# An optimization problem for one- dimensional Schrodinger operator

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?

• 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. – Christian Remling Nov 19 '19 at 17:59
• 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. – Christian Remling Nov 19 '19 at 18:15
• Agreed. It's more common with negative sign. – BigM Nov 20 '19 at 5:57
• 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. – Michael Engelhardt Nov 20 '19 at 6:25
• Have you solved this numerically? If not, I can do it pretty quickly, this problem is very simple. – Michael Engelhardt Nov 21 '19 at 0:29