# All Graphs with a Failed Zero-Forcing Number of Two

Luis Gomez, Karla Rubi, Jorden Terrazas  Luis Gomez  University of Arkansas
Karla Rubi  California State University - Dominguez Hills  Jorden Terrazas
Southern Methodist University  Darren Narayan  Rochester Institute of
Technology

(July 2021)

###### Abstract

Given a graph $G$, the zero-forcing number of $G$, $Z(G)$, is the smallest
cardinality of any set $S$ of vertices on which repeated applications of the
forcing rule results in all vertices being in $S$. The forcing rule is: if a
vertex $v$ is in $S$, and exactly one neighbor $u$ of $v$ is not in $S$, then
$u$ is added to $S$ in the next iteration. Zero-forcing numbers have attracted
great interest over the past 15 years and have been well studied. In this
paper we investigate the largest size of a set $S$ that does not force all of
the vertices in a graph to be in $S$. This quantity is known as the failed
zero-forcing number of a graphs and will be denoted by $F(G)$, and has
received attention in recent years. We present new results involving this
parameter. In particular, we completely characterize all graphs $G$ where
$F(G)=2$, solving a problem posed in 2015 by Fetcie, Jacob, and Saavedra.

##  1 Introduction

Given a graph $G$, the zero-forcing number of $G$, $Z(G)$, is the smallest
cardinality of any set $S$ of vertices on which repeated applications of the
forcing rule results in all vertices being in $S$. The forcing rule is: if a
vertex $v$ is in $S$, and exactly one neighbor $u$ of $v$ is not in $S$, then
$u$ is added to $S$ in the next iteration. Zero forcing numbers have attracted
great interest over the past 15 years and have been well studied [2]. In this
paper we investigate the largest size of a set $S$ that does not force all of
the vertices in a graph to be in $S$. This quantity is known as the failed
zero forcing number of a graphs and will be denoted by $F(G)$, and has
received attention in recent years [3]. Independently, a closely related
property called the zero blocking number of a graph was introduced by
Beaudouin-Lafona, Crawford, Chen, Karst, Nielsen, and Sakai Troxell [4] and
Karst, Shen, and Vu [5]. The zero blocking number of a graph $G$ equals
$|V(G)|-F(G)$.

We will use $K_{n}$, $P_{n}$, $C_{n}$, to denote the complete graph, path, and
cycle on $n$ vertices, respectively. The complete bipartite graph with $r$
vertices in one part and $t$ in the other part will be denoted $K_{r,t}$. The
disjoint union of graphs $G$ and $H$ will be denoted by $G+H$. Following the
terminology of [Erdos], we will refer an induced $K_{1,2}$ as a “cherry”. A
vertex $v$ is a cut-vertex of a graph $G$ if $G-v$ has more components than
$G$. We will refer to a subgraph that contains a cut-vertex $v$ of degree 3
that is adjacent to vertices $u$ and $w$, which are adjacent to each other and
have degree 2 as a pendant triangle.

Throughout the paper we will use colorings to describe our failed zero-forcing
sets where the vertex in $S$ will be colored blue (or referred to as filled)
and the vertices not in $S$ will be colored white (or referred to as
uncolored). A vertex will be called true blue if it is colored blue and all of
its neighbors are colored blue. A vertex that is not a true blue vertex will
be referred to as non true blue.

Failed zero-forcing numbers were introduced by Fetcie, Jacob, and Saavedra [3]
where they established numerous results. In particular they proved that for
any connected graph $0\leq F(G)\leq n-2$ and gave a characterization of all
graphs $G$ where $F(G)=0,1,n-1$ and $n-2$. They showed that the only graphs
where $F(G)=0$ are either $K_{1}$ or $K_{2}$ and the only graphs where
$F(G)=1$ are $2K_{1}$, $P_{3}$, $K_{3}$, or $P_{4}$. It was also shown that
$F(G)=n-1$ if and only if $G$ contains an isolated vertex. In addition,
$F(G)=n-2$ if and only if $G$ contains a module of order $2$, where a module
is defined as follows. A set $X$ of vertices in $V(G)$ is a module if all
vertices in $X$ have the same set of neighbors among vertices not in $X$. They
gave examples of some graphs where $F(G)=2$ and posed the problem of
characterizing all graphs with this property. We provide a complete solution
to this problem in Section 2.

