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Given the following evolution IVP/BVP: $\begin{cases} u_t - \Delta u = f &\text{in } U \times(0,\infty)\\ u=0 &\text{on } \partial U \times (0,\infty)\\ u= g &\text{on } U \times \{t=0\} \end{cases}$,

$g\in L^2(U) $, $f \in L^{\infty}(U_T)$ for each $T>0$ and $U_T:= U\times (0,T]$.

Suppose $\tau > 0$ and $f(x,t) = f(x,t+\tau)$ with $x \in U$, for each $t\geq 0$. Prove that there is a $L^2$ function $g$ whose associated $u$ solving the PDE has the same periodic property as $f$.

Showing uniqueness is pretty straightforward, but I am clueless for existence.

Any hint or suggestion is tremendously appreciated!

S_j
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    An alternative to Banach's theorem is to use an eigenfunction expansion $u(t,x)=\sum_1^\infty u_n(t)\phi_n(x)$. The $u_n$'s solve ODEs which are easy to solve, and imposing $\tau-$periodicity allows you to solve for the $g_n$'s in terms of the Fourier coefficients $(f(t,.),\phi_n)$. Then you just need to show that $G:=\sum g_n\phi_n(x)$ is a well defined sum in $L^2$. – user254433 Apr 15 '18 at 09:13
  • @user254433 what do you mean by "imposing $\tau$- periodicity allows you to solve for th $g_n$'s in terms of the Fourier coefficients"? What are the $g_n$ in this case? I'm sorry Im a bit confused since for each choice of the base function $g$, we have a (weak) solution for the PDE by 2nd-order parabolic PDE theory. It seems to me that you are assuming something for $u_n$ and then solve for $g_n$? – S_j Apr 15 '18 at 19:36
  • Sorry for being unclear, I'll elaborate more below. – user254433 Apr 15 '18 at 20:55

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This is an expanded version of my comment:

Let $\{\phi_n\}_1^\infty$ and $\{\lambda_n\}_1^\infty$ be the eigenfunctions and eigenvalues of $-\Delta$ on $U$, and project all functions involved onto the orthonormal basis:

\begin{align} u_n(t):=(u(t,x),\phi_n(x)),\\ g_n:=(g(x),\phi_n(x)),\\ f_n(t):=(f(t,x),\phi_n(x)). \end{align} The $f_n$'s are $\tau$-periodic, and by completeness of the basis, it suffices to consider whether the $u_n$'s are $\tau$-periodic as well.

The $u_n$'s solve the initial value problems $$ u_n'(t)+\lambda_n u_n(t)=f_n(t),\\ u_n(0)=g_n, $$ which have the solutions $$ u_n(t)=e^{-\lambda_n t}g_n+\int_0^te^{-\lambda_n(t-s)}f_n(s)ds. $$ We have $\tau$-periodicity if and only if $u_n(t+\tau)=u_n(t)$. Substituting the solution into this expression, extending $f_n$ to $\mathbb R$ using periodicity, and changing variables $s\to s+\tau$ in the integral yields the equation $$ e^{-\lambda_n(t+\tau)}g_n+\int_{-\tau}^te^{-\lambda_n(t-s)}f_n(s)ds=e^{-\lambda_nt}g_n+\int_{0}^te^{-\lambda_n(t-s)}f_n(s)ds, $$ which reduces to \begin{align} (e^{-\lambda_n\tau}-1)g_n&=-\int_{-\tau}^0e^{\lambda_ns}f_n(s)ds\\ &=-e^{-\lambda_n \tau}\int_0^\tau e^{\lambda_n s}f_n(s)ds. \end{align} Since $\lambda_n>0$ for all $n$, we conclude that $$ g_n=\frac{1}{e^{\lambda_n\tau}-1}\int_0^\tau e^{\lambda_n s}f_n(s)ds. $$

user254433
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  • Thank you very much for your help. I understand your approach now. One more thing, so for each $x \in U$, we look at the orthonormal basis provided above (the $u_n(t)$, $f_n(t)$ ,...). They become the basis for that particular point in $U$, which are functions from $(0,\infty) \rightarrow \mathbb{R}^2$. After those calculations involving integrating vector-valued functions, we obtain, for each $n$ and the corresponding eigenvalue $\lambda _ n$, the coefficient $g_n$ to be a function depending on $x$. Is this correct? The last step is to show $G$ converges on $U$. Is this correct? – S_j Apr 16 '18 at 00:27
  • How do we even show convergence for that? I can deduce that $|g_n| \leq \frac {||f_n||_{L^{\infty}}}{\lambda _n}$, but then it's not so clear that the series converges let alone well-definedness in $L^2$. – S_j Apr 16 '18 at 01:48
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    @S_j The orthonormal basis is the $\phi_n$'s, and $(u(t,x),\phi_n(x))$ denotes $\int_U u(t,x)\phi_n(x)dx$ (the inner product on $L^2$). The $g_n$'s are thus constants, and the $u_n(t)$'s are maps $(0,\infty)\to\mathbb R$.

    I have to think about convergence. If $f\in C^\infty$, then integrating the $g_n$ integral by parts gives series convergence (noting that $\lambda_n\sim n^{2/d}$ by Weyl's Law). Maybe after that approximation can be used to deduce the result for $f\in L^2$.

     – user254433 Apr 16 '18 at 02:16
  • @S_j Sorry I removed my edits since they didn't work. A tricky problem! – user254433 Apr 16 '18 at 04:46
  • No worries I was gonna mention something wasn't quite right. – S_j Apr 16 '18 at 08:56
  • $\lambda_{n} \sim n^{\frac{2}{d}}$ doesn't give convergence since $d>n$. – S_j Apr 17 '18 at 04:13