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AEM_3e_Chapter_15

AEM_3e_Chapter_15 - 15 15 Integral Transform Method...

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15 15 Integral Transform Method EXERCISES 15.1 Error Function 3. By the first translation theorem, e t erf( t ) = erf( t ) s s 1 = 1 s s + 1 s s 1 = 1 s ( s 1) . 6. We first 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 1 + e 2 s + e 4 s + · · · e ( a +1) s s 1 + e 2 s + e 4 s + · · · = 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 + · · · = n =0 e (2 n +1 a ) s s e (2 n +1+ a ) s s . Then sinh a s s sinh s = n =0 e (2 n +1 a ) s s e (2 n +1+ a ) s s = n =0 erfc 2 n + 1 a 2 t erfc 2 n + 1 + a 2 t = n =0 1 erf 2 n + 1 a 2 t 1 erf 2 n + 1 + a 2 t = n =0 erf 2 n + 1 + a 2 t erf 2 n + 1 a 2 t . 9. b a e u 2 du = 0 a e u 2 du + b 0 e u 2 du = b 0 e u 2 du 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 ) = 0 we have lim 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 t x a t x a . 6. Transforming the partial differential 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 , s ) = 0 and U (1 , s ) = 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 πx 1 ( s 2 + π 2 )( s 2 + ω 2 ) = ω ω 2 π 2 sin πx 1 π π s 2 + π 2 1 ω ω s 2 + ω 2 = ω π ( ω 2 π 2 ) sin πt sin πx 1 ω 2 π 2 sin ωt sin πx. 9. Transforming the partial differential 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 ) = 0 and U (0 , s ) = 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 ) = 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. 12. We use U ( x, s ) = c 1 e s x + c 2 e s x + u 1 x s .

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AEM_3e_Chapter_15 - 15 15 Integral Transform Method...

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