##  2 Graphs with a Failed Zero-forcing number of 2

In this paper we provide a characterization of all graphs with a failed zero-
forcing number of 2. We first examined all graphs with between 3 and 6
vertices and identified all graphs with $F(G)=2$. These are shown in Figure 1.
Later we show that no other graphs exist. We will show that if $G$ has at
least 7 vertices then $F(G)\geq 3$.

Figure 1: The 15 graphs with $F(G)=2$

It is not difficult to verify that each of these graphs have a failed zero-
forcing number of $2$. For the sake of completeness, we include the details.
The coloring of the vertices gives a lower bound for the failed zero-forcing
number of $2$. For the disconnected three graphs the upper bound for $F(G)=2$
follows since coloring any three vertices blue forces all of the vertices to
be colored blue. For the four connected graphs with four vertices the upper
bound follows since $F(G)\leq n-2$. For the two paths, the coloring of any
three vertices blue will result in either a vertex of degree 1 being colored
blue or two adjacent vertices colored blue, both of which will color all of
the vertices in the graph. For the graph of $C_{5}$ coloring any three
vertices blue will mean that two adjacent vertices were colored blue, which
will force the remaining vertices in the graph. For the graph of $C_{5}$ and a
chord note that coloring any three of the vertices in the graph blue will
either include two vertices of the triangle or three vertices in the $C_{4}$
both of which will force all vertices in the graph to be blue. For the wheel
graph with five vertices, coloring three vertices blue will either include the
vertex of degree four and two other vertices, or three vertices of degree two
or three, both of which will force all vertices in the graph to be blue. For
the graph consisting of a triangle and two pendant edges note that coloring
any three vertices blue will either include the three vertices of degree
greater than 1, one vertex of degree 1 and two vertices of degree greater than
1, or two vertices of degree 1, all of which will force all vertices to be
colored blue. Finally for the triangle with three pendant edges, coloring any
three vertices blue will either include all three vertices of degree greater
than 1, or at least one vertex of degree 1 and two other vertices that are
either degree 1 or greater than 1, all of which will force all vertices to be
blue.

We examined all connected graphs on six vertices [1] and found that all but
two graphs have a failed zero forcing number of at least 3. The two graphs
$P_{6}$ and the corona of $K_{3}$, (the last two graphs in Figure 1) have a
failed zero forcing number of $2$. Adding an additional vertex $v$ and any
number of edges between $v$ and other vertices of the graph results in a graph
with a failed zero forcing number of 3 (This is verified later in Section
2.1). The above suggests that any graph with 7 or more vertices has a failed
zero forcing number of at least 3, which we will prove.

The following theorem guarantees that our list of 15 graphs with $F(G)=2$ is
complete.

######  Theorem 2.1.

Let $G$ be a graph with at least $7$ vertices. Then $F(G^{\prime})\geq 3$.

It is tempting to think that if you have a graph $G$ where $F(G)=k$, then for
any supergraph $H$ of $G$ we have $F(H)\geq k$. However it can be the case
that $F(H)<F(G)$ and in fact we can construct graphs where $F(G)-F(H)$ is
arbitrarily large. In the figure below we give an example of a graph $G$ where
$F(G)=5$ and a supergraph $H$ of $G$ where $F(H)=4$.

Figure 2: Adding a single edge to $G$ to create a supergraph $H$

######  Theorem 2.2.

There exist graphs $G$ with a supergraph $H$ where $F(G)-F(H)$ is arbitrarily
large.

