SADDLE-NODE BIFURCATIONS OF MULTIPLE HOMOCLINIC SOLUTIONS IN ODES

We study codimension 3 degenerate homoclinic bifurcations under periodic perturbations. Assume that among the 3 bifurcation equations, one is due to the homoclinic tangecy along the orbital direction. To the lowest order, the bifurcation equations become 3 quadratic equations. Under generic conditions on perturbations of the normal and tangential directions of the homoclinic orbit, up to 8 homoclinic orbits can be created through saddlenode bifurcations. Our results generate the homoclinic tangency bifurcation in Guckenheimer and Holmes [8].

The variational equation of (2) along the homoclinic solution γ is Since γ is a bounded solution of (3), system (3) has d ≥ 1 linearly independent bounded solutions.
When µ = 0, (5) may have bifurcations near γ.The case d = 1 has been extensively studied.In this case breaking of the homoclinic orbit γ is restored by choosing the parameter τ , see [8].Hale [9] proposed to study the degenerate cases where d ≥ 2. The case d = 2 has been considered in [17].In this paper we treat the case d = 3.Using the method of Lyapunov-Schmidt reduction and exponential dichotomies, we derive a system of bifurcation functions H j , 1 ≤ j ≤ 3, the zeros of which correspond to the bifurcations of homoclinic solutions for (5) (For the definitions of H j , see ( 23)).To the lowest degree of H j , the bifurcation equations reduce to three quadratic equations M j = 0, 1 ≤ j ≤ 3 (For the definitions of M j , see (11)).
The last equation M 3 = 0 can be dealt with by selecting the parameter τ as usual, while M j = 0, j = 1, 2 can be simplified by the codiagonalization of quadratic forms.We show that M j = 0, j = 1, 2, can have 4 non-degenerate solutions.Substituting them into the last equation, we show M 3 = 0 undergoes the saddle-node bifurcation with respect to the parameter τ .Thus, each of the 4 solutions can generate 2 solutions, and 8 solutions can be obtained for M j = 0, 1 ≤ j ≤ 3. Finally, as perturbations to M j = 0, the system H j = 0, 1 ≤ j ≤ 3, can have up to 8 solutions.
Codiagonalization of matrices has been used by Jibin Li and Lin [15] to study systems of coupled KdV equations, and will be one of the main tool used in this paper.Given a symmetric real matrix B ∈ R 2×2 , then , where (x 1 , x 2 ) T = M (y 1 , y 2 ) T .The symmetric transformation described above is also called the congruence diagonalization.It should not confused with the similarity transformation of B which is defined by M −1 BM .For example the matrix diag(λ 1 , −λ 2 ), λ j > 0, can be reduced to diag(1, −1) by the matrix , which is a symmetric reduction, not similarity reduction.In §2, we introduce notations to be used in this paper.We also present the reduced bifurcation functions (11) which, to the lowest degree, represent the breaking of the homoclinc orbits under the periodic perturbations.In §3, using the Lyapunov-Schmidt reduction, we derive the bifurcation equations ( 23), which to the lowest degree, become three quadratic equations M j = 0, 1 ≤ j ≤ 3.In §4, we introduce conditional max/min problems to codiagonalize two quadratic forms, and obtain general conditions under which the two quadratic equations may have 4 real valued solutions.The case when one equation is elliptic is considered in §4.1 (Theorem 4.2).The other case when both equations are hyperbolic is considered in §4.2 (Theorem 4.2).Conditions for the existence of 4 real valued solutions to quadratic systems after codiagonalization are given in §4.3 (Theorems 4.3 and 4.4).In §5, we prove the existence of homoclinic solutions by solving the bifurcation equations using the contraction mapping principle (Theorems 5.1 and 5.2).In §6 (Theorem 6.1), we prove the transversality of the homoclinic solutions obtained in §5.

