Difference between revisions of "Unilateral Laplace transform"

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Let $\mathbb{T}$ be a [[time_scale | time scale]] and let $s \in \mathbb{T}$. If $f \in C_{rd}(\mathbb{T},\mathbb{C})$ ([[continuity | rd-continuous]]) then we define the Laplace transform of $f$ about $s$ by the formula <ref>Bohner, Martin ; Guseinov, Gusein Sh. ; Karpuz, Başak . Properties of the Laplace transform on time scales with arbitrary graininess. Integral Transforms Spec. Funct. 22 (2011), no. 11, pp.793.</ref>
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__NOTOC__
$$\mathscr{L}\{f\}(z;s) = \displaystyle\int_s^{\infty} f(t) e_{\ominus z}(\sigma(t),s) \Delta t,$$
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If $\mathbb{T}$ is a [[time_scale | time scale]], $s \in \mathbb{T}$, and $f$ is [[rd-continuous]], then we define the unilateral Laplace transform of $f$ about $s$ by the formula
where $z$ is in a domain $D \subset \mathbb{C}$ for which the integral converges. Let $\alpha$ be a non-negative [[regressive_function | regressive]] constant larger than $s$. We use the notation "$f(t;s)$" to denote we are thinking of $f$ as a function of $t$ with parameter $s$.
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$$\mathscr{L}_{\mathbb{T}}\{f\}(z;s) = \displaystyle\int_s^{\infty} f(t) e_{\ominus z}(\sigma(t),s) \Delta t,$$
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for all $z$ for which the integral converges, where $\displaystyle\int \ldots \Delta t$ denotes the [[delta integral]], $e_{\ominus z}$ denotes a [[delta exponential]] whose subscript is the [[forward circle minus]] of the constant $z$, and $\sigma$ is the [[forward jump]].  
  
 +
=Properties of Laplace Transforms=
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[[Unilateral Laplace transform is a linear operator]]<br />
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[[Unilateral Laplace transform of delta derivative]]<br />
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 +
=Table of Laplace transforms=
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<center>
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{| class="wikitable"
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|+Formula for unilateral Laplace transform
 +
|-
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|$\mathbb{T}=$
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|Unilateral Laplace transform
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|-
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|[[Real_numbers | $\mathbb{R}$]]
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|$\mathscr{L}_{\mathbb{R}}\{f\}(z;s)=\displaystyle\int_s^{\infty} f(\tau) e^{-z\tau} \mathrm{d}\tau$
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|-
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|[[Integers | $\mathbb{Z}$]]
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|
 +
|-
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|[[Multiples_of_integers | $h\mathbb{Z}$]]
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|
 +
|-
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| [[Square_integers | $\mathbb{Z}^2$]]
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|
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|-
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|[[Quantum_q_greater_than_1 | $\overline{q^{\mathbb{Z}}}, q > 1$]]
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|
 +
|-
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|[[Quantum_q_less_than_1 | $\overline{q^{\mathbb{Z}}}, q < 1$]]
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|
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|-
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|[[Harmonic_numbers | $\mathbb{H}$]]
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|
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|}
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</center>
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<center>
 
{| class="wikitable"
 
{| class="wikitable"
|+Laplace Transformations
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|+Laplace Transforms of special functions
 
|-
 
|-
|Function $f(t;s)$
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|$f(t;s)$
|Laplace Transformation $\mathscr{L}\{f(\cdot;s)\}(z)$
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|$\mathscr{L}\{f(\cdot;s)\}(z)$
 
|-
 
|-
 
|$e_{\alpha}(t;s)$
 
|$e_{\alpha}(t;s)$
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|-
 
|-
 
|}
 
|}
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</center>
  
=Properties of Laplace Transforms=
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=See also=
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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[[Bilateral Laplace transform]]<br />
<strong>Proposition:</strong> The Laplace transform is linear, i.e. for constants $\alpha, \beta$ and Laplace-transformable functions $f,g$, <ref>Bohner, Martin ; Guseinov, Gusein Sh. ; Karpuz, Başak . Properties of the Laplace transform on time scales with arbitrary graininess. Integral Transforms Spec. Funct. 22 (2011), no. 11, pp.795.</ref>
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[[Unilateral convolution]]<br />
$$\mathscr{L}\{\alpha f + \beta g\} = \alpha \mathscr{L}\{f\} + \beta \mathscr{L}\{g\}.$$
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █
 
