Let $\;y_1\left(x,\,\lambda\right),\: y_2\left(x,\,\lambda\right)\;$ be respective solutions of eigenvalue problem \begin{align}y\,'' &= \lambda \,y, & x&\in\big(\,0,\,\infty\,\big),&\lambda&\in\mathbb{C}\end{align} with boundary conditions \begin{align} &\begin{cases} y_{1}\big(0,\,\lambda\big)=1\\y_{1}'\big(0,\,\lambda\big)=0\end{cases} & \text{ and } & &\begin{cases} y_{2}\big(0,\,\lambda\big)=0\\y_{2}'\big(0,\,\lambda\big)=1\end{cases} \end{align}

I need to find a function $\,m\left(\lambda\right)\,$ such that the function \begin{align} \psi\big(x,\,\lambda\big) = y_{2}\big(x,\,\lambda\big) + m\left(\lambda\right)\, y_{1}\big(x,\,\lambda\big) \end{align} is in $\;L^{2}_{\left(0,\infty\right)}$.

The problem arose in context of solutions of Sturm-Liouville eigenproblems dependent on eigenvalue as parameter. I have no idea how to approach it, and will appreciate any relevant advice.

  • $\begingroup$ Aren't the solutions $sin$ and $cos$ or $sinh$ and $cosh$ depending on the sign of $\lambda$ ? $\endgroup$ – Keith McClary Nov 14 '15 at 3:48
  • $\begingroup$ Or am I misunderstanding your question? $\endgroup$ – Keith McClary Nov 14 '15 at 5:16

The solutions are $$ y_1(x,\lambda) = \cosh(\sqrt{\lambda}x), \;\;\; y_2(x,\lambda) = \frac{\sinh(\sqrt{\lambda}x)}{\sqrt{\lambda}}, $$ where $\sqrt{\lambda}$ has its branch cut along the negative real axis. Using this particular branch cut ensures that $\Re\sqrt{\lambda} > 0$ for $\mathbb{C}\setminus(-\infty,0]$. Then $e^{\sqrt{\lambda}x} \notin L^2[0,\infty)$ and $e^{-\sqrt{\lambda}x} \in L^2[0,\infty)$ for $\lambda\in\mathbb{C}\setminus(-\infty,0]$ Therefore, the unique $m(\lambda)$ for which $$ y_2 + my_1 \in L^2[0,\infty), \;\;\; \lambda\in\mathbb{C}\setminus(-\infty,0], $$ must be chosen so that the $e^{\sqrt{\lambda}x}$ term is absent. Hence, $$ \frac{1}{2\sqrt{\lambda}}+\frac{1}{2}m(\lambda) = 0 \\ \implies m = -1/\sqrt{\lambda}. $$ Note: It is more common to consider the operator with negative coefficient of $y''$ on the left. The reason for the branch along the negative real axis is that the spectrum of your operator is the negative real axis. One normally tries to arrange the spectrum to be infinite in the positive direction instead, which is arranged by considering $Lf=-(pf')'+qf$ instead of $Lf=(pf')'-qf$.

  • $\begingroup$ @Vlad : I did not assume $\lambda$ was real. I talked about the branch cut to use. $\endgroup$ – DisintegratingByParts Nov 14 '15 at 7:22
  • $\begingroup$ Sorry, my bad. I do not understand, however, where does the equation $$\frac{1}{2}+\frac{1}{2\sqrt{\lambda}}m(\lambda) = 0$$ come from. $\endgroup$ – Vlad Nov 14 '15 at 8:00
  • $\begingroup$ Same place the similar expression came from in my post: $$\frac{1}{\sqrt{\lambda}}\sinh(\sqrt{\lambda}x) +m\cosh(\sqrt{\lambda}x) = \frac{m + \frac{1}{\sqrt{\lambda}}}{2}e^{\sqrt{\lambda}x} + \frac{m - \frac{1}{\sqrt{\lambda}}}{2}e^{-\sqrt{\lambda}x}.$$ As he discussed, you need the coefficient of $e^{\sqrt{\lambda}x}$ to be $0$ to be in $L^2(0,\infty)$. $\endgroup$ – Paul Sinclair Nov 14 '15 at 14:33
  • $\begingroup$ @PaulSinclair According to your comment $\,m\,$ should be $\,-\dfrac{1}{\sqrt{\lambda}}\,$, not $\,-\sqrt{\lambda}\,$ as it is in the answer. $\endgroup$ – Vlad Nov 15 '15 at 23:37
  • $\begingroup$ Ahh. TrialAndError failed to catch that rather odd turn-around in your equation for $\psi$, where you placed $y_2$ first, and $y_1$ second. $\endgroup$ – Paul Sinclair Nov 16 '15 at 2:04

Keith is correct. for any given $\lambda > 0$, let $a^2 = \lambda$. Then $y_1 = \cosh ax$ and $y_2 = \sinh ax$. For any given $\lambda < 0$, letting $a^2 = -\lambda$ gives the solutions $y_1 = \cos ax$ and $y_2 = \sin ax$. And of course, if $\lambda = 0$, then $y_1 = 1$ and $y_2 = x$.

If $\lambda > 0$, $y_2 + my_1 = e^x\frac{m+1}{2} + e^{-x}\frac{m - 1}{2}$, which is in $L^2_{(0,\infty)}$ only when $m = -1$.

If $\lambda \le 0$, for all $m,\ y_2 + my_1 \notin L^2_{(0,\infty)}$

  • $\begingroup$ Thank you for the answer. However, in the original problem there was no assumption on $\,\lambda\,$ to be real. It is probably my fault that people misinterpreted problem, since I haven't pointed out explicitly that $\,\lambda\,$ can be complex. I have edited my post to make it clear. $\endgroup$ – Vlad Nov 14 '15 at 7:11

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