Elliptic Equations with Weight and Combined Nonlinearities

Theorem 1.1

Suppose that 1<q<2<p<2* and the potentials a, b and c satisfy (a1)–(a2), (b1)–(b2) and (c1)–(c2), respectively. Then, problem (P) has at least two nonnegative nontrivial solutions if |b|Lqσ⁢(Ω) is small.

Theorem 1.2

Suppose that 1<q<2<p=2* and that the potentials a, b and c satisfy (a1)–(a3), (b1), (c1) and (c3). Then, problem (P) has at least two nonnegative nontrivial solutions if |b|Lqσ⁢(Ω) is small and k>(N-2)/2.

Lemma 2.1

Suppose that b satisfies (b1) and |b|σq is small enough. Then, there exist ρ,α>0 such that I⁢(u)≥α>0 for any u∈H such that ∥u∥=ρ.

Proof.

From (a2), (2.2) and (2.1), it follows that

For B:=2⁢(p⁢a0)-1⁢|c|∞⁢Sp-p/2, the function f:(0,+∞)→ℝ, given by f⁢(t):=t2-q-B⁢tp-q, achieves its maximum value at

t 0 := [ ( 2 - q ) B ⁢ ( p - q ) ] 1 / ( p - 2 ) > 0 .

For M:=f⁢(t0) and ∥u∥=t0, we have that

I ⁢ ( u ) ≥ a 0 ⁢ t 0 q 2 ⁢ { M - 2 q ⁢ a 0 ⁢ | b | σ q ⁢ S q ⁢ σ q ′ - q / 2 } ≥ t 0 q 2 ⁢ M 2 > 0 ,

whenever

The lemma holds for α:=t0q⁢M/4, ρ:=t0 and |b|σq as above. ∎

Lemma 2.2

Suppose that b satisfies (b1), (b2) and (2.3). If ρ>0 is given by Lemma 2.1, then

- ∞ < I 0 := inf u ∈ B ρ ⁢ ( 0 ) ¯ ⁡ I ⁢ ( u ) < 0

is achieved at u0∈Bρ⁢(0), which is a nonnegative solution of (P).

Proof.

A straightforward calculation shows that I maps bounded sets into bounded sets, and therefore I0 is finite. Since Ωb+ has nonempty interior, there exists δ1>0 and x1∈Ω such that Bδ1⁢(x1) is contained in the set Ωb+. Hence, we can take a nonnegative function φ∈C0∞⁢(Bδ1⁢(x1)) such that ∫b⁢(x)⁢φq>0. Since q<2<p, we have that

lim sup t → 0 + ⁡ I ⁢ ( t ⁢ φ ) t q ≤ - 1 q ⁢ ∫ b ⁢ ( x ) ⁢ φ q < 0 .

So, for t>0 small, we have that I⁢(t⁢φ)<0, and therefore I0<0.

Let (un)⊂Bρ⁢(0)¯ be a minimizing sequence for I0. By Ekeland’s variational principle, we may assume that I⁢(un)→I0 and I′⁢(un)→0. Since (un) is bounded and 2≤q⁢σq′<2*, we have, up to a subsequence, that

Young’s inequality provides

| b ⁢ ( x ) ⁢ ( u n + ) q | ≤ 1 σ q ⁢ b ⁢ ( x ) σ q + 1 σ q ′ ⁢ ψ ⁢ ( x ) q ⁢ σ q ′   for a.e. ⁢ x ∈ Ω .

Since ψ∈Lq⁢σq′⁢(Ω) and b∈Lσq⁢(Ω), from (2.4) and the Lebesgue theorem it follows that

lim n → ∞ ⁡ ∫ b ⁢ ( x ) ⁢ ( u n + ) q = ∫ b ⁢ ( x ) ⁢ ( u 0 + ) q .

We now claim that I′⁢(u0)=0. Assuming the claim, the above equality and the weak convergence of (un) provide

and therefore I⁢(u0)=I0<0. By Lemma 2.1, ∥u0∥≠ρ. Hence, the infimum is achieved at u0∈Bρ⁢(0). Since I′⁢(u0)=0, the function u0 is nonnegative.

