pde4 - 48 BRUCE K. DRIVER 4. Cauchy Kovalevskaya Theorem As...

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48 BRUCE K. DRIVER 4. Cauchy — Kovalevskaya Theorem As a warm up we will start with the corresponding result for ordinary di f erential equations. Theorem 4.1 (ODE Version of Cauchy — Kovalevskaya, I.) . Suppose a> 0 and f :( a, a ) R is real analytic near 0 and u ( t ) istheun iqueso lu t iontotheODE (4.1) ˙ u ( t )= f ( u ( t )) with u (0) = 0 . Then u is also real analytic near 0 . We will give four proofs. However it is the last proof that the reader should focus on for understanding the PDE version of Theorem 4.1. Proof. (First Proof.)If f (0) = 0 , then u ( t )=0 for all t is the unique solution to Eq. (4.1) which is clearly analytic. So we may now assume that f (0) 6 =0 . Let G ( z ):= R z 0 1 f ( u ) du, another real analytic function near 0 . Then as usual we have d dt G ( u ( t )) = 1 f ( u ( t )) ˙ u ( t )=1 and hence G ( u ( t )) = t. We then have u ( t )= G 1 ( t ) which is real analytic near t =0 since G 0 (0) = 1 f (0) 6 =0 . Proof. (Second Proof.) For z C let u z ( t ) denote the solution to the ODE (4.2) ˙ u z ( t )= zf ( u z ( t )) with u z (0) = 0 . Notice that if u ( t ) is analytic, then t u ( tz ) satis f es the same equation as u z . Since G ( z,u )= zf ( u ) is holomorphic in z and u, it follows that u z in Eq. (4.2) depends holomorphically on z a scanbeseenbyshow ing ¯ z u z =0 , i.e. showing z u z satis f es the Cauchy Riemann equations. Therefore if ±> 0 is chosen small enough such that Eq. (4.2) has a solution for | t | and | z | < 2 , then (4.3) u ( t )= u 1 ( t )= X n =0 1 n n ! n z u z ( t ) | z =0 . Now when z R ,u z ( t )= u ( tz ) and therefore n z u z ( t ) | z =0 = n z u ( tz ) | z =0 = u ( n ) (0) t n . Putting this back in Eq. (4.3) shows u ( t )= X n =0 1 n ! u ( n ) (0) t n which shows u ( t ) is analytic for t near 0 . Proof. (Third Proof.) Go back to the original proof of existence of solutions, but now replace t by z C and R t 0 f ( u ( τ )) by R z 0 f ( u ( ξ )) = R 1 0 f ( u ( tz )) zdt. Then the usual Picard iterates proof work in the class of holomorphic functions to give a holomorphic function u ( z )
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PDE LECTURE NOTES, MATH 237A-B 49 learn ¨ u ( t )= f 0 ( u ( t )) ˙ u ( t )= f 0 ( u ( t )) f ( u ( t )) u (3) ( t )= f 00 ( u ( t )) f 2 ( u ( t )) + [ f 0 ( u ( t ))] 2 f ( u ( t )) . . . u ( n ) ( t )= p n ³ f ( u ( t )) ,...,f ( n 1) ( u ( t )) ´ where p n is a polynomial in n variables with all non-negative integer coe cients. The f rst few polynomials are p 1 ( x )= x, p 2 ( x, y )= xy, p 3 ( x, y, z )= x 2 z + xy 2 .
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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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pde4 - 48 BRUCE K. DRIVER 4. Cauchy Kovalevskaya Theorem As...

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