Chem Differential Eq HW Solutions Fall 2011 117

Chem Differential Eq HW Solutions Fall 2011 117 - Section...

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Section 7.4 The Heat Equation and Gauss’s Kernel 117 So u ( x, t )= 1 2 t 1 2 π ± -∞ s 2 e - ( x - s ) 2 t ds = - ² d 2 dw 2 F ³ 1 2 t e - ( x - s ) 2 t ´ ( w ) µ w =0 = - ² d 2 dw 2 e - ( iwx + w 2 t ) µ w =0 = - ² d dw - e - ( iwx + w 2 t ) ( ix +2 wt ) µ w =0 = - e - ( iwx + w 2 t ) ( ix +2 wt ) 2 +2 te - ( iwx + w 2 t ) · w =0 = x 2 +2 t. You can check the validity of this answer by plugging it back into the heat equation. The initial condition is also obviously met: u ( x, 0) = x 2 . The approach that we took can be used to solve the boundary value problem with f ( x ) x n as initial temperature distribution. See the end of this section for interesting applications. 9. Fourier transform the problem: du dt ¸ u ( w, t )= - e - t w 2 ¸ u ( w, t ) , ¸ u ( w, 0) = ¸ f ( w ) . Solve for ¸ u ( w, t ): ¸ u ( w, t )= ¸ f ( w ) e - w 2 (1 - e - t ) . Inverse Fourier transform and note that u ( x, t )= f *F - 1 ¹ e - w 2 (1 - e - t ) º . With the help of Theorem 5, Sec. 7.2 (take a =1 - e - t ), we ±nd F - 1 ¹ e - w 2 (1 - e - t )
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