It remains to prove that I′⁢(u0)=0. Let us denote by A the support of φ∈C0∞⁢(Ω). Since we know that σq>(p/q)′=p/(p-q), we can choose q0∈(2,p) such that

σ q > q 0 q 0 - q > q 0 ( q 0 + 1 ) - q .

Thus,

1 σ q + q - 1 q 0 < ( q 0 + 1 ) - q q 0 + q - 1 q 0 = 1 ,

and there exists θ>1 satisfying

1 σ q + 1 q 0 / ( q - 1 ) + 1 θ = 1 .

The above inequality implies that un→u0 in Lq0⁢(Ω) and provides ψq0 such that |un⁢(x)|≤ψq0⁢(x) a.e. in Ω. By Young’s inequality there exists C>0 such that

| b ⁢ ( x ) ⁢ ( u n + ) q - 1 ⁢ φ | ≤ C ⁢ ( | b ⁢ ( x ) | σ q + | ψ q 0 ⁢ ( x ) | q 0 + | φ | θ )   for a.e. ⁢ x ∈ A .

From the Lebesgue theorem, it follows that

lim n → + ∞ ⁡ ∫ b ⁢ ( x ) ⁢ ( u n + ) q - 1 ⁢ φ = ∫ b ⁢ ( x ) ⁢ ( u 0 + ) q - 1 ⁢ φ .

An analogous argument holds for ∫c⁢(x)⁢(un+)r-1⁢φ, and hence we conclude that 0=limn→+∞⁡I′⁢(un)⁢φ=I′⁢(u0)⁢φ for all φ∈C0∞⁢(Ω). The result follows by density. ∎

Lemma 2.3

Suppose that b and c satisfy (b1) and (c1)–(c2), respectively, and let Bδ1⁢(x1)⊂Ωc+. If u0 is given by Lemma 2.2 and φ∈C0∞⁢(Bδ1⁢(x1))∖{0} is nonnegative, then

lim t → + ∞ ⁡ I ⁢ ( u 0 + t ⁢ φ ) = - ∞ .

Proof.

Since φ=0 outside Bδ1⁢(x1)⊂Ωc+ and u0≥0 a.e. in Ω, we can easily compute

Since p>2, the result follows from the positivity of the last integral above. ∎

Lemma 2.4

If 2<p<2*, then the functional I satisfies the (PS)d condition for any d∈ℝ.

Proof.

Let (un)⊂H be such that I⁢(un)→d and I′⁢(un)→0. We have that

d + ∥ u n ∥ + o ⁢ ( 1 ) ≥ a 0 ⁢ ( 1 2 - 1 p ) ⁢ ∥ u n ∥ 2 - ( 1 q - 1 p ) ⁢ ∫ b ⁢ ( x ) ⁢ ( u n + ) q .

Hölder’s inequality and the embedding H↪Lτ⁢(Ω) provide C1>0 such that

d + ∥ u n ∥ + o ⁢ ( 1 ) ≥ a 0 ⁢ ( 1 2 - 1 p ) ⁢ ∥ u n ∥ 2 - ( 1 q - 1 p ) ⁢ C 1 ⁢ ∥ u n ∥ q ,

and therefore (un) is bounded in H. Up to a subsequence, we have that un⇀u weakly in H and un→u strongly in Lτ⁢(Ω) for 2≤τ<2*. By the definition of σq, we know that there exists 2≤p0<p such that σq=(p0/q)′=p0/(p0-q). So

1 σ q + 1 p 0 ⁢ ( q - 1 ) + 1 p 0 = p 0 - q p 0 + q - 1 p 0 + 1 p 0 = 1 .

From Hölder’s inequality, it follows that

| ∫ b ⁢ ( x ) ⁢ ( u n + ) q - 1 ⁢ ( u n - u ) | ≤ | b | σ q ⁢ | u n | p 0 q - 1 ⁢ | u n - u | p 0 = o ⁢ ( 1 )   as n → + ∞ .

