\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 243, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/243\hfil Lyapunov-type inequalities] {Lyapunov-type inequalities for odd order linear differential equations} \author[S. Dhar, Q. Kong \hfil EJDE-2016/243\hfilneg] {Sougata Dhar, Qingkai Kong} \address{Sougata Dhar \newline Department of Mathematics, Northern Illinois University, DeKalb, IL 60115, USA} \email{sdhar@niu.edu} \address{Qingkai Kong \newline Department of Mathematics, Northern Illinois University, DeKalb, IL 60115, USA} \email{qkong@niu.edu} \thanks{Submitted July 7, 2015. Published September 7, 2016.} \subjclass[2010]{34A40, 34B05, 34C11, 26D20} \keywords{Lyapunov-type inequalities; odd order; linear differential equations; \hfill\break\indent Green's function; boundary value problem} \begin{abstract} In this article, we obtain Lyapunov-type inequalities for certain odd order linear boundary-value problems. Our inequalities involve integrals of both $q_+(t)$ and $q_-(t)$ in addition to that of $|q(t)|$. The Green's function for even order boundary-value problems plays a key role in our proofs. Also, using the Fredholm alternative theorem, we obtain a criterion for the existence and uniqueness of solutions to the corresponding nonhomogeneous linear boundary-value problems. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{introduction} For the second-order linear differential equation \begin{equation} x''+q(t)x=0\label{1.1} \end{equation} with $q\in C([a,b],\mathbb{R})$, the following result is known as the Lyapunov inequality, see \cite{B,AML}. \begin{theorem} \label{t1.1} Assume \eqref{1.1} has a solution $x(t)$ satisfying $x(a)=x(b)=0$ and $x(t)\neq0$ for $t\in(a,b)$. Then \begin{equation} \int_a^b |q(t)|dt>\frac{4}{b-a}.\label{1.2} \end{equation} \end{theorem} It was first noticed by Wintner \cite{W} and later by several other authors that inequality \eqref{1.2} can be improved by replacing $|q(t)|$ by $q_{+}(t):=\max\{0,q(t)\}$, the nonnegative part of $q(t)$, to become \begin{equation} \int_a^b q_{+}(t)dt>\frac{4}{b-a}.\label{1.3} \end{equation} An extension of \eqref{1.3}, due to Hartman \cite[Chapter XI]{H}, to the more general equation \begin{equation} (r(t)x')'+q(t)x=0\label{1.4} \end{equation} with $q,r\in C([a,b],\mathbb{R})$ and $r(t)>0$ for $t\in[a,b]$, is as follows. \begin{theorem}\label{t1.2} Assume \eqref{1.4} has a solution $x(t)$ satisfying $x(a)=x(b)=0$ and $x(t)\neq0$ for $t\in(a,b)$. Then \begin{equation} \int_a^b q_{+}(t)dt>\frac{4}{\int_a^b r^{-1}(t)dt}.