# Exponential stability of a second order ordinary linear operator

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!!

• Is $\{T_t\}$ bounded and strongly continuous? – Pedro Mar 3 at 1:00
• @Pedro yes, I think so. – misakaczy Mar 3 at 20:40
• Then, in order to obtain (b), it is enough to put assumptions on $k(x)$ so that $\mathbf{i}\mathbb R\subset\rho(A)$. – Pedro Mar 3 at 21:09
• @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! – misakaczy Mar 3 at 22:32
• 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. – Pedro Mar 3 at 23:38

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.