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Unformatted text preview: Introduction to Nonlinear dispersive wave equations: Linares and Ponce Sudarshan Balakrishnan August 8, 2011 Chapter 1 Exercise 1.1 Let n 1 and f ( x ) = e 2  x  . Show that: f ( ) = [( n + 1) / 2] ( n +1) / 2 (1 +   2 ) ( n +1) / 2 Proof In order to prove this statement, we have that by example 1.5 Z  cos( ax ) dx x 2 + b 2 = b e ab . (1) For a = and b = 1, we notice that (1) becomes: e = 2 Z cos( x ) 1 + x 2 dx. (2) Combining (2) with 1 1 + x 2 = Z e (1+ x ) 2 d, (3) we get, e = Z e e 2 4 d. (4) The fourier transform of f is defined by, 1 f ( ) = Z f ( x ) e 2 ix . (5) Substituting f ( x ) in (5) we get: f ( ) = Z e 2  x  e 2 ix dx = Z e 2  x  2 ix dx. (6) The idea is to use the identity in (6) to obtain the desired result. But first, the Gamma function is defined by, ( n ) = Z y n 1 e x dx. (7) In (6), using identity (4) we get: Z Z  1 2 e e (2  x  +2 ix ) dxd (8) Integrating (8) by parts we get: Z Z  1 2 e e (2  x  +2 ix ) dxd = [( n + 1) / 2] ( n +1) / 2 (1 +   2 ) ( n +1) / 2 (9) The claim follows. Exercise 1.3 Prove Youngs inequality: Let f L p ( R n ), 1 p , and g L 1 ( R n ). Then f * g L p ( R n ) with: k f * g k k f k p k g k 1 . Proof I will try to prove a stronger version of Youngs inequality, the exercise follows almost immediately. For functions, f L p ( R n ) and g L 1 ( R n ) let us prove for p,q,r [1 , ] k f * g k r k f k p k g k q . (10) In order to see this, we need to use Holders Inequality . Holders inequality states that for p,q [1 , ] such that 1 q + 1 p = 1 we have: 2 k fg k 1 k f k p k g k q . (11) Using (11), we have :...
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This note was uploaded on 11/19/2011 for the course MATH 101 taught by Professor Wormer during the Spring '08 term at UCSD.
 Spring '08
 wormer
 Equations

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