1 iassignment i solutions fall 2012 1 9 z s z f xg

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Unformatted text preview: .i! )ˆ ( ˆt = + u lim AMATH 350 s 1 es e i s 1 = . Thus the Fourier Transform of f (Page 2 x) exists. 1 i!Assignment i! Solutions- Fall 2012 1 #9 Z s Z f (x)g (y )e i x dx = g (y ) f (x)e i x dx. Note: there is a simpler argument, using the result of part (a). If this Since the Fourier Transform of f (x) exists the integral on the right contransform exists, then verges, thus the integral on the left converges for any value of y , and we ˆ have u(! , y ) = Z (F){f !x) + g (y )} uy (F ,{f (x)Z+ y )(ˆ() ⇤ 1} ˆ g y f ( ( ). Clearly ˆ ! y ) = g ( g fy ! ). Further, = ix ˆ F{uy (x, y )} = f ( x) g ( y ) e dx = g (y ) f (x)e i x dx = g (y )f (! ). Z = F {f (x)} + g (y ) F {1} Z (since g (y )is constant Z with respect to x) (b) (f (x) + g (y ))e i x dx = f (x)e i x dx + g (y ) e i x dx. However, F {1} does not exist. The first integral on the right converges since the Fourier Transform of f (x) exists, however the second integral diverges (this was shown in class). Since g y ) following boundary value problem left diverges for Transform: 3. Solve (the6= 0, this means the integral on the using the Fourierany value of y and the Fourier Tranform of u(x, y ) does not exist. Z ux + ut + tu = 0 3. (a) F{ (x a)} = (x a)e i x dx = e i a , using the substitution property. lim u(x, t) = 0 ˆ (b) Let F{f (x)} = f (! ). Using1the result of part(a) and the Convolution x! ( Theorem we have ex , x 0 u(x, 0) = .Z ˆ F 1 {f (! )e i t } = F 1 {F{f (x)}F{ (xx >}} = t) 0 f (x z ) (z t) dz = f (x t). 0, 2. (a) Z 4. Let u(! , t) = F{u(x, t)} = ˆ u(x, t)e i x dx. Apply Fourier Transform to PDE to obtain i! u + ut + tu = 0 ˆˆ ˆ ut = (t + i! )ˆ ˆ u Z Z du ˆ 1 = u ˆ (t + i! ) dt t2 ˆ i! t + F (! ) 2 2 ˆ u ( ! , t ) = G ( ! ) e t /2 e i t ˆ ln |u| = ˆ ˆ where G is an arbitrary function. Apply Inverse Fourier Transform 1Z 2 ˆ u(x, t) = G(! )e t /2 e i t ei x d! 2 2 e t /2 Z ˆ = G( ! ) e i t e i x d ! 2 ˆ Let G(x) = F 1 {G(! )}. Then, using the Shifting Property, we have t2 /2 u(x, t) = e G(x t) where G is an arbitrary function. Apply BC:...
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