# Spectral characterization of the constant sign derivatives of Green’s
function related to two point boundary value conditions

Alberto Cabada, Lucía López-Somoza and Mouhcine Yousfi
1 CITMAga, 15782, Santiago de Compostela, Galicia, Spain
2 Departamento de Estatística, Análise Matemática e Optimización
Facultade de Matemáticas, Universidade de Santiago de Compostela, Spain.
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

In this paper we will study the set of parameters in which certain partial
derivatives of the Green’s function, related to a $n$-order linear operator
$T_{n}[M]$, depending on a real parameter $M$, coupled to different two-point
boundary conditions, are of constant sign. We will do it without using the
explicit expression of the Green’s function. The constant sign interval will
be characterized by the first eigenvalue related to suitable boundary
conditions of the studied operator. As a consequence of the main result, we
will be able to give sufficient conditions to ensure that the derivatives of
Green’s function cannot be nonpositive (nonnegative). These characterizations
and the obtained results can be used to deduce the existence of solutions of
nonlinear problems under additional conditions on the nonlinear part. To
illustrate the obtained results, some examples are given.

Keywords: Green’s function, Two-point boundary conditions, Spectral theory,
Comparison results, Constant sign.

##  1 Introduction

Many papers in the literature have studied the qualitative properties of
solutions of differential equations coupled with different boundary
conditions. Several techniques have been used to ensure the existence, non-
existence and multiplicity of solutions. Among them, stands out the method of
lower and upper solutions, coupled with monotone iterative techniques which,
as it is very well-known, are equivalent to the constant sign of the Green’s
function of the corresponding linear problem. Moreover the method of
constructing suitable cones in Banach spaces where to apply the well-known
Krasnoselskii’s fixed-point theorem, see for instance [7, 10, 9, 12] is a
powerful tool to deduce the existence of positive solutions of the nonlinear
problem. Such cones are constructed using constant sign properties of the
Green’s function and some of its derivatives.

To be concise, in this paper, we consider the following $n$-order linear
differential equation

$T_{n}[M]\,u(t)\equiv u^{(n)}(t)+M\,u(t)=0\,,\ t\in I:=[a,b]\,,$ (1.1)

where $M\in\mathbb{R}$ a is real parameter, coupled to the two-point boundary
conditions

$\displaystyle u^{(\sigma_{1})}(a)=$ $\displaystyle\cdots$
$\displaystyle=u^{(\sigma_{k})}(a)\hskip 8.5359pt=0,$ (1.2) $\displaystyle
u^{(\varepsilon_{1})}(b)=$ $\displaystyle\cdots$
$\displaystyle=u^{(\varepsilon_{n-k})}(b)=0,$ (1.3)

with $k\in\\{1,\dots,n-1\\}$ and the sets of indices
$\\{\sigma_{1},\dots,\sigma_{k}\\}$,
$\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}\subset\\{0,\dots,n-1\\}$,
satisfying that

$0\leq\sigma_{1}<\sigma_{2}<\cdots<\sigma_{k}\leq n-1\,,\quad
0\leq\varepsilon_{1}<\varepsilon_{2}<\cdots<\varepsilon_{n-k}\leq n-1\,.$

This paper is devoted to study the set of parameters $M$ in which certain
partial derivatives of the Green’s function $g_{M}$, related to problem
(1.1)–(1.3), are of constant sign without using the explicit expression of
Green’s function. It should be noted that, in general, it is not possible to
obtain the explicit expression of the Green’s function, specially for non
constant coefficients.

For the particular case of equation (1.1), together with boundary conditions
(1.2)–(1.3), the interval of parameters for which the Green’s function has
constant sign has been characterized in [5, Theorem 8.1]. Also for boundary
conditions of the type $(k,n-k)$, i.e., $\sigma_{j}=j-1$ for all
$j=1,\dots,k$, and $\varepsilon_{j}=j-1$ for all $j=1,\dots,n-k$ which are a
particular case of the previous ones, a characterization of the constant sign
intervals of the Green’s function is given in [3]. Moreover and the
calculation of the eigenvalues that characterizes these intervals up to $n=8$
is obtained.

The main assumption is the disconjugacy character of the operator $T_{n}[M]$,
which, as we know, is disconjugate for the value $M=0$ (see [3]) at any
arbitrary interval $[a,b]$. In the literature, sufficient conditions have been
given to ensure the disconjugacy character of linear operators, we refer to
[6, 11, 8, 13] and the references therein. A characterization of this property
is proved in [5], in which the optimal extremes of the intervals are given as
the eigenvalues of suitable $(k,n-k)$ problems.

It is in this context that the corresponding spectral theory arises: we will
prove that the constant sign interval is characterized by the first
eigenvalues of the studied operator, related to suitable boundary conditions.

The study starts from conditions (1.2)–(1.3) and constructs the conditions
satisfied by the derivatives of $g_{M}$, defining the necessary hypotheses
that we need to obtain the main results, and deducing the relationships
between the adjoint spaces of the respective related spaces. This way, we will
obtain results that characterize the constant sign interval of the
corresponding derivatives.

The novelty of the work is that the results generalize [5, Theorem 8.1] for
the particular operator $T_{n}[M]$ defined in (1.1). Also, as a consequence of
these results, we will deduce some cases in which the derivatives of the
Green’s function cannot be of constant sign. Furthermore, we will obtain a
necessary condition that allow us to ensure that the corresponding derivatives
are strictly positive (strictly negative) on the interior of the square of
definition.

The paper is organized as follows: in a preliminary section 2 we introduce the
fundamental concepts, the results and the main hypotheses that are needed in
the development of the article. The next section is devoted to prove the
results that characterize the constant sign of the derivatives of Green’s
function by distinguishing three cases. The main results are proved through
spectral theory. This section includes also a corollary that provides
sufficient conditions for derivatives to be nonpositive or nonnegative, a
necessary condition to ensure the negative (positive) sign of certain
derivatives with respect to $t$ of $g_{M}$, and examples of application of our
main results.

##  2 Preliminaries, hypotheses and main assumptions

In this section, we will make a survey of several results and properties that
will be used along the paper. The main part of these results are proven in
[5], and many of them are given in [3].

Let us consider the $n^{\rm th}$-order linear differential equation (1.1)
coupled to the two-point boundary conditions (1.2)–(1.3).

######  Definition 2.1.

Let us say that
$\\{\sigma_{1},\dots,\sigma_{k}\\}-\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}$
satisfy property $(N_{a})$ if

$\sum_{\sigma_{j}<h}1+\sum_{\varepsilon_{j}<h}1\geq h\,,\quad\forall
h\in\\{1,\dots,n-1\\}\,.$

######  Notation 2.2.

Let us denote

$\displaystyle\alpha=\min\left\\{i\in\\{0,\dots,n-1\\}:\;\;i\notin\\{\sigma_{1},\dots,\sigma_{k}\\}\right\\}\,,$
(2.1)
$\displaystyle\beta=\min\left\\{i\in\\{0,\dots,n-1\\}:\;\;i\notin\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}\right\\}\,.$
(2.2)

We introduce the following set of functions related to the boundary conditions
(1.2)–(1.3):

$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}=\left\\{u\in
C^{n}(I)\ \mid\
u^{(\sigma_{1})}(a)=\cdots=u^{(\sigma_{k})}(a)=u^{(\varepsilon_{1})}(b)=\cdots=u^{(\varepsilon_{n-k})}(b)=0\right\\}.$

Also, we define the following sets of functions:

$\begin{split}X_{\\{\sigma_{1},\dots,\sigma_{k-1}|\alpha\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}&=\left\\{u\in
C^{n}(I)\ \mid\ u^{(\sigma_{1})}(a)=\cdots=u^{(\sigma_{k-1})}(a)=0\,,\
u^{(\alpha)}(a)=0\,,u^{(\varepsilon_{1})}(b)=\cdots=u^{(\varepsilon_{n-k})}(b)=0\right\\},\\\
X_{\\{\sigma_{1},\dots,\sigma_{k}|\alpha\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k-1}\\}}&=\left\\{u\in
C^{n}(I)\ \mid\ u^{(\sigma_{1})}(a)=\cdots=u^{(\sigma_{k})}(a)=0\,,\
u^{(\alpha)}(a)=0\,,u^{(\varepsilon_{1})}(b)=\cdots=u^{(\varepsilon_{n-k-1})}(b)=0\right\\},\\\
X_{\\{\sigma_{1},\dots,\sigma_{k-1}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}|\beta\\}}&=\left\\{u\in
C^{n}(I)\ \mid\
u^{(\sigma_{1})}(a)=\cdots=u^{(\sigma_{k-1})}(a)=0\,,u^{(\varepsilon_{1})}(b)=\cdots=u^{(\varepsilon_{n-k})}(b)=0\,,\
u^{(\beta)}(b)=0\right\\},\\\
X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k-1}|\beta\\}}&=\left\\{u\in
C^{n}(I)\ \mid\
u^{(\sigma_{1})}(a)=\cdots=u^{(\sigma_{k})}(a)=0\,,u^{(\varepsilon_{1})}(b)=\cdots=u^{(\varepsilon_{n-k-1})}(b)=0\,,\
u^{(\beta)}(b)=0\right\\}.\end{split}$

######  Definition 2.3.

Let us say that the operator $T_{n}[M]$ satisfies the property $(T_{d})$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
if, and only if, there exists the following decomposition:

$T_{0}\,u(t)=u(t)\,,\quad
T_{l}\,u(t)=\dfrac{d}{dt}\left(\dfrac{T_{l-1}\,u(t)}{v_{l}(t)}\right)\,,\
l=1,\dots,n\,,\;t\in I,$ (2.3)

where $v_{l}>0$ on $I$, $v_{l}\in C^{n}(I)$ such that

$T_{n}[M]\,u(t)=v_{1}(t)\,\dots\,v_{n}(t)\,T_{n}\,u(t)\,,\ t\in I\,,$

and, moreover, such a decomposition satisfies, for every $u\in
X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$:

$\displaystyle T_{\sigma_{1}}\,u(a)\hskip 1.13791pt=$ $\displaystyle\cdots$
$\displaystyle=T_{\sigma_{k}}\,u(a)\hskip 8.5359pt=0\,,$ $\displaystyle
T_{\varepsilon_{1}}\,u(b)=$ $\displaystyle\cdots$
$\displaystyle=T_{\varepsilon_{n-k}}\,u(b)=0\,.$

######  Remark 2.4.

Note that decomposition (2.3) is not unique, it depends on the choice of
$v_{k}$ for $k=1,\dots,n$. Throughout this work, for the operator
$T_{n}[0]\,u(t)=u^{(n)}(t)$ we choose the following decomposition:

$T_{0}\,u(t)=u(t)\,,\quad
T_{k}\,u(t)=\dfrac{d}{dt}\left(\dfrac{T_{k-1}\,u(t)}{v_{k}(t)}\right)\,,\
k=1,\dots,n\,,\;\;t\in I,$

where $v_{k}\equiv 1$ on $I$. That is, $T_{k}\,u(t)=u^{(k)}(t)$, $t\in I$. In
particular:

$T_{n}[0]\,u(t)=v_{1}(t)\,\dots\,v_{n}(t)\,T_{n}\,u(t)\,,\ t\in I\,,$

and, moreover, this decomposition satisfies, for every $u\in
X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$:

$\displaystyle T_{\sigma_{1}}\,u(a)=u^{(\sigma_{1})}(a)=0\,,$
$\displaystyle\dots\,,$ $\displaystyle T_{\sigma_{k}}\,u(a)\hskip
8.5359pt=u^{(\sigma_{k})}(a)\hskip 8.5359pt=0\,,$ $\displaystyle
T_{\varepsilon_{1}}\,u(b)=u^{(\varepsilon_{1})}(b)=0\,,$
$\displaystyle\dots\,,$ $\displaystyle
T_{\varepsilon_{n-k}}\,u(b)=u^{(\varepsilon_{n-k})}(b)=0\,,$

or which is the same $T_{n}[0]$ satisfies property $(T_{d})$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.

######  Lemma 2.5.

[5, Lemma 3.8] Let $\bar{M}\in\mathbb{R}$ be such that $T_{n}[\bar{M}]$
satisfies the property $(T_{d})$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.
Then
$\\{\sigma_{1},\dots,\sigma_{k}\\}-\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}$
satisfy the property $(N_{a})$ if, and only if, $M=0$ is not an eigenvalue of
$T_{n}[\bar{M}]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.

Let us denote by $G_{M}$ the Green’s matrix associated with the equivalent
$n$-dimensional first order system associated to (1.1)–(1.3), which is given
by the expression

$G_{M}(t,s)=\left(\begin{array}[]{ccccc}g_{1}(t,s)&g_{2}(t,s)&\cdots&g_{n-1}(t,s)&g_{M}(t,s)\\\
&&&&\\\ \dfrac{\partial}{\partial t}\,g_{1}(t,s)&\dfrac{\partial}{\partial
t}\,g_{2}(t,s)&\cdots&\dfrac{\partial}{\partial
t}\,g_{n-1}(t,s)&\dfrac{\partial}{\partial t}\,g_{M}(t,s)\\\
\vdots&\vdots&\cdots&\vdots&\vdots\\\ \dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{1}(t,s)&\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{2}(t,s)&\cdots&\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{n-1}(t,s)&\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{M}(t,s)\end{array}\right),$ (2.4)

where $g_{M}$ is the scalar Green’s function related to operator $T_{n}[M]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
(see [1, Section 1.4] for details). Moreover, by the axiomatic definition of
$G_{M}$, we have that for all $s\in(a,b)$, the following equality holds

$\dfrac{\partial}{\partial t}\,G_{M}(t,s)=A\,G_{M}(t,s),\quad\text{for all
}t\in I\backslash\left\\{s\right\\},$ (2.5)

where $A=\left(\begin{array}[]{c|c}0&I_{n-1}\\\
\hline\cr-M&0\end{array}\right)$, with $I_{n-1}$ the $(n-1)$ dimensional
identity matrix, denotes the matrix that defines the $n$-dimensional system
equivalent to our problem (1.1)–(1.3) and

$B\,G_{M}(a,s)+C\,G_{M}(b,s)=0,\;\;\text{for all}\;\;s\in(a,b),$ (2.6)

with $B,\,\ C\in\mathcal{M}_{n\times n}$, defined by $b_{j,1+\sigma_{j}}=1$
for $j=1,\dots,k$ and $c_{j+k,1+\varepsilon_{j}}=1$ for ${j=1,\dots,n-k}$;
otherwise, $b_{ij}=0$ and $c_{ij}=0$.

In particular, it is easy to see, using the expressions (2.4) and (2.5) (see
[1, Section 1.4] for details), that $g_{M}\in C^{n-2}(I\times I)$. Moreover,
it is a $C^{n}$ function on the triangles $a\leq s<t\leq b$ and $a\leq t<s\leq
b$, it satisfies, as a function of $t$ for any $s\in(a,b)$ fixed, the two-
point boundary value conditions (1.2)-(1.3) and solves equation (1.1) for all
$t\in I\setminus\\{s\\}$.

Moreover, for any $t\in(a,b)$ it satisfies

$\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{M}(t^{+},t)=\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{M}(t^{-},t)+1.$ (2.7)

Studying the matrix Green’s function $G_{M}$, it is proved in [3, page 13]
that, for this particular case, $g_{n-j}(t,s)$ may be expressed as a function
of $g_{M}(t,s)$ for all $j=1,\dots,n-1$ as follows:

$g_{n-j}(t,s)=(-1)^{j}\dfrac{\partial^{j}}{\partial s^{j}}\,g_{M}(t,s)\,.$
(2.8)

The adjoint operator of $T_{n}[M]$ follows the expression

$T_{n}^{*}[M]v(t)\equiv(-1)^{n}\,v^{(n)}(t)+M\,v(t)\,,$ (2.9)

defined on the domain

$\footnotesize D(T_{n}^{*}[M])=\left\\{v\in C^{n}(I)\
\mid\sum_{i=0}^{n-1}(-1)^{n-1-i}\,v^{(n-1-i)}(b)\,u^{(i)}(b)=\sum_{i=0}^{n-1}(-1)^{n-1-i}\,v^{(n-1-i)}(a)\,u^{(i)}(a)\,\right\\},$

for all $u\in D(T_{n}[M])$.

We denote by $g_{M}^{*}(t,s)$ the Green’s function related to operator
$T_{n}^{*}[M]$. Furthermore, the following equality is satisfied (see [1,
section 1.4])

$g^{*}_{M}(t,s)=g_{M}(s,t),\,\;\;(t,s)\in I\times I.$ (2.10)

In such a case, if we define the following operator

$\widehat{T}_{n}[(-1)^{n}\,M]:=(-1)^{n}\,T_{n}^{*}[M]\,,$ (2.11)

we have, according to the above equality, that

$\widehat{g}_{(-1)^{n}\,M}(t,s)=(-1)^{n}\,g_{M}^{*}(t,s)=(-1)^{n}\,g_{M}(s,t)\,,$
(2.12)

where $\widehat{g}_{(-1)^{n}M}(t,s)$ is the scalar Green’s function related to
operator $\widehat{T}_{n}[(-1)^{n}M]$ in $D\left(T_{n}^{*}[M]\right)$.

