The purpose of this paper was the study of the dynamic behavior of impact oscillators:
{x″+a(x)x2n+1+m∑l=0pl(t)xl=0,for x(t)>0,x(t)≥0,x′(t+0)=−x′(t−0),if x(t0)=0,
where the positive function a(x) is a smooth T-periodic oscillator violating the monotone twist condition. We have proved that the above equation has an infinite number of bounded solutions as well as a solution that escapes to infinity in a finite amount of time.
Citation: Yanmei Sun. The coexistence of quasi-periodic and blow-up solutions for a class of impact oscillators without the twist condition[J]. AIMS Mathematics, 2025, 10(5): 10263-10282. doi: 10.3934/math.2025467
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The purpose of this paper was the study of the dynamic behavior of impact oscillators:
{x″+a(x)x2n+1+m∑l=0pl(t)xl=0,for x(t)>0,x(t)≥0,x′(t+0)=−x′(t−0),if x(t0)=0,
where the positive function a(x) is a smooth T-periodic oscillator violating the monotone twist condition. We have proved that the above equation has an infinite number of bounded solutions as well as a solution that escapes to infinity in a finite amount of time.
In [1], Littlewood proposed the Duffing equation
x″+G′(x)=p(t),p(t)=p(t+1) |
concerning Lagrangian stability and boundedness of its solutions. We use Moser's twist theorem to prove boundedness when the Poincaré map has a monotone twist at infinity.
However, Zhiguo Wang and Yiqian Wang [2] have proved that the differential equation of the second order is
x″+x2n+1a(x)+p(t)=0,with p(t)=p(t+1),a(x)=a(x+T) is smooth, and a(x)>0. |
Despite violating the monotone twist condition, the upper equation remains stable in Lagrangian mechanics. For the Duffing equation x″+G′(x)=p(t), violating the monotone twist condition means that the ratio G′(x)x is not monotone, and one can see [3] for more details. The author [3] defines
A+(k)=the annulus bounded byΔ2kπ−c0andΔ2kπ |
and
A−(k)=the annulus bounded byΔ2kπ+π−c0andΔ2kπ+π, |
where Δx is the closed orbit of the unperturbed system
x″+x2n+1(1+ccosx)=0, |
and which passes through the point (x,0) in the x−˙x plane, where c0 is a positive constant. The author pointed out that although the monotone twist condition is violated in the annulus A+(k)⋃A−(k) (see [3] for details), the sign of d2Idh2 would not change in each annulus A+(k) (A−(k)).
In [4], the authors showed that all solutions of the following equation can be found
x″+x2n+1+xlp(t)=0,if p(t)∈C1,0≤l≤n, |
and are bounded.
Zhiguo Wang and Yiqian Wang studied the following impact oscillators in [2]
{x″+x2n+1=p(t), for x(t)>0,x(t)≥0, x′(t+0)=−x′(t−0), if x(t0)=0. |
They found that there are an infinite number of periodic and quasi-periodic solutions.
From a mechanical point of view, the previous equation shows how particles that are attached to nonlinear springs will bounce off a fixed barrier (x=0). These types of systems are special cases of vibrating impact systems [5]. In addition, they are associated with the Fermi accelerator [6], the dual billiard [7], and certain models used in astronomy [8]. The application of many powerful mathematical tools is limited by the lack of smoothness caused by impact. However, the periodic and quasi-periodic motion of impact oscillators has been addressed in several recent papers, see refs. [9,10,11] and their references.
Let us start by studying the following impact system:
{x″+a(x)x2n+1+m∑l=0pl(t)xl=0, for x(t)>0,0≤m≤2n,x(t)≥0, x′(t+0)=−x′(t−0), if x(t0)=0, | (1.1) |
which has infinitely many bounded solutions. We suppose that the positive function a(x) is a smooth T-periodic oscillating function that violates the monotone twist condition and the coefficient pl(t) is 1-periodic and C5-smooth. Denote T=R/Z, and then we have pl(t)∈C5(T).
We can prove the first important result according to Moser's twist theorem.
Theorem 1.1. Suppose pl(t)∈C5(T),1<n∈N,0≤m≤2n−2. Then every solution x(t) of Eq (1.1) is bounded, i.e., it is in (−∞,+∞) and satisfies
supt∈R(|x(t)|+|x′(t)|)<+∞. |
In addition, for any positive integer m, there are an infinite number of m-periodic solutions, as well as an infinite number of quasi-periodic solutions of large amplitude, i.e., there exists a very large ω∗>0 such that for any irrational number ω>ω∗ which satisfies
|ω−pq|≥εq−52, | (1.2) |
there exists a smooth function F(θ1,θ2) periodic with period 1 for all integers p and q≠0 with constant ε>0. This function is such that x(t)=F(θ1+ωt,θ2+t) are solutions of Eq (1.1).
Remark 1.1. Here the existence of infinitely many periodic solutions can be obtained by the Aubry-Mather theory. We have chosen to omit it here because of the wealth of related results.
Remark 1.2. The ω set satisfying Eq (1.2) has a positive Lebesgue measure.