\label{1.5} \end{equation} \end{theorem} The above Lyapunov inequalities have been improved by replacing $\int_a^b q_{+}(t)dt$ by some integrals of $q(t)$ on parts of or the whole interval $[a,b]$, see Harris and Kong \cite{HK}, Brown and Hinton \cite{BH} for the details. Lyapunov-type inequalities have been further developed for higher order linear and half-linear differential equations by many authors. The reader is referred to Cakmak \cite{C1,C2}, He and Tang \cite{HT}, Pachpatte \cite{P1,P2}, Parhi and Panigrahi \cite{PP1,PP2}, Panigrahi \cite{PN}, Tiryaki, Unal and Cakmak \cite{TUC}, Yang \cite{Y1}, Yang and Lo \cite{YL}, and Zhang and He \cite{ZH} for the higher order linear case. Also, Pinasco \cite{JPP} provided an excellent survey on various Lyapunov-type inequalities. Among the above, Parhi and Panigrahi \cite{PP1} established the Lyapunov-type inequalities for the third-order linear differential equation \begin{equation} x'''+q(t)x=0\label{1.6} \end{equation} with $-\infty\frac{4}{(c-a)^{2}}.\label{1.7} \end{equation} \end{theorem} Recently, Dhar and Kong \cite{DK} obtained Lyapunov-type inequalities for third-order half-linear differential equations. Restricted to the linear equation \eqref{1.6}, inequality \eqref{1.7} becomes \begin{equation} \max_{\xi\in[a,c]}\Big\{ \int_a^{\xi}q_{-}(s)ds +\int_{\xi}^{c}q_{+}(s)ds\Big\} >\frac{4}{(c-a)^{2}}.\label{1.8} \end{equation} Clearly, \eqref{1.8} improves \eqref{1.7} by replacing $|q(t)|$ in the integral of the left-hand side by $q_{-}(t)$ and $q_{+}(t)$, the negative and positive pars of $q$, respectively. In a different direction, the constant $4$ on the right-hand side of \eqref{1.7} has been improved based on the Green's function for a corresponding second-order Dirichilet problem. In particular, motivated by the approach in Aktas, Cakmak, and Tiryaki \cite{ACT}, Dhar and Kong \cite{DK1} obtained the following result. \begin{theorem} \label{t1.4} Assume \eqref{1.6} has a solution $x(t)$ satisfying $x(a)=x(b)=x(c)=0$ and $x(t)\neq0$ for $t\in(a,b)\cup(b,c)$. Then one of the following holds: \begin{itemize} \item[(a)] $\int_a^{c}q_{-}(t)dt>\frac{8}{(c-a)^{2}},$ \item[(b)] $\int_a^{c}q_{+}(t)dt>\frac{8}{(c-a)^{2}},$ \item[(c)] $\int_a^b q_{-}(t)dt+\int_{b}^{c}q_{+}(t)dt>\frac{8}{(c-a)^{2}}.$ \end{itemize} As a result, \[ \int_a^{c}|q(t)| dt>\frac{8}{(c-a)^{2}}. \] \end{theorem} In this article, we use some ideas from \cite{ACT} and \cite{DK1} for third-order equations to derive Lyapunov-type inequalities for odd order equations. More specifically, we use the Green's function for even order linear boundary value problems (BVPs) to obtain Lyapunov-type inequalities for certain types of BVPs associated with odd order linear equations. Furthermore, by using the Fredholm alternative theorem, we obtain a criterion for the existence and uniqueness of solutions to nonhomogeneous linear boundary value problems of odd order. \section{Main results} We let $-\infty0$. \begin{theorem} \label{t2.1} Assume \eqref{2.1} has a nontrivial solution $x(t)$ satisfying \begin{equation} x^{(i+1)}(a)=x^{(i+1)}(b)=0,\quad i=0,1,\dots ,n-1\label{2.3} \end{equation} and $x(c)=0$ for $c\in[a,b]$. Then \begin{equation} \int_a^b |q(t)|dt>\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}.