Entire functions related to stationary solutions of the Kawahara equation

Abstract

In this study, we characterize the lengths of intervals for
which the linear Kawahara equation has a non-trivial solution,
whose energy is stationary. This gives rise to a family of complex
functions. Characterizing the lengths amounts to deciding which members
of this family are entire functions. Our approach is essentially
based on determining the existence of certain Mobius transformation.

Full text

Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 43, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ENTIRE FUNCTIONS RELATED TO STATIONARY SOLUTIONS
OF THE KAWAHARA EQUATION
ANDR´
E LUIZ C. DOS SANTOS, PATR´
ICIA N. DA SILVA,
CARLOS FREDERICO VASCONCELLOS
Abstract. In this study, we characterize the lengths of intervals for which the
linear Kawahara equation has a non-trivial solution, whose energy is stationary.
This gives rise to a family of complex functions. Characterizing the lengths
amounts to deciding which members of this family are entire functions. Our
approach is essentially based on determining the existence of certain M¨obius
transformation.
1. Introduction
In the Kawahara equation
ut+ux+κuxxx +ηuxxxxx +uux= 0,(1.1)
the conservative dispersive effect is represented by the term (κuxxx +ηuxxxxx).
This equation is a model for plasma wave, capilarity-gravity water waves and other
dispersive phenomena when the cubic KdV-type equation is weak. Kawahara [11]
pointed out that it happens when the coefficient of the third order derivative in
the KdV equation becomes very small or even zero. It is then necessary to take
into account the higher order effect of dispersion in order to balance the nonlinear
effect. Kakutani and Ono [10] showed that for a critical value of angle between the
magneto-acoustic wave in a cold collision-free plasma and the external magnetic
field, the third order derivative term in the KdV equation vanishes and may be
replaced by the fifth order derivative term. Following this idea, Kawahara [11]
studied a generalized nonlinear dispersive equation which has a form of the KdV
equation with an additional fifth order derivative term. This equation has also
been obtainded by Hasimoto [9] for the shallow wave near critical values of surface
tension. More precisely, in this work Hasimoto found these critical values when the
Bond number is near one third.
While analyzing the evolution of solutions of the water wave-problem, Schneider
and Wayne [17] also showed that the coefficient of the third order dispersive term
in nondimensionalized statements of the KdV equation vanishes when the Bond
number is equal to one third. The Bond number is proportional to the strength
2010 Mathematics Subject Classification. 30D20, 35Q53.
Key words and phrases. Entire functions; M¨obius transformations; stationary solutions;
Kawahara equation.
c
2016 Texas State University.
Submitted November 20, 2014. Published January 29, 2016.
1

2 A. L. C. DOS SANTOS, P. N. DA SILVA,C. F. VASCONCELLOS EJDE-2016/43
of the surface tension and in the KdV equation it is related to the leading order
dispersive effects in the water-waves problem. With its disappearance, the resulting
equation is just Burger’s equation whose solutions typically form shocks in finite
time. Thus, if we wish to model interesting behavior in the water-wave problem it
is necessary to include higher order terms. That is, it is necessary to consider the
Kawahara equation. In any case, the inclusion of the fifth order derivative term
takes into account the comparative magnitude of the coefficients of the third and
fifth power terms in the linearized dispersion relation.
Berloff and Howard [3] presented the Kawahara equation as the purely dispersive
form of the nonlinear partial differential equation
ut+urux+auxx +buxxx +cuxxxx +duxxxxx = 0.
The above equation describes the evolution of long waves in various problems in fluid
dynamics. The Kawahara equation corresponds to the choice a=c= 0 and r= 1
and describes water waves with surface tension. Bridges and Derks [6] presented
the Kawahara equation – or fifth-order KdV-type equation – as a particular case
of the general form
ut+κuxxx +ηuxxxxx =∂
∂x f(u, ux, uxx) (1.2)
where u(x, t) is a scalar real valued function, κand η6= 0 are real parameters
and f(u, ux, uxx) is some smooth function. The form (1.1) occurs most often in
applications and corresponds to the choice of fin (1.2) with the form f(u, ux, uxx) =
−u2/2.
As noted by Kawahara [11], we may assume without loss of generality that η < 0
in (1.1). In fact, if we introduce the following simple transformations
u→ −u, x → −x, t →t,
we can obtain an equation of the form of (1.1) in which κand ηare replaced,
respectively, by −κand −η.
Hagarus, Lombardi and Scheel [8] pointed out that the Kawahara equation
ut=uxxxxx −εuxxx +uux,(1.3)
in which εis a real parameter, models water waves in the long-wave regime for
moderate values of surface tension (Weber numbers close to 1/3). For such Weber
numbers, the usual description of long water waves via the Korteweg-de Vries (KdV)
equation fails since the cubic term in the linear dispersion relation vanishes and
fifth order dispersion becomes relevant at leading order, ω(k) = k5+εk3. Positive
(resp. negative) values of the parameter εin (1.3) correspond to Weber numbers
larger (resp. smaller) than 1/3.
Dispersive problems have been object of intensive research (see, for instance, the
classical paper of Benjamin, Bona and Mahoni [2], Biagioni and Linares [4], Bona
and Chen [5], Menzala et al. [13], Rosier [15], and references therein). Recently
global stabilization of the generalized KdV system have been obtained by Rosier and
Zhang [16] and Linares and Pazoto [12] studied the stabilization of the generalized
KdV system with critical exponents. For the stabilization of global solutions of the
Kawahara under the effect of a localized damping mechanism, see Vasconcellos and
Silva [19, 20, 21].
We consider the linear Kawahara equation
ut+βux+κuxxx +ηuxxxxx = 0 with (x, t)∈(0, L0)×(0,∞),(1.4)