Remark 1.3. Consider the impact system:
{x″+x2n+1(1+ccosx)=p(t),forx(t)>0, |c|<1,x(t)≥0,x′(t+0)=−x′(t−0),ifx(t0)=0, | (1.3) |
where 1<n∈N,p(t) are smooth, and a(x+2π)=a(x),p(t)=p(t+1). In this case, it is similar to statement 1.1, as Eq (1.1) is a generalization of Eq (1.3).
Next, for the boundedness problem, during the past years, people have obtained many results via some Kolmogorov-Arnold-Moser (KAM) theorems, see refs [12,13,14]. In addition, there are many references on the study of nonsmooth oscillators using the KAM theory, such as [15,16,17]. However, to the best of our knowledge, there are few results [6,18,19] related to the unbounded problem. In [20], Wang obtained an unbounded solution of the following equation:
¨x+x2n+1+p(t)xl=0, |
with p(t)∈C0(S1), n≥2,2n+1>l≥n+2.
Motivated by [2,3,14], we consider the blow-up solutions of Eq (1.1). For simplicity, we will only discuss the case where pl(t)=p(t), i.e.,
{¨x+a(x)x2n+1+p(t)xl=0,forx(t)>0,x(t)≥0,x′(t+0)=−x′(t−0),ifx(t0)=0, | (1.4) |
where p(t)=p(t+1),n+2≤l≤2n, a(x)=a(x+T)>0 are smooth.
From [3] we know that the monotone twist condition can be violated. Depending on the oscillation behavior of the potential G(x)=∫x0a(s)s2n+1ds of (1.4), the ratio G′(x)x may or may not be monotone. For example, it is monotone if a(x)=1, which corresponds to a non-oscillating potential, and not monotone when a(x)=1+ccosx,0<|c|<1, which corresponds to an oscillating potential [3]. The second main result is as follows.
Theorem 1.2. There exists p(t)∈C0(S1),n+2≤l≤2n, and a sufficiently large λ0 such that for the solution (x(t),x′(t)) with (x(0),x′(0))=((2λ0)1n+2,0) of Eq (1.4), there is a strictly increasing series {Ti}∞i=0 such that limi→∞Ti<1 with T0=0 and
minTi≤t≤Ti+1λ(x(t),x′(t))≥λMi0, |
where M>1 and λ(x,x′) is the new action variable after coordinate transformations Ψ1, Ψ2, Ψ3 in section 2.1 satisfying (x(t),x′(t))=Ψ1∘Ψ2∘Ψ3(ϕ(t),λ(t)).
Corollary 1.1. Equation (1.4) in Theorem 1.2 has a solution (x(t),x′(t)) which satisfies (|x(t)|+|x′(t)|)→+∞, as t→T∞<1.
First, let us introduce the idea to prove the boundedness of the solutions of Eq (1.1). Using transformation theory, we can first characterize Eq (1.1) in terms of a Hamiltonian function H(3)(ϕ,λ,t) (see Eq (2.8)) in action-angle variables, specified over the whole space T×R+×R. The action causes H(3)(ϕ,λ,t) to be C5 smooth in λ and t. It is only continuous in ϕ.
By changing the roles of the variables t and ϕ, we move away from a large disc in the space Dr={(ϕ,λ)∈T×R+,λ<r} in the (ϕ,λ) plane, and H(3)(ϕ,λ,t) is transformed into a perturbation of an integrable Hamiltonian system H(θ,ρ,τ) (see Eq (2.13)). This system is sufficiently smooth in θ and ρ. The Poincaré mapping concerning the new time parameter τ is closely related to a mapping called "twist mapping" in the region R2/Dr and exhibits area-preserving properties.
The KAM theorem ensures the existence of arbitrarily large invariant curves that are diffeomorphic to circles and lie in the (x,y) plane, as presented in reference [21]. One can return to the equivalent system Eq (1.1), where it is obvious that any such curve is a basis for a time-periodic and flow-invariant cylinder in the extended phase space (x,y,t)∈R+×R×R. This restriction of the solutions leads internally to bounds on these solutions. This is true as long as the uniqueness of the initial value problem persists.
Then the idea for the proof of the infinite solutions of Eq (1.4) is as follows. Indeed, we construct simultaneously the function p(t) and the solution x(t) of Theorem 1.2. The first thing we notice is that when the curve is spiraling once around the origin, the action variable λ is increasing at some points in time and decreasing at other points in time. So we have no idea whether the increase in λ will be positive or negative. But we can construct a time t1<<1 and modify p(t)≡1 at [0,t1] so that the increment is positive and O(1τλl−n+1n+20) if the starting point is (λ(0),ϕ(0))=(λ0,0) is far enough from the origin, where the "jump" τ(0<τ<1) is critical to modify p(t) and our estimate. Inductively, we can construct a series of times t1,t2,...,ti,ti+1,⋅⋅⋅ and modify p(t) on [ti,ti+1],i=1,2,⋅⋅⋅, so that on each such interval [ti,ti+1], the increment is positive and at least O(1τλl−n+1n+20). So we can build a time T1≤1τ′<1, so that the curve spirals at least O(1τ′λnn+20) times around the origin and λ1=λ(T1)>λ0+cττ′λl+1n+20 where l+1n+2>1 and τ′ are used to ensure that the time does not exceed 1. This completes an induction step: During the time interval [0,T1],λ increases from λ0 to λ1. Inductively, we can construct a series of times T1,T2,...,Ti,Ti+1,⋅⋅⋅, such that during the time interval [Tk,Tk+1],λ increases from λk to λk+1, where λk+1>λk+cτkτ′kλl+1n+2k with the jump 1τk, where Tk+1−Tk≤1τ′k. The reason why the jump is less and less is that we have to make sure that p(t) is continuous. Since the exponent is l+1n+2>1, the smaller and smaller jump will not be able to stop the rapid increase of λ. If 1τ′ is chosen small enough, we will find that Tk→T∞<1 as k→∞ and λt→+∞ as t→T∞.
