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d. None of the above (3) Based on part (2), we can deduce that: dw a. 1+ w2 Jo dw dw 1+ w? dw dw 1+ w? d. None of the above b. C.

d. None of the above (3) Based on part (2), we can deduce that: dw a. 1+ w2 Jo dw dw 1+ w? dw dw 1+ w? d. None of the above b. C.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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fourier transform part 3
Let f be the function defined on R by f(x) = e-lxl, and consider the following boundary problem (G): TE R, t> 0 (1) I u(x, 0) = S(x), r€R (2) (1) The Fourier Transform of f is given by: 2 1 11+ w² ' a. F(f(x)) = V w b. F(f(1)) = V + w² * 1 c. F(f(x)) = d. None of the above %3! T1+ w +oo (2) Given that iuz dw = -tuz dw and ew - 2 cos (wr), we obtain by applying +e the inverse Fourier Transform: w cos(wr) dw a. erl = 1+w Jo rtoo 2 w cos(wr) 1+ w² t∞ cos(wr) 1+ w² b. e-| dw T Jo 2 -dw * Jo d. None of the above (3) Based on part (2), we can deduce that: rtoo dw a. 1+ w² dw Jo mp 1+ w²' rtoo dw b. dw = 1+ w2 C. d. None of the above (4) Let v(w,t) be the Fourier Transform of u(r, t) acting to the variable r. By Applying the Fourier Transform to the first equation of (G), we obtain: a. vr(w,t) – iwv(w, t) = 0 b. vz(w, t) – w²v(uw,t) = 0 c. vr(w,t) – w*v(w, t) = 0 d. None of the above (5) The general solution of (G) is: rtoo ptwa ewt -dw 1+ w 1 a. u(x, t) =- / %3D b. u(r,t) = -dw 1+ w? rto0 piwa put -dw 1+ w3 -00 c. u(x, t) d. None of the above -00
Transcribed Image Text:Let f be the function defined on R by f(x) = e-lxl, and consider the following boundary problem (G): TE R, t> 0 (1) I u(x, 0) = S(x), r€R (2) (1) The Fourier Transform of f is given by: 2 1 11+ w² ' a. F(f(x)) = V w b. F(f(1)) = V + w² * 1 c. F(f(x)) = d. None of the above %3! T1+ w +oo (2) Given that iuz dw = -tuz dw and ew - 2 cos (wr), we obtain by applying +e the inverse Fourier Transform: w cos(wr) dw a. erl = 1+w Jo rtoo 2 w cos(wr) 1+ w² t∞ cos(wr) 1+ w² b. e-| dw T Jo 2 -dw * Jo d. None of the above (3) Based on part (2), we can deduce that: rtoo dw a. 1+ w² dw Jo mp 1+ w²' rtoo dw b. dw = 1+ w2 C. d. None of the above (4) Let v(w,t) be the Fourier Transform of u(r, t) acting to the variable r. By Applying the Fourier Transform to the first equation of (G), we obtain: a. vr(w,t) – iwv(w, t) = 0 b. vz(w, t) – w²v(uw,t) = 0 c. vr(w,t) – w*v(w, t) = 0 d. None of the above (5) The general solution of (G) is: rtoo ptwa ewt -dw 1+ w 1 a. u(x, t) =- / %3D b. u(r,t) = -dw 1+ w? rto0 piwa put -dw 1+ w3 -00 c. u(x, t) d. None of the above -00
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