EJDE-2016/43 ENTIRE FUNCTIONS RELATED TO STATIONARY SOLUTIONS 3
where the coefficients β, κ and ηare real numbers such that η < 0, κ6= 0, β∈ {0,1}.
Sometimes, while discussing the existence of solutions of certain partial differential
equations, it is necessary to establish when a certain quotient of entire functions
still turns out to be an entire function (see, for instance, Rosier [15], Vasconcellos
and Silva [19]). The problem of factoring an entire function is solved by the famous
Weierstrass factorization theorem and its corollaries. However, applying this result
may not be very practical or even viable in some cases.
Now we proceed to a general description of such kind of factoring problems. We
have a polynomial p:C→Cand a family of functions
Na:C×(0,∞)→C,
a∈C4∼{0}, whose restriction Na(·, L) is entire for each L > 0. We consider a
family of functions fa(·, L) defined by
Na(ξ, L) = fa(ξ , L)p(ξ) (1.5)
in its maximal domain. For a given polynomial p(·), the problem of characterizing
the set of values L0>0, for which it is possible to find a non null a0∈C4such that
the function fa0(·, L0) is entire, is a challenging problem and of particular interest
of the academic community.
Vasconcellos and Silva [19, Lemma 2.1] discussed the existence of non-zero so-
lutions for (1.4) whose energy is constant over time. Their results show that the
existence of such solutions is equivalent to determining the lengths of interval (0, L0)
for which it is possible to verify that the condition
(∃λ∈C, u0∈(H3
0(0, L0)∩H5(0, L0),C)⇒λu0+βu0
0+κu000
0+ηu00000
0= 0) (1.6)
is valid. Such condition in turn reduces to the problem of characterizing the set X
of L0>0 values, for which exist rand a0providing that function fa(·, L) is entire
for L=L0and a=a0. In this case, using (1.5), fa(·, L) is defined by
Na(ξ, L) = a1iξ −a2iξe−iξL +a3−a4e−iξL
p(ξ) = r+βξ −κξ3+ηξ5(1.7)
where a= (a1, a2, a3, a4) and r∈R. It follows from (1.6) that λis a pure imaginary
number. Thus, we only have to consider polynomials p(·) with r∈R.
For each r∈Rand a0∈C4∼{0}, let Xa0rbe the set of L0>0 values, for
which the function fa(·, L) is entire for L=L0and a=a0. The set Xis the union
of Xa0rfor r∈Rand a0∈C4∼{0}.
Here, we place emphasis on the following statements:
(S1) fa0(·, L0) is entire;
(S2) all the zeros, taking the respective multiplicities into account, of the poly-
nomial pare zeros of Na0(·, L);
(S3) the maximal domain of fa0(·, L0) is C;
which are, clearly, equivalent and will be widely used throughout this article. A
closer look shows that determining the solution to the problem guarantees the exis-
tence of a M¨obius transformation in some circumstances. Further, for the function
fa(·, L), defined by (1.5) and (1.7), to be entire, given the equivalence between
statements (S1) and (S2); informally, we must have
a1iξ0+a3
a2iξ0+a4
=e−iLξ0(1.8)

