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Let us consider the following linear system with damping: $$ \begin{cases} u_t - u_x = -\frac{1}{2} (u+v)\\ v_t + v_x = -\frac{1}{2} (u+v) \end{cases} $$ Let's write the solution as $w=(u,v)$ corresponding to initial data $w_0 := w(0,\cdot) \in L^1(\mathbb R) \cap L^2(\mathbb R)$. Can we prove that $$w=w_1+w_2$$ where $$\|w_1\|_{L^2} \lesssim e^{-\alpha t}\|w_0\|_{L^2}, \qquad \|w_2\|_{L^\infty} \lesssim t^{-1/2}\|w_0\|_{L^1}$$ or something similar? Does more hold?

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    $\begingroup$ Going to the Fourier side reduces the question to the system of two damped ODE. Thus what we really want is to investigate the decay of the solutions to the system $u_t=i\xi u-\frac 12(u+v)$, $v_t=-i\xi v-\frac12(u+v)$ with the hope that the decay rate is like $e^{-c\min(1,\xi^2)t}$ though I see no reason why it shouldn't be $\min(1,|\xi|)t$ in the exponent, which would give a better bound with $t^{-1}$ for the second (low frequency) component. OK, let's think about it a bit :-) $\endgroup$
    – fedja
    Commented Dec 29, 2021 at 3:38
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    $\begingroup$ Indeed, if $\Sigma=u+v$ and $\sigma=u-v$ for that ODE, then $|\Sigma|^2+|\sigma|^2-\lambda\Im(\Sigma\bar\sigma)$ with $\lambda\approx \xi$ for small $\xi>0$, say, gives the Lapunov function yielding the quadratic in $\xi$ bound I mentioned. The question whether the linear bound is possible remains open. $\endgroup$
    – fedja
    Commented Dec 29, 2021 at 4:01
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    $\begingroup$ No, the linear bound is impossible. Let's say we start with $\Sigma=0$ and $\sigma=1$. Then $\Sigma$ driven by the equation $\Sigma_t=i\xi\sigma-\Sigma$ remains imaginary and gets saturated at the size $\xi$, so $\sigma_t=i\xi\Sigma$ drives $\sigma$ down at the speed $\xi^2$ only, at least initially. So, yeah, what has been requested is true, but hardly much more. $\endgroup$
    – fedja
    Commented Dec 29, 2021 at 4:09
  • $\begingroup$ @fedja Thank you so much! I have a couple of questions on wht you wrote: (1) How does the Lyapunov function in your second comment give? (2) Can we get the same decay if we replace $\frac{1}{2}$ in both lines of the system with $\tfrac{1}{2} \mathbf{1}_{(-\infty,-1) \cup (1,\infty)}$? Or, even better, can we compare the two systems in any way? $\endgroup$
    – Riku
    Commented Dec 29, 2021 at 9:39
  • $\begingroup$ 1) It shows that the Fourier transform $|\widehat w(\xi,t)|\le C|\widehat w(\xi,0)|e^{-c\min(1,\xi^2)t}$, after which you can just split into the low frequency ($|\xi|\le 1$) and high frequency ($|\xi|>1$) parts. 2) I don't know at the moment. That seems to be a substantially more difficult question because the pure frequency split is no longer possible. I'll think of it if it is what you are really after :-) $\endgroup$
    – fedja
    Commented Dec 29, 2021 at 16:13

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