Multiplicity of Radial Solutions of Quasilinear Problems with Minimum and Maximum

Theorem 1.1

Let h satisfy the Carathéodory conditions on [a,b]×ℝ2 and

| h ⁢ ( t , x , y ) | ≤ M   for a.e. ⁢ t ∈ [ a , b ] , ( x , y ) ∈ ℝ 2

with some positive constant M. Let β-α>M8⁢(b-a)2. Then problem (1.4) has at least two different solutions.

Theorem 1.2

Let G:C1⁢[a,b]→L1⁢[a,b] be such that there exists a continuous nondecreasing function f:[0,∞)→[0,∞) satisfying

| G ⁢ ( u ) ⁢ ( t ) | ≤ f ⁢ ( | u ′ ⁢ ( t ) | ) , t ∈ [ a , b ] , u ∈ C 1 ⁢ [ a , b ] .

and

∫ 0 ∞ d ⁢ t f ⁢ ( t ) ≥ b - a , ∫ 0 ∞ t f ⁢ ( t ) ⁢ 𝑑 t = ∞ .

Let α,β satisfy

β - α > b - a 2 ⁢ P - 1 ⁢ ( b - a 2 ) , where ⁢ P ⁢ ( u ) = ∫ 0 u d ⁢ t f ⁢ ( t ) .

Then problem (1.5) has at least two different solutions.

Definition 2.1

A functional τ:X→ℝ is increasing if

x ⁢ ( r ) < y ⁢ ( r )   for all ⁢ r ∈ [ R 1 , R 2 ]   ⇒   τ ⁢ ( x ) < τ ⁢ ( y )

for x,y∈X. For every τ:X→ℝ, Im⁡(τ) denotes the range of τ. Denote

𝒜 = { τ ∣ τ : X → ℝ ⁢ is continuous and increasing } , 𝒜 0 = { τ ∣ τ ∈ 𝒜 , τ ⁢ ( 0 ) = 0 } .

Remark 2.2

If we take

τ ⁢ ( u ) = min ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } , ζ ⁢ ( u ) = max ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } ,

then the functionals τ,ζ belong to 𝒜0. It is easy to observe that the boundary condition (1.7) can reduce to

So, in the remaining part of the paper, we only need to deal with problem (1.6), (2.1).

Lemma 2.3

Let τ∈𝒜, h∈Im⁡(τ). Then there exists a unique d∈ℝ such that τ⁢(d)=h.

Proof.

Suppose that there exist x∈X and c1,c2∈ℝ satisfying τ⁢(x)=h and c1<x⁢(t)<c2. By the definition of 𝒜, we get τ⁢(c1)<τ⁢(x)<τ⁢(c2). Define the increasing function p∈C⁢([c1,c2]) by

p ⁢ ( c ) = τ ⁢ ( c ) .

Obviously, p⁢(c1)<h<p⁢(c2). Since p is continuous and strictly increasing, the equation p⁢(c)=h has a unique solution c=d. Of course, there exists a unique d∈ℝ such that τ⁢(d)=h. ∎

Lemma 2.4

Let τ∈𝒜 and e∈[0,1]. For each u∈X, let u satisfy the equality τ⁢(u)-e⁢τ⁢(-u)=0. Then there exists a ξ∈[R1,R2] such that u⁢(ξ)=0.

Proof.

Suppose on the contrary that u⁢(ξ)≠0. If u⁢(r)<0 on [R1,R2], then τ⁢(u)<τ⁢(0)=0, τ⁢(-u)>τ⁢(0)=0. Moreover, τ⁢(u)-e⁢τ⁢(-u)<0, this is a contradiction. Similarly, for u⁢(r)>0, we also can get a contradiction. ∎

Lemma 2.5

Let q:[0,+∞)→(0,+∞) be a nondecreasing continuous function, Q:[0,+∞)→[0,+∞) defined by Q⁢(u)=∫0ud⁢tq⁢(t) and let a∈[b,c]⊂ℝ. Let Q-1 denote the inverse function to Q. If y∈X satisfies the inequality

| y ⁢ ( r ) | ≤ | ∫ a r q ⁢ ( | y ⁢ ( s ) | ) ⁢ 𝑑 s |   for ⁢ r ∈ ( b , c ) ,

then

Lemma 2.6

Lemma 2.6 ([9, 18, 19])

For every u∈Y, the functional φ is continuous and

max ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } - min ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } = max ⁡ { φ ⁢ ( u ′ ) , φ ⁢ ( - u ′ ) } .

