On multivariate higher order Lyapunov-type inequalities

Abstract

In this paper, by using the best Sobolev constant method, we obtain some new Lyapunov-type inequalities for a class of even-order partial differential equations; the results of this paper are new which generalize and improve some early results in the literature.

1 Introduction

It is well known that the Lyapunov inequality for the second-order linear differential equation

$$x^{″}(t)+q(t)x(t)=0$$

(1)

states that if $q\in C[a,b]$, $x(t)$ is a nonzero solution of (1) such that $x(a)=x(b)=0$, then the following inequality holds:

$${\int }_a^b|q(t)|dt>\frac{4}{b-a}$$

(2)

and the constant 4 is sharp.

There have been many proofs and generalizations as well as improvements on this inequality. For example, the authors in [1–3] generalized the Lyapunov-type inequality to the partial differential equations or systems.

First let us recall some background and notations which are introduced in [1, 2].

Let A be a spherical shell $\subseteq {\mathbb{R}}^N$ for $N>1$, i.e. $A=B(0,R_2)-\overline{B(0,R_1)}$ for $0<R_1<R_2$, where $B(0,R)=\{x\in {\mathbb{R}}^N:\parallel x\parallel <R\}$ for $R>0$ and $\parallel \cdot \parallel$ is the Euclidean norm. Denote $S^{N-1}=\{x\in {\mathbb{R}}^N:\parallel x\parallel =1\}$, the unit sphere in ${\mathbb{R}}^N$ with surface area

$${\omega }_N=\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }(\frac{N}{2})},\mathit{\text{i.e. }}{\int }_{S^{N-1}}d\omega =\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }(\frac{N}{2})},$$

(3)

where $\mathrm{\Gamma }(\cdot )$ is the gamma function. Then every $x\in {\mathbb{R}}^N-\{0\}$ has a unique representation of the form $x=r\omega$, where $r=\parallel x\parallel >0$ and $\omega =\frac{x}{r}\in S^{N-1}$. Therefore, for any $f\in C(\overline{A})$, we have

$${\int }_{A}f(x)dx={\int }_{S^{N-1}}({\int }_{R_1}^{R_2}f(r\omega )r^{N-1}dr)d\omega .$$

In [1], Aktaş obtained the following results.

Theorem A If $f\in C^{2n}(\overline{A})$ is a nonzero solution of the following even-order partial differential equation:

$$\frac{{\partial }^{2n}f(x)}{\partial r^{2n}}+q(x)f(x)=0,x\in A,$$

(4)

where $n\in \mathbb{N}$ and $q\in C(\overline{A})$, with the boundary conditions

$$\frac{{\partial }^{2i}f}{\partial r^{2i}}(\partial B(0,R_1))=\frac{{\partial }^{2i}f}{\partial r^{2i}}(\partial B(0,R_2))=0,i=0,1,2,\dots ,n-1,$$

(5)

then the following inequality holds:

$${\int }_A|q(x)|dx>\frac{2^{3n-1}}{{(R_2-R_1)}^{2n-1}}\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }(\frac{N}{2})}{R}_1^{N-1}.$$

(6)

Theorem B If $f\in C^{2n}(\overline{A})$ is a nonzero solution of (4) with the boundary conditions

$$\frac{{\partial }^{i}f}{\partial r^i}(\partial B(0,R_1))=\frac{{\partial }^{i}f}{\partial r^i}(\partial B(0,R_2))=0,i=0,1,2,\dots ,n-1,$$

(7)

then the following inequality holds:

$${\int }_A|q(x)|dx>\frac{4^{2n-1}(2n-1){[(n-1)!]}^2}{{(R_2-R_1)}^{2n-1}}\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }(\frac{N}{2})}{R}_1^{N-1}.$$

(8)

In this paper, we generalize Theorem A and Theorem B to a more general class of even order partial differential equations. Moreover, as we shall see by the end of this paper, Theorem 1 improves Theorem A significantly.

2 Main results

Let us consider the following even-order partial differential equation:

$$\frac{{\partial }^{2n}y(x)}{\partial r^{2n}}+\sum _{k=0}^n{p}_k(x)\frac{{\partial }^{k}y(x)}{\partial r^k}=0,$$

(9)

where $y(x)\in C^{2n}(\overline{A})$, $p_k(x)\in C(\overline{A})$, $k=0,1,2,\dots ,n$, and $x\in {\mathbb{R}}^N$.

The main results of this paper are the following theorems.