As a particular case of the obtained results in [5, section 4], we can deduce
that

$D(T_{n}^{*}[M])\equiv X_{\
\,\\{\sigma_{1},\dots,\sigma_{k}\\}}^{*\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}=X_{\\{\tau_{1},\dots,\tau_{n-k}\\}}^{\\{\delta_{1},\dots,\delta_{k}\\}},$

being
$\\{\delta_{1},\dots,\delta_{k}\\}\,,\\{\tau_{1},\dots,\tau_{n-k}\\}\subset\\{0,\dots,n-1\\}$,
such that $\delta_{i}<\delta_{i+1}$ and $\tau_{j}<\tau_{j+1}$, for
$i=1,\dots,k-1$ and $j=1,\dots,n-k-1$, satisfying:

$\begin{split}\\{\sigma_{1},\dots,\sigma_{k},n-1-\tau_{1},\dots,\,n-1-\tau_{n-k}\\}=&\\{0,\dots,n-1\\},\\\
\\{\varepsilon_{1},\dots,\varepsilon_{n-k},n-1-\delta_{1},\dots,\,n-1-\delta_{k}\\}=&\\{0,\dots,n-1\\}.\end{split}$

######  Notation 2.6.

Let us define $\eta$, $\gamma$ in the following way:

$\displaystyle\eta=\min\left\\{i\in\\{0,\dots,n-1\\}:\;\;i\notin\\{\tau_{1},\dots,\tau_{n-k}\\}\right\\}\,,$
(2.13)
$\displaystyle\gamma=\min\left\\{i\in\\{0,\dots,n-1\\}:\;\;i\notin\\{\delta_{1},\dots,\delta_{k}\\}\right\\}\,.$
(2.14)

Analogously, let us denote by $\widehat{G}(t,s)$ the Green’s matrix associated
with the equivalent $n$-dimensional problem of
$\widehat{T}_{n}[(-1)^{n}\,M]\,v(t)=0$, $v\in
X_{\\{\tau_{1},\dots,\tau_{n-k}\\}}^{\\{\delta_{1},\dots,\delta_{k}\\}}$ given
by

$\widehat{G}(t,s)=\left(\begin{array}[]{llll}\widehat{g}_{1}(t,s)&\cdots&\widehat{g}_{n-1}(t,s)&\widehat{g}_{(-1)^{n}\,M}(t,s)\\\
&&&\\\ \dfrac{\partial}{\partial
t}\,\widehat{g}_{1}(t,s)&\cdots&\dfrac{\partial}{\partial
t}\,\widehat{g}_{n-1}(t,s)&\dfrac{\partial}{\partial
t}\,\widehat{g}_{(-1)^{n}\,M}(t,s)\\\
\qquad\vdots&\cdots&\qquad\vdots&\qquad\vdots\\\
\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,\widehat{g}_{1}(t,s)&\cdots&\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,\widehat{g}_{n-1}(t,s)&\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,\widehat{g}_{(-1)^{n}\,M}(t,s)\end{array}\right),$ (2.15)

where $\widehat{g}_{(-1)^{n}\,M}(t,s)$ is the scalar Green’s function related
to operator $\widehat{T}_{n}[(-1)^{n}\,M]$ in
$X_{\\{\tau_{1},\dots,\tau_{n-k}\\}}^{\\{\delta_{1},\dots,\delta_{k}\\}}$.

Similarly to (2.8) we can deduce that

$\widehat{g}_{n-j}(t,s)=(-1)^{j}\dfrac{\partial^{j}}{\partial
s^{j}}\,\widehat{g}_{(-1)^{n}M}(t,s)\,.$ (2.16)

By definition, $\widehat{G}(t,s)$ satisfies the equality

$\widehat{B}\,\widehat{G}(a,s)+\widehat{C}\,\widehat{G}(b,s)=0,\;\;\text{for
all}\;\;s\in(a,b),$ (2.17)

where $\widehat{B}$, $\widehat{C}\in\mathcal{M}_{n\times n}$, defined as
$(\widehat{B})_{i,\,\tau_{i}+1}=1$ for $i=1,\dots,n-k$ and
$(\widehat{C})_{j+n-k,\,1+\delta_{j}}=1$ for $j=1,\dots,k$; otherwise,
$(\widehat{B})_{i\,j}=0$ and $(\widehat{C})_{i\,j}=0$.

######  Remark 2.7.

From the previous definitions, as it is pointed out in [5, Remarks 4.1 and
4.2] we have that $\alpha=n-1-\tau_{n-k}$, $\beta=n-1-\delta_{k}$,
$\eta=n-1-\sigma_{k}$ and $\gamma=n-1-\varepsilon_{n-k}$.

######  Example 2.8.

Let us consider operator $T_{4}[M]$ coupled with the boundary conditions

$u(0)=u^{\prime}(0)=u^{\prime\prime}(0)=u^{\prime\prime}(1)=0.$

In this case,
$\\{\sigma_{1},\sigma_{2},\sigma_{3}\\}=\\{0,1,2\\},\\{\varepsilon_{1}\\}=\\{2\\},\\{\tau_{1}\\}=\\{0\\}$
and $\\{\delta_{1},\delta_{2},\delta_{3}\\}=\\{0,2,3\\}$. Thus, we deduce that
$X_{\ \,\\{0,1,2\\}}^{*\\{2\\}}=X_{\\{0\\}}^{\\{0,2,3\\}}.$ Moreover,
$\alpha=3\,,\beta=0,\,\eta=1$ and $\gamma=1$.

######  Definition 2.9.

Operator $T_{n}[M]$ is said to be inverse positive (negative) on
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
if every $u\in
X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
such that $T_{n}[M]\,u\geq 0$ on $I$, satisfies $u\geq 0$ ($u\leq 0$) on $I$.

The above definition is equivalent to the constant sign of Green’s function
$g_{M}$ related to $T_{n}[M]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
as it is shown in the following result whose proof is analogous the one made
in [5, Theorem 2.9].

######  Theorem 2.10.

Operator $T_{n}[M]$ is inverse positive (negative) in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
if, and only if, the Green’s function related to problem (1.1)–(1.3) is non-
negative (non-positive) on its square of definition.

######  Definition 2.11.

Operator $T_{n}[M]$ is said to be strongly inverse positive (negative) in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
if every $u\in
X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
such that $T_{n}[M]\,u\gneqq 0$ on $I$, satisfies that $u>0$ ($<0$) on $(a,b)$
and, moreover, $u^{(\alpha)}(a)>0$ $(<0)$ and $(-1)^{\beta}u^{(\beta)}(b)>0$
$(<0)$, where $\alpha$ and $\beta$ are defined in (2.1) and (2.2),
respectively.

Analogously to Theorem 2.10, the following characterization can be shown:

######  Theorem 2.12.

Operator $T_{n}[M]$ is strongly inverse positive (negative) in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
if, and only if, the Green’s function related to problem (1.1)–(1.3) satisfies
the following properties:

  * •

$g_{M}(t,s)>0\,(<0)$ a.e. on $(a,b)$.

  * •

$\frac{\partial^{\alpha}}{\partial t^{\alpha}}\,g_{M}(t,s)>0\,(<0)$ for a.e.
$s\in(a,b)$.

  * •

$(-1)^{\beta}\frac{\partial^{\beta}}{\partial t^{\beta}}\,g_{M}(t,s)>0\,(<0)$
for a.e. $s\in(a,b)$.

Let $u\in
X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
be such that $T_{n}[M]\,u(t)=0$, $t\in I$. Then, differentiating equation
(1.1) with respect to $t$, as a direct consequence of equations (2.4) and
(2.5) we deduce that, for any $s\in(a,b)$ given, the $q$-th derivative
$v_{s}^{q}[M](t):=\dfrac{\partial^{q}}{\partial t^{q}}\,g_{M}(t,s)$, with
$1\leq q\leq n-1$, satisfies the equation $T_{n}[M]\,v_{s}^{q}[M](t)=0$, for
all $t\in I\setminus\\{s\\}$, together with the new boundary conditions

$\displaystyle(v_{s}^{q})^{(\mu_{1}^{q})}[M](a)=$ $\displaystyle\cdots$
$\displaystyle=(v_{s}^{q})^{(\mu_{k}^{q})}[M](a)\hskip 8.5359pt=0\,,$ (2.18)
$\displaystyle(v_{s}^{q})^{(\rho_{1}^{q})}[M](b)=$ $\displaystyle\cdots$
$\displaystyle=(v_{s}^{q})^{(\rho_{n-k}^{q})}[M](b)=0\,.$ (2.19)

Here, the sets of indices
$\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\},\;\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}\subset\\{0,\dots,n-1\\}$,
satisfy that

$0\leq\mu_{1}^{q}<\mu_{2}^{q}<\cdots<\mu_{k}^{q}\leq n-1\,,\quad
0\leq\rho_{1}^{q}<\rho_{2}^{q}<\cdots<\rho_{n-k}^{q}\leq n-1\,,$

and are defined in the following way:

  * •

If $A_{q}=\\{i\in\\{1,\dots,k\\}\,/\;\;\sigma_{i}\geq q\\}\neq\varnothing$,
then

$\mu_{1}^{q}=\sigma_{j}-q,\,\mu_{2}^{q}=\sigma_{j+1}-q,\,\dots,\mu_{k-(j-1)}^{q}=\sigma_{k}-q,\,\mu_{k-(j-2)}^{q}=\sigma_{1}-q+n,\dots,\,\mu_{k}^{q}=\sigma_{j-1}-q+n\,,$

with

$j=\min A_{q}\geq 1.$ (2.20)

  * •

If $A_{q}=\varnothing$, then

$\mu_{i}^{q}=\sigma_{i}-q+n,\;\;i=1,\ldots,k.$

  * •

If $B_{q}=\\{i\in\\{1,\dots,n-k\\}\,/\;\;\varepsilon_{i}\geq
q\\}\neq\varnothing$, then

$\displaystyle\rho_{1}^{q}$
$\displaystyle=\varepsilon_{r}-q,\,\rho_{2}^{q}=\varepsilon_{r+1}-q,\,\dots,\rho_{n-k-(r-1)}^{q}=\varepsilon_{n-k}-q,\,\rho_{n-k-(r-2)}^{q}=\varepsilon_{1}-q+n,\dots,$
$\displaystyle\rho_{n-k}^{q}$ $\displaystyle=\varepsilon_{r-1}-q+n,$

with

$r=\min B_{q}\geq 1.$ (2.21)

  * •

If $B_{q}=\varnothing$, then

$\rho_{i}^{q}=\varepsilon_{i}-q+n,\;\;i=1,\ldots,n-k.$

######  Remark 2.13.

The above process of calculating the boundary conditions of $v_{s}^{q}[M]$ is
valid for both $M\neq 0$ and $M=0$.

Taking into account previous computations, we may define $\phi$ as the
function which maps the space
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
into the space of the boundary conditions that satisfies the first derivative,
that is,
$X_{\\{\mu_{1}^{1},\dots,\mu_{k}^{1}\\}}^{\\{\rho_{1}^{1},\dots,\rho_{n-k}^{1}\\}}$.
This way, we can extend this definition to the $q$-th derivative as

$\phi^{q}=\phi\circ\overset{q)}{\cdots}\circ\phi:X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}\longrightarrow
X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}.$

In particular, $\phi^{n}$ is the identity function.

######  Example 2.14.

Consider the operator $T_{5}[M]$ defined in $X_{\\{0,2,4\\}}^{\\{0,2\\}}$.
Applying the above definition of the spaces of the derivatives we obtain the
following sequence of spaces:

${X_{\\{0,2,4\\}}^{\\{0,2\\}}}$${X_{\\{1,3,4\\}}^{\\{1,4\\}}}$${X_{\\{0,2,3\\}}^{\\{0,3\\}}}$${X_{\\{1,2,4\\}}^{\\{2,4\\}}}$${X_{\\{0,1,3\\}}^{\\{1,3\\}}}$${X_{\\{0,2,4\\}}^{\\{0,2\\}}.}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$

Analogously to the Notation 2.2, let us define $\alpha^{q}$, $\beta^{q}$ as
follows:

$\displaystyle\alpha^{q}=\min\left\\{i\in\\{0,\dots,n-1\\}:\;\;i\notin\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}\right\\}\,,$
(2.22)
$\displaystyle\beta^{q}=\min\left\\{i\in\\{0,\dots,n-1\\}:\;\;i\notin\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}\right\\}\,.$
(2.23)

######  Remark 2.15.

It is important to point out that for any $q\geq 1$, the function
$v_{s}^{q}[M]$ is not the Green’s function related to the operator $T_{n}[M]$
in
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$.
This is due to the fact that $v_{s}^{q}[M]\in C^{n-2-q}(I\times I)$.

######  Remark 2.16.

The operator $T_{n}[0]$ satisfies the property $(T_{d})$ in
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$.
From Remark 2.4, we know that operator $T_{n}[0]\,u(t)=u^{(n)}(t)$ satisfies
the following decomposition:

$T_{0}\,u(t)=u(t)\,,\quad
T_{k}\,u(t)=\dfrac{d}{dt}\left(\dfrac{T_{k-1}\,u(t)}{v_{k}(t)}\right)\,,\
k=1,\dots,n\,,$

where $v_{k}\equiv 1$ on $I$ and

$T_{n}[0]\,u(t)=v_{1}(t)\,\dots\,v_{n}(t)\,T_{n}\,u(t)\,,\ t\in I\,,$

that is, $T_{k}\,u(t)=u^{(k)}(t)$.

Moreover, this decomposition satisfies, for every $u\in
X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$:

$\displaystyle T_{\mu_{1}^{q}}\,u(a)=u^{(\mu_{1}^{q})}(a)=0\,,$
$\displaystyle\dots\,,$ $\displaystyle T_{\mu_{k}^{q}}\,u(a)\hskip
8.5359pt=u^{(\mu_{k}^{q})}(a)\hskip 8.5359pt=0\,,$ $\displaystyle
T_{\rho_{1}^{q}}\,u(b)=u^{(\rho_{1}^{q})}(b)=0\,,$ $\displaystyle\dots\,,$
$\displaystyle T_{\rho_{n-k}^{q}}\,u(b)=u^{(\rho_{n-k}^{q})}(b)=0\,.$

We consider the general expression of the solution of equation

$T_{n}[0]\,u(t)+\lambda\,u(t)\equiv u^{(n)}(t)+\lambda\,u(t)=0,\;\;t\in I,$

to obtain the different eigenvalues.

Next we present a result concerning the spectrum of the spaces
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$.

######  Lemma 2.17.

$\bar{\lambda}\neq 0$ is an eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
if and only if $\bar{\lambda}$ is an eigenvalue of $T_{n}[0]$ in
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
for all $q\in\\{1,\dots,n-1\\}$.

###### Proof.

Suppose that $\bar{\lambda}\neq 0$ is an eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.
So, there is $u\not\equiv 0$ in $I$ such that
$u^{(n)}(t)+\bar{\lambda}\,u(t)=0$, $t\in I$ and $u\in
X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.
Since $u\not\equiv 0$ in $I$ we have that $u^{(n)}\not\equiv 0$ in $I$.
Therefore, $u^{(j)}\not\equiv 0$ in $I$ for all $j\in\\{1,\dots,n-1\\}$.

Taking into account that
$T_{n}[0]\,u^{(j)}(t)=u^{(j+n)}(t)=-\bar{\lambda}\,u^{(j)}(t)$ for all $t\in
I$ and $j\in{\mathbb{N}}$, we have that $u^{(j)}$ is an eigenvector related to
the eigenvalue $\bar{\lambda}$ such that $u^{(j)}\in
X_{\\{\mu_{1}^{j},\dots,\mu_{k}^{j}\\}}^{\\{\rho_{1}^{j},\dots,\rho_{n-k}^{j}\\}}$
for all $j\in\\{1,\dots,n-1\\}$.

For $j=n$, we have that $\mu_{i}^{n}=\sigma_{i}$ for all $i\in\\{1,\dots,k\\}$
and $\rho_{l}^{n}=\varepsilon_{l}$ for all $l\in\\{1,\dots,n-k\\}$. Therefore,
the result holds. ∎

######  Remark 2.18.

As a direct consequence of Remark 2.4, we have, from Lemma 2.5, that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$ if and only if $M=0$ is not an eigenvalue of
$u^{(n)}$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.

######  Remark 2.19.

Note that if $\lambda=0$ is not an eigenvalue of $T_{n}[0]$ in
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$,
then the spaces
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
and
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
have the same spectrum. However, it is very important to point out that
$\lambda=0$ can appear as an eigenvalue in some space
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$,
as it is shown in the following example.

######  Example 2.20.

Let us consider again the Example 2.14. As we know, the following sequence of
spaces is fulfilled:

${X_{\\{0,2,4\\}}^{\\{0,2\\}}}$${X_{\\{1,3,4\\}}^{\\{1,4\\}}}$${X_{\\{0,2,3\\}}^{\\{0,3\\}}}$${X_{\\{1,2,4\\}}^{\\{2,4\\}}}$${X_{\\{0,1,3\\}}^{\\{1,3\\}}}$${X_{\\{0,2,4\\}}^{\\{0,2\\}}.}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$

In this case, we have that $\lambda=0$ is not an eigenvalue of $T_{5}[0]$ in
$X_{\\{0,2,4\\}}^{\\{0,2\\}}$, $X_{\\{0,2,3\\}}^{\\{0,3\\}}$ and
$X_{\\{0,1,3\\}}^{\\{1,3\\}}$, while $\lambda=0$ is an eigenvalue of
$T_{5}[0]$ in $X_{\\{1,3,4\\}}^{\\{1,4\\}}$ and $X_{\\{1,2,4\\}}^{\\{2,4\\}}$.

Now, we will show how to determine the adjoint space of
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
from the adjoint space of
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.
We arrive to the next result.

######  Theorem 2.21.

The adjoint space of
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
is given by
$X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k}^{q}\\}}$,
where the indices
$\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}\\}\,,\\{\delta_{1}^{q},\dots,\delta_{k}^{q}\\}\subset\\{0,\dots,n-1\\}$
satisfy

$0\leq\tau_{1}^{q}<\tau_{2}^{q}<\cdots<\tau_{n-k}^{q}\leq n-1\,,\quad
0\leq\delta_{1}^{q}<\delta_{2}^{q}<\cdots<\delta_{k}^{q}\leq n-1\,,$

and are defined as follows:

  * •

If $C_{q}=\\{i\in\\{1,\dots,n-k\\}\,/\;\;\tau_{i}+q\leq
n-1\\}\neq\varnothing$, then

$\displaystyle\tau_{1}^{q}$
$\displaystyle=\tau_{l+1}+q-n,\,\tau_{2}^{q}=\tau_{l+2}+q-n,\,\dots,\tau_{n-k-l}^{q}=\tau_{n-k}+q-n,\,\tau_{n-k-l+1}^{q}=\tau_{1}+q,\dots,\,$
$\displaystyle\tau_{n-k}^{q}$ $\displaystyle=\tau_{l}+q,$

with

$l=\max C_{q}.$ (2.24)

  * •

If $C_{q}=\varnothing$, then

$\tau_{i}^{q}=\tau_{i}+q-n,\;\;i=1,\ldots,n-k.$

  * •

If $D_{q}=\\{i\in\\{1,\dots,k\\}\,/\;\;\delta_{i}+q\leq
n-1\\}\neq\varnothing$, then

$\delta_{1}^{q}=\delta_{p+1}+q-n,\,\delta_{2}^{q}=\delta_{p+2}+q-n,\,\dots,\delta_{k-p}^{q}=\delta_{k}+q-n,\,\delta_{k-p+1}^{q}=\delta_{1}+q,\dots,\delta_{k}^{q}=\delta_{p}+q,$

with

$p=\max D_{q}.$ (2.25)

  * •

If $D_{q}=\varnothing$, then

$\delta_{i}^{q}=\delta_{i}+q-n,\;\;i=1,\ldots,k.$

As a consequence, the following diagram

${X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}}$${X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}}$${X_{\\{\tau_{1},\dots,\tau_{n-k}\\}}^{\\{\delta_{1},\dots,\delta_{k}\\}}}$${X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k}^{q}\\}}}$$\scriptstyle{\phi^{q}}$$\scriptstyle{*}$$\scriptstyle{*}$$\scriptstyle{\phi^{q}}$
(2.26)

is commutative.