Lemma 2.7

Assume that (H0)–(H2) are satisfied. Let u be a solution of (1.6) on [R1,R2]. Then

where Q-1 denotes the inverse function to

Q ⁢ ( u ) = ∫ 0 u d ⁢ s g ⁢ ( ϕ 1 - 1 ⁢ ( s ) ) .

Proof.

Set

C + = { r ∣ u ′ ⁢ ( r ) > 0 , r ∈ ( R 1 , R 2 ) } , C - = { r ∣ u ′ ⁢ ( r ) < 0 , r ∈ ( R 1 , R 2 ) } .

Let μ⁢(C+) and μ⁢(C-) denote the Lebesgue measure of C+ and C-, respectively.

If C+=∅ (resp. C-=∅), then φ⁢(u′)=0 (resp. φ⁢(-u′)=0), and (2.3) is obviously satisfied. Suppose C+≠∅ and C-≠∅. Since u′∈X, we have that C+, C- are open subsets of [R1,R2] and therefore C+ (resp. C-) is a union of at most countably many disjoint open intervals (ai,bi), i∈I+⊂ℕ (resp. (ci,dj)∈I-⊂ℕ) without common elements, i.e.,

C + = ⋃ i ∈ I + ( a i , b i )   resp.   C - = ⋃ i ∈ I - ( c i , d i ) .

Of course, u′⁢(ai)≠0 or u′⁢(bi)≠0 for any i∈I+ (resp. u′⁢(cj)≠0 or u′⁢(dj)≠0 for any j∈I-) imply ai=R1 or bi=R2 (resp. ci=R1 or di=R2). Furthermore C+≠(R1,R2), since in the opposite case C-=∅, a contradiction. Similarly, C-≠(R1,R2).

From the inequality μ⁢(C+)+μ⁢(C-)≤R2-R1, it follows that

For our next consideration, recall that the interval [0,R2-R1) is contained in the domains of Q-1 by assumption (H2).

We now show the inequality

Fix i∈I+, let u′⁢(ε)=0, ε∈{ai,bi}. Integrating (1.6) from ε to r, we will easily get the following equality with ϕ1⁢(0)=0:

r N - 1 ⁢ ϕ 1 ⁢ ( u ′ ⁢ ( r ) ) = ∫ ε r s N - 1 ⁢ ( F ⁢ u ) ⁢ ( s ) ⁢ 𝑑 s , r ∈ [ a i , b i ] .

Since u′⁢(r)≥0 in r∈[ai,bi], it follows that ϕ1 is an increasing homeomorphism. Using (H0) and (H1), we have

From Lemma 2.5 with a=ε, b=ai, c=bi, y⁢(s)=ϕ1⁢(u′⁢(s)) and q⁢(y)=g⁢(ϕ1-1⁢(y)), it is easy to check that

ϕ 1 ⁢ ( u ′ ⁢ ( r ) ) ≤ Q - 1 ⁢ ( | ε - r | ) , r ∈ [ a i , b i ] .

Note that 0≤u′⁢(r)≤ϕ1-1⁢(Q-1⁢(bi-ai)) for r∈[ai,bi], i∈I+. We can see that

∫ a i b i u ′ ⁢ ( s ) ⁢ 𝑑 s ≤ ( b i - a i ) ⁢ ϕ 1 - 1 ⁢ ( Q - 1 ⁢ ( b i - a i ) ) .

Furthermore,

Consequently, (2.5) is fulfilled.

Next, we will prove that

Fix j∈I-, let u′⁢(η)=0, η∈{cj,dj}. Integrating (1.6) from η to r with ϕ1⁢(0)=0, we get

r N - 1 ⁢ ϕ 1 ⁢ ( u ′ ⁢ ( r ) ) = ∫ η r s N - 1 ⁢ ( F ⁢ u ) ⁢ ( s ) ⁢ 𝑑 s , r ∈ [ c j , d j ] .

Since u′⁢(r)≤0 in [cj,dj] and using the fact that ϕ1 is an odd increasing homeomorphism and (H0)–(H1), we obtain

Hence,

ϕ 1 ⁢ ( | u ′ ⁢ ( r ) | ) = - ϕ 1 ⁢ ( u ′ ⁢ ( r ) ) ≤ | ∫ η r g ⁢ ( ϕ 1 - 1 ⁢ ( ϕ 1 ⁢ ( | u ′ ⁢ ( s ) | ) ) ) ⁢ 𝑑 s | .