Theorem 1 If $y(x)$ is a nonzero solution of (9) satisfying boundary conditions (5), then the following inequality holds:

$$\begin{array}{rl}1<& \sqrt{\frac{(2^{2n}-1)\zeta (2n){(R_2-R_1)}^{2n-1}\mathrm{\Gamma }(\frac{N}{2})}{2^{2n}{\pi }^{2n+\frac{N}{2}}{R}_1^{N-1}}}{({\int }_A{p}_n^2(x)dx)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{(R_2-R_1)}^{2n-k-1}\mathrm{\Gamma }(\frac{N}{2})}{{(2\pi )}^{2n-k}{R}_1^{N-1}{\pi }^{\frac{N}{2}}}\sqrt{(2^{2n}-1)(2^{2(n-k)}-1)\zeta (2n)\zeta (2(n-k))}\\ \times {\int }_A|p_k(x)|dx,\end{array}$$

(10)

where $\zeta (s)={\sum }_{n=1}^{+\mathrm{\infty }}\frac{1}{n^s}$ is the Riemann zeta function.

Theorem 2 If $y(x)$ is a nonzero solution of (9) satisfying boundary conditions (7), then the following inequality holds:

$$\begin{array}{rl}1<& \frac{1}{(n-1)!2^{2n-1}}\sqrt{\frac{{(R_2-R_1)}^{2n-1}\mathrm{\Gamma }(\frac{N}{2})}{(2n-1)R_1^{N-1}2{\pi }^{\frac{N}{2}}}}{({\int }_A{p}_n^2(x)dx)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{(R_2-R_1)}^{2n-k-1}\mathrm{\Gamma }(\frac{N}{2})}{\sqrt{(2n-1)(2n-2k-1)}(n-1)!(n-k-1)!4^{2n-k-1}{R}_1^{N-1}2{\pi }^{\frac{N}{2}}}{\int }_A|p_k(x)|dx.\end{array}$$

(11)

3 Proofs of theorems

For the proofs of Theorem 1 and Theorem 2, let us consider first the following ordinary even-order linear ordinary differential equation:

$$x^{(2n)}(t)+\sum _{k=0}^n{p}_k(t)x^{(k)}(t)=0,$$

(12)

where $p_k(t)\in C[a,b]$, $k=0,1,2,\dots ,n$.

Proposition 3 If (12) has a nonzero solution $x(t)$ satisfying the following boundary value conditions:

$$x^{(2i)}(a)=x^{(2i)}(b)=0,i=0,1,2,\dots ,n-1,$$

(13)

then the following inequality holds:

$$\begin{array}{rl}1<& \sqrt{\frac{(2^{2n}-1)\zeta (2n){(b-a)}^{2n-1}}{2^{2n-1}{\pi }^{2n}}}{({\int }_a^b{p}_n^2(t)dt)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{(b-a)}^{2n-k-1}\sqrt{(2^{2n}-1)(2^{2n-2k}-1)\zeta (2n)\zeta (2(n-k))}}{2^{2n-k-1}{\pi }^{2n-k}}{\int }_a^b|p_k(t)|dt,\end{array}$$

(14)

where $\zeta (s)$ is the Riemann zeta function: $\zeta (s)={\sum }_{k=1}^{+\mathrm{\infty }}\frac{1}{k^s}$, $s>1$.

Proposition 4 If (12) has a nonzero solution $x(t)$ satisfying the following boundary value conditions:

$$x^{(i)}(a)=x^{(i)}(b)=0,i=0,1,2,\dots ,n-1,$$

(15)

then we have the following inequality:

$$\begin{array}{rl}1<& \frac{1}{(n-1)!2^{2n-1}}\sqrt{\frac{{(b-a)}^{2n-1}}{(2n-1)}}{({\int }_a^b{p}_n^2(t)dt)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{(b-a)}^{2n-k-1}}{(n-1)!(n-k-1)!4^{2n-k-1}\sqrt{(2n-1)(2n-2k-1)}}{\int }_a^b|p_k(t)|dt.\end{array}$$

(16)

In order to prove the above propositions, we need the following lemmas.

Lemma 5 ([[4], Proposition 2.1])

Let $M\in \mathbb{N}$ and

$$H_C=\{u|u^{(M)}\in L^2(a,b),u^{(2k)}(a)=u^{(2k)}(b)=0,0\le k\le [(M-1)/2]\}.$$

Then there exists a positive constant C such that, for any $u\in H_C$, the Sobolev inequality

$${\left(\underset{a\le t\le b}{sup}|u(t)|\right)}^2\le C{\int }_a^b|u^{(M)}(t){|}^{2}dt$$

holds. Moreover, the best constant $C=C(M)$ is as follows:

$$C(M)=\frac{(2^{2M}-1)\zeta (2M){(b-a)}^{2M-1}}{2^{2M-1}{\pi }^{2M}}.$$

Lemma 6 ([[5], Theorem 1.2 and Corollary 1.3])

