(A)periodicity and (In)dependence on the boundary condition for optimization problem related to ODE

The question is pair to MO117505 and translates some problem on error-correction codes to similar problem about differential operators. (See also If “force” is periodic does it imply “velocity” is periodic ?).

Setup Consider two given periodic functions $(r_1(x), r_2(x))$ (with same period "T"); two linear differential operators $D_1 (s(x)) = (\sum_i a_i s^{(i)}(x) )$ , and similar $D_2$ with some other coefficients (both coefficients are (say) constant, may be I also need something like positive definite - not sure)

Optimization problem: Find $(s(x) )$, such that $||D_1(s) - r_1||^2 + ||D_2(s) - r_2||^2 -> min.$ Where $||.||$ is the $L^2$-norm on the intervals specified below.

Boundary conditions Let us fix boundary condition at zero: $b_0 = s(0), b_1=s'(0) ...$.

Question 1 Consider optimization problem on interval $[0, Y]$. Does the result of optimization depends on the boundary conditions for large x>>T ? I mean take $Y=NT$ very large comparing with period "T" and look on $s(x)$ for $x$ very large, I would expect the difference between $s_{boundary1}(x)-s_{boundary2}(x)$ will be small, is it true ?

Question 2 Is true that function s(x) tends to be periodic for large "x"? I.e. $|s(x+T)-s(x)|-> 0$ for $x-> \infty$

Question 3 Look on s(x) for $x \in$ interval $[(N-1)T, NT]$ is it true that for $N-> \infty$ we get same $s(x)$ as if we solve the same optimization problem without boundary conditions ?

PS It might be that Question 3 is not clearly formulated.

Let me try again.

Question 3b Assume the answer on question 2 is positive i.e. s(x) does not depend on initial conditions for large x. Is it true in this case that the solution of the same optimization problem WITHOUT boundary conditions will be the same as s(x) for large x ?

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What is $|D(s)-r|$? Is this $L^2$-norm, $\| D(s)-r\|$ ? If yes, on what interval? – Alexandre Eremenko Dec 29 '12 at 17:13
@Alexandre Yes it is L^2 norm, on the intervals specified in questions: (0,NT). I corrected notations thank you ! – Alexander Chervov Dec 29 '12 at 19:16

The answer to Question 3 is yes, and to Questions 1,2 the answer is "no", unless the initial conditions are very special.

To address these questions, assume wlog that $T=2\pi$, and expand everything into Fourier series. Let $a_n, b_n$ be coefficients of $r_1,r_2$, and $s_n$ coefficients of $s$. Let $D_i=P_i(D)$, where $D=d/dx$. Let $c_{i,n}=P_i(n\sqrt{-1})$.

The quantity minimized is the sum of the squares of $L^2$ norms over $(0,T)$.

Let us address Question 3 first (no initial conditions). Then a simple computation shows that there is a minimum and its Fourier coefficients are $$s_n=\frac{a_nc_{1,n}+b_nc_{2,n}}{c_{1,n}^2+c_{2,n}^2}.$$ (Substitute the Fourier series, use Parseval, and solve the simple quadratic minimization problem). This solution is unique if $P_i$ have no common root (so the denominator is never $0$). If they have a common root, an exponential polynomial can be added to $s$ without affecting the minimum.

This gives a periodic solution. Well, we solved the problem on the interval $(0,T)$ and ANY function can be extended periodically. But one wants the extention to be smooth enough, so that operators $D_i$ make sense. The solution formula shows that the solution is in general has roughly $1$ more derivative in comparison with $r_i$. So if the $r_i$ are smooth enough than $s$ is smooth and periodic with the same period.

If we solve the same problem (without initial conditions) on an interval $(0,mT)$, but the $r_i$ still have period $T$, then the formula shows that the solution is the same (has period $T$) unless we are in the unlikely situation when the $P_i$ have a common root. But even in this situation, there always exists a solution with period $T$.

Now what happens when we add the initial condition (Questions 1,2). These conditions add a linear restriction on the $s_j$, so when we optimize on $(0,mT)$, in general harmonics $\exp(in/m)$ will be present.

And of course their contribution cannot tend to $0$ when $x\to\infty$ because they are periodic.

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Thank you very much ! I made some corrections to the question. – Alexander Chervov Dec 29 '12 at 19:26
And I made the corresponding correction to my answer:-) – Alexandre Eremenko Dec 29 '12 at 19:33