Since an analogous argument holds for ∫c⁢(x)⁢(un+)p-1⁢(un-u), we obtain

o ⁢ ( 1 ) = I ′ ⁢ ( u n ) ⁢ ( u n - u ) = ∥ u n ∥ a 2 - ∥ u ∥ a 2 + o ⁢ ( 1 ) .

Thus, ∥un∥2→∥u∥2 and from the weak convergence of (un), it follows that, along a subsequence, it converges. This completes the proof. ∎

Proof of Theorem 1.1.

If |b|σq is small, we can use Lemma 2.2 to obtain a nonnegative solution u0 such that I⁢(u0)<0. For the second one, we take ρ>0 and φ as in Lemmas 2.1 and 2.3, respectively. Let t0>0 be such that e:=u0+t0⁢φ satisfies I⁢(e)≤I⁢(u0). If we define

d ~ := inf γ ∈ Γ ⁡ max t ∈ [ 0 , 1 ] ⁡ I ⁢ ( γ ⁢ ( t ) ) ,

where

Γ := { γ ∈ C ⁢ ( [ 0 , 1 ] , H ) : γ ⁢ ( 0 ) = u 0 , γ ⁢ ( 1 ) = e } ,

then by Lemma 2.1 we get d~≥α>0. Moreover, the Mountain Pass Theorem provides (un)⊂H such that I⁢(un)→d~ and I′⁢(un)→0. By Lemma 2.4, un→u1 in H along a subsequence. Hence, I⁢(u1)=d~>0 and I′⁢(u1)=0, in such way that we have obtained a second (nonnegative) solution. ∎

Lemma 3.1

If p=2* and u0 is the only nontrivial critical point of I, then I satisfies (PS)d for

d < d * := I ⁢ ( u 0 ) + 1 N ⁢ ( a 0 ⁢ S ) N / 2 | c | ∞ ( N - 2 ) / 2 .

Proof.

Let (un)⊂H be such that I′⁢(un)→0 and I⁢(un)→d. As in the proof of Lemma 2.4, we can show that it is bounded. Hence, along a subsequence, we have that un⇀u weakly in H and un→u strongly in Lq⁢σq′⁢(Ω). Setting vn:=un-u, we can use the last convergence and the Brezies–Lieb lemma to get

As in the proof of Lemma 2.2, we have that I′⁢(u)=0. Hence, there exists l≥0, such that

lim n → ∞ ∥ v n ∥ a 2 = l = lim n → ∞ ∫ c ( x ) ( v n + ) 2 * .

If l=0, then un→u strongly in H and we are done. So, we may suppose that l>0. From the definition of S and a0, it follows that

( ∫ c ⁢ ( x ) ⁢ ( v n + ) 2 * ) 2 / 2 * ≤ | c | ∞ 2 / 2 * a 0 ⁢ S ⁢ ∫ a ⁢ ( x ) ⁢ | ∇ ⁡ v n | 2 .

Taking the limit, we obtain

On the other hand, arguing as in the beginning of the proof, we obtain

d + o ⁢ ( 1 ) = I ⁢ ( u n ) = I ⁢ ( u ) + 1 2 ⁢ ∥ v n ∥ a 2 - 1 2 * ⁢ ∫ c ⁢ ( x ) ⁢ ( v n + ) 2 * + o ⁢ ( 1 ) .

Taking the limit again and using (3.1), we get

d = I ⁢ ( u ) + ( 1 2 - 1 2 * ) ⁢ l = I ⁢ ( u ) + l N ≥ I ⁢ ( u ) + 1 N ⁢ ( a 0 ⁢ S ) N / 2 | c | ∞ ( N - 2 ) / 2 .

But we are assuming that the only critical points are u=0 and u=u0. Since max⁡{I⁢(0),I⁢(u0)}≤0, the above inequality contradicts d<d*. ∎

Proof of Theorem 1.2.