In the past, scholars studied either quasi-periodic solutions or blow-up solutions, but there were few research results on the coexistence of quasi-periodic solutions and blow-up solutions. In this paper, we obtained the result that quasi-periodic solutions and blow-up solutions coexist.
The article is structured as follows: In Section 2, the proof of Theorem 1.1 may be derived from Moser's twist theorem. To apply Moser's theorem, we provide the necessary estimates, which are proven in Section 3. These estimates are lengthy and complicated, but we think it is important to give the details because they are crucial for fitting the problem into the KAM theory framework. In Section 4, we construct the action-angle transformation and obtain some Lemmas. In Section 5, we construct p(t)∈C0(S1) and a series of time Tk. We then obtain an unbounded solution and finish the proof of Theorem 1.2.
Equation (1.1) of the second-order model, excluding impacts, is equivalent to the following system of the one-order model:
{x′=y,y′=−a(x)x2n+1−m∑l=0pl(t)xl,0≤m≤2n. | (2.1) |
Equation (2.1) is a Hamiltonian system defined in the entirety of the phase plane XOY of the Hamiltonian function
H(x,y,t)=12y2+G(x)+m∑l=0xl+1l+1pl(t), | (2.2) |
where
G(x)=∫x0a(s)s2n+1ds. | (2.3) |
Define I=I(h)=∫Γh√2h−2G(x)dx, where Γh represents the closed curve 12y2+G(x)=h. We also define
S(x,I)={∫x−x−√2h−2G(x)dx,y≥0,I−∫x−x−√2h−2G(x)dx,y<0, |
where G(−x−)=h.
Try changing the action angle like this:
y=∂S∂x,θ=∂S∂I, |
and we get
θ={H′0(I)∫x−x−dx√2h−2G(x),y≥0,1−H′0(I)∫x−x−dx√2h−2G(x),y<0, |
where H0 is the inverse function of I(h).
Denote
Ψ1:(θ,I)→(x,y). | (2.4) |
Subsequently, the Hamiltonian H of Eq (2.2) is transformed into the following expression:
H(1)(θ,I,t)=H∘Ψ1=H0(I)+m∑l=0xl+1(I,θ)l+1pl(t). | (2.5) |
The impact case transforms the phase space into a half-plane x≥0 of the original phase plane XOY. When x(t)=0, the smoothness and continuity of H(1) with variable θ is lost. Consequently, as outlined below, an alternative Hamiltonian, designated as H(3), will be pursued.
To normalize the angle variable at a later stage, the following definition is proposed:
Ψ2:(ϕ1,λ)→(θ,I):{θ=12ϕ1,I=2λ. | (2.6) |
We get
H(2)(ϕ1,λ,t)=H(1)∘Ψ2=H0(2λ)+m∑l=0xl+1l+1(2λ,12ϕ1)pl(t). | (2.7) |
The periodic extension of H(2) and ϕ1 to the interval [0,1) defines a new Hamiltonian, which we denote by
H(3)(ϕ,λ,t)=H(2)∘Ψ3=H0(2λ)+m∑l=0xl+1l+1(2λ,12(ϕ−[ϕ]))pl(t), | (2.8) |
where
Ψ3:(ϕ,λ)→(ϕ1,λ):{ϕ1=ϕ−[ϕ],λ=λ | (2.9) |
where [ϕ] represents the greatest integer that is no greater than ϕ. We know that
{ϕ′=∂H(3)∂λ, when ϕ∈(k,k+1),k∈Z,λ′=−∂H(3)∂ϕ, when ϕ∈(k,k+1),k∈Z,ϕ(t0)=k,λ(t0)=limt→t0λ(t), when limt→t0ϕ(t)=k,k∈Z, | (2.10) |
is the corresponding Hamiltonian system.
It is evident that H(3) is periodic in ϕ with a period of 1, and C∞ in ϕ when ϕ∉Z. However, it is only continuous at ϕ=ϕ0, ϕ0∈Z. In fact, H(3)|ϕ=0=H(3)|ϕ=1=H0(2λ) for x(2λ,0)=x(2λ,12)=0.
Then we provide the key lemma.
Lemma 2.1. For every solution (ϕ(t),λ(t)) of Eqs (2.8) and (2.10) with λ(t)≠0, the pair (x(t),x′(t))=Ψ1∘Ψ2∘Ψ3(ϕ(t),λ(t)) is a continuous solution of Eq (1.1) with (x(t),x′(t))≠(0,0), and vice verse.
