Chem Differential Eq HW Solutions Fall 2011 115

Chem Differential Eq HW Solutions Fall 2011 115 - Section...

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Unformatted text preview: Section 7.3 The Fourier Transform Method and d u(w, 0) = g(w) dt ⇒ −cw2 A(w) + cw2B (w) = g(w) g (w ) cw2 g (w ) ⇒ −f (w) + 2B (w) = cw2 1 g (w ) ; ⇒ B (w ) = f (w ) + 2 cw2 1 g (w ) ⇒ A(w) = . f (w ) − 2 cw2 ⇒ −A(w) + B (w) = Hence u(w, t) 1 2 = f (w ) − 2 = = g (w ) cw2 2 e−cw t + 2 1 2 f (w ) + g (w ) cw2 2 2 ecw t . 2 (ecw t + e−cw t) g(w) (ecw t − e−cw t ) f (w ) + 2 cw2 2 g (w ) f (w) cosh(cw2 t) + sinh(cw2 t) cw2 Taking inverse Fourier transforms, we get 1 u(x, t) = √ 2π ∞ f (w) cosh(cw2 t) + −∞ g (w ) sinh(cw2 t) cw2 eixw dw. 115 ...
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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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