Differential Equations Lecture Work Solutions 199

Differential Equations Lecture Work Solutions 199 - x = 1 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x = 1 2 , y = 1 x = 3 2 , y = 0 Use the formulae for A , B etc with for and for , 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...
View Full Document

Ask a homework question - tutors are online