Bi-Laplacians on graphs and networks

Abstract

We study the differential operator $A=\frac{d^4}{dx^4}$ acting on a connected network $\mathcal {G}$ along with $\mathcal L^2$, the square of the discrete Laplacian acting on a connected discrete graph $\mathsf {G}$. For both operators, we discuss well-posedness of the associated linear parabolic problems

$$\begin{aligned} \frac{\partial u}{\partial t}=-Au,\qquad \frac{df}{dt}=-{\mathcal {L}}^2 f, \end{aligned}$$

on $L^p(\mathcal {G})$ or $\ell ^p(\mathsf {V})$, respectively, for $1\le p\le \infty$. In view of the well-known lack of parabolic maximum principle for all elliptic differential operators of order 2N for $N>1$, our most surprising finding is that after some transient time, the parabolic equations driven by $-A$ may display Markovian features, depending on the imposed transmission conditions in the vertices. Analogous results seem to be unknown in the case of general domains and even bounded intervals. Our analysis is based on a detailed study of bi-harmonic functions complemented by simple combinatorial arguments. We elaborate on analogous issues for the discrete bi-Laplacian; a characterization of complete graphs in terms of the Markovian property of the semigroup generated by $-{\mathcal {L}}^2$ is also presented.

Appendix

We argue that in the context of parabolic problems associated with different realizations of the bi-Laplacian, eventual positivity and its relaxation, asymptotic positivity are more appropriate notions: in the context of operators on general Banach lattices, they have been thoroughly discussed in [18, 19]. For the sake of self-containedness, let us present a reminder of the most relevant results obtained so far in this area.

Recall that if X is a $\sigma$-finite measure space, then $L^2(X)$ is a Hilbert lattice [67]; if we denote its positive cone by $L^2(X)_+$, then for all $u\in L^2(X)_+$ the principal ideal generated by u is the set

$$\begin{aligned} L^2(X)_u:=\{f\in L^2(X):\exists c>0:|f|\le cu\}. \end{aligned}$$

We write $f\gg _u 0$ if f is real and there exists $\epsilon >0$ such that $f\ge \epsilon u$. Observe that if $f>0$ a.e., then $L^2(X)_u$ is dense in $L^2(X)$, i.e., u is a quasi-interior point of $L^2(X)_+$ in the language of Banach lattice theory; we can thus formulate the above-mentioned results as follows.

Definition 7.1

Let X be a $\sigma$-finite measure space and $L^2(X)_+$ be the positive cone of the Hilbert lattice $L^2(X)$. A strongly continuous semigroup $(T(t))_{t\ge 0}$ is said to be

• individually asymptotically positive if

  $$\begin{aligned} f\ge 0\Rightarrow \lim _{t\rightarrow \infty }\text {dist}\, (e^{-ts(-A)}T(t)f,L^2(X)_+)=0, \end{aligned}$$

  provided the spectral bound $s(-A)$ of its generator $-A$ is finite.

• uniformly eventually strongly positive with respect to $u\in L^2(X)_+$ if

  $$\begin{aligned} \exists t_0>0\hbox { s.t.}\quad f\ge 0,\ f\ne 0 \Rightarrow T(t)f\gg _u 0\quad \hbox {for all }t\ge t_0. \end{aligned}$$

If X has finite measure and thus u can be taken to be the constant function $\mathbf{1}$, then $f\gg _u 0$ is equivalent to $f\gg 0$ and so uniform eventual strong positivity with respect to$\mathbf{1}$ is nothing but eventual irreducibility, a rather strong property. At the other end of the spectrum, individual asymptotic positivity seems to be the weakest conceivable condition that still gives a scintilla of positivity to a semigroup.

Let us recall the following corollaries of [35, Thm. 10.2.1] and [19, Thm. 8.3].

