From the course notes page105 we have 7 from the

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

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

Unformatted text preview: 2)(1) + (8)(1)) + 2(16)(1)(1) = 36 (⌘ ⇠ )(8) = 3 ⇠ 3 ⌘ E= 6 1 1 1 6 3 3 ⇠ ⌘ D = 1 (⌘ ⇠ )(2) = 4 4 4 1 6 y ( ⇠ )(8) sin( E = sin((⌘ ⇠ , ⌘ )) = = 3⇠ ⇠ 3⌘ ⌘ ) ˆ f= 6 33 33 4 4 1 (⌘ ⇠ )(8) = 4 ⇠ ⌘ E= 1 ˆ 6 f = becomes = u 3 1 sin( ⇠ ⌘) u3 u Thus the PDEsin(y (⇠ , ⌘ )) 36ˆ + 3 (⇠ 33⌘ )ˆ + 4 (⇠ ⌘ )ˆ + sin( 4 ⇠ 1 ⌘ )ˆ = 0. 3 3u 3 4 1 1 4 1 ˆ form: u y (⇠ , ⌘1 (= sin( u⇠ 1 ⌘⇠ ⌘ )ˆ f )) ⇠ ⌘ ) 1 Canonical = sin( ˆ () u ⌘ )ˆ = 0. u 36 sin( 3 ⇠ 3u 3 ⌘ u Thus the PDE becomes108 36ˆ +3 (⇠ 27 )ˆ + 4 (⇠ ⌘ )ˆ + sin( 4 ⇠ 3 1 ⌘ )ˆ = 0. u 3 3 3u 1 1 4 11 CanonicalPDE becomes 1 (⇠6ˆ ⌘+ 1 (⇠ 27 ()ˆ + 4 (⇠ ⌘36 sin( 3 ⇠ 4 ⇠3 ⌘ )ˆ )ˆ 0. 0. ˆ u u== ⌘ ⇠ ⌘ )ˆ u )ˆ + sin( u ⌘u Thus the form: u 108 3 u )u Canonical form: u ˆ 1 108 (⇠ 3 ⌘ )u 1 27 (⇠ 3 ⌘ )ˆ u 1 36 4 sin( 3 ⇠ 3 3 1 ⌘ )ˆ 3u = 0. AMATH 350 Page 4 Assignment #8 Solutions - Fall 2012 5. From the course notes, page105 we have 7. From the Course Notes p. 105, we h...
View Full Document

This document was uploaded on 02/09/2014.

Ask a homework question - tutors are online