Invariant Tori for a Two-Dimensional Completely Resonant Beam Equation with a Quintic Nonlinear Term
Abstract
This paper focuses on a two-dimensional completely resonant beam equation with a quintic nonlinear term. This means studying , under periodic boundary conditions, where ε is a small positive parameter and f(u) is a real analytic odd function of the form It is proved that the equation admits small-amplitude, Whitney smooth, linearly stable quasiperiodic solutions on the phase-flow invariant subspace . Firstly, the corresponding Hamiltonian system of the equation is transformed into an angle-dependent block-diagonal normal form by using symplectic transformation, which can be achieved by selecting the appropriate tangential position. Finally, the existence of a class of invariant tori is proved, which implies the existence of quasiperiodic solutions for most values of frequency vector by an abstract KAM (Kolmogorov-Arnold-Moser) theorem for infinite dimensional Hamiltonian systems.
1. Introduction
It is proved that the equation admits the existence of a class of invariant tori, which implies the existence of quasiperiodic solutions for most values of frequency vector by an abstract KAM (Kolmogorov-Arnold-Moser) theorem for infinite dimensional Hamiltonian systems.
In nature, periodic phenomenon is an ideal state, but in practical problems, such as data observation, extraction, and operation, it always has errors and even some interference. In fact, quasiperiodic functions are always needed to be introduced in a system when there are two disturbance factors with incommensurability of periods; thus, quasiperiodic phenomenon is more common than periodic phenomenon, such as the celestial mechanics, ecology system, and economic volatility in many practical problems which often can be classified as quasiperiodic problem of differential equations. Generally speaking, there is more than one variable that causes the change of a phenomenon, so it is of great practical value to study the quasiperiodic problem of partial differential equations (PDEs), and the quasiperiodic solution problem of nonlinear Hamiltonian system is an important branch of nonlinear scientific research. As a kind of important Hamiltonian system, beam equation has also received corresponding attention.
The classical KAM theory, proposed by Kolmogorov [1, 2], Arnold [3], and Moser [4], is a theory about the long-term state of the solution of the integrable Hamiltonian system after it is perturbed, which is a significant progress of Newtonian mechanics in the 20th century and enables people to study the Hamiltonian system in a new way. In the late 1980s, in order to construct quasiperiodic solutions of one-dimensional Hamiltonian PDEs, the classical KAM theory was developed into infinite dimensional space by Wayne [5], Kuksin [6], and Pöschel [7]. Since then, KAM theory of Hamiltonian PDEs with one-dimensional spatial variables has been well developed and produced a lot of results, which we will not repeat here.
Although only the nonlinear term has changed from |u|2u to |u|4u, its normal form is completely different, so its corresponding KAM theory has also undergone essential changes. In recent years, more attention has been paid to the existence of quasiperiodic solutions of quintic Hamiltonian PDEs in high-dimensional space. Relevant results can be referred to references [24–26]. However, using the KAM theory to prove the existence of quasiperiodic solutions for two-dimensional completed resonant beam equations with higher order nonlinear terms remains to be solved.
This paper is focused on the study of (1) + (2). The nonlinear term of the Hamiltonian system corresponding to (6) is and that of (1) is , where τ1i + τ2j + τ3n + τ4m + τ5r + τ6s = 0, τ1 = ±, τ2 = ±, τ3 = ±, τ4 = ±, τ5 = ±, τ6 = ±, and . This difference leads to the essential difference of the normal form between the two, which leads to the fact that the KAM theory for (6) is not suitable for (1), and the phase flow invariant subspace where the quasiperiodic solution is located will also change. In this paper, we only discuss the existence of quasiperiodic solutions of (1) in the phase flow invariant subspace . By selecting phase flow invariant subspace , it can be ensured that the eigenvalue λr ≠ 0, and the normal form is simple and beautiful enough. However, due to the change of the phase flow invariant subspace where the quasiperiodic solution is located, we will have to reselect the tangential sites, that is, reconstruct the admissible set. Although this process does not require advanced mathematical knowledge and only involves elementary operations, it is cumbersome enough and requires strong skills. For the selected admissible set, (1) is turned into normal form by symplectic coordinate transformation, whose integrable terms only come from where i − j + n − m + r − s = 0; that is, the normal form of (1) is formally consistent with the normal form of (7). Of course, the two normal forms are only in the same form, and the coefficients of (1) are much more complex than (7). Therefore, we spend a lot of energy to prove that (1) meets the requirements of the KAM theory in [23] and then prove the existence of its quasiperiodic solution.