2.
Notations and preliminaries.Notations.Let X, Y be Banach spaces and L : X → Y be a linear operator.We use N (L) and R(L) to denote the null subspace and range subspace of L, respectively.Since y = 0 is a hyperbolic equilibrium, from [6,20], (3) has exponential dichotomies on J = R ± respectively.Let U (t) be the fundamental matrix of (3).Then there exist projections to the stable and unstable subspaces, P s + P u = I, and constants m > 0, K 0 ≥ 1 such that For the same m > 0, define the Banach space with the norm z = sup t∈R |z(t)|e m|t| .The linear variational system will be considered in Z.The adjoint operator for L is The domains of ( 7) and ( 8) are the dense subset of Z, defined as From the theory of homoclinic bifurcations [20], L : Z → Z is a Fredholm operator with index 0.The range of L is orthogonal to the null space of L * .That is From d = 3, N (L) is three dimensional.Note that γ ∈ N (L).Without loss in generality, let (u 1 , u 2 , u 3 ) be a basis of N (L), where we choose u 3 = γ.And let (ψ 1 , ψ 2 , ψ 3 ) be a basis of N (L * ).
We define some Melnikov types of integrals that will be used in the future.For integers p, q = 1, 2 and i = 1, 2, 3, define where ϕ(t, τ ) = K(I − P )g(γ(t), t + τ, 0) (See the definitions of operators K and P in Section 3).We look for conditions so that (5) can have homoclinic solutions near γ.

We now assume (H5):
Remark 1.In the special case d = 1, the term d ij and F 2 do not appear.And F 1 (τ ) = −a(τ ), where Then (H5) reduces to the following In this case bifurcations due to homoclinic tangencies may occur, [8].
By changing ψ i to −ψ i , we can change B (i) to −B (i) without altering the result of the paper.Hence we assume the following conditions are satisfied: 3. Derivation of the bifurcation equations using the Lyapunov-Schimidt reduction.In this section, we look for conditions such that for small µ = 0, (5) may have homoclinic solutions γ µ with γ − γ µ = O( |µ|).
Recall that L(u) = u − Df (γ)u in the Banach space Z.As in [6,20], we define the subspace of Z, which consists the range of L in Z.
Let Z ⊥ be the subspace of Z consisting of z(t) with z(0) ⊥ γ(0).If h ∈ Z, using the variation of constants, there exists an operator K : Z → Z ⊥ such that Kh is a solution of (16).Clearly, the general bounded solution of ( 16) is As in [20], the projection P satisfies the following properties: We now use the Lyapunov-Schmidt reduction to (13).Applying P and (I − P ) to (13), we have the following equivalent system First, we solve (17) for z ∈ Z ⊥ .Then the bifurcation equations are obtained by substituting z into (18).
We have proved the following important result.13) and hence the perturbed system (5) has a homoclinic orbit x = γ + φ.
Through direct calculations, we can prove the following Lemma.
Keeping up to quadratic terms of β and µ, We will see that µ 2 ζ i (τ ) can be dropped in the bifurcation analysis.Let M i : R 2 × R × R → R 3 be given by (11), which retains the quadratic terms of H i (β, τ, µ), except for µ 2 ζ i (τ ).Define a i (τ ) = −2ã i (τ ) and C i (τ ) = −2 Ci (τ ).We need to solve the quadratic system (12).and F (x, y) = z T Bz.We say that the quadratic equation F (x, y) = h, h = 0 is of elliptic (or hyperbolic, or line) type if b 2 − ac < 0 (or > 0, or = 0).In this case the graph of the equation is an ellipse (or two hyperbolas, or two lines).The graph of the line type equation is a special case of two hyperbolas, where the normal direction to two lines replaces the real axis of a hyperbola.If b 2 = ac, and a, c, h > 0, then F (x, y) = ( √ ax + √ cy) 2 .The solution represents two parallel lines √ ax + √ cy = ± √ h, symmetric about the origin.The hyperbolic rotation is well-known for its use in relativity theory [3].We shall define various transformations that keep a quadratic form F (x, y) = ax 2 +2bxy+cy 2 invariant.Consider the Hamiltonian system and its solution mapping T (t).The values of F (x, y) are invariant under T (t).
Definition 4.1.The solution mapping T (t) for (24) that maps the ray − − → OP 1 to − − → OP 2 , where P 2 = T (t)P 1 , will be called the quadratic rotation by the angle t.It will also be called the circular, elliptic or hyperbolic rotation if the graph of F (x, y) = h is a circle, ellipse or hyperbola.The angle θ from On the other hand, if there does not exist any t ∈ R with − − → OP 2 = T (t) − − → OP 1 , then the angle between the two rays is undefined.
We can pick any P 0 on a circle and define its angle coordinate to be θ(P 0 ) = 0.For other quadratic curves, if P 0 is a point on the major axis (or semi-real, or semiimaginary axis), then we define its angle coordinate to be θ(P 0 ) = 0. Then for any P ∈ R 2 , we define its angle coordinate θ(P ) to be the angle from Examples.Let a = c = 1, b = 0 in (24).The solution mapping defines the circular rotation in counter-clockwise direction.Let a = 1, c = −1, b = 0 in (24).The solution mapping defines the standard hyperbolic rotation in 4 invariant sectors of R 2 .More precisely, the lines y = ±x divide R 2 into 4 invariant sectors: If (x 0 , y 0 ) T ∈ S 1 or S 3 , then (x 0 , y 0 ) = r 0 (cosh(t 0 ), sinh(t 0 )), r 0 ∈ R.And (x(t), y(t)) T remains in S 1 or S 3 with x(t) y(t) = r 0 cosh(t) sinh(t) sinh(t) cosh(t) cosh(t 0 ) sinh(t 0 ) = r 0 cosh(t + t 0 ) sinh(t + t 0 ) .
(1) If the vector field (24) corresponding to F (x, y) satisfies x = 0 on the x-axis, or y = 0 on the y-axis, then matrix B is diagonal.
(2) In both cases, we have We consider the system of two quadratic equations (H7): Assume that the two quadratic forms F 1 (x, y), F 2 (x, y) are linearly independent, i.e., the two matrices B 1 , B 2 are linearly independent.Here is brief outline of the content in §4.1 - §4.3.In §4.1 and §4.2, we use conditional max/min problems to codiagonalize two quadratic equations.The max/min process also provides conditions for the existence of 4 real valued solutions without going through the codiagonalization.In §4.3 we give simple conditions for the existence of 4 real valued solutions on all the systems considered in §4.1and §4.2.However, these conditions are posed on the codiagonalized systems.

