1 Introduction

In [1], Lyapunov proved that, if \(p(x)\) is a nonnegative and continuous function and \(u(x)\in C(I,\mathbb{R})\), a necessary condition for the following boundary value problem:

$$ \left \{ \textstyle\begin{array}{l}u^{\prime\prime}(x)+p(x)u(x)=0, \quad u(x)\neq0,\forall x\in I, \\ u(a_{1})=0=u(b_{1}), \end{array}\displaystyle \right . $$
(1.1)

to have nontrivial solutions is

$$ 4/(b_{1}-a_{1})\leq \int_{a_{1}}^{b_{1}}p(x)\,\mathrm{d}x, $$
(1.2)

where \(I= [ a_{1},b_{1} ] \).

Since Lyapunov’s study, because the inequality of (1.2) plays a key role for the qualitative properties, such as oscillatory and disconjugacy etc., of differential equations’ solutions, several authors focused on the inequality of (1.2). Those authors improved and generalized the inequality of (1.2) in \(\mathbb{R}\). In this work the literature of the one-dimensional case is not studied in detail but it is listed in the references for the interested reader. See [1–17] and the references cited therein.

In addition to studies in \(\mathbb{R}\), several authors [18–24] have extended the inequality of (1.2) in \(\mathbb{R}^{n}\) recently. To the best of our knowledge, it was extended by Cañada, Montero, and Villegas [19] for the first time. In [19] Cañada et al. considered the linear elliptic problem as follows:

$$ \left \{ \textstyle\begin{array}{l}-\triangle u=a(x)u,\quad x\in\Omega, \\ \partial u/\partial n=0,\quad x\in\partial\Omega, \end{array}\displaystyle \right . $$
(1.3)

where \(\Omega\subset \mathbb{R}^{N}\) is a smooth bounded domain with \(N\geq2\) and the function \(a:\Omega \rightarrow \mathbb{R}\) belongs to the set

$$ \Lambda= \biggl\{ a\in L^{N/2}(\Omega)\setminus\{0\}: \int_{\Omega}a(x)\,\mathrm{d}x\geq0\text{ and (1.3) has a nontrivial solution} \biggr\} $$
(1.4)

if \(N\geq3\),

$$\begin{aligned} \Lambda =&\biggl\{ a:\Omega\rightarrow \mathbb{R} \text{ s.t. } \exists q\in( 1,\infty ] \text{ with }a\in L^{q}(\Omega)\setminus \{0\}: \int_{\Omega}a(x)\,\mathrm{d}x\geq0 \\ &\text{and (1.3) has a nontrivial solution}\biggr\} \end{aligned}$$
(1.5)

if \(N=2\), we define

$$ \beta_{q}:=\inf_{a\in\wedge\cap L^{q}(\Omega)}\Vert a\Vert _{L^{q}(\Omega)}, \quad 1\leq q\leq\infty. $$
(1.6)

Their main result is as follows.

Theorem A

The following statements hold.

  1. (1)

    If \(N=2\) then \(\beta_{q}>0 \Leftrightarrow 1< q\leq\infty\). If \(N\geq3\) then \(\beta _{q}>0 \Leftrightarrow\frac{N}{2}\leq q\leq\infty\).

  2. (2)

    If \(\frac{N}{2}< q\leq\infty\) then \(\beta_{q}\) is attained. In this case, any function \(a\in\wedge\cap L^{q}(\Omega)\) from which \(\beta_{q}\) is attained has one of the following forms:

    1. (i)

      \(a(x)=\lambda_{1}\), if \(p=\infty\), where \(\lambda_{1}\) is the first strictly positive eigenvalue of (1.3).

    2. (ii)

      \(a(x)=\vert u(x)\vert ^{{\frac{2}{p-1}}}\), if \(\frac{N}{2}< q<\infty\), where u is a solution of the problem as follows:

      $$ \left \{ \textstyle\begin{array}{l}-\triangle u=\vert u(x)\vert ^{2/(p-1)}u, \quad x\in\Omega, \\ \partial u/\partial n=0, \quad x\in\partial\Omega. \end{array}\displaystyle \right . $$
      (1.7)
  3. (3)

    The map \((\frac{N}{2},\infty)\rightarrow \mathbb{R}\), \(p\mapsto\beta_{p}\), is continuous and the map \([\frac{N}{2},\infty ) \rightarrow \mathbb{R}\), \(p\mapsto \vert \Omega \vert ^{\frac{-1}{p}}\beta_{p}\), is strictly increasing.

