Q 1030 q16 paper i 2000 by using a rectangular

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Q 10.30. (Q16, Paper I, 2000) By using a rectangular contour with corners at ± R , ± R + i and taking the limit as R → ∞ , or otherwise, show that if a is real and | a | , then integraldisplay 0 cosh ax cosh πx dx = 1 2 sec parenleftBig a 2 parenrightBig . Q 10.31. (Q16, Paper II, 1999) A real function y ( t ) satisfies the differential equation ¨ y + 4 ˙ y + 3 y = f ( t ) , where f ( t ) vanishes as | t | → ∞ and has Fourier transform F ( ω ). Assuming that y ( t ) and ˙ y ( t ) vanish as | t | → ∞ , find the Fourier transform Y ( ω ) of y ( t ) in terms of in terms of F ( ω ). Hence show that y ( t ) vanishes for t< 0 if f ( t ) vanishes for t< 0. Comment on the significance of this fact. Find the Fourier transform of the function f ( t ) = H ( t ) e t , where H ( t ) is the unit step function H ( t ) = braceleftBigg 1 if t> 0, 0 if t< 0. Hence determine the function y ( t ). Check, by direct substitution, that y is, indeed, the required solution. [ The examiner said that you could use the Fourier inversion theorem and Jordan’s lemma without proof. In spite of this invitation, I suggest you might be better off using the uniqueness of Fourier transforms. ] Q 10.32. (Q17, Paper IV, 1997) Use Fourier transforms to find g in the equation 5 e −| t | 8 e 2 | t | + 3 e 3 | t | = integraldisplay −∞ g ( τ ) e −| t τ | dτ. 32
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Q 10.33. (Q17, Paper IV, 2000) Assuming suitable decay of the function w ( x ) as x → ±∞ , express the Fourier transforms of w ( x ) and xw ( x ) in terms of the Fourier transform of w ( x ). Find the form of the Fourier transform of w ( x ), if x ( w ′′ ( x ) w ( x )) + w ( x ) = 0 . ( ) By using the inversion formula and a suitable change of variables, or other- wise, deduce that w ( x ) = 1 π integraldisplay 0 cos( x sinh u ) du is a solution of ( ). Part C Q 10.34. (Q16, Paper II, 2000) [This is not really very hard but is a bit out of the ordinary and a bit beyond the syllabus.] The complex plane is cut along the real axis from z = 1 to z = 1, and the branch of f ( z ) = ( z 2 1) 1 / 2 is chosen so that f ( z ) is real and positive when z is real and z > 1. Obtain expressions for f ( z ) just above and just below the cut and also when | z | ≫ 1. Show that integraldisplay 1 1 1 x 2 1 + x 2 dx = π ( 2 1) , where the square root gives positive values for 1 <x< 1. Q 10.35. (Q8(b), Paper IV, 1996) [This is very much on the hard side, involves fairly messy calculations and is beyond the syllabus so you may consider it as starred. Unfortunately the examiner did not.] In real life, Fourier integral techniques are most useful not for solving dif- ferential equations but for solving partial differential equations in the manner shown below. As a warm up, explain why the general solution of the partial differential equation ∂F ∂x ( x,y ) = 0 is F ( x,y ) = A ( y ) and the general solution of 2 F ∂x 2 ( x,y ) + F ( x,y ) = 0 is F ( x,y ) = A ( y ) sin x + B ( y ) cos x . 33
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A slab of material occupies the region { ( x,y ) : 0 y 1 } and moves with constant velocity U in the x -direction. The temperature T ( x,y ) in the slab satisfies the ‘advection diffusion equation’ U ∂T ∂x = 2 T ∂x 2 + 2 T ∂y 2 , and the boundary conditions are such that T ( x, 0) = T 0 e x 2 , T ( x, 1) = 0 .
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