###### Proof.

We do the proof only for $\tau_{i}^{q}$. For $\delta_{i}^{q}$, the proof is
done in a similar way. To prove it, we start from the definition of
$\tau_{i}$, we have that $\tau_{i}<\tau_{i+1}$ for all
$i\in\\{1,\dots,n-k-1\\}$ and

$\\{\sigma_{1},\dots,\sigma_{k},n-1-\tau_{n-k},\dots,\,n-1-\tau_{1}\\}\equiv\\{0,\dots,n-1\\}.$

Taking into account the definition of
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
we have that

${\\{\sigma_{1},\dots,\sigma_{k},n-1-\tau_{n-k},\dots,\,n-1-\tau_{1}\\}}$${\\{\mu_{1}^{q},\dots,\mu_{k}^{q},n-1-\tau_{n-k}^{q},\dots,\,n-1-\tau_{1}^{q}\\},}$$\scriptstyle{\phi^{q}}$

with
$\\{\mu_{1}^{q},\dots,\mu_{k}^{q},n-1-\tau_{n-k}^{q},\dots,\,n-1-\tau_{1}^{q}\\}\equiv\\{0,\dots,n-1\\}$
where $\tau_{i}^{q}$ has to fulfill that

$0\leq\tau_{1}^{q}<\tau_{2}^{q}<\cdots<\tau_{n-k}^{q}\leq n-1\,.$

To calculate them, we distinguish two cases:

  * 1)

If $C_{q}\neq\varnothing$, then:

    * –

If $n-1-\tau_{i}\geq q$ (i.e, $i\leq l$), $n-1-\tau_{i}$ becomes
$n-1-\tau_{i}-q$ by applying $\phi^{q}$.

    * –

If $n-1-\tau_{i}<q$ (i.e, $i>l$), $n-1-\tau_{i}$ becomes $n-1-\tau_{i}-q+n$ by
applying $\phi^{q}$.

Reordering the elements, we obtain that:

$\displaystyle 0$ $\displaystyle\leq
n-1-\tau_{l}-q<n-1-\tau_{l-1}-q<\cdots<n-1-\tau_{1}-q$
$\displaystyle<n-1-\tau_{n-k}-q+n<\cdots<n-1-\tau_{l+1}-q+n\leq n-1,$

or, equivalently,

$0\leq\tau_{l+1}+q-n<\tau_{l+2}+q-n<\cdots<\tau_{n-k}+q-n<\tau_{1}+q<\cdots<\tau_{l}+q\leq
n-1\,.$

Taking

$\displaystyle\tau_{1}^{q}$
$\displaystyle=\tau_{l+1}+q-n,\,\tau_{2}^{q}=\tau_{l+2}+q-n,\,\dots,\tau_{n-k-l}^{q}=\tau_{n-k}+q-n,\,\tau_{n-k-l+1}^{q}=\tau_{1}+q,\dots,\,$
$\displaystyle\tau_{n-k}^{q}$ $\displaystyle=\tau_{l}+q\,,$

the first statement yields.

  * 2)

If $C_{q}=\varnothing$, then $n-1-\tau_{i}$ becomes $n-1-\tau_{i}-q+n$ by
applying $\phi^{q}$ and we obtain that

$0\leq n-1-\tau_{n-k}-q+n<\cdots<n-1-\tau_{2}-q+n<n-1-\tau_{1}-q+n\leq n-1\,,$

and, reordering, we infer that

$0\leq\tau_{1}+q-n<\tau_{2}+q-n<\cdots<\tau_{n-k}+q-n\leq n-1\,.$

Taking $\tau_{i}^{q}=\tau_{i}+q-n,\;\;i=1,\ldots,n-k$, the second statement
yields.

Therefore, $X_{\
\,\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{*\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}=X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k}^{q}\\}}$.
∎

######  Remark 2.22.

Note that, unlike the direct arrows between the leading spaces, the arrows
between the adjoint spaces are in the inverse direction.

######  Example 2.23.

Consider the operator $T_{4}[M]$ coupled with the boundary conditions

$u(a)=u^{\prime\prime}(a)=u^{\prime}(b)=u^{\prime\prime}(b).$

In this case, $u\in X_{\\{0,2\\}}^{\\{1,2\\}}$ and $X_{\
\,\\{0,2\\}}^{*\\{1,2\\}}=X_{\\{0,2\\}}^{\\{0,3\\}}$. Using the diagram
(2.26), we get that

${X_{\\{0,2\\}}^{\\{1,2\\}}}$${X_{\\{1,3\\}}^{\\{0,1\\}}}$${X_{\\{0,2\\}}^{\\{0,3\\}}}$${X_{\\{1,3\\}}^{\\{2,3\\}}}$${X_{\\{0,2\\}}^{\\{0,3\\}}}$${X_{\\{1,3\\}}^{\\{0,1\\}}}$${X_{\\{0,2\\}}^{\\{1,2\\}}}$${X_{\\{1,3\\}}^{\\{2,3\\}}}$$\scriptstyle{\phi}$$\scriptstyle{*}$$\scriptstyle{\phi}$$\scriptstyle{*}$$\scriptstyle{\phi}$$\scriptstyle{*}$$\scriptstyle{*}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$$\scriptstyle{\phi}$

Using the fact that the adjoint operators have the same spectrum as the
original one, from Lemma 2.17, we arrive at the following result.

######  Lemma 2.24.

$\lambda^{*}\neq 0$ is an eigenvalue of $T_{n}[0]$ in
$X_{\\{\tau_{1},\dots,\tau_{n-k}\\}}^{\\{\delta_{1},\dots,\delta_{k}\\}}$ if
and only if $\lambda^{*}$ is an eigenvalue of $T_{n}[0]$ in
$X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k}^{q}\\}}$
for all $q\in\\{0,\dots,n-1\\}$. Moreover, for each fixed
$q\in\\{0,\dots,n-1\\}$, $\bar{\lambda}\in{\mathbb{R}}$ is an eigenvalue of
$T_{n}[0]$ in
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
if and only if $\bar{\lambda}$ is an eigenvalue of $T_{n}[0]$ in
$X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k}^{q}\\}}$.

Following the line of Definition 2.11 and the characterization of Theorem
2.12, we introduce the following concept for $v_{s}^{q}[M]$.

######  Definition 2.25.

Let $q\in\\{0,\dots,n-1\\}$ be fixed. We said that the function $v_{s}^{q}[M]$
is strongly positive (strongly negative) on $I\times I$ if it satisfies the
following properties:

  * •

$v_{s}^{q}[M]>0\,(<0)$ a.e. on $(a,b)$.

  * •

$(v_{s}^{q})^{(\alpha^{q})}[M](a)>0\,(<0)$ for a.e. $s\in(a,b)$, with
$\alpha^{q}$ defined in (2.22).

  * •

$(-1)^{\beta^{q}}(v_{s}^{q})^{(\beta^{q})}[M](b)>0\,(<0)$ for a.e.
$s\in(a,b)$, with $\beta^{q}$ defined in (2.23).

Let us consider the following two conditions on $g_{M}(t,s)$ introduced in [1,
pages 78 and 86] as follows:

  * $(P_{g}$)

Suppose that there is a continuous function $\phi(t)>0$ for all $t\in(a,b)$
and $k_{1},\ k_{2}\in L^{1}(I)$, such that $0<k_{1}(s)<k_{2}(s)$ for a.e.
$s\in I$, satisfying

$\phi(t)\,k_{1}(s)\leq g_{M}(t,s)\leq\phi(t)\,k_{2}(s)\,,\quad\text{for a.e.
}(t,s)\in I\times I\,.$

  * ($N_{g}$)

Suppose that there is a continuous function $\phi(t)>0$ for all $t\in(a,b)$
and $k_{1},\ k_{2}\in L^{1}(I)$, such that $k_{1}(s)<k_{2}(s)<0$ for a.e.
$s\in I$, satisfying

$\phi(t)\,k_{1}(s)\leq g_{M}(t,s)\leq\phi(t)\,k_{2}(s)\,,\quad\text{for a.e.
}(t,s)\in I\times I\,.$

As a particular case of [5, Theorem 5.1], the following result is attained.

######  Theorem 2.26.

Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$. Then the following properties are fulfilled:

  * •

If $n-k$ is even, then $T_{n}[0]$ is strongly inverse positive in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
and, moreover, the related Green’s function, $g_{0}(t,s)$, satisfies
$(P_{g})$.

  * •

If $n-k$ is odd, then $T_{n}[0]$ is strongly inverse negative in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
and, moreover, the related Green’s function, $g_{0}(t,s)$, satisfies
$(N_{g})$.

Next, for any $q\in\\{1,\dots,n-1\\}$ we determine the positive (negative)
sign of $v_{s}^{q}(t):=v_{s}^{q}[0](t)$ on the interval $I$.

Let us define
${\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}},{\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}}\subset\\{0,\dots,n-1\\}$,
in the following way:

$\displaystyle\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}$ $\displaystyle=$
$\displaystyle\\{\sigma_{1}-q,\dots,\sigma_{k}-q\\}\cap\\{0,\dots,n-q-1\\},$
(2.27) $\displaystyle\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}$
$\displaystyle=$
$\displaystyle\\{\varepsilon_{1}-q,\dots,\varepsilon_{n-k}-q\\}\cap\\{0,\dots,n-q-1\\}.$
(2.28)

If
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies the condition $(N_{a})$ then $c_{q}+d_{q}\leq n-q$. Moreover, if
$c_{q}+d_{q}<n-q$ then
${\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}}-{\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}}$
does not satisfy $(N_{a})$, and in this case $\lambda=0$ is an eigenvalue of
$X_{\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}}^{\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}}$
(we cannot study in this case the constant sign of its corresponding
derivative $v_{s}^{q}[M]$).

######  Remark 2.27.

Taking into account that function $v_{s}^{q}(t)=\frac{\partial^{q}}{\partial
t^{q}}\,g_{0}(t,s)$ is the Green’s function related to operator $T_{n-q}[0]$
in the space
$X_{\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}}^{\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}}$,
then operator $T_{n-q}[0]$ is strongly inverse positive (negative) in the
space
$X_{\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}}^{\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}}$
if and only if it satisfies the Definition 2.11 with the same $\alpha^{q}$ and
$\beta^{q}$ since

$\mu_{1}^{q}=\sigma_{1}^{q},\,\mu_{2}^{q}=\sigma_{2}^{q},\,\dots,\mu_{k-(j-1)}^{q}=\sigma_{c_{q}}^{q},$

and

$\rho_{1}^{q}=\varepsilon_{1}^{q},\,\rho_{2}^{q}=\varepsilon_{2}^{q},\,\dots,\rho_{n-k-(r-1)}^{q}=\varepsilon_{d_{q}}^{q}.$

######  Theorem 2.28.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$ and $c_{q}+d_{q}=n-q$. The following properties
are fulfilled:

  * (a)

If $c_{q}\geq 1$ and $d_{q}\geq 1$, then $v_{s}^{q}$ is strongly positive on
$I\times I$ if $n-q-c_{q}$ is even and $v_{s}^{q}$ is strongly negative on
$I\times I$ if $n-q-c_{q}$ is odd.

  * (b)

If $c_{q}=n-q$ and $d_{q}=0$, then $v_{s}^{q}(t)>0,\dots,v_{s}^{n-1}(t)>0$ in
$(s,b]$, ${v_{s}^{q}(t)=\dots=v_{s}^{n-2}(t)=0}$ in $[a,s]$ and
$v_{s}^{n-1}(t)=0$ in $[a,s)$.

  * (c)

If $c_{q}=0$ and $d_{q}=n-q$, then $v_{s}^{q}(t)>0$ in $[a,s)$ if $q=n-l$ with
$l$ even and $v_{s}^{q}(t)<0$ in $[a,s)$ if $q=n-l$ with $l$ odd. Moreover,
$v_{s}^{q}(t)=\dots=v_{s}^{n-2}(t)=0$ in $[s,b]$ and $v_{s}^{n-1}(t)=0$ in
$(s,b]$.

###### Proof.

  * (a)

In this case, it easy to verify that the space
$X_{\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}}^{\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}}$
satisfies the condition $(N_{a})$ since

$\sum_{\sigma_{j}-q<h}1+\sum_{\varepsilon_{j}-q<h}1\geq h\,,\quad\forall
h\in\\{1,\dots,n-q-1\\}\,.$

So, taking into account the fact that $v_{s}^{q}$ is the Green’s function
related to operator $T_{n-q}[0]$ in the space
$X_{\\{\sigma_{1}^{q},\dots,\sigma_{c_{q}}^{q}\\}}^{\\{\varepsilon_{1}^{q},\dots,\varepsilon_{d_{q}}^{q}\\}}$,
we apply Theorem 2.26, and we have proved the first statement.

  * (b)

Since $v_{s}^{(n)}(t)=0$ for all $t\neq 0$, we have, by (2.7) and
$v_{s}^{(n-1)}(a)=0$, that $\frac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{0}(t,s)=0$ if $t<s$ and $\frac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{0}(t,s)=1$ if $t>s$.

Since $v_{s}^{h}(a)=0$ for all $q\leq h\leq n-2$, and $g_{0}\in
C^{n-2}(I\times I)$, we deduce that $\frac{\partial^{h}}{\partial
t^{h}}\,g_{0}(t,s)=0$ if $t<s$ and $\frac{\partial^{h}}{\partial
t^{h}}\,g_{0}(t,s)>0$ if $t>s$. So, we conclude the proof of second claim.

  * (c)

In this case arguing as in the previous case, since $v_{s}^{(n)}(b)=0$, we
have that$\frac{\partial^{n-1}}{\partial t^{n-1}}\,g_{0}(t,s)=0$ if $t>s$ and
$\frac{\partial^{n-1}}{\partial t^{n-1}}\,g_{0}(t,s)=-1$ if $t<s$.

As consequence, since $v_{s}^{h}(b)=0$ for all $q\leq h\leq n-2$, and
$g_{0}\in C^{n-2}(I\times I)$, we deduce, by recurrence, that
$\frac{\partial^{h}}{\partial t^{h}}g_{0}(t,s)=0$ for all $t>s$ and
$(-1)^{n-h}\frac{\partial^{h}}{\partial t^{h}}g_{0}(t,s)>0$ for all $t<s$.
Thus, the third assertion holds.

∎

Let $q\in\\{0,\dots,n-1\\}$ be fixed. Suppose that $(-1)^{n-q-
c_{q}}\,v_{s}^{q}$ is strongly positive on $I\times I$. Then, for each
$s\in(a,b)$, we obtain the following limits:

$\displaystyle\ell_{1}^{q}(s)$ $\displaystyle:=$
$\displaystyle\lim_{t\rightarrow a^{+}}\dfrac{(-1)^{n-q-
c_{q}}\,v_{s}^{q}(t)}{(t-a)^{\alpha^{q}}\,(b-t)^{\beta^{q}}}=\dfrac{(-1)^{n-q-
c_{q}}\,(v_{s}^{q})^{(\alpha^{q})}(a)}{\alpha^{q}!\,(b-a)^{\beta^{q}}}\,,$
$\displaystyle\ell_{2}^{q}(s)$ $\displaystyle:=$
$\displaystyle\lim_{t\rightarrow b^{-}}\dfrac{(-1)^{n-q-
c_{q}}\,v_{s}^{q}(t)}{(t-a)^{\alpha^{q}}\,(b-t)^{\beta^{q}}}=\dfrac{(-1)^{n-q-
c_{q}-\beta^{q}}\,(v_{s}^{q})^{(\beta^{q})}(b)}{\beta^{q}!\,(b-a)^{\alpha^{q}}}.$

For each $s\in(a,b)$, let us consider the following function defined on $I$ by

$\displaystyle\tilde{u}_{s}^{q}(t)=\left\\{\begin{array}[]{ccc}\ell_{1}^{q}(s)\,,&t=a,\\\
\dfrac{(-1)^{n-q-
c_{q}}\,v_{s}^{q}(t)}{(t-a)^{\alpha^{q}}\,(b-t)^{\beta^{q}}}\,,&t\in(a,b),\\\
\ell_{2}^{q}(s)\,,&t=b.\end{array}\right.$

It is clear that $\tilde{u}_{s}^{q}>0$ on $[a,b]$ for all $s\in(a,b)$.

Since $g_{0}(t,s)\in C^{n-2}(I\times I)$, $\frac{\partial^{n-1}}{\partial
t^{n-1}}g_{0}\in C^{\infty}((I\times I)\setminus\\{(t,t)\;/\;t\in I\\})$ and
there exists $\lim\limits_{s\rightarrow t^{\pm}}\frac{\partial^{n-1}}{\partial
t^{n-1}}g_{0}(t,s)\in{\mathbb{R}}$, we deduce that there exists $K^{q}>0$ such
that $\tilde{u}_{s}^{q}(t)\leq K^{q}$ for every $(t,s)\in I\times I$ and
$q\in\\{1,\dots,n-1\\}$. Therefore, the following functions

$\displaystyle\tilde{k}_{1}^{q}(s)$ $\displaystyle=$ $\displaystyle\min_{t\in
I}\tilde{u}_{s}^{q}(t)\,,\quad s\in[a,b]\,,$
$\displaystyle\tilde{k}_{2}^{q}(s)$ $\displaystyle=$ $\displaystyle\max_{t\in
I}\tilde{u}_{s}^{q}(t)\,,\quad s\in[a,b]\,,$

are continuous on $I$ and positive in $(a,b)$.

Taking $\phi(t)=(t-a)^{\alpha^{q}}\,(b-t)^{\beta^{q}}>0$ on $(a,b)$, the
function $v_{s}^{q}(t)$ satisfies the condition $(P_{g})$ if $n-q-c_{q}$ is
even with $k_{1}(s)=\tilde{k}_{1}^{q}(s)$ and $k_{2}(s)=\tilde{k}_{2}^{q}(s)$
and condition $(N_{g})$ if $n-q-c_{q}$ is odd with
$k_{1}(s)=-\tilde{k}_{2}^{q}(s)$ and $k_{2}(s)=-\tilde{k}_{1}^{q}(s)$.

As for the set of parameters of $M$ in which the function $v_{s}^{q}[M]$,
$q\in\\{1,\ldots,n-1\\}$ maintains constant sign and monotony with respect to
$M$, we have the following results:

######  Lemma 2.29.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$. If there exists $M\in\mathbb{R}$ be such that
$\frac{\partial^{q}}{\partial t^{q}}g_{M}$ has constant sign, then $g_{M}$ has
constant sign.

###### Proof.

If $v_{s}^{q}[M](t)=\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)$ has
constant sign, then $v_{s}^{q-1}[M](t)=\frac{\partial^{q-1}}{\partial
t^{q-1}}g_{M}(t,s)$ is a monotone and continuous function. Thus,
$v_{s}^{q-1}[M]$ has at most one zero on $I$.

