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

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1302 J.M. Davis et al. / J. Math. Anal. Appl. 332 (2007) 1291–1307====∫ ∞0∫ ∞0∫ ∞0∫ ∞0f(τ)[ ∞∫σ(τ)g ( t,σ(τ) ) e σ ⊖z (t, 0)t ]τf(τ)L { u σ(τ) (t)g ( t,σ(τ) )} τf(τ) [ G(z)e σ ⊖z (τ, 0)] τf(τ)e⊖z σ (τ, 0)τG(z)= F (z)G(z).For the convergence of the integral, note that for f and g of exponential type II with constantsc f and c g , respectively, we have∣∣(f ∗ g)(t) ∣ ∫ t=f(τ)g ( t,σ(τ) ) τ∣∣0∫ t0∣∣f(τ) ∣ · ∣∣g ( t,σ(τ) )∣ ∣τ∫ t Me cg (t, 0)0∫ t Me cg (t, 0)0e cf (τ, 0)e cg(0,σ(τ))τe cf (τ, 0)e ⊖cg (t, 0)τM|c f − c g | e c g(t, 0) ( e cf ⊖c g(t, 0) − 1 )M (ecf (t, 0) + e cg (t, 0) )|c f − c g |2M|c f − c g | e ĉ(t, 0),so that f ∗ g is of exponential type II with constant ĉ.Example 2.1. Suppose f(t)= f(t,0) is one of the elementary functions: that is, f(t)is one ofh k (t, 0), e a (t, 0), sin a (t, 0), cos a (t, 0), cosh a (t, 0), or sinh a (t, 0). Direct calculations show thatthe delay f(t,σ(τ))of each of these functions is given by h k (t, σ (τ)), e a (t, σ (τ)),sin a (t, σ (τ)),cos a (t, σ (τ)) cosh a (t, σ (τ)), and sinh a (t, σ (τ)). Thus, convolving any two of these functionswill give the results obtained by Bohner and Peterson in [3].✷
J.M. Davis et al. / J. Math. Anal. Appl. 332 (2007) 1291–1307 1303Example 2.2. Now suppose g(t,σ(τ)) = 1. In this instance, we see that for any f ∈C prd-e2 (T, R), the <strong>transform</strong> of h(t) = ∫ t0f(τ)τ is given byL{h}=L{f ∗ 1}=F(z)L{1}= F(z)z ,another result obtained by direct calculation in Bohner and Peterson [3].Example 2.3. Letf(t)= 1 ∫ t2α sin α(t, 0)011 + (μ(τ)α) 2 τ − 1 ∫ t2 cos α(t, 0)0μ(τ)1 + (μ(τ)α) 2 τ,for α>0, and g(t) = e β (t, 0). We wish to compute (f ∗ g)(t). By the Convolution <strong>The</strong>orem, theconvolution product has <strong>transform</strong> F (z)G(z). By Example 1.2,HenceF(z)=z(z 2 + α 2 ) 2 .(f ∗ g)(t) = L −1{ F (z)G(z) }=3∑i=1Res z=zize z (t, 0)(z 2 + α 2 ) 2 (z − β)1=2α 2 (α 2 + β 2 )(α 4 + β 2 )( ∫t)(−βμ+ 1)(α 2 )(α 4 + β 2 )×1 + (μα) 2 τ + (α − 1)β 3 + 2α 3 β cos α (t, 0)0+( ∫t12α(α 2 + β 2 )(α 4 + β 2 )0+ α 4 + ( −α 2 + α + 1 ) β 2 )sin α (t, 0) +β(α 4 + β 2 ) + α 2 (α 4 + β 2 )μ1 + (μα) 2 τβ(α 2 + β 2 ) 2 e β(t, 0).It is worth noting that the convolution product is both commutative and associative. Indeed,the products f ∗ g and g ∗ f have the same <strong>transform</strong> as do the products f ∗ (g ∗ h) and(f ∗ g) ∗ h, and since the inverse is unique, the functions defined by these products must agreealmost everywhere.At first glance, one may think that the identity is vested in the Hilger delta. Unfortunately, thisis not the case. It can be easily shown that any identity for the convolution will of necessity have<strong>transform</strong> 1 by the Convolution <strong>The</strong>orem. But the Hilger delta does not have <strong>transform</strong> 1 sinceits <strong>transform</strong> is just the exponential. Thus, we develop an analogue of the Dirac delta in the nextsection in order to establish an identity element.
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