{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chem Differential Eq HW Solutions Fall 2011 58

Chem Differential Eq HW Solutions Fall 2011 58 - 58 Chapter...

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

View Full Document Right Arrow Icon
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 - u 2 ( r,θ ) = - r 3 sin 3 θ . reasoning as in Exercise 10, we find that the solution is u 2 ( r,θ,t ) = n =1 b 3 n J 3 ( α 3 n r ) sin(3 θ ) e - α 2 3 n t , where b 3 n = - 2 πJ 4 ( α 3 n ) 2 1 0 2 π 0 r 3 sin 2 3 θJ 3 ( α 3 n r ) rdθdr = - 2 J 4 ( α 3 n ) 2 1 0 r
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}