Existence and uniqueness of weak solutions for a class of nonlinear parabolic equations

In this paper, we study the Dirichlet boundary value problem of a class of nonlinear parabolic equations. By a priori estimates, difference and variation techniques, we establish the existence and uniqueness of weak solutions of this problem.


Introduction
The purpose of this paper is to investigate the existence and uniqueness of weak solutions to the following parabolic problem in Ω, (1.1) where Ω is a bounded, open domain of R N (N ≥ 2) with Lipschitz boundary ∂Ω, T is a positive number, and u 0 ∈ L 2 (Ω). The function a is such that φ : R → R defined by φ(s) = a(|s|)s s = 0, 0 s = 0 is an odd increasing homeomorphism from R onto itself. Problems of type (1.1) have been motivated by a various range of applications, such as motion of non-Newton fluids, image restoration, elastic materials, and mathematical biology. We refer to the bibliography (see, for example, [6,7,9,14,16,20]) for more detailed information on the physical situation.
Due to its prominent role in many modeling phenomena, equation (1.1) has been studied extensively. In 1990, Perona and Malik [5] proposed the Malik-Perona model u t − div(c(|∇u| 2 )∇u) = 0 in Ω × (0, T ], (1.2) where Ω is an image domain in R 2 and c(s) > 0. This model is well-known and has been widely used to de-noise and segment images.
with b(s) = c(s) + 2sc (s). Thus, the right-hand side of the equation (1.3) may be interpreted as a sum of a diffusion u TT in the tangent direction (T) plus a diffusion u NN in the normal (N = ∇u |∇u| ) direction. Although there are some results about the existence and uniqueness of weak solutions for the Malik-Perona model, the conditions to ensure these results are difficult to check (see [1]).
The difficulties in this paper are mainly two parts. First, we do not assume polynomial or exponential growth condition as in [19] and [13]. Second, C 1 (Ω) is not dense in W 1,1 0 (Ω), so we will structure new test functions. The novelty in this paper is that we provide an approximation argument to study this kind of problems by finding a weak limit for approximation solution sequence with bounded L 1 -norm under certain conditions and then proving this limit is a weak solution.
We assume that there exist l, m > 1 such that (ii) For any ϕ ∈ C 1 (Q) with ϕ(·, T ) = 0 and ϕ(·, t)| ∂Ω = 0, we have Remark 1.1. Let u be a weak solution in Definition 1.1. By using the approximation technique (see [6,Chapter 3] , The condition (i) in Definition 1.1 is crucial. It guarantees the uniqueness of weak solutions. Now we state our main result. This paper is organized as follows. In Section 2, we will list and prove some useful inequalities and lemmas. In Section 3, the proof of Theorem 1.2 will be given.

Preliminaries
In this section, we state some basic results that will be used later and utilize the properties of N -function to prove Lemma 2.7.
Inequality 2.1 is called the Young Inequality.
In this case, we write Φ ∈ 2 .
Lemma 2.6 ( [15]). The following statements are equivalent: For the case ξ = 0, η = 0 or ξ = 0, η = 0, the proof is similar to the case of Proof. For every pair of ξ 1 , ξ 2 ∈ R N and every λ ∈ [0, 1], we have Lemma 2.9 (The Biting Lemma [4,17]). Let Ω ⊂ R N be measurable with finite Lebesgue measure µΩ and suppose that {f n } is a bounded sequence in L 1 (Ω; R N ). Then there exist a subsequence {f nj } ⊂ {f n }, a function f ∈ L 1 (Ω; R N ), and a decreasing family of measurable sets E k such that µE k → 0 as k → ∞ and for Let Ω ⊂ R N be measurable with finite Lebesgue measure µΩ and suppose that where C is a positive constant. Then there exist a subsequence {f ni } ⊂ {f n }, and a function f ∈ L 1 (Ω; R N ) such that