###### Proof.

Let $G$ be a graph with vertices $v_{1},v_{2},...,v_{n}$ with edges
$\\{v_{i}v_{i+1}|$ $i=1,...,n-2\\}$ and $\\{v_{n-2}v_{n}\\}$. Here $F(G)=n-2$.

Now add the edge $v_{1}v_{n}$ to create a supergraph $H$. Since $H$ is a cycle
with a pendant vertex we can obtain a maximum-sized failed zero-forcing set by
selecting the pendant vertex and $\left\lfloor\frac{n}{2}\right\rfloor$
vertices along the cycle (including the neighbor of the pendant vertex). Hence
$F(H)=\left\lfloor\frac{n}{2}\right\rfloor+1$. Since
$F(G)-F(H)=n-2-\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)\geq\frac{n}{2}$
this difference can be made arbitrarily large, but choosing a graph with a
sufficiently large number of vertices. ∎

We next present a series of lemmas that will be used to prove Theorem 2.1.
A characterization of disconnected graphs with a failed zero-forcing number
$k$ was given by Fetcie, Jacob, and Saavedra [3].

######  Lemma 2.1.

Let $G$ be a disconnected graph with $k\geq 2$ maximal connected components.
Let $G_{1},G_{2},...,G_{k}$ be the $k$ maximal connected components of $G$.
Then $F(G)=\max\left\\{F(G_{k})+\sum\limits_{l\neq
k}\left|V(G_{l})\right|\right\\}$.

This theorem can be applied to determine all disconnected graphs with a
particular failed zero-forcing number. We show the result for the case where
$F(G)=2$.

######  Lemma 2.2.

For a disconnected graph $G$, $F(G)=2$ if and only if $G$ equals $3K_{1}$,
$K_{2}+K_{1}$, or $K_{2}+K_{2}$.

###### Proof.

If $G$ consists of three components the only possibility is for $G$ to equal
$3K_{1}$, since if any component contained $k>1$ vertices we would have
$F(G)\geq k+1=3$. If $G$ has two components one component must contain exactly
two vertices, and the other component must have a failed zero-forcing number
of 1, which are either $K_{1}$ or $K_{2}$. ∎

Now that we have identified all disconnected graphs with $F(G)=2$, we next
consider the case of connected graphs. We note that if a graph with 8 or more
vertices contains a cherry or a pendant triangle then $F(G)\geq 3$.

The approach will be as follows.

Connected graphswith pendant Vor a pendant triangle($F(G)\geq 3$)Connected
graphs with 8 or more verticesConnected graphs with 6vertices with $F(G)\geq
3$that can be extendedConnected graphs with 6 verticeswhich cannot be
extended(exceptions)Connected graphs with7 or 8 vertices with $F(G)\geq 3$that
can be extended Figure 3: Dotted arrows indicate extensions by Lemma 2.4

A pendant V is a subgraph containing a cut-vertex $v$ of degree at least $3$
whose removal results in at least two paths in different components.

We start with a connected graph with 8 or more vertices that does not contain
a pendant V or a pendant triangle and reduce it to a graph with 6 vertices by
removing vertices and incident edges. We note we can maintain a connected
graph by successively removing a leaf from a spanning tree of the graph. We
will show in Lemmas 2.6 and 2.7 that the reduced graphs will not contain a
pendant V or a pendant triangle. If the resulting graph on six vertices does
not have any true blue vertices, then we apply Lemma 2.3 add back vertices and
edges to obtain the original graph which will have a failed zero-forcing set
of size at least 3. If the resulting graph on six vertices has at least one
true blue vertex then we consider all possible extensions to supergraphs with
7 vertices. We show that all of these graphs have $F(G)\geq 3$. If these
graphs on 7 vertices do not have any true blue vertices then we can apply
Lemma 2.3 and extend the graph to the original graph which will a failed zero-
forcing set of size at least 3. If the extensions to 7 vertices have at least
one true blue vertex then we consider all possible extensions to 8 vertices.
We were able to show that all of these graphs do not have any true blue
vertices so they can be extended back to the original graph showing that
$F(G)\geq 3$.

