Difference between revisions of "Unilateral Laplace transform"
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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 $\beta > \alpha$. | <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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<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
Revision as of 14:53, 21 January 2023
Let $\mathbb{T}$ be a time scale and let $s \in \mathbb{T}$. If $f \in C_{rd}(\mathbb{T},\mathbb{C})$ ( rd-continuous) then we define the Laplace transform of $f$ about $s$ by the formula [pp.793] $$\mathscr{L}\{f\}(z;s) = \displaystyle\int_s^{\infty} f(t) e_{\ominus z}(\sigma(t),s) \Delta t,$$ where $z$ is in a domain $D \subset \mathbb{C}$ for which the integral converges. Let $\alpha$ be a non-negative 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 $s$.
Properties of Laplace Transforms
Proposition: The Laplace transform is linear, i.e. for constants $\alpha, \beta$ and Laplace-transformable functions $f,g$, [pp.795] $$\mathscr{L}\{\alpha f + \beta g\} = \alpha \mathscr{L}\{f\} + \beta \mathscr{L}\{g\}.$$
Proof: █
Proposition: The following formula holds: $$e_{\ominus z}(\sigma(t),s) = \dfrac{e_{\ominus z}(t,s)}{1+\mu(t)z} = -\dfrac{(\ominus z)(t)}{z} e_{\ominus z}(t,s).$$
Proposition: The Laplace transform of a delta derivative: $$\mathscr{L}\{f^{\Delta}\}(z;s) = -f(s) + z\mathscr{L}\{f\}(z).$$
Proof: Compute using integration by parts, $$\begin{array}{ll} \mathscr{L}\{f^{\Delta}\}(z) &= \displaystyle\int_0^{\infty} f^{\Delta}(\tau) e_{\ominus z}(\sigma(\tau),s) \Delta \tau \\ &= \end{array}$$ proving the claim. █
Assume there exist $M,\alpha > 0$ with $$|a_k| \leq M \alpha_k$$ for all $k=0,1,2,\ldots$. Then for all $z$ where it exists, [pp.796] $$\mathscr{L}\left\{ \displaystyle\sum_{k=0}^{\infty} a_k h_k(\cdot,s) \right\}(z;s) = \displaystyle\sum_{k=0}^{\infty} a_k \mathscr{L}\{h_k(\cdot,s)\}(z;s) = \displaystyle\sum_{k=0}^{\infty} \dfrac{a_k}{z^{k+1}},$$ where $h_k$ denotes the standard time scale polynomial.
Proposition: Let $m_z(t,s):=\displaystyle\int_s^t \dfrac{\Delta \tau}{1+\mu(\tau)z}$. Then [pp.797] $$\dfrac{d}{dz} \mathscr{L}\{f\}(z;s) = -\mathscr{L}\{m_z(\sigma(\cdot),s)f\}(z;s).$$
Proof: █
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.
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 $$\mathbb{C}_h(\lambda) = \left\{ z \in \mathbb{C}_h \colon \mathrm{Re}_h(z) > \lambda \right\}.$$
Theorem (Absolute convergence): Let $f \in C_{\mathrm{rd}}([s,\infty) \cap \mathbb{T},\mathbb{C})$ be of exponential order $\alpha$. Then $\mathscr{L}\{f\}(\cdot;s)$ exists on $\mathbb{C}_{\mu_*(s)}(\alpha)$ and converges absolutely.
Proof: █
Theorem (Uniform convergence): 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 $\beta > \alpha$.
Proof: █
Table of Laplace transforms
| $\mathbb{T}=$ | Unilateral Laplace transform |
| $\mathbb{R}$ | $\mathscr{L}_{\mathbb{R}}\{f\}(z;s)=\displaystyle\int_s^{\infty} f(\tau) e^{-z\tau} \mathrm{d}\tau$ |
| $\mathbb{Z}$ | |
| $h\mathbb{Z}$ | |
| $\mathbb{Z}^2$ | |
| $\overline{q^{\mathbb{Z}}}, q > 1$ | |
| $\overline{q^{\mathbb{Z}}}, q < 1$ | |
| $\mathbb{H}$ |
| $f(t;s)$ | $\mathscr{L}\{f(\cdot;s)\}(z)$ |
| $e_{\alpha}(t;s)$ | $\dfrac{1}{z-\alpha}$ |
| $h_n(t;s)$ | $\dfrac{1}{z^{n+1}}$ |
| $\sinh_{\alpha}(t;s)$ | $\dfrac{\alpha}{z^2-\alpha^2}$ |
| $\cosh_{\alpha}(t;s)$ | $\dfrac{z}{z^2-\alpha^2}$ |
| $\sin_{\alpha}(t;s)$ | $\dfrac{\alpha}{z^2+\alpha^2}$ |
| $\cos_{\alpha}(t;s)$ | $\dfrac{z}{z^2+\alpha^2}$ |
See also
Bilateral Laplace transform
Unilateral convolution
References
- Martin Bohner and Gusein Sh. Guseinov: The convolution on time scales (2007): (1.1)
- Martin Bohner and Başak Karpuz: The gamma function on time scales (2013)... (next): Section 3
