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Take $$ u_t(t) + A(t)u(t) = f(t), $$ $$ u(0) = u(T), $$ where $A$ is an linear elliptic operator and the first equation is an equality in $L^2(0,T;V^*)$ for $V \subset H \subset V^*$ Hilbert triple. (there is slight abuse of notation in the equality but never mind)

Under what conditions does a solution to this problem exist? By solution I mean $u \in L^2(0,T;V)$ with $u_t \in L^2(0,T;V^*)$ (or $u_t \in L^2(0,T;H)$ if data is smooth enough). Apart from requiring maybe $A(0) = A(T)$ and $f(0) = f(T)$.

How does one prove this via the Galerkin approach?

Thanks

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2 Answers 2

Step I. Solve the initial value problem $$ \left\{ \begin{array}{lll} X_t(t)=-A(t)X(t), \\ X(s)=I, \end{array} \right. $$ where $I : H\to H$ is the identity. Assume that $K(t,s)$ is the solution - this is a bounded operator from $H$ to $H$, for every $s,t$ - There is both theory and approximations for obtaining $K$, i.e., $$ K(t,s)=\lim_{n\to\infty}\prod_{k=1}^n \bigg(1-\frac{t-s}{n}A\Big(s+k\frac{t-s}{n}\Big)\bigg). $$

Step II. The solution of $$ X_t+A(t)X=f(t), \quad X(0)=X^0, $$ is then expressed as $$ X(t)=K(t,0)\,X^0+\int_0^t K(t,s)\,f(s)\,ds. $$

Step III. Check whether there exists a $X^0$, such that $$ X(T)=X(0) $$ or equivalently $$ K(T,0)\,X(0)+\int_0^T K(T,s)\,f(s)\,ds=X(0). $$ or $$ \big(K(T,0)-I\big)X(0)=-\int_0^T K(T,s)\,f(s)\,ds. $$ This only requires that the operator $K(T,0)-I :H\to H$ possesses an inverse or equivalently $$ 1\not\in \sigma\big(K(T,0)\big). $$

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Thanks for the answer! –  matt.x Feb 4 at 8:35

The conditions are the "usual" ones, see e.g. Brezis, Functional analysis, 2010:

o) $V \subset H$ is dense, $H$ and $H^*$ are identified, $V$ is separable.

o) $A(t) : V \to V^*$ linear, bounded and coercive (i.e. $A + \mu I$ is $V$-elliptic for a $\mu > 0$ large enough) uniformly in $t$, and $t \mapsto (A(t) u, v)$ is measurable for all $u, v \in V$.

o) $f \in L^2(0, T; V^*)$.

o) $u(0) = u(T)$ is understood as equality in $H$.

Then, there exists a unique solution $u \in H^1(0,T; V^*) \cap L^2(0,T; V)$ and it depends linearly and continuously on $f$.

Periodicity of data is not required; it doesn't even make sense.

The proof by the Faedo-Galerkin approximation method can be applied exactly as in the non-periodic case; see e.g. Evans, Partial differential equations, 1998; or Wloka Partial differential equations, 1987; and of course, Lions/Magenes and Dautray/Lions.

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Wow, thanks, I will check out that book. You say proof by Galerkin is the same. Do you mean to say that ODE system, the initial condition is replaced by $u_n(0) = u_n(T)$? If so, any idea where I can find existence result for that ODE system? I search long and hard but everything I find is too abstract/hard to apply for this case. –  matt.x Feb 4 at 8:34
    
I haven't checked out Dautray/Lions so apologies if that is covered in there... –  matt.x Feb 4 at 8:35
    
I'm sorry but I can't find anything in Brezis' book. Can you please recheck? –  matt.x Feb 4 at 8:41
    
@matt.x: I was referring to the comments on Chapter 10 (I chose that book because it might be available online). An yes, the ODE system with periodic initial/end condition. –  user66081 Feb 4 at 8:49
    
But I don't see anything about periodic ICs there. In fact I don't think what you wrote can be true: consider the equation $u_t - \Delta u = 0$ with zero Neumann BCs and periodic initial condition. Then it satisfies all your assumptions but there is no uniqueness: if $u$ is a solution so is $\lambda u$ for constant $\lambda$. I think in the weakly coercive case ($A+\mu I > ...$) one only gets existence. But I have yet to see a result for that. –  matt.x Feb 4 at 8:53

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