Bifurcation of Limit Cycles by Perturbing a Piecewise Linear Hamiltonian System

Theorem 1. Let (9) be satisfied. For any given n ≥ 1, one has Table 2.

Lemma 2. Suppose system (1) satisfies (9). Then,

•  

  (i) M⁺(h, δ) in (11) can be written as

  $$ ()

•  

   where

  $$ ()

•  

  with

  $$ ()

•  

  (ii)  (h, δ) in (11) can be expressed as

  $$ ()

•  

  where

  $$ ()

•  

  with

  $$ ()

Proof. We only prove (i) since (ii) can be verified in a similar way. By (11), we obtain

which follows that by Green formula and (3) where Then, by Green formula again where By (3), (4), and the above formulas, we have Combining (20)–(25) gives (13) and (14). Thus, the proof is ended.

Lemma 3. (i) If a₁ = 0,   a₀ > 0, then M⁺(h, δ) in (11) has form

where (ii)  If b₁ = 0,   b₀ < 0, then we have where

Proof. Note that along $$x = h/a₀ − (1/2a₀)y². Then, inserting it into (14) follows that

Let $$. Then we have $$ and the above integral can be carried into Thus, using (30) and the above equation we can write (13) as where which gives (i) by (15). Thus, (i) holds and we can prove (ii) in the same way by (16)–(18). This ends the proof.

Lemma 4. Let system (5a) satisfy (3) and (4). Then

•  

  (i) If a₁ > 0,   a₀ = 0,   M⁺(h, δ) has the expression

  $$ ()

•  

  where

  $$ ()

•  

  (ii) If a₁ > 0,   a₀ ≠ 0,   M⁺(h, δ) can be written as

  $$ ()

•  

  or

  $$ ()

•  

  where

  $$ ()

•  

  each $$ is a nonzero constant and $$ are polynomials of degree k + [(i + 1)/2], [n/2], [(n − 1)/2], respectively.

Proof. Since $$ along the curve $$ in (14) becomes

where Let $$. Then, we have $$ and the above equation becomes

For a₀ = 0, make the transformation $$. Then, we have by (41)

Substituting the above formula into (37), together with (13), gives that where Thus, by (15) and the above discussion we know that (i) holds.

For a₀ ≠ 0, (41) can be represented as

where Recall that Then, by (46) and the above equation we obtain that It follows that where which is a polynomial of degree k in h. For convenience, introduce Then, combining (49) and (51) gives that Further, by using the formula we have that It follows that where which is a polynomial of degree [(r − 1)/2] in h. Let Then, we have that by (55) and the above Substituting the above equation into (52), one can find that where $$ for r odd, $$ for r even, and which is a polynomial of degree k  +  [(r  +  1)/2] in h. Combining (37), (45), and (59) gives that which implies (35), together with (13) and (15).

Note that

Then, we have Inserting the above formula into (35), we can obtain (36). Hence, the proof is finished.

Lemma 5. Let system (5b) satisfy (3) and (4). Then

(i) If b₁ > 0, b₀ = 0, M⁻(h, δ) in (11) has the expression

where (ii) If b₁ > 0,   b₀ ≠ 0,   M⁻(h, δ) in (11) has the form or where each $$ is nonzero constant and $$ are polynomials of degree k + [(i + 1)/2], [n/2], [(n − 1)/2], respectively.

Subcase 1. a₁ = b₁ = 0,  a₀ > 0,  b₀ < 0. From (12) and Lemma 3, one can obtain that

which implies that M(h, δ) has at most n isolated positive zeros for h > 0. To show that this bound can be reached, take $$. Then, by (27) and (29), (69) has the form where Hence, using (70) we can take $$ as free parameters to produce n simple positive zeros of M(h, δ) near h = 0, which gives n limit cycles correspondingly near the origin. Thus, N(n) ≥ n in this case. This ends the proof.

Subcase 2. a₁ > 0,   b₁ > 0,   a₀ = b₀ = 0 Similar to the above and using (32) and (64), M(h, δ) in (12) has the expression of the form

where $$. Further, taking $$, then, by (34) and (65), M(h, δ) in (72) becomes where Thus, from (72) and (73), we can discuss similar to Subcase 1. This finishes the proof.