Remark 2.1. As stated in [22], a similar description is given for the equivalence of these systems. Therefore, the proof is omitted.
In order to prove the boundedness of every solution of Eq (1.1), i.e., |x(t)|+|x′(t)|<∞, by Lemma 2.1, we need to show the boundedness of λ(t). We will consider the Poincaré mapping of the system Eq (2.15) below which is equivalent to the system Eq (2.10) by exchanging the positions of variables (ϕ,λ) and (t,H(3)) to cope with the non-smoothness in ϕ. The next step is to use Moser's invariant curve theorem for similar considerations.
In this context, let a and A be suitable constants, without worrying about how big they are.
Lemma 2.2. Denote α=12n+2. For λ large enough, H(3) has the inverse function
λ=[H(3)(ϕ,⋅,t)]−1(ρ)=12H−10(ρ)+g(t,ρ,ϕ), |
and moreover,
|∂iρ∂jtg|<Aρmα+(2α−12)+(3α−1)i | (2.11) |
for 0≤i+j≤5,n≥1, provided ρ is large enough.
We will approve the proof in Section 3. Then we exchange the roles of (ϕ,λ) and (t,H(3)) by means of
Ψ4:(ϕ,λ,t)→(θ,ρ,τ):=(t,H(3)(λ,ϕ,t),ϕ),ϕ∉Z. | (2.12) |
This transformation again leads to a new Hamiltonian system, cf. [23], where the new Hamiltonian is
H(θ,ρ,τ)=[H(3)(τ,⋅,θ)]−1(ρ)=12H−10(ρ)+g(θ,ρ,τ). | (2.13) |
Therefore H(θ,ρ,τ) is C5 in θ, C∞ in ρ, and continuous in new time τ. This transformation is used by many authors, cf. [24,25], etc.
We know that the system (2.10) is the same as the Hamiltonian system
XH:{dθdτ=∂H∂ρ=12dH−10(ρ)dρ+∂g(θ,ρ,τ)∂ρ,τ∉Z,dρdτ=−∂H∂θ=−∂g(θ,ρ,τ)∂θ,τ∉Z,θ(k)=limτ→kθ(τ),ρ(k)=limτ→kρ(τ),k∈Z. | (2.14) |
Integrate the above system by τ from 0 to 1, and the Poincaré mapping is formed:
Φ1:{θ1=θ+γ(ρ)+g1(θ,ρ),ρ1=ρ+g2(θ,ρ) | (2.15) |
where γ(ρ)=12dH−10(ρ)dρ and g1,g2 satisfy the following lemma.
Lemma 2.3. If ρ is large enough, then
|∂iρ∂jtg1|<Aρmα+(2α−12)+(4α−1)(i+1),fori+j≤4,|∂iρ∂jtg2|<Aρmα+(2α−12)+(4α−1)i,fori+j≤4. | (2.16) |
Proof. Set
γ(ρ,t)=t2dH−10(ρ)dρ |
and set for the flow (θ(t),ρ(t))=Φt(θ,ρ) with Φ0=id:
{θt=θ+γ(ρ,t)+A(θ,ρ,t),ρt=ρ+B(θ,ρ,t). |
Then the integral equation
Φt(θ,ρ)=Φ0(θ,ρ)+∫t0XH∘Φsds |
for the flow is equivalent to the following equation for A and B:
A(θ,ρ,t)=t2∫t0∫10d2H−10(ρ+τB)dρ2Bdτds+∫t0∂g(θ+γ+A,ρ+B)∂ρds,B(θ,ρ,t)=−∫t0∂g(θ+γ+A,ρ+B)∂θds. | (2.17) |
It is easy to check that for any value of ρ≥ρ0, these equations have a unique solution in the space |A|,|B|≤1 using the construction principle. Also, A and B are smooth. The required estimates in Eq (2.16) can be verified from Eq (2.17) using induction.
Lemma 2.4. The mapping P=Φ1(see Eq (2.15)) has the intersection property, i.e., if an embedded circle Γ in R×[0,1] is homotopic to a circle ρ= const., then P(Γ)∩Γ≠∅.
Proof. One can see that the mapping Φ1 in (2.15) is the time-one map of the Hamiltonian system (2.10); recall that the time-one map of a Hamiltonian system is symplectic, thus Φ1 is symplectic, and hence, Φ1 has the intersection property.
Proof of Theorem 1.1. Through Lemma 3.3 in Section 3, denote G(x+)=h, and we can obtain an annulus kT−c0≤x+≤kT when a′(x)>0, or an annulus kT+T2−c0≤x+≤kT+T2 when a′(x)<0, respectively, such that γ(ρ) of Eq (2.15) satisfies γ′(ρ)≠0,∀ρ∈[a,b] and γ(ρ)∈C4[a,b]. When n>1,0≤m≤2n−2, by Lemma 2.3, we have ‖g1‖C4(A)+‖g2‖C4(A)<ε. Lemma 2.4 shows that Φ1 has the intersection property. Therefore, Φ1 meets the assumptions of Moser's twist theorem. Theorem 1.1 is complete.
