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Let's consider an unbounded second order linear differential operator $A := k(x)\frac{d^{2}}{dx^{2}}+\frac{d}{dx}$ defined over $L^{2}(0,1)$ whose domain is $H^{2}(0,1) \cap H_{0}^{1}(0,1)$. $k(x)$ is strictly positive over $(0,1)$, and smooth. Then this operator generates semigroup of operators $\{T_{t}\}_{t\geq0}$.

a. $\{T_{t}\}_{t\geq0}$ is called exponentially stable if there exist $M \geq 1$ and positive real number $\gamma$ such that $||T_{t}|| \leq Me^{-\gamma t}$.

b. $\{T_{t}\}_{t\geq0}$ is called asymptotically stable if for any element $h \in L^{2}(0,1)$, $lim_{t \to \infty}T_{t}(h) = 0$.

My problem is if we can put more assumptions on $k(x)$ so that operator $A$ is stable in either exponential or asymptoric sense.

Thank you so much!!

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  • $\begingroup$ Is $\{T_t\}$ bounded and strongly continuous? $\endgroup$ – Pedro Mar 3 at 1:00
  • $\begingroup$ @Pedro yes, I think so. $\endgroup$ – misakaczy Mar 3 at 20:40
  • $\begingroup$ Then, in order to obtain (b), it is enough to put assumptions on $k(x)$ so that $\mathbf{i}\mathbb R\subset\rho(A)$. $\endgroup$ – Pedro Mar 3 at 21:09
  • $\begingroup$ @Pedro Thank you for the comments. May I ask when you ask if $\{T_{t}\}$ is bounded you mean the usual bounded operator not uniformly bounded? Also, could you please be more accurate with " ... it is enough to put assumptions on $k(x)$ so that $\mathbf{i}\mathbb R\subset\rho(A)$ ... " please? Could you show an example? Thank you! $\endgroup$ – misakaczy Mar 3 at 22:32
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    $\begingroup$ In my comment I refer to the Arendt-Batty-Lyubich-Phong theorem (the meaning of bounded is given in the link). However, I don't know if this theorem can be applied in your case. $\endgroup$ – Pedro Mar 3 at 23:38
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I am sorry but I guess that for the problem I asked there may be an "easy" answer.

If $k(x)$ is just some positive real number $k/2$, then operator $A$ is self-adjoint and its eigenvalues form an orthnormal basis of $L^{2}((0,1),e^{2x/k}dx)$, thus, $A$ has pure point spectrum. Thus, all its spectrum are just its eigenvalues: $\{-(1+n^{2}\pi ^{2}k^{2})/k\}_{n=1}^{\infty}$. Denote the nth eigenfunction as $e_{n}$ and the nth eigenvalue as $-\lambda_{n}$, then $T_{t} = \sum_{n=1}^{\infty}exp(-\lambda t)e_{n}\otimes e_{n}$, thus, $T_{t}$ is bounded. Then we can apply Arendt-Batty-Lyubich-Phong theorem to conclude that $T_{t}$ is asymptotically stable.

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