Nonnegative Periodic Solutions of a Three-Term Recurrence Relation Depending on Two Real Parameters

Proposition 1. Let $$ be a solution of (7) defined by (φ₀, φ₁) = (α, β) and $$ be the “positive” orbit of φ. Then ω is a period of φ if and only if (x_ω, y_ω) = (α, β). Furthermore, if lim_k→∞|x_k| = ∞, then φ is aperiodic and if lim_k→∞x_ky_k < 0, then φ is neither a nonnegative nor a negative solution.

Proposition 2. Let $$ be a solution of (7) with (φ₀, φ₁) = (α, β).

• (i)

  Suppose ρ ≠ −2. Then φ is a constant solution if and only if α = β = μ/(2 + ρ).

• (ii)

  Suppose ρ = −2. Then φ is a constant solution if and only if μ = 0 and α = β.

Proof. Suppose ρ ≠ −2. If φ is a constant solution of (7), then α = β = φ₂ and

so that Accordingly, α = μ/(2 + ρ) = β. The converse can be verified by direct iteration. Next, suppose ρ = −2. If φ is a constant solution of (7), then α = β and by (7) which leads to μ = 0. Conversely, if μ = 0 and α = β, then by iteration φ is clearly a constant solution.

The proof is complete.

Corollary 3. Let $$ be a solution of (7) with φ₀ = φ₁ = γ. Let γ = μ/(2 + ρ) where ρ ≠ −2. Then φ is 1-periodic. φ is a nonnegative solution if and only if γ ≥ 0; and φ is a negative solution if and only if γ < 0.

Proposition 4. Let $$ be a solution of (5) with (φ₀, φ₁) = (α, β). Then φ is 2-periodic if and only if ρ = 2, α + β = μ/2, and α ≠ β.

Proof. Suppose φ is 2-periodic. Then φ₂ = α as well as φ₃ = β and by (5), φ₂ = μ − ρβ − α = α and φ₃ = μ − ρα − β = β which lead us to 2α + ρβ = μ and ρα + 2β = μ, respectively. Accordingly, we have (2 + ρ)(α + β) = 2μ and (2 − ρ)(α − β) = 0. Note that since φ is 2-periodic, we can be sure that α ≠ β. Hence, if ρ ≠ 2, then α − β = 0 which is a contradiction; if ρ = 2, then α + β = μ/2 and α ≠ β. The converse may be checked by direct substitution into (5) and this completes the proof.

Corollary 5. Let $$ be a solution of (7) with ρ = 2 and (φ₀, φ₁) = (α, β). Suppose α + β = μ/2. Then φ is periodic with period 2.  φ is nonnegative if and only if α, β, μ ≥ 0; and φ is negative if and only if α, β, μ < 0.

Proof. Suppose α + β = μ/2. First of all, if α = β = μ/4, then, by Proposition 2, φ is 1-periodic; otherwise, by Proposition 4, φ is 2-periodic. Hence, φ is periodic with period 2. If φ is a nonnegative solution, then α,  β ≥ 0 which implies μ ≥ 0. Furthermore, since φ is periodic with period 2, we can see that φ_k ≥ 0 for all k ∈ Z. If α, β, μ ≥ 0, then α + β = μ/2 is satisfied and φ is a nonnegative solution. The case where φ is a negative solution can be handled similarly and this completes the proof.

Proposition 6. Let $$ be a nonnull solution of (7). If μ ≥ 0 and ρ ≥ 0, then φ cannot be a nonpositive solution; if μ ≤ 0 and ρ ≤ 0, then φ cannot be a nonnegative solution.

Proof. Suppose μ ≥ 0 and ρ ≥ 0. If φ is nonpositive, then 0 ≥ φ₂ = −ρφ₁ − φ₀ + μ ≥ 0 and 0 ≥ φ₃ = −ρφ₂ − φ₁ + μ ≥ 0. Hence φ₂ = φ₃ = 0, which implies φ is null, contrary to our assumption.

Suppose μ ≤ 0 and ρ ≤ 0. If φ is nonnegative, then 0 ≤ φ₂ = −ρφ₁ − φ₀ + μ ≤ 0 and 0 ≤ φ₃ = −ρφ₂ − φ₁ + μ ≤ 0 so that φ₂ = φ₃ = 0. A contradiction is arrived and this completes the proof.

Proposition 7. φ is a solution of (5) if and only if −φ is a solution of (6).

Proof. Suppose that $$ is a solution of (5) with (φ₀, φ₁) = (α, β). By (5), we have φ₂ = μ − ρβ − α. Let ψ = −φ such that (ψ₀, ψ₁) = (−α, −β). Accordingly, ψ₂ = −φ₂ and we see that

which satisfies (6). By induction, it follows that ψ is a solution of (6) and this completes the proof.