\label{2.4} \end{equation} \end{theorem} \begin{proof} As shown in \cite{DV}, the Green's function for the BVP \begin{equation} \begin{gathered} y^{(2n)}+(-1)^{n-1}h(t)=0,\\ y^{(i)}(a)=y^{(i)}(b)=0,\quad i=0,1,\dots ,n-1 \end{gathered} \label{2.5} \end{equation} is \begin{equation} G(t,s) = \begin{cases} \frac{1}{(2n-1)!}\Big(\frac{(t-a)(b-s)}{b-a}\Big)^n \sum_{j=0}^{n-1}\binom{n+j-1}{j} (s-t)^{n-j-1} \big(\frac{(b-t)(s-a)}{b-a}\big)^j,\\ \quad a\le t\le s\le b;\\[4pt] \frac{1}{(2n-1)!}\Big(\frac{(s-a)(b-t)}{b-a}\Big)^n \sum_{j=0}^{n-1}\binom{n+j-1}{j} (t-s)^{n-j-1}\Big(\frac{(t-a)(b-s)}{b-a}\Big)^j, \\ \quad a\le s\le t\le b. \end{cases} \label{2.6} \end{equation} Hence the solution $y(t)$ of BVP \eqref{2.5} satisfies \begin{equation} y(t)=\int_a^b G(t,s)h(s)ds.\label{2.7} \end{equation} We note that for the solution $x(t)$ of \eqref{2.1}, $y(t):=x'(t)$ satisfies \eqref{2.5} with $h(t)=q(t)x(t)$. By \eqref{2.7} \begin{equation} x'(t)=\int_a^b G(t,s)q(s)x(s)ds.\label{2.8} \end{equation} Integrating \eqref{2.8} from $c$ to $t$ and noting that $x(c)=0$, we have \begin{equation} x(t)=\int_{c}^{t}\int_a^b G(\tau,s)q(s)x(s)dsd\tau =\int_a^b \Big(\int_{c}^{t}G(\tau,s)d\tau\Big)q(s)x(s)ds.\label{2.9} \end{equation} It is easy to see that $G(t,s)\ge 0$ on $[a,b]\times [a,b]$. It follows that \begin{equation} |x(t)|=\big|\int_a^b \Big(\int_{c}^{t}G(\tau,s)d\tau\Big)q(s)x(s)ds\big| \le\int_a^b \Big(\int_a^b G(\tau,s)d\tau\Big)|q(s)||x(s)|ds.\label{2.9*} \end{equation} We first show that for $s\in[a,b]$ \begin{equation} \int_a^b G(\tau,s)d\tau\le\frac{(b-a)^{2n}S_{n}}{2^{2n}(2n-1)!}\,, \label{2.9**} \end{equation} where $S_{n}$ is defined in \eqref{2.2}. In fact, \begin{equation} \int_a^b G(\tau,s)d\tau =\int_a^{s}G(\tau,s)d\tau+\int_{s}^b G(\tau,s)d\tau.\label{2.10} \end{equation} We consider each integral separately. To ease the notation, we denote \begin{equation} A_{j}(s)=\frac{1}{(2n-1)!}\binom{n+j-1}{j} \frac{(b-s)^{n}(s-a)^j}{(b-a)^{n+j}}.\label{2.11} \end{equation} Then from \eqref{2.6}, \begin{equation} \begin{aligned} \int_a^{s}G(\tau,s)d\tau &= \int_a^{s}(\tau-a)^{n}\sum_{j=0}^{n-1}A_{j}(s) (s-\tau)^{n-j-1}(b-\tau)^jd\tau \\ &= \sum_{j=0}^{n-1}A_{j}(s)\int_a^{s}(\tau-a)^{n} (s-\tau)^{n-j-1}(b-\tau)^jd\tau. \end{aligned}\label{2.12} \end{equation} We write $$ (b-\tau)^j=(b-s+s-\tau)^j=\sum_{k=0}^j\binom{j}{k} (b-s)^{j-k}(s-\tau)^{k}. $$ Substituting it into \eqref{2.12}, we obtain \begin{equation} \begin{aligned} \int_a^{s}G(\tau,s)d\tau &=\sum_{j=0}^{n-1}A_{j}(s)\int_a^{s}(\tau-a)^{n}(s-\tau)^{n-j-1} \sum_{k=0}^j\binom{j}{k} (b-s)^{j-k}(s-\tau)^{k}d\tau \\ &=\sum_{j=0}^{n-1}A_{j}(s)\sum_{k=0}^j\binom{j}{k} (b-s)^{j-k}\int_a^{s}(\tau-a)^{n}(s-\tau)^{n-j+k-1}d\tau. \end{aligned}\label{2.13} \end{equation} To evaluate the integral in \eqref{2.13}, we use the transformation $u=(\tau-a)/(s-a)$ which implies $1-u=(s-\tau)/(s-a)$. Hence \begin{align*} \int_a^{s}(\tau-a)^{n}(s-\tau)^{n-j+k-1}d\tau &= (s-a)^{2n-j+k}\int_{0}^{1}u^{n}(1-u)^{n-j+k-1}du \\ &= (s-a)^{2n-j+k}B(n+1,n-j+k). \end{align*} Then by \eqref{2.13}, \begin{equation} \int_a^{s}G(\tau,s)d\tau=\sum_{j=0}^{n-1}A_{j}(s)\sum_{k=0}^j \binom{j}{k} (b-s)^{j-k}(s-a)^{2n-j+k}B(n+1,n-j+k).