A Note on Sign-Changing Solutions to the NLS on the Double-Bridge Graph

Abstract

We study standing waves of the NLS equation posed on the double-bridge graph: two semi-infinite half-lines attached at a circle. At the two vertices, Kirchhoff boundary conditions are imposed. We pursue a recent study concerning solutions nonzero on the half-lines and periodic on the circle, by proving some existing results of sign-changing solutions non-periodic on the circle.

Full text

symmetry
S
S
Article
A Note on Sign-Changing Solutions to the NLS on the
Double-Bridge Graph
Diego Noja *, Sergio Rolando and Simone Secchi
Dipartimento di Matematica e Applicazioni, Università di Milano Bicocca, via R. Cozzi 55, 20126 Milano, Italy;
sergio.rolando@unimib.it (S.R.); simone.secchi@unimib.it (S.S.)
*Correspondence: diego.noja@unimib.it
Received: 15 January 2019; Accepted: 28 January 2019; Published: 1 February 2019


Abstract:
We study standing waves of the NLS equation posed on the double-bridge graph:
two semi-infinite half-lines attached at a circle. At the two vertices, Kirchhoff boundary conditions
are imposed. We pursue a recent study concerning solutions nonzero on the half-lines and periodic
on the circle, by proving some existing results of sign-changing solutions non-periodic on the circle.
Keywords: quantum graphs; non-linear Schrödinger equation; standing waves
MSC: 35Q55; 81Q35; 35R02
1. Introduction and Main Results
The study of nonlinear equations on graphs, especially the nonlinear Schrödinger equation
(NLS), is a quite recent research subject, which already produced a plenty of interesting results
(see [
1
–
3
]). The attractive feature of these mathematical models is the complexity allowed by the
graph structure, joined with the one dimensional character of the equations. While they are an
oversimplification in many real problems, they appear indicative of several dynamically interesting
phenomena that are atypical or unexpected in more standard frameworks. The most studied issue
concerning NLS is certainly the existence and characterization of standing waves (see, e.g., [
4
–
9
]).
More particularly, several results are known about ground states (standing waves of minimal energy
at fixed mass, i.e.,
L2
norm) as regard existence, non-existence and stability properties, depending on
various characteristics of the graph [2,10–13].
In this paper, we are interested in a special example, which reveals an unsuspectedly complex
structure of the set of standing waves. More precisely, we consider a metric graph
G
made up of two
half lines joined by two bounded edges, i.e., a so-called double-bridge graph (see Figure 1).
G
can
also be thought of as a ring with two half lines attached in two distinct vertices. The half lines are
both identified with the interval
[
0,
+∞)
, while the bounded edges are represented by two bounded
intervals of lengths L1>0 and L2≥L1, precisely [0, L1]and [L1,L]with L=L1+L2.
∞ ∞
L1
L2
Figure 1. The double-bridge graph.
Symmetry 2019,11, 161; doi:10.3390/sym11020161 www.mdpi.com/journal/symmetry

Symmetry 2019,11, 161 2 of 20
A function
ψ
on
G
is a Cartesian product
ψ(x1
, ...,
x4)=(ψ1(x1)
,
. . .
,
ψ4(x4))
with
xj∈Ij
for
j=
1,
. . .
, 4, where
I1= [
0,
L1]
,
I2= [L1
,
L]
and
I3=I4= [
0,
+∞)
. Then, a Schrödinger operator
HG
on Gis defined as
HGψ(x1, . . . , x4)=−ψ00
1(x1), . . . , −ψ00
4(x4),xj∈Ij, (1)
with domain
D(HG)
given by the functions
ψ
on
G
whose components satisfy
ψj∈H2(Ij)
together
with the so-called Kirchhoff boundary conditions, i.e.,
ψ1(0) = ψ2(L) = ψ3(0),ψ1(L1) = ψ2(L1) = ψ4(0), (2)
ψ0
1(0)−ψ0
2(L) + ψ0
3(0) = ψ0
1(L1)−ψ0
2(L1)−ψ0
4(0) = 0. (3)
As is well known (see [
14
] for general information on quantum graphs), the operator
HG
is
self-adjoint on the domain
D(HG)
, and it generates a unitary Schrödinger dynamics. Essential
information about its spectrum is given in ([
15
], Appendix A). We perturb this linear dynamics
with a focusing cubic term, namely we consider the following NLS on G
idψt
dt =HGψt−|ψt|2ψt(4)
where the nonlinear term
|ψt|2ψt
is a shortened notation for
(|ψ1,t|2ψ1,t
,
. . .
,
|ψ4,t|2ψ4,t)
. Hence,
Equation (4) is a system of scalar NLS equations on the intervals
Ij
coupled through the Kirchhoff
boundary conditions in Equations (2)–(3) included in the domain of
HG
. On rather general grounds, it
can be shown that this problem enjoys well-posedness both in strong sense and in the energy space
(see in particular ([2], Section 2.6)).
We are interested in standing waves of Equation (4), i.e., its solutions of the form
ψt=e−iωtU(x)
where
ω∈R
and
U(x1
, ...,
x4) = (u1(x1)
,
. . .
,
u4(x4))
is a purely spatial function on
G
, which may
also depend on
ω
. Such a problem has already been considered in [
11
,
12
,
15
,
16
]. In particular, in [
11
,
12
],
variational methods are used to show, among many other things, that Equation (4) has no ground
state, i.e., no standing wave exists that minimizes the energy at fixed
L2
-norm. In a recent paper [
16
],
information on positive bound states that are not ground states is given. The special example of tadpole
graph (a ring with a single half-line) is treated in detail in [17,18].
As for the results in [
15
], they can be summarized as follows. Writing the problem of standing
waves of Equation (4) component-wise, we get the following scalar problem:









−u00
j−u3
j=ωuj,uj∈H2(Ij)
u1(0) = u2(L) = u3(0),u1(L1) = u2(L1) = u4(0)
u0
1(0)−u0
2(L) + u0
3(0) = 0, u0
1(L1)−u0
2(L1)−u0
4(0) = 0.
(5)
Such a system has solutions with
u3=u4=
0 if and only if the ratio
L1/L2
is rational. In this
case, they form a sequence of continuous branches in the
(ω,kUkL2)
plane, bifurcating from the
linear eigenvectors of the Schrödinger operator
HG
(see Figure 2), and they are periodic on the ring
of
G
, that is,
u1
and
u2
are restrictions to
I1
and
I2
of a function
u
belonging to the second Sobolev
space of periodic functions
H2
per([
0,
L]) = u∈H2([0, L]) :u(0) = u(L),u0(0) = u0(L)
. In particular,
such function
u
is a rescaled Jacobi cnoidal function (see, e.g., [
19
,
20
] for a treatise on the Jacobian
elliptic functions). If
ω≥
0, no other nonzero standing waves exist, since the NLS on the unbounded
edges has no nontrivial solution. If
ω<
0, instead, the NLS on the half lines has soliton solutions,
so that standing waves with nonzero
u3
and
u4
are admissible. The general study of this kind of
solutions leads to a rather complicated system of equations, since, while
u3
and
u4
must be shifted
solitons, each of
u1
and
u2
can be (at least in principle) a cnoidal function, a dnoidal function or a
shifted soliton. To limit this complexity, the analysis in [
15
] is focused on the special case of standing

Symmetry 2019,11, 161 3 of 20
waves that are non-vanishing on the half lines but share the above-mentioned periodicity feature with
the bifurcation solutions. This amounts to study the following system:


−u00 −u3=ωu,u∈H2
per([0, L]),ω<0
u(0) = ±u(L1) = p2|ω|(6)
where the sign
±
distinguishes the cases of
u3
and
u4
with the same sign (which we may assume
positive, thanks to the odd parity of the equation) or with different signs. In [15], it is shown that:
(i)
If
L1/L2∈Q
, then the set of solutions to (6) is made up of a sequence of secondary
bifurcation branches
{(ω,e
un,ω):ω<0}n≥1
, originating at
ω=
0 from each of the previous
ones, together with a sequence {(ωn,un)}n≥1not lying on any branch (see Figure 2).
Figure 2. Bifurcation diagram for L1/L=p/qwith p,q∈Ncoprime.
(ii)
If
L1/L2/∈Q
, then the set of solutions to (6) reduces to two sequences
{(ω+
n,u+
n)}n≥1
and
{(ω−
n,u−
n)}n≥1
alone, solving the problem in Equation (6) with sign
±
, respectively, where the
frequency sequences
{ω±
n}n≥1
are unbounded below and have at least a finite nonzero cluster
point (see Figure 3). The functions u±
noscillate ntimes on the ring of the graph.
These results come rather unexpectedly, so the aim of this paper is to pursue the study begun
in [
15
] by deepening the understanding of such results in relation to the underlying physical model.
In particular, we ask the following questions: Does Equation (4) admit standing waves that are
non-periodic on ring of
G
? If so, do they form continuous branches to which the isolated periodic
solutions belong?

Symmetry 2019,11, 161 4 of 20
Figure 3.
The appearance of each of the sequences
{(ω+
n
,
u+
n)}n∈N
and
{(ω−
n
,
u−
n)}n∈N
for
L1/L∈R\Q.
With a view to especially answer the second question, we look for standing waves which include
the ones given by Equation (6) but still change sign on the bounded edges. More precisely, we look for
solutions to Equation (5) exhibiting the following features:
•u1,u2are sign-changing.
•u3,u4are nonzero.
The second feature implies ω<0 and
uj(x) = ±√2ηsech ηx+aj,aj∈R,j=3, 4 (7)
where we set η:=p|ω|for brevity. Then, the first feature implies
uj(x) = ηv
u
u
t2k2
j
2k2
j−1cn 
η
q2k2
j−1x+aj;kj
,kj∈1
√2, 1,aj∈0, Tj,j=1, 2 (8)
where
cn (·;k)
is the cnoidal function of parameter
k
and
Tj=Tjkj,η:=Skj/η
is the period of
the function cn η(·)/q2k2
j−1; kj. Here and in the rest of the paper, Sdenotes the function
S(k):=4p2k2−1K(k) = 4p2k2−1Z1
0
dt
q(1−t2)(1−k2t2)
, (9)
where
K(k)
is the so called complete elliptic integral of first kind. Notice that
S:(
1
/√2
, 1
)→R
is
strictly increasing, continuous and such that S(1/√2, 1)=(0, +∞).
Therefore, restricting ourselves for simplicity to the case with
u3
and
u4
of the same sign, which we
may assume positive thanks to the odd parity of the system in Equation (5), we are led to study
the existence of solutions
η>
0,
k1
,
k2∈1
√2, 1
,
a1∈[0, T1)
,
a2∈[0, T2)
,
a3
,
a4∈R
to the
following system:

Symmetry 2019,11, 161 5 of 20







































k1
√2k2
1−1cn ηa1
√2k2
1−1;k1=k2
√2k2
2−1cn η(L+a2)
√2k2
2−1;k2=sech (ηa3)
k1
√2k2
1−1cn η(L1+a1)
√2k2
1−1;k1=k2
√2k2
2−1cn η(L1+a2)
√2k2
2−1;k2=sech (ηa4)
tanh (ηa3)sech (ηa3)=
=−k1
2k2
1−1sn ηa1
√2k2
1−1;k1dn ηa1
√2k2
1−1;k1+k2
2k2
2−1sn η(L+a2)
√2k2
2−1;k2dn η(L+a2)
√2k2
2−1;k2
tanh (ηa4)sech (ηa4)=
=k1
2k2
1−1sn η(L1+a1)
√2k2
1−1;k1dn η(L1+a1)
√2k2
1−1;k1−k2
2k2
2−1sn η(L1+a2)
√2k2
2−1;k2dn η(L1+a2)
√2k2
2−1;k2.
(10)
This set of equations turns out to be still rather difficult to study in his full generality, and indeed
we have results only in the subcase where the two solitons in Equation (7) have the same height at
the vertices, i.e.,
sech (ηa3)=sech (ηa4)
(which corresponds to
θ1=θ2
in Section 2). More precisely,
in Section 2we reduce the system in Equation (10) to an equivalent one, which naturally splits into
different cases. Then, we study three of such cases, all with
sech (ηa3)=sech (ηa4)
, leading to our
existence results, which are the following three theorems.
The first two results only concern the case of irrational ratios
L1/L2
and give solutions with
k16=k2, i.e., non-periodic on the ring of the graph.
Theorem 1.
Assume that
L1/L2∈R\Q
. Then, there exists a sequence of positive integers
(nh)h∈N
such
that for every
ω<−
32
K(
1
/√2)2/(L1L2)
there exists
hω∈N
(also depending on
L1
and
L2
) such that for
all
h>hω
the problem in Equation (5) has two solutions
(u+
1,h
,
u+
2,h
,
u+
3,h
,
u+
4,h)
and
(u−
1,h
,
u−
2,h
,
u−
3,h
,
u−
4,h)
of
the form:
u±
j,h(x) = v
u
u
t2|ω|k2
j,h
2k2
j,h−1cn s|ω|
2k2
j,h−1x+a±
j,h;kj,h!,j=1, 2 (11)
u±
j,h(x) = q2|ω|sech q|ω|x+a±
j,h,j=3, 4 (12)
where u±
1,h(x)and u±
2,h(x)have periods T1,h=L1/[nhL1/L2+1]and T2,h=L2/nh, and for all h one has
1
√2
<k1,h<k2,h<1, a±
1,h∈0, T1,h
4,a±
2,h∈[0, T2,h),a±
3,h<0, a±
4,h>0, a+
j,h6=a−
j,h. (13)
Remark 1. More precisely, according to the proof, in Theorem 1, we have that
k1,h=S−1L1
[nhL1/L2+1]q|ω|,a±
1,h=γ1(k1,h,ω,θ±
h),
k2,h=S−1L2
nhq|ω|,a±
2,h=γ2(k2,h,ω,θ±
h)−L+pT2,h,−a±
3,h=a±
4,h=sech−1
|[0,+∞)(θ±
h),
where
p
is the unique positive integer such that
a±
2,h∈[0, T2,h)
,
θ±
h
are the two distinct solutions in
(0, 1]
of the
equation
θ21−θ2=tk1,h,k2,h
with
tk1,h,k2,h
given by Equation (17), and
γj(kj,h
,
ω
,
θ±
h)
is the unique preimage
in 0, Tj,h/4of θ±
hq2k2
j,h−1/kj,hby the function cn (·)p|ω|/q2k2
j,h−1; kj,h.
Theorem 2.
Assume that
L1/L2∈R\Q
. Then, there exists a sequence of positive integers
(nh)h∈N
such that for every
ω<−
32
K(
1
/√2)2/(L1L2)
there exists
hω∈N
(also depending on
L1
and
L2
)
such that for all
h>hω
the problem in Equation (5) has two solutions
(u±
1,h
,
u±
2,h
,
u±
3,h
,
u±
4,h)
of the form
of Equations (11)–(12), where
u±
1,h(x)
and
u±
2,h(x)
have periods
T1,h=L1/[nhL1/L2]
and
T2,h=L2/nh
,
the parameters a±
1,h,a±
2,h,a±
3,h,a±
4,hare as in Equation (13) and for all h one has

Symmetry 2019,11, 161 6 of 20
1
√2
<k2,h<k1,h<1.
Remark 2. More precisely, in Theorem 2we have that
k1,h=S−1L1
[nhL1/L2]q|ω|and k2,h=S−1L2
nhq|ω|,
whereas a±
j,hare exactly as in Remark 1.
The third result does not need
L1/L2
irrational and concerns the subcase of the system in
Equation (5) which, if
L1/L2∈R\Q
and
k1=k2
, is exactly the system in Equation (6) with plus sign
(see Remark 5).
Theorem 3.
Let
m
,
n∈N
be such that
n>m≥
1. Then, there exists
ωm,n<
0(also depending on
L1
) such
that for all
ω<ωm,n
the problem in Equation (5) has a solution
(u1,u2,u3,u4)
of the form of Equations (7)–(8),
with k1,k2∈√3/2, 1, a1∈(0, T1/4), a2∈[0, T2).
Remark 3.
According to the proof, in Theorem 3,
a1
,
a2
,
a3
,
a4
can be described in a similar way of
Theorems 1and 2
. On the contrary, the parameters
k1
,
k2
do exist, but are not explicit as in the previous theorems.
As already mentioned, Theorems 1–3do not exhaust the study of solutions to the problem in
Equation (5), and thus of standing waves of (NLS), as they only concern the case of solitons having
the same height at the vertices. In addition, they do not describe the whole family of this kind of
solutions, but only give existence results. However, they still provide some answer to the questions
raised above. Indeed, Theorems 1and 2answer in the affirmative to the first question, as they prove
existence of standing waves which are non-periodic on the ring of
G
. As to Theorem 3, for any
m
and
n
, it provides a family of solutions which depend on the continuous parameter
ω∈(−∞
,
ωm,n)
and, roughly speaking, make
m
oscillations on the edge of length
L1
and
n−m
oscillations on the
one of length
L2
(cf. the second and third equations of the system in Equation (33)). If
L1/L2
is
irrational and one of these families contain a solution with
k1=k2
, then such a solution is one of
the isolated solutions found in [
15
] in the irrational case and we can answer affirmatively also to the
second question. Unfortunately, the argument we used in proving Theorem 3does not allow us to
say wether we find solutions with
k1=k2
or not, and therefore we do not have a final answer to the
second question.
2. Preliminaries
In this section, we reduce the system in Equation (10) to a simpler equivalent one, which is the
system in Equation (14) with the last two equations replaced by the system in Equation (19).
For brevity, we set
X1=ηa1
q2k2
1−1
,X2=η(L+a2)
q2k2
2−1
,X3=η(L1+a1)
q2k2
1−1
,X4=η(L1+a2)
q2k2
2−1
,
and
σ1=sgn [sn (X1;k1)] ,σ2=sgn [sn (X2;k2)] ,σ3=sgn [sn (X3;k1)] ,σ4=sgn [sn (X4;k2)] .
Then, using well known identities (see [
20
]) and the first equation of the system in Equation (10),
we get