Lemmas 3.1, 3.2, 3.4–3.6 below are the direct results of [3]and [14].
Lemma 3.1. It holds that
ax2n+2<G(x)<Ax2n+2, |
and
|G(i)(x)|<A|x|2n+1, |
for any i≥1.
Lemma 3.2. Denote α=12n+2. There exist a,A∈R and 0<a<A, for large I>0, and we get
ah12+α≤I(h)≤Ah12+α,ahα−12≤I′(h)≤Ahα−12. |
If in the smooth T-periodic function a(x)≢d,d is a constant, then its maximum value points in (0,T) must exist, and we use x∗ to mark the largest point in (0,T). Then we have the following lemma.
Lemma 3.3. There exists c0>0, such that
1)−Ah32α−32≤I″(h)≤−ah32α−32,kT−c0≤x+≤kT,a′(x)>0,x∈(x∗,T);2)ah32α−32≤I″(h)≤Ah32α−32,kT+T2−c0≤x+≤kT+T2,a′(x)<0,x∈(x∗,T); |
where G(x+)=h.
Proof. We denote
a(x)=ˉa+b(x), |
where ˉa=1T∫T0a(s)ds,1T∫T0b(s)ds=0. Then we get
G(x)=∫x0a(s)s2n+1ds=∫x0(ˉa+b(s))s2n+1ds=ˉa2n+2x2n+2+∫x0b(s)s2n+1ds=ˉa2n+2x2n+2+B(x)x2n+1−(2n+1)∫x0B(s)s2nds, |
where B′(x)=b(x).
From the above equality, we have
GG″G′2=ˉa2n+2x2n+2+B(x)x2n+1−(2n+1)∫x0B(s)s2ndsa(x)x2n+1×(2n+1)ˉax2n+(2n+1)b(x)x2n+a′(x)x2n+1a(x)x2n+1=2n+12n+2ˉa2a2(x)+12n+2ˉaa′(x)xa2(x)+2n+12n+2ˉab(x)a2(x)+B(x)a′(x)a2(x)+O(1x)a2(x). |
Using the above equality, we get
I″(h)=4h∫x+0(12−GG″G′2)dx√2h−2G(x)=4h∫x+0(12−2n+12n+2ˉa2a2(x))dx√2h−2G(x)−4h∫x+012n+2ˉaa′(x)xa2(x)dx√2h−2G(x)−4h∫x+0(2n+12n+2ˉab(x)a2(x)+B(x)a′(x)a2(x))dx√2h−2G(x)−4h∫x+0O(1x)a2(x)dx√2h−2G(x)˙=J1+J2+J3+J4, | (3.1) |
with
J1=4h∫x+0(12−2n+12n+2ˉa2a2(x))dx√2h−2G(x),J2=−4h∫x+012n+2ˉaa′(x)xa2(x)dx√2h−2G(x),J3=−4h∫x+0(2n+12n+2ˉab(x)a2(x)+B(x)a′(x)a2(x))dx√2h−2G(x),J4=−4h∫x+0O(1x)a2(x)dx√2h−2G(x). |
Similarly to the proof of Lemma 2.1 in [3], it is easy to see that
J1=O(hα−32),J3=O(hα−32), |
and
J4=O(h−32). |
If the following inequality holds:
−Ah32α−32≤J2≤−ah32α−32,kT−c0≤x+≤kT,a′(x)>0,x∈(x∗,T), | (3.2) |
the proof of Lemma 3.3 can be reduced. Next, we will only prove (3.2). Due to 2) of the lemma, the proof is similar. Let x+=kT−δ∈[kT−c0,kT] with c0>0 to be determined later. Choose c1=c1(x+)∈(0,1) such that c1x+=kT−T+x∗.
We denote
J2=−4ˉa√2(2n+2)h˜J2 | (3.3) |
and
˜J2=∫x+0a′(x)xa2(x)dx√h−G(x)=∫c1x+0a′(x)xa2(x)dx√h−G(x)+∫x+c1x+a′(x)xa2(x)dx√h−G(x)˙=J21+J22 | (3.4) |
with
J21=∫c1x+0a′(x)xa2(x)dx√h−G(x)=−∫c1x+0x√h−G(x)d1a(x)=xa(x)√h−G(x)|0c1x++∫c1x+01a(x)(1√h−G(x)+xG′(x)√(h−G(x))3)dx=−c1x+a(c1x+)√h−G(c1x+)+∫c1x+01a(x)(1√h−G(x)+xG′(x)√(h−G(x))3)dx=O(hα−12), |
and
J22=∫x+c1x+a′(x)xa2(x)dx√h−G(x)=∫kT−δkT−T+x∗a′(x)xa2(x)dx√h−G(x)=∫T−x∗δa′(−x)(kT−x)a2(−x)√h−G(kT−x)dx=kT∫T−x∗δa′(−x)a2(−x)√h−G(kT−x)dx−∫T−x∗δa′(−x)xa2(−x)√h−G(kT−x)dx˙=I1+I2. |
After a simple calculation, we have
I2=O(hα2−12) |
and by δ<x<T−x∗, we get x∗<T−x<T−δ, and then a′(−x)=a′(T−x)>0. Therefore
ah3α2−12≤I1≤Ah3α2−12. |
From the fact that kT∈[x+,x++T]⊂[ahα,Ahα], it holds for small δ that
ah3α2−12≤J22≤Ah3α2−12. |
Combining (3.3), (3.4), and the last inequality, there exists c0>0, such that (3.2) holds, and thus the proof is complete.
