Differential Equations Lecture Work Solutions 201

Differential Equations Lecture Work Solutions 201 - x = 1 ,...

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5. b. 2 u ∂x 2 2 2 u ∂x∂y +5 2 u ∂y 2 + ∂u ∂y =0 A =1 ,B = 2 ,C =5 ,D =0 ,E =1 ,F = G =0 . The discriminant ∆ = ( 2) 2 4 · 1 · 5=4 20 = 16 < 0 Therefore the problem is elliptic. The characteristic equation is dy dx = B ± 2 A = 2 ± 4 i 2 = 1+2 i The solutions are y = x ± 2 ix + C The transformation is ξ = y + x +2 ix η = y + x 2 ix In elliptic problems we use another transformation (to stay with real functions) α = y + x β =2 x Since these are linear functions, we only need the Frst partials
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Unformatted text preview: x = 1 , y = 1 x = 2 , y = 0 Use the formulae for A , B etc with for and for , we have A = 1 1 2 2 1 1 + 5 1 2 = 1 2 + 5 = 4 B = 0 as should be for elliptic C = A = 4 D = 1 1 = 1 E = 1 0 = 0 F = G = 0 Thus the canonical form is 201...
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