Differential Equations Lecture Work Solutions 203

# Differential Equations Lecture Work Solutions 203 - A = 1 9...

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6. a. 2 u ∂x 2 6 2 u ∂x∂y +9 2 u ∂y 2 + ∂u ∂x e xy =1 A =1 ,B = 6 ,C =9 ,D =1 ,E = F =0 ,G =1+ e xy . The discriminant ∆ = ( 6) 2 4 · 1 · 9=36 36 = 0 Therefore the problem is parabolic. The characteristic equation is dy dx = B ± 2 A = 6 2 = 3 The solution is y = 3 x + C The transformation is ξ = y +3 x η = y arbitrary for parabolic Since these are linear functions, we only need the Frst partials ξ x =3 y =1 η x =0 y =1 Note that the Jacobian of the transformation is NOT zero. Use the formulae for A , B etc, we have
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Unformatted text preview: A = 1 9 6 3 1 + 9 1 = 0 B = 0 as should be for parabolic C = 0 0 + 9 = 9 D = 1 3 = 3 E = 0 F = 0 G = 1 + e xy Need to substitute for x, y into G . Note that from the transformation y = and 3 x = , so G = 1 + e ( ) / 3 Thus the canonical form is u + 1 3 u = 1 9 1 + e ( ) / 3 203...
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## This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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