Chem Differential Eq HW Solutions Fall 2011 130

Chem Differential Eq HW Solutions Fall 2011 130 - i e-iw w...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
130 Chapter 7 The Fourier Transform and its Applications The formula is good at w = 0 if we take the limit as w 0. You will get F ( x ( U - 1 -U 1 ) ) = lim w 0 i ± 2 π ² w cos w - sin w w 2 ³ = i ± 2 π lim w 0 - w sin w 2 w =0 . (Use l’Hospital’s rue.) Unlike the Fourier transform in Exercise 31, the transform here is a nice continuous function. There is a major difference between the trans- forms of the two exercises. In Exercise 31, the function is not integrable and its Fourier transform exists only as a generalized function. In Exercise 33, the function is integrable and its Fourier transform exists in the usual sense of Section 7.2. In fact, look at the transform in Exercise 31, it is not even de±ned at w =0 . An alternative way to do this problem is to realize that φ ± ( x )= - δ - 1 - δ 1 + U - 1 -U 1 . So F ( φ ± ( x ) ) = F ( - δ - 1 - δ 1 + U - 1 -U 1 ) = 1 2 π ´ - e iw - e - iw - i e iw w
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: i e-iw w . But F ( ( x ) ) = iw F ( ( x ) ) . So F ( ( x ) ) = i w F ( e iw + e-iw + i e iw w-i e-iw w ) = i 2 w 2 cos w + i w (2 i sin w ) = i 2 w cos w-sin w w 2 . 37. Write 2 ( x ) = x 2 U ( x ), then use the operational properties F ( 2 ( x ) ) = F ( x 2 U ( x ) ) =-d 2 dw 2 F ( U ( x )) =--i 2 d 2 dw 2 1 w = 2 i 2 1 w 3 = i 2 1 w 3 . 41. We have f ( x ) =-U-2 ( x ) + 2 U-1 ( x )- U 1 ( x ) + -1 ( x )-2 2 ( x ) . So F ( f ( x )) = F (-U-2 ( x ) + 2 U-1 ( x )- U 1 ( x ) + -1 ( x )-2 2 ( x )) =--i 2 e 2 iw w + 2-i 2 e iw w--i 2 e-iw w + 1 2 e iw-2 2 e-2 iw ....
View Full Document

Ask a homework question - tutors are online