By recurrence, we have that $v_{s}^{q-l}[M](t)=\frac{\partial^{q-l}}{\partial
t^{q-l}}g_{M}(t,s)$ has at most $l$ zeros on $I$. In particular,
$g_{M}(\cdot,s)$ has at most $q$ zeros on $I$.

Now, since condition $(N_{a})$ implies that $g_{M}(\cdot,s)$ has at least $q$
zeros of order smaller or equal to $q$ on the boundary, we deduce that
$g_{M}(\cdot,s)$ cannot have any zero on $(a,b)$. Thus, $g_{M}$ has constant
sign. ∎

Moreover, we recall the following facts:

  * •

If $n-k$ is even, then from Theorem 2.26 and [1, Lemma 1.8.33] we have that
$g_{M}(t,s)>0$ for all $(t,s)\in(a,b)\times(a,b)$ if and only if
$M\in(\lambda_{1},+\infty)$ or $M\in(\lambda_{1},\lambda_{2}]$, and it is
monotone decreasing with respect to $M$ on such interval, where
$\lambda_{1}<0$ is the first eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
and $\lambda_{2}>0$ (if the interval is bounded) is not an eigenvalue of
$T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.
The value of $\lambda_{2}$ is characterized in [5, Sect.7 and Sect.8].

  * •

If $n-k$ is odd, then from Theorem 2.26 and [1, Lemma 1.8.25] we have that
$g_{M}(t,s)<0$ for all $(t,s)\in(a,b)\times(a,b)$ if and only if
$M\in(-\infty,\lambda_{1})$ or $M\in[\overline{\lambda_{2}},\lambda_{1})$ and
it is monotone decreasing with respect to $M$ on such interval, where
$\lambda_{1}>0$ is the first eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
and $\overline{\lambda_{2}}<0$ (if the interval is bounded) is not an
eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.
The value of $\overline{\lambda_{2}}$ is characterized in [5, Sect. 7 and
Sect. 8].

Therefore, as a direct consequence of Theorem 2.26, Lemma 2.29 and [2, Theorem
3.6], the following result is deduced.

######  Lemma 2.30.

The following properties hold:

  * •

If $n-k$ is even and $\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)>0$ on
$(a,b)\times(a,b)$ for some $M\in\mathbb{R}$, then
$\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)>0$ if and only if
$M\in(\lambda_{1},M_{q}]$, with $M_{q}\leq\lambda_{2}$. Moreover,
$\frac{\partial^{q}}{\partial t^{q}}g_{M}$ is monotone decreasing with respect
to $M\in(\lambda_{1},M_{q}]$.

  * •

If $n-k$ is even and $\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)<0$ on
$(a,b)\times(a,b)$ for some $M\in\mathbb{R}$, then
$\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)<0$ if and only if
$M\in(\lambda_{1},M_{q}]$, with $M_{q}\leq\lambda_{2}$. Moreover,
$\frac{\partial^{q}}{\partial t^{q}}g_{M}$ is monotone increasing with respect
to $M\in(\lambda_{1},M_{q}]$.

  * •

If $n-k$ is odd and $\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)>0$ on
$(a,b)\times(a,b)$ for some $M\in\mathbb{R}$, then
$\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)>0$ if and only if
$M\in[M_{q},\lambda_{1})$, with $M_{q}\geq\overline{\lambda_{2}}$. Moreover,
$\frac{\partial^{q}}{\partial t^{q}}g_{M}$ is monotone increasing with respect
to $M\in[M_{q},\lambda_{1})$.

  * •

If $n-k$ is odd and $\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)<0$ on
$(a,b)\times(a,b)$ for some $M\in\mathbb{R}$, then
$\frac{\partial^{q}}{\partial t^{q}}g_{M}(t,s)<0$ if and only if
$M\in[M_{q},\lambda_{1})$, with $M_{q}\geq\overline{\lambda_{2}}$. Moreover,
$\frac{\partial^{q}}{\partial t^{q}}g_{M}$ is monotone decreasing with respect
to $M\in[M_{q},\lambda_{1})$.

##  3 Study of the constant sign of the derivatives of Green’s function

In this section for any $q\in\\{1,\dots,n-1\\}$ be fixed, we will give a proof
of the main result that characterizes the constant sign of the partial
derivative with respect to $t$ of order $q$ of the Green’s function related to
operator $T_{n}[M]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.

We distinguish three cases:

  * (a)

$c_{q}\geq 1\;\;\text{and}\;\;d_{q}\geq 1$,

  * (b)

$c_{q}=n-q\;\;\text{and}\;\;d_{q}=0$,

  * (c)

$c_{q}=0,\;\;\text{and}\;\;d_{q}=n-q$,

where $c_{q}$ and $d_{q}$ are defined in (2.27) and (2.28) and satisfy
$c_{q}+d_{q}=n-q$.

###  3.1 Case (a): $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and $d_{q}\geq 1$

######  Lemma 3.1.

Let $q\in\\{1,\dots,n-1\\}$ be fixed. If $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and
$d_{q}\geq 1$, then the indices $z=k-(j-1)$ and $h=n-k-(r-1)$ are such that

$\mu_{z}^{q}+q+\eta=n-1,\;\;\text{and}\;\;\rho_{h}^{q}+q+\gamma=n-1,$

with $j$ and $r$ defined in (2.20) and (2.21), $\eta$ and $\gamma$ defined in
(2.13) and (2.14).

###### Proof.

By the definition of $\mu_{z}^{q}=\mu_{k-(j-1)}^{q}$ and
$\rho_{h}^{q}=\rho_{n-k-(r-1)}^{q}$ we have that since $c_{q}\geq 1$,
$\mu_{z}^{q}=\sigma_{k}-q\geq 0$ and since $d_{q}\geq 1$,
$\rho_{h}^{q}=\varepsilon_{n-k}-q\geq 0$. Using Remark 2.7, we deduce that

$\mu_{z}^{q}+q+\eta=\sigma_{k}-q+q+n-1-\sigma_{k}=n-1,$

and

$\rho_{h}^{q}+q+\gamma=\varepsilon_{n-k}-q+q+n-1-\varepsilon_{n-k}=n-1.$

∎

######  Lemma 3.2.

Let $q\in\\{1,\dots,n-1\\}$ be fixed. If $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and
$d_{q}\geq 1$, then $z=p$ and $h=l$ with $z$ and $h$ defined in Lemma 3.1, $p$
and $l$ defined in (2.25) and (2.24).

###### Proof.

First, note that $z=|A_{q}|$ is the cardinal of the set
$A_{q}=\\{i\in\\{1,\dots,k\\}\,/\;\;\sigma_{i}\geq q\\}$ and $p=|D_{q}|$ is
the cardinal of the set $D_{q}=\\{i\in\\{1,\dots,k\\}\,/\;\;\delta_{i}+q\leq
n-1\\}$. Moreover, by (2.27), it is clear that $c_{q}$ corresponds with the
number of $\sigma$’s that are equal or greater than $q$, that is, $z=c_{q}$.

On the other hand, we have that

$\\{\varepsilon_{1},\dots,\varepsilon_{n-k},n-1-\delta_{k},\dots,\,n-1-\delta_{1}\\}\equiv\\{0,\dots,n-1\\}.$

Among these $n$ elements (all the numbers between $0$ and $n-1$) there are
$n-q$ elements that are equal or greater than $q$. By (2.28), we have that the
number of $\varepsilon$’s that are equal or greater than $q$ is $d_{q}$.
Therefore, the number of $\delta$’s such that $n-1-\delta_{i}\geq q$ is $n-q-
d_{q}$, that is, $p=n-q-d_{q}$. Since, $c_{q}+d_{q}=n-q$, we have that $z=p$.
In an analogous way, we can prove that $h=l$. ∎

######  Remark 3.3.

It should be noted that in the previous proof, since $j$ and $r$ depend on
$q$, we have that $z$ and $h$ also depend on $q$, but we will omit such
dependence in the notation for the sake of simplicity.

######  Remark 3.4.

Indices $z$ and $h$ depend on the space
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
and are unique. Indeed, for the rest of the components of space
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
we have that $(\mu_{i}^{q}+q+\eta)\bmod{n}<n-1$ for $i\neq z$ with
$i\in\\{1,\dots,k\\}$ and $(\rho_{l}^{q}+q+\gamma)\bmod{n}<n-1$ for $l\neq h$
with $l\in\\{1,\dots,n-k\\}$.

Now, we present some results that provide sufficient conditions to ensure the
constant sign of solutions of (1.1) in the spaces
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$.
Using Lemma 2.16, these results can be proved analogously to [5, Proposition
6.7] and [5, Proposition 6.9], respectively, and we omit them.

Let us consider the following spaces

$\displaystyle X_{2}$
$\displaystyle:=\left\\{\begin{array}[]{cc}X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{z-1}^{q},\,\mu_{z+1}^{q},\,\dots,\,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\,\dots,\,\rho_{n-k}^{q}|\beta^{q}\\}}\,,&\text{if
$z\neq 1$,}\\\ &\\\
X_{\\{\mu_{2}^{q},\,\dots,\,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\,\dots,\,\rho_{n-k}^{q}|\beta^{q}\\}}\,,&\text{if
$z=1$,}\end{array}\right.$ (3.4) $\displaystyle X_{3}$
$\displaystyle:=\left\\{\begin{array}[]{cc}X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{z-1}^{q},\,\mu_{z+1}^{q},\,\dots,\mu_{k}^{q}|\alpha^{q}\\}}^{\\{\rho_{1}^{q},\,\dots,\,\rho_{n-k}^{q}\\}}\,,&\text{if
$z\neq 1$,}\\\ &\\\
X_{\\{\mu_{2}^{q},\,\dots,\,\mu_{k}^{q}|\alpha^{q}\\}}^{\\{\rho_{1}^{q},\,\dots,\,\rho_{n-k}^{q}\\}}\,,&\text{if
$z=1$,}\end{array}\right.$ (3.8)
$X_{4}:=\left\\{\begin{array}[]{cc}X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{k}^{q}|\alpha^{q}\\}}^{\\{\rho_{1}^{q},\,\dots,\,\rho_{h-1}^{q},\,\rho_{h+1}^{q},\,\dots,\,\rho_{n-k}^{q}\\}}\,,&\text{if
$h\neq 1$,}\\\ &\\\
X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{k}^{q}|\alpha^{q}\\}}^{\\{\rho_{2}^{q},\,\dots,\,\rho_{n-k}^{q}\\}}\,,&\text{if
$h=1$,}\end{array}\right.$ (3.9)

and

$X_{5}:=\left\\{\begin{array}[]{cc}X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\,\dots,\,\rho_{h-1}^{q},\,\rho_{h+1}^{q},\,\dots,\,\rho_{n-k}^{q}|\beta^{q}\\}}\,,&\text{if
$h\neq 1$,}\\\ &\\\
X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{k}^{q}\\}}^{\\{\rho_{2}^{q},\,\dots,\,\rho_{n-k}^{q}|\beta^{q}\\}}\,,&\text{if
$h=1$,}\end{array}\right.$ (3.10)

where $z$ and $h$ are defined in Lemma 3.1.

######  Proposition 3.5.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$, $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and $d_{q}\geq
1$. If $u\in C^{n}(I)$ is a solution of (1.1) on $(a,b)$, satisfying the
boundary conditions:

$\displaystyle
u^{(\mu_{1}^{q})}(a)=\cdots=u^{(\mu_{z-1}^{q})}(a)=u^{(\mu_{z+1}^{q})}(a)=$
$\displaystyle\cdots$ $\displaystyle=u^{(\mu_{k}^{q})}(a)\hskip 8.5359pt=0\,,$
(3.11) $\displaystyle u^{(\rho_{1}^{q})}(b)=$ $\displaystyle\cdots$
$\displaystyle=u^{(\rho_{n-k}^{q})}(b)=0\,,$ (3.12)

then it does not have any zero on $(a,b)$ provided that one of the following
assertions is satisfied:

  * •

Let $n-k$ be even:

    * –

If $k>1$, $\mu_{z}^{q}\neq z-1$ and $M\in[\lambda_{3}^{q},\lambda_{2}^{q}]$,
where:

      * *

$\lambda_{3}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{3}$.

      * *

$\lambda_{2}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in $X_{2}$.

    * –

If $k>1$, $\mu_{z}^{q}=z-1$ and $M\in[\lambda_{1},\lambda_{2}^{q}]$, where:

      * *

$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

      * *

$\lambda_{2}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in $X_{2}$.

    * –

If $k=1$, $\mu_{1}^{q}\neq 0$ and $M\in[\lambda_{3}^{q},+\infty)$, where:

      * *

$\lambda_{3}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{\\{\alpha^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-1}^{q}\\}}$, where
$\alpha^{q}=0$.

    * –

If $k=1$, $\mu_{1}^{q}=0$ and $M\in[\lambda_{1},+\infty)$, where:

      * *

$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-1}\\}}.$

  * •

Let $n-k$ be odd:

    * –

If $k>1$, $\mu_{z}^{q}\neq z-1$ and $M\in[\lambda_{2}^{q},\lambda_{3}^{q}]$,
where:

      * *

$\lambda_{3}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in $X_{3}$.

      * *

$\lambda_{2}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{2}$.

    * –

If $k>1$, $\mu_{z}^{q}=z-1$ and $M\in[\lambda_{2}^{q},\lambda_{1}]$, where:

      * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$.

      * *

$\lambda_{2}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{2}$.

    * –

If $k=1$, $\mu_{1}^{q}\neq 0$ and $M\in(-\infty,\lambda_{3}^{q}]$, where:

      * *

$\lambda_{3}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\alpha^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-1}^{q}\\}}$, where
$\alpha^{q}=0$.

    * –

If $k=1$, $\mu_{1}^{q}=0$ and $M\in(-\infty,\lambda_{1}]$, where:

      * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-1}\\}}$.

######  Proposition 3.6.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$, $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and $d_{q}\geq
1$. If $u\in C^{n}(I)$ is a solution of (1.1) on $(a,b)$ satisfying the
boundary conditions:

$\displaystyle u^{(\mu_{1}^{q})}(a)=$ $\displaystyle\cdots$
$\displaystyle=u^{(\mu_{k}^{q})}(a)\hskip 8.5359pt=0\,,$ (3.13) $\displaystyle
u^{(\rho_{1}^{q})}(b)=$ $\displaystyle\cdots$
$\displaystyle=u^{(\rho_{h-1}^{q})}(b)=u^{(\rho_{h+1}^{q})}(b)=\cdots=u^{(\rho_{n-k}^{q})}(b)=0\,,$
(3.14)

then it does not have any zero on $(a,b)$ provided that one of the following
assertions is satisfied:

  * •

Let $n-k$ be even:

    * –

If $\rho_{h}^{q}\neq h-1$ and $M\in[\lambda_{5}^{q},\lambda_{4}^{q}]$, where:

      * *

$\lambda_{5}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{5}$.

      * *

$\lambda_{4}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in $X_{4}.$

    * –

If $\rho_{h}^{q}=h-1$ and $M\in[\lambda_{1},\lambda_{4}^{q}]$, where:

      * *

$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

      * *

$\lambda_{4}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in $X_{4}$.

  * •

Let $n-k$ be odd:

    * –

If $k<n-1$, $\rho_{h}^{q}\neq h-1$ and
$M\in[\lambda_{4}^{q},\lambda_{5}^{q}]$, where:

      * *

$\lambda_{5}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in $X_{5}$.

      * *

$\lambda_{4}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{4}$.

    * –

If $k<n-1$, $\rho_{h}^{q}=h-1$ and $M\in[\lambda_{4}^{q},\lambda_{1}]$, where:

      * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

      * *

$\lambda_{4}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{4}$.

    * –

If $k=n-1$, $\rho_{1}^{q}\neq 0$ and $M\in(-\infty,\lambda_{5}^{q}]$, where:

      * *

$\lambda_{5}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\mu_{1}^{q},\dots,\mu_{n-1}^{q}\\}}^{\\{\beta^{q}\\}}$, where
$\beta^{q}=0$.

    * –

If $k=n-1$, $\rho_{1}^{q}=0$ and $M\in(-\infty,\lambda_{1}]$, where:

      * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{n-1}\\}}^{\\{\varepsilon_{1}\\}}$.

######  Example 3.7.

Let us consider the operator $T_{4}[M]$ with the boundary conditions:

$u^{\prime}(0)=u^{\prime\prime\prime}(0)=u(1)=u^{\prime}(1)=0.$

The hypotheses of Propositions 3.5 and 3.6 are satisfied only for the value
$q=1$. Moreover, $z=2$, $h=1$, and $v_{s}^{1}[M]$ satisfies the conditions of
the space $X_{\\{0,2\\}}^{\\{0,3\\}}$.

From Proposition 3.5, we can affirm that any nontrivial solution of the
problem

$T_{4}[M]=0,\;t\in[0,1],\;\;u(0)=u(1)=u^{\prime\prime\prime}(1)=0\,,$ (3.15)

does not have any zero on $(0,1)$ for $M\in[\lambda_{3}^{1},\lambda_{2}^{1}]$,
where $\lambda_{3}^{1}<0$ is the biggest negative eigenvalue of $T_{4}[0]$ in
$X_{\\{0,1\\}}^{\\{0,3\\}}$ and $\lambda_{2}^{1}>0$ is the least positive
eigenvalue of $T_{4}[0]$ in $X_{\\{1\\}}^{\\{0,1,3\\}}$.

The eigenvalues of $T_{4}[0]$ in $X_{\\{0,1\\}}^{\\{0,3\\}}$ are given by
$-\lambda^{4}$, where $\lambda$ is a positive solution of
$\sin\left(\lambda\right)=0\,$. The smallest positive solution of this
equation is $\pi$, so $\lambda_{3}^{1}=-\pi^{4}$ is the biggest negative
eigenvalue of $T_{4}[0]$ in $X_{\\{0,1\\}}^{\\{0,3\\}}$.

Similarly, the eigenvalues of $T_{4}[0]$ in $X_{\\{1\\}}^{\\{0,1,3\\}}$ are
given by $\lambda^{4}$, where $\lambda$ is a positive solution of

$\left(-1+e^{\sqrt{2}\,m}\right)\,\cos\left(\frac{m}{\sqrt{2}}\right)+\left(1+e^{\sqrt{2}\,m}\right)\,\sin\left(\frac{m}{\sqrt{2}}\right)=0.$

Denoting by $m_{2}$ the smallest positive solution of this equation, we deduce
that$\lambda_{2}^{1}=m_{2}^{4}\approx 3.34^{4}$ is the least positive
eigenvalue of $T_{4}[0]$ in $X_{\\{0\\}}^{\\{0,1,3\\}}$.

