Delta derivative

Let $\mathbb{T}$ be a time scale and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$ and let $t \in \mathbb{T}^{\kappa}$. We define<ref>Bohner, Martin ; Peterson, Allan. Dynamic equations on time scales. An introduction with applications. Birkhäuser Boston, Inc., Boston, MA, 2001,p.5.</ref> the $\Delta$-derivative of $f$ at $t$ to be the number $f^{\Delta}(t)$ (if it exists) so that there exists a $\delta >0$ so that for all $s \in (t-\delta,t+\delta) \bigcap \mathbb{T}$, $$|[f(\sigma(t))-f(s)]-f^{\Delta}(t)[\sigma(t)-s]| \leq \epsilon |\sigma(t)-s|.$$ We sometimes use the notation $\dfrac{\Delta}{\Delta t} f(t)$ or $\dfrac{\Delta f}{\Delta t}$ for $f^{\Delta}(t)$.

Theorem

If $f$ is $\Delta$-differentiable at $t$, then $f$ is continuous at $t$.

Proof

Supposing \( f \) is differentiable, given that, by Definition 1.10, $ \epsilon^* > 0 $ there is a neighborhood \( U \) \( (U = (t - \delta, t + \delta) \cap \mathbb{T}, \text{ for some } \delta > 0) \) of \( t \) such that \( |t - s| < \delta \) and \( \left|f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]\right| < \epsilon^* \left|\sigma(t) - s\right| \).

Now, notice that

$$ |f(t) - f(s)| = \left|f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) - s] - f^\Delta(t)[\sigma(t) - s]\right| $$

Since \( \mu(t) = \sigma(t) - t \), then

$$ \begin{array}{rcl} |f(t) - f(s)| & < & \left|f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) - s] - f^\Delta(t)[\sigma(t) - s]\right| \\ & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\sigma(t) + t - t - s] - f^\Delta(t)[\sigma(t) - s]| \\ & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)[\mu(t) + t - s] - f^\Delta(t)[\sigma(t) - s]| \\ & = & |f(t) - f(s) + f(\sigma(t)) - f(\sigma(t)) + f^\Delta(t)\mu(t) + f^\Delta(t)[t - s] - f^\Delta(t)[\sigma(t) - s]| \\ & = & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s] - f(\sigma(t)) + f(t) + f^\Delta(t)\mu(t) + f^\Delta(t)[t - s]| \\ & \leq & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]| + |- f(\sigma(t)) + f(t) + f^\Delta(t)\mu(t)| + |f^\Delta(t)[t - s]| \\ & = & |f(\sigma(t)) - f(s) - f^\Delta(t)[\sigma(t) - s]| + |f(\sigma(t)) - f(t) - f^\Delta(t)\mu(t)| + |f^\Delta(t)[t - s]| \\ & \leq & \epsilon^*|\sigma(t) - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t -s| \end{array} $$

But, observe that

$$ \begin{array}{rcl} |\sigma(t) - s| & = & |\sigma(t) - s + t - t| \\ & = & |\sigma(t) - t + t - s| \\ & \leq & |\sigma(t) - t| + |t - s| \\ & = & |\mu(t)| + |t - s| \\ \end{array} $$

Thus, we have

$$ \begin{array}{rcl} |f(t) - f(s)| & \leq & \epsilon^*|\sigma(t) - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t -s| \\ & \leq & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t - s| \end{array} $$

However, taking \( \delta = \epsilon^* \) and knowing that \( |t - s| < \delta \) (by hypothesis), we have

$$ \begin{array}{rcl} |f(t) - f(s)| & \leq & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \cdot |t - s| \\ & < & \epsilon^*(|\mu(t)| + |t - s|) + \epsilon^*|\mu(t)| + |f^\Delta(t)| \epsilon^* \\ & = & \epsilon^*|\mu(t)| + \epsilon^*|t - s| + \epsilon^*|\mu(t)| + |f^\Delta(t)| \epsilon^* \\ & = & \epsilon^*[|\mu(t)| + |t - s| + |\mu(t)| + |f^\Delta(t)|] \\ & = & \epsilon^*[2|\mu(t)| + |t - s| + |f^\Delta(t)|] \end{array} $$