Existence and Uniqueness
To prove Theorem 1.2, as preparation we first study the existence and uniqueness of weak solutions of the following auxiliary elliptic problems. For h > 0 and u 0 ∈ L 2 (Ω), we consider Theorem 3.2. Assume that u 0 ∈ L 2 (Ω), then there exists a unique weak solution for problem (3.1).
Proof. We consider the variational problem and the functional J is We will establish that J(v) has a minimizer u 1 in V . As Since By Lemma 2.10 and the weak compactness of bounded set in reflexive Banach Space, we can find a subsequence {v mi } of {v m } and a function u This implies that u 1 ∈ V is a minimizer of the functional J(u) in V , i.e., Furthermore, we have J(u 1 ) ≤ J(λu 1 ), λ ∈ (0, 1). Since Φ is an N -function, by Lemma 2.8, we know that Φ(|ξ|) is a convex function with respect to ξ ∈ R N . Therefore we get .
Making use of the approximation argument, we conclude that v − u 1 can be a test function in (3.3). Therefore, Using inequality (2.6), we have which implies u 1 = v a.e. in Ω. Thus we complete the proof of the Theorem. on Ω.

PEIYING CHEN
By an approximation argument [8,18], we first extend solution u(x, t) to the initial value u 0 (x) when t < 0. We next mollify u in the spatial directions to have an approximation C ∞ sequence u ε , then introduce the time average of u ε (x, t), As as a test function in the above initial-boundary value problem to have By the approximation argument, we have Sending k → ∞, we conclude that which implies u = v a.e. in Q. Therefore we obtain the uniqueness of weak solutions. Now we prove the existence of weak solutions. Let n be a positive integer. Denote h = T n . In order to construct an approximation solution sequence {u h } for problem (1.1), we consider the following elliptic problems for k = 1, 2, . . . , n. As k = 1, it follows from Theorem 3.2 that there is a unique u 1 ∈ V satisfying (3.4). Following the same procedures, we can find weak solutions u k ∈ V of (3.4) for k = 2, 3, . . . , n. Moreover, for every ϕ ∈ C 1 c (Ω), we have and Ω u k − u k−1 h u k dx + Ω a(|∇u k |)∇u k · ∇u k dx = 0. Now for every h = T n , we define Choosing u k as a test function in (3.5), and using u k u k−1 ≤ For each t ∈ (0, T ], there exists some j ∈ {1, 2, . . . , n} such that t ∈ ((j − 1)h, jh].
Adding the inequality (3.7) from k = 1 to k = j, we get By the definition of u h (x, t), we obtain In particular, we get Therefore, after taking the supermum over [0, T ], we have We conclude that Thanks to Lemma 2.10, we may choose a subsequence (for simplicity, we also denote it by the original sequence) such that u h u weakly- * in L ∞ 0, T ; L 2 (Ω) , u h u weakly in L 1 0, T ; W 1,1 0 (Ω) . These yield that sup 0≤t≤T Ω u 2 (x, t)dx ≤ C and (3.10) It follows from (3.9) and (3.10) that Recalling (2.5), we have that Thus we can draw another subsequence {ξ h } (we also denote it by the original sequence for simplicity) such that We conclude from Lemma 2.10 that Recalling inequality (2.1), we have |ξ · ∇u| ≤ |ξ||∇u| ≤ Ψ(|ξ|) + Φ(|∇u|).
For every 0 < δ < T , we denote v δ (x, t) = u(x, t + δ). By the uniqueness of weak solutions, we conclude that v δ is a weak solution for the following problem in Ω.
Thanks to Lemma 2.7, it yields Hence, it follows from (3.21) that {u(x, δ i )} is bounded sequence in L 2 (Ω). Moreover, there exist a subsequence (for simplicity, we also denote it by the original sequence) and a u 0 ∈ L 2 (Ω) such that u(x, δ i ) u 0 (x) weakly in L 2 (Ω).
The above relation leads to a contradiction with (3.22). Therefore, we conclude that (3.19) is true and u ∈ C [0, T ]; L 2 (Ω) . Thus we complete the proof of the theorem.