Theorem 8. Let $$ be a solution of (5). If f = Γ_μ,ρ(φ_m, φ_m+1) for some m ∈ Z, then $$ Furthermore, if (α, β) ∈ Γ_μ,ρ,f, where f = Γ_μ,ρ(α, β), then $$

Proof. Without loss of generality, we let m = 0 and (φ₀, φ₁) = (α, β). Let f = Γ_μ,ρ(α, β). Then (α, β) ∈ Γ_μ,ρ,f. By (21), we see that

and by (7), it follows (φ₁, φ₂) = (β, μ − ρβ − α) so that which leads to (φ₁, φ₂) ∈ Γ_μ,ρ,f. By induction, for any n ∈ N, (φ_n, φ_n+1) ∈ Γ_μ,ρ,f which implies $$ Next, we consider (φ_n, φ_n+1), where n ∈ Z∖N. Let $$ such that (ψ₀, ψ₁) = (β, α) and φ_n = ψ_−n+1. Since Γ_μ,ρ,f is symmetric with respect to the x = y line, (β, α) ∈ Γ_μ,ρ,f and by the previous discussions, we can be sure that $$ which implies $$, where $$ Accordingly, $$ and by (15), it follows that $$ Furthermore, by the definitions, if (α, β) ∈ Γ_μ,ρ,f, where f = Γ_μ,ρ(α, β), then $$ and since $$ itself is symmetric with respect to the x = y line, it is clear $$ as desired. The proof is complete.

Theorem 9. Suppose |ρ| > 2 and let (α, β) ∈ R² such that f = Γ_μ,ρ(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Then Γ_μ,ρ,f is a hyperbola (or a degenerate one made up of two lines $$ and $$) and we have the following.

• (i)

  Suppose Γ_μ,ρ,f is a degenerate hyperbola. If $$, then $$ for all m ∈ N; if $$, then $$ for all n ∈ N.

• (ii)

  Suppose Γ_μ,ρ,f is a hyperbola. If ρ ≥ 0, then (x_m, x_m+1) and (x_m+1, x_m+2) are sitting on different branches for all m ∈ N; if ρ < 0, then (x_n, x_n+1) and (x_n+1, x_n+2) lie on the same branch for all n ∈ N.

Proof. Note that x₀ = α and x₁ = y₀ = β. By Theorem 8 and (7), we see that y₁ = μ − ρy₀ − x₀ = μ − ρβ − α = γ. Suppose $$ or $$ In the former case, $$ and by (31), as well as (32), $$ which implies $$; in the latter case, by similar arguments, we may see that $$ Note that if $$, then (α, β) = (μ/(2 + ρ), μ/(2 + ρ)) and by the Tracking Procedure, (x₁, y₁) = (μ/(2 + ρ), μ/(2 + ρ)) which implies (x_i, y_i) = (μ/(2 + ρ), μ/(2 + ρ)) for all i ∈ N. Suppose $$ Then Γ_μ,ρ,f is a hyperbola and (α, β) is on one branch of Γ_μ,ρ,f. By the Tracking Procedure, we may see that if ρ ≥ 0, then P₁ “jumps” to another branch (see Figure 3), while if ρ < 0, then P₁ will stay on the same branch (see Figure 4). By induction, we may see that (ii) holds. This completes our proof.

Theorem 10. Suppose |ρ| > 2 and let (α, β) ∈ R² such that f = Γ_μ,ρ(α, β). Let $$ be generated by the Tracking Procedure with (x₀, y₀) = (α, β). Then Γ_μ,ρ,f is a hyperbola (or a degenerate one made up of two lines $$ and $$) and we have the following.

• (i)

  Suppose $$ If (α, β) = (μ/(2 + ρ), μ/(2 + ρ)), then (x_k, y_k) = (α, β) for all k ∈ N; if $$ and (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)), then

  $$ (33)

•  

  and if $$ and (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)), then

  $$ (34)

• (ii)

  If ρ > 2 and Γ_μ,ρ,f is a hyperbola, then $$; and if ρ < −2 and Γ_μ,ρ,f is a hyperbola, then $$

Proof. For the sake of convenience, we let P_c = (μ/(2 + ρ), μ/(2 + ρ)) and P_k = (x_k, y_k) so that P₀ = (α, β) and f = Γ_μ,ρ(α, β). Let y₁ = γ = μ − ρβ − α. By Proposition 2, if P₀ = P_c, then φ is a constant solution of (7) which implies $$ Now, we suppose P₀ ≠ P_c. Since |ρ| > 2, Γ_μ,ρ,f    is a hyperbola (or a degenerate one). Hence, we discuss the two cases where P₀ is on a degenerate hyperbola or a hyperbola.

Suppose $$ and P₀ ≠ P_c. Then by Theorem 9, $$ and $$ are on the same line. In view of (24) and (25), we let m₊ = ρ + δ,  m₋ = ρ − δ, Δ₊ = μ + δ/(2 + ρ), and Δ₊ = μ − δ/(2 + ρ). For the sake of convenience, we also let

Since Γ_μ,ρ,f is symmetric with respect to the x = y line, we see that if ρ > 2, then κ < 1; if ρ < −2, then κ > 1. Suppose ρ > 2. If $$ and P₀ ≠ P_c, then $$ and $$ so that α + m₊β + Δ₊ = 0 and γ + m₋β + Δ₋ = 0, respectively. Also, by the symmetry of Γ_μ,ρ,f with respect to the x = y line, we can be sure that $$ so that On the other hand, we calculate and by (35) and (36), as well as (37), it follows that By induction, we have $$, where κ < 1 and i ∈ N. Accordingly, (34) holds. The case where ρ < −2 and $$ can be handled similarly with $$, where κ > 1 and j ∈ N. By similar arguments, (33) is also true.

