Chem Differential Eq HW Solutions Fall 2011 126

# Chem Differential Eq HW Solutions Fall 2011 126 - ± u c w...

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126 Chapter 7 The Fourier Transform and its Applications Solutions to Exercises 7.7 1. Fourier sine transform with respect to x : d dt ± u s ( w, t )= - w 2 ± u s ( w, t )+ ² 2 π w =0 ³ ´µ u (0 ,t ) d dt ± u s ( w, t )= - w 2 ± u s ( w, t ) . Solve the ±rst-order diﬀerential equation in ± u s ( w, t ) and get ± u s ( w, t )= A ( w ) e - w 2 t . Fourier sine transform the initial condition ± u s ( w, 0) = A ( w )= F s ( f ( x ))( w )= T 0 · 2 π 1 - cos bw w . Hence ± u s ( w, t )= · 2 π 1 - cos bw w e - w 2 t . Taking inverse Fourier sine transform: u ( x, t )= 2 π ¸ 0 1 - cos bw w e - w 2 t sin wxdw. 5. If you Fourier cosine the equations (1) and (2), using the Neumann type condition ∂u ∂x (0 ,t )=0 , you will get d dt ± u c ( w, t )= c 2 ¹ - w 2 ± u c ( w, t ) - ² 2 π =0 ³ ´µ d dx u (0 ,t ) º d dt ± u c ( w, t )= - c 2 w 2 ± u c ( w, t ) . Solve the ±rst-order diﬀerential equation in
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Unformatted text preview: ± u c ( w, t ) and get ± u c ( w, t ) = A ( w ) e-c 2 w 2 t . Fourier cosine transform the initial condition ± u c ( w, 0) = A ( w ) = F c ( f )( w ) . Hence ± u s ( w, t ) = F c ( f )( w ) e-c 2 w 2 t . Taking inverse Fourier cosine transform: u ( x, t ) = · 2 π ¸ ∞ F c ( f )( w ) e-c 2 w 2 t cos wxdw. 9. (a) Taking the sine transform of the heat equation (1) and using u (0 ,t ) = T for t > 0, we get d dt ± u s ( w,t ) = c 2 ¹-w 2 ± u s ( w,t ) + · 2 π wu (0 , t ) º ;...
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