Symmetry 2019,11, 161 7 of 20
sn (X1;k1)=σ1q1−cn2(X1;k1)=σ1s1−2k2
1−1
k2
1
sech2(ηa3),
dn (X1;k1)=q1−k2
1+k2
1cn2(X1;k1)=q1−k2
1+2k2
1−1sech2(ηa3)
and hence
k1
2k2
1−1sn (X1;k1)dn (X1;k1)=σ1sk2
1
2k2
1−1−sech2(ηa3)s1−k2
1
2k2
1−1+sech2(ηa3)
=σ1v
u
u
tk2
11−k2
1
2k2
1−12+sech2(ηa3)−sech4(ηa3).
Arguing similarly for the products
sn (X2;k2)dn (X2;k2)
,
sn (X3;k1)dn (X3;k1)
and
sn (X4;k2)dn (X4;k2), and defining
c(k):=k21−k2
(2k2−1)2,
we thus obtain that the system in Equation (10) is equivalent to

























k1
√2k2
1−1cn ηa1
√2k2
1−1;k1=k2
√2k2
2−1cn η(L+a2)
√2k2
2−1;k2=sech (ηa3)
k1
√2k2
1−1cn η(L1+a1)
√2k2
1−1;k1=k2
√2k2
2−1cn η(L1+a2)
√2k2
2−1;k2=sech (ηa4)
tanh (ηa3)sech (ηa3)=−σ1qc(k1)+sech2(ηa3)−sech4(ηa3)+σ2qc(k2)+sech2(ηa3)−sech4(ηa3)
tanh (ηa4)sech (ηa4)=σ3qc(k1)+sech2(ηa4)−sech4(ηa4)−σ4qc(k2)+sech2(ηa4)−sech4(ηa4).
(14)
Let us now focus on the last two equations. Setting
θ1=sech (ηa3),θ2=sech (ηa4),σ5=sgn (a3)=sgn (tanh (ηa3)) ,σ6=sgn (a4)=sgn (tanh (ηa4))
the couple of such equations is equivalent to





σ5q1−θ2
1θ1=−σ1qc(k1)+θ2
11−θ2
1+σ2qc(k2)+θ2
11−θ2
1
σ6q1−θ2
2θ2=σ3qc(k1)+θ2
21−θ2
2−σ4qc(k2)+θ2
21−θ2
2.
(15)
Squaring the equations, we get
c(k1)+θ2
11−θ2
1+c(k2)−2σ1σ2qc(k1)+θ2
11−θ2
1qc(k2)+θ2
11−θ2
1=0,
c(k1)+θ2
21−θ2
2+c(k2)−2σ3σ4qc(k1)+θ2
21−θ2
2qc(k2)+θ2
21−θ2
2=0,
which are impossible if
σ1σ2=−
1 or
σ3σ4=−
1. Hence, we can add the conditions
σ1=σ2
and
σ3=σ4to the system in Equation (15), and get











σ5q1−θ2
1θ1=σ1−qc(k1)+θ2
11−θ2
1+qc(k2)+θ2
11−θ2
1
σ6q1−θ2
2θ2=σ3qc(k1)+θ2
21−θ2
2−qc(k2)+θ2
21−θ2
2
σ2=σ1,σ4=σ3.
(16)

Symmetry 2019,11, 161 8 of 20
Moreover, both θ2
11−θ2
1and θ2
21−θ2
2must be solutions t∈[0, 1/4]of the equation
c(k1)+c(k2)+t−2qc(k1)+tqc(k2)+t=0.
Such equation has the unique nonnegative solution
t=tk1,k2=1
32qc(k1)2−c(k1)c(k2)+c(k2)2−c(k1)−c(k2), (17)
which belongs to [0, 1/4]if and only if (k1,k2)belongs to the set
A=((k1,k2)∈1
√2, 12
: 2qc(k1)2−c(k1)c(k2)+c(k2)2−c(k1)−c(k2)≤3
4),
i.e., as one can easily see after some computations,
A=

(k1,k2)∈R:k1∈1
√2, 1,q4k2
1−1
2k1≤k2≤1
2q1−k2
1
,k2<1


(the set Ais portrayed in Figure 4).
0.75 0.80 0.85 0.90 0.95 1.00
0.75
0.80
0.85
0.90
0.95
1.00
Figure 4. The set A. The point (√2/2, √2/2)and the straight lines of the boundary are not included.
In this case, the equation θ21−θ2=tk1,k2with θ∈(0, 1]has two distinct solutions
θ±
k1,k2=s1±p1−4tk1,k2
2(18)
if
tk1,k2∈(0, 1/4)
, two coincident solutions
θ+
k1,k2=θ−
k1,k2=
1
/√2
if
tk1,k2=
1
/
4, and a unique solution
θ+
k1,k2=
1 if
tk1,k2=
0 (i.e.,
k1=k2
). In this latter case, we still write
θ+
k1,k2=θ−
k1,k2=
1 for future
convenience. We also observe that the function
c(k)
is positive and strictly decreasing from
1/√2, 1
onto
(0, +∞)
, so that the terms within brackets on the right hand sides of the first two equations of
Equation (16) have a fixed sign according as
k1<k2
or
k1>k2
. Therefore, the system in Equation (15)
turns out to be equivalent to

Symmetry 2019,11, 161 9 of 20

























(k1,k2)∈A,θ1,θ2∈nθ+
k1,k2,θ−
k1,k2o
sech (ηa3)=θ1, sech (ηa4)=θ2





k1<k2
σ5=−σ1
σ6=σ3∨




k1>k2
σ5=σ1
σ6=−σ3∨(k1=k2
a3=a4=0
σ2=σ1,σ4=σ3.
(19)
As a conclusion, Equation (10) is equivalent to the system in Equation (14) with the last two
equations replaced by the system in Equation (19).
3. Case θ1=θ2,σ1=σ3and k1<k2. Proof of Theorem 1
We focus on the case
σ1=σ3=
1, which gives Theorem 1, leaving the analogous case
σ1=σ3=−1 to the interested reader. In such a case, condition (k1,k2)∈Abecomes
(k1,k2)∈A0=A∩{(k1,k2)∈R:k1<k2}=

(k1,k2)∈R:1
√2
<k1<k2≤1
2q1−k2
1
,k2<1


and, taking into account the equivalence between Equation (15) and Equation (19), the system in
Equation (14) becomes:











