4 A. L. C. DOS SANTOS, P. N. DA SILVA,C. F. VASCONCELLOS EJDE-2016/43
for each root ξ0of the polynomial p. We note that for asuch that a1a3−a2a46= 0,
the left side of (1.8) suggests that a M¨obius transformation is defined. Note that
we already have an indication that for a polynomial pwith at least two roots
differing by an integer multiple of 2π/L, we obtain L /∈ X. With this, a method for
solving the problem is revealed: we must verify for which structures of the roots of
the polynomial pis it possible to define a M¨obius transformation Mthat satisfies
M(ξ0) = e−iLξ0for each zero ξ0of polynomial p(See Lemma 2.6).
Taking (1.8), it is essential to define, for each non null a∈C4, the discriminant
of a, specifically, the complex number d(a) = a1a4−a2a3. It is natural, however,
to consider:
(i) d(a) = 0 or
(ii) d(a)6= 0.
The main result shown in this article guarantees that the existence of pairs (a0, L0)
that make fa(·, L) entire is intimately linked to whether or not the discriminant
is zero. In fact, when the discriminant of ais zero, such pairs do not exist for
any r∈R. On the other hand, if the discriminant of ais non-zero, we identify
situations where the pairs (a0, L0) can exist or not. Whereas case (i) has been
completely solved here, in case (ii) there are situations where the problem remains
to be solved, i.e., in some cases, we do not know whether or not it is possible to
satisfy (1.8). As far as we know, Rosier [15] was the first to analyze these kinds
of problems. In fact, he showed that the existence of non-trivial solutions for the
Kortweg de Vries equation, whose energies do not decay over time, is equivalent to
determining the set Uof values l0>0, for which there exists a non null k0∈C2
and s∈C, so that the function gk(·, L) with k= (k1, k2), defined by
Mk(ξ, l) = gk(ξ , l)q(ξ),(1.9)
is entire for k=k0and l=l0. Here, in particular, Mk(ξ, l) = k1−k2e−iLξ and
q(ξ) = ξ3−ξ+s. Then Rosier [15] proves that
U=n2πrm2+mn +n2
3:n, m ∈No.
Let us take case (i) from the same starting point as Rosier [15], i.e., the analysis of
zeros of Na(·, L). Here, it makes no sense to argue about the existence of a M¨obius
transformation. Case (ii) is completely based on equation (1.8). Our strategy is
quite efficient. It proved to be efficient in this situation, where using previously
established results, such as the Weierstrass factorization theorem, is not possible.
Notice that, for each choice of the coefficients β, κ and η, condition (1.6) asso-
ciates the Kawahara equation
ut+βux+κuxxx +ηuxxxxx = 0
to a family of polynomials
p(ξ) = r+βξ −κξ3+ηξ5, r ∈R.
Let Xbe the set of the lengths of interval (0, L0) for which exist non-zero solutions
for (1.4) whose energy is constant over time. Consider for each r∈Rand a∈C4∼
{0}, the set Xarof values L0>0 for which the function fa(·, L) defined by (1.5)
and (1.7) is entire for L=L0. We can decompose Xas the union of the sets Xar
for r∈Rand non null a∈C4

