Chem Differential Eq HW Solutions Fall 2011 40

Chem Differential Eq HW Solutions Fall 2011 40 - 40 Chapter...

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Unformatted text preview: 40 Chapter 3 Partial Differential Equations in Rectangular Coordinates To simplify the notation, replace x0 and t0 by x and y in the coordinates of the points Q1 and P2 and let φ(x, t) = −4t2 + x − x2 + 8tx. We have u(x, t) = u(Q1) − u(P2 ) x + 1 − 2t 1 − x + 2t =φ , 2 4 = 5 − 12t − 5x + 12tx, −φ 3 − x − 2t −1 + x + 2t , 2 4 where the last expression was derived after a few simplifications that we omit. It is interesting to note that the formula satisfies the wave equation and the boundary condition u(1, t) = 0 for all t > 0. Its restriction to the line x + 2t = 1 (part of the boundary of region I ) reduces to the formula for u(x, t) for (x, t) in region I . This is to be expected since u is continuous in (x, t). ...
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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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