Chem Differential Eq HW Solutions Fall 2011 134

Chem Differential Eq HW Solutions Fall 2011 134 - ) ds = 1...

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134 Chapter 7 The Fourier Transform and its Applications Solutions to Exercises 7.10 1. Apply Proposition 1 with f ( x, s )= e - ( x + s ) 2 , then dU dx = f ( x, x )+ ± x 0 ∂x f ( x, s ) ds = e - 4 x 2 - ± x 0 2( x + s ) e - ( x + s ) 2 ds = e - 4 x 2 - ± 2 x x 2 ve - v 2 dv (( v = x + s ) = e - 4 x 2 + e - v 2 ² ² ² 2 x x =2 e - 4 x 2 - e - x 2 . 5. Following Theorem 1, we frst solve φ t = φ xx , φ ( x, 0 ,s )= e - s x 2 , where s> 0 is fxed. The solution is φ ( x, t, s )= e - s (2 t + x 2 ) (see the solution o± Exercise 5, Section 7.4). The the desired solution is given by u ( x, t )= ± t 0 φ ( x, t - s, s ) ds = ± t 0 e - s (2( t - s )+ x 2 ) ds = - 2 te - s +2 se - s +2 e - s - x 2 e - s ² ² ² t 0 = - 2+2 t + x 2 + e - t (2 - x 2 ) . 9. Following Theorem 2, we frst solve φ tt = φ xx , φ ( x, 0 ,s )=0 , φ t ( x, 0 ,s )= cos( s + x ) where s> 0 is fxed. By d’Alembert’s method, the solution is φ ( x, t, s )= 1 2 ± x + t x - t cos( s + y ) dy = 1 2 ³ sin( s + x + t ) - sin( s + x - t ) ´ . The the desired solution is given by u ( x, t )= ± t 0 φ ( x, t - s, s )
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Unformatted text preview: ) ds = 1 2 ± t ³ sin( x + t )-sin( x-t + 2 s ) ´ ds = 1 2 ³ s sin( x + t ) + 1 2 cos( x-t + 2 s ) ´ ² ² ² t = 1 2 ³ t sin( x + t ) + 1 2 cos( x + t )-1 2 cos( x-t ) ´ . 13. start by solving φ tt = φ xx , φ ( x, , s ) = 0, φ t ( x, , s ) = δ ( x ) where s > 0 is fxed. By d’Alembert’s method, the solution is φ ( x, t, s ) = 1 2 ± x + t x-t δ ( y ) dy = 1 2 ³ U ( x + t )- U ( x-t ) ´ . By Theorem 2, the solution o± u tt = u xx + δ ( x ), u ( x, 0) = 0, φ t ( x, 0) = 0 is given...
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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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