inverse_lecture15

# inverse_lecture15 - 1 Lecture 15 From last time Z 1 K s t f...

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Unformatted text preview: 1 Lecture 15 From last time: Z 1 K ( s, t ) f ( t ) dt = g ( s ) ≤ s ≤ 1 (1) Given the SVE { s i , u i , v i } for the kernel K the solution to the IP is f ( t ) = ∞ X i =1 ( u i , g ) s i v i ( t ) For the existence of a solution to (1), g must satisfy ∞ X i =1 ( u i , g ) s i 2 < ∞ This is known as the Picard condition . This is a stronger requirement than g ∈ L 2 (0 , 1) since for g ∈ L 2 (0 , 1) we require ( u i , g ) decay faster than i- 1 / 2 , but here we require they decay faster than s i i- 1 / 2 . Note: The Picard condition is equivalent to the requirement g ∈ R ( K ). (Recall from last time ∑ ∞ i =1 s i ( v i , f ) u i ( s ) = ∑ ∞ i =1 ( u i , g ) u i ( s ). SO the right hand side is a projection of g onto span( { u 1 , . . . } ). If g has any arbitrary small component outside R ( K ), the IP has no L 2 solution. Suppose g 6∈ R ( K ). Let g k denote the approximation to g obtained by truncating its SVE after k terms: g k ( s ) = k X i =1 ( u i , g ) u i ( s ) Clearly, g k satisfies the Picard conditon for all k , and the corresponding solution is given by f k ( t ) = k X i =1 ( u...
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## This note was uploaded on 03/14/2011 for the course MATH 676 taught by Professor Staff during the Fall '08 term at Colorado State.

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inverse_lecture15 - 1 Lecture 15 From last time Z 1 K s t f...

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