# the asymptotic behaviour of function as $\lambda \to -\infty$

Let's consider the following differential equation on $\mathbb{R}$: $$-u''(x)+u(x)-V(x)u(x)=\lambda u(x),$$ where $\lambda<1$ and $V$ is a bounded. We consider only that solution $u(x) \in C^1$ which decays exponentially as $x\to +\infty$.

I am interested in the behaviour of $\frac{u'(x)}{u(x)}$ as $\lambda \to -\infty$, which, i guess, should diverge to $-\infty$ (and can be proved for some particular cases).

However, I am interested in general case, and would be very glad if someone could provide me with a good reference related to that question (which, i believe, exists in some classical literature about Sturm-Liouville problem).

• If $y(x)=u'(x)/u(x)$, then $y(x)$ satisfies $y'(x)=-y(x)^2-V(x)-\lambda+1$, apparently. – Brendan McKay Jan 13 '18 at 1:01
• @BrendanMcKay that's true, but it still seems to be hard to study the behaviour of $y(x)$ in this context, since as $\lambda \to -\infty$, $y'(x)$ depends on $-y(x)^2-\lambda$, and the latter expression is probably not good for analysis – Jane Jan 13 '18 at 1:06
• Maple explicitly solves $y'(x)=-y(x)^2-V(x)-\lambda+1$ for concrete $V(x)$, e.g. $\sin(x),\,\exp(x)$. This may be useful for you as model examples. – user64494 Jan 13 '18 at 5:14

First of all, I think you should really absorb the stray $u$ on the left-hand side by either $Vu$ or $\lambda u$, so consider $$-u''+Vu =\lambda u ,$$ under the same assumptions. Next, instead of talking about an exponentially decaying solution (whose existence you'd have to prove first), I think it would be much better to take the unique (up to a factor) solution $u(x,\lambda)\in L^2(0,\infty)$, which is guaranteed to exist for $\lambda\notin\mathbb R$ by basic theory; in more technical terms, a bounded $V$ is in the limit point case at infinity. (If there is an exponentially decaying solution, then it's of course the same $u$.)

Then the quotient $m(\lambda)=u'(x,\lambda)/u(x,\lambda)$ defines the Titchmarsh-Weyl $m$ function of the problem on $[x,\infty)$, with Dirichlet boundary conditions $v(x)=0$ at $x$. This has a holomorphic extension to the complement of the spectrum, so in your situation is defined for all small enough real $\lambda$. Its asymptotics have been studied extensively, and the answer to your question is that $$m(\lambda) = -\sqrt{-\lambda} + \textrm{error term}, \quad \lambda\to -\infty ,$$ with rather precise information on the error. In particular, indeed $m\to-\infty$, as you suspected.

See reference 271 here for much additional information (the asymptotic formula, as $\lambda\to -\infty$ is right at the beginning, in the abstract).

• Thank you for your help! This is a very good reference which contains another list of good references! I actually interested in $m$-function defined on $\mathbb{R}$ with the only boundary conditions converging to $0$ at $+\infty$. But the reference and the formula with "error term" you provided me with gave me kind of a confedence in my guess and, actually, seems to be very helpful for me, since I tried to get the result via Volterra's integral and got almost the same equations (but still need some corrections/proofs). Thank you! – Jane Jan 15 '18 at 0:30