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If I have a PDE $$u_t = Au + f$$ with conditions $$u(0,x) = u(T,x)$$ then if it has a solution, why is the solution called periodic? Isn't it only true that $u(0) = u(T)$? It does not follow that $u(0+\epsilon) = u(T+\epsilon)$, which I would have thought is what periodic should be.

Is that all that is required for the solution to be called that?

Finally, is there any literature that address weak periodic solutions of parabolic PDE via Galerkin method?

aere
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  • Actually, the condition $u(0,x)=u(T,x)$ is enough to guarantee periodicity due to the Cauchy-Lipschitz theorem. Indeed, if $v(t,x)=u(t+T,x)$, then $v$ verifies the same PDE and the same initial condition as $u$, which implies $u=v$. – zuggg Jul 15 '13 at 12:14
  • @zuggg why does it verify the IC? I get $v(0) = u(T)$ only – aere Jul 15 '13 at 12:29
  • Let's say $u$ is a solution of the PDE with no assumption on its initial conditions, then it satisfies both the PDE and the IC $u(0,x)=u(0,x)$, obviously. Then, with the same notation as above, $v$ satisfies the PDE and the IC $v(0,x)=u(T,x)=u(0,x)$. – zuggg Jul 15 '13 at 12:36

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The implicit assumption is that your PDE has a well-posed Cauchy problem, and that $A,f$ are either independent of time $t$ or periodic with period $T$.

Under the above two assumptions, the uniqueness of solutions for the Cauchy problem will mean that $$ u(0,x) = u(T,x) \implies u(t,x) = u(T+t,x) $$

Willie Wong
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  • Thanks a lot. I assume you are not aware of any text Galerkin method is used to address this.. – aere Jul 15 '13 at 12:44
  • No. Sorry. (I also didn't notice your final sentence in your OP. I would suggest in the future asking one question per post; but that's just my opinion.) – Willie Wong Jul 15 '13 at 12:49