Let $M\in \mathbb{N}$ and

$$H_D=\{u|u^{(M)}\in L^2(a,b),u^{(k)}(a)=u^{(k)}(b)=0,0\le k\le M-1\}.$$

Then there exists a positive constant D such that for any $u\in H_D$, the Sobolev inequality

$${\left(\underset{a\le t\le b}{sup}|u(t)|\right)}^2\le D{\int }_a^b|u^{(M)}(t){|}^{2}dt$$

holds. Moreover, the best constant $D=D(M)$ is as follows:

$$D(M)=\frac{{(b-a)}^{2M-1}}{(2M-1){[(M-1)!]}^2{4}^{2M-1}}.$$

(17)

We give the first seven values of $\zeta (2n)$, $C(n)$, and $D(n)$ in Table 1.

Table 1 The first seven values of $\mathit{\zeta }\mathbf{(}\mathbf{2}\mathit{n}\mathbf{)}$ , $\mathit{C}\mathbf{(}\mathit{n}\mathbf{)}$ and $\mathit{D}\mathbf{(}\mathit{n}\mathbf{)}$

Since the proof of Proposition 4 is similar to that of Proposition 3, we give only the proof of Proposition 3 below.

Proof of Proposition 3 Multiplying both sides of (12) by $x(t)$ and integrating from a to b by parts and using the boundary value condition (13), we can obtain

$${\int }_a^b{x}^{(2n)}(t)x(t)dt={(-1)}^n{\int }_a^b{(x^{(n)}(t))}^{2}dt=-\sum _{k=0}^n{\int }_a^b{p}_k(t)x^{(k)}(t)x(t)dt.$$

This yields

$$\begin{array}{rl}{\int }_a^b{(x^{(n)}(t))}^{2}dt& \le \sum _{k=0}^n{\int }_a^b|p_k(t)||x^{(k)}(t)x(t)|dt\\ ={\int }_a^b|p_n(t)||x^{(n)}(t)x(t)|dt+\sum _{k=0}^{n-1}{\int }_a^b|p_k(t)||x^{(k)}(t)x(t)|dt.\end{array}$$

(18)

Now, by using Lemma 5, we get for any $t\in [a,b]$, $k=1,2,\dots ,n-1$,

$$|x(t)|\le \sqrt{C(n)}{\left({\int }_a^b{(x^{(n)}(t))}^{2}dt\right)}^{\frac{1}{2}}$$

(19)

and

$$|x^{(k)}(t)|\le \sqrt{C(n-k)}{\left({\int }_a^b{(x^{(n)}(t))}^{2}dt\right)}^{\frac{1}{2}}.$$

(20)

Substituting (19) and (20) into (18), we obtain

$$\begin{array}{rl}{\int }_a^b{(x^{(n)}(t))}^{2}dt\le & \sqrt{C(n)}{\int }_a^b|p_n(t)||x^{(n)}(t)|dt{\left({\int }_a^b{(x^{(n)}(t))}^{2}dt\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\sqrt{C(n)C(n-k)}{\int }_a^b|p_k(t)|dt{\int }_a^b{(x^{(n)}(t))}^{2}dt.\end{array}$$

(21)

Now by applying Hölder’s inequality, we get

$${\int }_a^b|p_n(t)x^{(n)}(t)|dt\le {({\int }_a^b{p}_n^2(t)dt)}^{\frac{1}{2}}{\left({\int }_a^b{(x^{(n)}(t))}^{2}dt\right)}^{\frac{1}{2}}.$$

(22)

Substituting (22) into (21) and by using the fact that $x(t)$ is not a constant function, we obtain the following strict inequality:

$$\begin{array}{rl}{\int }_a^b{(x^{(n)}(t))}^{2}dt<& \sqrt{C(n)}{({\int }_a^b{p}_n^2(t)dt)}^{\frac{1}{2}}{\int }_a^b{(x^{(n)}(t))}^{2}dt\\ +\sum _{k=0}^{n-1}\sqrt{C(n)C(n-k)}{\int }_a^b|p_k(t)|dt{\int }_a^b{(x^{(n)}(t))}^{2}dt.\end{array}$$

(23)

Dividing both sides of (23) by ${\int }_a^b{(x^{(n)}(t))}^{2}dt$, which can be proved to be positive by using the boundary value condition (13) and the assumption that $x(t)\not\equiv 0$, we obtain

$$1<\sqrt{C(n)}{({\int }_a^b{p}_n^2(t)dt)}^{\frac{1}{2}}+\sum _{k=0}^{n-1}\sqrt{C(n)C(n-k)}{\int }_a^b|p_k(t)|dt.$$

