Complex calculus

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Let $\mathbb{T}_1$ and $\mathbb{T}_2$ be time scales. We define a distance function on $\mathbb{T}_1 \times \mathbb{T}_2$ by the formula $$d((x_1,y_1),(x_2,y_2))=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$ We let $\mathbb{T}_1 + i \mathbb{T}_2 = \{(t_1,t_2) \colon t_1 \in \mathbb{T}_1, t_2 \in \mathbb{T}_2\}.$ For brevity, we use the notation $\mathscr{T}=\mathbb{T}_1 + i\mathbb{T}_2$ to denote the time scale complex plane.

Proposition:The function $d$ is a metric.

Proof: $\blacksquare$

Propoisition:The metric $d$ is a complete metric.

Proof: $\blacksquare$

We say that a function $f \colon \mathbb{T} \rightarrow \mathbb{C}$ is $\Delta$-differentiable (or $\Delta$-analytic) at $z_0=x_0+iy_0 \in \mathbb{T}$ if there exists some $A_0 \in \mathbb{C}$ (depending possibly on $z_0$) and some functions $\alpha(\cdot,z_0), \beta(\cdot,z_0), \gamma(\cdot,z_0)$ such that $$\left\{ \begin{array}{ll} f(z_0)-f(z)=A_0(z_0-z) + \alpha(z_0-z) \\ f(z_0^{\sigma_1})-f(z) = A_0(z_0^{\sigma_1}-z)+\beta(z_0^{\sigma_1}-z) \\ f(z_0^{\sigma_2})-f(z) = A_0(z_0^{\sigma_2}-z)+\gamma(z_0^{\sigma_2}-z) \end{array} \right.$$ for all $z$ in an $\epsilon$ $\mathbb{T}$-neighborhood of $z_0$, $\alpha(z_0,z_0)=0, \beta(z_0,z_0)=0$, and $\gamma(z_0,z_0)=0$. Also $$\displaystyle\lim_{z \rightarrow z_0} \alpha(z_0,z) = \displaystyle\lim_{z \rightarrow z_0} \beta(z_0,z) = \displaystyle\lim_{z \rightarrow z_0} \gamma(z_0,z).$$ We will typically write a function $f \colon \mathscr{T} \rightarrow \mathbb{C}$ as $f(z)=f(x+iy)=u(x,y)+iv(x,y)$.

Theorem: The function $f$ is $\Delta$-differentiable ($\Delta$-analytic) as a function of the complex variable $z$ if and only if the functions $u$ and $v$ are completely $\Delta$-differentiable and satisfy the Cauchy-Riemann equations $$\left\{\begin{array}{ll} \dfrac{\partial u}{\Delta_1 x} = \dfrac{\partial v}{\Delta_2 y} \\ \dfrac{\partial u}{\Delta_2y y} = -\dfrac{\partial v}{\Delta_1 x}. \end{array} \right.$$

Moreover if these equations are satisfied, then $f^{\Delta}(z_0)$ can be represented in any of the forms $$f^{\Delta}(z_0)=\dfrac{\partial u}{\Delta_1 x} + i \dfrac{\partial v}{\Delta_1 x} = \dfrac{\partial v}{\Delta_2 y} - i\dfrac{\partial u}{\Delta_2 y} =\dfrac{\partial u}{\Delta_1 x} - i \dfrac{\partial u}{\Delta_2 y}=\dfrac{\partial v}{\Delta_2 y}+ i \dfrac{\partial v}{\Delta_1 x},$$ where the partial derivatives are evaluated at $(x_0,y_0)$.

Proof: $\blacksquare$

Let $\mathbb{T},\mathbb{T}_1,$ and $\mathbb{T}_2$ be time scales and let $a,b \in \mathbb{T}$. A complex function $z = \lambda(t) = \phi(t)+i\psi(t)$, $t \in [a,b] \cap \mathbb{T}$ where $\phi \colon [a,b] \cap \mathbb{T} \rightarrow \mathbb{T}_1$ and $\psi \colon [a,b] \cap \mathbb{T} \rightarrow \mathbb{T}_2$ are continuous (in the induced subspace topology on the time scale) is called a curve in the time scale complex plane $\mathbb{T}_1 + i\mathbb{T}_2$. The points of the curve $\Gamma$ are the elements of the image of $[a,b] \cap \mathbb{T}$ under $\lambda$. The initial point of $\Gamma$ is the point $z_0=\lambda(a)$ and the final point of $\Gamma$ is $z_1=\lambda(b)$.

