{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
48 BRUCE K. DRIVER 4. Cauchy — Kovalevskaya Theorem As a warm up we will start with the corresponding result for ordinary di ff 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 ) is the unique solution to the ODE (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 fi 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 as can be seen by showing ¯ z u z = 0 , i.e. showing z u z satis fi 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 ) solving Eq. (4.1). Proof. (Fourth Proof: Method of Majorants) Suppose for the moment we have an analytic solution to Eq. (4.1). Then by repeatedly di ff erentiating Eq. (4.1) we
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 fi rst few polynomials are p 1 ( x ) = x, p 2 ( x, y ) = xy, p 3 ( x, y, z ) = x 2 z + xy 2 . Notice that these polynomials are universal, i.e. are independent of the function f and ¯ ¯ ¯ u ( n ) (0) ¯ ¯ ¯ = ¯ ¯ ¯ p n ³ f (0) , . . . , f ( n 1) (0) ´¯ ¯ ¯ p n ³ | f (0) | , . . . , ¯ ¯ ¯ f ( n 1) (0) ¯ ¯ ¯ ´ p n ³ g (0) , . . . , g ( n 1) (0) ´ where g is any analytic function such that ¯ ¯ f ( k ) (0) ¯ ¯ g ( k ) (0) for all k Z + . (We will abbreviate this last condition as f ¿ g. ) Now suppose that v ( t ) is a solution to (4.4) ˙ v ( t ) = g ( v ( t )) with v (0) = 0 , then we know from above that v ( n ) (0) = p n ³ g (0) , . . . , g ( n 1) (0) ´ ¯ ¯ ¯ u ( n ) (0) ¯ ¯ ¯ for all n.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}