EJDE-2016/43 ENTIRE FUNCTIONS RELATED TO STATIONARY SOLUTIONS 5
We extend the results obtained by Vasconcellos and Silva [19, 20] for character-
izing the set Xfor the Kawahara equation (1.4). They have partially analyzed the
case κ= 0 in (1.4) and did not deal with the case κ= 1 in (1.4). In our proof, we
argue by exhaustion characterizing the sets Xar. In Part (I) of Theorem 1.1, we
see that if d(a) = 0, then Xar=∅for all r∈R. As a consequence of this result,
it follows that for any Kawahara equation (1.4), the set Xis given by the union
of the set Xarfor r∈Rand d(a)6= 0. Our results for d(a)6= 0 allow to partially
describe the set Xfor Kawahara equations (1.4) with β= 1 and κ6= 0 or β= 0
and κ < 0. For Kawahara equations (1.4) with β= 0, κ > 0, as a consequence of
Theorem 1.1, we obtain that Xis empty.
Now we summarize the results obtained in this article in the following theorem
guided by the roots of polynomial p, as we will shortly see.
Theorem 1.1. Let r∈R, a non null a∈C4and L > 0, and consider the function
fa(·, L)defined by the product
Na(ξ, L) = fa(ξ , L)p(ξ) (1.10)
in its maximal domain. Let us suppose that Na(ξ, L)and p(ξ)are as in (1.7). Let
Xarbe the set of values L0>0for which the function fa(·, L)defined by (1.10) is
entire for L=L0.
(I) If L0>0is such that fa0(·, L0)is entire for some non null a0∈C4, then
d(a0)6= 0. In other words, for any non null a, if d(a) = 0, we obtain Xar=∅, for
any r∈R. The reciprocal, however, is false.
(II) If ais such that d(a)6= 0 and one of following three items occurs:
(a) β= 1 and |r|> z −κz3+ηz5, where z=r3κ−√9κ2−20η
10η;
(b) β= 0,κ > 0and r∈R;
(c) β= 0,κ < 0and |r|>−κz3+ηz5, where z=q3κ
5η.
Then there is no L > 0that renders the function fa(·, L)entire. Therefore, Xar=∅.
(III)
(a) If β= 1 and r= 0, then there exist L0>0and non null a0such that
fa0(·, L0)is entire if and only if
L0∈nL∈R, k cotanh Lk
2=−ρcot Lρ
2o
where
ρ=sκ−pκ2−4η
2ηand k=v
u
u
t
κ+pκ2−4η
2η.
(b) If β= 0,κ < 0and r= 0, then there exist L0>0and non null a0such
that fa0(·, L0)is entire if and only if
L0∈nL > 0,tan ρL
2=ρL
2o,
where ρ=pκ/η.
The sets in (a) and (b) are enumerable.

6 A. L. C. DOS SANTOS, P. N. DA SILVA,C. F. VASCONCELLOS EJDE-2016/43
The knowledge of the zeros of gk(ξ, l) in (1.9) plays a key role in the Rosier’s
analysis of the existence of non-trivial solutions for the Kortweg de Vries equation,
whose energies do not decay over time. The function fa(ξ , L) related to Kawahara
equation does not resemble this fact and its structure together with the order of
the polynomial turn the analysis of the Kawahara case into a hard problem. Many
other authors have made efforts to tackle this problem (see for instance, Glass
and Guerrero [7], for a particular case of (III)(b); Araruna, Capistrano-Filho and
Doronin [1], for an example of a critical set). Our results take their contributions
into account. We show they can be presented and obtained in a systematic way
and we go a step further.
2. Proof of main results
First we establish some lemmas need for proving the three parts of Theorem 1.1.
Part (I). The main idea behind Part (I) of Theorem 1.1 is to find out whether
there is at least one zero of polynomial pthat is not a zero of Na(·, L). The following
lemma is a decisive factor in obtaining this result.
Lemma 2.1. Let non null a∈C4with d(a) = 0 and L > 0. Then the set of the
imaginary parts of the zeros of Na(·, L)has at most two elements.
Proof. Fix an arbitrarily L > 0 and a non null asuch that d(a) = 0. The nullity of
the discriminant of aguarantees that the vectors (a1, a3) and (a2, a4) are linearly
dependent. We can then assume that there exists some complex number λsuch
that (a1, a3) = λ(a2, a4). Thus (a2, a4) cannot be zero and, if (a1, a3) is zero, then
λ= 0. Therefore, we can write
Na(ξ, L)=(a2iξ +a4) (λ−e−iLξ) (2.1)
Finally, we see that in one of the factors of (2.1) there is at most one zero and, in
the other, an infinite number of zeros all with the same imaginary part. Thus, the
set of the imaginary parts of the zeros of Na(·, L) has no more than two elements. If
we assume that (a2, a4) = λ(a1, a3), the same conclusion about the zeros of Na(·, L)
is valid, proving the result. 
By quickly verifying the result shown in Lemma 2.6, we conclude the set of
the imaginary parts of the polynomial phas at least three elements, except for
r=β= 0 and κ < 0, when all the roots are real. With the aim of proving part (I)
of Theorem 1.1, we fix an arbitrary L > 0 and a non null asuch that d(a) = 0. We
initially consider any complementar case to r=β= 0 and κ < 0. The Lemmas 2.6
and 2.1 combined guarantee that there will always be a zero of the polynomial p
that is not a zero of Na(·, L). Consequently, the function fa(·, L) is not entire.
Finally, we suppose that r=β= 0 and κ < 0 and assume that fa(·, L) is entire.
Since 0 is a root of multiplicity three, given the equivalence between (S1) and (S2),
we have Na(0, L) = N0
a(0, L) = N00
a(0, L) = 0 (differentiating with respect to ξ).
These three equations imply that a=µ(−L
2,L
2,1,1) for a complex number µ6= 0.
Here, we obtain d(a) = −µL 6= 0, which contradicts the hypothesis of the nullity of
the discriminant of a. Therefore, fa(·, L) is not entire for any value of L > 0 and
non null asuch that d(a) = 0. This proves part (I) of Theorem 1.1.

