The weak solutions of a nonlinear parabolic equation from two-phase problem

A nonlinear parabolic equation from a two-phase problem is considered in this paper. The existence of weak solutions is proved by the standard parabolically regularized method. Different from the related papers, one of diffusion coefficients in the equation, b(x), is degenerate on the boundary. Then the Dirichlet boundary value condition may be overdetermined. In order to study the stability of weak solution, how to find a suitable partial boundary value condition is the foremost work. Once such a partial boundary value condition is found, the stability of weak solutions will naturally follow.


Introduction
In this paper, we study the following initial-boundary value problem: Let us give a brief review of the related works. We first noticed that the initial-boundary value problem of the equation u t = div |u| σ (x,t) + d 0 |∇u| p(x,t)-2 ∇u + c(x, t)b 0 u(x, t), (x, t) ∈ Q T , (1.5) has been considered in [14,19,23], where σ (x, t) > 1, d 0 > 0, c(x, t) ≥ 0, and b 0 > 0, ⊂ R N is a bounded domain with smooth boundary ∂ . This model may describe some properties of image restoration in space and time, u(x, t) represents a recovering image, p(x, t) reflects the corresponding observed noisy image. The authors of [14] obtained the existence and uniqueness of weak solutions with the assumption that the exponent σ (x, t) ≡ 0, 1 < p -< p + < 2. If σ (x, t) ≡ 0 and b 0 = 0, the existence of weak solutions was proved in [23] by Galerkin's method. Next, in [19], they proved the existence and uniqueness of weak solution when σ (x, t) ∈ (2, 2p + p + -1 ) or σ (x, t) ∈ (1, 2), 1 < p -< p + ≤ 1 + √ 2. Moreover, they applied energy estimates and Gronwall's inequality to obtain the extinction of solutions when the exponents pand p + belong to different intervals.
Secondly, the nonlinear parabolic equation from the double phase problems has been studied in [6][7][8][9][10] and [16,17,26] in recent years, where the diffusion coefficients a(x) and b(x) satisfy If f (x, t) = 0, the author of [6] studied the existence of weak solutions to equation (1.6) by the energy functional method. If f (x, t) ∈ L r (0, T; L s ( )) with some given positive constants r and s, by defining the local parabolic potential, the author of [12] obtained the local boundedness of weak solutions. In addition, there are many papers that worked on the double phase elliptic equations studied in the framework of the Musielak-Orlicz spaces, see [5,13,15,20,21,27,30].
In this paper, we use the parabolically regularized method to prove the existence of the weak solution to equation (1.1). If p(x) ≥ q(x), it is not difficult to show that the weak solution u is in L ∞ (0, T; W 1,p(x) 0 ( )). Then, based on the usual Dirichlet boundary value condition (1.3), the stability of weak solutions can be obtained in a simple way. So, in this paper, we assume that The greatest contribution of this paper lies in that, instead of using the usual boundary value condition (1.3), it proves the stability of weak solutions only under a partial boundary value condition u| 1T = 0, (x, t) ∈ 1T = 1 × (0, T), (1.8) and the uniqueness follows naturally. Here, 1 ⊂ ∂ is a relative open subset and will be specified below. The method used in this paper may be generalized to study the well-posedness problem of the following double phase equation with the variable exponents: (1.9) We are ready to study this problem in the future. For a general degenerate parabolic equation, the well-posedness of weak solutions based on a partial boundary value condition has been studied for a long time, relevant literature can be referred to [4,22,28,29,[31][32][33][34][35][36][37].

The definitions of weak solution and the main results
We assume that r(x) ∈ C( ), and quote some function spaces with variable exponents.
which is a separable, uniformly convex Banach space.
(ii) r(x)-Hölder's inequality. Let r 1 (x) and r 2 (x) be real functions with 1 r 1 (x) + 1 r 2 (x) = 1 and r 1 (x) > 1. Then the conjugate space of L r 1 If u L r(x) ( ) = 1, then |u| r(x) dx = 1, Besides this trivial embedding, it would be useful to know finer estimates of the type of Sobolev inequality.
But Zhikov [38] pointed out that unless r(x) ∈ C log ( ). Here, r(x) ∈ C log ( ) means that r(x) is a logarithmic Hölder continuity function, i.e., it satisfies where ω(s) is with the property Let ρ(x) be the Friedrichs mollifying kernel These two lemmas can be found in [2]. Certainly, for a constant p ≥ 1, it is well known that, if u ∈ L p ( ), then ρ ε * u ∈ L p ( ) and Let us give the definition of weak solution.
and for any function ϕ ∈ L ∞ (0, T; W Throughout this paper, we assume that q(x), p(x) both are logarithmic Hölder continuous functions and satisfy The main results are the following theorems.
The unusual thing is that, since b(x) satisfies (1.4), the stability of weak solutions can be proved under a partial boundary value condition (1.8) in which 1 has the form , then the partial boundary appearing in (1.8) (1.1) with the initial values u 0 (x) and v 0 (x) respectively, with the same partial boundary value condition (1.8) and 1 given by (2.7). If

Theorem 2.7 Let u(x, t) and v(x, t) be two solutions of equation
then the stability of (2.6) is true.