\label{2.15} \end{equation} Using the expression for $A_{j}(s)$ in \eqref{2.15} and rearranging terms we obtain \begin{equation} \begin{aligned} \int_a^{s}G(\tau,s)d\tau &= \frac{1}{(2n-1)!(b-a)^{n}}\sum_{j=0}^{n-1}\sum_{k=0}^j \binom{n+j-1}{j}\binom{j}{k} B(n+1,n-j+k) \\ &\quad \times \frac{(b-s)^{n+j-k}(s-a)^{2n+k}}{(b-a)^j}. \end{aligned}\label{2.16} \end{equation} Using the same technique, we also have \begin{equation} \begin{aligned} \int_{s}^b G(\tau,s)d\tau &= \frac{1}{(2n-1)!(b-a)^{n}}\sum_{j=0}^{n-1}\sum_{k=0}^j \binom{n+j-1}{j} \binom{j}{k} B(n+1,n-j+k) \\ &\quad \times \frac{(b-s)^{2n+k}(s-a)^{n+j-k}}{(b-a)^j}. \end{aligned} \label{2.17} \end{equation} Substituting \eqref{2.16} and \eqref{2.17} into \eqref{2.10} we obtain \begin{equation} \begin{aligned} \int_a^b G(\tau,s)d\tau &=\frac{1}{(2n-1)!(b-a)^{n}}\sum_{j=0}^{n-1}\sum_{k=0}^j \binom{n+j-1}{j} \binom{j}{k} B(n+1,n-j+k) \\ & \quad \times \Big\{ \frac{(b-s)^{n+j-k}(s-a)^{2n+k}}{(b-a)^j} +\frac{(b-s)^{2n+k}(s-a)^{n+j-k}}{(b-a)^j}\Big\}. \end{aligned} \label{2.18} \end{equation} Note that $\alpha\beta\le(\alpha+\beta)^{2}/4$ and ${ \alpha^{l}+\beta^{l}\le(\alpha+\beta)^{l}}$ for $\alpha,\beta>0$ and $l\in\mathbb{N}$. Letting $\alpha=b-s$, $\beta=s-a$ and $l=n-j+2k$ we have \begin{align*} & \frac{(b-s)^{n+j-k}(s-a)^{2n+k}}{(b-a)^j} +\frac{(b-s)^{2n+k}(s-a)^{n+j-k}}{(b-a)^j}\\ &= \frac{(b-s)^{n+j-k}(s-a)^{n+j-k}}{(b-a)^j}\Big[(s-a)^{n-j+2k}+(b-s)^{n-j+2k}\Big]\\ &\le \frac{(b-a)^{2n+2j-2k}}{2^{2n+2j-2k}(b-a)^j}(b-a)^{n-j+2k}\\ &=\frac{(b-a)^{3n}}{2^{2n+2j-2k}}. \end{align*} Then \eqref{2.9**} follows from \eqref{2.18}. We then show that \eqref{2.4} holds. Define $m:=\max\{|x(t)|:t\in[a,b]\}$. Then taking maximum of $|x(t)|$ in \eqref{2.9*} and using the fact that $x(t)\not\equiv m$ on $[a,b]$, we have \[ m\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}$, \item[(ii)] $\int_a^b q_{+}(t)dt>\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}$, \item[(iii)] $\int_a^{c}q_{-}(t)dt+\int_{c}^b q_{+}(t)dt >\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}$. \end{itemize} (b) Suppose $x(a)=0$ and $x(t)\neq0$ for $t\in(a,b]$. Then \[ \int_a^b q_{+}(t)dt>\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}. \] (c) Suppose $x(b)=0$ and $x(t)\neq0$ for $t\in[a,b)$. Then \[ \int_a^b q_{-}(t)dt>\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}. \] \end{theorem} \begin{proof} As in the proof of Theorem \ref{t2.1}, we see that \eqref{2.9} and \eqref{2.9**} hold. (a) Since $x(t)$ is continuous and $x(c)=0$ for $c\in(a,b)$, there exist $t_1\in(a,c)$ and $t_2\in(c,b)$ such that $|x(t_1)|=\max\{|x(t)|:t\in[a,c]\}$ and $|x(t_2)|=\max\{|x(t)|:t\in[c,b]\}$. Without loss of generality, we may assume $x(t)$ satisfies one of the following cases: \begin{itemize} \item{(I)} $x(t)>0$ on $(a,c)\cup(c,b)$ and $x(t_1)\geq x(t_2)$; \item{(II)} $x(t)>0$ on $(a,c)\cup(c,b)$ and $x(t_1)0$ on $(a,c)$ and $x(t)<0$ on $(c,b)$, and $x(t_1)\geq-x(t_2)$; \item{(IV)} $x(t)>0$ on $(a,c)$ and $x(t)<0$ on $(c,b)$, and $x(t_1)<-x(t_2)$. \end{itemize} In the sequel, we denote $m=\max\{|x(t_1)|,|x(t_2)|\}$. \smallskip \noindent\textbf{Case I: $m=x(t_1)$.