EJDE-2016/43 ENTIRE FUNCTIONS RELATED TO STATIONARY SOLUTIONS 7
Part (II). The following lemma essentially states that if the polynomial phas “too
many” complex roots, equation (1.8) cannot be satisfied.
Lemma 2.2. For any L > 0, there is no M¨obius transformation Msuch that
M(ξ) = e−iLξ, ξ ∈ {ξ1, ξ2, ξ1, ξ2}
with ξ1, ξ2, ξ1, ξ2all distinct in C.
Proof. Write ξj=xj+iyjand ωj=M(ξj) for j= 1,2 and note that M(ξj) = 1
ωj.
The M¨obius transformation Mthat satisfies
M(ξ) = e−iLξ, ξ ∈ {ξ1, ξ2, ξ1, ξ2}
exists if, and only if, the equality
(ξ1, ξ2;ξ1, ξ2) = ω1, ω2;1
ω1
,1
ω2
is valid [18, Theorem VI,p. 178]. (Note that (ξ1, ξ2;ξ3, ξ4) stands for the cross ratio
of pairs (ξ1, ξ2) and (ξ3, ξ4)) However, this does not occur, as
ω1, ω2;1
ω1
,1
ω2=4 sinh Ly1sinh Ly2
2 cosh L(y1+y2)−2 cos L(x2−x1)>4y1y2
K= (ξ1, ξ2;ξ1, ξ2),
where K=|ξ1−ξ2|. To prove this statement, let us assume, without loss of
generality, that y1, y2>0 and define the function:
F(t) = Ksinh(y1t) sinh(y2t)−2y1y2[cosh((y1+y2)t)−cos((x2−x1)t)]
for t∈R. A direct calculation (see the Appendix) shows that F(0) = F0(0) = 0
and F00(t)>0 for all t > 0. Using a second order Taylor expansion for the function
F, we conclude that F(t)>0 for all t > 0. In particular, this means that for L > 0,
the inequality
4Ksinh(Ly1) sinh(Ly2)>4y1y2(2 cosh(L(y1+y2)) −2 cos(L(x2−x1)))
is valid. This completes the proof. 
Thus, now we can prove part (II) of Theorem 1.1. Let us suppose there exists
asuch that d(a)6= 0 and that one of (a), (b) with r6= 0, or (c) of part (II) of the
theorem occurs (case (b) with r= 0 is proven in Lemma 2.5).
Here, Lemma 2.6 guarantees that polynomial phas a single real root, whose
multiplicity is equal to 1. This means that this polynomial has two pairs of complex
conjugate roots. Let us assume, in contradiction, that there exists L > 0 such that
the function fa(·, L) is entire. Then, all roots of polynomial pmust satisfy (1.8);
i.e., there exists a M¨obius transformation that takes each root ξ0of pinto e−iLξ0.
However, this contradicts Lemma 2.2 and proves part (II) of the theorem except
for the case where β= 0, κ > 0 and r= 0 that will be shown in Lemma 2.5.
Part (III). Lemma 2.3 below, unlike Lemma 2.2, guarantees the existence of a
M¨obius transformation in a case when the polynomial phas exactly three real roots
whose multiplicities are equal to 1. Lemma 2.5 below guarantees the existence of a
M¨obius transformation when all roots of polynomial pare real.

