Extinction in a finite time for solutions of a class of quasilinear parabolic equations

Abstract

We study the property of extinction in a finite time for nonnegative solutions of $\frac{1}{q} \frac{\partial}{\partial t} (u^{q}) - \nabla (| \nabla u |^{p - 2} \nabla u) + a (x) u^{λ} = 0$ for the Dirichlet Boundary Conditions when $q > λ > 0$ , $p ⩾ 1 + q$ , $p ⩾ 2$ , $a (x) ⩾ 0$ and Ω a bounded domain of $R^{N}$ ( $N ⩾ 1$ ). We prove some necessary and sufficient conditions. The threshold is for power functions when $p > 1 + q$ while finite time extinction occurs for very flat potentials $a (x)$ when $p = 1 + q$ .

Keywords

Extinction time semi-classical limit nonlinear eigenvalue problem

1. Introduction and formulations of main results

Let $q > λ > 0$ , $p ⩾ 1 + q$ , $p ⩾ 2$ and $Ω \subset R^{N}$ a bounded domain of $R^{N}$ ( $N ⩾ 1$ ). We consider a nonnegative solution u of $\begin{matrix} (1) & \{\begin{matrix} \frac{1}{q} \frac{\partial}{\partial t} (u^{q}) - \nabla (| \nabla u |^{p - 2} \nabla u) + a (x) u^{λ} = 0 in Ω \times (0, + \infty) \\ u (x, 0) = u_{0} (x) in Ω \end{matrix} \end{matrix}$ with homogeneous Dirichlet Boundary Condition. Here, we study conditions on absorption potential $a (x)$ , guaranteeing of extinction of solutions in a finite time or absence of such a extinction.

Definition 1.1 (See for example [1], where corresponding existence results were proved).

An energy solution of problem (1) is the function $u (x, t) \in L_{p} (0, T; W^{1, p} (Ω, \partial Ω)) : {(u^{q})}_{t} \in L_{p^{'}} (0, T; {(W^{1, p} (Ω, \partial Ω))}^{*})$ , $u (x, 0) = u_{0} (x)$ , satisfying the following integral identity: $\begin{matrix} (2) & \int_{0}^{T} ⟨ {(u^{q})}_{t}, φ ⟩ d t + \int_{Ω \times (0, T)} | \nabla_{x} u |^{p - 2} \nabla_{x} u \nabla_{x} φ d x d t + \int_{Ω \times (0, T)} a (x) u^{λ} φ d x d t = 0, \end{matrix}$ for arbitrary $φ \in L_{p} (0, T; W^{1, p} (Ω, \partial Ω))$ , $\forall T < \infty$ , where $p^{'} = \frac{p}{p - 1}$ , $W^{1, p} (Ω, Γ)$ is closure in the norm $W^{1, p} (Ω)$ of the set of smooth functions $v = 0$ on $Γ \subset \partial Ω$ .

Definition 1.2.
Let say that problem (1) has the extinction in finite time (EFT-property) if for arbitrary solution u, there exists some positive T such that $u (x, t) = 0$ a.e. in Ω, $\forall t ⩾ T$ .

The EFT-property has been intensively studied by many authors by differents methods for several types of equations [8,9,14] and [15]. In [11], the authors introduce a new energy method called semi-classical method. The large time behaviour for solutions of some parabolic equations is linked with the asymptotic behaviour of a family of first eigenvalues in the semi-classical limit. For $q = 1$ , $p = 2$ , and homogeneous Neumann Boundary Conditions, they introduce the fundamental states of the associated Schrödinger operator $\begin{matrix} μ_{n} = inf {\int_{Ω} (| \nabla u |^{2} + 2^{n} a (x) u^{λ + 1}) d x : u \in W^{1, 2} (Ω), \int_{Ω} u^{2} d x = 1} . \end{matrix}$ By iterations and using the maximum principle, they proved that if $\begin{matrix} \sum_{n = 0}^{\infty} μ_{n}^{- 1} ln (μ_{n}) < \infty, \end{matrix}$ then the EFT-property is satisfied for equation (1) ( $p = 2$ , $q = 1$ ). The Lieb–Thirring formula [12] gives estimates by below for eigenvalues of Schrödinger operator. In [5], they use it to give some concrete examples.

Moreover, in the same article, the authors have given a necessary condition using a Kaplan’s like-method [10,13] for some homothetic test-functions like in Lemma 2.

In [2], in only two cases for $p > 2$ , $q = 1$ and $p = 2$ , $q < 1$ , for the sufficient condition, the critical power $δ_{0} = \frac{p (q - λ)}{p - (1 + q)}$ appeared under the condition $N > p$ . The proof is also by iterations and uses maximum principle.

Some properties of the first eigenvalue and the first eigenfunction are also studied in [3].

In semilinear case ( $p = 2$ , $q = 1$ ), there was obtained in [6] the most sharp explicit sufficient condition of EFT-property for absorption potential $a (x) \approx exp (- \frac{ω (| x |)}{| x |^{2}})$ , $ω (s) \to 0$ as $s \to 0$ : $\begin{matrix} \int_{0}^{c} \frac{ω (s)}{s} d s < \infty . \end{matrix}$ Our conjecture (which is not proved up to now) is that mentioned Dini-like condition is also a necessary condition for EFT-property. Corresponding Dini-like sufficient condition for high-order operators was proved in [7].

One can mention [4] and [1] for some abstract results.

We introduce the following notation for $u_{0} \neq 0$ : $\begin{matrix} (3) & T (u_{0}) = sup {t > 0, \forall τ \in [0, t], {‖ u (τ) ‖}_{L^{2} (Ω)} > 0} . \end{matrix}$ We give a sufficient condition for $p > 1 + q$ and $p = 1 + q$ and a necessary condition for $p > 1 + q$ and $p = 2 = 1 + q$ , i.e., $q = 1$ .

There are two steps for the sufficient condition:
We transform the parabolic problem into a problem of behaviour for some kind of nonlinear first eigenvalues. There is extinction in a finite time for solutions of (1) if some integral is finite. For this purpose, we define $\begin{matrix} (4) & μ (h) = inf_{v \in W^{1, p} (Ω, \partial Ω) : ‖ v ‖_{L^{1 + q} (Ω)}^{1 + q} = h} (\int_{Ω} | \nabla v |^{p} d x + \int_{Ω} a (x) | v |^{1 + λ} d x) . \end{matrix}$ We have extinction in a finite time if $\begin{matrix} (5) & \int_{0}^{1} \frac{d h}{μ (h)} < + \infty . \end{matrix}$ This condition is valid for a large class of maximal monotone operators of Hilbert Spaces. But it seems that the genuine notion is a kind of uniform upper bound for the extinction time, i.e., we define for all $h_{0} > 0$ , $\begin{matrix} (6) & T^{+} (h_{0}) = sup {T (u_{0}) : u_{0} \in W^{1, p} (Ω, \partial Ω), 0 < ‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q} ⩽ h_{0}} . \end{matrix}$ The supremum is taken on the closed ball and not only on the sphere in order to get monotonicity for the function $h \mapsto T^{+} (h)$ . But in many cases, the supremum on the closed ball is equal to the supremum on the sphere.

With this new definition, condition (5) implies that $\begin{matrix} T^{+} (h_{0}) < + \infty \forall h_{0} > 0, and lim_{h_{0} \to 0} T^{+} (h_{0}) = 0 . \end{matrix}$

Now, the point is to find estimates of $μ (h)$ . The main tool is Sobolev injections, that’s why we have the restriction $N > p$ . The quantity $μ (h)$ is estimated in two different ways depending on the value of p with respect to $1 + q$ , i.e., $p > 1 + q$ or $p = 1 + q$ .
By using a Liapounov functional, we get the following estimate: $\begin{matrix} (7) & T^{+} (h_{0}) ⩾ sup_{0 < h ⩽ h_{0}} \frac{(1 + λ)}{p (q - λ)} \frac{h}{μ (h)} . \end{matrix}$ This estimate remains valid for a large class of maximal monotone operators of Hilbert Spaces. There are two important cases:
$\begin{matrix} (8) & lim_{h \to 0} \frac{h}{μ (h)} = + \infty ⟹ T^{+} (h_{0}) = + \infty, \end{matrix}$ means that for any $h_{0} > 0$ and $T > 0$ , there is an initial data $u_{0}$ such that $T (u_{0}) > T$ . But without any further assumption, it seems difficult to get a $u_{0}$ such that $T (u_{0}) = + \infty$ in the general case. When $p = 1 + q$ , (8) never happens.

$\begin{matrix} (9) & \underset{h \to 0}{lim inf} \frac{h}{μ (h)} > 0 ⟹ lim_{h \to 0} T^{+} (h) > 0, \end{matrix}$ implies that there exists $C > 0$ such that for any $h > 0$ , one can find an initial data $u_{0}$ such that $‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q} = h$ and $T (u_{0}) ⩾ C$ . An initial data with a positive norm as small as we want exists but with an extinction time greater than a fixed positive constant.

When $p = 1 + q$ , (9) happens only when $a (x) = 0$ a.e. in Ω which is a trivial case.
For $p > 1 + q$ , by means of “good” homothetic test-function, we estimate $μ (h)$ from above.

We have seen that for $p = 1 + q$ , condition (7) isn’t relevant. For this reason, when $p = 2$ (and so $q = 1$ ), we use some kind of asymptotic Kaplan’s method, i.e., we prove that under suitable condition on $a (x)$ , for all time $T > 0$ , a “good” first eigenfunction φ such that $\int_{Ω} u (x, T) φ (x) d x > 0$ exists.