We know that the problem has a nontrivial solution u0 such that I⁢(u0)<0. Arguing by contradiction, we suppose that this the only nontrivial critical point of I. As in the proof of Theorem 1.1, we set e:=u0+t⁢vε with t>0 large in such way that I⁢(e)≤I⁢(u0), and define the mountain pass level

where

Γ := { γ ∈ C ⁢ ( [ 0 , 1 ] , H ) : γ ⁢ ( 0 ) = u 0 , γ ⁢ ( 1 ) = e } .

We obtain (un)⊂H satisfying I′⁢(un)→0 and I⁢(un)→d~. From Proposition 3.2, which we state and prove in the sequel, it follows that d~<d*. Hence, Lemma 3.1 implies that, along a subsequence, (un) converges strongly to a solution with positive energy. But this contradicts the assumption that 0 and u0 are the only critical points of I. Hence, we conclude that there exist a nonzero solution u1 such that u1≠u0. As before, we have that u1≥0 a.e. in Ω. ∎

Proposition 3.2

Suppose that b and c satisfy (b1), (c1) and (c3). If N<(2⁢k+2), then

max t > 0 ⁡ I ⁢ ( u 0 + t ⁢ v ε ) < d * = I ⁢ ( u 0 ) + 1 N ⁢ ( a 0 ⁢ S ) N / 2 | c | ∞ ( N - 2 ) / 2

for ε>0 small enough. In particular, the minimax level defined in (3.2) satisfies d~<d*.

Proof.

Let δ>0 be given by hypothesis (c3), and consider ψ∈C0∞⁢(Ω) satisfying ψ⁢(x)=1 if x∈Bl/2⁢(x0) and ψ⁢(x)=0 if x∈Ω∖Bl⁢(x0), where 0<l<δ. Given ε>0, we define

u ε ⁢ ( x ) := ψ ⁢ ( x ) [ ε + | x - x 0 | 2 ] ( N - 2 ) / 2   and   v ε ⁢ ( x ) := u ε ⁢ ( x ) | u ε | 2 * .

For any ε>0, Lemma 2.3 implies that the function t→I⁢(u0+t⁢vε) achieves its maximum at tε>0. From I′⁢(u0)⁢vε=0, it follows that

for

A ε := ∫ B l ⁢ ( x 0 ) b ⁢ ( x ) ⁢ [ ( u 0 + t ε ⁢ v ε ) q - u 0 q - q ⁢ t ε ⁢ u 0 q - 1 ⁢ v ε ] ⁢ 𝑑 x

and

D ε := ∫ B l ⁢ ( x 0 ) c ⁢ ( x ) ⁢ [ ( u 0 + t ε ⁢ v ε ) 2 * - u 0 2 * - 2 * ⁢ t ε ⁢ u 0 2 * - 1 ⁢ v ε ] ⁢ 𝑑 x ,

where we also have used that vε≡0 outside Bl⁢(x0).

Since u0≥0 a.e. in Bl⁢(x0), we can apply the Mean Value Theorem to obtain η⁢(x)∈[0,1] such that

( u 0 ⁢ ( x ) + t ε ⁢ v ε ⁢ ( x ) ) q - u 0 ⁢ ( x ) q = q ⁢ ( u 0 ⁢ ( x ) + η ⁢ ( x ) ⁢ t ε ⁢ v ε ⁢ ( x ) ) q - 1 ⁢ t ε ⁢ v ε ≥ q ⁢ t ε ⁢ u 0 ⁢ ( x ) q - 1 ⁢ v ε ⁢ ( x )

for a.e. x∈Bl⁢(x0). Since b⁢(x)≥0 a.e. in Bl⁢(x0), we get Aε≥0.