For higher derivatives of I(h), from the definition of I(h), [26], and Lemma 3.1, we have:
Lemma 3.4. For large I>0, we have
|I(i)(h)|≤Ahα+12−i(1−α), i≥3. |
Lemma 3.5. For large I>0, we have
aI22α+1≤H0(I)≤AI22α+1,aI1−2α2α+1≤H′0(I)≤AI1−2α2α+1,|H″0(I)|≤AI−3α2α+1, |
and
|H(i)0(I)|≤AI2+3(i−1)α2α+1−i,foreachi≥3. |
Lemma 3.6. For i,j≥0, we have
|∂iI∂jθx(I,θ)|≤AIa(i,j)−i, | (3.5) |
where
a(i,j)={2α2α+1,i=0,j=0,1,(4i+2max{j−1,0})α2α+1,otherwise. | (3.6) |
Proof. The proof method is like the one in Lemma 3.4 in [3], so we will not go into detail about it here.
Lemma 3.7. For i,j≥0, we have
|∂iI∂jθxl+1(I,θ)|≤AIa(i,j)−i+2lα2α+1. | (3.7) |
Proof. We know ∂iI∂jθxl+1(I,θ) is the sum of terms
(xl+1)(ν)∂s1I∂t1θx(I,θ)⋅⋅⋅∂sνI∂tνθx(I,θ) |
with ν≥1,s1,t1,⋅⋅⋅,sν,tν≥0,s1+⋅⋅⋅+sν=i,t1+⋅⋅⋅+tν=j.
If you add the above inequality to Lemma 3.6, you can easily show that (3.7) is true.
The proof of Lemma 2.2. H(3)(ϕ,λ,t) is defined in R×R+×R, and is 1-periodic in ϕ and t. First, consider the existence of the inverse function of H(3)(ϕ,λ,t) with the second variable. Denote
ρ=H(3)(ϕ,λ,t)=H0(2λ)+m∑l=0xl+1l+1(2λ,12(ϕ−[ϕ]))pl(t). |
For λ large enough, by Lemma 3.5, we have ∂ρ∂λ>λ0.5−α0.5+α>0, where 0.5−α0.5+α>0, thus limλ→∞ρ=∞ and ρ has its inverse function
λ=[H(3)(ϕ,⋅,t)]−1(ρ)=12H−10(ρ)+g(t,ρ,ϕ). |
Next, we will provide the estimates in summary, for ρ large enough,
|∂iρ∂jtg|<Aρmα+(2α−12)+(3α−1)i, for 0≤i+j≤5. |
Rewrite
ρ=H0(2(12H−10(ρ)+g))+m∑l=0xl+1l+1(2(12H−10(ρ)+g),12(ϕ−[ϕ]))pl(t) | (3.8) |
into the following form:
2g∫10H′0(H−10(ρ)+2τg)dτ+m∑l=0xl+1l+1(H−10(ρ)+2g,12(ϕ−[ϕ]))pl(t)=0. | (3.9) |
If ρ is large, g is well determined by the contraction principle. Moreover, by the implicit function theorem, g is smooth in ρ, for large ρ. We can easily obtain |g(t,ρ,ϕ)|<Aρmα+2α−12.
By Eq (3.9), we have
g=−12m∑l=0xl+1l+1(H−10(ρ)+2g,12(ϕ−[ϕ]))pl(t)1∫10H′0(H−10(ρ)+2τg)dτ. | (3.10) |
Applying ∂iρ to Eq (3.10), the left-hand side is ∂iρg and the right-hand side is the algebraic sum of terms
−12∂i1ρxl+1l+1(H−10(ρ)+2g,12(ϕ−[ϕ]))pl(t)∂i2ρ1∫10H′0(H−10(ρ)+2τg)dτ, |
with i1+i2=i.
We know −12∂i1ρxl+1l+1(H−10(ρ)+2g,12(ϕ−[ϕ]))pl(t) is the algebraic sum of terms
−12(xl+1)(ν)l+1ν∏k=1∂νkρ(H−10(ρ)+2g)pl(t),ν≥1,ν∑k=1νk=i1 |
and ∂i2ρ1H′0(H−10(ρ)+2τg) is the algebraic sum of terms
∂μλ1H′0(λ)μ∏k=1∂μkρ(H−10(ρ)+2τg),μ≥1,μ∑k=1μk=i2,ν+μ≤i. |
Thus, by Lemma 3.7 and
|∂i2ρ1H′0(ρ)|<Aρ(i2+1)(2α−1)2α+1, |
we can get
|∂iρg|<Aρmα+(2α−12)+(3α−1)i. |
Next differentiating ∂iρg with respect to t, we have
|∂iρ∂jtg|<Aρmα+(2α−12)+(3α−1)i, for 0≤i+j≤5. |
Thus when n≥1, for 0≤i+j≤5, Eq (2.11) has been proved.