  4. (4)

    The limits \(\lim_{p\rightarrow\infty}\beta_{p}\) and \(\lim_{p\rightarrow(\frac{N}{2})^{+}}\beta_{p}\) always exist and take the values

    1. (i)

      \(\lim_{p\rightarrow\infty}\beta_{p}=\beta_{\infty}\), if \(N\geq2\),

    2. (ii)

      \(\lim_{p\rightarrow(\frac{N}{2})^{+}}\beta _{p}\geq \beta_{\frac{N}{2}}>0\), if \(N\geq3\), \(\lim_{p\rightarrow1^{+}}\beta_{p}=0\), if \(N=2\).

Here, we also note that in the study Cañada et al. they proved that the relation between the p and \(\frac{N}{2}\) plays a crucial role. They also considered the equation in (1.3) with zero Dirichlet boundary condition. They presented similar inequalities at their study. Then others established Lyapunov-type inequalities for different equations with boundary conditions. For more information about the studies in \(\mathbb{R}^{n}\), the interested reader can refer to [18–24] and the references cited therein.

The aim of this paper is to prove a Lyapunov-type inequality for the two-dimensional quasilinear elliptic problem as follows:

$$ \left \{ \textstyle\begin{array}{l}-(\Phi_{p}(u_{xy}))_{y}=r(x,y)\Phi_{p}(u),\quad (x,y)\in\Omega, \\ u(a_{1},y)=0=u(b_{1},y), \quad a_{2}\leq y\leq b_{2}, \\ u_{x}(x,a_{2})=0=u_{x}(x,b_{2}),\quad a_{1}\leq x\leq b_{1}, \end{array}\displaystyle \right . $$
(1.8)

where \(\Omega= [ a_{1},b_{1} ] \times [ a_{2},b_{2} ] \) and \(r(x,y)\) is a measurable function on Ω, and \(\Phi_{p}(u(x,y))=\vert u(x,y)\vert ^{p-2}u(x,y)\) for \(p>1\). In addition to this, we note that by a solution of the problem (1.8), we mean that \(u(x,y)\in W^{3,p}(\Omega)\) in that

$$ W^{3,p}(\Omega)= \bigl\{ u: u,u_{x},u_{xy}\text{ and }u_{xyy}\in L^{p}(\Omega) \bigr\} . $$
(1.9)

As usual, \(L^{p}(\Omega)\) is a space of Lebesgue measurable functions.

2 Main results

Now, we give a key lemma as a proof of our main conclusions.

Lemma 1

Assume that \(u(x,y)\in W^{3,p}(\Omega )\), it satisfies the boundary conditions in (1.8) and \(u(x,y)\neq0\) for \(\forall(x,y)\in\Omega^{0}\). Then

$$\begin{aligned}& \bigl(4\bigl\vert u(x,y)\bigr\vert \bigr)^{p}/(b_{1}-a_{1})^{p-1}(b_{2}-a_{2})^{p-1} \leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x, \end{aligned}$$
(2.1)
$$\begin{aligned}& \bigl( 2^{p}/(b_{2}-a_{2})^{p-1} \bigr) { \int_{a_{1}}^{b_{1}}} \vert u_{x}\vert ^{p}\,\mathrm{d}x\leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x, \end{aligned}$$
(2.2)

hold, respectively, where \(\Omega^{0}\) is the set of all interior points of Ω.

Proof

Let \((x,y)\in\Omega\). Since \(u(x,y)\) satisfies the boundary conditions in (1.8), it is easy to see

$$u(x,y)= { \int_{a_{1}}^{x}} { \int_{a_{2}}^{y}} u_{ts}\,\mathrm{d}s\, \mathrm{d}t, $$

taking the absolute value, we obtain

$$ \bigl\vert u(x,y)\bigr\vert \leq { \int_{a_{1}}^{x}} { \int_{a_{2}}^{y}} \vert u_{ts}\vert \, \mathrm{d}s\,\mathrm{d}t. $$
(2.3)

Similarly, we get

$$\begin{aligned}& \bigl\vert u(x,y)\bigr\vert \leq { \int_{a_{1}}^{x}} { \int_{y}^{b_{2}}} \vert u_{ts}\vert \, \mathrm{d}s\,\mathrm{d}t, \end{aligned}$$
(2.4)
$$\begin{aligned}& \bigl\vert u(x,y)\bigr\vert \leq { \int_{x}^{b_{1}}} { \int_{a_{2}}^{y}} \vert u_{ts}\vert \, \mathrm{d}s\,\mathrm{d}t, \end{aligned}$$
(2.5)

and

$$ \bigl\vert u(x,y)\bigr\vert \leq { \int_{x}^{b_{1}}} { \int_{y}^{b_{2}}} \vert u_{ts}\vert \, \mathrm{d}s\,\mathrm{d}t. $$
(2.6)