8 A. L. C. DOS SANTOS, P. N. DA SILVA,C. F. VASCONCELLOS EJDE-2016/43
Lemma 2.3. Let L, k and ρbe real numbers with L > 0and k6= 0. There is a
unique M¨obius transformation Mwhich satisfies M(0) = 1,M(±ρ) = e∓iLρ and
M(±ik) = e±Lk if and only if the following equality occurs
kcotanh Lk
2=−ρcot Lρ
2.
Proof. First we prove the necessity part. Assume that exists a unique M¨obius
transformation Mthat satisfies M(0) = 1, M(±ρ) = e∓iLρ and M(±ik) = e±Lk .
We know [18, Theorem V, p. 176] that the pairs (0,1), (ik, eLk) and (−ik, e−Lk)
unambiguously determine the M¨obius transformation of M1, whose expression is
M1(z) = −z+ik cotanh Lk
2
z−ik cotanh Lk
2(2.2)
and can be obtained by (5.1). Thus, we must obtain M=M1. (For the coefficients
M1(z) and M2(z), see the Appendix)
Similarly, the injectivity of Mimplies that Lis not an integer multiple of π
ρ. Thus
the pairs (0,1), (ρ, e−iLρ) and (−ρ, eiLρ ) determine a unique M¨obius transformation
M2. The formula (5.1) guarantees that M2is expressed by
M2(z) = −z−iρ cot Lρ
2
z+iρ cot Lρ
2.(2.3)
Consequently, we obtain M=M2. Thus, M1=M2. We know [14, Theorem 3, p.
26] that if M1and M2coincide, their coefficients must be such that
−1−ik cotanh Lk
2
1−ik cotanh Lk
2=−1iρ cot Lρ
2
1iρ cot Lρ
2.
The same result guarantees the sufficiency part. 
Lemma 2.4. Let kand ρbe real numbers with k6= 0. The positive solutions of the
equation
kcotanh Lk
2=−ρcot Lρ
2
form a countable set.
Proof. If ρ= 0, there is nothing to prove. Thus, suppose that ρ6= 0. For every
n∈N, let Jn= ((n−1)π, nπ) and J=∪n∈NJn. The continuous function g:J→R,
defined by g(t) = kcotanh(kt
2)+ ρcot( ρt
2), is strictly decreasing on each Jnand such
that g[Jn] = R. Therefore, there exists a unique Lnin each Jnsuch that g(Ln) = 0,
proving that the solutions of the equation form a countable set. This completes the
proof. 
We have all the ingredients to prove item (a) of part (III) of Theorem 1.1. If
β= 1 and r= 0, the polynomial phas exactly three real roots, whose multiplicities
are equal to 1. Now, it is sufficient to observe that Lemma 2.3 together with
equation (1.8), as well as Lemma 2.4, conclude this. Finally, we prove the final
lemma that supports item (b) of Theorem 1.1.
Lemma 2.5. Consider the family of functions fa(·, L)defined by (1.7) and let us
suppose that r=β= 0. Let Xarbe the set of values L0>0for which the function
fa(·, L)is entire for L=L0.

EJDE-2016/43 ENTIRE FUNCTIONS RELATED TO STATIONARY SOLUTIONS 9
(1) If κ < 0, there exist L0>0and non null a0such that fa0(·, L0)is entire if
and only if
L0∈nL > 0,tan ρL
2=ρL
2o
where ρ=pκ/η.
(2) If κ > 0, then fa(·, L)is not entire for any non null aand L > 0. That is,
Xa0=∅.
Proof. In the case κ < 0, the polynomial pallows the decomposition p(ξ) = ηξ3(ξ−
ρ)(ξ+ρ) where ρ=pκ/η. Assume L0is in
P=nL > 0,tan ρL
2=ρL
2o
and set a0= (−L0
2,L0
2,1,1). We obtain
Na0(0, L0) = N0
a0(0, L0) = N00
a0(0, L0)=0,
i.e., 0 is the root of Na0(0, L0) with multiplicity at least equal to 3. Now, if we
prove that Na0(±ρ, L0) = 0, then we will be done. To this end, observe that
Na0(ρ, L0) = 0 if and only if
iρL0
2+ 1e(iρL0
2+1)=iρL0
2+ 1e(iρL0
2+1) (2.4)
This equality, in turn, is equivalent to sin ρL0
2=ρL0
2cos ρL0
2, which is valid, since
we chose L0in P. The same argument proves that Na0(−ρ, L0) = 0. Then, the
function fa0(·, L0) is entire, proving that the condition is sufficient. Now, we will
prove that it is necessary. Suppose that fa0(·, L0) is entire for a certain pair (a0, L0).
This tells us that
Na0(0, L0) = N0
a0(0, L0) = N00
a0(0, L0) = Na0(±ρ, L0)=0.(2.5)
The three first equalities in (2.5) can be used to explicitly determine a0, i.e., a0=
a(L0) = γ(−L0
2,L0
2,1,1) for γ6= 0. As Na(L0)(±ρ, L0) = 0, we conclude from
argument (2.4), that L0must be in P, proving that the condition is also necessary.
Finally, we assume κ > 0. Here, the polynomial padmits the factorization
p(ξ) = ηξ3(ξ−eρ)(ξ+eρ) where eρ=iρ and ρ=p−κ/η. Suppose that function
fa(·, L) is entire for the pair (a, L). This implies
Na(0, L) = N0
a(0, L) = N00
a(0, L) = Na(±eρ, L)=0.(2.6)
As in the first part, we can obtain a(L) = γ(−iL
2,iL
2,1,1) for γ6= 0. Thus
Na(L)(ξ, L) = γh−iL
2ξ+ 1 −iL
2ξ+ 1e−iξLi.
On the other hand, as Na(L)(±eρ, L) = 0, the following equalities must be valid
±ρL
2+ 1 −∓ρL
2+ 1e±ρL = 0 (2.7)
However, this is absurd. To prove this, note that h:R→Rdefined by h(t) =
et(2 −t)−(2 + t) is such that h(0) = 0 and h0(t)<0 for all t∈R. This completes
the proof. 
Note that the sets obtained from the final lemma are clearly countable. Thus
Theorem 1.1 is proved.