The following are the conditions that the tangential sites need to meet, which are proposed by Zhang and Si [23].
Definition 1. A set is admissible, if
- (1)
arbitrary three of them are not vertices of a rectangle
- (2)
for any , if i − j + n − m + r − s = 0 and |i|2 − |j|2 + |n|2 − |m|2 + |r|2 − |s|2 = 0, then the intersection of {i, j, n, m, r, s} and contains at most four elements, that is,
- (3)
for any , there exists at most one triplet {i, j, s}, where such that i − j + r − s = 0 and |i|2 − |j|2 + |r|2 − |s|2 = 0. If such triplet exists, let us say that r, s are resonant in the first type and denote all such r by
- (4)
for any , there exists at most one triplet {i, j, s}, where such that i + j − r − s = 0 and |i|2 + |j|2 − |r|2 − |s|2 = 0. If such triplet exists, let us say that r, s are resonant in the second type and denote all such r by
- (5)
for any , there exists at most one quintuple {i, j, n, m, s}, where such that i − j + n − m + r − s = 0 and |i|2 − |j|2 + |n|2 − |m|2 + |r|2 − |s|2 = 0. If such quintuple exists, let us say that r, s are resonant in the third type and denote all such r by
- (6)
for any , there exists at most one quintuple {i, j, n, m, s}, where such that i − j + n + m − r − s = 0 and |i|2 − |j|2 + |n|2 + |m|2 − |r|2 − |s|2 = 0. If such quintuple exists, let us say that r, s are resonant in the fourth type and denote all such r by
- (7)
Any is not resonant of any two of the above four classes. Geometrically, any two of the above defined graphs cannot share vertex in
Remark 2. There are sets satisfying the above definition. For example, for any integer d ≥ 4, the first point is defined as and the second one is defined as ; the others are defined inductively by
The set given above is admissible, and the proof is shown in the appendix.
The main result of this paper is as follows.
Theorem 3 (main theorem). Let be an admissible set with d ≥ 4. There exist a Cantor set Ξ∗ with positive measure, such that for arbitrary (ζ1, ⋯, ζd) ∈ Ξ∗, the beam equation (1) + (2) has a solution
2. The Hamiltonian Setting and Birkhoff Normal Form
Before turning equation (1) into a Hamiltonian system, we first introduce the following notations which will appear later.
The vector function is function of the parameter ζ, and its norm is The vector function is functions of the parameter ζ, and its norm is
2.1. The Hamiltonian Setting
The regularity of the Hamiltonian system (15) is shown below, and its proof is similar to [21], which is omitted here.
Lemma 4. For a given a ≥ 0 and ϖ > 1/2, the gradients are real analytic as maps from some neighborhood of origin in la,ϖ × la,ϖ into la,ϖ and , .
2.2. Partial Birkhoff Normal Form
Let us introduce some partial Birkhoff form of order six.
Proposition 5. is an admissible set with d points; there exists a symplectic transformation that converts the Hamiltonian (16) into
Proof. Let
Then, converts H∗ into
Introduce the action-angle variable
Equation (37) converts the Hamiltonian into
Scaling through time
3. An Infinite-Dimensional KAM Theorem for PDEs
(θ, I) are d-dimensional angle-action coordinates, are infinite-dimensional coordinates, and the corresponding symplectic structure is . Frequencies and depend on the parameter ζ ∈ Ξ ⊂ ℝd, where Ξ is a closed bounded set with positive Lebesgue measure. In order to prove the existence of the invariant torus of small perturbation H = H⋄ + W⋄ of H⋄, we need the following assumptions.
