Chem Differential Eq HW Solutions Fall 2011 58

Chem Differential - 58 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates We now solve subproblem#2 which is a heat

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: 58 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates We now solve subproblem #2, which is a heat problem with 0 boundary data and initial temperature distribution given by −u2(r, θ) = −r3 sin 3θ. reasoning as in Exercise 10, we find that the solution is ∞ 2 b3nJ3 (α3nr) sin(3θ)e−α3n t, u2(r, θ, t) = n=1 where b3n = = = = = −2 πJ4 (α3n)2 1 2π r3 sin2 3θJ3 (α3nr)r dθ dr 0 1 0 −2 r4J3 (α3nr) dr J 4 ( α 3n ) 2 0 α3n −2 1 s4 J3(s) ds J 4 ( α 3n ) 2 α 5n 0 3 α3n −2 14 s J4 (s) 2 α5 J 4 ( α 3n ) 3n 0 −2 . α 3n J 4 ( α 3n ) Hence ∞ u(r, θ, t) = r3 sin 3θ − 2 sin(3θ) n=1 (where α3nr = s) J3 (α3nr) −α2n t e 3. α3nJ4 (α3n) ...
View Full Document

This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

Ask a homework question - tutors are online