Hartman and Lyapunov Inequalities

1. Introduction

Consider the second-order linear differential equation

$$x^{\prime \prime }\left(t\right)+q\left(t\right)x\left(t\right)=0; a\le t\le b,$$

(1)

where $q\left(t\right)\in L^1\left[a,b\right]$ is a real-valued function. If $x\left(t\right)\not\equiv 0$ is a solution of Eq. (1) having two consecutive zeros $a$ and $b$, where $a,b\in \mathbb{R}$ with $a<b$ and $x\left(t\right)\ne 0$ for $t\in \left(a,b\right)$, then the inequality

$$\underset{a}{\overset{b}{\int }}\left|q\left(t\right)\right|dt>{\displaystyle \frac{4}{b-a}}$$

(2)

holds. This striking inequality was first proved by Lyapunov [1] and in the literature known as “Lyapunov inequality”. Later Wintner [2] and thereafter some authors achieved to replace the function $\left|q\left(t\right)\right|$ in Ineq. (2) by the function $q^{+}\left(t\right)$ i.e., they obtained the following inequality:

$$\underset{a}{\overset{b}{\int }}{q}^{+}\left(t\right)dt>{\displaystyle \frac{4}{b-a}},$$

(3)

where $q^{+}\left(t\right)=\mathrm{m}\mathrm{a}\mathrm{x}\{q\left(t\right),0\}$ , and the constant 4 in the right hand side of Ineq. (3) (and Ineq. (2)) is the best possible largest number (see [1] and [3, Thm. 5.1]). In [3], Hartman has obtained the more general inequality than both (2) and (3):

$$\underset{a}{\overset{b}{\int }}\left(b-t\right)\left(t-a\right)q^{+}\left(t\right)dt>b-a.$$

(4)

Since

$$\left(b-t\right)\left(t-a\right)\le {\displaystyle \frac{{\left(b-a\right)}^2}{4}} \mathrm{for} \mathrm{all} t\in \left[a,b\right],$$

(5)

In eq. (4) implies Ineq. (3), see Theorem 3.2.

2. Preliminaries

he Green’s function method plays an important role for boundary value problems. Now,

we observe that the solution of the second-order boundary value problem

$${ x}^{\prime \prime }\left(t\right)=-f\left(t\right); a<t<b$$

(6)

satisfying the Dirichlet boundary conditions

$$x\left(a\right)=x\left(b\right)=0.$$

(7)

To express the solution of Prb. (6)–(7), we need the following lemma.

Lemma 2.1.

Solution of Eq. (6) satisfying the Dirichlet boundary conditions (7) can be expressed as the integral equation

$$x\left(t\right)=\underset{a}{\overset{b}{\int }}G\left(t,s\right)f\left(s\right)ds,$$

(8)

where the function $G(t, s)$ is called the Green’s function for the Prb. (6)-(7) and is defined as

$$G\left(t, s\right)={\displaystyle \frac{1}{b-a}}\left\{\begin{array}{c}\left(b-t\right)\left(s-a\right) if a\le s\le t;\\ \left(t-a\right)\left(b-s\right) if t\le s\le b.\end{array} \right.$$

(9)

Proof. 

Integrating the both sides of Eq. (6), we get

$${ x}^{\prime }\left(t\right)=x^{\prime }\left(a\right)-\underset{a}{\overset{t}{\int }}f\left(s\right)ds; a\le t\le b.$$

(10)

Again integrating the both sides of Eq. (10), we get

$$\underset{a}{\overset{t}{\int }}{x}^{\prime }\left(s\right)ds=\left(t-a\right)x^{\prime }\left(a\right)-\underset{a}{\overset{t}{\int }}\underset{a}{\overset{\tau }{\int }}f\left(s\right)dsd\tau ; a\le t\le b. \left(11\right)$$

(11)

Using Eq. (7) and Eq. (11), we obtain

$$x\left(t\right)=\left(t-a\right)x^{\prime }\left(a\right)-\underset{a}{\overset{t}{\int }}\underset{a}{\overset{\tau }{\int }}f\left(s\right)dsd\tau ; t\le b.$$

(12)

he double integral on the right hand side of Eq. (12) can be represented as

$$\underset{a}{\overset{t}{\int }}\underset{a}{\overset{\tau }{\int }}f\left(s\right)dsd\tau =\int \underset{\mathsf{\Omega }}{\int }f\left(s\right)dsd\tau \left(13\right)$$

