Differential Equations Lecture Work Solutions 204

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

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

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.
This is the end of the preview. Sign up to access the rest of the document.

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 ηη − 15 u ξ − 8 u η = 0 204...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online