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For a potential of the form $V(x)=ax^4+bx^2$, where $a,b>0$, let us consider the one dimensional Schrodinger operator $D=-\frac{d^2}{dx^2}+V$ with Dirichlet B.C on $[-L,L]$ and denote its first eigenvalue by $\lambda(a,b)$.

Q1. Is $\lambda(a,b)$ a differentiable function in $a$ and $b$?

Q2. For which $(a,b)$'s subject to $a+b=1$, the function $\lambda$ is maximized/minimized?

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    $\begingroup$ The answer to Q1 has to be yes, I believe, and a standard source for these things would be Kato's book. Also, you can make $V$ arbitrarily negative on $[-1/2,1/2]$, say, by taking $a\to\infty$, so $\lambda(a)$ will be unbounded below. $\endgroup$
    – Christian Remling
    Commented Nov 19, 2019 at 17:59
  • $\begingroup$ I just realize that the second part of my comment applies to the operator $-d^2/dx^2+V$ (I'm so used to this minus sign that it registers subconsciously). However, similar remarks can also be made in your case. $\endgroup$
    – Christian Remling
    Commented Nov 19, 2019 at 18:15
  • $\begingroup$ Agreed. It's more common with negative sign. $\endgroup$
    – BigM
    Commented Nov 20, 2019 at 5:57
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    $\begingroup$ For the case $L\leq 1$, one can partially answer Q2 using the variational principle: Given the ground state for a given $a,b$, with eigenvalue $\lambda (a,b)$, the expectation value of any Hamiltonian with larger $a$ and smaller $b$ in that state is lower than $\lambda (a,b)$, because $x^4 \leq x^2 $. Hence, the ground state energy of the latter Hamiltonian is even lower, by the variational principle. This means that the minimal $\lambda (a,b)$ is achieved for $a=1$, $b=0$. For $L>1$, it's not so clear - maybe there's a clever way to rescale. $\endgroup$
    – Michael Engelhardt
    Commented Nov 20, 2019 at 6:25
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    $\begingroup$ Have you solved this numerically? If not, I can do it pretty quickly, this problem is very simple. $\endgroup$
    – Michael Engelhardt
    Commented Nov 21, 2019 at 0:29

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