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!!
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.