(k1,k2)∈A0,θ∈nθ+
k1,k2,θ−
k1,k2o
sech (ηa3)=sech (ηa4)=θ,a3<0, a4>0
k1
√2k2
1−1cn ηa1
√2k2
1−1;k1=k2
√2k2
2−1cn η(L+a2)
√2k2
2−1;k2=θ
k1
√2k2
1−1cn η(L1+a1)
√2k2
1−1;k1=k2
√2k2
2−1cn η(L1+a2)
√2k2
2−1;k2=θ
sn ηa1
√2k2
1−1;k1>0, sn η(L1+a1)
√2k2
1−1;k1>0
sn η(L+a2)
√2k2
2−1;k2>0, sn η(L1+a2)
√2k2
2−1;k2>0.
(20)
We denote by
γj=γjkj,η,θ
the unique preimage in
0, Tj/4
of the value
q2k2
j−1
kjθ
by
the function cn η
q2k2
j−1(·);kj!. Then,









k1
√2k2
1−1cn ηa1
√2k2
1−1;k1=θ, sn ηa1
√2k2
1−1;k1>0
k1
√2k2
1−1cn η(L1+a1)
√2k2
1−1;k1=θ, sn η(L1+a1)
√2k2
1−1;k1>0
means
(a1=γ1
L1+a1=γ1+mT1for some m≥1, i.e., (a1=γ1
L1=mT1for some m≥1

Symmetry 2019,11, 161 10 of 20
while 








k2
√2k2
2−1cn η(L+a2)
√2k2
2−1;k2=θ, sn η(L+a2)
√2k2
2−1;k2>0
k2
√2k2
2−1cn η(L1+a2)
√2k2
2−1;k2=θ, sn η(L1+a2)
√2k2
2−1;k2>0
means
(L+a2=γ2+pT2for some p≥0
L1+a2=γ2+qT2for some 0 ≤q<p,i.e., (L+a2=γ2+pT2for some p≥0
L2=(p−q)T2for some 0 ≤q<p.
Hence, the system in Equation (20) becomes





























(k1,k2)∈A0,θ∈nθ+
k1,k2,θ−
k1,k2o
sech (ηa3)=sech (ηa4)=θ,a3<0, a4>0
L1=mT1(k1,η)for some m≥1
L2=nT2(k2,η)for some n≥1
a1=γ1(k1,η,θ)
a2=γ2(k2,η,θ)+pT2(k2,η)−Lfor some p≥n
(21)
(observe that θdepends on both k1and k2, and so do a1and a2according to the last two equations).
Remark 4.
The equivalence between the systems in Equation (20) and Equation (21) does not need assumption
k1<k2
. On the other hand, if
k1=k2
, then
T1(k1,η)=T2(k2,η)
and thus the third and fourth equations of
the system in Equation (21) imply
L1/L2∈Q
. This means that solutions to the system in Equation (10) with
k1=k2(which implies θ1=θ2=1) and σ1=σ3cannot exist if the ratio L1/L2is not rational.
Let us now focus on the following group of equations:







(k1,k2)∈A0
L1=mT1(k1,η), for some m≥1
L2=nT2(k2,η), for some n≥1.
(22)
Recalling that Tjkj,η=Skj/η, this system is equivalent to













1
√2<k1<k2≤1
2√1−k2
1
,k2<1
k1=S−1ηL1
mfor some m≥1
k2=S−1ηL2
nfor some n≥1.
(23)
and therefore, recalling that
S
is strictly increasing and continuous from
(
1
/√2
, 1
)
onto
(0, +∞)
,
we can obtain solutions by fixing η>0 and finding n,m≥1 such that







S−1ηL1
m<S−1ηL2
n
S−1ηL2
n≤1
2r1−hS−1ηL1
mi2,i.e., 








L1
m<L2
n
ηL2
n≤S
1
2r1−hS−1ηL1
mi2
.(24)

Symmetry 2019,11, 161 11 of 20
Lemma 1. One has
S
1
2q1−[S−1(t)]2
=t+1
32K2
0
t3+ot3as t →0+
(where, we recall, K0=K1/√2).
Proof. We have
lim
t→0+
S−1(t)−1
√2−t2
32K2
0√2
t4=lim
k→(1/√2)+
S−1(S(k)) −1
√2−S(k)2
32K2
0√2
S(k)4
=lim
k→(1/√2)+
k−1
√2−16K(k)2(2k2−1)
32K2
0√2
28K(k)4(2k2−1)2
=1
210K2
0
lim
k→(1/√2)+
2K2
0−K(k)2√2k+1
K(k)4√2k+12k−1/√2
where, setting K0
0=K01/√2, by De L’Hôpital’s rule, we get
lim
k→(1/√2)+
2K2
0−K(k)2√2k+1
k−1/√2=−4K0K0
0−K2
0√2.
Hence, we conclude
lim
t→0+
S−1(t)−1
√2−t2
32K2
0√2
t4=−K0+2√2K0
0
211√2K5
0
,
i.e.,
S−1(t)=1
√2+c1t2−c2t4+ot4as t→0+(25)
where c1=1
32√2K2
0
and c2=K0+2√2K0
0
211√2K5
0
. This implies
1
2q1−S−1(t)2=1
2r1
2−2
√2c1t2−c2
1−√2c2t4+o(t4)
=1
√2r1−2√2c1t2−2c2
1−√2c2t4+o(t4)
=1
√2+c1t2+2√2c2
1−c2t4+ot4.
Using De L’Hôpital’s rule again, we now compute
lim
k→(1/√2)+
S(k)−211/4K0k−1/√21/2
k−1/√23/2 =lim
k→(1/√2)+
S0(k)−27/4K0k−1/√2−1/2
3
2k−1/√21/2