Suppose $$ We first consider the case where ρ > 2. Then κ < 1 and by (24)

By induction, it follows that $$, where i ∈ N. Consequently, case (ii) is true when ρ > 2 and by similar arguments, we also see that ρ < −2 holds with $$, where j ∈ N and κ > 1.

The proof is complete.

Corollary 11. Suppose |ρ| > 2 and let (α, β) ∈ R² such that f = Γ_μ,ρ(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Then Γ_μ,ρ,f is a hyperbola (or a degenerate one made up of two lines $$ and $$). Suppose (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)). If $$, where ρ > 2, then x_kx_k+1 < 0 and |x_k| → ∞ when k → ∞; if $$, where ρ < −2, then x_kx_k+1 > 0 and |x_k| → ∞ when k → ∞.

Theorem 12. Suppose ρ = 2 and let (α, β) ∈ R² such that f = Γ_μ,2(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Then f ≥ −μ²/4 and Γ_μ,ρ,f is a degenerate parabola made up of $$ and $$. If f = −μ²/4, then $$ and (x_k, y_k) = (x_k+2, y_k+2) for all k ∈ N; otherwise, (x_k, y_k) as well as (x_k+1, y_k+1) are not on the same line. Furthermore, if f > −μ²/4, then $$ and x_ky_k < 0 when k → ∞.

Proof. For the sake of convenience, we let P_c = (μ/(2 + ρ), μ/(2 + ρ)) and P_k = (x_k, y_k). First of all, we see that

which implies f ≥ −μ²/4. Furthermore, which leads to Γ_μ,2,f being two parallel lines. Consequently, if f = −μ²/4, then $$; otherwise, $$ and $$ are two distinct parallel lines. Suppose f = −μ²/4. Then $$ In view of the Tracking Procedure, it is found that if P₀ = P_c, then P₁ = P₀ which implies P_m = P₀ for all m ∈ N; otherwise, $$ which leads to P_n = P_n+2 for all n ∈ N. Next, suppose f > −μ²/4.    By the Tracking Procedure, we can directly observe (see Figure 10) that for an arbitrary n ∈ N, if $$, then $$ and vice versa. Now, we discuss the asymptotic behaviors of $$ By the Tracking Procedure, we first observe that and clearly, the converses are also true. Note that d(P_n, Π) ≠ d(P_n+1, Π) for any n ∈ N. Furthermore, by the Tracking Procedure again, if $$, then for all k ∈ N otherwise, we see that for some j ∈ N, In view of (44), since $$,  $$, and $$ is a positive constant, there is i ∈ N such that $$ as well as $$ and, thus, by the previous discussions, $$ Accordingly, by the previous discussions again, $$ when k → ∞ and since the slopes of $$ and $$ are −1, x_ky_k < 0 as k → ∞. The proof is complete.

Theorem 13. Suppose ρ = −2 and let (α, β) ∈ R² such that f = Γ_μ,−2(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Then we have the following.

• (i)

  Suppose μ = 0. Then f ≥ 0 and Γ_0,−2,f is a degenerate parabola made up of two parallel lines $$ and $$ If f = 0, then $$ and α = β such that (x_k, y_k) = (α, α) for all k ∈ N; otherwise, (x_k, y_k) and (x_k+1, y_k+1) are on the same line and d((x_k, y_k), P₀) → ∞ when k → ∞.

• (ii)

  Suppose μ ≠ 0. Then Γ_μ,−2,f is a parabola and $$ when k → ∞.

Proof. Suppose μ = 0. By (20), Γ_0,−2(x, y) = x² − 2xy + y² = (x − y)² and, thus, f = Γ_0,−2(α, β) = (α − β)² ≥ 0. Accordingly, Γ_0,−2,f is a degenerate parabola which is made up of two parallel lines $$ and $$ as defined in (24) and (25), respectively. More precisely, if α = β, then f = 0 which leads to $$; if α ≠ β, then f > 0 which implies $$ and $$ are two distinct parallel lines. By the Tracking Procedure, if α = β, then (x_k, y_k) = (α, α) for all k ∈ N; if α ≠ β, then (x_k, y_k) and (x_k+1, y_k+1) are on the same line. Furthermore, if α ≠ β, then we have $$ and by the Tracking Procedure again, d((x_k, y_k), (α, β)) = kΔtan⁡(π/4) = kΔ, where k ∈ N. Accordingly, d((x_k, y_k), P₀) → ∞ when k → ∞. Next, suppose μ ≠ 0. Then Γ_μ,−2,f is parabola. By (7), for k ∈ N, we have y_k = x_k+1 such that x_k+2 − x_k+1 = μ + x_k+1 − x_k and thus

which leads to Hence, $$ when k → ∞. Note that, if μ > 0, then Γ_μ,−2,f is concave upward (see Figure 15), while if μ < 0, then Γ_μ,−2,f is concave downward. Accordingly, (ii) is true and this completes the proof.