Here, we have to split $p > 1 + q$ and $p = 1 + q$ (called critical case). Calculations and estimates are differents due to the nature of (1). Indeed, for $p > 1 + q$ , there are three different degrees of homogeneity while when $p = 1 + q$ , there are only two. It provides two radically different thresholds (power functions for $p > 1 + q$ and very flat functions for the critical case).

We define for $p > 1 + q$ , $\begin{matrix} δ_{0} = \frac{p (q - λ)}{p - (1 + q)} . \end{matrix}$ We assume $\begin{matrix} (10) & a (x) = \frac{| x |^{δ_{0}}}{ω (| x |)} for x small enough and otherwise a (x) ⩾ C > 0, \end{matrix}$ where $\begin{matrix} (11) & ω is a continuous positive function on (0, ρ_{0}), ρ_{0} > 0 . \end{matrix}$ For the necessary condition, we assume that for $ρ > 0$ small enough, $\begin{array}{l} (12) & ρ \mapsto \frac{ρ^{δ_{0}}}{ω (ρ)} is a continuous nondecreasing function, \\ (13) & \begin{array}{l} ρ \mapsto ω (ρ) is a nonincreasing function, \\ \forall γ_{1} \in (0, 1), \exists C > 0, \exists γ_{2} > 0, \forall h \in (0, 1], ω (h^{γ_{1}}) ⩾ C {(ω (h))}^{γ_{2}}, \\ \exists η > 0, \forall h \in (0, h_{0}], ω (h) ⩽ h^{- η} for some h_{0} > 0 . \end{array} \end{array}$ To some extend, the assumption above means that $ρ \mapsto ω (ρ)$ is less than a polynomial function. For the sufficient condition, we assume that $\begin{matrix} (14) & ω (ρ) = \frac{1}{{(- ln ρ)}^{α}} for α > \frac{p - (1 + λ)}{p - (1 + q)} + \frac{δ_{0}}{N} . \end{matrix}$
Theorem 1.1.
Let $p > 1 + q$ , $O \in Ω$ , potential $a (x)$ has radial structure ( 10 ) and ( 11 ) near to the point 0.
Let assumption ( 12 ) holds. Then condition ${lim inf}_{ρ \to 0} ω (ρ) > 0$ implies ${lim}_{h \to 0} T^{+} (h) > 0$ .

In particular, it happens when $ω (ρ) = \frac{1}{{(- ln ρ)}^{α}}$ for $α ⩽ 0$ .

Let additionally to ( 12 ), assumption ( 13 ) holds and ${lim}_{ρ \to 0} ω (ρ) = + \infty$ . Then for arbitrary $h_{0} > 0$ , $T^{+} (h_{0}) = + \infty$ .

In particular, it happens when $ω (ρ) = \frac{1}{{(- ln ρ)}^{α}}$ with $α < 0$ .

Let $N > p$ and function ω satisfies ( 14 ). Then all solutions of problem ( 1 ) vanish in a finite time.

More precisely, $\forall h_{0} > 0$ , $T^{+} (h_{0}) < + \infty$ and ${lim}_{h_{0} \to 0} T^{+} (h_{0}) = 0$ .

In the case $p = 1 + q$ , we will consider potential $a (x)$ of the form: $\begin{matrix} (15) & a (x) = exp (- \frac{ω (| x |)}{| x |^{p}}), \end{matrix}$ where the function ω satisfies conditions:
$(A_{1})$
$ω (r)$ is a continuous and nondecreasing function $\forall r ⩾ 0$ ,
$(A_{2})$
$ω (0) = 0$ , $ω (r) > 0$ , $\forall r > 0$ ,
$(A_{3})$
$ω (r) ⩽ ω_{0} < \infty$ , $\forall r > 0$ ,
$(A_{4})$
$ω (r) ⩾ r^{p - δ}$ , $\forall s \in (0, s_{0})$ , $s_{0} > 0$ , $p > δ > 0$ ,
$(A_{5})$
the function $\frac{ω (r)}{r^{p}}$ is nonincreasing on $(0, s_{0})$ .
Theorem 1.2.
Let $p = 1 + q$ , $O \in Ω$ , $N > p$ , potential $a (x)$ has structure ( 15 ) and function $ω (ρ)$ satisfies conditions $(A_{1})$ – $(A_{5})$ . If additionally, $\begin{matrix} \int_{0}^{c} \frac{ω (r)}{r} d r < + \infty, \end{matrix}$ then all solutions of problem ( 1 ) vanish in a finite time. More precisely, $\begin{matrix} \forall h > 0, T^{+} (h) < + \infty and lim_{h \to 0} T^{+} (h) = 0 . \end{matrix}$ Moreover, if $p = 2$ (i.e. $q = 1$ ) and potential $a (x)$ has structure ( 15 ) where $ω (\cdot)$ is arbitrary non-negative function, then the following properties take place:
If ${lim inf}_{ρ \to 0} ω (ρ) > 0$ then ${lim inf}_{h \to 0} T^{+} (h) > 0$ .

If ${lim}_{ρ \to 0} ω (ρ) = + \infty$ then $\forall h > 0$ , $T^{+} (h) = + \infty$ .

The proofs of these theorems are in §4. In §2, we formulate and prove theorem about extinction properties of solutions of some abstract Cauchy Problem. §3 containts proofs of some important technical and auxiliary lemma.
2. Abstract Cauchy problem

The next theorem is the key-stone for necessary condition and sufficient condition for $p > 1 + q$ and sufficient condition for $p = 1 + q$ . In what follows, we consider that $H = L^{2} (Ω)$ but it can be easily extend to all Hilbert Space when $q = 1$ . We always denote by J a proper convex lower semi-continuous functional on $L^{2} (Ω)$ . We define $A = \partial J$ as the subdifferential of J. Let’s denote the domain of A by $D (A)$ . Let’s remember that $D (J) = {u \in H, J (u) < + \infty}$ . For simplicity’s sake, we assume that A is a single-valued operator but it’s possible to extend to multivalued operators. Let u be a solution of the following Cauchy Problem: $\begin{matrix} (16) & \frac{1}{q} \frac{\partial}{\partial t} (| u |^{q - 1} u) + A u = 0, u (0) = u_{0} . \end{matrix}$ We define for all $u_{0} \in D (A) ∖ {0}$ , $\begin{matrix} (17) & T (u_{0}) : = sup {t > 0, \forall τ \in [0, t], u (τ) \neq 0} . \end{matrix}$ Moreover, for all $h_{0} > 0$ , define also: $\begin{matrix} (18) & T^{+} (h_{0}) = sup {T (u_{0}) : u_{0} \in D (A) \cap L^{1 + q} (Ω), 0 < ‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q} ⩽ h_{0}} . \end{matrix}$ We suppose that the following assumptions are fulfilled: $\begin{array}{l} H_{0} : inf_{v \in D (J)} J (v) = J (0) = 0 . \\ H_{1} : \forall h > 0, D (A) \cap {v \in L^{1 + q} (Ω), ‖ v ‖_{L^{1 + q} (Ω)}^{1 + q} = h} \neq \emptyset . \\ H_{2} : \forall h > 0, μ (h) > 0 . \\ H_{3} : \exists p > 1, \forall v \in D (A), ⟨ A v, v ⟩ ⩽ p J (v) . \\ H_{4} : \exists λ \in [0, p - 1), \forall v \in D (A), ⟨ A v, v ⟩ ⩾ (1 + λ) J (v) . \end{array}$

Remark 2.1.

Assumption $H_{0}$ implies $A 0 = 0$ .

Assumptions $H_{2}$ and $H_{3}$ imply $J (v) > 0$ for all $v \in D (J) ∖ {0}$ .

Assumption $H_{4}$ is always true for $λ = 0$ .

To some extend, $H_{3}$ means that J is some kind of polynome in u and of all its derivative.

Let’s define for all $h > 0$ , $\begin{matrix} (19) & μ (h) = inf_{v \in D (A) \cap {v \in L^{1 + q} (Ω), ‖ v ‖_{L^{1 + q} (Ω)}^{1 + q} = h}} ⟨ A v, v ⟩ . \end{matrix}$
Theorem 2.1.
Let $u_{0} \in D (A) ∖ {0}$ . Under $H_{0}$ – $H_{4}$ , for solutions of ( 16 ), we have the following results:
If $\int_{0}^{1} \frac{d h}{μ (h)} < + \infty$ then $\forall h > 0$ , $T^{+} (h) < + \infty$ and ${lim}_{h \to 0} T^{+} (h) = 0$ .

If $\forall h > 0$ , $T^{+} (h) < + \infty$ and ${lim}_{h \to 0} T^{+} (h) = 0$ then ${lim}_{h \to 0} \frac{h}{μ (h)} = 0$ .
If the previous conditions are fulfilled then we have the following estimates: $\begin{matrix} (20) & T (u_{0}) ⩽ T^{+} (‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}) ⩽ \frac{1}{1 + q} \int_{0}^{‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}} \frac{d h}{μ (h)}, \end{matrix}$ and $\begin{matrix} (21) & T^{+} (h_{0}) ⩾ sup_{0 < h < h_{0}} \frac{(1 + λ)}{p (q - λ)} \frac{h}{μ (h)} . \end{matrix}$
Proof.
For the sufficient condition, we assume that $\int_{0}^{1} \frac{d h}{μ (h)} < + \infty$ .

