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Consider the following minimization problem $$\inf \left\{ \int\limits_{-\infty}^0 \left[ (\psi')^2 + m(y)(\psi - F)^2 \right]\; : \; \psi \in H^1(\left(-\infty,0\right]) \right\}$$ where $m(y) > 0$, $\left\| F \right\|_{C^1(\left(-\infty,0\right])} < \infty$, $F$ is asymptotically 0, $F > 0$ for $y\in \left(-\infty,0\right)$, and $F(0) = 0$.

Let $\psi$ be a minimizer. My question is, can we say anything about $\psi(0)$?

Edit: I should also state that $c < m(y) < C$ for positive constants $c,C$.

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    $\begingroup$ A quick computation from the Euler equation gives $\psi'(0)=0$ and $\psi(0)=m\int_{-\infty}^0F(x)e^{\sqrt{m}x}dx$... $\endgroup$ Dec 8, 2014 at 18:54
  • $\begingroup$ I suppose I should have been clearer. $m = m(x)$. I think this makes a world of difference? Correct me if I'm wrong. $\endgroup$
    – k3thomps
    Dec 8, 2014 at 19:28
  • $\begingroup$ Then it is still true that $\psi''(x)-m(x)\psi(x)=-m(x)F(x)$ and $\psi'(0)=0$, and I think $\psi(0)=\int_{-\infty}^0 m(x)u(x)F(x)dx$, where $u$ is the solution of $u''(x)=m(x)u(x)$, $u'(0)=1$, $u(-\infty)=0$. $\endgroup$ Dec 8, 2014 at 21:04
  • $\begingroup$ I think this might be the way to wrap it up: you need the solution of $\psi''-m\psi=-mF$ with $\psi=0$ near $-\infty$ and $\psi'(0)=0$. This you can in principle express in terms of the solutions of the homogeneous equation $u''-mu=0$, by the variation of constants formula, and this gives a formula of sorts for $\psi(0)$. It will not be very explicit though since you don't know the solutions of $u''-mu=0$. $\endgroup$ Dec 8, 2014 at 21:34
  • $\begingroup$ Yes, you do understand the details of the comment :) and yes, the given value of $\psi(0)$ is not more explicit than the function $m(x)$, of course. $\endgroup$ Dec 8, 2014 at 22:15

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