Symmetry 2019,11, 161 12 of 20
=2
3lim
k→(1/√2)+
8k
√2k2−1K(k)+4√2k2−1K0(k)−27/4K0
(k−1/√2)1/2
k−1/√21/2
=215/4
3K0
0+2
3lim
k→(1/√2)+
8kK(k)
4
√2√√2k+1−27/4K0
k−1/√2
=215/4
3K0
0+2
3lim
k→(1/√2)+
8k(K(k)−K0)
4
√2√√2k+1+8k
4
√2√√2k+1−27/4K0
k−1/√2=25/4 K0+211/4 K0
0
where the result follows because K(k)−K0∼K0
0k−1/√2as k→1/√2+and
8k
4
√2p√2k+1−27/4 =27/4 2k−p√2k+1
p√2k+1=27/4 4k2−√2k−1
p√2k+12k+p√2k+1
=27/4 4k+√2k−1/√2
p√2k+12k+p√2k+1.
This means
S(k)=211/4K0k−1/√21/2 +25/4 K0+211/4 K0
0k−1/√23/2 +ok−1/√23/2 (26)
as k→1/√2+and therefore we deduce that as t→0+one has (note that 211/4 K0√c1=1)
S
1
2q1−S−1(t)2
=211/4K0√c1t 1+2√2c2
1−c2
c1
t2+ot2!1/2
+
+25/4K0+211/4K0
0c1√c1t3 1+2√2c2
1−c2
c1
t2+ot2!3/2
+ot3
=t 1+1
2
2√2c2
1−c2
c1
t2+ot2!+
+25/4K0+211/4K0
0c1√c1t3 1+3
2
2√2c2
1−c2
c1
t2+ot2!+ot3
=t+ 211/4K0√c1
1
2
2√2c2
1−c2
c1
+25/4K0+211/4K0
0c1√c1!t3+ot3.
Simplifying the coefficient of t3, this gives the result.
Thanks to Lemma 1, the system in Equation (24) becomes
0<m
n−L1
L2≤L3
1η2
32K2
0L2
1
m2+ζm(27)
where
(ζm)m
is a suitable sequence (also dependent on
L1
,
L2
,
η
) such that
ζm=om−2
as
m→∞
.
Notice that, according to systems (23) and (24), the equality sign in the second inequality amounts to
k2=1
2√1−k2
1
.

Symmetry 2019,11, 161 13 of 20
Proof of Theorem 1.
Since
L1/L2∈R\Q
, by ([
21
], Corollary 1.9) there exist infinitely many rational
numbers m/nsuch that
0<m
n−L1
L2
<1
n2. (28)
This implies
nL1/L2<m<nL1/L2+
1 and thus
m=[nL1/L2+1]
. Since the denominators
of such rationals
m/n
must be infinite, we may arrange them in a diverging sequence
(nh)⊂N
;
accordingly, the corresponding numerators are
mh=[nhL1/L2+1]
. Now, let
η>
4
√2K0(L1L2)−1/2
and fix ε>0 such that
η2>L1
L2
+ε232K2
0L2
L3
1
.
Since Equation (28) implies that
mh/nh→L1/L2
as
h→∞
, for
h
large enough, we have that
mh/nh<L1/L2+ε, so that
1
n2
h
<L1
L2
+ε21
m2
h
<L3
1η2
32K2
0L2
1
m2
h
.
Hence, up to further enlarging h, Equation (28) gives
0<mh
nh−L1
L2
<L1
L2
+ε21
m2
h
<L3
1η2
32K2
0L2
1
m2
h
+ζmh, (29)
so that
nh
and
mh
satisfy Equation (27). For every
h
, this provides solutions to the system in
Equation (22) by taking
k1=k1,h=S−1(ηL1/mh)
and
k2=k2,h=S−1(ηL2/nh)
, and thus solutions
to the system in Equation (21) by choosing
θ=θh∈ {θ+
k1,h,k2,h
,
θ−
k1,h,k2,h}
, taking
p
as the unique integer
such that
0≤γ2(k2,h,η,θh)+pT2(k2,h,η)−L<T2(k2,h,η)
(where
T2(k2,h,η)=L2/nh
), which turns out to be greater than or equal to
nh
, and defining
a1
,
a2
,
a3
,
a4
according to the second, fifth and sixth equation of the system. Note that
θ+
k1,h,k2,h
and
θ−
k1,h,k2,h
are
different for all
h
, since
tk1,h,k2,h6=
0 (because
k1,h6=k2,h
) and
tk1,h,k2,h6=
1
/
4 (because of the strict
inequality signs in Equation (29)). Up to discarding a finite number of terms of the sequence
(nh)
,
the proof is complete.
4. Case θ1=θ2,σ1=σ3and k1>k2. Proof of Theorem 2
As in the previous section, we focus on the case
σ1=σ3=
1. In this case, the system
in Equation (14) becomes again the system in Equation (21), but with
(k1,k2)∈A0
replaced by
(k1,k2)∈A00, where
A00 =A∩{(k1,k2)∈R:k1>k2}=

(k1,k2)∈R:q4k2
1−1
2k1≤k2<k1<1

.
Then, Equation (22) is now equivalent to the system













q1−1
4k2
1≤k2<k1<1
k1=S−1ηL1
mfor some m≥1
k2=S−1ηL2
nfor some n≥1,

Symmetry 2019,11, 161 14 of 20
i.e., 














L2
n<L1
m
ηL2
n≥S r1−1
4S−1ηL1
m2!
k1=S−1ηL1
m,k2=S−1ηL2
n
(30)
with η>0 and n,m∈N.
Lemma 2. One has
S s1−1
4S−1(t)2!=t−1
32K2
0
t3+ot3as t →0+
(where, we recall, K0=K1/√2).
Proof. Since S−1(t)=1
√2+c1t2−c2t4+ot4as t→0+(see Equation (25)), we have
1−1
2S−1(t)2=1−1
2S−1(t)−1/√2+1/√22
=1−1
2
1
S−1(t)−1/√22+1/2 +2S−1(t)−1/√2/√2
=1−1
2
1
(c1t2−c2t4+o(t4))2+1/2 +2(c1t2−c2t4+o(t4)) /√2
=1−1
1+2c1√2t2+2c2
1−c2√2t4+o(t4)
=2c1√2t2−23c2
1+c2√2t4+ot4
and therefore
s1−1
4S−1(t)2=1
√2v
u
u
t1+ 1−1
2S−1(t)2!
=1
√2
1+1
2 1−1
2S−1(t)2!−1
8 1−1
2S−1(t)2!2
+o
 1−1
2S−1(t)2!2


=1
√2+c1t2−2√2c2
1+c2t4+ot4.

