Bounds for Roots of Functions fv0 and fz0

April 29, 2016

Series expansion of Jacobian is given as:

J =16

3k3u21[−12(−3 + 6u21 + 2u41) + (−36 + 33u41 − 8u61) cos(2u2)

+ cos(4u1)(−12(−1 + u21

)−

(12− 24u21 + u41) cos(2u2)

)−

−2u1(21− 68u21 + 18u41 + 10(−3 + 2u21) cos(2u2)

)sin(2u1)

+4 cos(2u1)(−12 + 21u21 − 24u41 + 2u61

+cos(2u2)(12− 6u21 + 4u41 + u1(−15 + 4u21) sin(2u1)

))

+u1(21− 2u21) sin(4u1)] +O(k3). (1)

where u1 = am(p, k) and u2 = am(τ, k).Similary series expansion of the functions fz(p, k) and fv(p, k) in variables u1and u2 can be written as:

fz0(u1, k) = −u1 cosu1 + sinu1, (2)

fv0(u1, k) = 4u21 cosu1 − 4u1 sinu1 +4

3u31 sinu1. (3)

It turns out that (1) in terms of fz0 and fv0 can be written as:

J = 1(u1)fz0(u1, k), u2 =π

2,

J = 2(u1)fv0(u1, k), u2 = 0,

where1(u1) = −2u1(27−85u21+8u41) cos(u1)−2u1(−27+u21) cos(3u1)+4u21(−45+14u21+9 cos(2u1)) sin(u1)+192 sin3(u1) and 2(u1) =

32 (−3(1 + 4u21) cos(u1) + 3 cos(3u1) + 2u1(11− 4u21 + cos(2u1) sin(u1)).

Lemma 1. The first root of the function fz0(u1) lies in the interval u1 ∈(π, 3π2

).

Proof. Clearly, function fz0(u1) is continuous with its derivative given as f ′z0(u1) = u1 sinu1and f ′z0(u1) >0, u1 ∈ (0, π) where function is increasing and f ′z0(u1) < 0, u1 ∈ (π, 2π) where function is decreasing.Also,

fz0(0) = 0,

fz0 (π) = π,

fz0

(3π

2

)= −1.

Now by intermediate function theorem, fz0(u1) = 0 for some u1 ∈(π, 3π2

).

Lemma 2. The first root of the function fv0(u1) lies in the interval u1 ∈(3π2 , 2π

).

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Proof. Consider the function gv0(u1) which has same roots as fv0(u1) and has following properties:

gv0(u1) =fv0(u1)

sin(u1)=

4

3u1

(−3 + u21 + 3u1 cotu1

),

limu1→0

gz0(u1) = 0,

g′z0(u1) = −4 (−1 + u1 cot(u1))2< 0,

The function gv0(u1) has vertical asymptotes at u1 = ±nπ, n ∈ N such that:

limu1→0+

gv0(u1) = 0,

limu1→π−

gv0(u1) = −∞,

limu1→π+

gv0(u1) = +∞,

limu1→2π−

gv0(u1) = −∞.

Hence, the function gv0(u1) and therefore fv0(u1) has its first positive root in the interval u1 ∈ (π, 2π). Also,fv0

(3π2

)< 0 and fz0 (2π) > 0 and therefore by intermediate value theorem the function fv0(u) has root in

the interval u1 ∈(3π2 , 2π

).

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