Corollary 14. Suppose ρ = −2 and let (α, β) ∈ R² such that f = Γ_μ,−2(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Then we have the following.

• (i)

  Suppose μ = 0. Then f ≥ 0 and Γ_0,−2,f is a degenerate parabola with two parallel lines, $$ and $$ If α = β, then f = 0 such that $$ and x_k = α for all k ∈ N; otherwise, f > 0 such that $$ as well as $$ are two distinct parallel lines and

  $$ (47)

• (ii)

  Suppose μ ≠ 0. Then Γ_μ,−2,f is a parabola. If μ > 0, then Γ_μ,−2,f is concave upward and x_k → ∞ when k → ∞; otherwise, Γ_μ,−2,f is concave downward and x_k → −∞ when k → ∞.

Theorem 15. Suppose |ρ| < 2 and (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)) such that f = Γ_μ,ρ(α, β). Let $$ be generated by the Tracking Procedure with input (α, β) = (x(θ₀), y(θ₀)). For any m ∈ N,

Proof. In view of (49), we first consider the case where ϑ = π/4. By (49), it follows that

as well as Note that x_c = y_c = μ/(2 + ρ). By (7) and (52), we also have In view of (52), with respect to x_k+2, it follows that which leads to The case where ϑ = −π/4 can be handled similarly. Hence, (50) is true and this completes the proof.

Corollary 16. Suppose |ρ| < 2 and (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)) such that f = Γ_μ,ρ(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Then $$ is periodic if and only if

Furthermore, suppose π⁻¹arccos(−ρ/2) = (m/n) ∈ Q such that m/n is irreducible and 0 < m < n. If m is even, then $$ is n-periodic; otherwise, $$ is 2n-periodic.

Proof. Note that if ρ ≠ −2 and (α, β) = (μ/(2 + ρ), μ/(2 + ρ)), then by the Tracking Procedure and Proposition 1, (x₁, y₁) = (α, β) which implies (x_k, y_k) = (α, β) for all k ∈ N. Suppose |ρ| < 2 and (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)). By (49), we have

By Theorem 15, for any k ∈ N, θ_Δ = θ_k+1 − θ_k is constant such that cos⁡(θ_Δ) = −ρ/2. Hence θ_k = θ₀ + kθ_Δ. If $$ is periodic with period n ∈ Z⁺, then we see that there is some m ∈ Z⁺ such that θ_n = θ₀ + 2mπ. Accordingly, nθ_Δ = 2mπ which implies On the other hand, if (56) holds, then, by the previous discussions and (58), $$ is periodic with period 2n. Note that by Propositions 2 and 4, if $$ is ω-periodic with (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)), then ω > 2. Next, suppose r ∈ Q, where r = π⁻¹arccos⁡(−ρ/2). Since 2jπ where j ∈ Z are periods of cos⁡(x), we can first restrict ourselves to 0 < r < 2, where r ≠ π. Also, if π < x < 2π, then cos⁡(x) = cos⁡(−x) = cos⁡(2π − x) and this allows us to further concentrate on 0 < r < 1. Note that cos⁡(x) is bijective on (0, π). Hence, for each ρ such that r ∈ Q, where 0 < r < 1, there is a unique pair (m, n) such that r = (m/n), where m and n are coprime. In view of (58), it is easy to see that if m is even, then n is a period of $$; if m is odd, then $$ is periodic with period 2n. In the former case, we claim that $$ is q-periodic. To this end, if q^′ < q is a period of $$, then, by the previous discussions, there is p^′ ∈ Z⁺ such that p^′ as well q^′ are coprime and r = p^′/q^′. Hence, (p^′/q^′) = (p/q) which is not possible. The latter case can be handled similarly and this completes the proof.

Theorem 17. Let r ∉ Q. Suppose ϕ_k ≡ 2πkr mod 2π where k ∈ N. Then {ϕ_k∣k ∈ N} is dense in [0,2π].

Corollary 18. Suppose −2 < ρ < 0 and (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)) such that f = Γ_μ,ρ(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). The set {(x_m, y_m)∣m ∈ N} cannot lie entirely in one of the half planes divided by $$

Proof. Since −2 < ρ < 0 and (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)), Γ_μ,ρ,f is an ellipse. By Corollary 16 and Propositions 2 and 4, if π⁻¹arccos⁡(−ρ/2) ∈ Q, then φ is ω-periodic, where ω ≥ 3; otherwise, φ is aperiodic. Suppose π⁻¹arccos⁡(−ρ/2) ∈ Q. Then $$ is ω-periodic, where ω ≥ 3 such that (x_ω, y_ω) = (α, β). First of all, by the Tracking Procedure, we see that (x_k, y_k) moves on Γ_μ,ρ,f in a clockwise manner. Secondly, Γ_μ,ρ,f is symmetric with respect to x = y. Hence, without loss of generality, we suppose Π(α, β) ≥ 0. Since (x_k, y_k) move on Γ_μ,ρ,f in a clockwise manner; by the Tracking Procedure, we can be sure that for some i ∈ {0,1, 2, …, ω − 1}, H(Π_⊥(x_i, y_i))H(Π_⊥(α, β)) < 0. Note that if H(Π_⊥(α, β)) = 0, then we may investigate the orbit $$ or $$ so that H(Π_⊥(β, μ − ρβ − α)) ≠ 0 or H(Π_⊥(μ − ρα − β, α)) ≠ 0, respectively. Suppose π⁻¹arccos⁡(−ρ/2) ∉ Q. Without loss of generality, we suppose Π_⊥(α, β) < 0 and (s, t) ∈ Γ_μ,ρ,f such that Π_⊥(s, t) > 0. Then by Theorem 17, there is a subsequence $$ such that $$ as required and this completes the proof.