(13)

where $\Omega$ is the region defined as

$$\Omega ≔\left\{\left(t,s\right):a\le s\le t, a\le \tau \le t\right\}.$$

By changing the order of integration in the double integral (13), we get

$$\int \underset{\mathsf{\Omega }}{\int }f\left(s\right)dsd\tau =\underset{a}{\overset{t}{\int }}f\left(s\right)\underset{s}{\overset{t}{\int }}d\tau ds=\underset{a}{\overset{t}{\int }}\left(t-s\right)f\left(s\right)ds.$$

(14)

By using (14), Eq. (12) turns out to be

$$x\left(t\right)=\left(t-a\right)x^{\prime }\left(a\right)-\underset{a}{\overset{t}{\int }}\left(t-s\right)f\left(s\right)ds.$$

(15)

On the other hand, if we take $t = b$ in (15), then we obtain;

$$x\left(b\right)=\left(b-a\right)x^{\prime }\left(a\right)-\underset{a}{\overset{b}{\int }}\left(b-s\right)f\left(s\right)ds. \left(16\right)$$

(16)

which implies that

$${ x}^{\prime }\left(a\right)={\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}\left(b-s\right)f\left(s\right)ds.$$

(17)

Using (15) and (17), we obtain

$$\begin{gathered} x\left(t\right)={\displaystyle \frac{t-a}{b-a}}\underset{a}{\overset{b}{\int }}\left(b-s\right)f\left(s\right)ds-\underset{a}{\overset{t}{\int }}\begin{array}{c}\left(t-s\right)f\left(s\right)ds \\ \\ \end{array} \\ ={\displaystyle \frac{t-a}{b-a}}\underset{a}{\overset{t}{\int }}\left(b-s\right)f\left(s\right)ds+{\displaystyle \frac{t-a}{b-a}}\underset{t}{\overset{b}{\int }}\left(b-s\right)f\left(s\right)ds-\underset{a}{\overset{t}{\int }}\left(t-s\right)f\left(s\right)ds \\ =\underset{a}{\overset{t}{\int }}{\displaystyle \frac{\left(t-a\right)\left(b-s\right)}{\left(b-a\right)}}f\left(s\right)ds-\underset{a}{\overset{t}{\int }}{\displaystyle \frac{\left(t-s\right)\left(b-a\right)}{b-a}}f\left(s\right)ds+\underset{t}{\overset{b}{\int }}{\displaystyle \frac{\left(t-a\right)\left(b-s\right)}{b-a}}f\left(s\right)ds \end{gathered}$$

(18)

Since

$${\displaystyle \frac{\left(t-a\right)\left(b-s\right)}{\left(b-a\right)}}-{\displaystyle \frac{\left(t-s\right)\left(b-a\right)}{b-a}}={\displaystyle \frac{\left(b-t\right)\left(s-a\right)}{b-a}},$$

(18) turns to

$$x\left(t\right)=\underset{a}{\overset{t}{\int }}{\displaystyle \frac{\left(b-t\right)\left(s-a\right)}{b-a}}f\left(s\right)ds+\underset{t}{\overset{b}{\int }}{\displaystyle \frac{\left(t-a\right)\left(b-s\right)}{b-a}}f\left(s\right)ds=\underset{a}{\overset{b}{\int }}G\left(t,s\right)f\left(s\right)ds,$$

(19)

where the function $G(t, s)$ is defined as in (9). 󠇈󠇈

3. Main Results

3.1. Hartman’s Inequality

In what follows we shall assume that $p\left(t\right),q\left(t\right)\in L^1\left[a,b\right].$ To motivate the formulation of our main results, we state the Hartman and Lyapunov inequalities for the linear Eq. (1).

Theorem 3.1. (Hartman’s Inequality):

If $x\left(t\right)$ is a non-trivial solution of Eq. (1) satisfying the Dirichlet boundary conditions (7), where $a, b \in \mathbb{R}$ with $a < b$ are consecutive zeros, then Ineq. (4) holds.

Proof. 