Taking the scalar product with u leads to $\begin{matrix} (22) & \frac{1}{1 + q} \frac{d}{d t} (‖ u ‖_{L^{1 + q} (Ω)}^{1 + q}) + ⟨ A u, u ⟩ = 0 . \end{matrix}$ But by definition of $μ (h)$ , $⟨ A u, u ⟩ ⩾ μ (‖ u ‖_{L^{1 + q} (Ω)}^{1 + q})$ . So, (22) yields $\begin{matrix} \frac{1}{1 + q} \frac{d}{d t} (‖ u ‖_{L^{1 + q} (Ω)}^{1 + q}) + μ (‖ u ‖_{L^{1 + q} (Ω)}^{1 + q}) ⩽ 0 . \end{matrix}$ Since $μ (h) > 0$ for all $h > 0$ , $\frac{\frac{d}{d t} (‖ u ‖_{L^{1 + q} (Ω)}^{1 + q})}{μ (‖ u ‖_{L^{1 + q} (Ω)}^{1 + q})} ⩽ - (1 + q)$ .

Integrating between 0 and $t < T (u_{0})$ , we get: $\begin{matrix} \int_{‖ u ‖_{L^{1 + q} (Ω)}^{1 + q}}^{‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}} \frac{d h}{μ (h)} ⩾ (1 + q) t . \end{matrix}$ This inequality yields $\begin{matrix} \int_{0}^{‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}} \frac{d h}{μ (h)} ⩾ (1 + q) t, \forall t < T (u_{0}) . \end{matrix}$ Passing to the limit $t \to T (u_{0})$ gives $\begin{matrix} \int_{0}^{‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}} \frac{d h}{μ (h)} ⩾ (1 + q) T (u_{0}), \end{matrix}$ and therefore $T (u_{0}) < + \infty$ . Moreover, $T^{+} (h) ⩽ \frac{1}{1 + q} \int_{0}^{h} \frac{d θ}{μ (θ)}$ and $T^{+} (h) \to 0$ as $h \to 0$ .

For the necessary condition, we have for the scalar product in $L^{2} (Ω)$ , $\begin{matrix} \frac{d}{d t} J (u (t)) = ⟨ u_{t}, A u ⟩, \end{matrix}$ from the theory of Maximum Monotone Operators. But $A u = u_{t} | u |^{q - 1}$ so, $\begin{matrix} \frac{d}{d t} J (u (t)) = - ⟨ u_{t}, u_{t} | u |^{q - 1} ⟩ . \end{matrix}$ Since $L^{2} (Ω)$ is an algebra, $\begin{matrix} \frac{d}{d t} J (u (t)) = - \int_{Ω} u_{t} u_{t} | u |^{q - 1} d x = - \int_{Ω} u_{t} | u |^{\frac{q - 1}{2}} u_{t} | u |^{\frac{q - 1}{2}} d x . \end{matrix}$ So, $\frac{d}{d t} J (u (t)) = - ‖ u_{t} | u |^{\frac{q - 1}{2}} ‖_{L^{2} (Ω)}^{2}$ .

According to Cauchy–Schwarz–Buniakowsky inequality, $\begin{matrix} {| ⟨ u_{t}, u | u |^{q - 1} ⟩ |}^{2} = {| ⟨ u_{t} | u |^{\frac{q - 1}{2}}, u | u |^{\frac{q - 1}{2}} ⟩ |}^{2} ⩽ {‖ u_{t} | u |^{\frac{q - 1}{2}} ‖}_{L^{2} (Ω)}^{2} ‖ u ‖_{L^{1 + q} (Ω)}^{1 + q} . \end{matrix}$ Hence, $\begin{matrix} {(⟨ A u (t), u (t) ⟩)}^{2} ⩽ (- \frac{d}{d t} J (u (t))) {‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q} . \end{matrix}$ So, $\begin{matrix} (23) & \frac{d}{d t} J (u (t)) ⩽ - \frac{{(⟨ A u (t), u (t) ⟩)}^{2}}{‖ u (t) ‖_{L^{1 + q} (Ω)}^{1 + q}} = \frac{1}{1 + q} \frac{d}{d t} ({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q}) \frac{⟨ A u (t), u (t) ⟩}{‖ u (t) ‖_{L^{1 + q} (Ω)}^{1 + q}} . \end{matrix}$ Then, using $H_{4}$ , $\begin{matrix} \frac{d}{d t} J (u (t)) ⩽ \frac{1 + λ}{1 + q} \frac{d}{d t} ({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q}) \frac{J (u (t))}{‖ u (t) ‖_{L^{1 + q} (Ω)}^{1 + q}} . \end{matrix}$ By integration, we obtain $\begin{matrix} (24) & J (u (t)) ⩽ \frac{J (u_{0})}{{(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}}} {({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}} . \end{matrix}$ By $H_{3}$ , $\begin{matrix} \frac{1}{1 + q} \frac{d}{d t} ({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q}) + p J (u (t)) ⩾ 0 . \end{matrix}$ Therefore, using additionally (24), we get $\begin{matrix} \frac{1}{1 + q} \frac{d}{d t} ({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q}) + \frac{p J (u_{0})}{{(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}}} {({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}} ⩾ 0 . \end{matrix}$ So, $\begin{matrix} \frac{\frac{d}{d t} (‖ u (t) ‖_{L^{1 + q} (Ω)}^{1 + q})}{{(‖ u (t) ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}}} + \frac{(1 + q) p J (u_{0})}{{(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}}} ⩾ 0 . \end{matrix}$ By integration between 0 and t, $\begin{array}{l} \frac{1 + q}{q - λ} {({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{q - λ}{1 + q}} - \frac{1 + q}{q - λ} {(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{q - λ}{1 + q}} + \frac{(1 + q) p J (u_{0})}{{(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}}} t ⩾ 0, \\ \frac{1}{q - λ} {({‖ u (t) ‖}_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{q - λ}{1 + q}} ⩾ \frac{1}{q - λ} {(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{q - λ}{1 + q}} - \frac{p J (u_{0})}{{(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}}} t . \end{array}$ We take $t = T (u_{0})$ . Hence, $‖ u (T (u_{0})) ‖_{L^{1 + q} (Ω)}^{1 + q} = 0$ . Therefore, $\begin{array}{l} \frac{p J (u_{0})}{{(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{1 + λ}{1 + q}}} T (u_{0}) ⩾ \frac{1}{q - λ} {(‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q})}^{\frac{q - λ}{1 + q}}, \end{array}$ which yields $\begin{matrix} (25) & T (u_{0}) ⩾ \frac{1}{p (q - λ)} \frac{‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}}{J (u_{0})} ⟹ T (u_{0}) ⩾ \frac{(1 + λ)}{p (q - λ)} \frac{‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}}{⟨ A u_{0}, u_{0} ⟩} . \end{matrix}$ Due to (25), it follows from definition (6) with $h_{0} > 0$ : $\begin{matrix} (26) & T^{+} (h_{0}) ⩾ sup_{0 < ‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q} ⩽ h_{0}} \frac{(1 + λ)}{p (q - λ)} \frac{‖ u_{0} ‖_{L^{1 + q} (Ω)}^{1 + q}}{⟨ A u_{0}, u_{0} ⟩} . \end{matrix}$ Let $h \in (0, h_{0}]$ . By definition (4), there exists a sequence $(u_{n})$ such that $‖ u_{n} ‖_{L^{1 + q} (Ω)}^{1 + q} = h$ and $⟨ A u_{n}, u_{n} ⟩ \to μ (h)$ . Passing to the limit in (26), we get finally $\begin{matrix} (27) & T^{+} (h_{0}) ⩾ sup_{0 < h ⩽ h_{0}} \frac{(1 + λ)}{p (q - λ)} \frac{h}{μ (h)} . \end{matrix}$ □

Let $Ω \subset R^{N}$ a bounded domain with $O \in Ω$ . In $L^{2} (Ω)$ , let’s take $\begin{matrix} J (u) = \frac{1}{p} \int_{Ω} | \nabla u |^{p} d x + \frac{1}{1 + λ} \int_{Ω} a (x) | u |^{1 + λ} d x, \end{matrix}$ for $0 < λ < q ⩽ p - 1$ with $D (J) = W^{1, p} (Ω, \partial Ω)$ for $p ⩾ 2$ . Proposition 1.
The functional J satisfies assumptions $H_{0}$ – $H_{4}$ .
Proof.
Assumptions $H_{0}$ and $H_{1}$ are satisfied when $D (J) = W^{1, p} (Ω, \partial Ω)$ . Using Poincaré’s inequality and Hölder’s inequality, we get $\begin{matrix} ⟨ A v, v ⟩ ⩾ C \int_{Ω} | v |^{p} d x ⩾ C^{'} {(\int_{Ω} | v |^{1 + q} d x)}^{\frac{p}{1 + q}}, \end{matrix}$ which means $μ (h) ⩾ C^{'} h^{\frac{p}{1 + q}} > 0$ for all $h > 0$ . So $H_{2}$ is valid. Moreover, it is clear that $H_{3}$ and $H_{4}$ are also valid. □

3. Auxiliary statements

The following lemma gives an upper bound for $μ (h)$ :

Lemma 1.
Let $p > 1 + q$ , $O \in Ω$ , and conditions ( 10 ) and ( 12 ) are satisfied. Then $\begin{matrix} (28) & μ (h) ⩽ C (p, q, λ, v) h (1 + \frac{1}{ω (h^{ε_{0}})}) . \end{matrix}$
Proof.
Since $O \in Ω$ then there exists $ρ_{0} > 0$ such that $B_{ρ_{0}}$ , the ball of center O and radius $ρ_{0}$ , is included in Ω. Let $v \in C_{0}^{\infty} (B_{1})$ be a nonnegative function such that $\int_{B_{1}} v^{1 + q} = 1$ .

