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- fi eld on U X. In this section we will consider the ordinary di ff 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 ff 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 ff 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 fi 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 fi 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 Z ( x ) = | x | α with 0 < α < 1 and we now consider the ordinary di ff erential equation (5.6) ˙ y ( t ) = Z ( y ( t )) = | y ( t ) | α with y (0) = x R . Working as above we fi nd, if x 6 = 0 that t = Z t 0 ˙ y ( τ ) | y ( t ) | α = Z y ( t ) 0 | u | α du = [ y ( t )] 1 α x 1 α 1 α , where u 1 α := | u | 1 α sgn( u ) . Since sgn( y ( t )) = sgn( x ) the previous equation im- plies sgn( x )(1 α ) t = sgn( x ) h sgn( y ( t )) | y ( t ) | 1 α sgn( x ) | x | 1 α i = | y ( t ) | 1 α | x | 1 α and therefore, (5.7) y ( t, x ) = sgn( x ) ³ | x | 1 α + sgn( x )(1 α ) t ´ 1 1 α is uniquely determined by this formula until the fi rst time t
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