Proposition 19. For μ, ρ ∈ R, the lines y = 0 and x = 0 intercept with $$ at (μ/2,0) and (0, μ/2), respectively. Furthermore, if ρ < 2 and μ ≠ 0, then $$ is a nondegenerate conic section and the lines y = 0 and x = 0 are tangents of $$ at the points (μ/2,0) and (0, μ/2), respectively.

Proof. First of all, we show that $$ meets the y = 0 line at (μ/2,0). To this end, note that

and, thus, x = μ/2 is the unique solution to $$ and y = 0. Accordingly, the y = 0 line intercept with $$ at (μ/2,0). By the previous discussions, if ρ = 2, then $$ is a degenerate parabola made up of the x + y − (1/2)μ = 0 line; if ρ = −2 and μ = 0, then Γ_0,−2,0  is a degenerate parabola made up of the x + y = 0 line. Note that the $$ is a parabola with μ ≠ 0. Accordingly, if ρ < 2 and μ ≠ 0, then $$ is not a degenerate conic section which implies $$ is tangent to x = 0 and y = 0 at (0, μ/2) as well as (μ/2,0), respectively. The proof is complete.

Proposition 20. Let $$ and $$ be solutions of (7) with (φ₀, φ₁) = (α, β) and (c₀, c₁) = (0,0). Let $$ and $$ be solutions of (7) with μ = 0, with (a₀, a₁) = (0,1), and with (b₀, b₁) = (1,0). Then we have

Proof. For the sake of convenience, we suppose k ∈ N. First of all, if (a₀, b₀, c₀) = (0,1, 0) and (a₁, b₁, c₁) = (1,0, 0), then (φ₀, φ₁) = (α, β) which implies (63) is true when k ≤ 1. Next, we show that (63) holds for k ≥ 2. To this end, by (7), it follows that

as required. Note that in view of (66), as well as (67), we have Next, we consider the case where k ∈ Z∖N. Let ψ = UE⁻¹φ such that (ψ₀, ψ₁) = (β, α). Thus, we may reformulate (63) for ψ_m, where m ∈ N as Furthermore, in view of (68), UE⁻¹a = b and UE⁻¹c = c which implies (63) is true and this completes the proof.

Theorem 21. Given ρ,   μ ∈ R, where ρ < 2 and μ > 0, then we have

Proof. First of all, we consider the case where k ≥ 0 and we show that h_μ,ρ,k(x, y) = rl_μ,ρ,k(x, y), where r = μ(ρ − 2)/2 < 0. To this end, we show the following results,

are true for all k ∈ Z. Let $$ be $$ with $$ Since (a₀, b₀, c₀) = (0,1, 0)    and (a₁, b₁, c₁) = (1,0, 0), by (70) and (74), it follows that (75) holds for m ≤ 1. Now, we suppose that for some j ∈ N such that (75) holds for j and j + 1. Then we have seen that for i ∈ {j, j + 1} For c_j+2, by (68), (76), and (7), it follows that as required. Next we consider a_j+2 and b_j+2. By (68), (77), and (7) again, in the former case, as desired; the latter case can be handled similarly. By induction, we see that for all m ∈ N, h_μ,ρ,k(x, y) = rl_μ,ρ,k(x, y) is true. Now, we consider the case where k < 0. By the reflection and translation invariances, we transform ψ = UE⁻¹φ in Proposition 20 and ζ = UE⁻¹(_ρ^μχ) and by similarly applied arguments for k ≥ 0, we see that h_μ,ρ,k(x, y) = rl_μ,ρ,k(x, y), for all k < 0. Note that U(_ρ^μχ) = E⁻²(_ρ^μχ) which implies $$, where k < 0. Accordingly, if k ≥ 0, then h_μ,ρ,k(x, y) = rl_μ,ρ,k(x, y); if k < 0, then $$ Finally, since r < 0, we see that (75) holds and this completes the proof.

Corollary 22. Let $$ be a solution of (7) with (φ_m, φ_m+1) = (σ, τ), for some m ∈ Z. If $$, then there is some n ∈ Z such that φ_n = 0.

Proof. By the definitions of $$ in (75), if $$, then (σ, τ) is on the tangent line $$ at $$, for some j ∈ Z. By (75) again, without loss of generality, we may further suppose that $$, for some i ∈ N. For the sake of convenience, we let δ = μ − ρτ − σ and $$ such that $$ Since $$, by (74), we have

Next, we show that $$ To this end, we first have We apply (x, y) = (τ, δ) = (τ, μ − ρτ − σ) and u = μ − ρt − s to (81) and by simplifying, it is found that l_i+1(τ, δ) = l_i(σ, τ) = 0 which implies $$ as desired. Since $$ and $$ are the y = 0 and x = 0 lines, respectively, by the previous discussions, there is some n ∈ Z such that φ_n = 0 and this completes the proof.

