Chem Differential Eq HW Solutions Fall 2011 119

# Chem Differential - Section 7.4 The Heat Equation and Gausss Kernel 119 29 Parts(a(c are obvious from the denition of gt(x(d The total area under

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Section 7.4 The Heat Equation and Gauss’s Kernel 119 29. Parts (a)-(c) are obvious from the deFnition of g t ( x ). (d) The total area under the graph of g t ( x ) and above the x -axis is ± -∞ g t ( x ) dx = 1 c 2 t ± -∞ e - x 2 / (4 c 2 t ) dx = 2 c t c 2 t ± -∞ e - z 2 dz ( z = x 2 c t ,dx =2 c tdz ) 2 ± -∞ e - z 2 dz = 2 π, by (4), Sec. 7.2. (e) To Fnd the ±ourier transform of g t ( x ), apply (5), Sec. 7.2, with a = 1 4 c 2 t , 1 2 a c 2 t, 1 4 a = c 2 t. We get ² g t ( w )= 1 c 2 t F ³ e - x 2 / (4 c 2 t ) ´ dx = 1 c 2 t × 2 c 2 te - c 2 2 = e - c 2 2 . (f) If f is an integrable and piecewise smooth function, then at its points of conti- nuity, we have lim t 0 g t * f ( x f ( x ) . This is a true fact that can be proved by using properties of Gauss’s kernel. If we interpret f ( x ) as an initial temperature distribution in a heat problem, then the solution of this heat problem is given by u ( x, t g t * f ( x ) . If t 0, the temperature u ( x, t ) should approach the initial temperature distribu- tion f ( x ). Thus lim t 0 g t * f ( x f ( x ) . Alternatively, we can use part (e) and argue as follows. Since lim t 0 F ( g t )( ω ) = lim t 0 e - c 2 2 =1 , So lim t 0 F ( g t * f ) = lim t 0 F ( g
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## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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