{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chem Differential Eq HW Solutions Fall 2011 116

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

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

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 , dv = - 1 t ds ) = 20 π x +1 t x - 1 t e - v 2 ds = 10 erf( x + 1 t ) - erf( x - 1 t ) . 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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}