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Chem Differential Eq HW Solutions Fall 2011 112

# Chem Differential Eq HW Solutions Fall 2011 112 - 2 u x 0 =...

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112 Chapter 7 The Fourier Transform and its Applications Solutions to Exercises 7.3 1. 2 u ∂t 2 = 2 u ∂x 2 , u ( x, 0) = 1 1+ x 2 , ∂u ∂t ( x, 0) = 0 . Follow the solution of Example 1. Fix t and Fourier transform the problem with respect to the variable x : d 2 dt 2 ± u ( w, t )= - w 2 ± u ( w, t ) , ± u ( w, 0) = F ( 1 1+ x 2 ) = ² π 2 e -| w | , d dt ± u ( w, 0) = 0 . Solve the second order diﬀerential equation in ± u ( w, t ): ± u ( w, t )= A ( w ) cos wt + B ( w ) sin wt. Using d dt ± u ( w, 0) = 0, we get - A ( w ) w sin wt + B ( w ) w cos wt ³ ³ ³ t =0 =0 B ( w ) w =0 B ( w )=0 . Hence ± u ( w, t )= A ( w ) cos wt. Using ± u ( w, 0) = ´ π 2 e -| w | , we see that A ( w )= ´ π 2 e -| w | and so ± u ( w, t )= ² π 2 e -| w | cos wt. Taking inverse Fourier transforms, we get u ( x, t )= µ -∞ e -| w | cos wte ixw dw. 5. 2 u ∂t 2 = c 2 2 u ∂x
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Unformatted text preview: 2 , u ( x, 0) = ² 2 π sin x x , ∂u ∂t ( x, 0) = 0 . Fix t and Fourier transform the problem with respect to the variable x : d 2 dt 2 ± u ( w, t ) =-c 2 w 2 ± u ( w, t ) , ± u ( w, 0) = F ( ² sin x x ) ( w ) = ± f ( w ) = ¶ 1 if | w | < 1 0 if | w | > 1 , d dt ± u ( w, 0) = 0 . Solve the second order diﬀerential equation in ± u ( w, t ): ± u ( w, t ) = A ( w ) cos cwt + B ( w ) sin cwt. Using d dt ± u ( w, 0) = 0, we get ± u ( w, t ) = A ( w ) cos cwt....
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