If we take a=η, b=cj, c=dj, y⁢(s)=ϕ1⁢(|u′⁢(s)|) and q⁢(y)=g⁢(ϕ1-1⁢(y)), then by Lemma 2.5 we deduce that

ϕ 1 ⁢ ( | u ′ ⁢ ( r ) | ) ≤ Q - 1 ⁢ ( | r - η | ) , r ∈ [ c j , d j ] .

Hence we obtain 0≤-u′⁢(r)≤ϕ1-1⁢(Q-1⁢(dj-cj)) for r∈[cj,dj], j∈I-. As a consequence,

- ∫ c j d j u ′ ⁢ ( s ) ⁢ 𝑑 s ≤ ( d j - c j ) ⁢ ϕ 1 - 1 ⁢ ( Q - 1 ⁢ ( d j - c j ) ) .

Moreover,

Thus, (2.6) is fulfilled.

From (2.4), (2.5) and (2.6), we can conclude inequality (2.3). ∎

Lemma 2.8

Suppose that u is a solution of (2.7) for any λ∈[0,1] and satisfies boundary condition (1.7) with A=0. Then the following two conclusions are satisfied:

Proof.

According to τ⁢(u)=A=0 and Lemma 2.4, there exists a ξ∈[R1,R2] such that u⁢(ξ)=0. It is easy to observe that

min ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } ≤ 0 , max ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } ≥ 0 ,

and consequently the boundary condition

max ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } - min ⁡ { u ⁢ ( r ) ∣ r ∈ [ R 1 , R 2 ] } = B

implies (2.8). Using the fact that ϕ1:(-1,1)→ℝ is an increasing homeomorphism and (2.8), we deduce that

∥ u ∥ C 1 = ∥ u ∥ 0 + ∥ u ′ ∥ 0 < B + 1 . ∎

Lemma 2.9

Let B be a positive constant, τ∈𝒜 and φ be defined in (2.2). Set

Ω = { ( u , μ , ν ) ∣ ( u , μ , ν ) ∈ Y × ℝ 2 , ∥ u ∥ C 1 < ρ , ∥ u ′ ∥ 0 < 1 , | μ | < ρ , | ν | < R 2 N - 1 ⁢ ϕ 1 ⁢ ( 1 ) } .

Define Γi:Ω¯→Y×ℝ2 for i=1,2 by

Then

d LS ⁢ ( I - Γ i , Ω , 0 ) ≠ 0 , i = 1 , 2 ,

where ρ=B+1 and ρ<R2-R1, and I denotes the identity operator on Y×ℝ2.

Proof.

Observe that Ω is a bounded open subset of the Banach space Y×ℝ2 with usual norm and symmetric with respect to θ∈Ω. Define Gi:[0,1]×Ω→Y×ℝ2 for i=1,2 by

For every (u,μ,ν)∈Ω¯, it is easy to check that

G i ⁢ ( 1 , u , μ , ν ) = Γ i ⁢ ( u , μ , ν ) , i = 1 , 2 .

By the Borsuk theorem [12, Theorem 8.3], to prove dLS⁢(I-Γi,Ω,0)≠0, it is sufficient to show that the following hypotheses hold:

1. Gi⁢(0,⋅,⋅,⋅) is an odd operator on Ω¯, that is,

   G i ⁢ ( 0 , - u , - μ , - ν ) = - G i ⁢ ( 0 , u , μ , ν ) , i = 1 , 2 , ( u , μ , ν ) ∈ Ω ¯ ;

2. Gi is a completely continuous operator;

3. Gi⁢(λ,u,μ,ν)≠(u,μ,ν) for (λ,u,μ,ν)∈[0,1]×∂⁡Ω.

At a first step, we take (u,μ,ν)∈Ω¯, for i=1,

A similar argument shows that G2⁢(0,-u,-μ,-ν)=-G2⁢(0,u,μ,ν). So (1) is satisfied.

Moreover, to prove (2), we need to show that Gi is continuous and that Gi takes bounded sets of [0,1]×Ω into relatively compact sets of Y×ℝ2.

Let (λn,un,μn,νn) be a sequence in [0,1]×Ω¯. For each n∈ℤ+ and r∈[R1,R2], |λn|≤1, ∥un∥0<ρ, |μn|≤ρ and |νn|<R2N-1⁢ϕ1⁢(1), at the same time, we have that {τ⁢(un)}, {τ⁢(-un)}, {φ⁢(un)} and {φ⁢(-un)} are bounded. From the Bolzano–Weierstrass theorem and the Arzelà–Ascoli theorem, it is not difficult to see that {λn}, {μn}, {νn}, {un} have convergent subsequences. Without loss of generality, we also denote the subsequences by {λn}, {μn}, {νn}, {un}. Let

lim n → ∞ ⁡ λ n = λ 0 , lim n → ∞ ⁡ u n = u 0 , lim n → ∞ ⁡ μ n = μ 0 , lim n → ∞ ⁡ ν n = ν 0 .