######  Lemma 2.3.

Let $G$ be a graph with a failed zero-forcing set $S$. If every vertex in $S$
is adjacent to two vertices outside of $S$, then $S$ is a failed-zero forcing
set for any connected supergraph $H$ of $G$, and thus $F(H)\geq F(G)$.

###### Proof.

When adding vertices and edges to $G$ to obtain the graph $H$ we can color all
of the vertices in $V(H)-V(G)$ white. Since each member of $S$ is adjacent to
two vertices not in $S$, $S$ will not force any vertices in $H$. ∎

In our next lemma we show that a if a graph $G$ can be extended to contain an
additional vertex $v$ and if the extended graph has a failed zero-forcing set
$S$ where $v$ is not in $S$ then any number of edges can be added between $v$
and vertices of $G$ and $S$ will still be a failed zero-forcing set of any of
these new graphs.

######  Lemma 2.4.

Let $G$ be a connected graph with a set of $S$ of blue vertices. If each
vertex in $S$ is adjacent to two vertices not in $S$, then for any supergraph
$H$ of $G$, we have that $F(H)\geq F(G)$.

###### Proof.

We can add any number of vertices and edges $G$ to create a supergraph $H$.
Then color all of the vertices in $V(G)-V(H)$ white. Then since white vertices
cannot force any other vertices and the blue vertices are still adjacent to
two white vertices, no vertices in the graph will be forced. Hence $F(H)\geq
F(G)$. ∎

######  Lemma 2.5.

If a connected graph $G$ has a pendant $K_{3}$. Then $F(G)\geq n-2$.

###### Proof.

All of the vertices except the two vertices of degree $2$ in the $K_{3}$ form
a failed zero forcing set of $|V(G)|-2$ which is the largest possible for a
connected graph. ∎

The following lemma shows that if $G$ is a graph with at least $5$ vertices
then $F(G)\geq 3$.

######  Lemma 2.6.

Let $G$ be a graph with $5$ or more vertices and contains a cut-vertex $v$
such that $G-v$ has at least three components, at least two of which are
paths. Then $F(G)\geq 3$.

###### Proof.

We consider two cases.

Case (i) $G$ has a cut-vertex $v$ such that $G-v$ consists of three disjoint
paths $P_{i}$, $P_{j}$, and $P_{k}$. Then either $i$, $j$, or $k$ must be at
least $2$. Without loss of generality suppose $i\geq 3$. Then $v$ along with
the vertices of $P_{i}$ form a failed zero forcing set of at least $3$.

Case (ii) $G$ is a graph with $5$ or more vertices and has a cut-vertex $v$
such that $G-v$ consists of a subgraph $H$ with at least two vertices and two
disjoint paths. Then $v$ along with the vertices of $H$ form a failed zero
forcing set of at least $3$. ∎

In our next two lemmas we prove that if a graph $G$ is a graph does contains a
pendant V or a pendant $K_{3}$ then it can be reduced to a subgraph $H$ by
removing vertices and edges, so that $H$ does not contain a pendant V or a
pendant $K_{3}$. If all reductions from a graph $G$ lead to a graph $H$ with
one of these two pendant subgraphs, then $G$ contained one of these two
pendant subgraphs.

######  Lemma 2.7 (pendant V).

Let $G$ be a connected graph with at least five vertices that does not contain
a pendant V. Then $\left|V(G)\right|-5$ vertices of $G$ can be successively
removed (along with incident edges) to create a graph $G^{\prime}$ that does
not have a pendant V.

