Chem Differential Eq HW Solutions Fall 2011 127

Chem Differential Eq HW Solutions Fall 2011 127 - Section...

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Unformatted text preview: Section 7.7 Problems Involving Semi-Infinite Intervals 127 or d 2 us (w, t) + c2 ω2 us(w, t) = c2 wT0 . dt π Taking the Fourier sine transform of the boundary condition u(x, 0) = 0 for x > 0, we get us (w, 0) = 0. (b) A particular solution of the differential equation can be guessed easily: us (w, t) = 2 T0 π w. The general solution of the homogeneous differential equation: d us (w, t) + c2 ω2 us(w, t) = 0 dt is us (w, t) = A(w)e−c ential equation is 2 w2 t . So the general solution of the nonhomogeneous differ- us(w, t) = A(w)e−c 2 T0 πw Using us(w, 0) = A(w) 2 w2 t 2 T0 . πw 2 T0 π w. = 0, we find A(w) = − 2 T0 − πw us ( w t ) = So 2 T0 −c2 w2 t . e πw Taking inverse sine transforms, we find u(x, t) = ∞ 2 π 0 T0 T0 −c2 w2 t sin wx dw −e w w =sgn(x)=1 sin wx 2T0 ∞ sin wx −c2 w2 t dw dw − e w π0 w 0 2T0 ∞ sin wx −c2 w2 t = T0 − dw e π0 w = T0 ∞ 2 π 13. Proceed as in Exercise 11 using the Fourier sine transform instead of the cosine transform and the condition u(x 0) = 0 instead of uy (x, 0) = 0. This yields =0 d2 dx2 us (x, w) − w2 us(x, w) + d2 dx2 us (x, 2 π u(x, 0) = 0 w) = w2us(x, w). The general solution is us (x, w) = A(w) cosh wx + B (w) sinh wx. Using us (0, w) = 0 and us(1, w) = Fs(e−y ) = 2w , π 1 + w2 we get A(w) = 0 and B (w) = 2w 1 · . π 1 + w2 sinh w Hence us (x, w) = 2 w sinh wx . π 1 + w2 sinh w Taking inverse sine transforms: u(x, y) = 2 π ∞ 0 w sinh wx sin wy dw. 1 + w2 sinh w ...
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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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