Adding (2.3)-(2.6), we have

$$ 4\bigl\vert u(x,y)\bigr\vert \leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert \, \mathrm{d}y\,\mathrm{d}x. $$
(2.7)

Then, applying Hölder’s inequality

$$ \int_{a_{1}}^{b_{1}}\bigl\vert f(t)g(t)\bigr\vert \, \mathrm{d}t\leq \biggl( \int_{a_{1}}^{b_{1}}\bigl\vert f(t)\bigr\vert ^{q}\,\mathrm{d}t \biggr) ^{1/q} \biggl( \int_{a_{1}}^{b_{1}}\bigl\vert g(t)\bigr\vert ^{p}\,\mathrm{d}t \biggr) ^{1/p} $$
(2.8)

to the right-hand side of (2.7), we get

$$ \bigl( 4\bigl\vert u(x,y)\bigr\vert \bigr) ^{p}\leq ( b_{1}-a_{1} ) ^{p-1}{ \int_{a_{1}}^{b_{1}}} \biggl[ { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert \, \mathrm{d}y \biggr] ^{p}\,\mathrm{d}x. $$
(2.9)

Applying Hölder’s inequality to the right hand side of (2.9) again, we obtain

$$\frac{(4\vert u(x,y)\vert )^{p}}{(b_{1}-a_{1})^{p-1}(b_{2}-a_{2})^{p-1}}\leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x. $$

Thereby, the proof of (2.1) is completed.

Similarly we have

$$ \bigl\vert u_{x}(x,y)\bigr\vert \leq { \int_{a_{2}}^{y}} \vert u_{xs}\vert \, \mathrm{d}s $$
(2.10)

and

$$ \bigl\vert u_{x}(x,y)\bigr\vert \leq { \int_{y}^{b_{2}}} \vert u_{xs}\vert \, \mathrm{d}s. $$
(2.11)

Adding (2.10) and (2.11), we get

$$ 2\bigl\vert u_{x}(x,y)\bigr\vert \leq { \int_{a_{2}}^{b_{2}}} \vert u_{xs}\vert \, \mathrm{d}s. $$
(2.12)

Applying Hölder’s inequality to the right hand side of (2.12) and integrating from \(a_{1}\) to \(b_{1}\), we have

$$\frac{2^{p}}{(b_{2}-a_{2})^{p-1}}{ \int_{a_{1}}^{b_{1}}} \vert u_{x}\vert ^{p}\,\mathrm{d}x\leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x. $$

Consequently, the proof of (2.2) is completed. □

Theorem 1

If \(u(x,y)\in W^{3,p}(\Omega )\) is a nontrivial solution of the problem (1.8), then the following inequality:

$$ 2^{2p+1}/(b_{1}-a_{1})^{p-1}(b_{2}-a_{2})^{p} \leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert ^{q}\,\mathrm{d}y\,\mathrm{d}x $$
(2.13)

holds, where q is the Hölder conjugate of p.

Proof

Let \(u(x,y)\in W^{3,p}(\Omega)\) is a nontrivial solution of the problem (1.8). Multiplying the equation in (1.8) by \(u_{x}\) and integrating on Ω,we obtain

$$ { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} -\bigl(\vert u_{xy} \vert ^{p-2}u_{xy}\bigr)_{y}u_{x}\, \mathrm{d}y\,\mathrm{d}x= { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} r(x,y)\vert u\vert ^{p-2}uu_{x}\,\mathrm{d}y\,\mathrm{d}x. $$
(2.14)

Then, applying partial integration in \({ \int_{a_{2}}^{b_{2}}} -(\vert u_{xy}\vert ^{p-2}u_{xy})_{y}u_{x}\, \mbox{d}y\) and using the boundary conditions in (1.8), we have

$$ { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x={ \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} r(x,y)\vert u\vert ^{p-2}uu_{x}\,\mathrm{d}y\,\mathrm{d}x. $$
(2.15)

By taking the absolute value on right hand side of (2.15), we get

$$ { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x\leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert \vert u\vert ^{p-1}\vert u_{x}\vert \,\mathrm{d}y\, \mathrm{d}x. $$
(2.16)