Subcase 3. a₁ > 0,   b₀ < 0,   a₀ = b₁ = 0 By Lemmas 3, and 4 and (12), we can have that

Let us prove that M^*(h, δ) has at most n + [(n + 1)/2] zeros on the open interval (0, +∞). For the purpose, let $$. Then, for n = 2l,   l ≥ 1, M^*(h, δ) in (75) has the expression where To prove M^*(h, δ) has at most n + [(n + 1)/2] zeros, it suffices to prove $$ has at most n + [(n + 1)/2] = 3l zeros for λ > 0. By Rolles theorem we need only to prove that $$ has at most l zeros for λ ∈ (0, +∞). From (76), we can have that which shows that $$ has at most l zeros for λ > 0. Thus, M(h, δ) has at most 3l zeros for h > 0. To prove 3l zeros can appear, we only need to prove that M^*(h, δ) in (75) can appear 3l zeros for h > 0 small. Let $$, and $$. Then M^*(h, δ) in (75) can be expressed as by (29) and (34) where Thus, by changing the sign of $$, $$ in turn such that we can find 3l simply positive zeros h₁, h₂, …, h_3l with 0 < h_3l < h_3l−1 < ⋯<h₁ ≪ 1. For n = 2l + 1, l = 0,1, …, we can discuss in a similar way. Thus, this bound can be reached and N(n) ≥ n + [(n + 1)/2]. The proof is finished.

Subcase 4. a₁ > 0,   a₀ > 0,   b₁ = 0,   b₀ < 0 From (12) and Lemmas 3 and 4, we get that

where which is a polynomial of degree n in h. Let $$. Then h = λ²,   λ ∈ (0, +∞), and (82) becomes where One can see that $$, where Denote by #{λ ∈ (0, +∞)∣f(λ) = 0} the number of zeros of the function in the interval (0, +∞) taking into account their multiplicities. Note that and they are even functions in λ. Therefore, Then, from [8], we can obtain that which implies that Z(n) ≤ n + 2[(n + 1)/2]. Now, we verify Z(n) ≥ n + [(n + 1)/2].

Make the transformation $$. Then $$ in (35) becomes

which follows that as h > 0 small Note that Inserting the above formula into (91) gives that where Take $$. Then, by (28), (35), and (93), we can obtain that for h > 0 small where with L_i,  i = 0,1, …, n being linear combination, L_i(0,0, …, 0) = 0. One can find that where A₂ is a (n + 1)×[(n + 1)/2] matrix, with β_i are [(n + 1)/2] × 1 matrix satisfying Hence, we can obtain that from (97) where and β_i are given in (99). We claim that |A | ≠ 0. We only need to prove |B | ≠ 0 by the above formula. Using elementary transformations to |B| by multiplying ith column by $$, we can obtain that by (99) and (101) where Now we will use elementary transformations to B₁ as follows.

• (1)

  Add the first column multiplying by −1 to ith column, i = 2,3, …, [(n + 1)/2].

• (2)

  Add the second column multiplying by $$ to ith column, i = 3,4, …, [(n + 1)/2].

• (3)

  Add the third column multiplying by $$ to ith column, i = 4,5, …, [(n + 1)/2]

•  

  ⋮

[(n − 1)/2]. Add the [(n − 1)/2]th column multiplying by $$ to [(n + 1)/2]th column,

[(n + 1)/2]. multiply ith column by (−1) ⁱ⁻¹, i = 2,3, …, [(n + 1)/2].

