A Lyapunov-type inequality for problem (S)-(DBC) is established in this section, and some particular cases are discussed.
Theorem 2.1
If (S)-(DBC) admits a nontrivial solution
\((u,v)\in C^{2}[a,b]\times C^{2}[a,b]\), then
$$\begin{aligned} & \biggl[\min \biggl\{ \frac{2^{p_{1}}}{(b-a)^{p_{1}-1}},\frac{2^{q_{1}}}{(b-a)^{q_{1}-1}} \biggr\} \biggr]^{\frac{2\alpha}{p_{1}+q_{1}}} \biggl[\min \biggl\{ \frac{2^{p_{2}}}{(b-a)^{p_{2}-1}}, \frac{2^{q_{2}}}{(b-a)^{q_{2}-1}} \biggr\} \biggr]^{\frac{2\beta}{p_{2}+q_{2}}} \\ &\quad \leq \biggl(\frac{1}{2} \int_{a}^{b} f(x)\,dx \biggr)^{\frac{2\alpha}{p_{1}+q_{1}}} \biggl( \frac{1}{2} \int_{a}^{b} g(x)\,dx \biggr)^{\frac{2\beta}{p_{2}+q_{2}}}. \end{aligned}$$
(7)
Proof
Let \((u,v)\in C^{2}[a,b]\times C^{2}[a,b]\) be a nontrivial solution to (S)-(DBC). Let \((x_{0},y_{0})\in (a,b)\times (a,b)\) be such that
$$\bigl\vert u(x_{0}) \bigr\vert =\max\bigl\{ \bigl\vert u(x) \bigr\vert :\, a\leq x\leq b\bigr\} $$
and
$$\bigl\vert v(y_{0}) \bigr\vert =\max\bigl\{ \bigl\vert v(x) \bigr\vert :\, a\leq x\leq b\bigr\} . $$
From the boundary conditions (DBC), we can write that
$$2u(x_{0})= \int_{a}^{x_{0}} u'(x)\,dx - \int_{x_{0}}^{b} u'(x)\,dx, $$
which yields
$$2 \bigl\vert u(x_{0}) \bigr\vert \leq \int_{a}^{b} \bigl\vert u'(x) \bigr\vert \,dx. $$
Using Hölder’s inequality with parameters \(p_{1}\) and \(p_{1}'=\frac{p_{1}}{p_{1}-1}\), we get
$$2 \bigl\vert u(x_{0}) \bigr\vert \leq (b-a)^{\frac{1}{p_{1}'}} \biggl( \int_{a}^{b} \bigl\vert u'(x) \bigr\vert ^{p_{1}}\,dx \biggr)^{\frac{1}{p_{1}}}, $$
that is,
$$ \frac{2^{p_{1}}}{(b-a)^{p_{1}-1}} \bigl\vert u(x_{0}) \bigr\vert ^{p_{1}} \leq \int_{a}^{b} \bigl\vert u'(x) \bigr\vert ^{p_{1}}\,dx. $$
(8)
Similarly, using Hölder’s inequality with parameters \(q_{1}\) and \(q_{1}'=\frac{q_{1}}{q_{1}-1}\), we get
$$ \frac{2^{q_{1}}}{(b-a)^{q_{1}-1}} \bigl\vert u(x_{0}) \bigr\vert ^{q_{1}} \leq \int_{a}^{b} \bigl\vert u'(x) \bigr\vert ^{q_{1}}\,dx. $$
(9)
By repeating the same argument for the function v, we obtain
$$ \frac{2^{p_{2}}}{(b-a)^{p_{2}-1}} \bigl\vert v(y_{0}) \bigr\vert ^{p_{2}} \leq \int_{a}^{b} \bigl\vert v'(x) \bigr\vert ^{p_{2}}\,dx $$
(10)
and
$$ \frac{2^{q_{2}}}{(b-a)^{q_{2}-1}} \bigl\vert v(y_{0}) \bigr\vert ^{q_{2}} \leq \int_{a}^{b} \bigl\vert v'(x) \bigr\vert ^{q_{2}}\,dx. $$
(11)
Now, multiplying the first equation of (S) by u and integrating over \((a,b)\), we obtain
