Week 15 - Solution

# Therefore y y cn e n y a 2cn e cn e bn a e 2bn a n

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Unformatted text preview: n sin nπ a and λn = 2 , where n = 1, 2, . . . III. Solving for the Y ′′ (y ) − λY (y ) = 0 with λ = α2 > 0, λn = nπ 2 a and Y (y ) = emy , Y ′′ (y ) − λY (y ) = 0 =⇒ m2 emy − λemy = 0 =⇒ m2 − α2 = 0 =⇒ m = ±α. Hence, Y (y ) = cn e nπ y a + dn e− nπ y a , where cn and dn are constants. Again, from the boundary condition, u (x, b) = X (x) Y (b) = 0 =⇒ Y (b) = 0 and Y (b) = cn e nπ b a + dn e− nπ b a = 0 =⇒ dn = −cn e 2bnπ a . Therefore, Y (y ) =cn e nπ y a =2cn e + −cn e bnπ a e 2bnπ a nπ ( y − b) a where e− − e− 2 nπ y a = cn e bnπ a e nπ ( y − b) a = 2cn e nπ ( y − b) a bnπ a sinh − e− nπ ( y − b) a nπ (y − b) , a ex − e−x sinh x = . 2 IV. From Step II and III, ∞ u (x, y ) = ∞ Xn (x) Yn (y ) = n=1 ∞ = 2βn cn e bnπ a sin n=1 Bn sin n=1 nπ nπ x sinh (y − b) a a bnπ nπ nπ x sinh (y − b) , where Bn = 2βn cn e a . a a V. Using the last boundary condition u (x, 0) = g (x), ∞ u (x, 0) = Bn sin n=1 nπ nπ x sinh (−b) = g (x) . a a VI. Finally, we have to ﬁnd the coeﬃcients Bn by a g (x) sin Bn = 0 sinh − bnπ a a 0 nπ x dx a nπ x sin a where Kn = 5 a = Kn 2 dx 2 . a sinh − bnπ a g (x) sin 0 nπ x dx, a (a) g (x) = x (a − x) a nπ x dx a 0 a2 nπ nπ a x + (a − 2x) x cos s...
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## This document was uploaded on 01/23/2014.

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