Question

State and prove that the convolution theorem for inverse laplace theorem

Answer 1

Let F ( s ) = L { f ( t ) } = ∫ + ∞ 0 e − s t d t f ( t ) and G ( s ) = L { g ( t ) } = ∫ + ∞ 0 e − s t g ( t ) d t F(s)=L{f(t)}=∫0+∞e−stdtf(t)andG(s)=L{g(t)}=∫0+∞e−stg(t)dt be the two Laplace transormations and let's denote ( f ⋆ g ) ( t ) = ∫ + ∞ 0 f ( τ ) g ( t − τ ) d τ (f⋆g)(t)=∫0+∞f(τ)g(t−τ)dτ the convolution of f , g f,g . Thus the Laplace transform for the convolution would be L { ( f ⋆ g ) ( t ) } = ∫ + ∞ 0 e − s t ( ∫ + ∞ 0 f ( τ ) g ( t − τ ) d τ ) d t = ∫ + ∞ 0 ∫ + ∞ 0 e − s t f ( τ ) g ( t − τ ) d τ d t ( 1 ) L{(f⋆g)(t)}=∫0+∞e−st(∫0+∞f(τ)g(t−τ)dτ)dt=∫0+∞∫0+∞e−stf(τ)g(t−τ)dτdt(1) The above double integral is to be evaluated in the domain D = { ( t , τ ) : t ∈ ( 0 , + ∞ ) , τ ∈ ( 0 , + ∞ ) } D={(t,τ):t∈(0,+∞),τ∈(0,+∞)} . Performing the change of variables ( t , τ ) → ( u , v ) : t = u + v , τ = v (t,τ)→(u,v):t=u+v,τ=v with u ∈ ( 0 , + ∞ ) , v ∈ ( 0 , + ∞ ) u∈(0,+∞),v∈(0,+∞) we have d τ d t = d u d v dτdt=dudv since the Jacobian equals to 1. Thus (1) becomes L { ( f ⋆ g ) ( t ) } = ∫ + ∞ 0 ∫ + ∞ 0 e − s ( u + v ) f ( u ) g ( v ) d u d v = ∫ + ∞ 0 e − s u f ( u ) d u ⋅ ∫ + ∞ 0 e − s v g ( v ) d v L{(f⋆g)(t)}=∫0+∞∫0+∞e−s(u+v)f(u)g(v)dudv=∫0+∞e−suf(u)du⋅∫0+∞e−svg(v)dv yielding to L { ( f ⋆ g ) ( t ) } = L { f ( t ) } ⋅ L { g ( t ) }