10 A. L. C. DOS SANTOS, P. N. DA SILVA,C. F. VASCONCELLOS EJDE-2016/43
2.1. Describing the roots. The following lemma separates the roots into groups
according to their algebraic structure. To characterize a group, we consider the
quantity of real roots and their respective multiplicities. It is worth noting that for
each polynomial p, the relation of its roots with these groups determines whether
or not a solution exists for the problem of determining an entire member of family
fa(·, L).
Lemma 2.6. Consider the polynomial p(ξ) = r+βξ −κξ3+ηξ5where r, β, κ and
ηare real such that β∈ {0,1},κ6= 0 and η < 0.
(1) If β= 0,
(a) κ > 0and
(i) r= 0, then 0is the only real root of pwith multiplicity 3.
(ii) r6= 0, then phas a single real root, with multiplicity 1.
(b) κ < 0and
(i) r= 0, then 0and ±ρare roots of p, where ρ=qκ
η, being 0the root
of multiplicity 3.
(ii) 0 <|r|<−κz3+ηz5, with z=q3κ
5η, then phas exactly three real
roots, all of which are of multiplicity 1.
(iii) |r|=−κz3+ηz5, with z=q3κ
5η, then phas exactly three real roots,
one of which has multiplicity 2.
(iv) |r|>−κz3+ηz5, with z=q3κ
5η, then phas exactly one real root, with
multiplicity 1.
(2) If β= 1 and
(a) r= 0, then the roots of pare 0,±ρ, ±ik where
ρ=sκ−pκ2−4η
2ηand k=v
u
u
t
κ+pκ2−4η
2η.
(b) 0 <|r|< z −κz3+ηz5, where z=r3κ−√9κ2−20η
10η, then phas exactly three
non null real roots, with multiplicity 1.
(c) |r|=z−κz3+ηz5, where z=r3κ−√9κ2−20η
10η, then phas exactly three real
roots, one of which has multiplicity 2.
(d) |r|> z −κz3+ηz5, where z=r3κ−√9κ2−20η
10η, then phas exactly one real
root with multiplicity 1.
3. Final Remarks
It is worth noting that when d(a) = 0, the sets Xarwere completely characterized
for any r∈R. The same happens when we consider d(a)6= 0, β= 0, κ > 0 and
r∈R; i.e., when p(ξ) = r−κξ3+ηξ5. In particular, in this case, Theorem 1.1
tells us that the sets Xarare empty for all non null a∈C4and r∈R. Thus
the set Xis empty and the problem of the initial and boundary value, analyzed
by Vasconcellos and Silva [19] and associated with the linear Kawahara equation
ut+κuxxx +ηuxxxxx = 0, does not admit non-trivial solutions whose energies do
not decay over time. Note that for p(ξ) = r+βξ −κξ3+ηξ5, the case d(a)6= 0

