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527solns12

# 527solns12 - 642:527 SOLUTIONS ASSIGNMENT 12 FALL 2009...

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642:527 SOLUTIONS: ASSIGNMENT 12 FALL 2009 Section 17.10: 2. If Re a > 0 then lim x →−∞ e ( a ) x = 0 for ω real, so F{ H ( x ) e ax } = integraldisplay −∞ H ( x ) e ax e iωx dx = integraldisplay 0 −∞ e ( a ) x dx = e ( a ) x a vextendsingle vextendsingle vextendsingle 0 −∞ = 1 a . 3. If a > 0 then all the integrals below converge: F{ e a | x | } = integraldisplay −∞ e a | x | e iωx dx = integraldisplay 0 −∞ e ( a ) x dx + integraldisplay 0 e ( a ) x dx = 1 a + 1 a + = 2 a a 2 + ω 2 . 4. (c) f ( x ) = 1 2 π integraldisplay −∞ [ H ( ω ) H ( ω 1)] ωe iωx = 1 2 π integraldisplay 1 0 ωe iωx = e ix 1 ixe ix x 2 . (f) f ( x ) = 1 2 π integraldisplay −∞ δ ( ω a ) e iωx = e iax 2 π . 6. (a) Using formulas 4, 17, and 18 of Appendix D we have F{ 4 x 2 e 3 | x | } = 4 i 2 d 2 2 6 9 + ω 2 = 144 3 ω 2 (9 + ω 2 ) 3 . (c) By formula 1 of Appendix D, F{ 1 / ( x 2 + 2) } = πe 2 | ω | / 2, so by formula 13, F braceleftbigg cos 3 x ( x 2 + 2) bracerightbigg = π 2 2 bracketleftBigg e 2 | ω 3 | 2 + e 2 | ω +3 | 2 bracketrightBigg (g) By formulas 7 and 9, F 1 braceleftbigg 4 sin ω ω 1 ω bracerightbigg = 2( H ( x + 1) H ( x 1)) + 1 radicalbig 2 π | x | . Section 18.3: 14. Discussed below, after problem 29. 15. The problem is Equation : α 2 u xx = u t F, 0 < x < L, t > 0; Boundary conditions : u (0 , t ) = 0 , u ( L, t ) = 50 Initial condition : u ( x, 0) = f ( x ) . (12.1) We follow the suggestion of the text and first look for a function v ( x ) which solves the PDE and the BC, but not necssarily the IC: that is, we need α 2 v ′′ = F, v (0) = 0 , v ( L ) = 50 . 1

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642:527 SOLUTIONS: ASSIGNMENT 12 FALL 2009 The solution is easily found to be v ( x ) = 50 F ( x 2 L 2 ) / (2 α 2 ). Now we write u ( x, t ) = v ( x ) + w ( x, t ), and by putting this into (12.1) find that w ( x, t ) must satisfy Equation : α 2 w xx = w t , 0 < x < L, t > 0; Boundary conditions : w (0 , t ) = 0 , w ( L, t ) = 0 Initial condition : w ( x, 0) = f ( x ) v ( x ) . (12.2) This problem is easily solved by a half range sine series, and so finally u ( x, t ) = v ( x ) + w ( x, t ) = 50 F 2 α 2 ( x 2 L 2 ) + summationdisplay n =1 b n sin nπx L , with b n = 2 L integraldisplay L 0 ( f ( x ) v ( x ) ) sin nπx L dx. (12.3) Notice that the steady state v ( x ) appears in the initial condition for problem (12.2) for w ( x, t ) and therefore in the formula (12.3) for b n . 19. Suppose that we are looking for a function u ( x, t ) satisfying Equation : α 2 u xx = u t , 0 < x < L, t > 0; Boundary conditions : u x (0 , t ) = Q 1 , u x ( L, t ) = Q 2 Initial condition : u ( x, 0) = f ( x ) . (12.4) The book takes f ( x ) = 0, Q 1 = 1, and Q 2 = 0, but that is not really necessary; however, we do assume that Q 1 negationslash = Q 2 . Since Q 1 negationslash = Q 2 these boundary conditons are precisely those for which our usual method does not work, because we cannot find a stationary solution u = A + Bx which satisfies the boundary conditions; this is covered in 10(c). The text suggests that we instead use a quadratic function z ( x ) which satisfies the boundary condtions: z (0) = Q 1 , z ( L ) = Q 2 . This means that we must take z ( x ) = A + Q 1 x +( Q 2 Q 1 ) x 2 / 2 L , with A some arbitrary constant; suppose in the future that we have chosen one such z ( x ).
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527solns12 - 642:527 SOLUTIONS ASSIGNMENT 12 FALL 2009...

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