Difference between revisions of "Derivation of delta exponential T=hZ"

Revision as of 02:27, 27 July 2015

Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that $$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s}{h}} 1+hp(hk),$$ $$y(s+2h)=(1+hp(s+h))y(s+h)=(1+hp(s+h))(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+h}{h}-1} 1+hp(hk),$$ $$\vdots$$ $$y(s+nh)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+hn}{h}-1} 1+hp(hk),$$ so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} 1+hp(hk).$ Now rearrange the IVP to get $y(t) = \dfrac{y(\sigma(t))}{1+hp(t)}$, from which we see $$y(s-h)=\dfrac{y(s)}{1+hp(s-h)}=\displaystyle\prod_{k=\frac{s-h}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)},$$ $$y(s-2h)=\dfrac{y(s-h)}{1+hp(s-2h)} = \dfrac{1}{(1+hp(s-2h))(1+hp(s-h))}=\displaystyle\prod_{k=\frac{s-2h}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)},$$ $$\vdots$$ $$y(s-nh)=\displaystyle\prod_{k=\frac{s-nh}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)},$$ so that if $t=s-nh$ then $y(t)=\displaystyle\prod_{k=\frac{t}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)}.$ Hence we have derived $$e_p(t,s)=\left\{ \begin{array}{ll} \displaystyle\prod_{k=\frac{t}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)} &; t < s \\ 1 &; t=s \\ \displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} 1+hp(hk) &; t>s. \end{array} \right.$$