Let u t fux t ux te i x dx apply fourier

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Unformatted text preview: u(x, 0) = ex , x 0 0, x > 0 ) ex , x 0 0, x > 0 G( x ) = Thus the solution of the BVP is t2 /2 x t e u(x, t) = e ,x 0, x t0 t>0 Z 5. Let u(! , t) = F{u(x, t)} = ˆ u(x, t)e i x dx. Apply Fourier Transform to find to general 4. Use theFourier Transform to PDEthe obtain solution of the PDE ut + t ! 2 u ˆ ˆ ut ut ˆ Z du ˆ u ˆ =0 tuxx = 0. = t! 2 u ˆ = Z t! 2 dt 12 2 ˆ t ! + F (! ) 2 22 ˆ u(! , t) = G(! )e t /2 ˆ ln |u| = ˆ ˆ where G(! ) is an arbitrary function. Let G(x) = F on p. 121 of the Course Notes we have 2 1 ˆ {G(! )}. Using a result 2 e x /(2t ) F {e }= = f ( x) . |t| 2 Thus, applying the Convolution Theorem we have t2 1 u(x, t) = F 1 {u(! , t)} = ˆ Z 2 /2 G( x where G is an arbitrary function. z )f (z ) dz = Z 2 G( x 2 e z /(2t ) z) dz |t| 2 Thus the solution of the BVP is Page 3 2 AMATH 350 Assignment #9 Solutions- tFall 2012 0 e t /2 ex , x t u(x, t) = 0, x t>0 Z 5. Let u(! , t) = F{u(x, t)} = ˆ u(x, t)e i x dx. Apply Fourier Transform to PDE to obtain ut + t ! 2 u = 0 ˆ ˆ ut = t ! 2 u ˆ ˆ Z du Z ˆ = t! 2 dt u ˆ 12 2 ˆ ln |u| = ˆ t ! + F (! ) 2 22 ˆ u(! , t) = G(! )e t /2 ˆ ˆ where G(! ) is an arbitrary function. Let G(x) = F on p. 121 of the Course Notes we have 2 1 ˆ {G(! )}. Using a result 2 e x /(2t ) F {e }= = f ( x) . |t| 2 Thus, applying the Convolution Theorem we have t2 1 u(x, t) = F 1 {u(! , t)} = ˆ Z 2 /2 G( x z )f (z ) dz = Z 2 G( x where G is an arbitrary function. 5. a) This should be easy: 2 e z /(2t ) z) dz |t| 2 2 1 ˘ f ( x) = [f ( x) + f (x)] 2 1 = [f (x) + f ( x)] 2 ˘ = f ( x) 1 e f ( x) = [f ( x) f (x)] 2 1 = [f (x) f ( x)] 2 e = f ( x) b) Following the hint, ˆ f (! ) = = = Z 1 1 h Z 1 Z 1 = i Z 1 1 1 f (x) [cos (! x) 1 ˘ f (x) cos (! x) dx + ˘ f (x) sin (! x) dx 1 dx 1 i ˘ e f (x) + f (x) cos (! x) dx Z i! x f ( x) e i i sin (! x)] dx Z Z i Z 1 1 1 h i ˘ e f (x) + f (x) sin (! x) dx e f (x) cos (! x) dx 1 1 1 e f (x) sin (! x) dx Now, the 2nd and 3rd integrands are odd (as functions of x), so those integrals are zero, and Z1 Z1 ˆ ˘ e f (! ) = f (x) cos (! x) dx i f (x) sin (! x) dx. 1 1 ⇣ ⌘ ˆ Viewing this now as a function of ! , we can see that Re f (! ) is even ⇣ ⌘ ˆ (! ) is odd. (since cos ( ! x) = cos (! x)), and similarly, Im f AMATH 350 Assignment #9 Solutions- Fall 2012 Page 4 ˘ e Furthermore, if f is even, then f = f , and f is zero, so Z1 ˆ (! ) = f f (x) cos (! x) dx, 1 which is real and even. Simlarly, if f is odd, then it’s the second integral which survives instead, ˆ so f is imaginary and odd....
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