Theorem 23. Let $$ be a solution of (7) with ρ > 2. Suppose (φ_m, φ_m+1) ≠ (μ/(2 + ρ), μ/(2 + ρ)), for some m ∈ Z. Then φ is neither a nonnegative solution nor a negative solution.

Proof. Without loss of generality, we let m = 0. Let (φ₀, φ₁) = (α, β) such that f = Γ_μ,ρ(α, β). Let $$ and $$ be generated by the Tracking Procedure with inputs (x₀, y₀) = (α, β) and (s₀, t₀) = (β, α), respectively. Then Γ_μ,ρ,f is a hyperbola (or a degenerate one made up of two lines $$ and $$) such that $$ For the sake of convenience, we let P_c = (μ/(2 + ρ), μ/(2 + ρ)). If (α, β) = P_c, then, by Proposition 2, (x_k, y_k) = (s_k, t_k) = P_c, for all k ∈ N. Suppose (α, β) ≠ P_c. By the previous discussions, we first consider the case where $$ If $$, then $$ and by Corollary 11, it is found that x_kx_k+1 = φ_kφ_k+1 < 0 when k → ∞; if $$, then we take $$ and we see that $$ so that s_ks_k+1 < 0 when k → ∞    as required. Next, suppose Γ_μ,ρ,f is a hyperbola. Then by Corollary 11 again, x_kx_k+1 = φ_kφ_k+1 < 0 when k → ∞. Accordingly, if (α, β) ≠ P_c, then φ is neither a nonnegative nor a negative solution and this completes the proof.

Theorem 24. Let $$ be a solution of (7) with ρ < −2 and μ = 0. Suppose (φ_m, φ_m+1) = (α, β), for some m ∈ Z such that f = Γ_μ,ρ(α, β). Then φ is a nonnegative solution if and only if (α, β) ∈ Θ_0,ρ; and φ is a negative solution if and only if (α, β) ∈ Υ_0,ρ.

Proof. Note that in view of (82), Θ_0,ρ is symmetric with respect to the x = y line and Θ_0,ρ∖{(0,0)} and Υ_0,ρ are symmetric with respect to the x + y = 0 line. Let $$ and $$ be generated by the Tracking Procedure with inputs (x₀, y₀) = (α, β) and (s₀, t₀) = (β, α), respectively. Here, Γ_0,ρ,f, where f = Γ_0,ρ(α, β), is a hyperbola (or a degenerate one made up of two lines $$ and $$). Note that Θ_0,ρ is a cone with two rays $$ and $$ such that u, v ≥ 0, where (u, v) ∈ Θ_0,ρ. First of all, we show that Φ_0,ρ = Θ_0,ρ. To this end, we investigate (x_k, y_k) and (s_k, y_k) directly, and by Theorem 10, it is found that (x_k, y_k), (s_k, t_k) ≥ 0 for all k ∈ N if and only if (x₀, y₀) ∈ Θ_0,ρ. Accordingly, we have Φ_0,ρ = Θ_0,ρ and the argument about Υ_0,ρ can be handled similarly. Note that both Θ_0,ρ and Θ_0,ρ are symmetric with respect to the x = y line. Hence, the cases where k ∈ Z∖N can be handled by the invariances of translation and reflection. The proof is complete.

Theorem 25. Let $$ be a solution of (7) with ρ < −2 and μ ≠ 0. Suppose (φ_m, φ_m+1) = (α, β) for some m ∈ Z such that f = Γ_μ,ρ(α, β).

• (i)

  Suppose μ > 0. Then φ is a nonnegative solution if and only if $$ which is the intersection of the closed half planes defined in (75). Furthermore, φ is a negative solution if and only if (α, β) ∈ Υ_μ,ρ.

• (ii)

  Suppose μ < 0.  φ is a nonnegative solution if and only if (α, β) ∈ Θ_μ,ρ; and φ is a negative solution if and only if $$, where $$ is the intersection of the closed half planes defined in (75).

Proof. Without loss of generality, we let m = 0 and P_c = (μ/(2 + ρ), μ/(2 + ρ)). Since f = Γ_μ,ρ(α, β), Γ_μ,ρ,f is a hyperbola (or a degenerate one, two lines $$ and $$). Let $$ and $$ be generated by the Tracking Procedure with inputs (x₀, y₀) = (α, β) and (s₀, t₀) = (β, α), respectively. Suppose μ > 0. Then μ/(2 + ρ) < 0. By Theorem 21, it follows that $$ as desired. Next, by Corollary 11, if (α, β) ∈ Υ_μ,ρ, then x_k, s_k < 0, for all k ∈ N. Conversely, if x_k, s_k < 0 for all k ∈ N, then, by Theorem 10 and Corollary 11, (α, β) ∈ Υ_μ,ρ. Now, we consider the case where μ < 0. From the previous discussions, Proposition 7 and Theorem 8, we can be sure that Φ_μ,ρ = Θ_μ,ρ. Next, we claim that $$, where $$ is given by (75). From the previous discussions and Corollary 22, we can be sure that φ is a positive solution if and only if $$ Thus, from Proposition 7 and Theorem 8, it follows that $$ and this completes the proof.

