Real Inversion Formulas of the Laplace Transform on Weighted - PDF document

(1) Real Inversion Formulas of the Laplace Transform on Weighted Function Spaces Hiroshi Fujiwara, Saitoh Saburoh and Yoshihiro Sawano August, 2008

(2) 1 Introduction The aim of this talk is to consider a numerical inver- sion formula of the Laplace transform ∫ ∞ e − p t f ( t ) dt. L f ( p ) = 0

(3) 2 Our new approach We define ∫ ∞ f ( t ) e − p t dt. Lf ( p ) := p 0 This operator is rather easy to deal with. Indeed, assuming some integrability conditions and smooth- ness, we obtain ∫ ∞ f ′ ( t ) e − p t dt. Lf ( p ) = 0

(4) 2.1 Model case Here we shall present our model case. First we define } 1 {∫ ∞ | f ′ ( t ) | 2 e t 2 ∥ f : H K ∥ := t dt 0 when f : R → R is absolutely continuous. Below we postulate the function on normalization conditions. Namely, we assume f (0) = 0 .

(5) With this in mind, we define H K to be the set of all absolutely continuous funtions f : R → R for which f (0) = 0 and ∥ f : H K ∥ < ∞ .

(6) We formulate our main theorem. Theorem 1 The mapping L : f ∈ H K �→ Lf ∈ L 2 ([0 , ∞ )) is an injective and compact linear operator.

(7) The injectivity of L is clear because L itself is injec- tive. As for the compactness, we argue by approxi- mation. We define ∫ ∞ e − p t f ( t ) dt. L R f ( p ) := p R

(8) Note that ∫ R e − p t f ( t ) dt Lf ( p ) − L R f ( p ) = p 0 and that ∫ R e − p t f ( t ) dt f ∈ H K → p 0 is compact.

(9) Therefore, to establish that L itself is compact, it suffices to show that L R shrinks to 0 in the operator topology as R → ∞ .

(10) We proceed as follows : An integration by parts yields ∫ ∞ e − p t f ′ ( t ) dt + e − R p f ( R ) . L R f ( p ) = R Hence it follows that ∫ ∞ ∥ e − p t ∥ L 2 ([0 , ∞ ) p ) | f ′ ( t ) | dt ∥ L R f ∥ 2 ≤ R + ∥ e − R p ∥ L 2 ([0 , ∞ ) p ) · | f ( R ) | ∫ ∞ | f ′ ( t ) | dt + | f ( R ) | = t R R by the triangle inequality.

(11) Equality ∫ R f ′ ( t ) dt f ( R ) = 0 yields ∫ ∞ | f ′ ( t ) | dt ∥ L R f ∥ 2 ≤ max( t, R ) 0 √∫ ∞ ∫ ∞ | f ′ ( t ) | 2 e t dt t e − t dt ≤ max( t, R ) 2 t 0 0 √∫ ∞ t e − t dt ≤ ∥ f : H K ∥ max( t, R ) 2 . 0

(12) Therefore, it follows that √∫ ∞ t e − t dt ∥ L R ∥ H K → L 2 ([0 , ∞ )) ≤ max( t, R ) 2 . 0 As a consequence we obtain L R shrinks to 0 in the norm topology of L 2 .

(13) Now that L is established to be a compact operator, we can take { v n } n ∈ N and { u n } n ∈ N such that Lv n = λ n u n , L ∗ u n = λ n v n , where λ n is a singular value of L and we assume that λ 1 ≥ λ 2 ≥ · · · ≥ 0 .

(14) Under this notation, we establish the following : Suppose that F ∈ L 2 is a function Theorem 2 that can be written as F = L f . Then we have (∫ ∞ ∞ 1 ) ∑ L − 1 F ( t ) = F ( p ) p u n ( p ) dp v n ( t ) . λ n 0 n =1

(15) We prove our main theorem by calculating the right- hand side. (∫ ∞ ∞ 1 ) ∑ F ( p ) p u n ( p ) dp v n ( t ) λ n 0 n =1 (∫ ∞ ∞ 1 ) ∑ = L f ( p ) p u n ( p ) dp v n ( t ) λ n 0 n =1 (∫ ∞ ∞ 1 ) ∑ = Lf ( p ) u n ( p ) dp v n ( t ) λ n 0 n =1

(16) ∞ 1 ∑ = ⟨ Lf, u n ⟩ L 2 ([0 , ∞ )) v n ( t ) λ n n =1 ∞ 1 ∑ ⟨ f, L ∗ u n ⟩ H K v n ( t ) = λ n n =1 ∞ ∑ ⟨ f, v n ⟩ H K v n ( t ) = f ( t ) = L − 1 F ( t ) . = n =1 This is the desired result.

(17) We shall consider the original question. Lf = F, F ∈ L 2 ([0 , ∞ )) . In general, as we have seen before, this problem does not admit a solution unless we assume that F ∈ L ( H K ) .

