Chem Differential Eq HW Solutions Fall 2011 73

Chem Differential Eq HW Solutions Fall 2011 73 - Section...

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

View Full Document Right Arrow Icon
Section 4.7 Bessel’s Equation and Bessel Functions 73 As t →∞ , u ( t ) 0, J 0 ( u ) 1, and y ( t ) c 1 . In this case the solution is bounded. (ii) If c 1 ± =0and c 2 ± = 0, as t , u ( t ) 0, J 0 ( u ) 1, Y 0 ( u ) →-∞ . Since Y 0 will dominate, the solution will behave like case (i). It makes sense to have unbounded solutions because eventually the spring wears out and does not affect the motion. Newton’s laws tell us the mass will continue with unperturbed momentum, i.e., as t , y ±± = 0 and so y ( t )= c 1 t + c 2 ,a linear function, which is unbounded if c 1 ± =0 . 33. (a) In (13), let u 2 = t ,2 udu = dt , then Γ( x ± 0 t x - 1 e - t dt = ± 0 u 2( x - 1) e - u 2 (2 u ) du =2 ± 0 u 2 x - 1 e - u 2 du. (b) Using (a) Γ( x )Γ( y )=2 ± 0 u 2 x - 1 e - u 2 du 2 ± 0 v 2 y - 1 e - v 2 dv =4 ± 0 ± 0 e - ( u 2 + v 2 ) u 2 x - 1 v 2 y - 1 dudv. (c) Switching to polar coordinates: u = r cos θ , v = r sin θ , u 2 + v 2 = r 2 , dudv = rdrdθ ; for ( u, v ) varying in the Frst quadrant (0 u< and 0 v< ), we have 0 θ π 2 , and 0 r< , and the double integral in (b) becomes Γ( x )Γ(
Background image of page 1
This is the end of the preview. Sign up to access the rest of the 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