This is equivalent to (14). Thus we finished the proof of Proposition 3. □

Lemma 7 For any $f\in C(A)$, we have

$${\int }_A|f(x)|dx\ge \frac{R_1^{N-1}2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }(\frac{N}{2})}{\int }_{R_1}^{R_2}|f(r\omega )|dr.$$

(24)

Proof Similar to the proofs given in [1] and [2], we have

$${\int }_{R_1}^{R_2}|f(r\omega )|dr={\int }_{R_1}^{R_2}{r}^{1-N}{r}^{N-1}|f(r\omega )|dr\le \left({\int }_{R_1}^{R_2}{r}^{N-1}|f(r\omega )|dr\right)R_1^{1-N},$$

which implies that

$$\begin{array}{rl}{\int }_A|f(x)|dx& ={\int }_{S^{N-1}}\left({\int }_{R_1}^{R_2}{r}^{N-1}|f(r\omega )|dr\right)d\omega \\ \ge {\int }_{S^{N-1}}\left(R_1^{N-1}{\int }_{R_1}^{R_2}|f(r\omega )|dr\right)d\omega \\ =\left({\int }_{R_1}^{R_2}|f(r\omega )|dr\right)\frac{R_1^{N-1}2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }(\frac{N}{2})}.\end{array}$$

 □

Proof of Theorem 1 It follows from (14) and Lemma 7 that for any fixed $\omega \in S^{N-1}$, we have

$$\begin{array}{rl}1<& \sqrt{\frac{(2^{2n}-1)\zeta (2n){(R_2-R_1)}^{2n-1}}{2^{2n-1}{\pi }^{2n}}}{({\int }_{R_1}^{R_2}{p}_n^2(r\omega )dr)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{(R_2-R_1)}^{2n-k-1}\sqrt{(2^{2n}-1)(2^{2n-2k}-1)\zeta (2n)\zeta (2(n-k))}}{2^{2n-k-1}{\pi }^{2n-k}}{\int }_a^b|p_k(r\omega )|dr\\ \le & \sqrt{\frac{(2^{2n}-1)\zeta (2n){(R_2-R_1)}^{2n-1}\mathrm{\Gamma }(\frac{N}{2})}{2^{2n}{\pi }^{2n+\frac{N}{2}}{R}_1^{N-1}}}{({\int }_A{p}_n^2(x)dx)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{(R_2-R_1)}^{2n-k-1}\mathrm{\Gamma }(\frac{N}{2})}{{(2\pi )}^{2n-k}{R}_1^{N-1}{\pi }^{\frac{N}{2}}}\sqrt{(2^{2n}-1)(2^{2(n-k)}-1)\zeta (2n)\zeta (2(n-k))}{\int }_A|p_k(x)|dx,\end{array}$$

which is (10). This finishes the proof of Theorem 1. □

The proof of Theorem 2 is similar to that of Theorem 1, so we omit it for simplicity.

Let us compare Theorem 1 and Theorem 2 with Theorem A and Theorem B. It is evident that Theorem 2 is a natural generalization of Theorem B. If we let $p_n(x)=p_{n-1}(x)=\cdots =p_1(x)\equiv 0$, $p_0(x)=q(x)$, $\mathrm{\forall }x\in A$, then (10) reduces to the following inequality:

$${\int }_A|q(x)|dx>\frac{2^{2n-1}{\pi }^{2n}}{(2^{2n}-1)\zeta (2n){(R_2-R_1)}^{2n-1}}\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }(\frac{N}{2})}{R}_1^{N-1}.$$

(25)

Let us compare the right sides of inequalities (6) and (25): if we denote ${\delta }_n=\frac{2^{2n-1}{\pi }^{2n}}{(2^{2n}-1)\zeta (2n)2^{3n-1}}$, then we have

$${\delta }_n=\frac{{\pi }^{2n}}{2^n(2^{2n}-1)\zeta (2n)}>\frac{{\pi }^{2n}}{2^{3n}\zeta (2n)}={\left(\frac{{\pi }^2}{8}\right)}^n\frac{1}{\zeta (2n)}\to \mathrm{\infty },\text{as }n\to \mathrm{\infty },$$

since $\zeta (2n)\to 1$ as $n\to \mathrm{\infty }$. Table 2 gives the first eight values of ${\delta }_n$.

Table 2 The first eight values of ${\mathit{\delta }}_{\mathit{n}}$

From Table 2 we see that ${\delta }_n$ increases very quickly, so Theorem 1 improves Theorem A significantly even in the special case of (4).