The proof of Theorem 2.5
Let q(x) ≥ q -≥ p + ≥ p(x). Consider the following regularized problem: where u 0ε ∈ C ∞ 0 ( ) and (b(x) + ε)|∇u 0ε | q(x) ∈ L 1 ( ) are uniformly bounded, and u 0ε converges to u 0 in W If p(x) and q(x) are with logarithmic Hölder continuous property, similar to [1,3,19,23], by constructing suitable function spaces and applying Galerkin's method, we can prove that there is a weak solution to problem (3.1)- In what follows, we shall show that the constant c 1 in (3.4) is independent of ε.
Then the following properties hold: 1. For any s ∈ R, we have We introduce a function space endowed with the norm u V = |∇u| L p(x) (Q T ) , or equivalent norm u V = |u| L p -(0,T;W 1,p(x) 0 ( )) + |∇u| L p(x) (Q T ) , and the equivalence follows from the p(x)-Poincare inequality. Then V is a separable and reflexive Banach space. We denote by V * its dual space.
is a weak solution of (3.1), then there is a constant c (independent of ε ) that depends on p -, N, T , let Proof In the proof, we simply denote that u ε = u. If k is a real number and u 0 L ∞ ( ) ≤ k, as a test function (where χ A is an eigenfunction on the set A ). At the same time, we know that v (3.12) Substituting (3.12) into (3.11), we can deduce that which implies so the measure μ(A k (τ )) = 0, and the conclusion follows naturally. Lemma 3.3 implies that one can choose a subsequence of u ε (we still denote it as u ε ) such that u ε * u, weakly star in L ∞ (Q T ), (3.14) where u(x, t) ∈ L ∞ (Q T ). Now, we can show that u(x, t) is a weak solution of equation (1.1) with the initial value (1.2) in the sense of Definition 2.4.
Proof of Theorem 2.5 First, for any t ∈ [0, T), we multiply (3.1) by u ε to obtain and so we have Secondly, by condition (2.4), Bögelein, Duzaar, and Marcellini [9][10][11] proved where the constant c is independent of ε. By u ε → u a.e. in Q T , and u εt u t , in L 2 (Q T ), In order to prove that u satisfies equation (2.1), we have to show that for any ϕ 1 ∈ C 1 0 (Q T ). In the first place, for any ϕ ∈ C 1 0 (Q T ), we have In the second place, let 0 ≤ ψ ∈ C ∞ 0 (Q T ) and If we choose ψu ε as the test function of equation (3.1), then By the facts Let ε → 0 in (3.23). By (3.24) and using the Hölder inequality, we can deduce that In the third place, let ϕ = ψu in (3.21). We get Combining (3.25) with (3.26), we have At last, when we choose v = uλϕ 1 , λ > 0, we have (3.28) Simultaneously, if we choose v = uλϕ 1 , λ < 0, then λ → 0 similarly yields Since ψ = 1 on supp ϕ 1 , namely we know that (3.19) is true, for any ϕ 1 ∈ C 1 0 (Q T ), we have is true for any given r > 1. For any given t ∈ (0, T), if we denote by t the compact support set of ϕ 2 (x, t), for any 1t satisfying t ⊂⊂ 1t ⊂⊂ , by (3.31), we get Here, we have used the assumption that q(x) satisfies the logarithmic Hölder continuity condition, then Since q(x) ≥ p(x), using the p(x)-Hölder inequality, we know By choosing a subsequence of ϕ n2 (x, t) (we still denote it as ϕ n2 (x, t)), we may think that ϕ n2 (x, t) satisfies Then, by (3.30), we have Letting n → ∞, we get for any ϕ 2 ∈ L r (0, T; W 1,p(x) 0 ( )) ∩ L r (0, T; W 1,q(x) loc ( )). As for the initial value, (2.3) can be showed as in [1], the proof of Theorem 2.5 ends.
Proof of Theorem 2.6 By Definition 2.4, for any Thus, if we choose S η (uv) as the test function, then we have Let η → 0 + in (4.2). Then, by (4.3), we have Theorem 2.6 is proved.

The partial boundary value condition
Proof of Theorem 2.7 If u(x, t) and v(x, t) are two weak solutions of equation (1.1) with the partial homogeneous boundary value condition and with the different initial values u(x, 0) and v(x, 0) respectively. For small η > 0, let Then ∇φ η = ∇b(x) η when x ∈ \ η , and in the other place, it is identically zero. Choosing φ η S η (uv) as the test function, we have In the first place, following [9, Lemma 3.1] we find In the second place, it is easy to see that In the third place, to evaluate the third term on the left-hand side of (5.3), in consideration of (5.2), by a straightforward calculation we obtain where p 1 = p + or paccording to (iii) of Lemma 2.2, q(x) = p(x) p(x)-1 and q 1 = q + or q -. If we denote 2 = ∂ \ 1 and define η1 = x ∈ \ η : dist(x, 2 ) > dist(x, 1 ) , η2 = x ∈ \ η : dist(x, 2 ) ≤ dist(x, 1 ) , In the fourth place, to evaluate the fourth term on the left-hand side of (5.3), by a direct calculation, we have By (