Let ρ be in $(0, ρ_{0})$ . We define $v_{ρ} (x) = v (\frac{x}{ρ})$ . Then, $\begin{matrix} \int_{Ω} v_{ρ}^{1 + q} (x) d x = \int_{B_{ρ}} v_{ρ}^{1 + q} (x) d x = \int_{B_{ρ}} v^{1 + q} (\frac{x}{ρ}) d x = ρ^{N} \int_{B_{1}} v^{1 + q} (y) d y = ρ^{N} . \end{matrix}$ As a consequence, $‖ \frac{v_{ρ}}{ρ^{\frac{N}{1 + q}}} ‖_{L^{1 + q} (Ω)} = 1$ . On the other hand, $\nabla_{x} v_{ρ} (x) = ρ^{- 1} \nabla_{y} v (\frac{x}{ρ})$ . Hence, since the support of v is in $B_{1}$ , $\begin{matrix} \int_{Ω} | \nabla v_{ρ} |^{p} d x = \int_{B_{ρ}} | \nabla v_{ρ} |^{p} d x = ρ^{N - p} \int_{B_{1}} | \nabla v |^{p} d y . \end{matrix}$ By using $h^{\frac{1}{1 + q}} \frac{v_{ρ}}{ρ^{\frac{N}{1 + q}}}$ in the definition of $μ (h)$ , we derive $\begin{matrix} μ (h) ⩽ h^{\frac{p}{1 + q}} ρ^{- \frac{p N}{1 + q}} \int_{Ω} | \nabla v_{ρ} |^{p} d x + h^{\frac{1 + λ}{1 + q}} ρ^{- N \frac{1 + λ}{1 + q}} \int_{Ω} a (x) v_{ρ}^{1 + λ} d x . \end{matrix}$ Therefore, $\begin{matrix} (29) & μ (h) ⩽ h^{\frac{p}{1 + q}} ρ^{N - p - \frac{p N}{1 + q}} \int_{B_{1}} | \nabla v |^{p} d y + h^{\frac{1 + λ}{1 + q}} ρ^{N - N \frac{1 + λ}{1 + q}} \int_{B_{1}} a (ρ y) v^{1 + λ} (y) d y . \end{matrix}$ Since $ρ \mapsto \frac{ρ^{δ_{0}}}{ω (ρ)}$ is a nondecreasing function for h small enough, $\begin{matrix} a (ρ y) = \frac{{(ρ ‖ y ‖)}^{δ_{0}}}{ω (ρ ‖ y ‖)} ⩽ \frac{ρ^{δ_{0}}}{ω (ρ)}, \end{matrix}$ for $y \in B_{1}$ . Therefore, $\begin{array}{l} μ (h) & ⩽ C (p, q, λ, v) (h^{\frac{p}{1 + q}} ρ^{N - p - \frac{p N}{1 + q}} + h^{\frac{1 + λ}{1 + q}} ρ^{N - N \frac{1 + λ}{1 + q} + δ_{0}} \frac{1}{ω (ρ)}) . \\ (30) & = C (p, q, λ, v) (h^{\frac{p}{1 + q}} ρ^{- \frac{N (p - (1 + q))}{1 + q} - p} + h^{\frac{1 + λ}{1 + q}} ρ^{N \frac{q - λ}{1 + q} + δ_{0}} \frac{1}{ω (ρ)}) . \end{array}$ We take $ρ = h^{ε_{0}}$ for $ε_{0} > 0$ . Then estimate (30) yields: $\begin{matrix} μ (h) ⩽ C (p, q, λ, v) (h^{\frac{p}{1 + q} - ε_{0} \frac{N (p - (1 + q))}{1 + q} - ε_{0} p} + h^{\frac{1 + λ}{1 + q} + ε_{0} N \frac{q - λ}{1 + q} + ε_{0} δ_{0}} \frac{1}{ω (h^{ε_{0}})}) . \end{matrix}$ We choose $ε_{0}$ such that $\begin{matrix} \frac{p}{1 + q} - ε_{0} \frac{N (p - (1 + q))}{1 + q} - ε_{0} p = \frac{λ + 1}{1 + q} + ε_{0} N \frac{q - λ}{1 + q} + ε_{0} δ_{0}, \end{matrix}$ i.e., $\begin{matrix} ε_{0} = \frac{p - (1 + λ)}{N (p - (1 + λ)) + (δ_{0} + p) (1 + q)} . \end{matrix}$ Hence, simple computations lead to $\begin{array}{l} \frac{1 + λ}{1 + q} + ε_{0} N \frac{q - λ}{1 + q} + ε_{0} δ_{0} \\ = \frac{(1 + q) N (p - (1 + λ)) + p (1 + λ) (1 + q) + p (1 + q) δ_{0}}{(1 + q) (N (p - (1 + λ)) + (δ_{0} + p) (1 + q))} . \end{array}$ But according to the definition of $δ_{0}$ , $p (1 + λ) + p δ_{0} = (δ_{0} + p) (1 + q)$ which leads to $\begin{matrix} \frac{p}{1 + q} - ε_{0} \frac{N (p - (1 + q))}{1 + q} - ε_{0} p = \frac{λ + 1}{1 + q} + ε_{0} N \frac{q - λ}{1 + q} + ε_{0} δ_{0} = 1 . \end{matrix}$ With this $ε_{0}$ , we obtain the result. □
Lemma 2.
Let $p > 1 + q$ , $O \in Ω$ , and conditions ( 10 ) and ( 12 ) are satisfied.
If ${lim}_{ρ \to 0} ω (ρ) = 0$ then $\begin{matrix} (31) & μ (h) ⩽ C \frac{h}{ω (h^{ε_{0}})} \end{matrix}$ for some $C < \infty$ , where $ε_{0} = \frac{p - (1 + λ)}{N (p - (1 + λ)) + (δ_{0} + p) (1 + q)}$ .

If ${lim inf}_{ρ \to 0} ω (ρ) > 0$ then $\begin{matrix} (32) & \underset{h \to 0}{lim inf} \frac{h}{μ (h)} > 0 . \end{matrix}$

If additionally conditions ( 13 ) and ${lim}_{ρ \to 0} ω (ρ) = + \infty$ are satisfied, then $\begin{matrix} (33) & lim_{h \to 0} \frac{h}{μ (h)} = + \infty . \end{matrix}$

Proof.
We start with the estimate (28): $\begin{matrix} μ (h) ⩽ C (p, q, λ, v) h (1 + \frac{1}{ω (h^{ε_{0}})}) . \end{matrix}$ If ${lim}_{ρ \to 0} ω (ρ) = 0$ , then for h small enough, $μ (h) ⩽ C \frac{h}{ω (h^{ε_{0}})}$ , $C = const < + \infty$ .

If ${lim inf}_{ρ \to 0} ω (ρ) > 0$ , then ${lim inf}_{h \to 0} \frac{h}{μ (h)} > 0$ . Let’s go back to (30). We take $ρ = h^{ε_{0}} ω {(h)}^{α}$ for $α > 0$ . This value of $ε_{0}$ implies that $\begin{matrix} \frac{μ (h)}{h} ⩽ C ({(ω (h))}^{- α (\frac{N (p - (1 + q))}{1 + q} + p)} + {(ω (h))}^{α (N \frac{q - λ}{1 + q} + δ_{0})} \frac{1}{ω (h^{ε_{0}} ω {(h)}^{α})}), \end{matrix}$ where $C = C (p, q, λ, v) < + \infty$ . Using (13), we can find $h_{0} > 0$ such that for all $0 < h ⩽ h_{0}$ , $h^{ε_{0}} ω {(h)}^{α} ⩽ h^{ε_{0} - α η}$ . Moreover, since ω is a nonincreasing function, $ω (h^{ε_{0}} ω {(h)}^{α}) ⩾ ω (h^{ε_{0} - α η})$ . It is always possible to assume that $ε_{0} - α η ⩾ \frac{ε_{0}}{2}$ .