Hence, applying Hölder’s inequality to the right hand side of (2.16), we find

$$\begin{aligned}& { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x \\& \quad \leq \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert ^{q}\vert u\vert ^{(p-1)q}\,\mathrm{d}y\, \mathrm{d}x \biggr) ^{1/q} \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{x}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x \biggr) ^{1/p}. \end{aligned}$$
(2.17)

Now, considering only the second term of right hand side in (2.17) from Fubini’s theorem, we can rewrite the inequality (2.17) as follows:

$$\begin{aligned}& { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x \\& \quad \leq \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert ^{q}\vert u\vert ^{(p-1)q}\,\mathrm{d}y\, \mathrm{d}x \biggr) ^{1/q} \biggl( { \int_{a_{2}}^{b_{2}}} \biggl[ { \int_{a_{1}}^{b_{1}}} \vert u_{x}\vert ^{p}\,\mathrm{d}x \biggr] \,\mathrm{d}y \biggr) ^{1/p}. \end{aligned}$$
(2.18)

Hence, using the inequality (2.2) in (2.18), we obtain

$$\begin{aligned}& \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x \biggr) ^{(p-1)/p} \\& \quad \leq \bigl( (b_{2}-a_{2})/2 \bigr) \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert ^{q}\vert u\vert ^{(p-1)q}\,\mathrm{d}y\, \mathrm{d}x \biggr) ^{1/q}. \end{aligned}$$
(2.19)

Then, replacing the point of \((x,y)\), which is used in Lemma 1, with the maximum point of \(\vert u(x,y)\vert \), from (2.1), we get

$$\begin{aligned}& \bigl(4\max\bigl\vert u(x,y)\bigr\vert \bigr)^{(p-1)}/(b_{1}-a_{1})^{(p-1)/q}(b_{2}-a_{2})^{(p-1)/q} \\& \quad \leq \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \vert u_{xy}\vert ^{p}\,\mathrm{d}y\,\mathrm{d}x \biggr) ^{(p-1)/p}. \end{aligned}$$
(2.20)

Then, using the inequality (2.20) in the inequality (2.19), we have

$$\begin{aligned}& \bigl(4\max\bigl\vert u(x,y)\bigr\vert \bigr)^{(p-1)}/(b_{1}-a_{1})^{(p-1)/q}(b_{2}-a_{2})^{(p-1)/q} \\& \quad \leq \bigl( (b_{2}-a_{2})/2 \bigr) \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert ^{q}\vert u\vert ^{(p-1)q}\,\mathrm{d}y\, \mathrm{d}x \biggr) ^{1/q} \\& \quad \leq \bigl( (b_{2}-a_{2})/2 \bigr) \biggl( { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert ^{q}\,\mathrm{d}y\,\mathrm{d}x \biggr) ^{1/q}\bigl( \max\bigl\vert u(x,y)\bigr\vert \bigr)^{(p-1)}. \end{aligned}$$
(2.21)

Since \(u(x,y)\) is a nontrivial solution, we have \(\max \vert u(x,y)\vert \neq0\). Therefore, we obtain

$$ 2^{2p+1}/(b_{1}-a_{1})^{p-1}(b_{2}-a_{2})^{p} \leq { \int_{a_{1}}^{b_{1}}} { \int_{a_{2}}^{b_{2}}} \bigl\vert r(x,y)\bigr\vert ^{q}\,\mathrm{d}y\,\mathrm{d}x. $$
(2.22)

Thus, the proof is completed. □

Corollary 1

Let \(\lambda_{1}\) be the first eigenvalue of the equation that is defined on Ω as follows:

$$-\bigl(\Phi_{p}(u_{xy})\bigr)_{y}= \lambda_{1}r(x,y)\Phi_{p}(u), $$

where Ω is a domain, which is defined in the beginning of the paper, and with the boundary conditions in (1.8). Then we have

$$ 2^{2p+1}/(b_{1}-a_{1})^{p-1}(b_{2}-a_{2})^{p} \bigl\Vert r(x,y) \bigr\Vert _{L^{q}(\Omega)}^{q}\leq \lambda_{1}. $$
(2.23)

Remark 1

If we take the Dirichlet boundary conditions, which are \(u(a_{1},y)=0=u(b_{1},y)\) and \(u(x,a_{2})=0=u(x,b_{2})\), instead of the Robin boundary conditions in the problem (1.8), then we obtain the identical conclusions given above.

Remark 2

The result, which is obtained in this study, is also the necessary condition for the problem of (1.8) to have a nontrivial solution.