Mathematical Methods for Physics and Engineering - Matematica.NET

13.2 LAPLACE TRANSFORMSWe may now consider the effect of multiplying the Laplace transform ¯f(s) bye −bs (b>0). From the definition (13.53),∫ ∞e −bs ¯f(s) = e −s(t+b) f(t) dt=0∫ ∞0e −sz f(z − b) dz,on putting t + b = z. Thus e −bs ¯f(s) is the Laplace transform of a function g(t)defined by{0 for 0 b.In other words, the function f has been translated to ‘later’ t (larger values of t)by an amount b.Further properties of Laplace transforms can be proved in similar ways andare listed below.(i)L [f(at)] = 1 a ¯f( s, (13.61)a)(ii)(iii)L [t n n dn ¯f(s)f(t)] =(−1)ds n , for n =1, 2, 3,..., (13.62)[ ] ∫ f(t) ∞L = ¯f(u) du, (13.63)t sprovided lim t→0 [f(t)/t] exists.Related results may be easily proved.◮Find an expression for the Laplace transform of td 2 f/dt 2 .From the definition of the Laplace transform we have[ ] ∫L t d2 f ∞= e −st t d2 fdt 2 dt dt 20∫ ∞= − d e −st d2 fds 0 dt dt 2= − d ds [s2 ¯f(s) − sf(0) − f ′ (0)]= −s 2 d ¯fds − 2s ¯f + f(0). ◭Finally we mention the convolution theorem for Laplace transforms (which isanalogous to that for Fourier transforms discussed in subsection 13.1.7). If thefunctions f and g have Laplace transforms ¯f(s) andḡ(s) then[∫ t]L f(u)g(t − u) du = ¯f(s)ḡ(s), (13.64)0457