ch7 - Chapter 7: Fourier transform 7.1. Preliminaries The...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 7: Fourier transform 7.1. Preliminaries The goal of this chapter is to provide tools on how to solve some PDE using the Fourier transform. The idea is the following: the Fourier transform allows us to replace a given PDE by a simplier, merely algebraic problem. After solving that problem, the Inverse Fourier transform gives us the solution of the original PDE problem. In contrast to the previous chapters, we will deal now with functions f : IR IC which are Riemann integrable on bounded intervals of IR. We will say that | f | p is Riemann integrable on IR - in short | f | p ∈ R (IR), ( p 1) - if Z IR | f ( x ) | p dx < . In the case p = 2 we work with the following interior product and norm: h f, g i := Z R f ( x ) g ( x ) dx, k f k := q h f, f i . A certain rˆ o le, both in proofs and in applications of our results, will play the convolution of two functions f, g which is defined as ( f * g )( x ) := Z IR f ( x - y ) g ( y ) dy, provided that the integral exists. Notice that this is the case, if both | f | 2 and | g | 2 are Riemann integrable on IR, by the Cauchy-Schwarz inequality, or if f is bounded and | g | ∈ R (IR). For instance, if f is piecewise continuous and g is bounded and vanishes on a finite interval, then f * g ( x ) exists for all x . The convolution obeys the following operations, ( f, g, h : IR IC, a, b IR): f * ( ag + bh ) = a ( f * g ) + b ( f * h ) , (1) f * g = g * h, (2) f * ( g * h ) = ( f * g ) * h, and (3) ( f * g ) 0 = f 0 * g, if f is differentiable . (4) 59
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The following result is basic. We omit its proof. Theorem 7.1. Let f : IR IC be Riemann integrable on each bounded interval, and assume that g C ( IR ) has the following properties: g ( x ) = g ( - x ) 0 x IR , (5) + Z -∞ g ( x ) dx = 1 . (6) For each ε > 0 , define g ε ( x ) := 1 ε g ± 1 ε x ² . Then f * g ε is continuous ε > 0 , and lim ε 0 ( f * g ε )( x ) = 1 2 ( f ( x - ) + f ( x +)) x IR . (7) Moreover, if | f | 2 ∈ R ( IR ) , we also have | f * g ε | 2 ∈ R ( IR ) , and f * g ε converges to f in norm. Finally, if g C k ( IR ) , ( k 1 ), then also f * g ε C k ( IR ) . 7.2. Fourier transform If | f | ∈ R (IR), its Fourier transform is the function ˆ f on IR defined by ˆ f ( ξ ) := Z IR e - iξx f ( x ) dx. (8) We will also use the following notation: F [ f ( x )] = ˆ f ( ξ ) . Since e - iξx has absolute value 1, we have, | ˆ f ( ξ ) | ≤ Z IR | f ( x ) | dx. (9) Theorem 7.2. Suppose | f | ∈ R ( IR ) . Then ( a ) F [ f ( x - a )] = e - iaξ ˆ f ( ξ ) and F [ e iax f ( x )] = ˆ f ( ξ - a ) a IR; ( b ) if δ > 0 and f δ ( x ) := (1 ) f ( x/δ ) then d ( f δ )( ξ ) = ˆ f ( δξ ) and F [ f ( δx )] = ³ ˆ f ´ δ ( ξ ); 60
Background image of page 2
( c ) if f is continuous and piecewise smooth, and | f 0 | ∈ R (IR), then F [ f 0 ( x )] = ˆ f ( ξ ); if f 0 is continuous and f 0 is piecewise smooth, and | f 0 | , | f 00 | ∈ R (IR), then F [ f 00 ( x )] = - ξ 2 ˆ f ( ξ ); if | xf ( x ) | ∈ R
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 11

ch7 - Chapter 7: Fourier transform 7.1. Preliminaries The...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online