Difference between revisions of "Delta Lyapunov inequality"

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__NOTOC__
<strong>Theorem:</strong> Let $p \colon \mathbb{T} \rightarrow \mathbb{R}^+$ be positive-valued and [[continuity | rd-continuous]]. If the [[Sturm-Liouville dynamic equation]]
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==Theorem==
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Let $p \colon \mathbb{T} \rightarrow \mathbb{R}^+$ be positive-valued and [[continuity | rd-continuous]]. If the [[Sturm-Liouville dynamic equation]]
 
$$x^{\Delta^2} + p(t) x^{\sigma} = 0$$
 
$$x^{\Delta^2} + p(t) x^{\sigma} = 0$$
 
has a nontrivial solution $x$ with $x(a)=x(b)=0$, then the Lyapunov inequality  
 
has a nontrivial solution $x$ with $x(a)=x(b)=0$, then the Lyapunov inequality  
 
$$\displaystyle\int_a^b p(t) \Delta t \geq \dfrac{b-a}{f(d)}$$
 
$$\displaystyle\int_a^b p(t) \Delta t \geq \dfrac{b-a}{f(d)}$$
holds, where $f \colon \mathbb{T} \rightarrow \mathbb{R}$ is defined by
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holds, where $f \colon \mathbb{T} \rightarrow \mathbb{R}$ is defined by $f(t) = (t-a)(b-t)$ and $d \in \mathbb{T}$ is such that
$$f(t) = (t-a)(b-t)$$
 
and $d \in \mathbb{T}$ is such that
 
 
$$\left| \dfrac{a+b}{2} - d \right| = \min \left\{ \left| \dfrac{a+b}{2} -s \right| \colon s \in [a,b] \cap \mathbb{T} \right\}.$$
 
$$\left| \dfrac{a+b}{2} - d \right| = \min \left\{ \left| \dfrac{a+b}{2} -s \right| \colon s \in [a,b] \cap \mathbb{T} \right\}.$$
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<strong>Proof:</strong> █
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==Proof==
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==References==
 
==References==
 
[http://www.math.unl.edu/~apeterson1/pub/ineq.pdf R. Agarwal, M. Bohner, A. Peterson - Inequalities on Time Scales: A Survey]
 
[http://www.math.unl.edu/~apeterson1/pub/ineq.pdf R. Agarwal, M. Bohner, A. Peterson - Inequalities on Time Scales: A Survey]
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{{:Delta inequalities footer}}
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 14:29, 22 January 2023

Theorem

Let $p \colon \mathbb{T} \rightarrow \mathbb{R}^+$ be positive-valued and rd-continuous. If the Sturm-Liouville dynamic equation $$x^{\Delta^2} + p(t) x^{\sigma} = 0$$ has a nontrivial solution $x$ with $x(a)=x(b)=0$, then the Lyapunov inequality $$\displaystyle\int_a^b p(t) \Delta t \geq \dfrac{b-a}{f(d)}$$ holds, where $f \colon \mathbb{T} \rightarrow \mathbb{R}$ is defined by $f(t) = (t-a)(b-t)$ and $d \in \mathbb{T}$ is such that $$\left| \dfrac{a+b}{2} - d \right| = \min \left\{ \left| \dfrac{a+b}{2} -s \right| \colon s \in [a,b] \cap \mathbb{T} \right\}.$$

Proof

References

R. Agarwal, M. Bohner, A. Peterson - Inequalities on Time Scales: A Survey

$\Delta$-Inequalities

Bernoulli Bihari Cauchy-Schwarz Gronwall Hölder Jensen Lyapunov Markov Minkowski Opial Tschebycheff Wirtinger