It is easy to verify that the solutions of problem (3.15) are given as
multiples of the following expression:

  * •

If $M=-m^{4}<0$,

$\begin{split}u(t)=&\cosh(m-m\,t)\,\sin(m)-\sin(m\,t)-\cosh(m)\,\sin(m-m\,t)-\cos(m-m\,t)\,\sinh(m)\\\
&+\sinh(m\,t)+\cos(m)\,\sinh(m-m\,t).\end{split}$

  * •

If $M=0$,

$u(t)=t^{2}-t.$

  * •

If $M=m^{4}>0$,

$\begin{split}u(t)=&\,e^{-\frac{m\,t}{\sqrt{2}}}\left(e^{\sqrt{2}\,m}\left(-1+e^{\sqrt{2}\,m\,t}\right)\,\cos\left(\frac{m(-2+t)}{\sqrt{2}}\right)-e^{\sqrt{2}\,m}\left(-1+e^{\sqrt{2}\,m}\right)\,\cos\left(\frac{m\,t}{\sqrt{2}}\right)\right)\\\
&+e^{-\frac{m\,t}{\sqrt{2}}}\,\left(-1+e^{\sqrt{2}\,m}\right)\left(-e^{\sqrt{2}\,m}+e^{\sqrt{2}\,m\,t}\right)\,\sin\left(\frac{m\,t}{\sqrt{2}}\right).\end{split}$

Analogously, from Proposition 3.6, we conclude that any solution of the
problem

$T_{4}[M]\,u(t)=0,\;t\in[0,1],\;\;u(0)=u^{\prime\prime}(0)=u^{\prime\prime\prime}(1)=0,$
(3.16)

does not have any zero on $(0,1)$ for $M\in[\lambda_{1},\lambda_{4}^{1}]$,
where $\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{4}[0]$ in
$X_{\\{1,3\\}}^{\\{0,1\\}}$ and $\lambda_{4}^{1}>0$ is the least positive
eigenvalue of $T_{4}[0]$ in $X_{\\{0,1,2\\}}^{\\{3\\}}$.

The eigenvalues of $T_{4}[0]$ in $X_{\\{1,3\\}}^{\\{0,1\\}}$ are given by
$-\lambda^{4}$, where $\lambda$ is a positive solution of

$\sin(\lambda)+\cos(\lambda)\,\tanh(\lambda)=0\,.$

Now, if $m_{1}$ is the smallest positive solution of this equation, then
$\lambda_{1}=-m_{1}^{4}\approx-2.36^{4}$ is the biggest negative eigenvalue of
$T_{4}[0]$ in $X_{\\{0,1\\}}^{\\{0,3\\}}$.

On the other hand, the eigenvalues of $T_{4}[0]$ in
$X_{\\{0,1,2\\}}^{\\{3\\}}$ are given by $\lambda^{4}$, where $\lambda$ is a
positive solution of $\cos\left(\frac{m}{\sqrt{2}}\right)=0.$ The smallest
positive solution of this equation is $\frac{\sqrt{2}}{2}\pi$, so
$\lambda_{4}^{1}=\left(\frac{\sqrt{2}}{2}\pi\right)^{4}$ is the least positive
eigenvalue of $T_{4}[0]$ in $X_{\\{0,1,2\\}}^{\\{3\\}}$.

It is immediate to show that solutions of problem (3.16) are given as
multiples of:

  * •

If $M=-m^{4}<0$,

$u(t)=\sin(m\,t)+\cos(m)\,\operatorname{sech}(m)\sinh(m\,t).$

  * •

If $M=0$,

$u(t)=t.$

  * •

If $M=m^{4}>0$,

$\begin{split}u(t)=&\,e^{-\frac{m\,t}{\sqrt{2}}}\left(\left(-1+e^{\sqrt{2}\,m\,(1+t)}\right)\,\cos\left(\frac{m(-1+t)}{\sqrt{2}}\right)+\left(-e^{\sqrt{2}\,m}+e^{\sqrt{2}\,m\,t}\right)\,\cos\left(\frac{m(1+t)}{\sqrt{2}}\right)\right)\\\
&+e^{-\frac{m\,t}{\sqrt{2}}}\left(\left(1+e^{\sqrt{2}\,m\,(1+t)}\right)\,\sin\left(\frac{m(-1+t)}{\sqrt{2}}\right)+\left(e^{\sqrt{2}\,m}+e^{\sqrt{2}\,m\,t}\right)\,\sin\left(\frac{m(1+t)}{\sqrt{2}}\right)\right).\end{split}$

######  Remark 3.8.

It should be noted that in the above propositions we have used the fact proven
in Lemma 2.17 that the first nonzero eigenvalues of the spaces
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
and
$X_{\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}\\}}$
coincide.

Let us define the following spaces

$X_{2}^{*}:=\left\\{\begin{array}[]{cc}X_{\\{\tau_{1},\,\dots,\,\tau_{n-k}|\eta\\}}^{\\{\delta_{1},\,\dots,\,\delta_{z-1},\,\delta_{z+1},\,\dots,\,\delta_{k}\\}}\,,&\text{if
$z\neq 1$,}\\\ &\\\
X_{\\{\tau_{1},\,\dots,\,\tau_{n-k}|\eta\\}}^{\\{\delta_{2},\,\dots,\,\delta_{k}\\}}\,,&\text{if
$z=1$,}\end{array}\right.$ (3.17)

and

$X_{4}^{*}:=\left\\{\begin{array}[]{cc}X_{\\{\tau_{1},\,\dots,\,\tau_{h-1},\,\tau_{h+1},\,\dots,\,\tau_{n-k}\\}}^{\\{\delta_{1},\,\dots,\,\delta_{k}|\gamma\\}}\,,&\text{if
$h\neq 1$,}\\\ &\\\
X_{\\{\tau_{2},\,\dots,\,\tau_{n-k}\\}}^{\\{\delta_{1},\,\dots,\,\delta_{k}|\gamma\\}}\,,&\text{if
$h=1$.}\end{array}\right.$ (3.18)

######  Lemma 3.9.

$\lambda\neq 0$ is an eigenvalue of $T_{n}[0]$ in $X_{2}^{*}$ defined in
(3.17) if and only if $\lambda$ is an eigenvalue of $T_{n}[0]$ in $X_{2}$. In
particular, $(\lambda_{2}^{*})^{q}=\lambda_{2}^{q}$ where
$(\lambda_{2}^{*})^{q}\neq 0$ is the least positive (biggest negative)
eigenvalue of $T_{n}[0]$ in $X_{2}^{*}$ and $\lambda_{2}^{q}\neq 0$ is the
least positive (biggest negative) eigenvalue of $T_{n}[0]$ in $X_{2}$.

###### Proof.

Suppose that $z\neq 1$ (the case $z=1$ is proved similarly). By the definition
of adjoint space of
$X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\,\dots,\,\rho_{n-k}^{q}\\}}$
we have that

$\\{\mu_{1}^{q},\,\dots,\,\mu_{z-1}^{q},\,\mu_{z}^{q},\,\mu_{z+1}^{q},\,\dots,\,\mu_{k}^{q},\,n-1-\tau_{n-k}^{q},\,\dots,\,n-1-\tau_{1}^{q}\\}=\\{0,\,\dots,\,n-1\\},$

and

$\\{\rho_{1}^{q},\,\dots,\,\rho_{n-k}^{q},\,n-1-\delta_{k}^{q},\,\dots,\,n-1-\delta_{1}^{q}\\}=\\{0,\,\dots,\,n-1\\}.$

Since, $\mu_{z}^{q}=\sigma_{k}-q$ and $\beta^{q}=n-1-\delta_{k}^{q}$ we deduce
that

${X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{z-1}^{q},\,\mu_{z+1}^{q},\,\dots,\,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}|\beta^{q}\\}}}$${X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}|n-1-\sigma_{k}+q\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k-1}^{q}\\}}.}$$\scriptstyle{*}$$\scriptstyle{*}$
(3.19)

Using that $\eta=n-1-\sigma_{k}$, the definition of
$X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k}^{q}\\}}$
and the diagram (2.26) we infer that

${X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}|n-1-\sigma_{k}+q\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k-1}^{q}\\}}}$${X_{\\{\tau_{1},\,\dots,\,\tau_{n-k}|\eta\\}}^{\\{\delta_{1},\,\dots,\,\delta_{p-1},\,\delta_{p+1},\,\dots,\,\delta_{k}\\}}.}$$\scriptstyle{\phi^{q}}$

From Lemma 3.2 we know that $p=z$ and therefore

${X_{\\{\tau_{1}^{q},\dots,\tau_{n-k}^{q}|n-1-\sigma_{k}+q\\}}^{\\{\delta_{1}^{q},\dots,\delta_{k-1}^{q}\\}}}$${X_{\\{\tau_{1},\,\dots,\,\tau_{n-k}|\eta\\}}^{\\{\delta_{1},\,\dots,\,\delta_{z-1},\,\delta_{z+1},\,\dots,\,\delta_{k}\\}}.}$$\scriptstyle{\phi^{q}}$
(3.20)

Considering Lemma 2.17 and the fact that the adjoint spaces have the same
eigenvalues, from (3.19) and (3.20), we obtain that the spaces
$X_{\\{\mu_{1}^{q},\,\dots,\,\mu_{z-1}^{q},\,\mu_{z+1}^{q},\,\dots,\,\mu_{k}^{q}\\}}^{\\{\rho_{1}^{q},\dots,\rho_{n-k}^{q}|\beta^{q}\\}}$
and
$X_{\\{\tau_{1},\,\dots,\,\tau_{n-k}|\eta\\}}^{\\{\delta_{1},\,\dots,\,\delta_{z-1},\,\delta_{z+1},\,\dots,\,\delta_{k}\\}}$
have the same eigenvalues. In particular,
$(\lambda_{2}^{*})^{q}=\lambda_{2}^{q}$. ∎

Using similar argument to the previous result, we arrive to the next one.

######  Lemma 3.10.

$\bar{\lambda}\neq 0$ is an eigenvalue of $T_{n}[0]$ in $X_{4}^{*}$ defined in
(3.18) if and only if $\bar{\lambda}$ is an eigenvalue of $T_{n}[0]$ in
$X_{4}$. In particular, $(\lambda_{4}^{*})^{q}=\lambda_{4}^{q}$ where
$(\lambda_{4}^{*})^{q}\neq 0$ is the least positive (biggest negative)
eigenvalue of $T_{n}[0]$ in $X_{4}^{*}$ and $\lambda_{4}^{q}\neq 0$ is the
least positive (biggest negative) eigenvalue of $T_{n}[0]$ in $X_{4}$.

Next, we state and prove the main result, which gives the characterization of
the set of parameter $M$ where $v_{s}^{q}[M]$ is of constant sign.

######  Theorem 3.11.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$, $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and $d_{q}\geq
1$. The following properties are fulfilled:

  * •

Let $2\leq k\leq n-2$ and $n-k$ even:

    * –

If $n-q-c_{q}$ is even, then $v_{s}^{q}[M]$ is strongly positive on $I\times
I$ if, and only if, $M\in(\lambda_{1},\lambda^{q}]$, and if $n-q-c_{q}$ is
odd, then $v_{s}^{q}[M]$ is strongly negative on $I\times I$ if, and only if,
$M\in(\lambda_{1},\lambda^{q}]$ where:

      * *

$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

      * *

$\lambda^{q}>0$ is the minimum between:

        * ·

$\lambda_{2}^{q}>0$, the least positive eigenvalue of $T_{n}[0]$ in $X_{2}$
defined in (3.4).

        * ·

$\lambda_{4}^{q}>0$, the least positive eigenvalue of $T_{n}[0]$ in $X_{4}$
defined in (3.9).

  * •

Let $2\leq k\leq n-2$ and $n-k$ odd:

    * –

If $n-q-c_{q}$ is even, then $v_{s}^{q}[M]$ is strongly positive on $I\times
I$ if, and only if, $M\in[\lambda^{q},\lambda_{1})$, and if $n-q-c_{q}$ is
odd, then $v_{s}^{q}[M]$ is strongly negative on $I\times I$ if, and only if,
$M\in[\lambda^{q},\lambda_{1})$ where:

      * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

      * *

$\lambda^{q}<0$ is the maximum between:

        * ·

$\lambda_{2}^{q}<0$, the biggest negative eigenvalue of $T_{n}[0]$ in $X_{2}.$

        * ·

$\lambda_{4}^{q}<0$, the biggest negative eigenvalue of $T_{n}[0]$ in $X_{4}$.

  * •

Let $k=1$ and $n>2$ odd:

    * –

If $n-q-1$ is even, then $v_{s}^{q}[M]$ is strongly positive on $I\times I$
if, and only if, $M\in(\lambda_{1},\lambda_{2}^{q}]$, and if $n-q-1$ is odd,
then $v_{s}^{q}[M]$ is strongly negative on $I\times I$ if, and only if,
$M\in(\lambda_{1},\lambda_{2}^{q}]$ where:

      * *

$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-1}\\}}.$

      * *

$\lambda_{2}^{q}>0$ is the least positive eigenvalue of $T_{n}[0]$ in $X_{2}$.

  * •

Let $k=1$ and $n>2$ even:

    * –

If $n-q-1$ is even, then $v_{s}^{q}[M]$ is strongly positive on $I\times I$
if, and only if, $M\in[\lambda_{2}^{q},\lambda_{1})$, and if $n-q-1$ is odd,
then $v_{s}^{q}[M]$ is strongly negative on $I\times I$ if, and only if,
$M\in[\lambda_{2}^{q},\lambda_{1})$, where:

      * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-1}\\}}$.

      * *

$\lambda_{2}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{2}$.

  * •

If $k=n-1$, $n>2$ and $n-q-c_{q}=1$, then $v_{s}^{q}[M]$ is strongly negative
on $I\times I$ if, and only if, $M\in[\lambda_{2}^{q},\lambda_{1})$, where

    * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{n-1}\\}}^{\\{\varepsilon_{1}\\}}.$

    * *

$\lambda_{2}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{2}$.

###### Proof.

First, it has been shown in Theorem 2.28 that $v_{s}^{q}$ satisfies the
property $(P_{g})$ if $n-q-c_{q}$ is even and the property $(N_{g})$ if $n-q-
c_{q}$ is odd. In addition, from Lemma 2.30 we know that if $v_{s}^{q}[M]$ has
constant sign then the first eigenvalue $\lambda_{1}$ of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
is one of the extremes of the interval where $v_{s}^{q}[M]$ holds that sign.

Again, taking into account Lemma 2.30 on the monotony of $v_{s}^{q}[M]$ with
respect to $M$, the constant sign starts/ends at the first eigenvalue
$\lambda_{1}$ of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$.
Thus, from Theorems 2.26 and 2.28 we conclude that:

  * •

If $n-k$ and $n-q-c_{q}$ are even, and $M\leq 0$, then $v_{s}^{q}[M]$ is
strongly positive on $I\times I$ if, and only if, $M\in(\lambda_{1},0]$.

  * •

If $n-k$ is even, $n-q-c_{q}$ is odd and $M\leq 0$, then $v_{s}^{q}[M]$ is
strongly negative on $I\times I$ if, and only if, $M\in(\lambda_{1},0]$.

  * •

If $n-k$ is odd, $n-q-c_{q}$ is even and $M\geq 0$, then $v_{s}^{q}[M]$ is
strongly positive on $I\times I$ if, and only if, $M\in[0,\lambda_{1})$.

  * •

If $n-k$ and $n-q-c_{q}$ are odd, and $M\geq 0$, then $v_{s}^{q}[M]$ is
strongly negative on $I\times I$ if, and only if, $M\in[0,\lambda_{1})$.

We now need to determine the other extreme of the constant sign interval of
$v_{s}^{q}[M]$ using Definition 2.25, Propositions 3.5 and 3.6. We divide the
proof into five steps assuming that $n-k$ and $n-q-c_{q}$ are even, and $2\leq
k\leq n-2$. For the rest of cases the proof is done in an analogous way.

  * Step 1.

Study of $v_{s}^{q}[M]$ at $s=a$.

  * Step 2.

Study of $v_{s}^{q}[M]$ at $s=b$.

  * Step 3.

Study of $v_{s}^{q}[M]$ at $t=a$.

  * Step 4.

Study of $v_{s}^{q}[M]$ at $t=b$.

  * Step 5.

Study of $v_{s}^{q}[M]$ on $(a,b)\times(a,b)$.

Let us denote

$g_{M}(t,s)=\left\\{\begin{array}[]{cc}g_{M}^{1}(t,s)\,,&a\leq s\leq t\leq
b\,,\\\ \\\ g_{M}^{2}(t,s)\,,&a<t<s<b\,,\end{array}\right.$

as the related Green’s function of $T_{n}[M]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

  * Step 1.

Study of the function $v_{s}^{q}[M]$ at $s=a$.

Let us consider the function

$w_{M,q}(t)=\dfrac{\partial^{\eta}}{\partial
s^{\eta}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{1}(t,s)\Big{)}_{\mid s=a},$

where $\eta$ has been defined in (2.13).

If $\eta>0$, since
$\\{0,\dots,\eta-1\\}\subset\\{\tau_{1},\dots,\tau_{n-k}\\}$ using the
equality (2.10) we obtain that

$g_{M}^{1}(t,a)=\dfrac{\partial}{\partial
s}g_{M}^{1}(t,a)=\cdots=\dfrac{\partial^{\eta-1}}{\partial
s^{\eta-1}}g_{M}^{1}(t,a)=0\,,\;\;\text{for all}\;t\in(a,b].$

Therefore, for all $t\in(a,b]$ and $q\in\\{1,\dots,n-1\\}$ we infer that

$\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{1}(t,a)=\dfrac{\partial^{q}}{\partial
t^{q}}\Big{(}\dfrac{\partial}{\partial s}g_{M}^{1}(t,s)\Big{)}_{\mid
s=a}=\cdots=\dfrac{\partial^{q}}{\partial
t^{q}}\Big{(}\dfrac{\partial^{\eta-1}}{\partial
s^{\eta-1}}g_{M}^{1}(t,s)\Big{)}_{\mid s=a}=0\,,$

equivalently,

$\dfrac{\partial^{q}}{\partial t^{q}}g_{M}^{1}(t,a)=\dfrac{\partial}{\partial
s}\Big{(}\dfrac{\partial^{q}}{\partial t^{q}}g_{M}^{1}(t,s)\Big{)}_{\mid
s=a}=\cdots=\dfrac{\partial^{\eta-1}}{\partial
s^{\eta-1}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{1}(t,s)\Big{)}_{\mid s=a}=0\,.$

Note that it is necessary that $w_{M,q}>0$ to guarantee the strongly positive
sign of the function $v_{s}^{q}[M]$. Indeed, if there exists $t^{*}\in[a,b]$,
such that $w_{M,q}(t^{*})<0$, then there exists $\rho(t^{*})>a$ such that
$v_{s}^{q}[M](t^{*})<0$ for all $s\in(a,\rho(t^{*}))$, which contradicts the
strongly positive sign.

