Chem Differential Eq HW Solutions Fall 2011 146

# Chem Differential Eq HW Solutions Fall 2011 146 - 146...

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Unformatted text preview: 146 Chapter 8 The Laplace and Hankel Transforms with Applications Solutions to Exercises 8.3 1. The solution is the same as Example 2. Simply take T0 = 70 in that example. 5. Using the formula from Example 3, we get t u(x, t) = (t − τ )[τ − (t − x) U 0 (τ − x)] dτ 0 t = t (t − τ )τ dτ − (t − τ )(τ − x)( U 0 (τ − x) dτ 0 = = 0 t2 13 τ−τ 2 3 t3 − 3! t t − 0 (t − τ )(τ − x) U 0 (τ − x) dτ 0 t (t − τ )(τ − x) U 0 (τ − x) dτ 0 Note that U 0(τ − x) = 1 if τ > x and 0 if τ < x. So the integral is 0 if t < x (since in this case τ ≤ t < x). If x < τ < t, then t (t − τ )(τ − x) U 0(τ − x) dτ 0 t = (t − τ )(τ − x) dτ x t (tτ − tx − τ 2 + τ x) dτ = x = 12 1 1 tτ − txτ − τ 3 + τ 2 x 2 3 2 t x = 13 1 1 1 1 1 t − t2x − t3 + t2 x − tx2 + tx2 + x3 − x3 2 3 2 2 3 2 = 13 12 1 1 t − t x + tx2 − x3 6 2 2 6 = 1 ( t − x) 3 6 Hence u(x, t) = 13 6t 13 6t if t < x − 1 ( t − x) 3 6 if t > x, or u(x, t) = 13 1 t − ( t − x) 3 U 0 ( t − x) . 6 6 ...
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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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