So, if h belongs to $(0, min (h_{0}, 1))$ , $h^{ε_{0} - α η} ⩽ h^{\frac{ε_{0}}{2}}$ . Always for the same reason, $ω (h^{ε_{0} - α η}) ⩾ ω (h^{\frac{ε_{0}}{2}})$ , which leads to $ω (h^{ε_{0}} ω {(h)}^{α}) ⩾ ω (h^{\frac{ε_{0}}{2}})$ . The previous inequality is true for all $0 < h ⩽ min (h_{0}, 1)$ and for all $α \in (0, \frac{ε_{0}}{2 η})$ . So, we use assumption (13) with $γ_{1} = \frac{ε_{0}}{2}$ . There exist $C > 0$ and $γ_{2} > 0$ such that $ω (h^{\frac{ε_{0}}{2}}) ⩾ C {(ω (h))}^{γ_{2}}$ . Therefore, $ω (h^{ε_{0}} ω {(h)}^{α}) ⩾ C {(ω (h))}^{γ_{2}}$ . $\begin{matrix} \frac{{(ω (h))}^{α (N \frac{q - λ}{1 + q} + δ_{0})}}{ω (h^{ε_{0}} ω {(h)}^{α})} ⩽ \frac{1}{C} \frac{1}{{(ω (h))}^{γ_{2} - α (N \frac{q - λ}{1 + q} + δ_{0})}} . \end{matrix}$ We can take $α > 0$ small enough such that $\begin{matrix} γ_{2} - α (N \frac{q - λ}{1 + q} + δ_{0}) > 0 . \end{matrix}$ Finally, ${lim}_{h \to 0} \frac{μ (h)}{h} = 0$ , since ${lim}_{ρ \to 0} ω (ρ) = 0$ . □
Lemma 3.
Let $p > 1 + q$ , $N > p$ , $O \in Ω$ , and $a (x)$ is a radial function which satisfies ( 10 ) and ( 14 ). Then $\begin{matrix} (34) & \int_{0}^{1} \frac{d h}{μ (h)} < + \infty . \end{matrix}$
Proof.
For all $h > 0$ , there is $v_{h} \in W^{1, p} (Ω, \partial Ω)$ such that $‖ v_{h} ‖_{L^{1 + q} (Ω)}^{1 + q} = h$ and $\begin{matrix} 2 μ (h) ⩾ \int_{Ω} | \nabla v_{h} |^{p} d x + \int_{Ω} a (x) v_{h}^{1 + λ} d x . \end{matrix}$ To simplify, we drop h for $v_{h} : = v$ . So, $\begin{matrix} \int_{Ω} | \nabla v |^{p} d x ⩽ \frac{2 μ (h)}{h} \int_{Ω} v^{1 + q} d x - \int_{Ω} a (x) v^{1 + λ} d x . \end{matrix}$ Since $p > 1 + q$ , $\begin{matrix} \int_{Ω} | \nabla v |^{p} d x ⩽ \int_{Ω} v^{p} (\frac{2 μ (h)}{h} \frac{1}{v^{p - (1 + q)}} - \frac{a (x)}{v^{p - (1 + λ)}}) d x . \end{matrix}$ We define $Γ (x, h) : = \frac{2 μ (h)}{h} \frac{1}{v^{p - (1 + q)}} - \frac{a (x)}{v^{p - (1 + λ)}}$ . Sobolev injection and Hölder inequality when $N > p$ give $\begin{matrix} C_{S} {(\int_{Ω} v^{p^{}} d x)}^{\frac{p}{p^{}}} ⩽ {(\int_{Ω} v^{p^{}} d x)}^{\frac{p}{p^{}}} {(\int_{Ω} {({[Γ (x, h)]}^{+})}^{\frac{N}{p}} d x)}^{\frac{p}{N}}, C_{S} > 0, \end{matrix}$ when $p^{} = \frac{p N}{p - N}$ . Hence, $C_{S}^{\frac{N}{p}} ⩽ \int_{Ω} {({[Γ (x, h)]}^{+})}^{\frac{N}{p}} d x = \int_{{x : a (x) ⩽ \frac{2 μ (h)}{h} v^{q - λ} (x)}} Γ {(x, h)}^{\frac{N}{p}} d x$ .