In order to estimate Dε, we shall use the following inequality (see [4]):

( m + n ) s ≥ m s + n s + s ⁢ m s - 1 ⁢ n + s ⁢ m ⁢ n s - 1 - C μ ⁢ n μ ⁢ m s - μ

for m,n≥0, s>2, 1<μ<s-1 and for some Cμ>0. If we choose m=u0, n=tε⁢vε and s=2*, we obtain

( u 0 + t ε ⁢ v ε ) 2 * - u 0 2 * - 2 * ⁢ t ε ⁢ u 0 2 * - 1 ⁢ v ε ≥ t ε 2 * ⁢ v ε 2 * + 2 * ⁢ t ε 2 * - 1 ⁢ u 0 ⁢ v ε 2 * - 1 - C μ ⁢ t ε μ ⁢ v ε μ ⁢ u 0 2 * - μ .

Since c⁢(x)≥0 a.e. in Bl⁢(x0), we obtain

D ε ≥ ∫ B l ⁢ ( x 0 ) c ⁢ ( x ) ⁢ [ t ε 2 * ⁢ v ε 2 * + 2 * ⁢ t ε 2 * - 1 ⁢ u 0 ⁢ v ε 2 * - 1 - C μ ⁢ t ε μ ⁢ v ε μ ⁢ u 0 2 * - μ ] ⁢ 𝑑 x .

Hence, we can use (3.3) to get

where we have used ∫Bl⁢(x0)vε2*⁢𝑑x=1.

In what follows we shall suppose that (tε) is bounded. The other case will be considered later. We compute

As proved in [5, (5.10)], the Lσ-norms of vε are such that

∫ | v ε | σ = O ( ε ( N ⁢ ( 2 - σ ) + 2 ⁢ σ ) / 4 ) , N / ( N - 2 ) < σ < 2 * ,

as ε→0+. Recalling that the functions u0 and c are bounded in Bl⁢(x0), we can choose μ=(N+1)/(N-2) to get

In the same way,

for some A0>0.

On the other hand, a known estimate from the paper of Brezis and Nirenberg [3] states that, for some A1>0,

| u ε | 2 * 2 * = ε - N / 2 ⁢ A 1 + O ⁢ ( 1 ) .

Thus, we can use the inequality in (c3) to obtain

Hence, setting y:=(x-x0)/ε, we get

where ωN is the area of the unit sphere in ℝN. All together, the last estimates provide

If we now substitute (3.5)–(3.8) in inequality (3.4), we obtain

Without loss of generality we may suppose that γ/2<(N-1)/4, and therefore the above expression becomes

We now recall a key estimate, which is a consequence of [9, (3.13)]:

We first consider the last case, that is, N≥5 and N>k+2. By using the Mean Value Theorem, we have that

∥ v ε ∥ a N = ( ∥ v ε ∥ a 2 ) N / 2 = ( a 0 ⁢ S ) N / 2 + O ⁢ ( ε k / 2 ) .

Hence,

Since we are supposing that max⁡{γ,k}>(N-2)/2, the above expression implies that mε<d*, if ε>0 is small enough. The other three cases in (3.9) can be handled with the same kind of argument. Actually, in all of them that are no extra restrictions on k.

It remains to consider the case where lim supε→0+⁡tε=+∞. Recalling that the support of vε is contained in Bl⁢(x0) and arguing as in the proof of Lemma 2.3, we obtain

m ε ≤ t ε 2 2 ⁢ ∥ v ε ∥ a 2 + t ε ⁢ ∫ p ⁢ ( x ) ⁢ ( ∇ ⁡ u 0 ⋅ ∇ ⁡ v ε ) - t ε 2 * 2 * ⁢ ∫ B l ⁢ ( x 0 ) c ⁢ ( x ) ⁢ v ε 2 * ⁢ 𝑑 x + O ⁢ ( 1 ) .

Since ∥vε∥a2=a0⁢S+o⁢(1), Bl⁢(x0)⊂Ωc+ and ∫Bl⁢(x0)vε2*⁢𝑑x=1, we infer from the above inequality that mε→-∞ as ε→0+. Hence, for small ε>0, we have that mε<d*, and we are done. ∎