Symmetry 2019,11, 161 15 of 20
Hence, using the expansion in Equation (25), we deduce that
S s1−1
4S−1(t)2!=211/4K0√c1t 1−2√2c2
1+c2
c1
t2+ot2!1/2
+
+25/4K0+211/4K0
0c1√c1t3 1−2√2c2
1+c2
c1
t2+ot2!3/2
+ot3
=t 1−1
2
2√2c2
1+c2
c1
t2+ot2!+
+25/4K0+211/4K0
0c1√c1t3 1−3
2
2√2c2
1+c2
c1
t2+ot2!+ot3
=t+ 25/4K0+211/4K0
0c1√c1−211/4K0√c1
1
2
2√2c2
1+c2
c1!t3+ot3.
Simplifying the coefficient of t3, the result ensues.
By Lemma 2, the first two conditions of the system in Equation (24) become
0>m
n−L1
L2≥ − L3
1η2
32K2
0L2
1
m2+ζm
where
(ζm)m
is a suitable sequence such that
ζm=om−2
as
m→∞
. Notice that the equality sign in
the second inequality amounts to k2=√4k2
1−1
2k1.
Proof of Theorem 2.
Since
L1/L2∈R\Q
, by ([
21
], Corollary 1.9) there exist infinitely many rational
numbers m/nsuch that
0>m
n−L1
L2
>−1
n2.
This implies
nL1/L2−
1
<m<nL1/L2
and thus
m=[nL1/L2]
. Proceeding exactly as in the
proof of Theorem 1, the result follows.
5. Case θ1=θ2and σ1=−σ3. Proof of Theorem 3
We focus on the case
θ1=θ2=θ+
k1,k2
and
σ1=−σ3=
1, which gives Theorem 3, leaving the
analogous cases
θ1=θ2=θ−
k1,k2
or
σ1=−σ3=−
1 to the interested reader. In such a case, the system in
Equation (14) becomes

































(k1,k2)∈A
k1
√2k2
1−1cn ηa1
√2k2
1−1;k1=k2
√2k2
2−1cn η(L+a2)
√2k2
2−1;k2=sech (ηa3)=θ+
k1,k2
k1
√2k2
1−1cn η(L1+a1)
√2k2
1−1;k1=k2
√2k2
2−1cn η(L1+a2)
√2k2
2−1;k2=sech (ηa4)=θ+
k1,k2
σ2=−σ4=1
(k1<k2
σ5=σ6=−1∨(k1>k2
σ5=σ6=1∨(k1=k2
a3=a4=0

Symmetry 2019,11, 161 16 of 20
that is 














































(k1,k2)∈A
k1
√2k2
1−1cn ηa1
√2k2
1−1;k1=k2
√2k2
2−1cn η(L+a2)
√2k2
2−1;k2=sech (ηa3)=θ+
k1,k2
k1
√2k2
1−1cn η(L1+a1)
√2k2
1−1;k1=k2
√2k2
2−1cn η(L1+a2)
√2k2
2−1;k2=sech (ηa4)=θ+
k1,k2
sn ηa1
√2k2
1−1;k1>0, sn η(L1+a1)
√2k2
1−1;k1<0
sn η(L+a2)
√2k2
2−1;k2>0, sn η(L1+a2)
√2k2
2−1;k2<0
(k1<k2
σ5=σ6=−1∨(k1>k2
σ5=σ6=1∨(k1=k2
a3=a4=0
(31)
Defining γjkj,η,θas in Section 3, we have that









k1
√2k2
1−1cn ηa1
√2k2
1−1;k1=θ+
k1,k2, sn ηa1
√2k2
1−1;k1>0
k1
√2k2
1−1cn η(L1+a1)
√2k2
1−1;k1=θ+
k1,k2, sn η(L1+a1)
√2k2
1−1;k1<0
means 


a1=γ1k1,η,θ+
k1,k2
L1=mT1(k1,η)−2γ1k1,η,θ+
k1,k2for some m≥1(32)
and 








k2
√2k2
2−1cn η(L+a2)
√2k2
2−1;k2=θ+
k1,k2, sn η(L+a2)
√2k2
2−1;k2>0
k2
√2k2
2−1cn η(L1+a2)
√2k2
2−1;k2=θ+
k1,k2, sn η(L1+a2)
√2k2
2−1;k2<0
means 


L2=(n−m)T2(k2,η)+2γ2k2,η,θ+
k1,k2for some n≥m
a2=γ2k2,η,θ+
k1,k2−L+pT2(k2,η)for some p≥n−m+1
where
m
is the same integer of the system in Equation (32). Hence, the system in Equation (31)
amounts to













































(k1,k2)∈A
L1=mT1(k1,η)−2γ1k1,η,θ+
k1,k2for some m≥1
L2=(n−m)T2(k2,η)+2γ2k2,η,θ+
k1,k2for some n≥m
a1=γ1k1,η,θ+
k1,k2
a2=γ2k2,η,θ+
k1,k2−L+pT2(k2,η)for some p≥n−m+1
sech (ηa3)=sech (ηa4)=θ+
k1,k2
(k1<k2
a3,a4<0∨(k1>k2
a3,a4>0∨(k1=k2
a3=a4=0.
(33)
Remark 5.
Suppose
L1/L2/∈Q
. If we assume
k1=k2
in the system in Equation (14), then we have
θ1=θ2=
1and
σ1=−σ3
(see Remark 4). Hence, a solution to the problem in. Equation (6) with plus

Symmetry 2019,11, 161 17 of 20
sign gives rise to a solution to the system in Equation (33). On the other hand, a solution to the system in
Equation (33) with
k1=k2
is such that
L=L1+L2=nT
and
a2=a1−L+pT =a1+ (p−n)T
,
where
T=T1(k1,η)=T2(k2,η)
,
a1∈(
0,
T/
4
)
and
a2∈[
0,
T)
. This forces
p=n
and thus
a1=a2
, so that
the corresponding solution to the problem in Equation (6) is periodic on the circle.
Now, recall that Tjkj,η:=Skj
η. By the definition of γj=γjkj,η,θ+
k1,k2, one has
cn 
η
q2k2
j−1
γj;kj
=q2k2
j−1
kj
θ+
k1,k2(34)
with γj∈0, Tj/4. This implies
0<η
q2k2
j−1
γj<η
q2k2
j−1
Skj
4η=Skj
4q2k2
j−1
=Kkj
and therefore Equation (34) yields that
γjkj,η,θ+
k1,k2=q2k2
j−1
ηarccn 
q2k2
j−1
kj
θ+
k1,k2;kj
.
Hence, defining
γ(k1,k2):=q2k2
1−1 arccn 
q2k2
1−1
k1
θ+
k1,k2;k1
=q2k2
1−1Z1
√2k2
1−1
k1θ+
k1,k2
dt
q(1−t2)1−k2
1(1−t2)
and observing that θ+
k1,k2=θ+
k2,k1, one has
γ1k1,η,θ+
k1,k2=1
ηγ(k1,k2)and γ2k2,η,θ+
k1,k2=1
ηγ(k2,k1).
Thus, the first three equations of the system in Equation (33) are equivalent to









