Difference between revisions of "Unilateral Laplace transform"

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__NOTOC__
 
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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 <sup>[[#potltots|[pp.793]]]</sup>
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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
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$$\mathscr{L}\{f\}(z;s) = \displaystyle\int_s^{\infty} f(t) e_{\ominus z}(\sigma(t),s) \Delta t,$$
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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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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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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=
 
=Properties of Laplace Transforms=
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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[[Unilateral Laplace transform is a linear operator]]<br />
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<strong>Proposition:</strong> The Laplace transform is linear, i.e. for constants $\alpha, \beta$ and Laplace-transformable functions $f,g$, <sup>[[#potltots|[pp.795]]]</sup>
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[[Unilateral Laplace transform of delta derivative]]<br />
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$$\mathscr{L}\{\alpha f + \beta g\} = \alpha \mathscr{L}\{f\} + \beta \mathscr{L}\{g\}.$$
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> █
 
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</div>
 
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</div>
 
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
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<strong>Proposition:</strong> The following formula holds:
 
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$$e_{\ominus z}(\sigma(t),s) = \dfrac{e_{\ominus z}(t,s)}{1+\mu(t)z} = -\dfrac{(\ominus z)(t)}{z} e_{\ominus z}(t,s).$$
 
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<div class="mw-collapsible-content">
 
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</div>
 
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</div>
 
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
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<strong>Proposition:</strong> The Laplace transform of a [[delta derivative]]:
 
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$$\mathscr{L}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z).$$
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> Compute using integration by parts,
 
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$$\begin{array}{ll}
 
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\mathscr{L}\{f^{\Delta}\}(z) &= \displaystyle\int_0^{\infty} f^{\Delta}(\tau) e_{\ominus z}(\sigma(\tau),s) \Delta \tau \\
 
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&=
 
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\end{array}$$
 
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proving the claim. █
 
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</div>
 
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</div>
 
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Assume there exist $M,\alpha > 0$ with
 
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$$|a_k| \leq M \alpha_k$$
 
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for all $k=0,1,2,\ldots$. Then for all $z$ where it exists, <sup>[[#potltots|[pp.796]]]</sup>
 
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$$\mathscr{L}\left\{ \displaystyle\sum_{k=0}^{\infty} a_k h_k(\cdot,s) \right\}(z;s) = \displaystyle\sum_{k=0}^{\infty} a_k \mathscr{L}\{h_k(\cdot,s)\}(z;s) = \displaystyle\sum_{k=0}^{\infty} \dfrac{a_k}{z^{k+1}},$$
 
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where $h_k$ denotes the standard time scale [[Polynomials | polynomial]].
 
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
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<strong>Proposition:</strong> Let $m_z(t,s):=\displaystyle\int_s^t \dfrac{\Delta \tau}{1+\mu(\tau)z}$. Then <sup>[[#potltots|[pp.797]]]</sup>
 
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$$\dfrac{d}{dz} \mathscr{L}\{f\}(z;s) = -\mathscr{L}\{m_z(\sigma(\cdot),s)f\}(z;s).$$
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> █
 
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</div>
 
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</div>
 
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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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==Convergence==
 
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We define the minimal graininess function
 
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$$\mu_*(s)=\inf_{\tau \in [s,\infty) \cap \mathbb{T}} \mu(\tau).$$
 
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Let $h\geq 0$. We also define the Hilger real part of $z \in \mathbb{C}$ by
 
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$$\mathrm{Re}_h(z)=\dfrac{1}{h}(|1+hz|-1)$$
 
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and the Hilger imaginary part of $z \in \mathbb{C}$ by
 
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$$\mathrm{Re}_h(z)=\mathrm{Arg}(1+hz),$$
 
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where $\mathrm{Arg}$ denotes the principal argument of $1+hz$. We let
 
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$$\mathbb{C}_h = \left\{ z \in \mathbb{C} \colon z \neq -\dfrac{1}{h} \right\}.$$
 
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Finally given some $\lambda \in \mathbb{R}$ we define
 
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$$\mathbb{C}_h(\lambda) = \left\{ z \in \mathbb{C}_h \colon \mathrm{Re}_h(z) > \lambda \right\}.$$
 
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
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<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.
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> █
 
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</div>
 
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</div>
 
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
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<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 $\beta > \alpha$.
 
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<div class="mw-collapsible-content">
 
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<strong>Proof:</strong> █
 
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</div>
 
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</div>
 
  
 
=Table of Laplace transforms=
 
=Table of Laplace transforms=
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=References=
 
=References=
 
*{{PaperReference|The convolution on time scales|2007|Martin Bohner|author2=Gusein Sh. Guseinov|prev=|next=}}: (1.1)
 
*{{PaperReference|The convolution on time scales|2007|Martin Bohner|author2=Gusein Sh. Guseinov|prev=|next=}}: (1.1)
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* {{PaperReference|The gamma function on time scales|2013|Martin Bohner|author2=Başak Karpuz|prev=|next=Gamma function}}: Section 3
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*{{PaperReference|The gamma function on time scales|2013|Martin Bohner|author2=Başak Karpuz|prev=|next=Gamma function}}: Section 3
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[[Category:Definition]]

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

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