} Then \eqref{2.9} with $t=t_1$ shows that \[ m=\int_a^b \Big(\int_{t_1}^{c}G(\tau,s)d\tau\Big)(-q(s))x(s)ds\,. \] Using that $0\leq x(t)\leq m$ and $x(t)\not\equiv m$, and $-q(t)\leq q_{-}(t)$, we have \[ m\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}\,; \] i.e., conclusion (i) in Part (a) holds. \smallskip \noindent\textbf{Case II: $m=x(t_2)$.} Then \eqref{2.9} with $t=t_2$ shows that \[ m=\int_a^b \Big(\int_{c}^{t_2}G(\tau,s)d\tau\Big)q(s)x(s)ds. \] Again using the facts that $0\leq x(t)\leq m$ and $x(t)\not\equiv m$, and $q(t)\leq q_{+}(t)$, we have \[ m\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}. \] i.e., conclusion (ii) in Part (a) holds. \smallskip \noindent\textbf{Case III: $m=x(t_1)$.} Then \eqref{2.9} with $t=t_1$ shows that \begin{align*} m & =\int_a^b \Big(\int_{t_1}^{c}G(\tau,s)d\tau\Big)(-q(s))x(s)ds\\ & =\int_a^{c}\Big(\int_{t_1}^{c}G(\tau,s)d\tau\Big)(-q(s))x(s)ds +\int_{c}^b \Big(\int_{t_1}^{c}G(\tau,s)d\tau\Big)q(s)(-x(s))ds. \end{align*} Note that $x(t)>0$ on $[a,c)$ and $x(t)<0$ on $(c,b]$. Then by a similar argument to Cases I and II, we see that \[ \int_a^{c}q_{-}(s)ds+\int_{c}^b q_{+}(s)ds>\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}\,; \] i.e., conclusion (iii) in Part (a) holds. \smallskip \noindent\textbf{Case IV.} The same argument as in Case III shows that conclusion (iii) in Part (a) holds. We omit the details. (b) Note that $x(a)=0$. Then it follows from \eqref{2.9} that \begin{equation} x(t)=\int_a^b \Big(\int_a^{t}G(\tau,s)d\tau\Big)q(s)x(s)ds.\label{2.21} \end{equation} Without loss of generality, we may assume $x(t)>0$ in $(a,b]$. Then there exists $t_2\in(a,b]$ such that $m=x(t_2)=\max\{x(t):t\in[a,b]\}$. Using $t=t_2$ in \eqref{2.21} we obtain $$ m=\int_a^b \Big(\int_a^{t_2}G(\tau,s)d\tau\Big)q(s)x(s)ds \le \int_a^b \Big(\int_a^b G(\tau,s)d\tau\Big)q_+(s)x(s)ds. $$ Using \eqref{2.9**} and a similar technique as before, we see that conclusion in Part (b) holds. (c) In this case, a similar argument as Part (b) holds with $m=x(t_1)=\max\{|x(t)|:t\in[a,b]\}|$. We omit the details. \end{proof} Now we interpret the results in Theorems \ref{t2.1} and \ref{t2.2} to the special case with $n=1$, i.e., the third-order linear differential equation \begin{equation} x'''+q(t)x=0. \label{a1} \end{equation} From \eqref{2.2}, \[ S_1=\sum_{j=0}^{0}\sum_{k=0}^j2^{2k-2j}\binom{j}{j} \binom{j}{k} B(2,1+k-j)=B(2,1)=\frac{1}{2}. \] \begin{corollary}\label{c2.1} Assume \eqref{a1} has a nontrivial solution $x(t)$ satisfying \[ x'(a)=x'(b)=0 \] and $x(c)=0$ for $c\in[a,b]$. Then \[ \int_a^b |q(t)|dt>\frac{8}{(b-a)^{2}}. \] \end{corollary} \begin{corollary} \label{c2.2} Assume \eqref{a1} has a solution $x(t)$ satisfying \[ x'(a)=x'(b)=0. \] (a) Suppose $x(c)=0$ for $c\in(a,b)$ and $x(t)\neq0$ for $t\in[a,c)\cup(c,b]$. Then one of the following holds: \begin{itemize} \item[(i)] $\int_a^b q_{-}(t)dt>\frac{8}{(b-a)^{2}}$, \item[(ii)] $\int_a^b q_{+}(t)dt>\frac{8}{(b-a)^{2}}$, \item[(iii)] $\int_a^{c}q_{-}(t)dt+\int_{c}^b q_{+}(t)dt>\frac{8}{(b-a)^{2}}$. \end{itemize} (b) Suppose $x(a)=0$ and $x(t)\neq0$ for $t\in(a,b]$. Then \[ \int_a^b q_{+}(t)dt>\frac{8}{(b-a)^{2}}. \] (c) Suppose $x(b)=0$ and $x(t)\neq0$ for $t\in[a,b)$. Then \[ \int_a^b q_{-}(t)dt>\frac{8}{(b-a)^{2}}. \] \end{corollary} We observe that the inequalities in Corollaries \ref{c2.1} and \ref{c2.2} supplement those in \cite[Corollary 2.1]{DK1} for different boundary conditions. \section{Applications to boundary-value problems} In the final section, we apply the results on the Lyapunov-type Inequalities obtained in Section 2 to study the nonexistence, uniqueness, and existence-uniqueness for solutions of certain BVPs. Consider the BVP consisting of \eqref{2.1} and the boundary conditions \begin{equation} \begin{gathered} x^{(i+1)}(a)=x^{(i+1)}(b)=0, \quad i=0,1,\dots ,n-1;\\ x(c)=0, \quad c\in[a,b]. \end{gathered} \label{3.1} \end{equation} In the following, we let $S_{n}$ be defined by \eqref{2.2}. The first result is on the nonexistence of solutions of the boundary-value problem \eqref{2.1} \eqref{3.1}. \begin{theorem} \label{t3.1} Assume \begin{equation} \int_a^b |q(t)|dt\le\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}.\label{3.1*} \end{equation} Then BVP \eqref{2.1} \eqref{3.1} has no nontrivial solution for any $c\in[a,b]$. \end{theorem} \begin{proof} Assume the contrary, i.e., \eqref{2.1} \eqref{3.1} has a nontrivial solution $x(t)$. Then by Theorem \ref{t2.1}, inequality \eqref{2.4} holds. This contradicts assumption \eqref{3.1*}. \end{proof} As a direct application of Theorem \ref{t2.2}, we present the following result. \begin{theorem} \label{t3.2} Assume \begin{equation} \max_{\xi\in[a,b]}\Big\{ \int_a^{\xi}q_{-}(t)dt+\int_{\xi}^b q_{+}(t)dt\Big \} \le\frac{2^{2n}(2n-1)!}{(b-a)^{2n}S_{n}}.\label{3.1**} \end{equation} Then every nontrivial solution of BVP \eqref{2.1}, \eqref{3.1} has at least two zeros in $[a,b]$. \end{theorem} \begin{proof} Assume the contrary, i.e., \eqref{2.1}, \eqref{3.1} has a nontrivial solution $x(t)$ with only one zero $c_1$ in $[a,b]$. Then $c_1=c$ and $x(t)\neq0$ for $t\in[a,c)\cup(c,b]$. It follows that one of the conclusions in Part (a) of Theorem \ref{t2.2} holds. This contradicts \eqref{3.1**}. \end{proof} Next we consider the odd order nonhomogeneous linear BVPs consisting of the equation \begin{equation} x^{(2n+1)}+(-1)^{n-1}q(t)x=f(t)\quad \text{on }(A,B)\label{3.2} \end{equation} with $-\infty