Difference between revisions of "Delta derivative"
(→Properties of the $\Delta$-derivativeBohner, Martin ; Peterson, Allan. Dynamic equations on time scales. An introduction with applications. Birkhäuser Boston, Inc., Boston, MA, 2001,p.8.) |
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| − | This function is clearly not continuous at $t=1$ and hence it is not $\Delta$-differentiable at $t=1$. | + | This function is clearly not continuous at $t=1$ and hence by [[Delta derivative implies continuous|this theorem]] it is not $\Delta$-differentiable at $t=1$. |
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== References == | == References == | ||
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Revision as of 18:47, 21 March 2015
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 $f^{\Delta}(t)$.
Contents
Properties of the $\Delta$-derivative<ref>Bohner, Martin ; Peterson, Allan. Dynamic equations on time scales. An introduction with applications. Birkhäuser Boston, Inc., Boston, MA, 2001,p.8.</ref>
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: █
Interesting Examples
- The jump operator $\sigma$ is not $\Delta$-differentiable on all time scales. Consider $\mathbb{T}=[0,1] \bigcup \{2,3,4,\ldots\}$, then we see
$$\sigma(t) = \left\{ \begin{array}{ll} 0 &; t \in [0,1) \\ 1 &; t \in \{1,2,3,\ldots\}. \end{array}\right.$$ This function is clearly not continuous at $t=1$ and hence by this theorem it is not $\Delta$-differentiable at $t=1$.
References
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