We define $f (v) : = \frac{β}{v^{p - (1 + q)}} - \frac{α}{v^{p - (1 + λ)}}$ , $v > 0$ , where $α, β > 0$ . It is easy to see that ${max}_{v > 0} f (v) = f (v_{M})$ , $v_{M} = {(\frac{α}{β} \frac{p - (1 + λ)}{p - (1 + q)})}^{\frac{1}{q - λ}}$ . Therefore $\begin{matrix} f (v_{M}) = \frac{β}{{(\frac{α}{β} \frac{p - (1 + λ)}{p - (1 + q)})}^{\frac{p - (1 + q)}{q - λ}}} - \frac{α}{{(\frac{α}{β} \frac{p - (1 + λ)}{p - (1 + q)})}^{\frac{p - (1 + λ)}{q - λ}}} = C (p, q, λ) \frac{β^{\frac{p - (1 + λ)}{q - λ}}}{α^{\frac{p - (1 + q)}{q - λ}}}, \end{matrix}$ where $C (p, q, λ) = [{(\frac{p - (1 + q)}{p - (1 + λ)})}^{\frac{p - (1 + q)}{q - λ}} - {(\frac{p - (1 + q)}{p - (1 + λ)})}^{\frac{p - (1 + λ)}{q - λ}}]$ . Since $meas {x : a (x) = 0} = 0$ , we deduce that for almost all $x \in {x : a (x) ⩽ \frac{2 μ (h)}{h} v^{q - λ} (x)}$ , $Γ (x, h) ⩽ \frac{C (p, q, λ)}{a {(x)}^{\frac{p - (1 + q)}{q - λ}}} {(\frac{2 μ (h)}{h})}^{\frac{p - (1 + λ)}{q - λ}}$ . Moreover, let $ε > 0$ . We have $\begin{matrix} h = ‖ v ‖_{L^{1 + q} (Ω)}^{1 + q} = \int_{Ω} v^{1 + q} d x ⩾ \int_{v^{1 + q} ⩾ ε} v^{1 + q} d x ⩾ ε meas {x : v^{1 + q} ⩾ ε} . \end{matrix}$ By taking $ε = ε (h)$ , we obtain $\begin{matrix} \frac{h}{ε (h)} ⩾ meas {x : v^{1 + q} ⩾ ε (h)} = meas {x : v ⩾ ε {(h)}^{\frac{1}{1 + q}}} . \end{matrix}$ So, $\begin{array}{l} C_{S}^{\frac{N}{p}} ⩽ & \int_{{x : a (x) ⩽ \frac{2 μ (h)}{h} v^{q - λ} (x)} \cap {x : v ⩾ ε {(h)}^{\frac{1}{1 + q}}}} {({[Γ (x, h)]}^{+})}^{\frac{N}{p}} d x \\ + \int_{{x : a (x) ⩽ \frac{2 μ (h)}{h} v^{q - λ} (x)} \cap {x : v < ε {(h)}^{\frac{1}{1 + q}}}} {({[Γ (x, h)]}^{+})}^{\frac{N}{p}} d x . \end{array}$ It follows that $C_{S}^{\frac{N}{p}} ⩽ \int_{{x : v ⩾ ε {(h)}^{\frac{1}{1 + q}}}} {({[Γ (x, h)]}^{+})}^{\frac{N}{p}} d x + \int_{{x : a (x) ⩽ \frac{2 μ (h)}{h} ε {(h)}^{\frac{q - λ}{1 + q}}}} {({[Γ (x, h)]}^{+})}^{\frac{N}{p}} d x$ . Hence, $\begin{array}{l} C_{S}^{\frac{N}{p}} ⩽ & \int_{{x : v ⩾ ε {(h)}^{\frac{1}{1 + q}}}} {(\frac{2 μ (h)}{h} \frac{1}{ε {(h)}^{\frac{p - (1 + q)}{1 + q}}})}^{\frac{N}{p}} d x \\ + C {(p, q, λ)}^{\frac{N}{p}} \int_{{x : a (x) ⩽ \frac{2 μ (h)}{h} ε {(h)}^{\frac{q - λ}{1 + q}}}} {({(\frac{2 μ (h)}{h})}^{\frac{p - (1 + λ)}{q - λ}} \frac{1}{a {(x)}^{\frac{p - (1 + q)}{q - λ}}})}^{\frac{N}{p}} d x . \end{array}$ Thus, it follows estimate: $\begin{array}{l} 0 < C = & C (Ω, N, p, q, λ) ⩽ {(\frac{μ (h)}{h} \frac{1}{ε {(h)}^{\frac{p - (1 + q)}{1 + q}}})}^{\frac{N}{p}} \frac{h}{ε (h)} \\ + {(\frac{μ (h)}{h})}^{\frac{N (p - (1 + λ))}{p (q - λ)}} \int_{{x : a (x) ⩽ \frac{2 μ (h)}{h} ε {(h)}^{\frac{q - λ}{1 + q}}}} \frac{d x}{a {(x)}^{\frac{N}{δ_{0}}}}, δ_{0} = \frac{p (q - λ)}{p - (1 + q)} . \end{array}$ This inequality yields $\begin{array}{l} 0 < C ⩽ & {(\frac{μ (h)}{h} \frac{1}{ε {(h)}^{\frac{p - (1 + q)}{1 + q}}})}^{\frac{N}{p}} \frac{h}{ε (h)} \\ (35) & + {(\frac{μ (h)}{h})}^{\frac{N (p - (1 + λ))}{p (q - λ)}} \int_{{x : a (x) ⩽ \frac{2 μ (h)}{h} ε {(h)}^{\frac{q - λ}{1 + q}}}} \frac{ω {(| x |)}^{\frac{N}{δ_{0}}}}{| x |^{N}} d x . \end{array}$ For $ρ > 0$ small enough, we define functions $\begin{matrix} ϕ : ρ \mapsto \frac{ρ}{ω {(ρ)}^{\frac{N}{δ_{0}}}} and ψ : ρ \mapsto \frac{ρ^{N}}{ω {(ρ)}^{\frac{N}{δ_{0}}}} . \end{matrix}$ Both functions are continuous, increasing and $ψ (ρ) = ρ^{N - 1} ϕ (ρ)$ . Since the function $a (x) = ψ {(| x |)}^{\frac{δ_{0}}{N}}$ is radial, we use a change of variables in radial coordinates. As results, estimate (35) yields $\begin{matrix} 0 < C ⩽ & {(\frac{μ (h)}{h} \frac{1}{ε {(h)}^{\frac{p - (1 + q)}{1 + q}}})}^{\frac{N}{p}} \frac{h}{ε (h)} \\ + {(\frac{μ (h)}{h})}^{\frac{N (p - (1 + λ))}{p (q - λ)}} \int_{0}^{ψ^{- 1} ({(\frac{2 μ (h)}{h} ε {(h)}^{\frac{q - λ}{1 + q}})}^{\frac{N}{δ_{0}}})} \frac{1}{ϕ (ρ)} d ρ . \end{matrix}$ Using (31), we have $μ (h) ⩽ C \frac{h}{ω (h^{ε_{0}})}$ which leads to, $\begin{array}{l} 0 < C ⩽ & {(\frac{1}{ω (h^{ε_{0}})} \frac{1}{ε {(h)}^{\frac{p - (1 + q)}{1 + q}}})}^{\frac{N}{p}} \frac{h}{ε (h)} \\ + {(\frac{μ (h)}{h})}^{\frac{N (p - (1 + λ))}{p (q - λ)}} \int_{0}^{ψ^{- 1} ({(\frac{2 μ (h)}{h} ε {(h)}^{\frac{q - λ}{1 + q}})}^{\frac{N}{δ_{0}}})} \frac{1}{ϕ (ρ)} d ρ . \end{array}$ Let us define $γ_{0} : = \frac{p (1 + q)}{N (p - (1 + q)) + p (1 + q)}$ . We take $0 < γ < γ_{0}$ and $ε (h) = h^{γ}$ . So, since $ω (ρ) ≫ ρ^{ε}$ for all $ε > 0$ , $\begin{matrix} {(\frac{1}{ω (h^{ε_{0}})} \frac{1}{ε {(h)}^{\frac{p - (1 + q)}{1 + q}}})}^{\frac{N}{p}} \frac{h}{ε (h)} \to 0 . \end{matrix}$ As a consequence, $\begin{matrix} 0 < C ⩽ {(\frac{μ (h)}{h})}^{\frac{N (p - (1 + λ))}{p (q - λ)}} \int_{0}^{ψ^{- 1} ({(\frac{2 μ (h)}{h} ε {(h)}^{\frac{q - λ}{1 + q}})}^{\frac{N}{δ_{0}}})} \frac{1}{ϕ (ρ)} d ρ . \end{matrix}$ By definition, for $ρ > 0$ small enough, $ψ {(ρ)}^{\frac{1}{N}} = \frac{ρ}{ω {(ρ)}^{\frac{1}{δ_{0}}}}$ . We deduce that for $r > 0$ small enough, $ψ^{- 1} (r) = r^{\frac{1}{N}} ω {(ψ^{- 1} (r))}^{\frac{1}{δ_{0}}} ⩽ r^{\frac{1}{N}} ω_{0}^{\frac{1}{δ_{0}}}$ . Hence, $\begin{matrix} 0 < C ⩽ {(\frac{μ (h)}{h})}^{\frac{N (p - (1 + λ))}{p (q - λ)}} \int_{0}^{K h^{\frac{γ (q - λ)}{(1 + q) δ_{0}}}} \frac{1}{ϕ (ρ)} d ρ, \end{matrix}$ for $γ \in (0, γ_{0})$ , $K > 0$ and $h > 0$ small enough. For $ω (ρ) = {(- ln ρ)}^{- α}$ , inequality $α > \frac{δ_{0}}{N}$ guarantees $\int_{0}^{1} \frac{ω {(ρ)}^{\frac{N}{δ_{0}}}}{ρ} d ρ < + \infty$ . So, for $C > 0$ and $h > 0$ small enough, $0 < C ⩽ {(\frac{μ (h)}{h})}^{\frac{N (p - (1 + λ))}{p (q - λ)}} {(- ln h)}^{\frac{δ_{0} - α N}{δ_{0}}}$ . Hence, $\begin{matrix} \frac{C}{μ (h)} ⩽ \frac{1}{h {(- ln h)}^{\frac{α N - δ_{0}}{δ_{0}} \frac{p (q - λ)}{N (p - (1 + λ))}}} = \frac{1}{h {(- ln h)}^{\frac{(α N - δ_{0}) (p - (1 + q))}{N (p - (1 + λ)}}} . \end{matrix}$ Thus, for $α > \frac{p - (1 + λ)}{p - (1 + q)} + \frac{δ_{0}}{N}$ , we obtain $\int_{0}^{1} \frac{d h}{μ (h)} < + \infty$ . □
Lemma 4.
Let* $p = 1 + q = 2$ , $O \in Ω$ and $a (x)$ has the structure ( 15 ). Then for arbitrary $h_{0} > 0$ , there exist $u_{0} \in W_{0}^{1, 2} (Ω)$ with $0 < ‖ u_{0} ‖_{L^{2} (Ω)} ⩽ h_{0}$ and $C > 0$ such that for all $r > 0$ small enough, $T (u_{0}) ⩾ C ω (r)$ . Moreover, if additionally, ${lim}_{r \to 0} ω (r) = + \infty$ , then $T^{+} (h) = + \infty$ , for all $h > 0$ .
Proof.
The proof is a simpler version of corresponding proof in [5]. Let’s take $α > 1$ , $r > 0$ and $u_{0} \in W_{0}^{1, 2} (Ω)$ such that $‖ u_{0} ‖_{L^{\infty} (Ω)} = 1$ and $u_{0} = 1$ in a neighbourhood of O. We define by $φ_{r}$ the first eigenfunction for $- Δ$ in the ball of centre 0 and radius r for the Dirichlet boundary condition. We normalize it to 1, i.e., $‖ φ_{r} ‖_{L^{2} (Ω)} = 1$ . The first eigenvalue is denoted by $λ_{1, r}$ . The function $φ_{r}^{α}$ and its normal derivative is equal to zero on the sphere of radius r. Therefore, $\begin{matrix} \frac{d}{d t} \int_{B_{r}} u (x, t) φ_{r}^{α} (x) d x - \int_{B_{r}} u (x, t) Δ (φ_{r}^{α}) (x) d x + \int_{B_{r}} a (x) u^{λ} (x, t) φ_{r}^{α} (x) d x = 0 . \end{matrix}$ But $\begin{matrix} Δ (φ_{r}^{α}) = α (α - 1) φ_{r}^{α - 2} | \nabla φ_{r} |^{2} + α φ_{r}^{α - 1} Δ φ_{r} = α (α - 1) φ_{r}^{α - 2} | \nabla φ_{r} |^{2} - α λ_{1, r} φ_{r}^{α} . \end{matrix}$ Therefore, $\begin{array}{l} \frac{d}{d t} \int_{B_{r}} u (x, t) φ_{r}^{α} (x) d x + α λ_{1, r} \int_{B_{r}} a u (x, t) φ_{r}^{α} (x) d x + \int_{B_{r}} a (x) u^{λ} (x, t) φ_{r}^{α} (x) d x \\ = α (α - 1) \int_{B_{r}} u (x, t) φ_{r}^{α - 2} (x) | \nabla φ_{r} |^{2} (x) d x ⩾ 0 . \end{array}$ Using Hölder’s inequality, we have $\begin{matrix} \int_{B_{r}} a (x) u^{λ} (x, t) φ_{r}^{α} (x) d x ⩽ {(\int_{B_{r}} u (x, t) φ_{r}^{α} (x) d x)}^{λ} {(\int_{B_{r}} a {(x)}^{\frac{1}{1 - λ}} φ_{r}^{α} (x) d x)}^{1 - λ} . \end{matrix}$ We define $\begin{matrix} y (t) = \int_{B_{r}} u (x, t) φ_{r}^{α} (x) d x, \end{matrix}$ and $λ_{1, 1} = λ_{1}$ the first eigenvalue of the unit ball. Hence, $\begin{matrix} (36) & y^{'} (t) + \frac{α λ_{1}}{r^{2}} y (t) + {(\int_{B_{r}} a {(x)}^{\frac{1}{1 - λ}} φ_{r}^{α} (x) d x)}^{1 - λ} y {(t)}^{λ} ⩾ 0 . \end{matrix}$ Since $r \mapsto a (r)$ is a non decreasing function, $\begin{matrix} {(\int_{Ω} a {(x)}^{\frac{1}{1 - λ}} φ_{r}^{α} (x) d x)}^{1 - λ} ⩽ a (r) {(\int_{B_{r}} φ_{r}^{α} (x) d x)}^{1 - λ} = a (r) r^{N (1 - λ)} {(\int_{B_{1}} φ_{1}^{α} (x) d x)}^{1 - λ} . \end{matrix}$ We define $C_{1} = {(\int_{B_{1}} φ_{1}^{α} (x) d x)}^{1 - λ}$ . Then, it follows from (36), $\begin{matrix} (37) & y^{'} (t) + \frac{α λ_{1}}{r^{2}} y (t) + C_{1} a (r) r^{N (1 - λ)} y {(t)}^{λ} ⩾ 0 . \end{matrix}$ We define $z (t) = y {(t)}^{1 - λ}$ . Then, it follows from (37), $\begin{matrix} \frac{z^{'} (t)}{1 - λ} + \frac{α λ_{1}}{r^{2}} z (t) + C_{1} a (r) r^{N (1 - λ)} ⩾ 0 . \end{matrix}$ Hence, $z^{'} (t) + \frac{α (1 - λ) λ_{1}}{r^{2}} z (t) + C_{1} (1 - λ) a (r) r^{N (1 - λ)} ⩾ 0$ , $\begin{matrix} (38) & \frac{d}{d t} (z (t) exp (\frac{α (1 - λ) λ_{1}}{r^{2}} t)) + C_{1} (1 - λ) a (r) r^{N (1 - λ)} exp (\frac{α (1 - λ) λ_{1}}{r^{2}} t) ⩾ 0 . \end{matrix}$ We define $ζ (t) = z (t) exp (\frac{α (1 - λ) λ_{1}}{r^{2}} t)$ . With this new notation, we derive from (38), $\begin{matrix} ζ^{'} (t) + C_{1} (1 - λ) a (r) r^{N (1 - λ)} exp (\frac{α (1 - λ) λ_{1}}{r^{2}} t) ⩾ 0 . \end{matrix}$ Integrating this inequality, we get: $\begin{array}{l} ζ (t) - ζ (0) + C_{1} \frac{a (r) r^{N (1 - λ) + 2}}{α λ_{1}} (exp (\frac{α (1 - λ) λ_{1}}{r^{2}} t) - 1) ⩾ 0, \forall t ⩾ 0 . \end{array}$ But, $\begin{matrix} ζ (0) = z (0) = y {(0)}^{1 - λ} = {(\int_{B_{r}} u_{0} (x) φ_{r}^{α} (x) d x)}^{1 - λ} = {(\int_{B_{r}} φ_{r}^{α} (x) d x)}^{1 - λ} = C_{1}, \end{matrix}$ implies that $\begin{matrix} ζ (t) ⩾ C_{1} (1 - \frac{a (r) r^{N (1 - λ) + 2}}{α λ_{1}} (exp (\frac{α (1 - λ) λ_{1}}{r^{2}} t) - 1)) . \end{matrix}$ We have $ζ (t) > 0$ for all $t \in [0, T_{0} (r))$ where $\begin{matrix} T_{0} (r) = \frac{r^{2}}{α (1 - λ) λ_{1}} ln (1 + \frac{α λ_{1}}{a (r) r^{N (1 - λ) + 2}}) . \end{matrix}$ Hence, for $r > 0$ small enough, $T_{0} (r) ⩾ C_{1} r^{2} ln (\frac{1}{a (r) r^{N (1 - λ) + 2}}) ⩾ C_{2} r^{2} ln (\frac{1}{a (r)})$ so $T_{0} (r) ⩾ C ω (r)$ for some $C, C_{1}, C_{2} > 0$ . Now, it is clear that if ${lim}_{r \to 0} ω (r) = + \infty$ , then $T (u_{0}) ⩾ {lim}_{r \to 0} T_{0} (r) = + \infty$ , which implies that $T^{+} (h) = + \infty$ for all $h > 0$ . □
Lemma 5.
We assume that $p = 1 + q$ , $O \in Ω$ , $a (x)$ has the structure ( 15 ) and the function ω satisfies conditions $(A_{1})$ – $(A_{5})$ . If $\int_{0}^{c} \frac{ω (r)}{r} d r < + \infty$ then $\begin{matrix} (39) & \int_{0}^{1} \frac{d h}{μ (h)} < + \infty . \end{matrix}$
Proof.
We start with (29) for $p = 1 + q$ so, for all $ρ > 0$ , $\begin{matrix} μ (h) ⩽ h ρ^{- p} \int_{B_{1}} | \nabla v |^{p} d y + h^{\frac{1 + λ}{p}} ρ^{N - N \frac{1 + λ}{p}} \int_{B_{1}} a (ρ y) v^{1 + λ} (y) d y . \end{matrix}$ We estimate $a (ρ y)$ by $exp (\frac{- ω (ρ)}{ρ^{p}})$ , so for h small enough, $\begin{matrix} (40) & μ (h) ⩽ C (p, λ, v) (h ρ^{- p} + h^{\frac{1 + λ}{p}} ρ^{N - N \frac{1 + λ}{p}} exp (\frac{- ω (ρ)}{ρ^{p}})) . \end{matrix}$ Using assumption $(A_{5})$ , for $h > 0$ small enough, we can find a function $ρ (h)$ such that $\begin{matrix} (41) & h = exp (\frac{- ω (ρ (h))}{ρ^{p} (h)}) . \end{matrix}$ Hence, $ln h = - ω (ρ (h)) ρ {(h)}^{- p} ⩾ - ω_{0} ρ {(h)}^{- p}$ , which gives $ρ (h) ⩽ {(\frac{ω_{0}}{- ln h})}^{\frac{1}{p}}$ . Moreover, $ln h = \frac{- ω (ρ (h))}{ρ {(h)}^{p}} ⩽ - \frac{1}{ρ^{δ} (h)}$ . Therefore, $ρ (h) ⩾ {(- ln h)}^{- \frac{1}{δ}}$ . Definition (41) yields $ρ {(h)}^{- p} = \frac{- ln h}{ω (ρ (h))} ⩽ - ln h {(ω ({(\frac{1}{- ln h})}^{\frac{1}{δ}}))}^{- 1}$ . Therefore, $\begin{matrix} μ (h) ⩽ C (p, λ, v) (h (- ln h) {(ω ({(\frac{1}{- ln h})}^{\frac{1}{δ}}))}^{- 1} + h^{\frac{1 + λ}{p}} {(\frac{ω_{0}}{- ln h})}^{\frac{N}{p} - N \frac{1 + λ}{p^{2}}} h) . \end{matrix}$ From this estimate, is follows for $h > 0$ small enough, $\begin{matrix} (42) & μ (h) ⩽ C h (- ln h) {(ω ({(\frac{1}{- ln h})}^{\frac{1}{δ}}))}^{- 1} . \end{matrix}$ Now after an upper bound, we derive the lower bound. For all $h > 0$ , there is $v_{h} \in W^{1, p} (Ω, \partial Ω)$ such that $‖ v ‖_{L^{1 + q} (Ω)}^{1 + q} = h$ and $\begin{matrix} 2 μ (h) ⩾ \int_{Ω} | \nabla v |^{p} d x + \int_{Ω} a (x) v^{1 + λ} d x . \end{matrix}$ To simplify, we drop h for $v_{h} : = v$ . So, since $p = 1 + q$ , we get $\begin{matrix} \int_{Ω} | \nabla v |^{p} d x ⩽ \frac{2 μ (h)}{h} \int_{Ω} v^{1 + q} d x - \int_{Ω} a (x) v^{1 + λ} d x, \end{matrix}$ and $\int_{Ω} | \nabla v |^{p} d x ⩽ \int_{Ω} v^{p} (\frac{2 μ (h)}{h} - \frac{a (x)}{v^{p - (1 + λ)}}) d x$ . Sobolev injection and Hölder inequality when $N > p$ give $\begin{matrix} C_{S} {(\int_{Ω} v^{p^{}} d x)}^{\frac{p}{p^{}}} ⩽ {(\int_{Ω} v^{p^{}} d x)}^{\frac{p}{p^{}}} {(\int_{Ω} {({[\frac{2 μ (h)}{h} - \frac{a (x)}{v^{p - (1 + λ)}}]}^{+})}^{\frac{N}{p}} d x)}^{\frac{p}{N}} . \end{matrix}$ Hence, $C_{S}^{\frac{N}{p}} ⩽ \int_{Ω} {({[\frac{2 μ (h)}{h} - \frac{a (x)}{v^{p - (1 + λ)}}]}^{+})}^{\frac{N}{p}} d x$ . Therefore, $C_{S}^{\frac{N}{p}} ⩽ {(\frac{2 μ (h)}{h})}^{\frac{N}{p}} meas {x : \frac{2 μ (h)}{h} ⩾ \frac{a (x)}{v^{p - (1 + λ)}}}$ . Let $ε > 0$ . We have $\begin{matrix} h = ‖ v ‖_{L^{1 + q} (Ω)}^{1 + q} = \int_{Ω} v^{1 + q} d x ⩾ \int_{v^{1 + q} ⩾ ε} v^{1 + q} d x ⩾ ε meas {x : v^{1 + q} ⩾ ε} . \end{matrix}$ By taking $ε = h^{γ}$ with $0 < γ < 1$ , we obtain $\begin{matrix} h^{1 - γ} ⩾ meas {x : v^{1 + q} ⩾ h^{γ}} = meas {x : v^{p - (1 + λ)} ⩾ h^{\frac{γ (p - (1 + λ))}{1 + q}}}, \end{matrix}$ which yields $\begin{array}{l} meas {x : \frac{2 μ (h)}{h} ⩾ \frac{a (x)}{v^{p - (1 + λ)}}} \\ = meas ({x : \frac{2 μ (h)}{h} ⩾ \frac{a (x)}{v^{p - (1 + λ)}}} \cap {x : v^{p - (1 + λ)} ⩾ h^{\frac{γ (p - (1 + λ))}{1 + q}}}) \\ + meas ({x : \frac{2 μ (h)}{h} ⩾ \frac{a (x)}{v^{p - (1 + λ)}}} \cap {x : v^{p - (1 + λ)} < h^{\frac{γ (p - (1 + λ))}{1 + q}}}) . \end{array}$ It follows that $\begin{array}{l} meas {x : \frac{2 μ (h)}{h} ⩾ \frac{a (x)}{v^{p - (1 + λ)}}} \\ ⩽ h^{1 - γ} + meas ({x : \frac{2 μ (h)}{h} ⩾ \frac{a (x)}{v^{p - (1 + λ)}}} \cap {x : v^{p - (1 + λ)} < h^{\frac{γ (p - (1 + λ))}{1 + q}}}) . \end{array}$ We obtain $\begin{matrix} meas {x : \frac{2 μ (h)}{h} ⩾ \frac{a (x)}{v^{p - (1 + λ)}}} ⩽ h^{1 - γ} + meas {x : \frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}} ⩾ a (x)} . \end{matrix}$ Now, from (42), for $h > 0$ small enough, $μ (h) ⩽ C h (- ln h) {(ω ({(\frac{1}{- ln h})}^{\frac{1}{δ}}))}^{- 1}$ . By $(A_{4})$ , $ω (r) ⩾ r^{p - δ}$ for $r > 0$ small enough. It leads to $\begin{matrix} (43) & μ (h) ⩽ C h \frac{- ln h}{({(\frac{1}{- ln h})}^{\frac{p - δ}{δ}})} = C h {(- ln h)}^{\frac{p}{δ}} . \end{matrix}$ There are two consequences: ${(\frac{2 μ (h)}{h})}^{\frac{N}{p}} h^{1 - γ} \to 0$ , and $\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}} \to 0$ , when $h \to 0$ . We deduce that there exists $C > 0$ such that for $h > 0$ small enough, $\begin{matrix} C ⩽ {(\frac{2 μ (h)}{h})}^{\frac{N}{p}} meas {x : \frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}} ⩾ exp (- \frac{ω (| x |)}{| x |^{p}})} . \end{matrix}$ So, $\begin{matrix} C ⩽ {(\frac{2 μ (h)}{h})}^{\frac{N}{p}} meas {x : ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}}) ⩾ - \frac{ω (| x |)}{| x |^{p}}} . \end{matrix}$ We take x such that $ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}}) ⩾ - \frac{ω (| x |)}{| x |^{p}}$ . Since ω is bounded, it satisfies $\begin{matrix} ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}}) ⩾ - \frac{ω_{0}}{| x |^{p}}, \end{matrix}$ i.e., $| x | ⩽ {(\frac{ω_{0}}{- ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}})})}^{\frac{1}{p}}$ . By monotonicity of ω (assumption $(A_{1})$ ), $\begin{matrix} ω (| x |) ⩽ ω ({(\frac{ω_{0}}{- ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}})})}^{\frac{1}{p}}) . \end{matrix}$ Therefore, $\begin{array}{l} C ⩽ & {(\frac{2 μ (h)}{h})}^{\frac{N}{p}} \\ \times meas {x : ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}}) ⩾ - ω ({(\frac{ω_{0}}{- ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}})})}^{\frac{1}{p}}) | x |^{- p}} . \end{array}$ Consequently, $\begin{matrix} C ⩽ {(\frac{2 μ (h)}{h})}^{\frac{N}{p}} meas {x : | x |^{p} ⩽ \frac{ω ({(\frac{ω_{0}}{- ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}})})}^{\frac{1}{p}})}{- ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}})}} . \end{matrix}$ Last inequality leads to $\begin{matrix} C ⩽ (\frac{2 μ (h)}{h}) \frac{ω ({(\frac{ω_{0}}{- ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}})})}^{\frac{1}{p}})}{- ln (\frac{2 μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}})}, \end{matrix}$ for some other $C > 0$ . Now, by (43), for $h > 0$ small enough, $\begin{matrix} \frac{μ (h)}{h} h^{\frac{γ (p - (1 + λ))}{1 + q}} ⩽ h^{ε}, \end{matrix}$ for some $ε > 0$ . We obtain $\begin{matrix} C ⩽ (\frac{μ (h)}{h}) \frac{ω ({(\frac{K}{- ln h})}^{\frac{1}{p}})}{- ln h}, \end{matrix}$ for $C, K > 0$ . Therefore, $\begin{matrix} μ (h) ⩾ C h \frac{- ln h}{ω ({(\frac{K}{- ln h})}^{\frac{1}{p}})} . \end{matrix}$ So, $\begin{matrix} \int_{0}^{\frac{1}{2}} \frac{d h}{μ (h)} ⩽ C \int_{0}^{\frac{1}{2}} \frac{ω ({(\frac{K}{- ln h})}^{\frac{1}{p}})}{- h ln h} d h . \end{matrix}$ We use the change of variables $r = {(\frac{K}{- ln h})}^{\frac{1}{p}}$ , i.e., $ln r = \frac{1}{p} (ln K - ln (- ln h))$ , $d (ln r) = \frac{1}{p} \frac{d h}{h (- ln h)}$ . This last equality leads to $\begin{matrix} \int_{0}^{\frac{1}{2}} \frac{d h}{μ (h)} ⩽ C \int_{0}^{c} \frac{ω (r)}{r} d r, \end{matrix}$ and (39) is proved. □

