Differential Equations Lecture Work Solutions 199

# Differential Equations Lecture Work Solutions 199 - α x =...

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5. a. 2 u ∂x 2 + 2 u ∂x∂y + 2 u ∂y 2 =0 A =1 ,B =1 ,C =1 . The discriminant ∆ = 1 2 4 · 1 · 1=1 4= 3 < 0 Therefore the problem is elliptic. The characteristic equation is dy dx = B ± 2 A = 1 ± i 3 2 The solutions are y = 1 2 x ± i 3 2 x + C The transformation is ξ = y 1 2 x i 3 2 x η = y 1 2 x + i 3 2 x In elliptic problems we use another transformation (to stay with real functions) α = y 1 2 x β = 3 2 x Since these are linear functions, we only need the Frst partials
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Unformatted text preview: α x = − 1 2 , α y = 1 β x = √ 3 2 , β y = 0 Use the formulae for A ∗ , B ∗ etc with α for ξ and β for η , we have A ∗ = 1 · ( − 1 2 ) 2 + 1 · ( − 1 2 ) · 1 + 1 · 1 2 = 3 4 B ∗ = 0 as should be for elliptic C ∗ = A ∗ = 3 4 The rest are zero (since they were zero and the transformation is linear) 199...
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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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