</div>
 
</div>
 
  
Assume there exist $M,\alpha > 0$ with
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=References=
$$|a_k| \leq M \alpha_k$$
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*{{PaperReference|The convolution on time scales|2007|Martin Bohner|author2=Gusein Sh. Guseinov|prev=|next=}}: (1.1)
for all $k=0,1,2,\ldots$. Then for all $z$ where it exists, <ref>Bohner, Martin ; Guseinov, Gusein Sh. ; Karpuz, Başak . Properties of the Laplace transform on time scales with arbitrary graininess. Integral Transforms Spec. Funct. 22 (2011), no. 11, pp.796.</ref>
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*{{PaperReference|The gamma function on time scales|2013|Martin Bohner|author2=Başak Karpuz|prev=|next=Gamma function}}: Section 3
$$\mathscr{L}\left\{ \displaystyle\sum_{k=0}^{\infty} a_k h_k(\cdot,s) \right\}(z;s) = \displaystyle\sum_{k=0}^{\infty} a_k \mathscr{L}\{h_k(\cdot,s)\}(z;s) = \displaystyle\sum_{k=0}^{\infty} \dfrac{a_k}{z^{k+1}},$$
 
where $h_k$ denotes the standard time scale [[Polynomials | polynomial]].
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Proposition:</strong> Let $m_z(t,s):=\displaystyle\int_s^t \dfrac{\Delta \tau}{1+\mu(\tau)z}$. Then <ref>Bohner, Martin ; Guseinov, Gusein Sh. ; Karpuz, Başak . Properties of the Laplace transform on time scales with arbitrary graininess. Integral Transforms Spec. Funct. 22 (2011), no. 11, pp.797.</ref>
 
$$\dfrac{d}{dz} \mathscr{L}\{f\}(z;s) = -\mathscr{L}\{m_z(\sigma(\cdot),s)f\}(z;s).$$
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █
 
</div>
 
</div>
 
  
It is known that $\dfrac{d}{dz} e_z(t,t_0) = m_z(t,t_0)e_z(t,t_0)$ and $\dfrac{d}{dz} e_{\ominus z}(t,t_0)=-m_z(t,t_0)e_{\ominus z}(t,t_0)$. These formulas are analogues of the formulas $\dfrac{d}{dz} e^{z(t-t_0)}=(t-t_0)e^{z(t-t_0)}$ and $\dfrac{d}{dz} e^{-z(t-t_0)}=-(t-t_0)e^{-z(t-t_0)}$ which occur in the case $\mathbb{T}=\mathbb{R}$. An important difference from the classical case is that $t-t_0$ has no dependence on the variable $z$, while $m_z(t,t_0)$ does.
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[[Category:Definition]]
 
 
==Convergence==
 
We define the minimal graininess function
 
$$\mu_*(s)=\inf_{\tau \in [s,\infty) \cap \mathbb{T}} \mu(\tau).$$
 
Let $h\geq 0$. We also define the Hilger real part of $z \in \mathbb{C}$ by
 
$$\mathrm{Re}_h(z)=\dfrac{1}{h}(|1+hz|-1)$$
 
and the Hilger imaginary part of $z \in \mathbb{C}$ by
 
$$\mathrm{Re}_h(z)=\mathrm{Arg}(1+hz),$$
 
where $\mathrm{Arg}$ denotes the principal argument of $1+hz$. We let
 
$$\mathbb{C}_h = \left\{ z \in \mathbb{C} \colon z \neq -\dfrac{1}{h} \right\}.$$
 
Finally given some $\lambda \in \mathbb{R}$ we define
 
$$\mathbb{C}_h(\lambda) = \left\{ z \in \mathbb{C}_h \colon \mathrm{Re}_h(z) > \lambda \right\}.$$
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem (Absolute convergence):</strong> Let $f \in C_{\mathrm{rd}}([s,\infty) \cap \mathbb{T},\mathbb{C})$ be of [[exponential_order | exponential order $\alpha$]]. Then $\mathscr{L}\{f\}(\cdot;s)$ exists on $\mathbb{C}_{\mu_*(s)}(\alpha)$ and converges absolutely.
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █
 
</div>
 
</div>
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem (Uniform convergence):</strong> Let $f \in C_{\mathrm{rd}}([s,\infty)\cap\mathbb{T},\mathbb{C})$ be of exponential order $\alpha$. Then the Laplace transform $\mathscr{L}\{f\}$ converges uniformly in the half-plane $C_{\mu_*(s)}(\beta)$ for any $\beta > \alpha$.
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █
 
</div>
 
</div>
 
 
 
=Inverse Transform on isolated time scales=
 
The following information is from <ref>Davis, John M. ;  Gravagne, Ian A. ;  Jackson, Billy J. ;  Marks, Robert J., II ;  Ramos, Alice A.  The Laplace transform on time scales revisited.
 