4. Proof of main results

Proof of Theorem 1.1.

Let u be a non negative solution of (1), the initial problem. Assume (10) and (11), that is $\begin{matrix} a (x) = \frac{| x |^{δ_{0}}}{ω (| x |)} for x small enough and otherwise a (x) ⩾ C > 0, \end{matrix}$ and $\begin{matrix} ω is a continuous positive function on (0, ρ_{0}), ρ_{0} > 0 . \end{matrix}$ For the necessary condition, assume additionally (12), i.e, $\begin{matrix} ρ \mapsto \frac{ρ^{δ_{0}}}{ω (ρ)} is a continuous nondecreasing function . \end{matrix}$ First, we suppose ${lim inf}_{ρ \to 0} ω (ρ) > 0$ . By Lemma 2, ${lim inf}_{h \to 0} \frac{h}{μ (h)} > 0$ . Moreover, if we set $\begin{matrix} J (u) = \frac{1}{p} \int_{Ω} | \nabla u |^{p} d x + \frac{1}{1 + λ} \int_{Ω} a (x) | u |^{1 + λ} d x, \end{matrix}$ then by Proposition 1, the functional J satisfies $H_{0}$ – $H_{4}$ . We use the contraposition in Theorem 2.1. Hence, ${lim}_{h \to 0} T^{+} (h) \neq 0$ . But the function $h \mapsto T^{+} (h)$ is a nondecreasing function which gives ${lim}_{h \to 0} T^{+} (h) > 0$ . Now, let assume further (13), i.e., $\begin{array}{l} ρ \mapsto ω (ρ) is a nonincreasing function, \\ \forall γ_{1} \in (0, 1), \exists C > 0, \exists γ_{2} > 0, \forall h \in (0, 1], ω (h^{γ_{1}}) ⩾ C {(ω (h))}^{γ_{2}}, \\ \exists η > 0, \forall h \in (0, h_{0}], ω (h) ⩽ h^{- η} for some h_{0} > 0 \end{array}$ and ${lim}_{ρ \to 0} ω (ρ) = + \infty$ . By Lemma 2, ${lim}_{h \to 0} \frac{h}{μ (h)} = + \infty$ . The functional J always satisfies $H_{0}$ – $H_{4}$ and by estimate (27) in the proof of Theorem 2.1, $\forall h_{0} > 0$ , $T^{+} (h_{0}) = + \infty$ .

For the sufficient condition (case 3 in Theorem 1.1), assume $N > p$ , (10), (11) and (14), i.e., $\begin{matrix} ω (ρ) = \frac{1}{{(- ln ρ)}^{α}} for α > \frac{p - (1 + λ)}{p - (1 + q)} + \frac{δ_{0}}{N} . \end{matrix}$ By Lemma 3, $\begin{matrix} \int_{0}^{1} \frac{d h}{μ (h)} < + \infty . \end{matrix}$ By Theorem 2.1, ${lim}_{h \to 0} T^{+} (h) = 0$ . For the sufficient condition, we don’t need assumptions $H_{0}$ – $H_{4}$ . □

Proof of Theorem 1.2.

Let u be a non negative solution of (1) with $p = 1 + q$ , $N > p$ , potential $a (x)$ has structure (15), function $ω (ρ)$ satisfies conditions $(A_{1})$ – $(A_{5})$ , i.e., $\begin{matrix} a (x) = exp (- \frac{ω (| x |)}{| x |^{p}}), \end{matrix}$ $(A_{1})$

$ω (r)$ is a continuous and nondecreasing function $\forall r ⩾ 0$ ,

(A_{2})

$ω (0) = 0$ , $ω (r) > 0$ , $\forall r > 0$ ,

(A_{3})

$ω (r) ⩽ ω_{0} < \infty$ , $\forall r > 0$ ,

(A_{4})

$ω (r) ⩾ r^{p - δ}$ , $\forall s \in (0, s_{0})$ , $s_{0} > 0$ , $p > δ > 0$ ,

(A_{5})

the function $\frac{ω (r)}{r^{p}}$ is nonincreasing on $(0, s_{0})$ , and $\int_{0}^{c} \frac{ω (r)}{r} d r < + \infty$ . By Lemma 5, $\int_{0}^{1} \frac{d h}{μ (h)} < + \infty$ . By Theorem 2.1, ${lim}_{h \to 0} T^{+} (h) = 0$ . For the necessary condition, we assume that $p = 1 + q = 2$ and ${lim}_{r \to 0} ω (r) = + \infty$ . By Lemma 4, $T^{+} (h) = + \infty$ for all $h > 0$ . □

Footnotes

Acknowledgement

This paper has been supported by the RUDN University Strategic Academic Leadership Program.

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