Chem Differential Eq HW Solutions Fall 2011 3

Chem Differential Eq HW Solutions Fall 2011 3 - Section 1.2...

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Unformatted text preview: Section 1.2 Solving and Interpreting a Partial Differential Equation For m = 1, 3, 5, and 7, the second term is 0, because cos times, we have, for, m = 1, 3, 5, and 7, u(x, mπ 2 3 = 0. Hence at these m 1 ) = sin πx cos πt + sin 3πx cos 3πt. 4 3 To say that the graph of this function is symmetric about x = 1/2 is equivalent to the assertion that, for 0 < x < 1/2, u(1/2 + x, m ) = u(1/2 − x, m ). Does this 4 4 equality hold? Let’s check: u(1/2 + x, m )= 4 = mπ 1 3mπ + sin 3π(x + 1/2) cos 4 3 4 mπ 1 3mπ cos πx cos − cos 3πx cos , 4 3 4 sin π(x + 1/2) cos where we have used the identities sin π(x + 1/2) = cos πx and sin 2π(x + 1/2) = − cos 3πx. Similalry, u(1/2 − x, m )= 4 = So u(1/2 + x, m 4) mπ 1 3mπ + sin 3π(1/2 − x) cos 4 3 4 mπ 1 3mπ cos πx cos − cos 3πx cos . 4 3 4 sin π(1/2 − x) cos = u(1/2 − x, m 4 ), as expected. 17. Same reasoning as in the previous exercise, we find the solution u(x, t) = 1 πx cπt 1 3πx 3cπt 2 7πx 7cπt sin cos + sin cos + sin cos . 2 L L 4 L L 5 L L 21. (a) We have to show that u( 1 , t) is a constant for all t > 0. With c = L = 1, 2 we have u(x, t) = sin 2πx cos 2πt ⇒ u(1/2, t) = sin π cos 2πt = 0 for all t > 0. (b) One way for x = 1/3 not to move is to have u(x, t) = sin 3πx cos 3πt. This is the solution that corresponds to the initial condition u(x, 0) = sin 3πx and ∂u (x, 0) = 0. For this solution, we also have that x = 2/3 does not move for ∂t all t. 25. The solution (2) is u(x, t) = sin Its initial conditions at time t0 = u(x, 3L 2c 3L πx ) = sin cos 2c L πx πct cos . L L are πc 3L · L 2c = sin πx 3π cos = 0; L 2 and ∂u 3L πc πx (x, ) = − sin sin ∂t 2c L L πc 3L · L 2c =− πc πx 3π πc πx sin sin = sin . L L 2 L L ...
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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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