J. Math. Anal. Appl.  332  (2007),  no. 2, 1291--1307.</ref>. Define the notation
 
$$\mathrm{Re}_h(z) := \dfrac{1}{h} (|1+hz|-1).$$
 
Let $\mathbb{T}$ be a time scale with $0 &lt; \mu_{\mathrm{min}} \leq \mu(t) \leq \mu_{\mathrm{max}} &lt; \infty.$ Let $\mu_* := \mu_{\mathrm{min}}$ and $\mu^* := \mu_{\mathrm{max}}$. Define in $\mathbb{C}$ the Hilger circles
 
$$\mathbb{H}_{\mu(t)} = \left\{ z \in \mathbb{C} \colon 0 < \left| z + \dfrac{1}{\mu(t)} \right| < \dfrac{1}{\mu(t)} \right\},$$
 
which are disks centered at $\left(-\dfrac{1}{\mu(t)},0 \right)$ with radius $\dfrac{1}{\mu(t)}$.
 
[[File:Hilger_circles.png|thumb|250px|Here $H_{\mathrm{min}}=\mathbb{H}_{\mu^*}$ and $H_{\mathrm{max}}=\mathbb{H}_{\mu_*}$.]]
 
These circles define the region of convergence for the Laplace transform $\mathscr{L}_\mathbb{T}$. If $\mu_*=0$, then the region of convergence is the right half plane, but if $\mu_*>0$ the region of convergence is outside of $\mathbb{H}_{\mu_*}$.
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> Suppose that $F(z)$ is analytic in the region $\mathrm{Re}_{\mu}(z)>\mathrm{Re}_{\mu}(c)$ and $F(z) \rightarrow 0$ uniformly as $|z| \rightarrow \infty$ in this region. Suppose $F(z)$ has finitely many regressive poles of finite order $\{z_1,\ldots,z_n\}$ and $\tilde{F}_{\mathbb{R}}(z)$ is the transform of the function $\tilde{f}(t)$ on $\mathbb{R}$ that corresponds to the transform $F(z)=F_{\mathbb{T}}(z)$ of $f(t)$ on $\mathbb{T}$. If
 
$$\displaystyle\int_{c-i\infty}^{c+i\infty} |\tilde{F}_{\mathbb{R}}(z)||dz|<\infty,$$
 
then
 
$$f(t)=\displaystyle\sum_{i=1}^n \mathrm{Res}_{z=z_i} e_z(t,0) F(z)$$
 
has transform $F(z)$ for all $z$ with $\mathrm{Re}(z)>c$.
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █
 
</div>
 
</div>
 
 
 
=References=
 
<references />
 

Latest revision as of 15:21, 21 January 2023

If $\mathbb{T}$ is a time scale, $s \in \mathbb{T}$, and $f$ is rd-continuous, then we define the unilateral Laplace transform of $f$ about $s$ by the formula $$\mathscr{L}_{\mathbb{T}}\{f\}(z;s) = \displaystyle\int_s^{\infty} f(t) e_{\ominus z}(\sigma(t),s) \Delta t,$$ for all $z$ for which the integral converges, where $\displaystyle\int \ldots \Delta t$ denotes the delta integral, $e_{\ominus z}$ denotes a delta exponential whose subscript is the forward circle minus of the constant $z$, and $\sigma$ is the forward jump.

Properties of Laplace Transforms

Unilateral Laplace transform is a linear operator
Unilateral Laplace transform of delta derivative

Table of Laplace transforms

Formula for unilateral Laplace transform
$\mathbb{T}=$ Unilateral Laplace transform
$\mathbb{R}$ $\mathscr{L}_{\mathbb{R}}\{f\}(z;s)=\displaystyle\int_s^{\infty} f(\tau) e^{-z\tau} \mathrm{d}\tau$
$\mathbb{Z}$
$h\mathbb{Z}$
$\mathbb{Z}^2$
$\overline{q^{\mathbb{Z}}}, q > 1$
$\overline{q^{\mathbb{Z}}}, q < 1$
$\mathbb{H}$
Laplace Transforms of special functions
$f(t;s)$ $\mathscr{L}\{f(\cdot;s)\}(z)$
$e_{\alpha}(t;s)$ $\dfrac{1}{z-\alpha}$
$h_n(t;s)$ $\dfrac{1}{z^{n+1}}$
$\sinh_{\alpha}(t;s)$ $\dfrac{\alpha}{z^2-\alpha^2}$
$\cosh_{\alpha}(t;s)$ $\dfrac{z}{z^2-\alpha^2}$
$\sin_{\alpha}(t;s)$ $\dfrac{\alpha}{z^2+\alpha^2}$
$\cos_{\alpha}(t;s)$ $\dfrac{z}{z^2+\alpha^2}$

See also

Bilateral Laplace transform
Unilateral convolution

References