Proposition 7.2

Let X be a $\sigma$-finite measure space and $(T(t))_{t\ge 0}$ be a strongly continuous semigroup on the complex Hilbert lattice $L^2(X)$ with self-adjoint generator $-A$. Assume that $s(-A)>-\infty$, that $(e^{-t(A+s(-A))})_{t\ge 0}$ is bounded, and that $s(-A)$ is a pole of the resolvent.

Then, $(T(t))_{t\ge 0}$ is individually asymptotically positive if and only if $s(-A)$ is a dominant spectral value of A and the associated spectral projection P is positive.

Proposition 7.3

Let X be a $\sigma$-finite measure space and $(T(t))_{t\ge 0}$ be a real, strongly continuous semigroup on the complex Hilbert lattice $L^2(X)$ with self-adjoint generator $-A$. Let $u>0$ a.e. and assume that $\bigcap _{k=1}^\infty D(A^k)\subset L^2(X)_u$.

Then, $(T(t))_{t\ge 0}$ is uniformly eventually strongly positive with respect to u if and only if the spectral bound $s(-A)$ is a simple eigenvalue and the associated eigenspace contains a vector v such that $v\gg _u 0$.

It is remarkable that under the assumption of Proposition 7.3, the semigroup is necessarily of trace class and hence, its generator has purely point spectrum, whereas Proposition 7.2 may apply also in the case of a generator whose spectrum is not purely discrete, but which does have a dominant eigenvalue.

We can deduce the following result for semigroups associated with forms from the Beurling–Deny criteria.

Corollary 7.4

Let X be a finite measure space and a be a closed symmetric form on $L^2(X)$; denote by A the associated operator. Then, the following assertions hold.

• (1) $(e^{-tA})_{t\ge 0}$ is individually asymptotically positive if and only if $s(-A)$ is a dominant spectral value of $-A$ and the associated spectral projection is positive.

• (2) Let additionally $D(a)\subset L^\infty (X)$ and ${{\,\mathrm{Re}\,}}u\in D(a)$ and $a({{\,\mathrm{Re}\,}}u, {{\,\mathrm{Im}\,}}u)\in \mathbb {R}$ for all $u\in D(a)$. Then, $(e^{-tA})_{t\ge 0}$ is eventually irreducible if and only if the spectral bound $s(-A)$ is a simple eigenvalue and the associated eigenspace contains a vector v such that $v\gg 0$.

In the same spirit one may ask whether a given semigroup is eventually$L^\infty$-contractive. This issue was not investigated in [18, 19] but an answer can be given in a simple case that covers, in particular, finite graphs and networks.

Our result seems to be of independent interest. It has been observed by the second author together with Jochen Glück (Ulm).

Proposition 7.5

Let X be a finite measure space and A a self-adjoint operator with compact resolvent on $L^2(X)$. Let 0 be the spectral bound of A, and let the associated eigenspace $E_0$ be a one-dimensional space spanned by $\mathbf 1$. Also assume that $\bigcap _{k\in {\mathbb {N}}}D(A^k)\hookrightarrow L^\infty (X)$. Then, the semigroup generated by A is uniformly eventually sub-Markovian, i.e., uniformly eventually positive and uniformly eventually $L^\infty$-contractive.

Proof

First of all, let us observe that the eigenprojector onto $E_0$, which is the rank-one operator

$$\begin{aligned} P:f\mapsto \frac{1}{|X|}\int _X |f|\ {\hbox {d}}x \end{aligned}$$

is also a positive, bounded linear operator on $L^\infty (X)$ with norm 1. It follows from Proposition 7.3 that the semigroup is uniformly eventually strongly positive—say it is positive for all $t>t_0$.

Let now $t>t_0$ and hence $e^{tA}$ be positive: in order to show that $e^{tA}$ is $L^\infty$-contractive and hence sub-Markovian, it suffices to check that $e^{t_0 A}{} \mathbf{1}= \mathbf{1}$, i.e., $A\mathbf{1}=0$. This holds by assumption. $\square$