(18) The Tikhonov regularization is a method to over- come this difficulty. Instead of considering the above quesion directly, we consider α ∥ f : H K ∥ 2 + ∥ Lf − F : L 2 ([0 , ∞ )) ∥ 2 ) ( min , f ∈ H K where α is the smoothing parameter.

(19) It is not so hard to solve the minimizing problem α ∥ f : H K ∥ 2 + ∥ Lf − F : L 2 ([0 , ∞ )) ∥ 2 ) ( min f ∈ H K because we can complete the square.

(20) After completing the square, we see that f α = ( α + L ∗ L ) − 1 L ∗ [ p · F ] is the desired element.

(21) If we expand f α = ( α + L ∗ L ) − 1 L ∗ F, then we obtain ∞ 1 ∑ α + λ n 2 ⟨ L ∗ [ p · F ] , v n ⟩ H K v n f α = j =1 ∞ λ n ∑ = α + λ n 2 ⟨ p · F, u n ⟩ L 2 ([0 , ∞ )) v n j =1 ∞ λ n (∫ ) ∑ = p · F ( p ) u n ( p ) dp v n . α + λ n 2 j =1

(22) To realize our scheme numerically, we use ∫ ∞ N ∑ f ( x ) dx ≃ f ( x i ) w i , 0 i =0 where { x i } N i =1 and { w i } N i =1 is a collection of points and a correction of weights respectively.

(23) Finally, to conclude our first part of the talk, we present two examples. Example 1  t 0 ≤ t < 1     f ( t ) := 3 / 2 − t/ 2 1 ≤ t < 3   0 otherwise   is the first example of our result.

(24) This is our numerical result.

(25) The second example is f ( t ) = δ ( t − 1) . Observe that f does not belong to H K . However, Lf ( p ) = p e − p does belong to L 2 . So we are in the position of applying our result of the Tikhonov regularization.

(26) This is our numerical result.

(27) ラプラス変換の実逆変換への再生核空間の応用 Hiroshi Fujiwara, Naotaka Kajino and Yoshihiro Sawano August, 2008

(28) Based on the previous talk, we consider the above result in the generalized framework.

(29) Below we shall consider weighted function space L 2 ([0 , ∞ ) , u ) whose norm is given by (∫ ∞ ) 1 2 ∥ f : L 2 ([0 , ∞ ) , u ) ∥ := | f ( t ) | 2 u ( t ) dt . 0

(30) We define H 0 K ( w ) f ∈ C 1 ([0 , ∞ )) : f ′ ∈ L 2 ( w − 1 ) , f (0) = 0 { } := and H K ( w ) := the completion of H 0 K ( w ) .

(31) We want to consider the following problem. What is the condition for L to be compact ? L : H K ( w ) → L 2 ( u ) . Here L is given by ∫ ∞ f ( t ) e − p t dt. Lf ( p ) := p 0

(32) In this problem, it is tacitly included that we need to give a sufficient condition for Lf, f ∈ H K ( w ) to make sense.

(33) A closer look at the proof, which we gave in our earlier paper, makes us notice that the quantity M is a key quantity, where ∫ ∞ ∫ ∞ e − t p u ( p ) w ( t ) dp dt < ∞ . M := 0 0

(34) If M < ∞ , then we proved that L : H K ( w ) → L 2 ( u ) is a Hilbert-Schmidt class operator and that the Hilbert- Schmit norm is M . Under the assumption M < ∞ , we see that L is well-defined on H K ( w ) .

(35) To show that L : H K ( w ) → L 2 ( u ) belongs to the Hilbert-Schmidt class, it suffices to show that LL ∗ : L 2 ( u ) → L 2 ( u ) is a trace class operator.

(36) First, we obtain L ∗ g ( t ) = ⟨ L ∗ g, K w ( · , t ) ⟩ H K ( w ) = ⟨ g, L [ K w ( · , t )] ⟩ L 2 ( u ) ∫ ∞ = g ( p ) L [ K w ( · , t )]( p ) u ( p ) dp 0 ∫ ∞ (∫ t ) = g ( p ) exp( − s p ) w ( s ) dp u ( p ) dt. 0 0

(37) Hence it follows that LL ∗ g ( p ) = L [( L ∗ g ) ′ ]( p ) ∫ ∞ ∫ ∞ = g ( q ) exp( − t ( p + q )) w ( t ) u ( p ) dq dt. 0 0

(38) Let ϕ j , j = 1 , 2 , · · · be eigenvectors of LL ∗ . Then the integral kernel can be written as ∫ ∞ ∞ ∑ λ j ϕ j ( p ) ϕ j ( q ) = w ( t ) exp( − t ( p + q )) dt. 0 j =1

(39) As a result we obtain ∫ ∞ ∞ ∞ ∑ ∑ ϕ j ( p ) 2 u ( p ) dp λ j = λ j 0 j =1 j =1 ∫ ∞ ∫ ∞ = w ( t ) u ( p ) exp( − 2 p t ) dp dt. 0 0