Since the function $v_{s}^{q}[M]$ is a solution of
$T_{n}[M]\,v_{s}^{q}[M](t)=0$ for $t\neq s$, we have that

$\dfrac{\partial^{\eta}}{\partial
s^{\eta}}\left(T_{n}[M]\,v_{s}^{q}[M](t)\right)_{\mid
s=a}=T_{n}[M]\,w_{M,q}(t)=0\,,\quad t\in(a,b]\,,$

that is, the function $w_{M,q}$ solves the equation $T_{n}[M]\,u(t)=0$,
$t\in(a,b]$.

It remains to obtain the boundary conditions satisfied by the function
$w_{M,q}$.

Using the matrix of the Green’s function given in (2.4), the equality (2.6),
the expression of $g_{n-j}$, $j=1,\dots,n-1$ given in (2.8) and the following
identity

$\dfrac{\partial}{\partial t}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}\,G_{M}(t,s)\Big{)}=A\,\dfrac{\partial^{q}}{\partial
t^{q}}G_{M}(t,s),\quad\text{for all }t\in
I\backslash\left\\{s\right\\},\,q\in\\{1,\dots,n-1\\},$

it follows that

$\dfrac{\partial^{\mu_{1}^{q}+q}}{\partial
t^{\mu_{1}^{q}+q}}g_{M}^{2}(t,s)_{\mid
t=a}=0,\,-\dfrac{\partial^{\mu_{1}^{q}+q+1}}{\partial
t^{\mu_{1}^{q}+q}\partial s}g_{M}^{2}(t,s)_{\mid
t=a}=0,\,\dots,\,(-1)^{\eta}\dfrac{\partial^{\mu_{1}^{q}+q+\eta}}{\partial
t^{\mu_{1}^{q}+q}\partial s^{\eta}}g_{M}^{2}(t,s)_{\mid t=a}=0\,.$

Since $(\eta+\mu_{1}^{q}+q)\mod{n}<n-1$, then we are not in an element of the
diagonal of $\dfrac{\partial^{q}}{\partial t^{q}}G_{M}(t,s)$, and taking in
account that the above equalities are satisfied for $s=a$, by continuity we
deduce that

$\left\\{\begin{array}[]{r}\dfrac{\partial^{\mu_{1}^{q}+q}}{\partial
t^{\mu_{1}^{q}+q}}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=0\,,\\\ \\\
-\dfrac{\partial^{\mu_{1}^{q}+q+1}}{\partial t^{\mu_{1}^{q}+q}\partial
s}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=0\,,\\\ \\\ \vdots\\\ \\\
(-1)^{\eta}\dfrac{\partial^{\mu_{1}^{q}+q+\eta}}{\partial
t^{\mu_{1}^{q}+q}\partial
s^{\eta}}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=0\,.\end{array}\right.$

Therefore,

$w_{M,q}^{(\mu_{1}^{q})}(a)=\dfrac{\partial^{\mu_{1}^{q}+q+\eta}}{\partial
t^{\mu_{1}^{q}+q}\partial s^{\eta}}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=0\,.$

Making a similar reasoning we deduce that

$w_{M,q}^{(\mu_{2}^{q})}(a)=\cdots=w_{M,q}^{(\mu_{z-1}^{q})}(a)=w_{M,q}^{(\mu_{z+1}^{q})}(a)=\cdots=w_{M,q}^{(\mu_{k}^{q})}(a)=0\,.$

For $\mu_{z}^{q}$, we have the following equalities

$\dfrac{\partial^{\mu_{z}^{q}+q}}{\partial
t^{\mu_{z}^{q}+q}}g_{M}^{2}(t,s)_{\mid
t=a}=0,\,-\dfrac{\partial^{\mu_{z}^{q}+q+1}}{\partial
t^{\mu_{z}^{q}+q}\partial s}g_{M}^{2}(t,s)_{\mid
t=a}=0,\,\dots,\,(-1)^{\eta}\dfrac{\partial^{\mu_{z}^{q}+q+\eta}}{\partial
t^{\mu_{z}^{q}+q}\partial s^{\eta}}g_{M}^{2}(t,s)_{\mid t=a}=0\,.$

By assumption, we have that $\eta+\mu_{z}^{q}+q=n-1$, so we are in an element
of the diagonal of $\dfrac{\partial^{q}}{\partial t^{q}}G_{M}(t,s)$, and
taking into account again that the previous equalities are satisfied for $s=a$
and that $\dfrac{\partial^{q}}{\partial t^{q}}G_{M}(t,s)$ has a jump equal to
$1$ on the diagonal we infer that

$\left\\{\begin{array}[]{r}\dfrac{\partial^{\mu_{z}^{q}+q}}{\partial
t^{\mu_{z}^{q}+q}}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=0\,,\\\ \\\
-\dfrac{\partial^{\mu_{z}^{q}+q+1}}{\partial t^{\mu_{z}^{q}+q}\partial
s}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=0\,,\\\ \\\ \vdots\\\ \\\
(-1)^{\eta}\dfrac{\partial^{\mu_{z}^{q}+q+\eta}}{\partial
t^{\mu_{z}^{q}+q}\partial
s^{\eta}}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=1\,.\end{array}\right.$

Therefore,

$w_{M,q}^{(\mu_{z}^{q})}(a)=\dfrac{\partial^{\mu_{z}^{q}+q+\eta}}{\partial
t^{\mu_{z}^{q}+q}\partial
s^{\eta}}g_{M}^{1}(t,s)_{\mid(t,s)=(a,a)}=(-1)^{\eta}\,.$

Let’s see what conditions $w_{M,q}$ satisfies at $t=b$. To do this, proceeding
with the previous argument we obtain that

$\dfrac{\partial^{\rho_{1}^{q}+q}}{\partial
t^{\rho_{1}^{q}+q}}g_{M}^{1}(t,s)_{\mid
t=b}=0,\,-\dfrac{\partial^{\rho_{1}^{q}+q+1}}{\partial
t^{\rho_{1}^{q}+q}\partial s}g_{M}^{1}(t,s)_{\mid
t=b}=0,\,\dots,\,(-1)^{\eta}\dfrac{\partial^{\rho_{1}^{q}+q+\eta}}{\partial
t^{\rho_{1}^{q}+q}\partial s^{\eta}}g_{M}^{1}(t,s)_{\mid t=b}=0\,.$

The above equalities are satisfied at $s=a$ and since $b\neq a$ we have that

$w_{M,q}^{(\rho_{1}^{q})}(b)=\dfrac{\partial^{\rho_{1}^{q}+q+\eta}}{\partial
t^{\rho_{1}^{q}+q}\partial s^{\eta}}g_{M}^{1}(t,s)_{\mid(t,s)=(b,a)}=0\,.$

Analogously we obtain that

$w_{M,q}^{(\rho_{2}^{q})}(b)=\dots=w_{M,q}^{(\rho_{n-k}^{q})}(b)=0\,.$

If $\eta=0$, then it follows from the matrix argument we made earlier that the
function $w_{M,q}(t)=\frac{\partial^{q}}{\partial t^{q}}g_{M}^{1}(t,a)$
satisfies the same boundary conditions (3.11)–(3.12).

Then from Proposition 3.5 and Theorem 2.28 we conclude that $w_{M,q}>0$ on
$(a,b)$ for all $M\in[0,\lambda_{2}^{q}]$.

Now, let us see that $v_{s}^{q}[M]$ cannot be positive for
$M>\lambda_{2}^{q}$.

Suppose that there exists $\widehat{M}>\lambda_{2}^{q}$ such that
$v_{s}^{q}[\widehat{M}]$ is positive. Then, by monotony of $v_{s}^{q}[M]$ we
have for every $M\in[\lambda_{2}^{q},\widehat{M}]$ that $w_{\widehat{M},q}\leq
w_{M,q}\leq w_{\lambda_{2}^{q},q}$.

In particular, $w_{\widehat{M},q}^{(\beta^{q})}(b)\geq
w_{M,q}^{(\beta^{q})}(b)\geq w_{\lambda_{2}^{q},q}^{(\beta^{q})}(b)=0$ if
$\beta^{q}$ is even and $w_{\widehat{M},q}^{(\beta^{q})}(b)\leq
w_{M,q}^{(\beta^{q})}(b)\leq w_{\lambda_{2}^{q},q}^{(\beta^{q})}(b)=0$ if
$\beta^{q}$ is odd.

If $w_{\widehat{M},q}^{(\beta^{q})}(b)\neq 0$, then there exists $\rho>0$ such
that $w_{\widehat{M},q}(t)<0$ for all $t\in(b-\rho,b)$, which contradicts our
assumption. Hence,

$0=w_{\widehat{M},q}^{(\beta^{q})}(b)=w_{M,q}^{(\beta^{q})}(b)=w_{\lambda_{2}^{q},q}^{(\beta^{q})}(b)\,,\quad\forall
M\in[\lambda_{2}^{q},\widehat{M}],$

and this fact contradicts the discrete character of the spectrum.

Thus, we conclude that, if $M\in[0,\lambda_{2}^{q}]$, then

$\forall t\in(a,b)\,,\quad\exists\rho^{q}(t)>0\ \mid\ v_{s}^{q}[M](t)>0\
\forall s\in(a,a+\rho^{q}(t))\,.$

Moreover, if $M>\lambda_{2}^{q}$, then $v_{s}^{q}[M]$ is not positive.

  * Step 2.

Study of the function $v_{s}^{q}[M]$ at $s=b$.

We study the following function by applying similar reasoning to Step 1. In
this case we consider

$y_{M,q}(t)=\dfrac{\partial^{\gamma}}{\partial
s^{\gamma}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(t,s)\Big{)}_{\mid s=b}\,,$

where $\gamma$ has been defined in (2.14).

In this case, using arguments analogous to Step 1 we obtain that if
$\gamma>0$, then

$\dfrac{\partial^{q}}{\partial t^{q}}g_{M}^{2}(t,b)=\dfrac{\partial}{\partial
s}g_{M}^{2}(t,s)_{\mid s=b}=\cdots=\dfrac{\partial^{\gamma-1}}{\partial
s^{\gamma-1}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(t,s)\Big{)}_{\mid s=b}=0\,.$

Moreover, if $\gamma$ is even, then $y_{M,q}\geq 0$ and if $\gamma$ is odd,
then $y_{M,q}\leq 0$.

Again, we have that

$T_{n}[M]\,y_{M,q}(t)=0\,,\quad\forall\,t\in[a,b)\,.$

In the same way than above, studying $\dfrac{\partial^{q}}{\partial
t^{q}}G_{M}(t,s)$ to determine the boundary conditions of $y_{M,q}$ we infer
that

$\begin{split}y_{M,q}^{(\mu_{1}^{q})}(a)=\cdots=y_{M,q}^{(\mu_{k}^{q})}(a)=&\,0\,,\\\
y_{M,q}^{(\rho_{1}^{q})}(b)\hskip
8.5359pt=\cdots=y_{M,q}^{(\rho_{h-1}^{q})}(b)=y_{M,q}^{(\rho_{h+1}^{q})}(b)=\cdots=y_{M,q}^{(\rho_{n-k}^{q})}(b)=&\,0\,,\\\
\hskip 85.35826pty_{M,q}^{(\rho_{h}^{q})}(b)=&\,(-1)^{\gamma+1}\,.\end{split}$

Analogously, if $\gamma=0$, then $y_{M,q}(t)=\frac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(t,b)$ satisfies the conditions (3.13)–(3.14) by an argument
similar to the case $\gamma>0$.

Then from Proposition 3.6 and Theorem 2.28 we conclude that $y_{M,q}>0$ if
$\gamma$ is even and $y_{M,q}<0$ if $\gamma$ is odd on $(a,b)$ for all
$M\in[0,\lambda_{4}^{q}]$.

As in the previous step it can be seen that for $M>\lambda_{4}^{q}$, the
function $v_{s}^{q}[M]$ is never positive. Therefore, we conclude from this
step that, if $M\in[0,\lambda_{4}^{q}]$, then

$\forall t\in(a,b)\,,\quad\exists\rho^{q}(t)>0\ \mid\ v_{s}^{q}[M](t)>0\
\forall s\in(b-\rho^{q}(t),b)\,.$

Moreover, if $M>\lambda_{4}^{q}$, then $v_{s}^{q}[M]$ is cannot be positive.

  * Step 3.

Study of the function $v_{s}^{q}[M]$ at $t=a$.

Let us denote

$\widehat{g}_{(-1)^{n}M}(t,s)=\left\\{\begin{array}[]{cc}\widehat{g}_{(-1)^{n}M}^{1}(t,s)\,,&a\leq
s\leq t\leq b\,,\\\ \\\
\widehat{g}_{(-1)^{n}M}^{2}(t,s)\,,&a<t<s<b\,,\end{array}\right.$

as the related Green’s function of $\widehat{T}_{n}[(-1)^{n}M]$ in $X_{\
\,\\{\sigma_{1},\dots,\sigma_{k}\\}}^{*\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}=X_{\\{\tau_{1},\dots,\tau_{n-k}\\}}^{\\{\delta_{1},\dots,\delta_{k}\\}}$.

In this case, we study the following function by applying similar reasoning to
Step 1:

$\widehat{w}_{M,q}(t)=(-1)^{n}\dfrac{\partial^{\alpha^{q}+q}}{\partial\,s^{\alpha^{q}+q}}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid
s=a}\,,$

which is equivalent, using equality (2.12), to study the function

$\widehat{w}_{M,q}(s)=\dfrac{\partial^{\alpha^{q}+q}}{\partial\,t^{\alpha^{q}+q}}g_{M}^{2}(t,s)_{\mid
t=a}=\dfrac{\partial^{\alpha^{q}}}{\partial\,t^{\alpha^{q}}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(t,s)\Big{)}_{\mid t=a}\,,\quad s\in I\,.$

In this case, if $\alpha^{q}>0$ using that
$\\{0,\dots,\alpha^{q}-1\\}\subset\\{\mu_{1}^{q},\dots,\mu_{k}^{q}\\}$ we have

$\dfrac{\partial^{q}}{\partial t^{q}}g_{M}^{2}(a,s)=\dfrac{\partial}{\partial
t}\Big{(}\dfrac{\partial^{q}}{\partial t^{q}}g_{M}^{2}(t,s)\Big{)}_{\mid
t=a}=\cdots=\dfrac{\partial^{\alpha^{q}-1}}{\partial
t^{\alpha^{q}-1}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(t,s)\Big{)}_{\mid t=a}=0\,,\quad\forall s\in(a,b).$

As in the Steps 1 and 2 we can deduce that

$\widehat{T}_{n}[(-1)^{n}M]\,\widehat{w}_{M,q}(t)=0,\;\;t\in(a,b].$

Using the arguments of Step 1, we can affirm that if there exists
$t^{*}\in(a,b)$ such that $\widehat{w}_{M,q}(t^{*})<0$, then $v_{s}^{q}[M]$ is
not positive.

By the first row of (2.17) and (2.16) we deduce that:

$\dfrac{\partial^{\tau_{1}}}{\partial
t^{\tau_{1}}}\widehat{g}_{(-1)^{n}M}^{2}(t,s)_{\mid
t=a}=0,\,-\dfrac{\partial^{\tau_{1}+1}}{\partial t^{\tau_{1}}\partial
s}\widehat{g}_{(-1)^{n}M}^{2}(t,s)_{\mid
t=a}=0,\,\dots,(-1)^{\alpha^{q}+q}\dfrac{\partial^{\tau_{1}+\alpha^{q}+q}}{\partial
t^{\tau_{1}}\partial s^{\alpha^{q}+q}}\widehat{g}_{(-1)^{n}M}^{2}(t,s)_{\mid
t=a}=0\,.$

Since $(\tau_{1}+\alpha^{q}+q)\mod{n}<n-1$, we do not reach any diagonal
element, hence the previous equalities are satisfied for $s=a$, and we obtain:

$\left\\{\begin{array}[]{rl}\dfrac{\partial^{\tau_{1}}}{\partial
t^{\tau_{1}}}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid(t,s)=(a,a)}&=0\,,\\\ \\\
-\dfrac{\partial^{\tau_{1}+1}}{\partial t^{\tau_{1}}\partial
s}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid(t,s)=(a,a)}&=0\,,\\\ \\\ \vdots\\\
\\\ (-1)^{\alpha^{q}+q}\dfrac{\partial^{\tau_{1}+\alpha^{q}+q}}{\partial
t^{\tau_{1}}\partial
s^{\alpha^{q}+q}}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid(t,s)=(a,a)}&=0\,.\end{array}\right.$

So, we conclude that $\widehat{w}_{M,q}^{(\tau_{1})}(a)=0\,.$

Analogously we obtain that

$\widehat{w}_{M,q}^{(\tau_{2})}(a)=\cdots=\widehat{w}_{M,q}^{(\tau_{h-1})}(a)=\widehat{w}_{M,q}^{(\tau_{h+1})}(a)=\cdots=\widehat{w}_{M,q}^{(\tau_{n-k})}(a)=0.$

For $\tau_{h}$, we have the following equalities

$\dfrac{\partial^{\tau_{h}}}{\partial
t^{\tau_{h}}}\widehat{g}_{(-1)^{n}M}^{2}(t,s)_{\mid
t=a}=0,\,-\dfrac{\partial^{\tau_{h}+1}}{\partial t^{\tau_{h}}\partial
s}\widehat{g}_{(-1)^{n}M}^{2}(t,s)_{\mid
t=a}=0,\,\dots,(-1)^{\alpha^{q}+q}\dfrac{\partial^{\tau_{h}+\alpha^{q}+q}}{\partial
t^{\tau_{h}}\partial s^{\alpha^{q}+q}}\widehat{g}_{(-1)^{n}M}^{2}(t,s)_{\mid
t=a}=0\,.$

In this case, since $\tau_{h}+\alpha^{q}+q=n-1$, we reach a diagonal element
of $\widehat{G}(t,s)$ given in (2.15), and as consequence we obtain the
following equalities for $s=a$:

$\left\\{\begin{array}[]{rl}\dfrac{\partial^{\tau_{h}}}{\partial
t^{\tau_{h}}}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid(t,s)=(a,a)}&=0\,,\\\ \\\
-\dfrac{\partial^{\tau_{h}+1}}{\partial t^{\tau_{h}}\partial
s}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid(t,s)=(a,a)}&=0\,,\\\ \\\ \vdots\\\
\\\ (-1)^{\alpha^{q}+q}\dfrac{\partial^{\tau_{h}+\alpha^{q}+q}}{\partial
t^{\tau_{h}}\partial
s^{\alpha^{q}+q}}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid(t,s)=(a,a)}&=1\,.\end{array}\right.$

Therefore, $\widehat{w}_{M,q}^{(\tau_{h})}(a)=(-1)^{n-\alpha^{q}-q}\,.$

Now, let us study the behavior of $\widehat{w}_{M,q}$ at $t=b$. Studying the
$(n-k+1)^{\mathrm{th}}$ row of (2.17), we have for all $s\in(a,b)$:

$\dfrac{\partial^{\delta_{1}}}{\partial
t^{\delta_{1}}}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid
t=b}=0,-\dfrac{\partial^{\delta_{1}+1}}{\partial t^{\delta_{1}}\partial
s}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid
t=b}=0,\dots,(-1)^{\alpha^{q}+q}\dfrac{\partial^{\delta_{1}+\alpha^{q}+q}}{\partial
t^{\delta_{1}}\partial s^{\alpha^{q}+q}}\widehat{g}_{(-1)^{n}M}^{1}(t,s)_{\mid
t=b}=0\,.$

The above equalities are satisfied at $s=a$ and since $b\neq a$ we have that
$\widehat{w}_{M,q}^{(\delta_{1})}(b)=0$.