Codiagonalization and solutions of (25) if one equation is elliptic.
It is well-known that two symmetric matrices can be simultaneously diagonalized if one of the matrices is positive definite, [11,12].However, it is not clear if the resulting matrices are real valued. If From (H6), we find that a 2 > 0, c 2 > 0 and h 2 > 0. We shall use the elliptic rotation T 2 (θ) defined by (24) with B = B 2 .First we shall find two points P j = (x j , y j ) T , j = 1, 2, from the conditional maximum/minimum problems.
We look for the critical points (x, y) of the following Lagranginan: Notice our definition of the Lagrangian is slightly different from those in standard literatures.
To find critical points of the Lagrangian, we look for the generalized eigenvectors of the following system We consider three types of systems.(i) (EE) type: In this case, (28) has two eigenvalues (λ 1 , λ 2 ) with (nonunique) eigenvectors (P 1 , P 2 ) = ((x 1 , y 1 ) T , (x 2 , y 2 ) T ).Then after rescaling of P 1 and P 2 , we assume that on the curve F 1 = h 1 , F 2 reaches the minimum r 1 at P 1 and the maximum r 2 at P 2 .There exists an appropriate angle θ 0 such that T 2 (−θ 0 )P 2 coincides with the major axis of F 2 (x, y) = h 2 .

Using the property
in all the three cases, the image of T 2 (−θ 0 )P 1 should coincide with the minor axis of F 2 = h 2 .Also, under the rotation T 2 (θ 0 ), the quadratic form F 1 (x, y) = h 1 becomes F 3 (x, y) = h 1 while F 2 (x, y) = h 2 is unchanged.Now apply a circular rotation R(−θ 0 ) to both F 3 (x, y) = h 1 and F 2 (x, y) = h 2 so the major axis of F 2 (x, y) = h 2 is mapped to the x-axis.The matrices that represent the two quadratic forms are We have proved the following results: Lemma 4.2.Assume (H1)-(H7) are satisfied.If one equation of the quadratic system is elliptic, then the two quadratic forms can always be codiagonalized by the real valued matrices.More specifically, the codiagonalized graphs of F 1 = h 1 are as follows.In the Case (EE) type, the major axis of the ellipse F 1 = h 1 is on the x-axis.In the case (HE) type, the real axis of the hyperbola F 1 = h 1 is on the x-axis.In the Case (LE) type, the normal direction of the two parallel lines is on the x-axis.
Theorem 4.1.The (EE) type of system has 4 solutions if r 1 < h 2 < r 2 .The (HE) type of system has 4 solutions if The (LE) type of system has 4 solutions if r 2 < h 2 .
Proof.Case (EE) type.It is given that F 2 (P 1 ) = r 1 < h 2 < r 2 = F 2 (P 2 ).Let the angle of P i be θ i .Then there exists an angle θ 0 between θ 1 and θ 2 such that the corresponding point is P 0 on the graph of F 1 = h 1 with F 2 (P 0 ) = h 2 .There exist 4 pairs of such (P 1 , P 2 ) so the total number of solutions is 4.
Case (HE) type.If F 2 (P 2 ) = r 2 < h 2 , then as θ → ±∞, the values of F 2 on the graph of F 1 = h 1 approach ∞ that is greater than h 2 .So there exists two points P ± on each of the two branches of F 1 = h 1 such that F 2 (P ± ) = h 2 .The other case h 1 < 0 can be proved similarly.
Case (LE) type.At each of the two parallel lines, there exists a point P 2 such that F 2 (P 2 ) = r 2 < h 2 .Moving away from P 2 on the line F 1 = h 1 , the value of F 2 gets greater than h 2 in two opposite directions.So on two opposite directions of each line, there exist P ± such that F 2 (P ± ) = h 2 .

