Q 0 otherwise.
Then, when 0 < |h| < T , similarly to (5.23) we have Z
0
T
2 w ¯m (x, t + h) − w ˜m (x, t + h) ≤
Ω(t)
Z
T
m 2 w ¯ (x, t + h) − w ˜m (x, t + h)
Ω(t)\Ω(t+h)
0
+
Z
0
≤c
dt
Z
0
T
T
dt (5.27)
m 2 w ¯ (x, t + h) − w ˜m (x, t + h)
Ω(t+h)\Ω(t)
dt
m
2 1 1
w ¯ (x, t + h) Ω dt · |h| q ≤ c|h| q .
From (5.25) and (5.27) we have Z
T
m 2 w ˜ (x, t + h) − wm (x, t)
1
dt ≤ c|h| q
for 0 < |h| < T.
Ωj (t) = {x ∈ Ω(t) : dist(x, ∂Ω(t)) > 2/j},
j = 1, 2, 3, · · · .
0
˜ ∈ RN and Let h
Ω(t)
14
(5.28)
Then, by (5.6) and (5.8) Z
Ω(t)\Ωj (t)
|wm (x, t)|2 dx = =
hZ
Ω(t)\Ωj (t)
i p2 1 |wm (x, t)|p dx · mes Ω(t) \ Ωj (t) q dt
(5.29)
q1 ≤ ckwm (x, t)k2Ω(t) · 1/j ,
where c depends only on Q. ˜ < 1/j, then x + sh ˜ ∈ Ω2j (t) provided x ∈ Ωj (t) and s ∈ [0, 1]. For wm ∈ C ∞ (Ω(t)), Now, if |h| Z
Ωj (t)
˜ t) − wm (x, t)|2 dx ≤ |wm (x + h, ≤
Z
Z
Ωj (t)
˜2 ≤ |h| ˜2 ≤ |h|
1
d m i2 ˜ t) ds w (x + sh, ds Ωj (t) 0 hZ 1 i2 ˜ t) · ˜ ∇wm (x + sh, dx h ds hZ dx
Z
0
1
0
Z
Ωj (t)
Z
Ω2j (t)
(5.30)
˜ t) 2 dxds ∇wm (x + sh,
|∇wm |2 dx ≤ (1/j)2 kwm k2Ω(t) .
Since C ∞ (Ω(t)) is dense in H 1 (Ω(t)), (5.30) is valid for any wm ∈ H 1 (Ω(t)). By (5.29), (5.30) and (5.8) we have that Z m 2 ˜ t) − w ˜ < 1/j w ˜ (x + h, ˜m (x, t) dxdt ≤ c(1/j)1/q for |h| (5.31) Q
where c is independent of m. From (5.28) and (5.31) we get Z m 2 ˜ h) → 0 in RN +1 . w ˜ (x + ˜ h, t + h) − w ˜m (x, t) dxdt → 0 as (h,
(5.32)
Q
From (5.26) and (5.29), we get
¯ε ⊂ Q : ∀ε, ∃Qε such that Q
Z
Q\Qε
m w (x, t) 2 dxdt < ε.
(5.33)
By (5.32) and (5.33) we know that the set {wm } is relatively compact in L2 (Q) (cf. Theorem 2.32 in [1]). Thus, we can choose a subsequence, which is still denoted by {wm }, such that wm → w ∈ L2 (Q). Therefore C(x, t, um ) → C(x, t, u) in L2 (Ω(t)) for a.e. t, where u ≡ w + u¯. Therefore, using (5.4) and passing to the limit in (4.1), we have (5.5) which shows that u is a solution to problem (3.2)-(3.4). References [1] R. A. Adams, J. J. F. Fournier, Sobolev Spaces, Academic Press, 2003. [2] M. L. Bernardi, G. A. Pozzi and G. Savar´e, Variational equations of Schr¨odinger-type in non-cylindrical domains, J. Differential Equations 171 (1) (2001) 63-87. [3] S. Bonaccorsi, G. Guatteri, A variational approach to evolution problems with variable domains, Journal of Differential Equations 175 (2001) 51-70.
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