Then, |B₁| becomes, together with (94)

where with i = 0,1, …, [(n − 1)/2]. For B₂ in (104) by adding a column on the left and a row on the above, we can obtain that, together with adding ith column to (i + 1)th column with i = 1,2, …, [(n + 1)/2] which implies that |A₃ | ≠ 0 by (102) and (104) if |B₂ | ≠ 0. We claim that |B₂ | ≠ 0 and Now, we prove them by induction on n. For n = 1,2 we have which means that (106) and (107) hold for n = 1,2. Suppose (106) and (107) hold for n = 2l − 1,   2l,   l ≥ 1. That is, we have for n = 2l Then for n = 2l + 1, we have Note that by the first equation of (109) there only exist α₁, α₂, …, α_l+1 such that since α₁ + α₂ + ⋯+α_l+1 = 1, where C is given in (109). If |B₂ | = 0, then we can obtain that from (109) and (111) Note that which follows that, together with (112) By the second equation in (109) and the above formula, we have This is a contradiction with α₁ + α₂ + ⋯+α_l+1 = 1. Hence α₁/(6l + 3) + 2α₂/(6l + 1) + ⋯+(l + 1)α_l+1/(4l + 3) ≠ 0, which means that (106) holds for n = 2l + 1. Since |B₂ | ≠ 0 in (110), there only exist β₁, β₂, …, β_l+1 such that with β₁ ≠ 0 since the last row in the second formula of (110) is linearly independent with all rows in the first formula of (110), which means that (107) holds for n = 2l + 1. In a similar way, we can prove (106) and (107) hold for n = 2l + 2. Hence, the claim holds. So, from (99) we can know that $$ can be taken as free parameters. So we can choose these values such that which yields that by (97) and (96) M(h, δ) can appear n + [(n + 1)/2] positive zeros for h > 0 small. We also can know that N(n) ≥ n + [(n + 1)/2]. Hence, the conclusion is proved.

Subcase 5. a₁ > 0,   a₀ > 0,   b₁ > 0,   b₀ = 0 By (36) and (67), one can see that

where For convenience, we denote by g_n any polynomial of degree n although its coefficients may be different when it appears in different place. Then, we claim that for any 2 ≤ k ≤ [(n − 1)/2] Now, we verify this claim by induction on k. For k = 2, by (118) we can obtain that which follows that Hence, (120) holds for k = 2. Suppose (120) holds for  k, 2 ≤ k ≤ [(n − 1)/2] − 1. Then for k + 1, we have which implies that (120) holds for k + 1. Thus, the claim is proved. Then, taking k = [(n − 1)/2], we can obtain that by differentiating it One can find that where where F is a polynomial of degree [n/2]+[(n + 1)/2]. Since M(0, δ) = 0 from (118), it is easy to see that M(h, δ) has at most [n/2] + 2[(n + 1)/2] = n + [(n + 1)/2] zeros for h > 0 by Rolle theorem. As the above discussion, we only prove Z(n) ≥ n + [(n + 1)/2] as h > 0 small, which implies N(n) ≥ n + [(n + 1)/2]. For the purpose, take $$. Then using (49), (51), we can write M(h, δ) in (12) as

For n = 2l,  l ≥ 1, (127) can be written as

where which implies that M(h, δ) can appear n + [(n + 1)/2] zeros in h > 0 small by using the same method with the Subcase 4. For n = 2l + 1,   l = 0,1, …, we can discuss by (127) in a similar way. Hence, the conclusion holds.

Subcase 6. a₁ > 0,   a₀ > 0,   b₁ > 0,   b₀ < 0. We have as the above

Similar to the Subcase 5, we can prove that for any 2 ≤ k ≤ [(n − 1)/2] Taking k = [(n − 1)/2] and differentiating the above twice follow that If $$, then it is easy to see that (132) has the same form with (125). Hence, we can know that M(h, δ) has at most [n/2] + 2[(n + 1)/2] zeros for h ∈ (0, +∞). If $$, then (132) can be written as where where $$ is a polynomial of degree [n/2] + 2[(n + 1)/2] in h. By Rolle theorem, we can obtain that M(h, δ) has at most [n/2] + 3[(n + 1)/2] zeros for h > 0 since M(0, δ) = 0. Now, we only need to prove Z(n) ≥ n. Take $$. Then, by Lemmas 4, 5, one can see that for h > 0 small Similarly, we can discuss the above formula such that such that M(h, δ) can appear n zeros for h > 0 small. Hence, the conclusion is proved.