In this section, we are concerned with the blow-up solutions of the system (1.4). Recall that by similar variable transformations as in Section 2.1, the system (1.4) can be determined by
{dϕdt=∂H3∂λ=2H′0(2λ)+xl(2λ,12(ϕ−[ϕ]))∂x(2λ,12(ϕ−[ϕ]))∂λ(p(t)−1), when ϕ∈(k,k+1),dλdt=−∂H3∂ϕ=−xl(2λ,12(ϕ−[ϕ]))∂x(2λ,12(ϕ−[ϕ]))∂ϕ(p(t)−1), when ϕ∈(k,k+1),ϕ(t0)=k,λ(t0)=limt→t0λ(t), when limt→t0ϕ(t)=k,k∈Z, | (4.1) |
where
H3(λ,ϕ,t)=H2∘Ψ3=H0(2λ)+xl+1l+1(2λ,12(ϕ−[ϕ]))(p(t)−1). | (4.2) |
Similarly to the Lemmas 3.1 and 3.2 in [14], it is easy to imply that there exist some constants ai>0,i=1,2,3,4,5, such that
a1Inn+2<H′0(I)<a2Inn+2, | (4.3) |
|xl(I,θ)|<a3Iln+2, | (4.4) |
|∂x∂I(I,θ)|<a4I−nn+2, | (4.5) |
|∂x∂θ(I,θ)|<a5I1n+2, | (4.6) |
∂x∂θ≥0,when y≥0;∂x∂θ<0,when y<0. | (4.7) |
We assume that l is an even number, and then define p0(t) to be a piecewise continuous function, where tk is the unique time which satisfies θ(tk)=k−12, Ik=Itk for the solution of the new system corresponding with p0(t).
p0(t)={1,[0,t12],1−σ,(t12,t1],1,(t1,1]. | (4.8) |
Remark 4.1. By (4.1) and (4.8), it is easy to imply that λ12=λ0.
In the following, all ci are independent of the steps in the induction process.
Lemma 4.1. If λ0 is sufficiently large, then
t1−t12≤c1λ−nn+20. |
Proof. By (4.7), we have ∂x∂ϕ≥0, and when t∈[t12,t1],λ is an increasing function on this interval. We know that ϕ(t12)=0,ϕ(t1)=1. By (4.1), (4.3), (4.4), and (4.5), if λ0 is sufficiently large, we get
t1−t12=∫10dϕ2H′0(2λ)−σxl∂x∂λ≤∫10dϕ2a1(2λ)nn+2−σa3a4(2λ)l−nn+2≤∫10dϕ(2λ0)nn+2(2a1−σa3a4(2λ0)l−2nn+2)≤(2λ0)−nn+2a1≤c1λ−nn+20. |
Lemma 4.2. If λ0 is sufficiently large, then
λ1≤λ0(1+c2σλl−1−2nn+20). |
Proof. By (4.1), (4.4), and (4.6), one has
t1−t12=∫t1t12dt=∫λ1λ12dλσxl∂x∂ϕ≥∫λ1λ12dλσa3a5(2λ)l+1n+2=(2λ1)n+1−ln+2−(2λ12)n+1−ln+22σa3a5n+1−ln+2. |
Because n+1−l<0, by the above inequality, Remark 4.1, and Lemma 4.1, we have
λn+1−ln+21−λn+1−ln+212n+1−ln+2≤σa3a5a1λ−nn+20, |
i.e.,
λn+1−ln+21≥λn+1−ln+20+σa3a5a1(n+1−l)n+2λ−nn+20=λn+1−ln+20(1+σa3a5a1(n+1−l)n+2λl−1−2nn+20). |
Then
λ1≤λ0(1+σa3a5a1(n+1−l)n+2λl−1−2nn+20)n+2n+1−l≤λ0(1+2σa3a5a1λl−1−2nn+20)≤λ0(1+c2σλl−1−2nn+20). |
Theorem 4.1.
λ1≥λ0(1+c3σλl−1−2nn+20). |
Proof. If t∈[t12,t1], by (4.1)–(4.5), there exists a6>0, such that
ϕ′(t)<a6λnn+20. | (4.9) |
Denote
xl=λln+2q1(λ,ϕ),∂x∂ϕ=λ1n+2q2(λ,ϕ), |
and then if ϕ∈(34,54), we have q2(λ,ϕ)>0 and if q1(λ,ϕ)=0, then ϕ=1. Thus
∫5434q1(λ,ϕ)q2(λ,ϕ)dϕ>0. | (4.10) |
By (4.1), (4.9), and (4.10), we get
λ1−λ12=∫λ1λ12dλ=∫t1t12σxl∂x∂ϕdt≥σ(2λ0)l+1n+2∫t1t12q1(λ,ϕ)q2(λ,ϕ)dt=σ(2λ0)l+1n+2∫5434q1(λ,τ)q2(λ,τ)ϕ′(t)dτ≥σ(2λ0)l+1−nn+2a6∫5434q1(λ,τ)q2(λ,τ)dτ≥c3σλl−n+1n+20. |
Lemma 4.3. If I0 is sufficiently large, then
∃ η>nn+2suchthatt1−t12>2λ−η0. |
Proof. By (4.1)–(4.5), and Lemma 4.2, we get
t1−t12=∫5434dϕH′0(2λ)−σxl∂x∂λ>121a2(2λ1)nn+2+σa3a4(2λ1)l−nn+2≥121(2λ1)nn+2(a2+σa3a4λl−2nn+21)>116a2λ−nn+20>2λ−η0, |
where η>nn+2.
