L61 - Fall 2003 Math 308/501502 6 Theory of Higher-Order...

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Fall 2003 Math 308/501–502 6 Theory of Higher-Order Linear ODEs 6.1 Basic Theory of Linear ODEs Wed, 24/Sep c ± 2003, Art Belmonte Summary We give an overview of the theory of n th order linear differential equations; here n 1. Of course, the case where n = 1 was dealt with in Section 2.3, whereas n = 2 is dealt with in Chapter 4. Using summation and matrix notation along with a smattering of linear algebra concepts, we treat the general case once and for all. In the following, I = ( a , b ) is an open real interval. TERMINOLOGY A linear ODE of order n has the form n X k = 0 a k ( x ) y ( k ) ( x ) = b ( x ) . Here the a k depend only on x (the independent variable), not on y (the dependent variable). The equation has constant coefficients if the a k are constants; otherwise, it has variable coefficients .If b ( x ) or g ( x ) is zero on I , then the equation is homogeneous ; otherwise it is nonhomogeneous . With the a k and b continuous on I and a n ( x ) 6= 0on I ,divideto obtain the standard form y ( n ) ( x ) + n X j = 1 p j ( x ) y ( n - j ) ( x ) = g ( x ) or L [ y ] ( x ) = g ( x ) ,where L = D n + n j = 1 p j D n - j is called a linear differential operator ; i.e., L " m X i = 1 y i # = m X i = 1 L [ y i ]and L [ cy ] = cL [ y ]. (This follows from properties of differentiation.) DEFINITIONS Let f 1 ,..., f n be n functions that are differentiable ( n - 1 ) times. The Wronskian matrix is n × n square array of derivatives M = f 1 f 2 ··· f n f 0 1 f 0 2 f 0 n . . . . . . . . . . . . f ( n - 1 ) 1 f ( n - 1 ) 2 f ( n - 1 ) n , Its determinant W is called the Wronskian [determinant]. The m functions f 1 f m are linearly dependent on I if there exist constants c 1 ,... c m , not all zero, such that m X k = 1 c k f k ( x ) = 0 for all x I . Otherwise, the functions are linearly independent . (Note that in the linearly dependent case, one function is a linear combination of the other m - 1 functions.) THEOREMS Existence and Uniqueness Let g ( x ) and p k ( x ) be continuous on I , an interval containing x 0 . For any choice of constants γ k , there exists a unique solution y ( x ) on the entire interval I to the initial value problem y ( n ) ( x ) + n X j = 1 p j ( x ) y ( n - j ) ( x ) = g ( x ), y ( k ) ( x 0 ) = γ k , k = 0 n - 1 . Representation of Solutions (Homogeneous Case) Suppose that y 1 y n are n solutions on I of y ( n ) ( x ) + n X j = 1 p j ( x ) y ( n - j ) ( x ) = 0 ,( * ) where the p k are continuous on I . If the Wronskian of the y k is nonzero at some point x 0 I ,then every solution of (*) on I may be expressed as y ( x ) = n X k = 1 c k y k ( x ) ,wherethe c k are constants.
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This note was uploaded on 03/29/2008 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.

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L61 - Fall 2003 Math 308/501502 6 Theory of Higher-Order...

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