4.2.
Codiagonalization and solutions of (25) if both equations are hyperbolic.In this subsection we consider the system F j (x, y) = h j , j = 1, 2, where both equations are of hyperbolic type, denoted by (HH).To simplify the illustration, assume h 1 > 0, h 2 > 0. Let T 2 (θ) be the hyperbolic rotation defined by (24) with B = B 2 .Unlike the cases studied in §4.1, a general conditional max/min problem is not well posed for (HH) type systems.We can divide the (HH) type system into two sub-cases, and find a suitable conditional max/min problem for each sub-case.
For the (HH) type system, b 2 j − a j c j > 0, j = 1, 2, so with (a, b, c) = (a j , b j , c j ), the equilibrium (0, 0) of ( 24) is hyperbolic and there exist stable and unstable eigenspaces for the equilibrium (0, 0).Definition 4.2.Let L (i) j , i = 1, 2, be the stable and unstable eigenspaces of the equilibrium for (24), where (a, b, c, ) = (a j , b j , c j ).They are called the asymptotes for F j (x, y) = h j , and four sectors.We say (x, y) is in the positive (or negative) sector if F j (x, y) > 0 ( or F j (x, y) < 0).Consider 4 cases, as depicted in Fig. 1: (i) The two sectors of F 1 > 0 are in the interior of the sectors of F 2 > 0.
(ii) The two sectors of F 1 < 0 are in the interior of the sectors of F 2 > 0. (iii) The two sectors of F 1 > 0 are in the interior of the sectors of F 2 < 0. (iv) The two sectors of F 1 < 0 are in the interior of the sectors F 2 < 0.
Finally, using the hyperbolic rotation T 2 (θ) defined by (24) with B = B 2 , and the method that proves Lemma 4.2, the system F j = h j , j = 1, 2, can be codiagonalized.
Proof.Case (i).Along the asymptotes of F 1 = h 1 or −h 1 , we find that F 2 (x, y) → ∞ as x 2 + y 2 → ∞.The curves for F 1 = ± h 1 have 4 continuous branches.Let (x(t), y(t)) be the orbit of ( 24) that is on one of such branches.Then F 2 (x(t), y(t)) → ∞ as t → ±∞.The search for minimum can be restrict to a compact subinterval of t ∈ R, on which the continuous function F 2 (x(t), y(t)) must reach a minimum.Hence there are at least 4 solutions for the max/min problem (29).
It remain to show that (29) cannot have more than 4 solutions.To this end, let P = (x, y) T be a point where the minimum is reached.Then (x, y) T is an critical point for the Lagrangian (27) and satisfies the generalized eigenvalue problem (28).There can only be 2 linearly independent vectors (x, y) T .Using the symmetry about the origin, we find that there are exactly 4 such critical points.The assertion r 2 < 0 < r 1 is obvious and the proof shall be omitted.
The proofs for cases (ii)-(iv) are similar and will not be given here.
From a counter example at the end of this subsection, if the asymptotes of F 1 = h 1 and F 2 = h 2 are alternating, then the system cannot be codiagonalized.So results in Lemma 4.3 are the best we can obtain.Theorem 4.2.Let r j , 1 ≤ j ≤ 4 be defined case by case as in Lemma 4.3.Then the conditions for the following system of quadratic equations to have 4 solutions are determined by the asymptotes and positions of positivenegative sectors separated by the asymptotes as follows: Case (i).The system has 4 solutions provided that h 1 > 0, r 1 < h 2 , see Fig. 2; or −h 1 < 0 for any r 2 < 0.
Case (iv).The system has 4 solutions provided that Proof.We will prove case (i) only.Let (x(t), y(t)) T be the point on a branch of F 1 (x, y) = h 1 , and the minimum of Hence there exist exactly two points t 1 < 0 < t 2 where F 2 (x(t j ), y(t j )) = h 2 .Obviously F 1 (x(t j ), y(t j )) = h 1 .Therefore (x j , y j ) T , j = 1, 2 are the solutions for There are two more solutions on the other branch of F 1 (x, y) = h 1 .So the total number of solutions is 4. We can similarly consider the (x(t), y(t)) T on a branch of F 1 (x, y) = −h 1 .This time using r 2 < 0, we can find two solutions of F 1 = −h 1 , F 2 = h 2 on each of the two branches.So the total number of solutions is 4. In case (4.6-1), if h 1 > 0 and r 1 < r 2 , or in case (4.6-2), if h 1 < 0 and r 1 < r 2 , then the system has 4 solutions.In case (4.6-3), if h 1 < 0 and r 1 > r 2 , or in case (4.6-4), if h 1 > 0 and r 1 > r 2 , then the system has 4 solutions.
Before ending this subsection, we present an example from [17] showing that not all the (HH) case can be codiagonalized.
A Counter Example.Assume that the asymptotes of two hyperbolas are alternating.It means that none of the positive or negative sectors of F 1 are inside the positive or negative sectors of F 2 and vise versa.See Figure .4. In this case the two quadratic forms cannot be codiagonalized.