Assumption 6 (nondegeneracy). Suppose that ∀ζ ∈ Ξ,
Assumption 7 (asymptotics of normal frequencies).
Assumption 8 (Melnikov’s nondegeneracy). Let for
Assumption 9 (regularity). is real analytic in and in ζ and
Assumption 10 (special form). has the following special form:
Assumption 11 (Töplitz-Lipschitz property). For given , the limits
The following KAM theorem comes from Zhang and Si [23].
Theorem 12 ([23] Theorem 2.1). If satisfies the Assumptions (6)–(11), and γ′ > 0 is small enough. Then, there exists a positive constant ε = ε(d, L, K∗, τ, γ′, r∗, s∗, a, ϖ), so that if , the following results true: there exists a Cantor subset Ξγ′ ⊂ Ξ with and two transformations (analytic in θ and in ζ) , where Ψ is -close to the trivial embedding and is ε-close to the unperturbed frequency η, so that for ζ ∈ Ξγ′ and , the curve is a quasiperiodic solution of the Hamiltonian equations corresponding to H∗.
4. Proof of the Main Theorem
Let us show that the Hamiltonian (23) satisfies the Assumptions (6)–(11).
From , we have when 0 < ε ≪ 1, that is, . Hence, Assumptions (6) is verified.
Verifying Assumption (7): take ς = 4; the proof is obvious.
It has been proved that none of the eigenvalues of are zero in [23]. Moreover, when or or and so on, the situations are similar, so omit the proofs. That is, none of the eigenvalues of are zero for k ≠ 0. From Lemma 3.1 in [19], then is a polynomial function in the components of ζ with order at most eight and By excluding some parameter set with measure , then (A3) is verified.
Verifying Assumption (9): similar to [19], from Lemma 4, it is obvious that (A4) holds, so omit its proof.
Verifying Assumption (10): similar to [19], the result is obvious.
Verifying Assumption (11): similar to [23], the result is obvious.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This paper is partially supported by the Fundamental Research Funds for the Central Universities (Grant Nos. 22CX03008A and 19CX02054A) and the Natural Science Foundation of Shandong Province (Grant Nos. ZR2021MA053 and ZR2022MA078).
Appendix
A. Proof of Non Empty Admissible Set
Lemma 13. Set as the given set with d points in Remark 2. For , the following conclusion is true:
- (i)
n − m = 0, n = m, and |n| = |m| are equivalent
- (ii)
n − m + r − s = 0, {n, r} = {m, s}, and {|n|, |r|} = {|m|, |s|} are equivalent
- (iii)
, , and are equivalent
- (iv)
, , and are equivalent
- (v)
<n, r > = <m, s>, {n, r} = {m, s}, and {|n|, |r|} = {|m|, |s|} are equivalent
- (vi)
<n − m, r − s > = 0 is equivalent to n = m or r = s
Lemma 14. The set given in Remark 2 is admissible.
Proof. According to i − j + r − s = 0 and |i|2 − |j|2 + |r|2 − |s|2 = 0, <r − j, i − j > = 0. By i − j + n − m + r − s = 0 and |i|2 − |j|2 + |n|2 − |m|2 + |r|2 − |s|2 = 0, then <r − m, i − j + n − m > = <i − j, j − n>. In order to prove that has the properties (3)–(12) in Definition 1, we need to prove that
Case 1.1. Only one element of {|i|, |j|, |n|, |m|, |r|} gets the maximum value.
Case 1.1.1. Suppose that |r| = max{|i|, |j|, |n|, |m|, |r|}. We write <r − m, i − j + n − m > = <i − j, j − n> in terms of the following components:
This is contradictory to .
Case 1.1.2. Suppose that |j| = max{|i|, |j|, |n|, |m|, |r|}; then, we have
This is contradictory to the hypothesis <i − j, j − n > = <r − m, i − j + n − m>.
Case 1.2. Two elements of {|i|, |j|, |n|, |m|, |r|} get the maximum value.