(k1,k2)∈A
ηL1=mS (k1)−2γ(k1,k2)for some m≥1
ηL2=(n−m)S(k2)+2γ(k2,k1)for some n≥m.
(35)
To prove Theorem 3, we use the following lemma, concerning the existence of a globally defined
implicit function. Its proof is classical, so we leave it to the interested reader.
Lemma 3.
Let
bi∈R
for
i=
1, ..., 4 and let
G:(b1
,
b2)×(b3
,
b4)→R
be a continuous function such that
for all x ∈(b1,b2)the following properties hold:
•the mapping G(x,·)is strictly increasing on (b3,b4);
•lim
y→b+
3
G(x,y)<0and lim
y→b−
4
G(x,y)>0.
Then, the set of solutions to the equation
G(x
,
y) =
0is the graph of a continuous function
g:(b1
,
b2)→
(b3,b4).

Symmetry 2019,11, 161 18 of 20
Proof of Theorem 3.Let n>m≥1 and for (k1,k2)∈Adefine the continuous functions
Fm(k1,k2):=mS (k1)−2γ(k1,k2)and Fm,n(k1,k2):=(n−m)S(k2)+2γ(k2,k1).
We also define
Fm
and
Fm,n
on the segments
n(k1, 1):√3/2 ≤k1<1o
and
n(1, k2):√3/2 ≤k2<1o
of the boundary of
A
, respectively, where the above definitions
also make sense.
Fix
√3/
2
<λ<
1 such that the square
Q= [λ
, 1
]×[λ
, 1
]
is contained into the closure of
A
and
the partial derivatives
∂F1/∂k1
and
∂F1,2/∂k2
are strictly positive on
Q
. The existence of such a square
can be checked by using the explicit expressions
F1(k1,k2)=2q2k2
1−1
2K(k1)−Z1
√2k2
1−1
k1θ+
k1,k2
dt
q(1−t2)1−k2
1(1−t2)
, (36)
F1,2 (k1,k2)=2q2k2
2−1
2K(k2)+Z1
√2k2
2−1
k2θ+
k1,k2
dt
q(1−t2)1−k2
2(1−t2)
, (37)
where
θ+
k1,k2
is given by Equation (18). Similarly, one checks that also
F1
is strictly positive on
Q
,
while
F1,2
obviously is. Consequently,
∂Fm/∂k1
,
∂Fm,n/∂k2
,
Fm
and
Fm,n
are also strictly positive on
Q
(recall that the function Sis strictly increasing and positive). Define
µm:=max
λ≤k2≤1Fm(λ,k2),µm,n:=max
λ≤k1≤1Fm,n(k1,λ)and ηm,n:=max {µm,µm,n}
L1
,
and let
η>ηm,n
, so that
ηL2>ηL1>max {µm,µm,n}
. By continuity of
Fm
and
Fm,n
, and using
again the explicit expressions in Equations (36)–(37) (with general
m
and
n
inserted) as
k1
,
k2→
1, we
have that
lim
k1→λ+Fm(k1,k2)=Fm(λ,k2)≤µm<ηL1and lim
k1→1−Fm(k1,k2)= +∞
for every fixed k2∈[λ, 1], and
lim
k2→λ+Fm,n(k1,k2)=Fm,n(k1,λ)≤µm,n<ηL2and lim
k2→1−Fm,n(k1,k2)= +∞
for every fixed k1∈[λ, 1]. Then, Lemma 3ensures that the level sets
{(k1,k2)∈Q:Fm(k1,k2)=ηL1}and {(k1,k2)∈Q:Fm,n(k1,k2)=ηL2}
respectively, are the graphs
k1=f(k2)
and
k2=g(k1)
of two continuous functions
f
,
g
defined on
[λ, 1]
. The first graph joins a point on the segment
[λ, 1]×{1}
to a point on
[λ, 1]×{λ}
, the latter one
joins a point on
{λ}×[λ, 1]
to a point on
{1}×[λ, 1]
, and therefore the two level sets must intersect in
the interior of
Q
at a point
(k1,k2)
, which thus solves the system in Equation (35). Then, Lines 4–7 of
the system in Equation (33) fix the values of
a1
,
a2
,
a3
,
a4
, by taking
p
as the unique integer such that
the corresponding a4belongs to (0, T2]. This completes the proof.
Remark 6.
In the proof of Theorem 3, the sign of the function
F1
can be easily checked. Indeed, taking into
account that θ+
k1,k2≥1/√2, one has
F1(k1,k2)>2q2k2
1−1Z1
√2k2
1−1
k1√2
1
√1−t2
1
q1−k2
1t2−1
q1−k2
1(1−t2)
>0.

Symmetry 2019,11, 161 19 of 20
On the contrary, the analysis of the sign of
∂F1/∂k1
and
∂F1,2/∂k2
over the set
A
is rather involved and we
could not perform it exactly. Therefore, we based our argument concerning the existence of the square
Q
on the
numerical evidence given by the plots of their graphs (see Figure 5), for which we used the software
Wolfram
MATHEMATICA 10.4.1.
Figure 5. The functions ∂F1/∂k1and ∂F1,2/∂k2over the square [λ, 1 ]2with λ=0.88.
Author Contributions: All the authors contribute equally to this work
Funding:
The research of D.N. and S.R. was funded in part by Departmental Project 2018-CONT-0127, University
of Milano Bicocca.
Conflicts of Interest:
The authors declare no conflict of interest. The funders had no role in the design of the
study; in the collection, analyses, or interpretation of data and in the writing of the manuscript, or in the decision
to publish the results.
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2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access
article distributed under the terms and conditions of the Creative Commons Attribution
(CC BY) license (http://creativecommons.org/licenses/by/4.0/).