Chem Differential Eq HW Solutions Fall 2011 113

Chem Differential Eq HW Solutions Fall 2011 113 - just have...

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Section 7.3 The Fourier Transform Method 113 Using ± u ( w, 0) = ± f ( w ), we see that ± u ( w, t )= ± f ( w ) cos wt. Taking inverse Fourier transforms, we get u ( x, t )= 1 2 π ² -∞ ± f ( w ) cos cwte ixw dw = 1 2 π ² 1 - 1 cos cwte ixw dw. 17. 2 u ∂t 2 = 4 u ∂x 4 u ( x, 0) = ³ 100 if | x | < 2, 0 otherwise. Fourier transform the problem with respect to the variable x : d 2 dt 2 ± u ( w, t )= w 4 ± u ( w, t ) , ± u ( w, 0) = ± f ( w ) = 100 ´ 2 π sin 2 w w . Solve the second order differential equation in ± u ( w, t ): ± u ( w, t )= A ( w ) e - w 2 t + B ( w ) e w 2 t . Because a Fourier transform is expected to tend to 0 as w →±∞ ,i fwe±x t> 0 and let w →∞ or w →-∞ , we see that one way to make ± u ( w, t ) 0 is to take B ( w ) = 0. Then ± u ( w, t )= A ( w ) e - w 2 t , and from the initial condition we obtain B ( w )= ± f ( w ). So ± u ( w, t )= ± f ( w ) e - w 2 t = 100 ´ 2 π sin 2 w w e - w 2 t . Taking inverse Fourier transforms, we get u ( x, t )= 1 2 π ² -∞ 100 ´ 2 π sin 2 w w e - w 2 t e ixw dw = 100 π ² -∞ sin 2 w w e - w 2 t e ixw dw. 21. (a) To verify that u ( x,t )= 1 2 [ f ( x - ct )+ f ( x + ct )] + 1 2 c ² x + ct x - ct g ( s ) ds is a solution of the boundary value problem of Example 1 is straightforward. You
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Unformatted text preview: just have to plug the solution into the equation and the initial and boundary con-ditions and see that the equations are veried. The details are sketched in Section 3.4, following Example 1 of that section. (b) In Example 1, we derived the solution as an inverse Fourier transform: u ( x, t ) = 1 2 - f ( w ) cos cwt + 1 cw g ( w ) sin cwt e iwx dx. Using properties of the Fourier transform, we will show that 1 2 - f ( w ) cos cwte iwx dw = 1 2 [ f ( x-ct ) + f ( x + ct )]; (1) 1 2 - 1 w g ( w ) sin cwte iwx dw = 1 2 x + ct x-ct g ( s ) ds. (2)...
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