The Laplace transform on time scales revisited - ECS - Baylor ...

1298 J.M. Davis et al. / J. Math. Anal. Appl. 332 (2007) 1291–1307Apply θ to f(t)to ˜ retrieve the time scale functionn∑f(t)= Res z=zi e z (t, 0)F T (z),i=1whereby (γ ◦ L R ◦ θ −1 )(f (t)) = F T (z) as claimed.✷Before looking at a few examples, some remarks are in order. First, it is reasonable to ask ifthere is a contour in the complex plane around which it is possible to integrate to obtain the sameresults that we have obtained here through a more operational approach. At present, we leavethis as a very interesting although nontrivial open problem. It is well known that there are suchcontours when T = R or T = Z. In fact, it can easily be shown that if T is completely discrete,then if we choose any circle in the region of convergence which encloses all of the singularitiesof F(z), we will obtain the inversion formula. However, in general, we do not know whether ornot there exists a contour which gives the formula, and if so, what it is.Second, it is possible to use the technique we have presented here to define and find inversesfor any time scale <strong>transform</strong>. <strong>The</strong>se would include the Fourier, Mellin, and many other <strong>transform</strong>s.Once the inverse is known for T = R and the appropriate time scale integral is developedto give the correct <strong>transform</strong> analogues for any T, the diagram becomes completed and the inversionformula for any T is readily obtained.Finally, notice in our construction, for any <strong>transform</strong>able function f(t)on T, there is a shadowfunction f(t) ˜ defined on R. That is, to determine the appropriate time scale analogue of thefunction f(t) ˜ in terms of the <strong>transform</strong>, we use the diagram to map its <strong>Laplace</strong> <strong>transform</strong> on Rto its <strong>Laplace</strong> <strong>transform</strong> on T.Example 1.1 (New representation for h k (t, 0)). Suppose F(z)= 1 . <strong>The</strong> hypotheses of <strong>The</strong>orem1.4 for F(z)are readily verified, so that F(z)has an inverse given 2 byzL −1 e z (t, 0){F }=f(t)= Res z=0z 2 = t.For F(z)= 1 ,wehavez 3L −1 e z (t, 0){F }=f(t)= Res z=0z 3 = t2 − ∫ t0μ(τ) τ= h 2 (t, 0).2<strong>The</strong> last equality is justified since the functionf(t)= t2 − ∫ t0μ(τ) τ,2is the unique solution to the initial value problem f (t) = h 1 (t, 0), f(0) = 0. In a similar manner,we can use an induction argument coupled with <strong>The</strong>orem 1.4 to show that the inverse<strong>transform</strong> of F(z) = 1 ,fork a positive integer, is hz k+1 k (t, 0). In fact, it is easy to show thatthe inversion formula gives the claimed inverses for any of the elementary functions that Bohnerand Peterson have in their table in [3]. <strong>The</strong>se elementary functions become the proper time scaleanalogues or shadows of their continuous counterparts.Example 1.2. Suppose we would like to determine the appropriate shadow of the functionf(t)= ˜1 t sin at, a > 0,2a