Corollary 26. Let $$ be a solution of (7) with |ρ| > 2.  φ is a nonnegative and periodic solution if and only if μ ≥ 0 and (φ_m, φ_m+1) = (μ/(2 + ρ), μ/(2 + ρ)) for some m ∈ Z; φ is a negative and periodic solution if and only if μ < 0 and (φ_n, φ_n+1) = (μ/(2 + ρ), μ/(2 + ρ)) for some n ∈ Z.

Theorem 27. Let $$ be a solution of (7) with ρ = 2. Suppose (φ_m, φ_m+1) = (α, β) for some m ∈ Z.

• (i)

  Suppose μ ≥ 0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if

  $$ (83)

• (ii)

  Suppose μ < 0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if

  $$ (84)

Proof. From the previous discussions, Γ_μ,2,f, where f = Γ_μ,2(α, β) ≥ −μ²/4, is a degenerate parabola made up of two lines, $$ and $$ defined by (24) and (25), respectively. Moreover, if f = −μ²/4, then $$; otherwise, $$ and $$ are two distinct parallel lines. We first consider the case where μ ≥ 0. By Theorem 12 and the Tracking Procedure, if α, β ≥ 0 such that f = Γ_μ,2(α, β) = −μ²/4, then φ is periodic with period 2 and φ₂ ≥ 0 which implies φ is a nonnegative solution. Conversely, if φ is a nonnegative solution, then α, β ≥ 0 and by Theorem 12, f = −μ²/4 which implies $$ Accordingly, we see that

On the other hand, by Theorem 12 again, we can be sure that φ cannot be a nonpositive solution which implies Ω_μ,ρ is empty. Next, we suppose μ < 0. From the previous discussions and Proposition 7, if φ is a nonnegative solution then −φ is a solution of (7) with −μ > 0 which is not possible. Also, $$ and $$ are symmetric with respect to line x + y = 0 and from Proposition 7 and Theorem 8, it follows that and this completes the proof.

Theorem 28. Let $$ be a solution of (7) with ρ = −2. Suppose (φ_m, φ_m+1) = (α, β) for some m ∈ Z.

• (i)

  Suppose μ = 0. Then φ is a nonnegative solution if and only if α, β ≥ 0 and $$ which is the x = y line; furthermore, φ is negative solution if and only if $$ and α, β < 0.

• (ii)

  Suppose μ > 0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if $$ which is the intersection of the closed half planes defined by (75).

• (iii)

  Suppose μ < 0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if

  $$ (87)

•  

  where $$ is the intersection of the closed half planes defined by (75).

Proof. Recall that $$ is the line x = y which is one principal axis of Γ_μ,ρ,f, where f = Γ_μ,ρ(α, β). Suppose μ = 0. By Theorem 13, if $$, then φ is neither a nonnegative nor a nonpositive solution. If $$, then φ is 1-periodic and we see that (i) holds by direct verification. Suppose μ > 0. Then by Theorem 21, we can be sure that $$ as required and $$ is depicted in Figure 19. For the case of (iii), from Proposition 7 and Theorem 8, it follows that (87) is true and this completes the proof.

Theorem 29. Let $$ be a solution of (7) with |ρ| < 2 and μ = 0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if φ is the null solution.

Proof. Let (φ₀, φ₁) = (α, β) such that f = Γ_0,ρ(α, β). Let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Suppose (α, β) = (0,0). From the previous discussions, f = Γ_0,ρ(0,0) = 0 which implies Γ_0,ρ,0 is a point ellipse. On the other hand, by Proposition 2, (φ₀, φ₁) = (0,0) which leads to φ_k = 0, for all k ∈ N and, thus, φ is a nonnegative solution with least period 1. Now, we suppose (α, β) ≠ (0,0). Then Γ_0,ρ,f is an ellipse and we show that φ is neither a nonnegative nor a negative solution. To this end, we consider 0 ≤ ρ < 2 and −2 < ρ < 0. Suppose 0 ≤ ρ < 2. By Proposition 6, φ cannot be a nonpositive solution which implies φ is not negative. If φ is a nonnegative solution, then by Proposition 7, −φ is a nonpositive solution which is a contradiction. The case where −2 < ρ < 0 can be handled similarly and this completes the proof.

Theorem 30. Let $$ be a solution of (7) with |ρ| < 2 and μ ≠ 0. Suppose π⁻¹arccos⁡(−ρ/2) ∉ Q and let (φ_m, φ_m+1) = (α, β) for some m ∈ Z.

