Differential Equations Lecture Work Solutions 204

Differential Equations Lecture Work Solutions 204 - Use the...

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6. b. 2 u ∂x 2 +2 2 u ∂x∂y + 2 u ∂y 2 +7 ∂u ∂x 8 ∂u ∂y =0 A =1 ,B =2 ,C =1 ,D =7 ,E = 8 ,F = G =0 . The discriminant ∆ = 2 2 4 · 1 · 1=4 4=0 Therefore the problem is parabolic. The characteristic equation is dy dx = B ± 2 A = 2 2 =1 The solution is y = x + C The transformation is ξ = y x η = y arbitrary for parabolic Since these are linear functions, we only need the Frst partials ξ x = 1 y =1 η x =0 y =1 Note that the Jacobian of the transformation is NOT zero.
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Unformatted text preview: Use the formulae for A , B etc, we have A = 0 B = 0 as should be for parabolic C = 0 + 0 + 1 = 1 D = 7 ( 1) 8 1 = 15 E = 8 1 = 8 F = 0 G = 0 Thus the canonical form is u 15 u 8 u = 0 204...
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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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