periodic-solns

periodic-solns - a and f are periodic with period P : u 00...

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Math 425/525, Spring 2011 Jerry L. Kazdan Periodic Solutions of ODEs In class we discussed some aspects of periodic solutions of ordinary differential equations. From the questions I received, my presentation was not so clear. Here I’ll give a detailed formal proof for the first order equation u 0 ( x ) + a ( x ) u ( x ) = f ( x ) (1) where both a ( x ) and f ( x ) are periodic with period P , so, for instance, a ( x + P ) = a ( x ) for all x . Theorem 1 If u ( x ) is a solution of (1) with u ( P ) = u ( 0 ) , then u is periodic with period P, that is, u ( x + P ) = u ( x ) for all x. Proof The key ingredient is the uniqueness assertion: If u ( x ) and v ( x ) both satisfy (1) with u ( 0 ) = v ( 0 ) , then u ( x ) = v ( x ) for all x . We take this as a known fact. Since u is a solution of (1) for all x . then u 00 ( x + P ) + a ( x + P ) u ( x + P ) = f ( x + P ) for all x . Therefore, because both
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Unformatted text preview: a and f are periodic with period P : u 00 ( x + P ) + a ( x ) u ( x + P ) = f ( x ) for all x . Thus, if we let v ( x ) : = u ( x + P ) , then v ( x ) + a ( x ) v ( x ) = f ( x ) . But if u ( P ) = u ( ) , then v ( ) = u ( ) . Therefore by the uniqueness assertion, v ( x ) = u ( x ) for all x , that is, u ( x + P ) = u ( x ) for all x . The related assertion for a solution of a second order equation is essentially identical except there we need to assume that both u ( ) = u ( P ) and u ( ) = u ( P ) , since the corresponding uniqueness assertion for second order equations requires that. 1...
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