Analogously we obtain that

$\widehat{w}_{M,q}^{(\delta_{2})}(b)=\cdots=\widehat{w}_{M,q}^{(\delta_{k})}(b)=0.$

Note that if $\alpha^{q}=0$, then from the same matrix argument above we
deduce that the function $\widehat{w}_{M,q}(s)=\frac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(a,s)$ verifies the same boundary conditions:

$\displaystyle\widehat{w}_{M,q}^{(\tau_{2})}(a)=\cdots=\widehat{w}_{M,q}^{(\tau_{h-1})}(a)=\widehat{w}_{M,q}^{(\tau_{h+1})}(a)=$
$\displaystyle\cdots$ $\displaystyle=\widehat{w}_{M,q}^{(\tau_{n-k})}(a)=0\,,$
$\displaystyle\widehat{w}_{M,q}^{(\delta_{2})}(b)=$ $\displaystyle\cdots$
$\displaystyle=\widehat{w}_{M,q}^{(\delta_{k})}(b)\hskip 8.5359pt=0\,.$

From (2.11) and (2.9) we deduce that

$\widehat{T}_{n}[0]\,v(t):=(-1)^{n}\,T_{n}^{*}[0]\,v(t)=T_{n}[0]\,v(t).$

Hence, using arguments similar to Step 1, we conclude that
$\widehat{w}_{M,q}>0$ on $(a,b)$ for all $M\in[0,(\lambda_{4}^{*})^{q}]$,
where $(\lambda_{4}^{*})^{q}>0$ is the least positive eigenvalue of
$\widehat{T}_{n}[0]\equiv T_{n}[0]$ in $X_{4}^{*}$ defined in (3.18).

From Lemma 3.10, we know that $(\lambda_{4}^{*})^{q}=\lambda_{4}^{q}$. Thus,
from this step we have that, if $M\in[0,\lambda_{4}^{q}]$, then

$\forall s\in(a,b)\,,\quad\exists\rho^{q}(s)>0\ \mid\ v_{s}^{q}[M](t)>0\
\forall t\in(a,a+\rho^{q}(s))\,.$

  * Step 4.

Study of the function $v_{s}^{q}[M]$ at $t=b$.

In this case, we study the following function by applying similar reasoning to
Step 3

$\widehat{y}_{M,q}(t)=(-1)^{n}\dfrac{\partial^{\beta^{q}+q}}{\partial\,s^{\beta^{q}+q}}\widehat{g}_{(-1)^{n}M}^{2}(t,s)_{\mid
s=b}\,,$

which is equivalent using equality (2.12) to studying the function

$\widehat{y}_{M,q}(s)=\dfrac{\partial^{\beta^{q}+q}}{\partial\,t^{\beta^{q}+q}}g_{M}^{1}(t,s)_{\mid
t=b}=\dfrac{\partial^{\beta^{q}}}{\partial\,t^{\beta^{q}}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(t,s)\Big{)}_{\mid t=b}\,.$

Since $\\{0,\dots,\beta^{q}\\}\subset\\{\rho_{1}^{q},\dots,\rho_{n-k}\\}$ if
$\beta^{q}>0$ we obtain that

$\dfrac{\partial^{q}}{\partial t^{q}}g_{M}(b,t)=\dfrac{\partial}{\partial
t}\Big{(}\dfrac{\partial^{q}}{\partial t^{q}}{g}_{M}(t,s)\Big{)}_{\mid
t=b}=\cdots=\dfrac{\partial^{\beta^{q}-1}}{\partial
t^{\beta^{q}-1}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}{g}_{M}(t,s)\Big{)}_{\mid t=b}=0\,.$

Moreover, using arguments analogous to Step 1 we obtain that if $\beta^{q}$ is
even, then $\widehat{y}_{M,q}\geq 0$ and if $\beta^{q}$ is odd, then
$\widehat{y}_{M,q}\leq 0$.

Again, we have that

$\widehat{T}_{n}[(-1)^{n}M]\,\widehat{y}_{M,q}(t)=0\,,\quad\forall\,t\in[a,b)\,.$

A study analogous to Step 3 leads us to the fact that $\widehat{y}_{M,q}$
satisfies the boundary conditions

$\begin{split}\widehat{y}_{M,q}^{(\tau_{1})}(a)=\cdots=\widehat{y}_{M,q}^{(\tau_{n-k})}(a)=&\,0\,,\\\
\widehat{y}_{M,q}^{(\delta_{1})}(b)=\cdots=\widehat{y}_{M,q}^{(\delta_{z-1})}(b)=\widehat{y}_{M,q}^{(\delta_{z+1})}(b)=\cdots=\widehat{y}_{M,q}^{(\delta_{k})}(b)=&\,0\,,\\\
y_{M,q}^{(\delta_{z})}(b)=&\,(-1)^{n-\beta^{q}-q+1}\,.\end{split}$ (3.21)

Analogously, if $\beta^{q}=0$, then
$\widehat{y}_{M,q}(s)=\frac{\partial^{q}}{\partial t^{q}}g_{M}^{2}(b,s)$
satisfies the conditions (3.21) by an argument similar to the case
$\beta^{q}>0$.

Thus, we deduce that $\widehat{y}_{M,q}>0$ on $(a,b)$ if $\beta^{q}$ is even
and $\widehat{y}_{M,q}<0$ on $(a,b)$ if $\beta^{q}$ is odd for all
$M\in[0,(\lambda_{2}^{*})^{q}]$ , where $(\lambda_{2}^{*})^{q}>0$ is the least
positive eigenvalue of $\widehat{T}_{n}[0]\equiv T_{n}[0]$ in $X_{2}^{*}$
defined in (3.17).

Again, from Lemma 3.9 we have that $(\lambda_{2}^{*})^{q}=\lambda_{2}^{q}$.
Therefore, we conclude from this step that, if $M\in[0,\lambda_{2}^{q}]$, then

$\forall s\in(a,b)\,,\quad\exists\rho^{q}(s)>0\ \mid\ v_{s}^{q}[M](t)>0\
\forall t\in(b-\rho^{q}(t),b)\,.$

  * Step 5.

Study of the function $v_{s}^{q}[M]$ on $(a,b)\times(a,b)$.

In this step, we verify that the function $v_{s}^{q}[M](t)>0$ for all
$(t,s)\in I\times I$ if $M$ belongs to the given intervals. Let us denote
$u_{M}^{s}(t)=g_{M}(t,s)$, for all $s\in(a,b)$.

The function $v_{s}^{q}[M](t)$ is in $C^{n-2-q}(I)$ and it satisfies the
boundary conditions (2.18)-(2.19).

Using the definition of $v_{s}^{q}[M]$, the following equality holds for all
$s\in(a,b)$:

$T_{n-q}[0]\,v_{s}^{q}[M](t)=-M\,u_{M}^{s}(t)\,,\quad\forall\,t\neq s\,,\ t\in
I\,.$ (3.22)

Let’s see that for the values of the parameter $M$ for which $v_{s}^{q}[M](t)$
has constant sign in $I$, it cannot have a double zero in $(a,b)$, and this
implies that the change of the sign must be at $t=a$ or $t=b$, and the result
is proven.

We do the proof for $n-k$ even (the case $n-k$ odd is similarly proved). In
this case, we study the behavior of $v_{s}^{q}[M](t)$ for $M>0$ and
$u_{M}^{s}\geq 0$. From (3.22), we have that

$\dfrac{d^{n-q}}{dt^{n-q}}\,v_{s}^{q}[M](t)=\dfrac{\partial^{n}}{\partial
t^{n}}\,g_{M}(t,s)\leq 0.$

Therefore, using the condition $(T_{d})$, since $v_{1}=\dots=v_{n}=1$, we have
that

$\dfrac{d^{n-q-1}}{dt^{n-q-1}}\,v_{s}^{q}[M](t)=\dfrac{\partial^{n-1}}{\partial
t^{n-1}}\,g_{M}(t,s)$

is a decreasing function, with two continuous components and a jump $+1$ at
$t=s$. Then, it has at most two zeros on $I$ (see Figure 1).

Figure 1:  $\dfrac{d^{n-q-1}}{dt^{n-q-1}}\,v_{s}^{q}[M](t)$, maximal
oscillation with $t\in I=[0,1].$

Now, we have that

$\dfrac{d^{n-q-2}}{dt^{n-q-2}}\,v_{s}^{q}[M](t)=\dfrac{\partial^{n-2}}{\partial
t^{n-2}}\,g_{M}(t,s)$

is a continuous function with at most four zeros on $I$ (See Figure 2).

Figure 2:  $\dfrac{d^{n-q-1}}{dt^{n-q-1}}\,v_{s}^{q}[M](t)$, maximal
oscillation with $t\in I=[0,1].$

By recurrence, we have that

$\dfrac{d^{n-q-l}}{dt^{n-q-l}}\,v_{s}^{q}[M](t)=\dfrac{\partial^{n-l}}{\partial
t^{n-l}}\,g_{M}(t,s)$

has at most $l+2$ zeros on $I$. In particular, $v_{s}^{q}[M]$ has at most
$n-q+2$ zeros on $I$. Moreover, since

$\displaystyle(v_{s}^{q})^{(\sigma_{1}^{q})}[M](a)=$ $\displaystyle\cdots$
$\displaystyle=(v_{s}^{q})^{(\sigma_{c_{q}}^{q})}[M](a)=0\,,$
$\displaystyle(v_{s}^{q})^{(\varepsilon_{1}^{q})}[M](b)=$
$\displaystyle\cdots$
$\displaystyle=(v_{s}^{q})^{(\varepsilon_{d_{q}}^{q})}[M](b)\hskip
1.13791pt=0\,,$

$v_{s}^{q}[M]$ has $c_{q}+d_{q}=n-q$ zeros on the boundary. Thus,
$v_{s}^{q}[M]$ has at most two zeros on $(a,b)$.

Now, making an argument similar to the one made in [5, Theorem 5.1], we deduce
that the maximal oscillation (that is, the maximal number of zeros) is
possible if and only if

$\left\\{\begin{array}[]{cc}(v_{s}^{q})^{(\alpha^{q})}[M](a)\leq 0\,,&\text{if
$n-q-c_{q}$ is even,}\\\ &\\\ (v_{s}^{q})^{(\alpha^{q})}[M](a)\geq
0\,,&\text{if $n-q-c_{q}$ is odd.}\end{array}\right.$ (3.23)

Since we are in the case $n-q-c_{q}$ even, the maximal oscillation is only
possible if$(v_{s}^{q})^{(\alpha^{q})}[M](a)\leq 0$. However, by definition,
$(v_{s}^{q})^{(\alpha^{q})}[M](a)\neq 0$ and, since $v_{s}^{q}[M]$ is
nonnegative and $\alpha^{q}$ denotes the order of the smallest derivative at
$t=a$ which is different from zero, necessarily
$(v_{s}^{q})^{(\alpha^{q})}[M](a)>0$. Therefore the maximal oscillation can
not happen (that is, $v_{s}^{q}[M]$ cannot have two zero on $(a,b)$). Thus,
$v_{s}^{q}[M]$ has at most one simple zero on $(a,b)$, which is not possible
due to the constant sign of $v_{s}^{q}[M]$, and we conclude that
$v_{s}^{q}[M]$ does not have any zero on $(a,b)$.

From this step we deduce that if $M>0$ and $u_{M}^{s}\geq 0$, then
$v_{s}^{q}[M]>0$ on $(a,b)$.

In conclusion, from the previous steps the result is proved. ∎

In the sequel, we present some examples of application of the previous result.

######  Example 3.12.

Consider again the fourth order operator $T_{4}[M]$ coupled with the boundary
conditions $X_{\\{0,1,2\\}}^{\\{2\\}}$ defined in Example 2.8.

The functions $\frac{\partial}{\partial t}{g}_{M}$ and
$\frac{\partial^{2}}{\partial t^{2}}{g}_{M}$ satisfy the conditions of spaces
$X_{\\{0,1,3\\}}^{\\{1\\}}$ and $X_{\\{0,2,3\\}}^{\\{0\\}}$, respectively.

In this case $k=3$ and $n-k=1$ is odd. For $q=1$, we have that $c_{q}=2$,
$d_{q}=1$, $z=2\neq 1$ and $\beta^{1}=0$. By applying Theorem 3.11 it follows
that $\frac{\partial}{\partial t}{g}_{M}$ is strongly negative on
$[0,1]\times[0,1]$ if, and only if $M\in[\lambda_{2}^{1},\lambda_{1})$, where

  * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{4}[0]$ in
$X_{\\{0,1,2\\}}^{\\{2\\}}.$

  * *

$\lambda_{2}^{1}<0$ is the biggest negative eigenvalue of $T_{4}[0]$ in
$X_{\\{0,3\\}}^{\\{0,1\\}}.$

The eigenvalues of $T_{4}[0]$ in $X_{\\{0,1,2\\}}^{\\{2\\}}$ are given by
$\lambda^{4}$, where $\lambda$ is a positive solution of

$\tanh\Big{(}\lambda\sqrt{2}\Big{)}+\tan\Big{(}\frac{\lambda}{\sqrt{2}}\Big{)}=0.$

Let us denote $m_{1}$ as the smallest positive solution of this equation.
Then, $\lambda_{1}=m_{1}^{4}$ is the least positive eigenvalue of $T_{4}[0]$
in $X_{\\{0,1,2\\}}^{\\{2\\}}.$

In the same way, the eigenvalues of $T_{4}[0]$ in $X_{\\{0,3\\}}^{\\{0,1\\}}$
are given by $-\lambda^{4}$, where $\lambda$ is a positive solution of
$\sin(\lambda)=0$.

In this case, $\lambda_{min}^{1}=\pi$ is the smallest positive solution of
this equation. Then, $\lambda_{2}^{1}=-(\lambda_{min}^{1})^{4}=-\pi^{4}$ is
the biggest negative eigenvalue of $T_{4}[0]$ in $X_{\\{0,3\\}}^{\\{0,1\\}}.$

By numerical approach, it can be seen that $\lambda_{1}=m_{1}^{4}\approx
3.34^{4}$.

For $q=2$, we have that $c_{q}=n-q-1=1$, $d_{q}=1$, $z=1$ and $\beta^{2}=1$.
Again, by applying Theorem 3.11 it follows that $\frac{\partial^{2}}{\partial
t^{2}}{g}_{M}$ is strongly negative on $[0,1]\times[0,1]$ if, and only if
$M\in[\lambda_{2}^{2},\lambda_{1})$, where

  * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{4}[0]$ in
$X_{\\{0,1,2\\}}^{\\{2\\}}.$

  * *

$\lambda_{2}^{2}<0$ is the biggest negative eigenvalue of $T_{4}[0]$ in
$X_{\\{2,3\\}}^{\\{0,1\\}}.$

The eigenvalues of $T_{4}[0]$ in $X_{\\{2,3\\}}^{\\{0,1\\}}$ are given by
$-\lambda^{4}$, where $\lambda$ is a least positive solution of

$2e^{2\lambda}+(e^{2\lambda}+1)\,\cos(\lambda)=0.$

By denoting $\lambda_{min}^{2}$ as the smallest positive solution of this
equation, then $\lambda_{2}^{2}=-(\lambda_{min}^{2})^{4}$ is the biggest
negative eigenvalue of $T_{4}[0]$ in $X_{\\{2,3\\}}^{\\{0,1\\}}.$

In this case, we have that
$\lambda_{2}^{2}=-(\lambda_{min}^{2})^{4}\approx-1.87^{4}$.

######  Example 3.13.

We consider the fourth order operator $T_{4}[M]$ defined in the space of the
boundary conditions $X_{\\{0,1\\}}^{\\{1,3\\}}$.

In this case $k=2$ and $n-k$ is even. For $q=1$ we have that $c_{q}=1$,
$d_{q}=n-q-c_{q}=2$, $z=1$, $h=2$ and $\alpha^{1}=\beta^{1}=1$. Moreover, the
function $\frac{\partial}{\partial t}{g}_{M}$ satisfies the conditions of
space $X_{\\{0,3\\}}^{\\{0,2\\}}$.

By applying Theorem 3.11 we obtain that $\frac{\partial}{\partial t}{g}_{M}$
is strongly positive on $[0,1]\times[0,1]$ if, and only if
$M\in(\lambda_{1},\lambda^{1}]$, where

  * *

$\lambda_{1}<0$ is the least positive eigenvalue of $T_{4}[0]$ in
$X_{\\{0,1\\}}^{\\{1,3\\}}.$

  * *

$\lambda^{1}>0$ is the minimum between:

    * ·

$\lambda_{2}^{1}>0$, the least positive eigenvalue of $T_{4}[0]$ in
$X_{\\{3\\}}^{\\{0,1,2\\}}.$

    * ·

$\lambda_{4}^{1}>0$ is the least positive eigenvalue of $T_{4}[0]$ in
$X_{\\{0,1,3\\}}^{\\{0\\}}.$

By numerical approximation, we obtain that $\lambda_{1}\approx-2.36^{4}$,
$\lambda_{2}^{1}\approx 2.22^{4}$ and $\lambda_{4}^{1}\approx 4.44^{4}$.
Therefore, $\frac{\partial}{\partial t}{g}_{M}$ is strongly positive on
$[0,1]\times[0,1]$ if, and only if, $M\in(-2.36^{4},2.22^{4}]$.

As a consequence of the Theorem 3.11 we arrive at the following result.

######  Corollary 3.14.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$, $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and $d_{q}\geq
1$. If either $\sigma_{k}=q+c_{q}-1$ or $\varepsilon_{n-k}=q+d_{q}-1$, we have
the following properties:

  * •

If $n-q-c_{q}$ is even, then there is not any $M\in\mathbb{R}$ such that
$v_{s}^{q}[M]$ is strongly negative on $I\times I$.