Hence,

lim n → ∞ ⁡ G ⁢ ( λ n , u n , μ n , ν n ) = G ⁢ ( λ 0 , u 0 , μ 0 , ν 0 ) .

Further, by the continuity of ϕ1-1,τ and φ, we obtain that G1 and G2 are continuous. So G1 and G2 are completely continuous.

To prove (3), suppose on the contrary that Gi⁢(λ0,u0,μ0,ν0)=(u0,μ0,ν0) for some (λ0,u0,μ0,ν0)∈[0,1]×∂⁡Ω. Observe that

By Lemma 2.4 (take u=u0,e=1-λ0) and (2.10) we infer that, there exists δ∈[R1,R2] such that u0⁢(δ)=0. Together with (2.9), it is easy to see that

From (2.9), by simple computation, we get

Next, we will divide the proof into three cases.

Case 1. Let ν0=0. It follows from (2.12), (2.13) that μ0=0,u0=0, so

( 0 , 0 , 0 ) = ( u 0 , μ 0 , ν 0 ) ∈ ∂ ⁡ Ω ,

which is a contradiction.

Case 2. If ν0>0, then

ϕ 1 - 1 ⁢ ( r 1 - N ⁢ ν 0 ) - ( 1 - λ 0 ) ⁢ ϕ 1 - 1 ⁢ ( - r 1 - N ⁢ ν 0 ) > 0 .

By simple computation, from the definition of φ in (2.2), we deduce that

φ ⁢ ( u 0 ′ ) - φ ⁢ ( ( λ 0 - 1 ) ⁢ u 0 ′ ) = ( ϕ 1 - 1 ⁢ ( r 1 - N ⁢ ν 0 ) - ( 1 - λ 0 ) ⁢ ϕ 1 - 1 ⁢ ( - r 1 - N ⁢ ν 0 ) ) ⁢ ( R 2 - R 1 ) .

Together with (2.11), we get

and

ϕ 1 - 1 ⁢ ( r 1 - N ⁢ ν 0 ) ≤ λ 0 ⁢ B R 2 - R 1 ,

since

- ( 1 - λ 0 ) ⁢ ϕ 1 - 1 ⁢ ( - r 1 - N ⁢ ν 0 ) ≥ 0 .

Thus,

ν 0 ≤ r N - 1 ⁢ ϕ 1 ⁢ ( λ 0 ⁢ B R 2 - R 1 ) ≤ r N - 1 ⁢ ϕ 1 ⁢ ( λ 0 ⁢ ρ R 2 - R 1 ) ≤ r N - 1 ⁢ ϕ 1 ⁢ ( ρ R 2 - R 1 ) < R 2 N - 1 ⁢ ϕ 1 ⁢ ( 1 ) .

Furthermore, by (2.12)–(2.14), for every r∈[R1,R2], we conclude that

Consequently, (u0,α0,β0)∉∂⁡Ω, a contradiction.

Case 3. If ν0<0, taking into account the inequality

ϕ 1 - 1 ⁢ ( r 1 - N ⁢ ν 0 ) - ( 1 - λ 0 ) ⁢ ϕ 1 - 1 ⁢ ( - r 1 - N ⁢ ν 0 ) < 0

and the definition of φ in (2.2), we have

Combining this with (2.11), we deduce that

If λ0=0, then (2.15) implies

ϕ 1 - 1 ⁢ ( r 1 - N ⁢ ν 0 ) - ( 1 - 0 ) ⁢ ϕ 1 - 1 ⁢ ( - r 1 - N ⁢ ν 0 ) = 0 ,

which contradicts ϕ1-1⁢(r1-N⁢ν0)-(1-0)⁢ϕ1-1⁢(-r1-N⁢ν0)<0.

If λ0=1, then λ0⁢B=0, i.e. B=0, which is impossible.

If λ0∈(0,1), then

( 1 - λ 0 ) ⁢ ( ϕ 1 - 1 ⁢ ( r 1 - N ⁢ ν 0 ) - ( 1 - λ 0 ) ⁢ ϕ 1 - 1 ⁢ ( - r 1 - N ⁢ ν 0 ) ) ⁢ ( R 2 - R 1 ) < 0 , also ⁢ λ 0 ⁢ B > 0 .

This is a contradiction. The proof is completed. ∎