Duality of delta and nabla

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Let $\mathbb{T}$ be a time scale. Define the dual time scale $\mathbb{T}^*=\{s \in \mathbb{R} \colon -s \in \mathbb{T} \}$, which is a time scale. We label the basic time scale operations of $\mathbb{T}$ as $\sigma, \mu, \rho, $ and $\nu$ and we label the operations of $\mathbb{T}^*$ as $\hat{\sigma}, \hat{\mu}, \hat{\rho},$ and $\hat{\nu}$.

Properties[edit]

Lemma: If $a,b \in \mathbb{T}$ with $a<b$ then $$[a,b]^*=[-b,-a].$$

Proof:

Lemma: The following formulas hold: $$\hat{\sigma}(s)=-\rho(-s)$$ and $$\hat{\rho}(s)=-\sigma(-s)$$ for all $s \in \mathbb{T}^*$.

Proof:

Proposition: If $\mathbb{T}$ is a time scale, then $$(\mathbb{T}^{\kappa})^* = (\mathbb{T}^*)_{\kappa}$$ and $$(\mathbb{T}_{\kappa})^*=(\mathbb{T}^*)^{\kappa}.$$

Proof:

Lemma: The following formulas hold: $$\hat{\nu}(s)=\mu^*(s)$$ and $$\hat{\mu}(s)=\nu^*(s)$$ for all $s \in \mathbb{T}^*$.

Proof:

Lemma: Given $f \colon \mathbb{T} \rightarrow \mathbb{R}$ with $f$ is rd continuous (respectively, ld continuous) if and only if $f^* \colon \mathbb{T}^* \rightarrow \mathbb{R}$ is ld continuous (respectively, rd continuous).

Proof:

Theorem: The following formulas hold: $$f^{\Delta}(t_0)=-(f^*)^{\hat{\nabla}}(-t_0),$$ $$f^{\nabla}(t_0)=-(f^*)^{\hat{\Delta}}(-t_0),$$ $$f^{\Delta}(t_0)=-((f^*)^{\hat{\nabla}})^*(t_0),$$ $$f^{\nabla}(t_0)=-((f^*)^{\hat{\Delta}})^*(t_0),$$ $$(f^{\Delta})^*(-t_0)=((f^*)^{\hat{\nabla}})(-t_0),$$ and $$(f^{\nabla})^*(-t_0)=-(f^*)^{\hat{\Delta}})(-t_0).$$

Proof:

Theorem: If $f \colon [a,b]\cap \mathbb{T} \rightarrow \mathbb{R}$ is rd continuous, then $$\displaystyle\int_a^b f(t) \Delta t = \displaystyle\int_{-a}^{-b}f^*(s) \hat{\nabla}s.$$

Proof:

Theorem: If $f \colon [a,b] \rightarrow \mathbb{R}$ is ld continuous, then $$\displaystyle\int_a^b f(t) \nabla t = \displaystyle\int_{-a}^{-b} f^*(s) \hat{\Delta}s.$$

Proof:

References[edit]

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