AEM_3e_Chapter_15

# AEM_3e_Chapter_15 - 15 15 Integral Transform Method EXERCISES 15.1 Error Function 3 By the rst translation theorem et erf t = 6 We rst compute erf

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15 15 Integral Transform Method EXERCISES 15.1 Error Function 3. By the frst translation theorem, n e t erF( t ) o = n erF( t ) o ¯ ¯ ¯ ¯ s s 1 = 1 s s +1 ¯ ¯ ¯ ¯ s s 1 = 1 s ( s 1) . 6. We frst compute sinh a s s sinh s = e a s e a s s ( e s e s ) = e ( a 1) s e ( a +1) s s (1 e 2 s ) = e ( a 1) s s h 1+ e 2 s + e 4 s + ··· i e ( a +1) s s h e 2 s + e 4 s + i = e (1 a ) s s + e (3 a ) s s + e (5 a ) s s + # e (1+ a ) s s + e (3+ a ) s s + e (5+ a ) s s + # = X n =0 e (2 n +1 a ) s s e (2 n +1+ a ) s s # . Then ½ sinh a s s sinh s ± = X n =0 ( e (2 n +1 a ) s s ) ( e (2 n +1+ a ) s s )# = X n =0 ² erFc µ 2 n a 2 t erFc µ 2 n +1+ a 2 t ¶¸ = X n =0 µ² 1 erF µ 2 n a 2 t ¶¸ ² 1 erF µ 2 n a 2 t ¶¸¶ = X n =0 ² erF µ 2 n a 2 t erF µ 2 n a 2 t ¶¸ . 9. Z b a e u 2 du = Z 0 a e u 2 du + Z b 0 e u 2 du = Z b 0 e u 2 du Z a 0 e u 2 du = π 2 erF( b ) π 2 erF( a )= π 2 [erF( b ) erF( a )] 255

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EXERCISES 15.2 Applications of the Laplace Transform 15.2 Applications of the Laplace Transform 3. The solution of a 2 d 2 U dx 2 s 2 U =0 is in this case U ( x, s )= c 1 e ( x/a ) s + c 2 e ( x/a ) s . Since lim x →∞ u ( x, t )=0wehavel im x →∞ U ( x, s )=0.Thus c 2 = 0 and U ( x, s c 1 e ( x/a ) s . If { u (0 ,t ) } = { f ( t ) } = F ( s ) then U (0 ,s F ( s ). From this we have c 1 = F ( s ) and U ( x, s F ( s ) e ( x/a ) s . Hence, by the second translation theorem, u ( x, t f ³ t x a ´³ t x a ´ . 6. Transforming the partial di±erential equation gives d 2 U dx 2 s 2 U = ω s 2 + ω 2 sin πx. Using undetermined coeﬃcients we obtain U ( x, s c 1 cosh sx + c 2 sinh sx + ω ( s 2 + π 2 )( s 2 + ω 2 ) sin πx. The transformed boundary conditions U (0 )=0and U (1 ) = 0 give, in turn, c 1 = 0 and c 2 = 0. Therefore U ( x, s ω ( s 2 + π 2 )( s 2 + ω 2 ) sin πx and u ( x, t ω sin ½ 1 ( s 2 + π 2 )( s 2 + ω 2 ) ± = ω ω 2 π 2 sin ½ 1 π π s 2 + π 2 1 ω ω s 2 + ω 2 ± = ω π ( ω 2 π 2 ) sin πt sin 1 ω 2 π 2 sin ωt sin πx. 9. Transforming the partial di±erential equation gives d 2 U dx 2 s 2 U = sxe x . Using undetermined coeﬃcients we obtain U ( x, s c 1 e sx + c 2 e sx 2 s ( s 2 1) 2 e x + s s 2 1 xe x . The transformed boundary conditions lim x →∞ U ( x, s U (0 ) = 0 give, in turn, c 2 = 0 and 256
15.2 Applications of the Laplace Transform c 1 =2 s/ ( s 2 1) 2 . Therefore U ( x, s )= 2 s ( s 2 1) 2 e sx 2 s ( s 2 1) 2 e x + s s 2 1 xe x . From entries (13) and (26) in the Table of Laplace transforms we obtain u ( x, t ½ 2 s ( s 2 1) 2 e sx 2 s ( s 2 1) 2 e x + s s 2 1 xe x ± =2( t x )sinh( t x )( t x ) te x sinh t + xe x cosh t.

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## This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.

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AEM_3e_Chapter_15 - 15 15 Integral Transform Method EXERCISES 15.1 Error Function 3 By the rst translation theorem et erf t = 6 We rst compute erf

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