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# chap5 - A NALYSIS TOOLS W ITH APPLICATIONS 55 5. Ordinary...

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ANALYSIS TOOLS WITH APPLICATIONS 55 5. Ordinary Differential Equations in a Banach Space Let X be a Banach space, U o X, J =( a, b ) 3 0 and Z C ( J × U, X ) Z is to be interpreted as a time dependent vector- f eld on U X. In this section we will consider the ordinary di f erential equation (ODE for short) (5.1) ˙ y ( t )= Z ( t, y ( t )) with y (0) = x U. The reader should check that any solution y C 1 ( J, U ) to Eq. (5.1) gives a solution y C ( J, U ) to the integral equation: (5.2) y ( t )= x + Z t 0 Z ( τ,y ( τ )) and conversely if y C ( J, U ) solves Eq. (5.2) then y C 1 ( J, U ) and y solves Eq. (5.1). Remark 5.1 . For notational simplicity we have assumed that the initial condition for the ODE in Eq. (5.1) is taken at t =0 . There is no loss in generality in doing this since if ˜ y solves d ˜ y dt ( t )= ˜ Z ( t, ˜ y ( t )) with ˜ y ( t 0 )= x U i f y ( t ):=˜ y ( t + t 0 ) solves Eq. (5.1) with Z ( t, x )= ˜ Z ( t + t 0 ,x ) . 5.1. Examples. Let X = R ,Z ( x )= x n with n N and consider the ordinary di f erential equation (5.3) ˙ y ( t )= Z ( y ( t )) = y n ( t ) with y (0) = x R . If y solves Eq. (5.3) with x 6 =0 , then y ( t ) is not zero for t near 0 . Therefore up to the f rst time y possibly hits 0 , we must have t = Z t 0 ˙ y ( τ ) y ( τ ) n = Z y ( t ) 0 u n du = [ y ( t )] 1 n x 1 n 1 n if n> 1 ln ¯ ¯ ¯ y ( t ) x ¯ ¯ ¯ if n =1 and solving these equations for y ( t ) implies (5.4) y ( t )= y ( t, x )= ( x n 1 1 ( n 1) tx n 1 if n> 1 e t x if n =1 . The reader should verify by direct calculation that y ( t, x ) de f ned above does in- deed solve Eq. (5.3). The above argument shows that these are the only possible solutions to the Equations in (5.3). Notice that when n =1 , the solution exists for all time while for n> 1 , we must require 1 ( n 1) tx n 1 > 0 or equivalently that t< 1 (1 n ) x n 1 if x n 1 > 0 and t> 1 (1 n ) | x | n 1 if x n 1 < 0 .

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56 BRUCE K. DRIVER Moreover for n> 1 ,y ( t, x ) blows up as t approaches the value for which 1 ( n 1) tx n 1 =0 . The reader should also observe that, at least for s and t close to 0 , (5.5) y ( t, y ( s, x )) = y ( t + s, x ) for each of the solutions above. Indeed, if n =1 Eq. (5.5) is equivalent to the well know identity, e t e s = e t + s and for n> 1 , y ( t, y ( s, x )) = y ( s, x ) n 1 p 1 ( n 1) ty ( s, x ) n 1 = x n 1 1 ( n 1) sx n 1 n 1 s 1 ( n 1) t · x n 1 1 ( n 1) sx n 1 ¸ n 1 = x n 1 1 ( n 1) sx n 1 n 1 q 1 ( n 1) t x n 1 1 ( n 1) sx n 1 = x n 1 p 1 ( n 1) sx n 1 ( n 1) tx n 1 = x n 1 p 1 ( n 1)( s + t ) x n 1 = y ( t + s, x ) . Now suppose
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## This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.

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chap5 - A NALYSIS TOOLS W ITH APPLICATIONS 55 5. Ordinary...

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