$$ \int_{a}^{b} \bigl\vert u'(x) \bigr\vert ^{p_{1}}\,dx+ \int_{a}^{b} \bigl\vert u'(x) \bigr\vert ^{q_{1}}\,dx= \int_{a}^{b} f(x) \bigl\vert u(x) \bigr\vert ^{\alpha}\bigl\vert v(x) \bigr\vert ^{\beta}\,dx. $$
(12)
Multiplying the second equation of (S) by v and integrating over \((a,b)\), we obtain
$$ \int_{a}^{b} \bigl\vert v'(x) \bigr\vert ^{p_{2}}\,dx+ \int_{a}^{b} \bigl\vert v'(x) \bigr\vert ^{q_{2}}\,dx= \int_{a}^{b} g(x) \bigl\vert u(x) \bigr\vert ^{\alpha}\bigl\vert v(x) \bigr\vert ^{\beta}\,dx. $$
(13)
Using (8), (9) and (12), we obtain
$$\bigl\vert u(x_{0}) \bigr\vert ^{\alpha}\bigl\vert v(y_{0}) \bigr\vert ^{\beta}\int_{a}^{b} f(x) \,dx\geq \frac{2^{p_{1}}}{(b-a)^{p_{1}-1}} \bigl\vert u(x_{0}) \bigr\vert ^{p_{1}}+\frac{2^{q_{1}}}{(b-a)^{q_{1}-1}} \bigl\vert u(x_{0}) \bigr\vert ^{q_{1}}, $$
which yields
$$\bigl\vert u(x_{0}) \bigr\vert ^{\alpha}\bigl\vert v(y_{0}) \bigr\vert ^{\beta}\int_{a}^{b} f(x) \,dx\geq \min \biggl\{ \frac{2^{p_{1}}}{(b-a)^{p_{1}-1}},\frac{2^{q_{1}}}{(b-a)^{q_{1}-1}} \biggr\} \bigl( \bigl\vert u(x_{0}) \bigr\vert ^{p_{1}}+ \bigl\vert u(x_{0}) \bigr\vert ^{q_{1}} \bigr). $$
Using the inequality
$$A+B\geq 2\sqrt{A}\sqrt{B} $$
with \(A= \vert u(x_{0}) \vert ^{p_{1}}\) and \(B= \vert u(x_{0}) \vert ^{q_{1}}\), we get
$$ \min \biggl\{ \frac{2^{p_{1}+1}}{(b-a)^{p_{1}-1}},\frac{2^{q_{1}+1}}{(b-a)^{q_{1}-1}} \biggr\} \leq \bigl\vert u(x_{0}) \bigr\vert ^{\alpha-\frac{p_{1}+q_{1}}{2}} \bigl\vert v(y_{0}) \bigr\vert ^{\beta} \int_{a}^{b} f(x)\,dx. $$
(14)
Similarly, using (10), (11) and (13), we obtain
$$ \min \biggl\{ \frac{2^{p_{2}+1}}{(b-a)^{p_{2}-1}},\frac{2^{q_{2}+1}}{(b-a)^{q_{2}-1}} \biggr\} \leq \bigl\vert u(x_{0}) \bigr\vert ^{\alpha} \bigl\vert v(y_{0}) \bigr\vert ^{\beta-\frac{p_{2}+q_{2}}{2}} \int_{a}^{b} g(x)\,dx. $$
(15)
Raising inequality (14) to a power \(e_{1}>0\), inequality (15) to a power \(e_{2}>0\), and multiplying the resulting inequalities, we obtain
$$\begin{aligned} & \biggl[\min \biggl\{ \frac{2^{p_{1}+1}}{(b-a)^{p_{1}-1}},\frac{2^{q_{1}+1}}{(b-a)^{q_{1}-1}} \biggr\} \biggr]^{e_{1}} \biggl[\min \biggl\{ \frac{2^{p_{2}+1}}{(b-a)^{p_{2}-1}},\frac{2^{q_{2}+1}}{(b-a)^{q_{2}-1}} \biggr\} \biggr]^{e_{2}} \\ &\quad \leq \bigl\vert u(x_{0}) \bigr\vert ^{ (\alpha-\frac{p_{1}+q_{1}}{2} )e_{1}+\alpha e_{2}} \bigl\vert v(y_{0}) \bigr\vert ^{\beta e_{1}+ (\beta-\frac{p_{2}+q_{2}}{2} )e_{2}} \biggl( \int_{a}^{b} f(x)\,dx \biggr)^{e_{1}} \biggl( \int_{a}^{b} g(x)\,dx \biggr)^{e_{2}}. \end{aligned}$$