It is clear from (9) that

$$0\le G\left(t,s\right)\le {\displaystyle \frac{\left(b-s\right)\left(s-a\right)}{b-a}}; s\in \left(a,b\right).$$

(20)

Now let $x\left(t\right)$ be a non-trivial solution of Eq. (1) satisfying Dirichlet boundary conditions (7), where $a, b \in \mathbb{R}$ with $a < b$ are consecutive zeros. Without loss of generality, we may assume that $x\left(t\right) > 0$ for $t \in (a, b).$ In fact, if $x\left(t\right) < 0$ for $t \in (a, b),$ then we can consider $x\left(t\right)$, which is also a solution. Then by using (1) and (8), $x\left(t\right)$can be expressed as,

$$x\left(t\right)=\underset{a}{\overset{b}{\int }}G\left(t,s\right)q\left(s\right)x\left(s\right)ds.$$

(21)

Let $x\left(c\right)={max}_{t\in \left(a,b\right)}x\left(t\right).$ hen by (20) and (21),

$$x\left(c\right)=\underset{a}{\overset{b}{\int }}G\left(c,s\right)q\left(s\right)x\left(s\right)ds\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}\left(b-s\right)\left(s-a\right)q\left(s\right)x\left(s\right)ds.$$

(22)

Since $q^{+}\left(t\right)\ge q\left(t\right),$(22) turns out to be

$$\left(b-a\right)x\left(c\right)\le x\left(c\right)\underset{a}{\overset{b}{\int }}\left(b-s\right)\left(s-a\right)q^{+}\left(s\right)ds.$$

Dividing both sides of the inequality above by $x\left(c\right) > 0,$ we obtain the desired inequality (4). ☐

3.2. Lyapunov’s Inequality

Theorem 3.2. (Lyapunov’s Inequality)

If $x\left(t\right)$ is a non-trivial solution of Eq. (1) satisfying the Dirichlet boundary conditions (7), where $a, b \in \mathbb{R}$ with  $a<b$ are consecutive zeros, then Ineq. (3) holds.

Proof. 

Using In eq. (5), Ineq. (4) immediately turns out to be

$$b-a<\underset{a}{\overset{b}{\int }}\left(b-t\right)\left(t-a\right)q^{+}\left(t\right)dt<{\displaystyle \frac{{\left(b-a\right)}^2}{4}}\underset{a}{\overset{b}{\int }}{q}^{+}\left(t\right)dt.$$

(23)

Multiplying both sides of the inequality above by $4/{\left(b-a\right)}^2$, we obtain the desired inequality

$$\underset{a}{\overset{b}{\int }}{q}^{+}\left(t\right)dt>{\displaystyle \frac{4}{b-a}}.$$

☐

3.3. An Example

Example 3.3.:Consider the linear differential equation

$${ x}^{\prime \prime }\left(t\right)+{\lambda }^{2}x\left(t\right)=0, \lambda \in {\mathbb{R}}^{+}.$$

(24)

Let $x\left(t\right)$ be a non-trivial solution of Eq. (24) satisfying the Dirichlet boundary conditions

$$x\left(0\right)=x\left(\pi /\lambda \right)=0.$$

(25)

Then we have $a=0, b=\pi /\lambda$ and $q\left(t\right)={\lambda }^2.$ Now we look for the integral,

$$\begin{gathered} \underset{a}{\overset{b}{\int }}\left(b-t\right)\left(t-a\right)q^{+}\left(t\right)dt=\underset{0}{\overset{\pi /\lambda }{\int }}\left({\displaystyle \frac{\pi }{\lambda }}-t\right)(t-0){\lambda }^{2}dt \\ ={\lambda }^2\underset{0}{\overset{\pi /\lambda }{\int }}\left({\displaystyle \frac{\pi t}{\lambda }}-t^2\right)dt \\ ={\lambda }^2\left({\displaystyle \frac{\pi t^2}{2\lambda }}-{\displaystyle \frac{t^3}{3}}\right)\left|{\displaystyle \genfrac{}{}{0pt}{}{\pi /\lambda }{0}}\right. \\ ={\lambda }^2\left({\displaystyle \frac{{\pi }^3}{2{\lambda }^3}}-{\displaystyle \frac{{\pi }^3}{3{\lambda }^3}}\right) \\ ={\displaystyle \frac{{\pi }^3}{6\lambda }} \end{gathered}$$

Since, ${\displaystyle \frac{{\pi }^3}{\left(6\lambda \right)}}>{\displaystyle \frac{\pi }{\lambda }}-0=b-a,$ Hartman’s inequality is verified for Eq. (24).

Now, if we evaluate the integral

$$\underset{a}{\overset{b}{\int }}{q}^{+}\left(t\right)dt=\underset{0}{\overset{\pi /\lambda }{\int }}{\lambda }^{2}dt=\lambda \pi >{\displaystyle \frac{4\lambda }{\pi }}={\displaystyle \frac{4}{\frac{\pi }{\lambda }-0}}={\displaystyle \frac{4}{b-a}},$$

then we see that Lyapunov inequality is verified for Eq. (24). We note that $x\left(t\right)=sin\left(\lambda x\right)$ is the unique solution of Eq. (24).