Let $P$ be a partition of $[a,b] \cap \mathbb{T}$. Define $$\ell(\Gamma,P) = \displaystyle\sum_{k=1}^n |z_k-z_{k-1}| \sqrt{[\phi(t_k)-\phi(t_{k-1})]^2+[\psi(t_k)-\psi(t_{k-1})]^2}.$$ The curve $\Gamma$ is said to be rectifiable with length $\ell(\Gamma)$ if $$\ell(\Gamma) = \sup \left\{ \ell(\Gamma,P) \colon P \mathrm{\hspace{2pt}is\hspace{2pt}a\hspace{2pt}partition\hspace{2pt}of\hspace{2pt}}[a,b] \right\} < \infty.$$

Theorem: Let $\phi$ and $\psi$ be continuous on $[a,b] \cap \mathbb{T}$ and $\Delta$-differentiable on $[a,b) \cap \mathbb{T}$. If $\phi^{\Delta}$ and $\psi^{\Delta}$ are bounded and $\Delta$-integrable over $[a,b] \cap \mathbb{T}$, then the curve $\Gamma$ is rectifiable and its length $\ell(\Gamma)$ can be evaluated by the formula $$\ell(\Gamma) = \displaystyle\int_a^b \left| \lambda^{\Delta}(t) \right| \Delta t = \displaystyle\int_a^b \sqrt{[\phi^{\Delta}(t)]^2 + [\psi^{\Delta}(t)]^2} \Delta t.$$

Proof:

Let $-\Gamma$ denote the curve $\Gamma$ traversed in the opposite direction.

Proposition $\displaystyle\int_{-\Gamma} f(z) \Delta z=-\displaystyle\int_{\Gamma} f(z) \Delta z$

Proof:

A $\Delta$-smooth curve $\Gamma$ is a curve that satisfies the hypothesis of the above theorem. Let $\Gamma$ be a $\Delta$-smooth curve and suppose that $$f(z)=u(x,y)+iv(x,y)$$ be a complex function defined and continuous on $\Gamma$. Let $P$ be a partition of $[a,b] \cap \mathbb{T}$ and $$z_k = \lambda(t_k) = \phi(t_k) + i \psi(t_k), k \in \{0,1,\ldots,n\}.$$ Let $\tau_k \in [t_{k-1},t_k) \cap \mathbb{T}$ for $k \in \{1,2,\ldots,n\}$ and set $$\xi_k = \lambda(\tau_k) = \phi(\tau_k) + i \psi(\tau_k), k \in \{1,2,\ldots,n\}.$$ Define the sum $$S_P = \displaystyle\int{k=1}^n f(\xi_k) (z_k-z_{k-1}).$$ We say that the complex number $I$ is the line $\Delta$-integral of $f$ along $\Gamma$ if for every $\epsilon > 0$ there is $\delta > 0$ such that $|S-I|<\epsilon$ for every $S_P$ where $P \in P_{\delta}([a,b])$ (the set of partitions of $[a,b]\cap\mathbb{T}$ where $t_k$ is within $\delta$ of $t_{k+1}$ if it is possible). We use the notation $$\displaystyle\int_{\Gamma} f(z) \Delta z$$ to represent the value $I$. We can also write $\displaystyle\lim_{\delta \rightarrow 0} = I$.

Proposition: Suppose that $\Gamma$ is comprised of subcurves $\Gamma_1,\ldots,\Gamma_n$ in the sense that the initial point of $\Gamma_{j+1}$ is the final point of $\Gamma_j$ for all $j=0,1,\ldots,n-1$. Then $$\displaystyle\int_{\Gamma} f(z) \Delta z = \displaystyle\sum_{k=1}^n \displaystyle\int_{\Gamma_k} f(z)\Delta z.$$

Proof:

References[edit]

<bibtex>
@inproceedings{MR2215851,
  title="An introduction to complex functions on products of two time scales",
  author="Bohner, Martin and Guseinov, Gusein Sh.",
  booktitle="Journal of Difference Equations and Applications 2006",
  doi="10.1080/10236190500489657",
  url="http://dx.doi.org/10.1080/10236190500489657"
}

</bibtex>