  * •

If $n-q-c_{q}$ is odd, then there is not any $M\in\mathbb{R}$ such that
$v_{s}^{q}[M]$ is strongly positive on $I\times I$.

###### Proof.

If $\sigma_{k}=q+c_{q}-1$, then $\eta=n-1-\sigma_{k}=n-q-c_{q}$.

We consider

$w_{M,q}(t)=\dfrac{\partial^{\eta}}{\partial
s^{\eta}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{1}(t,s)\Big{)}_{\mid s=a},$

defined in Step 1 of the proof of Theorem 3.11.

We know that this function satisfies the following boundary conditions:

$\begin{split}w_{M,q}^{(\mu_{1}^{q})}(a)=\cdots=w_{M,q}^{(\mu_{z-1}^{q})}(a)=w_{M,q}^{(\mu_{z+1}^{q})}(a)=\cdots=w_{M,q}^{(\mu_{k}^{q})}(a)=&\,0\,,\\\
w_{M,q}^{(\rho_{1}^{q})}(b)=\cdots=w_{M,q}^{(\rho_{n-k}^{q})}(b)=&\,0\,,\\\
w_{M,q}^{(\mu_{z}^{q})}(a)=&\,(-1)^{\eta}=(-1)^{n-q-c_{q}}\,.\end{split}$

Hence, if $n-q-c_{q}$ is even, then there exists $\rho>0$, such that
$w_{M,q}(t)>0$ for all $t\in(a,a+\rho)$. So, $v_{s}^{q}[M]$ cannot be negative
for any real $M$.

Now, if $n-q-c_{q}$ is odd, then there exists $\rho>0$, such that
$w_{M,q}(t)<0$ for all $t\in(a,a+\rho)$. Thus, $v_{s}^{q}[M]$ cannot be
positive for any $M\in\mathbb{R}$.

Analogously, if $\varepsilon_{n-k}=q+d_{q}-1$, then
$\gamma=n-\varepsilon_{n-k}-1=n-q-d_{q}=c_{q}$. In this case, we consider the
function

$y_{M,q}(t)=\dfrac{\partial^{\gamma}}{\partial
s^{\gamma}}\Big{(}\dfrac{\partial^{q}}{\partial
t^{q}}g_{M}^{2}(t,s)\Big{)}_{\mid s=b}\,,$

defined in Step 2 of the proof of Theorem 3.11.

We know that for all $M\in\mathbb{R}$, the function $y_{M,q}$ satisfies the
following boundary conditions:

$\begin{split}y_{M,q}^{(\mu_{1}^{q})}(a)=\cdots=y_{M,q}^{(\mu_{k}^{q})}(a)=&\,0\,,\\\
y_{M,q}^{(\rho_{1}^{q})}(b)=\cdots=y_{M,q}^{(\rho_{h-1}^{q})}(b)=y_{M,q}^{(\rho_{h+1}^{q})}(b)=\cdots=y_{M,q}^{(\rho_{n-k}^{q})}(b)=&\,0\,,\\\
y_{M,q}^{(\rho_{h}^{q})}(b)=&\,(-1)^{c_{q}+1}\,.\end{split}$

Hence, if $n-q-c_{q}$ and $c_{q}$ are even, then there exists $\rho>0$, such
that $y_{M,q}(t)>0$ for all $t\in(b-\rho,b)$. So, $v_{s}^{q}[M]$ cannot be
negative for any real $M$.

Moreover, if $n-q-c_{q}$ is even and $c_{q}$ odd, then there exists $\rho>0$,
such that $y_{M,q}(t)<0$ for all $t\in(b-\rho,b)$. So, $v_{s}^{q}[M]$ cannot
be positive for any real $M$.

Now, if $n-q-c_{q}$ is odd and $c_{q}$ even, then there exists $\rho>0$, such
that $y_{M,q}(t)<0$ for all $t\in(b-\rho,b)$. So, $v_{s}^{q}[M]$ cannot be
positive for any real $M$.

Finally, if $n-q-c_{q}$ and $c_{q}$ are odd, then there exists $\rho>0$, such
that $y_{M,q}(t)>0$ for all $t\in(b-\rho,b)$.

As consequence, $v_{s}^{q}[M]$ cannot be negative for any real $M$. ∎

Now, we give a necessary condition of the nonpositive (nonnegative) sign of
$v_{s}^{q}[M]$. This characterization will be an interval in which
$v_{s}^{q}[M]$ has a nonpositive (nonnegative) sign whose infimum or supremum
is given by the first eigenvalue $\lambda_{1}$ of $T_{n}[M]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

######  Theorem 3.15.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$, $c_{q}+d_{q}=n-q$, $c_{q}\geq 1$ and $d_{q}\geq
1$. If $\mu_{z}^{q}\neq z-1$ and $\rho_{h}^{q}\neq h-1$, then the following
properties are fulfilled:

  * •

Let $n-k$ be even:

    * –

If $n-q-c_{q}$ is even and $v_{s}^{q}[M]$ is nonpositive on $I\times I$, then
$M\in[\lambda_{*}^{q},\lambda_{1})$, and if $n-q-c_{q}$ is odd and
$v_{s}^{q}[M]$ is nonnegative on $I\times I$, then
$M\in[\lambda_{*}^{q},\lambda_{1})$, where

      * *

$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

      * *

$\lambda_{*}^{q}<0$ is the maximum between:

        * ·

$\lambda_{3}^{q}<0$ is the biggest negative eigenvalue of $T_{n}[0]$ in
$X_{3}$ defined in (3.8).

        * ·

$\lambda_{5}^{q}<0$, the biggest negative eigenvalue of $T_{n}[0]$ in $X_{5}$
defined in (3.10).

  * •

Let $n-k$ be odd:

    * –

If $n-q-c_{q}$ is even and $v_{s}^{q}[M]$ is nonpositive on $I\times I$, then
$M\in(\lambda_{1},\lambda_{*}^{q}]$, and if $n-q-c_{q}$ is odd and
$v_{s}^{q}[M]$ is nonnegative on $I\times I$, then
$M\in(\lambda_{1},\lambda_{*}^{q}]$ where

      * *

$\lambda_{1}>0$ is the least positive eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

      * *

$\lambda_{*}^{q}>0$ is the minimum between:

        * ·

$\lambda_{3}^{q}>0$, the biggest negative eigenvalue of $T_{n}[0]$ in $X_{3}$.

        * ·

$\lambda_{5}^{q}>0$, the biggest negative eigenvalue of $T_{n}[0]$ in $X_{5}$.

###### Proof.

Since $\mu_{z}^{q}\neq z-1$ and $\rho_{h}^{q}\neq h-1$ we have from
Propositions 3.5 and 3.6 that there are eigenvalues $\lambda_{3}^{q}$ and
$\lambda_{5}^{q}$ with corresponding related eigenfunctions of constant sign.
We suppose that $n-k$ and $n-q-c_{q}$ are even. For the other cases the proof
is done in an analogous way.

Let us assume that there exists $M^{*}\notin[\lambda_{*}^{q},\lambda_{1})$,
such that $v_{s}^{q}[M^{*}]$ is nonpositive on $I\times I$. From Theorem 3.11,
we can affirm that $M^{*}<\lambda_{1}$. So, $M^{*}<\lambda_{*}^{q}$ and for
all $M\in[M^{*},\lambda_{1})$ the function $v_{s}^{q}[M]$ is nonpositive on
$I\times I$ because the set where its takes nonpositive values is an interval.
By the monotone decreasing character of $v_{s}^{q}[M]$ we infer that

$v_{s}^{q}[\lambda_{*}^{q}](t)\leq v_{s}^{q}[M](t)\leq v_{s}^{q}[M^{*}](t)\leq
0.$

So, in particular

$w_{\lambda_{*}^{q},q}(t)\leq w_{M,q}(t)\leq w_{M^{*},q}(t)\leq 0\,,$

and

$\left\\{\begin{array}[]{cc}y_{\lambda_{*}^{q},q}(t)\leq y_{M,q}(t)\leq
y_{M^{*},q}(t)\leq 0\,,&\text{if }\gamma\text{ is even,}\\\ \\\ 0\leq
y_{M^{*},q}(t)\leq y_{M,q}(t)\leq y_{\lambda_{*}^{q},q}(t)\,,&\text{if
}\gamma\text{ is odd.}\end{array}\right.$

If $\lambda_{*}^{q}=\lambda_{3}^{q}$, then
$w_{\lambda_{*}^{q},q}^{(\alpha^{q})}(a)=0$. So, we conclude that, for all
$M\in[M^{*},\lambda_{*}^{q})$, $w_{M,q}^{(\alpha^{q})}(a)=0$, which
contradicts the discrete character of the spectrum of $T_{n}[0]$ in $X_{3}$.

If $\lambda_{*}^{q}=\lambda_{5}^{q}$, then
$y_{\lambda_{*}^{q},q}^{(\beta^{q})}(b)=0$. So, we conclude that, for all
$M\in[M^{*},\lambda_{*}^{q})$, $y_{M,q}^{(\beta^{q})}(b)=0$, which contradicts
the discrete character of the spectrum of $T_{n}[0]$ in $X_{5}$.

So, we arrive to a contradiction, thus the result is proved. ∎

Next, we present an example to illustrate the previous result.

######  Example 3.16.

Consider the operator $T_{5}[M]$ defined in the space of the boundary
conditions $X_{\\{0,2,3\\}}^{\\{1,3\\}}$.

For $q=1$ we have that $c_{q}=2$, $n-q-c_{q}=d_{q}=2$ is even, $z=h=2$,
$\alpha^{1}=0$ and $\beta^{1}=1$. Moreover, the function
$\frac{\partial}{\partial t}{g}_{M}$ satisfies the conditions of space
$X_{\\{1,2,4\\}}^{\\{0,2\\}}$ that fulfills the assumptions of Theorem 3.15.
Therefore, if $\frac{\partial}{\partial t}{g}_{M}$ is nonpositive on
$[0,1]\times[0,1]$, then $M\in[\lambda_{*}^{1},\lambda^{1})$, where

  * *

$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{5}[0]$ in
$X_{\\{0,2,3\\}}^{\\{1,3\\}}.$

  * *

$\lambda_{*}^{1}<0$ is the maximum between:

    * ·

$\lambda_{3}^{1}<0$, the biggest negative eigenvalue of $T_{5}[0]$ in
$X_{\\{0,1,4\\}}^{\\{0,2\\}}.$

    * ·

$\lambda_{5}^{1}<0$ is the biggest negative eigenvalue of $T_{5}[0]$ in
$X_{\\{1,2,4\\}}^{\\{0,1\\}}.$

By numerical approximation, we obtain that $\lambda_{1}\approx-2.23^{5}$,
$\lambda_{3}^{1}\approx-3.67^{5}$ and $\lambda_{5}^{1}\approx-2.88^{5}$. Thus,
if $\frac{\partial}{\partial}{g}_{M}$ is nonpositive on $[0,1]\times[0,1]$,
then $M\in[-2.88^{5},-2.23^{5})$.

###  3.2 Case (b): $c_{q}=n-q$ and $d_{q}=0$

In this subsection we deal with the constant sign of $v_{s}^{q}[M]$ for the
case $c_{q}=n-q$ and $d_{q}=0$. We arrive to the next theorem.

######  Theorem 3.17.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$, $c_{q}+d_{q}=n-q$, $c_{q}=n-q$ and $d_{q}=0$.
The following properties are fulfilled:

  * •

If $n-k$ is even, then $v_{s}^{q}[M]$ is nonnegative on $I\times I$ if, and
only if, $M\in(\lambda_{1},0]$, where $\lambda_{1}<0$ is the biggest negative
eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

  * •

If $n-k$ is odd, then $v_{s}^{q}[M]$ is nonnegative on $I\times I$ if, and
only if, $M\in[0,\lambda_{1})$, where $\lambda_{1}>0$ is the least positive
eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

###### Proof.

Let us assume that $n-k$ is even. For the case $n-k$ odd the proof is done in
an analogous way. We know from the Theorem 2.28 that $v_{s}^{q}\in
X_{\\{0,\dots,c_{q}\\}}$ satisfies an initial problem and it is nonnegative on
$[a,b]$. Moreover, $v_{s}^{q}(t)=\frac{\partial^{q}}{\partial
t^{q}}g_{0}(t,s)>0$ if $s<t$ and $v_{s}^{q}(t)=\frac{\partial^{q}}{\partial
t^{q}}g_{0}(t,s)=0$ if $s>t$.

Taking into account Lemma 2.30, we obtain that
$v_{s}^{q}(t)=\frac{\partial^{q}}{\partial t^{q}}g_{0}(t,s)>0$ on
$(a,b)\times(a,b)$ if and only if $M\in(\lambda_{1},M_{q}],$ with
$M_{q}>\lambda_{1}$. Moreover, $\frac{\partial^{q}}{\partial t^{q}}g_{M}$ is
monotone decreasing with respect to $M\in(\lambda_{1},M_{q}]$.

Let’s see that $M_{q}=0$. Using the monotone decreasing character of
$v_{s}^{q}[M]$ with respect to $M\in(\lambda_{1},M_{q}]$ and since
$v_{s}^{q}(t)=\frac{\partial^{q}}{\partial t^{q}}g_{0}(t,s)=0$ if $s>t$ we
have that $v_{s}^{q}[M_{q}](t)<v_{s}^{q}(t)=0$ if $M_{q}>0$. On the other
hand, if $s<t$ then there exists $(t^{*},s^{*})\in(a,b)\times(a,b)$ with
$s^{*}<t^{*}$ such that $0<v_{s^{*}}^{q}[M_{q}](t^{*})<v_{s^{*}}^{q}(t^{*})$
if $M_{q}>0$. Thus, $v_{s}^{q}[M]$ changes sign for $M_{q}>0$ and this
completes the proof. ∎

######  Example 3.18.

Consider the operator $T_{4}[M]$ coupled with the boundary conditions in the
space $X_{\\{2,3\\}}^{\\{0,1\\}}.$

In this case $n-k$ is even. For $q=2$ we have that $c_{q}=n-q=2$ and
$d_{q}=0$. Then, by Theorem 3.17 we have that $\frac{\partial^{2}}{\partial
t^{2}}{g}_{M}$ is nonnegative on $[0,1]\times[0,1]$ if, and only if,
$M\in(\lambda_{1},0]$, where $\lambda_{1}<0$ is the biggest negative
eigenvalue of $T_{4}[0]$ in $X_{\\{2,3\\}}^{\\{0,1\\}}.$

By numerical approximation, we obtain that $\lambda_{1}\approx-1.8751^{4}$.
Thus, $\frac{\partial^{2}}{\partial t^{2}}{g}_{M}$ is nonnegative on
$[0,1]\times[0,1]$ if, and only if, $M\in(-1.8751^{4},0]$.

For $q=3$ we have that $c_{q}=n-q=1$ and $d_{q}=0$. Again, by Theorem 3.17 we
have that $\frac{\partial^{3}}{\partial t^{3}}{g}_{M}$ is nonnegative on
$[0,1]\times[0,1]$ if, and only if, $M\in(-1.8751^{4},0]$.

###  3.3 Case (c): $c_{q}=0$ and $d_{q}=n-q$

In this section, arguing in a similar manner than in previous one, we can
characterize the constant sign of $v_{s}^{q}[M]$ for the case $c_{q}=0$ and
$d_{q}=n-q$.

Using Theorem 2.28, Lemma 2.30, and reasoning in an analogous way to the proof
of Theorem 3.17, we get to the next result.

######  Theorem 3.19.

Let $q\in\\{1,\dots,n-1\\}$. Suppose that
${\\{\sigma_{1},\dots,\sigma_{k}\\}}-{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}$
satisfies condition $(N_{a})$, $c_{q}+d_{q}=n-q$, $c_{q}=0$ and $d_{q}=n-q$.
Then, the following properties are fulfilled:

  * •

Let $n-k$ be even:

    * –

If $n-q$ is even, then $v_{s}^{q}[M]$ is nonnegative on $I\times I$ if, and
only if, $M\in(\lambda_{1},0]$, where $\lambda_{1}<0$ is the biggest negative
eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

    * –

If $n-q$ is odd, then $v_{s}^{q}[M]$ is nonpositive on $I\times I$ if, and
only if, $M\in(\lambda_{1},0]$, where $\lambda_{1}<0$ is the biggest negative
eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

  * •

Let $n-k$ be odd:

    * –

If $n-q$ is even, then $v_{s}^{q}[M]$ is nonnegative on $I\times I$ if, and
only if, $M\in[0,\lambda_{1})$, where $\lambda_{1}>0$ is the least positive
eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

    * –

If $n-q$ is odd, then $v_{s}^{q}[M]$ is nonpositive on $I\times I$ if, and
only if, $M\in[0,\lambda_{1})$, where $\lambda_{1}>0$ is the least positive
eigenvalue of $T_{n}[0]$ in
$X_{\\{\sigma_{1},\dots,\sigma_{k}\\}}^{\\{\varepsilon_{1},\dots,\varepsilon_{n-k}\\}}.$

######  Example 3.20.

Consider the operator $T_{6}[M]$ coupled with the boundary conditions in the
space $X_{\\{0,2\\}}^{\\{1,3,4,5\\}}.$

In this case $n-k=4$ is even. For $q=3$ we have that $c_{q}=0$ and
$d_{q}=n-q=3$ is odd. Then, by Theorem 3.19 we have that
$\frac{\partial^{3}}{\partial t^{3}}{g}_{M}$ is nonpositive on
$[0,1]\times[0,1]$ if, and only if, $M\in(\lambda_{1},0]$, where
$\lambda_{1}<0$ is the biggest negative eigenvalue of $T_{6}[0]$ in
$X_{\\{0,2\\}}^{\\{1,3,4,5\\}}.$

By numerical approximation, we obtain that $\lambda_{1}\approx-1.953^{6}$.
Thus, $\frac{\partial^{3}}{\partial t^{3}}{g}_{M}$ is nonpositive on
$[0,1]\times[0,1]$ if, and only if, $M\in(-1.953^{6},0]$.

For $q=4$ we have that $c_{q}=0$ and $d_{q}=n-q=2$ is even. Again, from
Theorem 3.19 we have that $\frac{\partial^{4}}{\partial t^{4}}{g}_{M}$ is
nonnegative on $[0,1]\times[0,1]$ if, and only if, $M\in(-1.953^{6},0]$.

Finally, for $q=5$ we have that $c_{q}=0$ and $d_{q}=n-q=1$ is odd. Then,
$\frac{\partial^{5}}{\partial t^{5}}{g}_{M}$ is nonpositive on
$[0,1]\times[0,1]$ if, and only if, $M\in(-1.953^{6},0]$.

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