Next, we take \((e_{1},e_{2})\) any solution of the homogeneous linear system
$$\begin{aligned} \textstyle\begin{cases} (\alpha-\frac{p_{1}+q_{1}}{2} )e_{1}+\alpha e_{2}= 0,\\ \beta e_{1}+ (\beta-\frac{p_{2}+q_{2}}{2} )e_{2}=0. \end{cases}\displaystyle \end{aligned}$$
Using (1), we may take
$$\begin{aligned} \textstyle\begin{cases} e_{1}= \alpha,\\ e_{2}=\frac{\beta(p_{1}+q_{1})}{p_{2}+q_{2}}. \end{cases}\displaystyle \end{aligned}$$
Therefore, we obtain
$$\begin{aligned} &2^{\alpha+\frac{\beta(p_{1}+q_{1})}{p_{2}+q_{2}}} \biggl[\min \biggl\{ \frac{2^{p_{1}}}{(b-a)^{p_{1}-1}},\frac{2^{q_{1}}}{(b-a)^{q_{1}-1}} \biggr\} \biggr]^{\alpha} \biggl[\min \biggl\{ \frac{2^{p_{2}}}{(b-a)^{p_{2}-1}}, \frac{2^{q_{2}}}{(b-a)^{q_{2}-1}} \biggr\} \biggr]^{\frac{\beta(p_{1}+q_{1})}{p_{2}+q_{2}}} \\ &\quad \leq \biggl( \int_{a}^{b} f(x)\,dx \biggr)^{\alpha} \biggl( \int_{a}^{b} g(x)\,dx \biggr)^{\frac{\beta(p_{1}+q_{1})}{p_{2}+q_{2}}}. \end{aligned}$$
Using again (1), we get
$$\begin{aligned} &2 \biggl[\min \biggl\{ \frac{2^{p_{1}}}{(b-a)^{p_{1}-1}},\frac{2^{q_{1}}}{(b-a)^{q_{1}-1}} \biggr\} \biggr]^{\frac{2\alpha}{p_{1}+q_{1}}} \biggl[\min \biggl\{ \frac{2^{p_{2}}}{(b-a)^{p_{2}-1}},\frac{2^{q_{2}}}{(b-a)^{q_{2}-1}} \biggr\} \biggr]^{\frac{2\beta}{p_{2}+q_{2}}} \\ &\quad \leq \biggl( \int_{a}^{b} f(x)\,dx \biggr)^{\frac{2\alpha}{p_{1}+q_{1}}} \biggl( \int_{a}^{b} g(x)\,dx \biggr)^{\frac{2\beta}{p_{2}+q_{2}}}, \end{aligned}$$
which proves Theorem 2.1. □
As a consequence of Theorem 2.1, we deduce the following result for the case of a single equation.
Corollary 1
Let us assume that there exists a nontrivial solution of
$$\begin{aligned} \textstyle\begin{cases} -( \vert u'(x) \vert ^{p-2}u'(x))'-( \vert u'(x) \vert ^{q-2}u'(x))'=f(x) \vert u(x) \vert ^{\frac{p+q}{2}-2}u(x),\quad x\in (a,b),\\ u(a)=u(b)=0, \end{cases}\displaystyle \end{aligned}$$
where
\(p>1\), \(q>1\), \(f\geq 0\), and
\(f\in L^{1}(a,b)\). Then
$$\min \biggl\{ \frac{2^{p}}{(b-a)^{p-1}},\frac{2^{q}}{(b-a)^{q-1}} \biggr\} \leq \frac{1}{2} \int_{a}^{b} f(x)\,dx. $$
Proof
An application of Theorem 2.1 with
$$p_{1}=p_{2}=p,\qquad q_{1}=q_{2}=q,\qquad \alpha=\frac{p+q}{2},\qquad \beta=0,\qquad v=u,\qquad g=f, $$
yields the desired result. □
Remark 1
Taking \(f=2h\) and \(q=p\) in Corollary 1, we obtain Lyapunov-type inequality (3) for the one-dimensional p-Laplacian equation.
Remark 2
Taking \(p_{1}=q_{1}=p\) and \(p_{2}=q_{2}=q\) in Theorem 2.1, we obtain Lyapunov-type inequality (6).