• (i)

  Suppose μ > 0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if

  $$ (88)

•  

  which is the interior of the ellipse $$

• (ii)

  Suppose μ < 0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if $$, where

  $$ (89)

Proof. Without loss of generality, we let m = 0. Also, for the sake of convenience, let P_c = (μ/(2 + ρ), μ/(2 + ρ)) and r = π⁻¹arccos⁡(−ρ/2). Also, let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Then we consider two cases where μ > 0 and μ < 0. Here, if (α, β) = P_c, then, by the previous discussions, $$ is a point ellipse so that (x_k, y_k) = P_c, for all k ∈ N. Note that $$ is tangent to lines x = 0 and y = 0 at (0, μ/2) and (μ/2,0), respectively. Suppose μ > 0. We first show that φ cannot be a nonpositive solution. If ρ ≥ 0, then, by Proposition 6, φ is not a nonpositive solution. If φ is nonpositive with ρ < 0, then, by the previous discussions, (α, β) ≠ P_c and by Corollary 18, there is some m ∈ N such that x_m ≥ 0 or y_m ≥ 0. Furthermore, if α = β = 0, then, by (7), φ₂ = μ > 0. Accordingly, φ is not a nonpositive solution when μ > 0.

Now, we show

where ρ > 0. Note that s, t ≥ 0, where (s, t) ∈ {(x, y) ∈ R²∣Γ_μ,ρ(x, y) ≤ (1/4)μ²}. If (88) holds, then, by Theorem 8, we can be sure that Γ_μ,ρ,f⊆Φ_μ,ρ, where f = Γ_μ,ρ(α, β). Accordingly, φ_m ≥ 0 for all m ∈ Z. On the other hand, suppose (88) does not hold. Since r ∉ Q, by Theorem 17, there is some i ∈ N such that $$ (or some j ∈ N such that $$) which implies φ is not nonnegative. Hence, (90) is true. Next, suppose μ < 0. First of all, by the previous discussions and Corollary 22, we see that φ is a positive solution if and only if (α, β) ∈ Λ_−μ,ρ. Thus, from Proposition 7 and Theorem 8, it follows that $$ as required.

The proof is complete.

Theorem 31. Let $$ be a solution of (7) with |ρ| < 2 and μ ≠ 0. Suppose π⁻¹arccos⁡(−ρ/2) ∈ Q and (φ_m, φ_m+1) = (α, β), for some m ∈ Z. Then φ is periodic and we have the following.

• (i)

  Suppose μ > 0. Then φ cannot be a negative solution; furthermore, φ is a nonnegative solution if and only if $$ which is the intersection of the closed half planes defined by (75).

• (ii)

  Suppose μ < 0. Then φ cannot be a nonnegative solution; furthermore, φ is a negative solution if and only if

  $$ (91)

•  

  where $$ is the intersection of the closed half planes defined by (75).

Proof. Without loss of generality, we let m = 0. Also, let $$ be generated by the Tracking Procedure with input (x₀, y₀) = (α, β). Suppose μ > 0. From the previous discussions, if (α, β) = (μ/(2 + ρ), μ/(2 + ρ)), then φ is a nonnegative solution with least period 1. If (α, β) ≠ (μ/(2 + ρ), μ/(2 + ρ)), then, by Corollary 16, φ is periodic and Γ_μ,ρ,f is an ellipse, where f = Γ_μ,ρ(α, β). Also, by Theorem 21, we can be sure that $$ as desired. Next, we show that φ is not a nonpositive solution. First of all, by Proposition 6, φ is not nonpositive when ρ ≥ 0. Secondly, suppose φ is a nonpositive solution of (7) with ρ < 0. Then by Corollary 18, there is some m ∈ N such that x_m or y_m ≥ 0. If x_m = y_m = 0, then, by (7), x_m+1 = μ > 0 which is a contradiction. Accordingly, φ is not a nonpositive solution.

Suppose μ < 0. By Corollary 22, Proposition 7, and Theorem 8, it follows that (91) holds. Also, if φ is a nonnegative solution of (7) with μ < 0, then, by Proposition 7, −φ is a solution of (7) with −μ > 0 which is not possible and this completes the proof.

Corollary 32. Let $$ be a solution of (7) with |ρ| < 2 and μ > 0. Suppose r = π⁻¹arccos⁡(−ρ/2) = (n/m) ∈ Q such that 0 < r < 1 and r is irreducible. Then Φ_μ,ρ is a closed polygon with ω sides which is circumscribed around the ellipse $$ and ω is either 2n or n depending on whether m is even or odd, respectively.

Proof. In view of Theorem 21, $$ which is defined by the tangent lines of $$ Furthermore, all the lines are tangent to $$ at the points which are generated by a specific solution _ρ^μχ of (7) with $$ Since π⁻¹arccos⁡(−ρ/2) ∈ Q, by Corollary 16, _ρ^μχ is ω-periodic where ω is dependent on n and the parity of m. Note that ω > 2. Accordingly, we see that Φ_μ,ρ is defined by ω distinct lines. To see Φ_μ,ρ is a closed polygon with ω sides which is circumscribed around $$, it is sufficient to show that Φ_μ,ρ is closed. If ρ = 0, then _ρ^μχ is 4-periodic such that

if 0 < ρ < 2, then, by (70), h_μ,ρ,2(x, y) = μ − ρx − y such that Accordingly, Φ_μ,ρ is closed, where 0 ≤ ρ < 2. Suppose −2 < ρ < 0. To see Φ_μ,ρ is closed, it is sufficient to show that there is some i ∈ N such that $$ Since $$, in view of Corollary 18, there is some j ∈ N such that $$ as required and this completes the proof.