4.3.
Existence of 4 real valued solutions of (25) for all the possible cases.After the codiagonalization of two quadratic equations, it is simple to list conditions for the coupled system to have 4 real valued solutions, including all the cases studied in §4.1 and §4.2.After codiagonalization, we have b 1 = b 2 = 0, then (25) becomes In the first quadrant, the graphs of F j (x, y) = h j , j = 1, 2, may intersect with the boundaries of the first quadrant at certain points, which will simply be called the x-intercept and/or y-intercept, and denoted by jx and/or jy respectively.A hyperbola has two continuous branches.The opening angle of a hyperbola C, denoted by Θ(C), is defined to be the angle between the two asymptotic lines of any branch of the hyperbola.
More precisely, an ellipse has x and y intercepts with the coordinate lines.A hyperbola has x or y-intercepts but not both.A vertical line has a x-intercept, and a horizontal line has a y-intercept with the coordinate axes.Based on those observations, we list the intersection of curves defined by (32) as follows: (EE): (i) Both ellipses have x and y-intercepts.All the possible cases, except for case (LL), are plotted in Figures 5-7.Also if two graphs are related by flipping the horizontal and vertical axes, then only one is plotted.Proof.The proof can be done by the Intermediate Value Theorem.For example, in the (EE) case, let us follow the C 1 in the first quadrant from the x-intercept to the y-intercept.If 1x < 2x , 1y > 2y , then C 1 started below, but ended above C 2 .Thus the two curves must intersect somewhere in the first quadrant.The (HE) and (LE) cases can be proved similarly.
Alternatively, each case listed in the theorem can be identified with a unique case in Theorems 4.1 and 4.2.And it can be shown that the conditions given there are the same as some conditions in Theorem 4.6.Hence the system has 4 solutions.See Remark 2 for two such examples.Remark 2. The conditions listed above look different but are equivalent to the conditions in Theorems 4.1 and 4.2.This can be checked case by case.For example, in the case (HH) (i), assume that C 1 has y-intercept, C 2 has x-intercept, and Θ(C 1 )+ Θ(C 2 ) > π, as in Theorem 4.3.Then the condition on the opening angles implies that the two asymptotes of F 1 = h 1 are in the sector there F 2 > 0. And among them, the sectors of F 1 < 0 are in the interior of the sectors of F 2 > 0. This is exactly the case (ii) considered in Lemma 4.5 and Theorem 4.6.From Theorem 4.6, the system F j = h j , j = 1, 2, has 4 solutions.
As a second example, in the case (HH) (ii), assume that 1x < 2x and Θ(C 1 ) < Θ(C 2 ) as in 4.3.Then the condition on the opening angles implies that the sectors of F 1 > 0 are in the sector F 2 > 0. This case is considered in as case (i) in Lemma 4.5 and Theorem 4.6.Now the condition 1x < 2x implies that r 1 < h 2 as in Theorem 4.6.Therefore the coupled system has 4 solutions.
In following we show the four solutions obtained in Theorem 4.3 are simple.Theorem 4.4.Assume that the two quadratic forms are linearly independent as in (H7).Then the four solutions of (25) obtained in Theorem 4.3 are simple.
Proof.We only give the proof for (EE) type systems for the other cases can be proved similarly.It suffices to consider the two equations after codiagonalization.
The normal directions of F 1 and F 2 at (x (i) , y (i) ) are (a 1 x (i) , c 1 y (i) ) and (a 2 x (i) , c 2 y (i) ), respectively.Due to the linear independence of the two quadratic forms, and nonzeroness of (x (i) , y (i) ), which implies that (x (i) , y (i) ), 1 ≤ i ≤ 4, are simple zeros of (33).
Finally, using the change of variables u = x 2 , v = y 2 , the quadratic system F j (x, y) = h j , j = 1, 2 becomes the following linear system in (u, v): Under the conditions a := a 1 c 2 − c 1 a 2 = 0, the system has a unique solution Thus F j (x, y) = h j , j = 1, 2 can have 4, possibly complex valued, solutions.
Observe that if jx , jy are the x and/or y-intercepts for the jth quadratic equation (32), then L ju = 2 jx , L j v = 2 jy are the u and/or v-intercepts for (34).If k j , j = 1, 2 are the slopes for the asymptote of the jth hyperbola, then K j = k 2 j are the slopes for corresponding lines for (34).
If the system on (u, v) has a solution in the first quadrant-u > 0, v > 0 then the original system in (x, y) has 4 solutions.Now, under the conditions as in Theorem 4.3 (or equivalently, Theorems 4.1 and 4.2), the 4 solutions are real valued solutions.Under the same conditions (34) has a positive valued solution u > 0, v > 0.
In the following, we will prove that the solutions obtained above are simple.From (36), we have 2 ) = 0 for small µ.
is a quadratic form associated to B. If B is diagonalized by a nonsingular matrix M : M T BM = diag(d 1 , d 2 ), then

