Chem Differential Eq HW Solutions Fall 2011 116

Chem Differential Eq HW Solutions Fall 2011 116 - 116...

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116 Chapter 7 The Fourier Transform and its Applications Solutions to Exercises 7.4 1. Repeat the solution of Example 1 making some adjustments: c = 1 2 , g t ( x )= 2 t e - x 2 t , u ( x, t f * g t ( x ) = 1 2 π ± -∞ f ( s ) 2 t e - ( x - s ) 2 t ds = 20 ± 1 - 1 e - ( x - s ) 2 t ds ( v = x - s t ,d v = - 1 t ds ) = 20 π ± x +1 t x - 1 t e - v 2 ds =1 0 ² erf( x +1 t ) - erf( x - 1 t ) ³ . 5. Apply (4) with f ( s s 2 : u ( x, t f * g t ( x ) = 1 2 t 1 2 π ± -∞ s 2 e - ( x - s ) 2 t ds. You can evaluate this integral by using integration by parts twice and then appealing to Theorem 5, Section 7.2. However, we will use a different technique based on the operational properties of the Fourier transform that enables us to evaluate a much more general integral. Let n be a nonnegative integer and suppose that
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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