Unilateral Laplace transform
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 $\displaystyle\int$ 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
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
