Timeline for Sturm-Liouville Problem: When does $w y^2$ vanish at a singular boundary point?
Current License: CC BY-SA 4.0
20 events
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| Aug 29, 2020 at 7:51 | comment | added | Giorgio Metafune | It is one of the results of the theory which gives solvability in $H^1$; basically is is the Riesz representation theorem applied to the weak formulation of the problem or, in other words, the existence of a minimizer. You are right that self-adjointness (and semiboundedness) is preserved under the transformation. But $q$ bounded below is a sufficient condition for the domain; the tranfsormed domain could be self-adjoint, positive but with a form domain not contained in H^1. | |
| Aug 29, 2020 at 1:56 | comment | added | stewori | Another thing puzzles me: If I know from another criterion (reference in the question) that the problem (1), (2) is self-adjoint, wouldn't that self-adjointness be preserved by the transformation? Let's call the original operator $\tilde{D}$, then I think the transformed one can be written as $D u = \sqrt{p} \tilde{D} \tfrac{u}{\sqrt{p}}$. Then one has $(Du, v) = (u, Dv)$ given that $(\tilde{D}y, z) = (y, \tilde{D}z)$. Would $q$ bounded below still be required then? Maybe I get something wrong here with self-adjoint vs self-adjoint extension. Maybe I should raise this as a separate question... | |
| Aug 28, 2020 at 20:30 | comment | added | stewori | How do we know that there exists $v \in D(A)$ (rather than in $L^2 \setminus D(A)$) such that $\lambda v - Av = f$? | |
| Aug 27, 2020 at 21:50 | comment | added | Giorgio Metafune | In principle, yes. But if you have a look at the very definition of Schroedinger operators with form methods, then it is in the definition of $D(A). | |
| Aug 27, 2020 at 19:38 | comment | added | stewori | Why is $D(A)$ contained in $H^1$? Is it because $u$ would be a minimizer of $\int_a^b q u^2 + (u')^2 dt$ and that expression would be infinite if $u \notin H^1$? | |
| Aug 27, 2020 at 7:33 | comment | added | Giorgio Metafune | Essential selfadjoiness means that the closure of $A$, initially defined on smooth functions with comapct support, is self adjoint and can be restated by saying that $\lambda -A$ is injective on the maximal domain if $\lambda$ is sufficiently large. Now that $u$ in the maximal domain and let $f=\lambda u-Au$; then you find $v \in D(A)$ such that $\lambda v-Av=f$ and then $w=u-v$ is in the maxiamal domain and $\lambda w-Aw=0$ which gives $w=0$ and $u=v \in D(A)$. | |
| Aug 27, 2020 at 7:30 | comment | added | Giorgio Metafune | At this point I can explain. Assume that $q$ is bounded below, then you define the Schroedinger operator $A$ through a form and $D(A)$ will be contained in $H^1$ and $\lambda -A$ will be invertible from $D(A)$ to $L^2$ for large $\lambda$. You can also define the maximal domain $D_{max}$ as the set of all functions $u\in L^2$ such that $Au \in L^2$ (usually you need some local regularity) and this set is larger than $D(A)$ since you do not require global integrability on $\nabla u$. | |
| Aug 27, 2020 at 0:38 | comment | added | stewori | Literature search based on "essential self-adjointness" leads to the criteria, but is then usually concerned with applying the spectral theorem and discussing properties of the spectrum. I also looked "randomly" into some books on the topic (difficult to search in a library), (e.g. into "Schrödinger Operators" by Simon Barry) but did not find this particular property discussed. | |
| Aug 27, 2020 at 0:34 | comment | added | stewori | Thanks for the hints. That (5) follows from $u \in H^1$ is straight forward. I also found various statements (based on the Rellich-Kato Theorem) that assert essential self-adjointness of Schrödinger operator if $q$ is bounded below. However, I struggle to find references for coincidence of the form domain with the maximal one. Some reference or keyword on this topic would be helpful. | |
| Aug 23, 2020 at 17:35 | comment | added | Giorgio Metafune | If you refer to the behavior at $\infty$ of $H^1$ functions in $1d$, then you can find it in any book on Sobolev spaces. Concerning the fact that $L^2$ eigenfunctions are in $H^1$, under condition on the potential $q$, this is not so difficult but requires more space (the key word is essential self-adjointness of Schr\"odinger operators; if this is true, the domain of the operator defined through a form coincides with the maximal one, hence if $u, D^2u-qu \in L^2$, then $u \in H^1$. | |
| Aug 23, 2020 at 17:04 | comment | added | stewori | Okay, then $u$ tending to zero means also $u^2$ tends to zero, so $p y^2$ tends to zero, i.e. (5) holds. Do you know a reference for what you explained above? Is it a standard argument in Sobolev spaces theory? | |
| Aug 23, 2020 at 6:12 | comment | added | Giorgio Metafune | Sorry, It Is a misprint. It should be $u=y\sqrt p$. | |
| Aug 23, 2020 at 2:17 | comment | added | stewori | Is there a reason to write $u=y \sqrt{u}$ instead of $u = y^2$? | |
| Aug 22, 2020 at 23:19 | comment | added | Giorgio Metafune | Actually, if $q$ Is bounded below, each $L^2$ eigenfunction is in $H^1$. | |
| Aug 22, 2020 at 15:44 | comment | added | Giorgio Metafune | If $q$ is bounded below you can assume that it is positive and the operator $D^2-q$ can be defined through a form with domain $H^1$. I mean that the equation $\lambda u-D^2 u+qu=f$ has a solution $u \in H^1$ for every $f \in L^2$, so the whole theory starts in $H^1$, eigenfunctions are in $H^1$ and so on. Now it depends on your problem; you could work directly with the transformed one in $H^1$ and your problem disappears...or stay with the original one and show that an eigenfunction $y$ produces $u \in H^1$ (or even that all $L^2$ eigenfunctions of the transformed problem are in $H^1$). | |
| Aug 22, 2020 at 14:01 | comment | added | stewori | Thanks, this looks promising! If the $q$ in your equation is unbounded below I assume that $u$ is in $H^1$ follows from some theorem, but this is not obvious to me. Could you give a hint or reference on why this follows? Is the criterion ($q$ unbounded below) a necessary one or "just" a sufficient one? | |
| Aug 22, 2020 at 9:29 | comment | added | Giorgio Metafune | Consider $q=0$, $p=w$ and $b=\infty$ and $u=y \sqrt{u}$. Then (1) is equivalent to $$u''-\frac{u}{2} \left (\frac{p''}{p}-\frac{p'^2}{p^2} \right )=-\lambda u.$$ Now everything depends on the potential $q=p''/p-p'^2/p^2$. If this is bounded from below, then $u$ is in the Sobolev space $H^1$ and then tends to 0 at $\infty$. | |
| Aug 21, 2020 at 18:38 | comment | added | stewori | In principle: yes. However, I read that (2) must not be applied at LP endpoints and Hermite Polynomials are an example that shows that (5) can still hold at LP endpoints, so (2) should not be required in every case. $y$ shall be an eigenfunction. I specified $b = \infty$ because my reason to believe that (5) holds is (4) under sort of Barbalat's lemma. | |
| Aug 21, 2020 at 18:29 | comment | added | Giorgio Metafune | To check if I have understood the question. Do you ask whether (5) holds when $y$ is an eigenfunction subject to the boundary condition (2)? And you assume that $b=\infty$? | |
| Aug 21, 2020 at 17:33 | history | asked | stewori | CC BY-SA 4.0 |