Now we change the piecewise continuous function p0(t) of (4.8) into a continuous function:
p0(t)={1,[0,t12],σ(t12−t)λη0+1,(t12,t12+λ−η0],1−σ,(t12+λ−η0,t1−λ−η0],1+(t−t1)λη0σ,(t1−λ−η0,t1]1,(t1,1]. | (4.11) |
It is easy to check that Lemmas 4.1–4.3 and Theorem 4.1 still hold with ~ci after this modification in view of λ−η0≪λ−nn+20.
We will modify p0(t) inductively and denote the function obtained and the corresponding solution with (λ0,ϕ0) as the initial point by pi and ϕi(t),λi(t) with ϕi(ti)=i,λi(ti)=λi.
Suppose we have obtained p0,p1,…,pi. pi+1 is constructed by modifying pi on the interval [ti,ti+1], where ti+1 satisfies ϕi+1(ti+1)=i+1 in the same way as above if we regard λi,ti as λ0,t0. All the lemmas are true after the modification.
In the process of constructing pi, we keep the jump σ=1/τ(τ>2) unchanged until i=j1. Then we let σ=1/τ2 and keep it unchanged until i=j2. Inductively, we choose σ=1/τk when ϕ∈[jk−1,jk], where j0=0, j1,j2,… satisfies
T0=0;t∈[0,tj1],σ=1τ,j1=[O(λnn+20τ′)],T1=tj1;t∈(tj1,tj2],σ=1τ2,j2−j1=[O(λnn+2j1τ′)],T2=tj2;t∈(tj2,tj3],σ=1τ3,j3−j2=[O(λnn+2j2τ′2)],T3=tj3;⋅⋅⋅⋅⋅⋅t∈(tjk−1,tjk],σ=1τk,jk−jk−1=[O(λnn+2jk−1τ′k)],Tk=tjk;⋅⋅⋅⋅⋅⋅ | (5.1) |
Then we can imply that
λj1>λ0+c′j1ττ′λl+1n+20;λj2>λj1+c′j2ττ′λl+1n+2j1;λj3>λj2+c′j3τ2τ′2λl+1n+2j2;⋅⋅⋅⋅⋅⋅λjk>λjk−1+c′jkτk−1τ′k−1λl+1n+2jk−1;⋅⋅⋅⋅⋅⋅ | (5.2) |
Lemma 5.1.
limk→∞Tk<1,if τ′islargeenough. |
Proof. First, we will show
Tk+1−Tk<c1jk+1τ′k,k=0,1,⋅⋅⋅ | (5.3) |
In fact, by (4.1), we have
t12<λ−nn+202a1 |
and
T1=tj1<j1⋅2t12<[O(λnn+20τ′)]⋅λ−nn+20a1<c11τ′. |
If Tk−Tk−1<c1jkτ′k−1, then
Tk+1−Tk=tjk+1−tjk<(jk+1−jk)⋅2(tjk+12−tjk)<[O(λnn+2jkτ′k)]⋅2cjk+1λ−nn+2jk<c1jk+1τ′k, |
and therefore, if τ′ is large enough, we have
limk→∞Tk=limk→∞[T1+(T2−T1)+(T3−T2)+⋅⋅⋅+(Tk−Tk−1)+⋅⋅⋅]≤c11τ′+c1j2τ′+c1j3τ′2+⋅⋅⋅+c1jkτ′k−1+⋅⋅⋅<1. |
Lemma 5.2.
λjk>2λMk0,M>1. |
Proof. By (5.2), one has
λj1>λ0+c′j1ττ′λl+1n+20>2λM0,M=1+l−n−12(n+2). |
If
λjk−1>2λMk−10, |
then
λjk>λjk−1+c′jkτk−1τ′k−1λl+1n+2jk−1>c′jkτk−1τ′k−1(2λMk−10)l+1n+2=2λMk0c′jkτk−1τ′k−12l+1n+2−1λMk−1⋅l+1n+2−Mk0=2λMk0c′jkτk−1τ′k−12l−n−1n+2λl−n−12(n+2)⋅Mk0>2λMk0. |
Proof of Theorem 1.2. According to Lemma 5.2, one can see that
mint∈[Ti,Ti+1]λ(t)≥12λji≥λMi0. |
Thus Theorem 1.2 has been proved.
Since M>1, we imply that λ(t)→+∞ as i→+∞. Therefore, Eq (1.4) in Theorem 1.2 possesses an unbounded solution (x(t),x′(t)) satisfying (|x(t)|+|x′(t)|)→+∞, as t→T∞<1.
The author declares she has not used Artificial Intelligence (AI) tools in the creation of this article.
This work is supported by the National Natural Science Foundation of China (Grant Nos. 12371256 and 11971475).
The author declares no conflict of interest.
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