4 .
Codiagonalization and solutions of two quadratic equations.Let z = x y , B = a b b c

1 Figure 4 .
Figure 4.If the asymptotes of F 1 = 0 and F 2 = 0 are alternating, then there always exist exactly two solutions.
(HE): (i) The hyperbola has x-intercept.(ii) The hyperbola has y intercept.(LE): (i) The line has x-intercept.(ii) The line has y-intercept.(HH): (i) One hyperbola has x, the other has y-intercept.(ii) Both hyperbolas have x-intercept.(iii) Both hyperbolas have y-intercept.(LH): (i) The hyperbola has y and the line has x-intercept.(ii) Both have yintercepts.(iii) The hyperbola has x and the line has y-intercept.(iv) both have x-intercepts.(LL): (i) One line has x-intercept and the other line has y-intercept.

Figure 5 .
Figure 5.Figure (1) is about the (EE) case.Figure (2) is about the (HE) case.The (HE) case where the hyperbola has y intercept is not shown.Figure (3) is about the (LE) case where the lines have x-intercepts.The (LE) case where the lines have y intercepts is not shown.

Figure ( 1 )
Figure 5.Figure (1) is about the (EE) case.Figure (2) is about the (HE) case.The (HE) case where the hyperbola has y intercept is not shown.Figure (3) is about the (LE) case where the lines have x-intercepts.The (LE) case where the lines have y intercepts is not shown.

Figure 6 .
Figure 6.Figure (1) is about the (HH) case where one curve has x and the other has y intercepts.Figure (2) is about the (HH) case where both graphs have x-intercepts.The (HH) case where both have y intercepts in not shown.

Figure ( 1 )
Figure 6.Figure (1) is about the (HH) case where one curve has x and the other has y intercepts.Figure (2) is about the (HH) case where both graphs have x-intercepts.The (HH) case where both have y intercepts in not shown.

Figure 7 .
Figure 7.Figure (1) is about the (LH) case where the lines have x-intercepts and the hyperbola has y-intercept.Figure (2) is about the (LH) case where both the lines and the hyperbola have y-intercepts.The two (LH) cases where the hyperbola has x-intercepts are not shown.

Figure ( 1 )
Figure 7.Figure (1) is about the (LH) case where the lines have x-intercepts and the hyperbola has y-intercept.Figure (2) is about the (LH) case where both the lines and the hyperbola have y-intercepts.The two (LH) cases where the hyperbola has x-intercepts are not shown.