{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chem Differential Eq HW Solutions Fall 2011 57

# Chem Differential Eq HW Solutions Fall 2011 57 - u 1 rr 1 r...

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

Section 4.3 Vibrations of a Circular Membrane: General Case 57 u = r 2 l , dv = r k +1 J k +1 ( r ) dr , and hence du = 2 lr 2 l - 1 dr and v = r k +1 J k +1 ( r ): r k +1+2 l J k ( r ) dr = r 2 l [ r k +1 J k ( r )] dr = r 2 l r k +1 J k +1 ( r ) - 2 l r 2 l - 1 r k +1 J k +1 ( r ) dr = r k +1+2 l J k +1 ( r ) - 2 l r k +2 l J k +1 ( r ) dr = r k +1+2 l J k +1 ( r ) - 2 l r ( k +1)+1+2( l - 1) J k +1 ( r ) dr and so, by the induction hypothesis, we get r k +1+2 l J k ( r ) dr = r k +1+2 l J k +1 ( r ) - 2 l l - 1 n =0 ( - 1) n 2 n ( l - 1)! ( l - 1 - n )! r k +2 l - n J k + n +2 ( r ) + C = r k +1+2 l J k +1 ( r ) + l - 1 n =0 ( - 1) n +1 2 n +1 l ! ( l - ( n + 1))! r k +1+2 l - ( n +1) J k +( n +1)+1 ( r ) + C = r k +1+2 l J k +1 ( r ) + l m =1 ( - 1) m 2 m l ! ( l - m )! r k +1+2 l - m J k + m +1 ( r ) + C = l m =0 ( - 1) m 2 m l ! ( l - m )! r k +1+2 l - m J k + m +1 ( r ) + C, which completes the proof by induction for all integers k 0 and all l 0. 13. The proper place for this problem is in the next section, since its solution invovles solving a Dirichlet problem on the unit disk. The initial steps are similar to the solution of the heat problem on a rectangle with nonzero boundary data (Exercise 11, Section 3.8). In order to solve the problem, we consider the following two subproblems: Subproblem #1 (Dirichlet problem)
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( u 1 ) rr + 1 r ( u 1 ) r + 1 r 2 ( u 1 ) θθ = 0 , < r < 1 , ≤ θ < 2 π, u 1 (1 , θ ) = sin 3 θ, ≤ θ < 2 π. Subproblem #2 (to be solved after Fnding u 1 ( r, θ ) from Subproblem #1) ( u 2 ) t = ( u 2 ) rr + 1 r ( u 2 ) r + 1 r 2 ( u 2 ) θθ , < r < 1 , ≤ θ < 2 π, t > , u 2 (1 , θ, t ) = 0 , ≤ θ < 2 π, t > , u 2 ( r, θ, 0) =-u 1 ( r, θ ) , < r < 1 , ≤ θ < 2 π. You can check, using linearity (or superposition), that u ( r, θ, t ) = u 1 ( r, θ ) + u 2 ( r, θ, t ) is a solution of the given problem. The solution of subproblem #1 follows immediately from the method of Sec-tion 4.5. We have u 2 ( r, θ ) = r 3 sin 3 θ....
View Full Document

{[ snackBarMessage ]}