This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 Perturbation Techniques In this series of lectures, we will like to be introduced to the basics of asymptotic expansions and perturbation techniques. The most elementary application of perturbation techniques is to algebraic equations which depend on a small parameter. We will study three examples, each with increasing complexity. They will clearly demonstrate the ideas, the problems that arise under certain conditions, the reasons for these difficulties (nonuniformities in solutions), and the manner in which these nonuniformities are removed. Algebraic Equations: Example 1 : Consider the quadratic equation (or, a polynomial of any degree) 2 (3 2 ) 2 0, 1 x x  + + + = < << (1) When =0, the equation reduces to the solvable quadratic equation 2 3 2 ( 2)( 1) 0. x x x x + = = (2) Equation (2) is called the reduced equation corresponding to the original problem in equation (1). This equation has the zeros (or, roots) given by 1 2 1, 2. x x = = (3) Equation (1) is called the perturbed equation , and it can be thought of as a function ( , ) f x whose zeros are being sought. Before we give a general result for the function ( , ) f x , let us assume that the zeros of the quadratic equation, ( ), 1,2, i i x x i = = are analytical functions of the small parameter . Thus, there exist a Taylor series expansion in powers of which is convergent for every value of within some radius of convergence . Therefore, let & & & & & 4 2 3 2 3 1 2 3 1 sec ( ) ( ) ( ) ( ) ( ) ( ) . . . . . i i i first order ond zeroth third O terms O order order order O O O x x x x x x = = + + + + = & (4) The aim is now to be able to find the coefficients , 0,1,2,3,4,...... i x i = . Substituting equation (4) into the equation (1) and expanding the resulting expression as an infinite series in gives 2 3 2 2 3 1 2 3 1 2 3 ( ...) (3 2 )( ...) 2 0. x x x x x x x x + + + + + + + + + + + = Now, 2 2 3 2 2 2 3 2 3 2 1 2 3 1 2 3 1 2 3 2 2 2 3 1 2 1 3 1 2 ( ...) 2 ( ...) ( ...) 2 (2 ) (2 2 ) . . . . . . x x x x x x x x x x x x x x x x x x x x x x + + + + = + + + + + + + + = + + + + + + and 2 3 2 1 2 3 1 2 1 (3 2 )( . . . ) 3 (3 2 ) (3 2 ) . . . x x x x x x x x x + + + + + = + + + + + Thus, equation (1) becomes 2 2 2 1 1 2 1 2 1 ( 3 2) (2 3 2 1) (2 3 2 ) . . . . x x x x x x x x x x x  + + + + + + = (5) Setting the coefficient of each power of to zero seperately gives 2 3 2 x x  + = (6) 1 1 1 2 3 2 1 x x x x  + = (7) 2 2 2 1 2 1 2 3 2 x x x x x + = (8) Similar equations are obtained at higher orders in . Thus, we get an infinite sequence of algebraic equations which need to be solved sequentially to determine the unknowns , 0,1,2,3,4,.........
View
Full
Document
This note was uploaded on 12/28/2011 for the course ME 580 taught by Professor Na during the Fall '10 term at Purdue UniversityWest Lafayette.
 Fall '10
 NA

Click to edit the document details