Existence of Solutions for Degenerate Elliptic Problems in Weighted Sobolev Space

Lemma 1. Let a and b be two nonnegative real numbers, and let

with θ = b²/4a². Then,

Lemma 2. Let g ∈ L^r(Ω, γ) and g_n ∈ L^r(Ω, γ), with $$, 1 < r < ∞. If g_n → g  a.e. in Ω, then g_n⇀g weakly in L^r(Ω, γ), where γ is a weight function on Ω.

Lemma 3 (assume (H1)). Let $$ be uniformly Lipschitzian, with G(0) = 0. Let $$. Moreover, if the set D of discontinuity points of G^′ is finite, then

Lemma 4 (assume (H1)). Let $$ and T_k(u), $$, be the usual truncation. Then, $$. Moreover, one has

Lemma 5 (assume (H1) and (H2)). Let {u_n} be a sequence of functions in $$ such that u_n⇀u weakly in $$ and

Then, u_n → u strongly in $$.

Definition 6. One says $$ is a weak solution to problem (1), provided that

Theorem 7. Let f be in L¹(Ω) and $$. Then, there exists at least one solution u to problem (1).

Proof. The proof will be divided into 5 steps.

Step 1 (the approximation equation). We introduce the following approximation equation of problem (1). Let f_n be a sequence of L^∞(Ω) functions that converges to f strongly in L¹(Ω) and let n ∈ N,

then g_n(x, s, ξ) is bounded and satisfies (10) and for almost every x in Ω, for every ξ in $$, and for every s in $$ with |s | ≥ h. By the results of [2], there exists a solution $$ of which satisfies for every $$.

Step 2 (the weak convergence u_n⇀u in $$). Take v = φ(T_h(u_n)) as a test function in (27), where h > 0 is defined in (9) and φ(s) is as in (19). Writing $$ and φ_h = φ(T_h(u_n)) for simplicity, we have

Thanks to Young’s inequality and (7), we have Since {f_n} is bounded in L¹(Ω) and $$, it follows from the above inequality that where C is independent of n. Splitting the second term on the left-hand side where |u_n | < h and |u_n | ≥ h, we can write Using (9) and (10), we get Hence, Recalling (19) in Lemma 1, let a = c₀/2, b = b(h); we then obtain which implies or equivalently where C is some positive constant. Therefore, we can extract a subsequence, still denoted by itself, such that By (5) and (37), we have for a subsequence u_n → u strongly in L^q(Ω, ω) and a.e. in Ω. Then, T_k(u_n) is bounded in $$. Hence, by the results of [8], we have

Step 3 (the strong convergence u_n → u in $$). For every k ≥ h, we will prove that T_k(u_n) converges strongly to T_k(u) in $$. We first prove that

Here, we denote by N^∗ the set of natural numbers. Choosing v = T_k(u_n) − T_k−1(u_n) as a test function in (27) with k ≥ h + 1, using (7) and Young’s inequality, we obtain Noticing (9) and that T_k(u_n) − T_k−1(u_n) has the same sign as g_n(x, u_n, ∇u_n) if |u_n | > h and is zero if |u_n | ≤ h, we get Dropping the nonnegative term, we have Since we obtain Taking into account the fact that {f_n} is compact in L¹(Ω) and $$, we deduce that Hence, Noticing that k ≥ h and (9), this completes the proof of assertion (39).

Let k ≥ h be fixed, 0 < ɛ < k, and choose v = φ(T_ɛ(u_n − T_k(u))) as a test function in (27), where φ(s) is defined in Lemma 1 (refer to [8–10]). We thus obtain

In the following, δ(ɛ, n) represents a quantity which converges to zero as firstly n → ∞ and secondly ɛ → 0. For convenience, we write Observe that, in the weak ^∗ topology of L^∞(Ω) and almost everywhere in Ω, we have Now, as {f_n} is compact in L¹(Ω) and (49), we have Thanks to T_k+ɛ(u_n)⇀T_k(u) weakly in $$, $$, and (49), we obtain where χ_ɛ{|u_n − T_k(u)|} = χ{|u_n − T_k(u)| ≤ ɛ}. We can decompose (A) as Owing to {|u_n − T_k(u)| ≤ ɛ}⊂{|u_n | ≤ k + ɛ}, we get Since ∇T_k+ɛ(u_n) is zero whenever ∇G_k+ɛ(u_n) is not zero, hence, Since ∇T_k(u) ≡ 0 on the set {|u | > k + ɛ}, we see that, as n → ∞, As $$, Lebesgue’s dominated convergence theorem guarantees that By (6), (49), and the fact that a(x, u_n, ∇G_k+ɛ(u_n)) is bounded in $$, we obtain Now we split (E) into We will prove that In fact, where c₁, c₂ are positive constants. Since T_k+ɛ(u_n)⇀T_k+ɛ(u) weakly in $$ and $$ is compact, then T_k+ɛ(u_n) → T_k+ɛ(u) strongly in L^q(Ω, σ) and a.e. in Ω. Hence, Then, by the generalized Lebesgue dominated convergence theorem, we deduce (59). By ∂T_k+ɛ(u_n)/∂x_i⇀∂T_k(u)/∂x_i weakly in L^p(Ω, ω_i) and (49), we have Using (57) and (62), we have As for the term (B), we decompose it as It is clear that on the set {u_n ≥ k + ɛ} we get while on the set {u_n < −k − ɛ} we get By (9) and the fact that k ≥ h, we obtain Using (7) and (10) and noticing that ɛ < k, we have By the weak convergence of ∂T_k+ɛ(u_n)/∂x_i⇀∂T_k(u)/∂x_i in L^p(Ω, ω_i), (49), and (59), we have Since a(x, T_k+ɛ(u_n), ∇T_k+ɛ(u_n)) is bounded in $$ and (49), We have Invoking (63), we have Consequently, by (19) and letting a = 1 and b = b(2k)/c₀, it yields Since {|u_n | ≤ k}⊂{|u_n | ≤ k + ɛ}, we have Together with Lemma 5 and the assumptions on a, we obtain which in turn implies

For any measurable set E of Ω, we have

Let ɛ > 0. Thanks to by (39), there exists k ≥ h such that While k is fixed, we get Owing to the strong compactness of {T_k(u_n)} in $$, there exists δ^′ > 0 such that if meas⁡(E) < δ^′, then Hence, Thus, the sequence {|∇u_n|^p} is equi-integrable. Thanks to Vitali theorem, the equi-integrability together with (76) implies that u_n converges strongly to u in $$.

Step 4 (the strong convergence g_n(x, u_n, ∇u_n) → g(x, u, ∇u) in L¹(Ω)). Note that (76) implies that

On the other hand, for any measurable set E of Ω, we have Let ɛ > 0 be fixed. We have Choose k ≥ h in (46) such that By (10), we have Since d(x) belongs to L¹(Ω) and T_k(u_n) is compact in $$, there exists δ^′′ > 0 such that if meas⁡(E) < δ^′′, then Thus, we have proved that {g_n(x, u_n, ∇u_n)} is equi-integrable. Invoking (86) and (88) and by Vitali theorem,

Step 5 (passing to the limit). Now, by passing to the limit in (27), we obtain

that is, u is a weak solution to problem (1). The proof is complete.