Case 1.2.1. Suppose that |m| = |j| = max{|i|, |j|, |n|, |m|, |r|}, then
But we have
It is a contradiction.
Case 2.1. Only one of gets the maximum value.
Case 2.1.1. Suppose that . According to the calculation,
To prove β211 ≠ 0 by contradiction, suppose that β211 = 0, then
From the system above, we get . And by , then <i − j, i − m > = 0 is obtained. In view of Lemma 13, then i = j or i = m. It is contradictory to . That is, β211 ≠ 0. Because the order of the numerator β211 is no more than n1 and the order of the divisor α211 is n2, we have .
Case 2.1.2. Suppose that . By the calculation, then
Let
Thus, .
Other situations are similar to the above cases.
Case 2.2. Two elements of get the maximum value.
Case 2.2.1. Suppose that , then i = n. By the calculation, then
Denote
Hence, .
Case 2.2.2. Suppose that , then . By the calculation, then
We prove β222 ≠ 0 by contradiction. Suppose that β222 = 0, then
From the system above, we have . And from , then we have <j − i, j − n > = 0. From Lemma 13, then j = i or j = n. It is contradictory to . That is, β222 ≠ 0. Due to the order of the numerator β222 which is no more than m1 and the order of the divisor α222 which is m2, we have .
Other situations are similar to the above cases.
Case 2.3. Three elements of get the maximum value. It can be seen by [23] that such situation is similar to those mentioned above; thus, omit the proof.
Case 2.4. Four elements of get the maximum value. It can be seen by [23] that such situation is similar to those mentioned above; thus, omit the proof. It is shown that equation (A.3) has no solution in ℤ2,⋄. That is, satisfies the property (6) in Definition 1. Similarly, satisfies the property (3) in Definition 1 and
- (III)
Let us show that equation (A.4) has no solution in ℤ2,⋄. The proof for (A.2), (A.5), and (A.7)–(A.10), is similar and simpler than the proof for (A.4)
By the calculation, equation (A.4) is equivalent to
We assert |j| and will not be the maximum. Suppose that . According to |m|2 + |n|2 + |i|2 = |r|2 + |s|2 + |j|2 ≥ |j|2 and the definition of , then there is one element of the set {m, n, i} that is identical to j. If i = j, then r + s − n − m = 0 and |r|2 + |s|2 − |n|2 − |m|2 = 0. That is, . This is contradictory to . That is, . Similarly, we have .
Case 3.1. Only one element of gets the maximum value. Suppose that , then
By the calculation to (A.32),
We prove β31 ≠ 0 by contradiction. Suppose that β31 = 0, then
That is, i − j + m − r = 0. Thus, we have n = s by i − j + n + m − r − s = 0. This is contradictory to and s ∈ ℤ2,⋄. That is, β31 ≠ 0. Due to the order of the numerator β31 which is no more than n1 and the order of the divisor α31 which is n2, we have .
Other situations are similar to the above cases.
Case 3.2. Two of get the maximum value.
Case 3.2.1. Suppose that , then . By equation (A.32), then
Therefore,
Due to σ321 which is of order m1 which is far less than m2, we have
Therefore,
Due to , we have .
Other situations are similar to the above cases.
Case 3.3. Three elements of get the maximum value. It can be seen by [23] that such situation is similar to that mentioned above; thus, omit the proof.
Case 3.4. Four elements of get the maximum value.
Case 3.4.1. Suppose that , then . By the calculation to (A.32),
It is shown that equation (A.32) has no solution in ℤ2,⋄ because
Other situations are similar to the above cases.
Case 3.5. Five elements of get the maximum value. It can be seen by [23] that such situation is similar to that mentioned above; thus, omit the proof.
Case 3.6. Six elements of get the maximum value. It can be seen by [23] that such situation is similar to that mentioned above; thus, omit the proof.
It is shown that equation (A.4) has no solution in ℤ2,⋄. That is, satisfies the property (7) in Definition 1. Similarly, satisfies the properties (5) and (12) in Definition 1.
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