Furthermore, being aware that \( |\mu(t)| = \mu(t) \) (because \( \mu(t) \) is a function with non negative image) and considering (without loss of generality) \( \epsilon^* < 1 \), it follows that

$$ \begin{array}{rcl} |f(t) - f(s)| & < & \epsilon^*[2|\mu(t)| + |t - s| + |f^\Delta(t)|] \\ & = & \epsilon^*[2\mu(t) + |t - s| + |f^\Delta(t)|] \\ & < & \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] \end{array} $$

Thus, since \( 2\mu(t) \geq 0 \), \( |t - s| \geq 0 \) and \( |f^\Delta(t)| \geq 0 \), we can let \( \epsilon = \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] \). Hence, we get

$$ |f(t) - f(s)| < \epsilon^*[2\mu(t) + 1 + |f^\Delta(t)|] = \epsilon $$

Therefore, \( f \) is continuous at \( t \).

References

• Martin Bohner and Allan Peterson: Dynamic Equations on Time Scales (2001)... (previous)... (next): Theorem $1.16 (i)$

Theorem: If $f$ is continuous at $t$ and $t$ is right-scattered, then $$f^{\Delta}(t) = \dfrac{f(\sigma(t))-f(t)}{\mu(t)}.$$

Proof: █

Theorem: If $t$ is right-dense, then (if it exists), $$f^{\Delta}(t) = \displaystyle\lim_{s \rightarrow t}\dfrac{f(t)-f(s)}{t-s}.$$

Proof: █

Theorem: If $f$ is differentiable at $t$, then $$f(\sigma(t))=f(t)+\mu(t)f^{\Delta}(t).$$

Proof: █

Theorem (Sum rule): $$(f+g)^{\Delta}(t)=f^{\Delta}(t)+g^{\Delta}(t).$$

Proof: █

Theorem (Constant rule): If $\alpha$ is constant with respect to $t$, then $$(\alpha f)^{\Delta}(t) = \alpha f^{\Delta}(t).$$

Proof: █

Theorem (Product rule,I): The following formula holds: $$(fg)^{\Delta}(t)=f^{\Delta}(t)g(t)+f(\sigma(t))g^{\Delta}(t)).$$

Proof: █

Theorem (Product rule,II): The following formula holds: $$(fg)^{\Delta}(t) = f(t)g^{\Delta}(t)+ f^{\Delta}(t)g(\sigma(t)).$$

Proof: █

Theorem (Quotient rule): The following formula holds: $$\left( \dfrac{f}{g} \right)^{\Delta}(t) = \dfrac{f^{\Delta}(t)g(t)-f(t)g^{\Delta}(t)}{g(t)g(\sigma(t))}.$$

Proof: █

Theorem

Let $\mathbb{T}$ be a time scale and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is $\Delta$-differentiable and $f^{\Delta}$ is rd continous on $\mathbb{T}^{\kappa}$, then $f$ is $\nabla$-differentiable on $\mathbb{T}_{\kappa}$ and $$f^{\nabla}(t) = f^{\Delta}(\rho(t)).$$

Theorem

Let $\mathbb{T}$ be a time scale and let $f \colon \mathbb{T} \rightarrow \mathbb{R}$. If $f$ is $\nabla$-differentiable on $\mathbb{T}_{\kappa}$ and $g^{\nabla}$ is ld continuous on $\mathbb{T}_{\kappa}$, then $f$ is $\Delta$-differentiable on $\mathbb{T}^{\kappa}$ and $$g^{\Delta}(t) = g^{\nabla}(\sigma(t)).$$

References

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