prelim2001 - Table of Contents Complete Partially Complete...

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Table of Contents Complete Partially Complete Not Complete Preliminary Examination 2001 2 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Problem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Problem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Problem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1
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Preliminary Examination September , 2001 1. Consider the integral equation f ( x ) = e −| x | + λ ´ −∞ e −| x y | f ( y ) d y where f ( x ) is finite at x = ±∞ . (a) Compute f ( x ) to first order in λ , for x> 0 . (b) Compute f ( x ) exactly. Hint: One possible approach is to consider the Fourier transform. Assume λ< 1 / 2 . Solution (a) This is a recursive definition. f ( x ) = e −| x | + λ ˆ −∞ e −| x y | ( e −| y | + λ ˆ −∞ e −| y z | f ( z ) d z ) d y = e −| x | + λ ˆ −∞ e −| x y | e −| x | d y + λ 2 ˆ −∞ ˆ −∞ e −| x y | e −| y z | f ( z ) d z d y To first order and for x> 0 , f ( x ) = e −| x | + λ ˆ −∞ e −| x y | e −| y | d y = e x + λ ˆ 0 −∞ e y x e y d y + λ ˆ x 0 e y x e y d y + λ ˆ x e x y e y = e x + λ 1 2 e x + λx e x + λ 1 2 e x = e x (1 + λ + λx ) (b) We make use of convolution theorem by realizing that the integral is a convolution. F [ f ( x )] = F [ e −| x | ] + λ F [ ˆ −∞ e −| x y | f ( y ) d y ] = F [ e −| x | ] + λ F [ e −| x | f ( x ) ] = F [ e −| x | ] + λ F [ e −| x | ] F [ f ( x )] Thus, f ( x ) = F 1 [ F [ e −| x | ] 1 λ F [ e −| x | ] ] Now, F [ e −| x | ] = ˆ −∞ e −| x | e i kx d x = ˆ 0 −∞ e x (1 i k ) d x + ˆ 0 e x (1+i k ) d x = 2 1 + k 2 x ̸ = 0 2 πδ ( k ) x = 0 So for x ̸ = 0 , f ( x ) = 1 2 π ˆ −∞ 2 1+ a 2 1 λ 2 1+ a 2 e i kx d k = 1 π ˆ −∞ e i kx k 2 + (1 2 λ ) 2
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Preliminary Examination September , 2001 I don’t know what the answer is when x = 0 . My guess is that f ( x ) doesn’t converge at x = 0 , and thus, it isn’t defined. When we proceed assuming x ̸ = 0 , we obtain results that seem to imply a discontinuity at x = 0 . From here on out, we assume x ̸ = 0 . So, why must λ < 1 / 2 ? Briefly, λ < 1 / 2 guarantees that the roots are imaginary and the results converge. A detailed explanation follows. Also, one must be careful when choosing a Fourier transform pair. In the work above, I am using the definition where a 1 / 2 π is present only on the inversion integral. If you want to use the following Fourier transform pair, then use must use a corresponding definition of the convolution. ˜ f ( k ) = 1 2 π ˆ −∞ f ( x )e i kx d x f ( x ) = 1 2 π ˆ −∞ f ( x )e i kx d x f g ( x ) = 1 2 π ˆ −∞ f ( x y ) g ( y ) d y and ˜ f ( k ) = ˆ −∞ f ( x )e i kx d x f ( x ) = 1 2 π ˆ −∞ f ( x )e i kx d x f g ( x ) = ˆ −∞ f ( x y ) g ( y ) d y So, let us look λ = 1 / 2 . In this case, we have f ( x ) = 1 π ˆ −∞ e i kx k 2 d k For x > 0 , we can do this integral using a contour on the upper half-plane. The integral over the infinite semi-circle will disappear by Jordan’s lemma, but the integral over the vanishing semi-circle about the origin blows up as the radius goes to zero. Thus, f ( x ) does not converge when λ = 1 / 2 . A
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