Supersolutions to nonautonomous Choquard equations in general domains

Definition 2.1

Let 1 < m < ∞ , β ≥ 0 , 0 < α < N and assume that Ω ⊂ R N is a domain. A function u ∈ W loc 1 , m ( Ω ) is a positive weak supersolution to (1.1)

1. if u > 0 and

   H ( x ) ∫ Ω Q ( y ) f ( u ( y ) ) ∣ x − y ∣ N − α d y β g ( u ( x ) ) ∈ L loc 1 ( Ω ) ;

2. if β > 0 , then u satisfies

   ∫ Ω Q ( y ) f ( u ( y ) ) 1 + ∣ y ∣ N − α d y < ∞ ;

3. for any ϕ ∈ C c ∞ ( Ω ) with ϕ ≥ 0 , we have

   ∫ Ω ∣ ∇ u ( x ) ∣ m − 2 ∇ u ( x ) ⋅ ∇ ϕ ( x ) d x ≥ ∫ Ω H ( x ) ( ( I α * ( Q f ( u ) ) ) ( x ) ) β g ( u ( x ) ) ϕ ( x ) d x .

Remark 2.2

We note that if u ∈ W loc 1 , m ( Ω ) is a solution to (1.1), then it is a weak supersolution to the m -Laplace equation. By [31, Theorem 3.63], we conclude that u is locally essentially bounded from below and that there exists a lower semicontinuous representative of u that satisfies

for every x ∈ Ω . Here,

In particular,

Weak supersolutions to (1.1) are, a priori, defined only up to a set of Lebesgue measure zero, but the aforementioned lower semicontinuous representative allows us to discuss pointwise defined supersolutions. When it is useful, we may replace u by its lower semicontinuous representative denoted again by u .

Theorem 2.3

Let 1 < m < ∞ , β ≥ 0 , 0 < α < N and assume that Ω ⊂ R N is a domain. Assume that f , g , H , and Q satisfy ( C ) and let u be a positive supersolution to (1.1) in Ω . Then,

for every x ∈ Ω and 0 < r < d Ω ( x ) , with

Here, ω N = π N 2 Γ N 2 + 1 is the volume of the unit ball in R N .

Proof

Let x ∈ Ω with 0 < r < d Ω ( x ) . From Definition 2.1, we obtain

for every ϕ ∈ C c ∞ ( B r ( x ) ) with ϕ ≥ 0 . For short, we write

For z ∈ B r ( x ) , we set r z = r − ∣ x − z ∣ . Then, we have ∣ y − x ∣ ≤ ∣ y − z ∣ + ∣ x − z ∣ < r z + ∣ x − z ∣ = r for y ∈ B r z ( z ) , which implies that y ∈ B r ( x ) . We observe that

for almost every z ∈ B r ( x ) . A simple calculation gives

which implies that

for every ϕ ∈ C c ∞ ( B r ( x ) ) with ϕ ≥ 0 , i.e.,

We consider an auxiliary function

which is the unique radial solution of

By a scaling and translation argument, we observe that the function

is a solution to

From (2.4), we obtain

Let v ( z ) = u ( z ) − m x ( r ) , z ∈ B r ( x ) , and

with

By the facts that Δ m is positively homogeneous of order m − 1 and that constants can be added to a solution, we obtain

i.e.,

for every ϕ ∈ C c ∞ ( B r ( x ) ) with ϕ ≥ 0 . Since v ≥ 0 on ∂ B r ( x ) and w = 0 on ∂ B r ( x ) , by a comparison result, see Tolksdorf [45] or [43, Corollary 3.4.2], we have v ≥ w in B r ( x ) . It follows that

in B r ( x ) . Since Φ is decreasing, we have Φ ∣ x − z ∣ r ≥ Φ h r for z ∈ B h ( x ) with 0 < h < r . It follows that

By taking essential infimum on the left-hand side over z ∈ B h ( x ) , we obtain

and then dividing both sides by r − h , we arrive at

By letting h → r and using the fact that

we obtain the following ordinary differential inequality with an initial value condition:

Dividing through by g ( m x ( r ) ) 1 m − 1 f ( m x ( r ) ) β m − 1 , we may rewrite (2.6) as:

where J : ( 0 , d Ω ( x ) ) → R is defined by:

Since m x ( r ) is nonincreasing and f and g are the positive functions, the function J is nondecreasing. By the Lebesgue differentiation theorem,

Thus, integrating (2.7) from 0 to r yields

which proves (2.2).□

Corollary 2.4

Let 1 < m < ∞ and 0 < α < N and assume that Ω ⊂ R N is a domain. Assume that g and H satisfy ( C ) and let u be a positive supersolution to

Then,

for every x ∈ Ω and 0 < r < d Ω ( x ) . In particular, if H ≡ 1 , then

for every x ∈ Ω and 0 < r < d Ω ( x ) .

Theorem 2.5

Let 1 < m < ∞ , γ , β ≥ 0 , 0 < α < N , Ω a domain in R N , and x 0 ∈ Ω . Assume that f and g satisfy ( C ) and let u be a positive supersolution to

Then,

for every 0 < r < d Ω ( x 0 ) , with

Proof

As in the proof of Theorem 2.3, by (2.8), we obtain

for every ϕ ∈ C c ∞ ( B r ( x 0 ) ) , 0 < r < d Ω ( x 0 ) , with ϕ ≥ 0 , i.e.,