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I recently came across a problem in research, and I'm asking about it here after trying in math stack exchange with no luck.

Suppose I have a metric graph $G$ (or even a closed interval, to make things simpler if needed) and a Schrödinger operator $L=-\frac{\partial ^2}{\partial x^2}+V$ (you can even ignore $V$ if that helps). Suppose also that I have a finite set of points $N\subset G$ (you can even think of $N$ as a single point if that helps). On each $x_0\in N$, I impose the following boundary condition - $$f'(x_0^+)=f'(x_0^-)=:f'(x_0)$$ $$f(x_0^+)-f(x_0^-)=\alpha f'(x_0)$$ This means that $f'$ is 'continuous' at $x_0$ (not really continuous in the usual sense, just that the two sided limits exist and are equal, the function itself doesn't have to be differentiable), and the jump of $f$ at $x_0$ is proportional by a factor $\alpha$ to the value of the one sided limits of $f'(x_0)$ (although $f'$ need not exist).

The eigenvalue problem $Lf=\lambda f$ with the imposed boundary condition has a unique spectrum - $\{\lambda_n\}_{n=1}^{\infty}$.

Now, I choose to change the boundary conditions to something similar, but different: $$f'(x_0^+)=f'(x_0^-)=:f'(x_0)$$ $$f(x_0^+)+f(x_0^-)=\alpha f'(x_0)$$ The change is in the minus sign in the second formula.

My question is – can I say something meaningful about how the spectrum of the problem changes due to the change in boundary conditions? The boundary conditions are (at least intuitively) very similar, so can I maybe somehow find a relation between the spectra of the two problems?

You can try to suggest ideas for the more general case of $G,L,N$, or even just think about some of the simplified cases I suggested. Ideally, I'd like an answer in the style of 'the change in boundary condition translates the spectrum by a fixed number' or something of this type - something which gives me a concrete relation between the two problems. But I'm aware that the answer might be much more complicated (if there even is an answer), so feel free to share any idea.

One thing I thought of is – given an eigenfunction $f$ of the first problem, maybe I can do some 'procedure' to it in order to turn it into an eigenfunction of the second problem (I thought about adding a constant to it, to match the new boundary condition, but then it won't necessarily be an eigenfunction of $L$, but maybe something else will work). But I don't know if this is a good way to go.

Thanks in advance!

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  • $\begingroup$ If you change $-$ with $+$ the problem is no longer self-adjoint and the spectrum can be the whole complex plane. For example take $+$ and $\alpha=0$ in $I=[0,1]$ and $c=1/2$ and consider also Dirichlet bc at $0,1$. For a given $\lambda$, let $u$ solve $\lambda u-u''=0$ in $I$ and $u(0)=0$. Then $v(x)=-u(1-x)$ solves the same equation with $v(1)=0$. Now take $w=u$ in $[0,1/2]$ and $w=v$ in $[1/2,1]$. Left and right derivatives are equal at $1/2$ but $w$ has opposite values from both sides of $1/2$. $\endgroup$
    – Giorgio Metafune
    Sep 20, 2020 at 10:18
  • $\begingroup$ @GiorgioMetafune Why does $v$ solve the eigenvalue equation $v''=-\lambda u$? $$v''=(-u+ux)''=(-u'+u'x+u)'=u''(-1+x)+2u'$$ But: $$-\lambda v = \lambda u(1-x) = -u''(1-x)$$ And so if I'm not mistaken, the two expressions differ by $2u'$ which is generally non-zero. I think that the $+$ problem is indeed self adjoint since it can be represented by the symmetric bilinear form: $B(u,v)=\int (u'v'+Vuv)+\frac{1}{\alpha}\sum_{x\in N}(u(x_+) + u(x_-))(v(x_+)+v(x_-))$ $\endgroup$
    – GSofer
    Sep 20, 2020 at 11:15
  • $\begingroup$ $v$ Is $u$ evaluated at the point $1-x$ and I mean that $v"=\lambda v$. $\endgroup$
    – Giorgio Metafune
    Sep 20, 2020 at 11:44
  • $\begingroup$ @GiorgioMetafune Oh, that makes more sense, thanks. But why do you conclude that the problem is not self adjoint? I don't see why it says anything about the spectrum. Doesn't the representation of the boundary condition via the symmetric bilinear form automatically gives that the problem is self adjoint? From what I know, this boundary condition is related to what's known in the literature as $\delta '$ boundary conditions, which are indeed self adjoint. $\endgroup$
    – GSofer
    Sep 20, 2020 at 12:02
  • $\begingroup$ Actually I took only $\alpha=0$ which Is excluded in your form. I only used that the spectrum of a selfadjoint operatori Is real. $\endgroup$
    – Giorgio Metafune
    Sep 20, 2020 at 12:09

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