EJDE-2016/43 ENTIRE FUNCTIONS RELATED TO STATIONARY SOLUTIONS 11
remains to be solved in two situations: (a) when r6= 0 and phas exactly three real
roots, with all the multiplicities being equal to 1 and (b) phas exactly three real
roots with one of them having a multiplicity of 2.
4. Appendix: Derivatives of F(t)
Consider the function
F(t) = Ksinh y1tsinh y2t−2y1y2(cosh((y1+y2)t)−cos((x2−x1)t))
for t∈R. Note that F(0) = 0 and that
F0(t) = K(y1cosh y1tsinh y2t+y2sinh y1tcosh y2t)
−2y1y2((y1+y2) sinh((y1+y2)t)+(x2−x1) sin((x2−x1)t))
F00(t)=(x2−x1)2(y2
1+y2
2) sinh y1tsinh y2t
+ (y1+y2)2(y1−y2)2sinh y1tsinh y2t+ 2(x2−x1)2y1y2(cosh y1tcosh y2t
−cos((x2−x1)t))
Note that F0(0) = 0 and that, for y1, y2>0, F00 (t)>0 for all t > 0.
5. Appendix: Coefficients of M1and M2
Let ξ1, ξ2and ξ3be three distinct points that are mapped by Minto three distinct
points w1, w2and w3. Since there is always one and only one linear fractional
transformation that transforms any three distinct points into three given distinct
points (see, for instance, [18, Theorem V p. 176], if we take
a=
w1ξ1w11
w2ξ2w21
w3ξ3w31, b =
w1ξ1ξ1w1
w2ξ2ξ2w2
w3ξ3ξ3w3,
c=
ξ1w11
ξ2w21
ξ3w31, d =
w1ξ1ξ11
w2ξ2ξ21
w3ξ3ξ31,
(5.1)
then M(ξ) = aξ+b
cξ+dis the M¨obius transformation M(ξ) such that M(ξj) = wj
(j= 1,2,3).
5.1. Coefficients of M1.Let M1(ξ) be the linear fractional transformation such
that M1(0) = 1, M1(ik) = eLk,M1(−ik) = e−Lk . By (5.1), the coefficients of M1
may be determined by
aI=−ikwI+1
wI−2=−2ik(cosh(Lk)−1)
bI=k2wI−1
wI=dI= 2k2sinh(Lk)
cI=ikwI+1
wI−2= 2ik(cosh(Lk)−1)
Therefore,
M1(ξ) = −2ik(cosh(Lk)−1)ξ+ 2k2sinh(Lk)
2ik(cosh(Lk)−1)ξ+ 2k2sinh(Lk)=−ξ+ik cotanh Lk
2
ξ−ik cotanh Lk
2.

12 A. L. C. DOS SANTOS, P. N. DA SILVA,C. F. VASCONCELLOS EJDE-2016/43
5.2. Coefficients of M2.Let M2(ξ) be the linear fractional transformation such
that M2(0) = 1, wρ=M2(ρ) = e−iLρ ,M2(−ρ) = eiLρ. By (5.1), the coefficients of
M2may be determined by
aρ=−ρ(wρ+wρ−2) = −2ρ(cos(Lρ)−1)
bρ=−ρ2(wρ−wρ) = dρ=−2iρ2sin(Lρ)
cρ=ρ(wρ+wρ−2) = 2ρ(cos(Lρ)−1) .
Therefore,
M2(ξ) = −2ρ(cos(Lρ)−1)ξ−2iρ2sin(Lρ)
2ρ(cos(Lρ)−1)ξ−2iρ2sin(Lρ)=−ξ−iρ cot Lρ
2
ξ+iρ cot Lρ
2.
Acknowledgments. This work was partially supported by FAPERJ under grant
E26/010.002646/2014.
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Andr´
e Luiz C. dos Santos
DEMAT/CEFET/Maracan˜
a; PPG-EM/UERJ - Rua Fonseca Teles, 121, 1o. andar, S˜
ao
Crist´
ov˜
ao, Rio de Janeiro, RJ. CEP: 20940-903, Brazil
E-mail address:andreluiz.cordeiro@gmail.com
Patr
´
ıcia N. da Silva
IME/UERJ, Rua S˜
ao Francisco Xavier, 524, 6o. andar, Rio de Janeiro, RJ, CEP 20550-
900, Brazil
E-mail address:nunes@ime.uerj.br
Carlos Frederico Vasconcellos
IME/UERJ, Rua S˜
ao Francisco Xavier, 524, 6o. andar, Rio de Janeiro, RJ, CEP 20550-
900, Brazil
E-mail address:cfredvasc@ime.uerj.br