Chapt_15 - 15 15 Integral Transform Method EXERCISES 15.1...

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15 15 Integral Transform Method EXERCISES 15.1 Error Function 1. (a) The result follows by letting τ = u 2 or u = τ in erf( t )= 2 π Z t 0 e u 2 du . (b) Using { t 1 / 2 } = π s 1 / 2 and the Frst translation theorem, it follows from the convolution theorem that n erf( t ) o = 1 π ½ Z t 0 e τ τ ¾ = 1 π { 1 } n t 1 / 2 e t o = 1 π 1 s n t 1 / 2 o ¯ ¯ ¯ ¯ s s +1 = 1 π 1 s π s +1 = 1 s s . 2. Since erfc( t )=1 erf( t )wehave n erfc( t ) o = { 1 }− n erf( t ) o = 1 s 1 s s = 1 s · 1 1 s ¸ . 3. By the Frst translation theorem, n e t erf( t ) o = n erf( t ) o ¯ ¯ ¯ ¯ s s 1 = 1 s s ¯ ¯ ¯ ¯ s s 1 = 1 s ( s 1) . 4. By the Frst translation theorem and the result of Problem 2, n e t erfc( t ) o = n erfc( t ) o ¯ ¯ ¯ ¯ s s 1 = µ 1 s 1 s s ± ¯ ¯ ¯ ¯ s s 1 = 1 s 1 1 s ( s 1) = s 1 s ( s 1) = s 1 s ( s + 1)( s 1) = 1 s ( s +1) . 793
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15.1 Error Function 5. From entry 3 in Table 15.1 and the frst translation theorem we have ( e Gt/C er± à x 2 ± RC t ) = ( e Gt/C 1 er±c à x 2 ± RC t ) = n e Gt/C o ( e Gt/C er±c à x 2 ± RC t ) = 1 s + G/C e x RC s s ¯ ¯ ¯ ¯ s s + G/C = 1 s + G/C e x RC s + G/C s + G/C = C Cs + G ³ 1 e x RCs + RG ´ . 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 ¾ = X n =0 ( e (2 n +1 a ) s s ) ( e (2 n +1+ a ) s s ) = X n =0 · er±c µ 2 n +1 a 2 t ² er±c µ 2 n +1+ a 2 t ²¸ = X n =0 µ· 1 er± µ 2 n a 2 t ²¸ · 1 er± µ 2 n a 2 t ²¸² = X n =0 · er± µ 2 n a 2 t ² er± µ 2 n a 2 t ²¸ . 7. Taking the Laplace trans±orm o± both sides o± the equation we obtain { y ( t ) } = { 1 }− ½ Z t 0 y ( τ ) t τ ¾ Y ( s )= 1 s Y ( s ) π s s + π s Y ( s 1 s Y ( s 1 s ( s + π ) . 794
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-10 -5 5 10 x -2 -1 1 2 y erf H x L erfc H x L 15.1 Error Function Thus y ( t )= ½ 1 s ( s + π ) ¾ = e πt erfc( ) . By entry 5 in Table 15.1 8. Using entries 3 and 5 in Table 15.1, we have ( e ab e b 2 t erfc µ b t + a 2 t ± + erfc µ a 2 t ± ) = ½ e ab e b 2 t erfc µ b t + a 2 t ±¾ + ½ erfc µ a 2 t ±¾ = e a s s ( s + b ) + e a s s = e a s · 1 s 1 s ( s + b ) ¸ = e a s · 1 s s s ( s + b ) ¸ = e a s · s + b s s ( s + b ) ¸ = be a s s ( s + b ) .
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This note was uploaded on 02/17/2010 for the course MATHEMATIC MAS201 taught by Professor Xingqin during the Spring '10 term at Korea Advanced Institute of Science and Technology.

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Chapt_15 - 15 15 Integral Transform Method EXERCISES 15.1...

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