###### Proof.

We will show that if a graph $G$ does not have a pendant V, then there is some
vertex we can remove from $G$ so that the resulting graph does not have a
pendant V. Let $G$ be a graph that is not a cycle and does not have a pendant
$V$ subgraph. We want to show there is always some vertex $v$ whose removal
does not leave a pendant $V$ subgraph. We can assume that $\delta(G)\geq 2$
since we could remove a vertex of degree $1$ and not create a pendant $V$
subgraph. Since $\delta(G)\geq 2$, $G$ must contain a cycle $C$ that has no
pendant paths. Since $G\neq C$ the cycle $C$ must contain a vertex $w$ with
degree at least $3$. Removing a vertex on $C$ that has degree 2 is adjacent to
$w$ will not leave a pendant $V$. ∎

######  Lemma 2.8 (pendant triangle).

Let $G$ be a connected graph with at least five vertices that does not contain
a pendant $K_{3}$. Then $\left|V(G)\right|-5$ vertices of $G$ can be
successively removed (along with incident edges) to create a graph
$G^{\prime}$ that does not have a pendant $K_{3}$.

###### Proof.

We will show that if $G$ does not contain a pendant triangle, there is always
some vertex we can remove so that the resulting graph does not contain a
pendant triangle. Let $G$ be a graph without a pendant triangle, but $G-v$
contains a pendant triangle with vertices $a$,$b$,$c$, and $d$ where $a$ and
$b$ are adjacent, and $b$,$c$, and $d$ are all mutually adjacent, and $c$ and
$d$ are not adjacent to $a$. Since $G$ has more than $5$ vertices we can
assume that either $d(a)>2$ or $d(v)>2$. We will consider various cases and
show in each case that a different vertex can be removed from $G$ that will
not result in a pendant triangle.

Case (i): $v$ is adjacent to $a$,$b$,$c$, and $d$. Removing $d$ leaves an
induced $K_{4}-e$ on vertices $a$,$v$,$b$, and $c$.

Case (ii): $v$ is adjacent to $a$,$b$, and $c$ only. Then removing $b$ results
in a path on vertices $a$,$v$,$c$, and $d$.

Case (iii): $v$ is adjacent to $c$ and $d$ only. Then removing $c$ results in
a path on vertices $v$,$d$,$b$, and $a$.

Case (iv): $v$ is adjacent to $c$ only. Then removing $d$ results in a path on
vertices $v$,$c$,$b$, and $a$.

Case (v): $v$ is adjacent to $d$ only. Then removing $d$ results in a path on
vertices $v$,$c$,$b$, and $a$. ∎

###  2.1 Graphs with six vertices

We will consider the 112 connected graphs with six vertices. We will refer to
the table and numbering found in [1]. For convenience, we explicitly show each
graph in each of the individual cases.

Out of $112$ different graphs, there were 16 cases where there does not exist
three blue vertices with at most one true blue vertex. We will show that show
that all possible extensions of these exception cases have three blue vertices
with none being a true blue vertex. This will not only show that they have a
failed zero-forcing number of at least 3, but they can be extended
indefinitely by adding vertices and edges and the resulting supergraph will
also have a failed zero-forcing number of at least 3.

These graphs are the exceptions:
54,58,59,60,76,77,80,85,86,87,96,98,102,103,105,112, each of which we will
present individually.

$v_{1}$$v_{2}$$v_{5}$$v_{4}$$v_{6}$$v_{3}$ Figure 4: Graph 54

Graph 54: We begin by extending graph 54 by one vertex, $v$ and connecting it
to each vertex in G. Notice there will be no case where $v$ is adjacent to
$v_{2}$ or $v_{5}$. This is because connecting a vertex to either one will
create a cherry structure and that case is covered by our cherry lemma. Our
goal with extending is to find no true blue vertex within the graph. It is
also important to notice that in each case, we are able to find a failed zero-
forcing number of three.

If $v$ is adjacent to $v_{1}$, then $v_{1},v_{3},$ and $v_{4}$ can be filled.

If $v$ is adjacent to $v_{6}$, then $v_{3},v_{4},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{4}$, then $v_{2},v_{5},$ and $v_{6}$ can be filled.
Notice that $v_{6}$ is true blue, so this case must be extended further.
Suppose we add a new vertex called $u$.

If $u$ is adjacent to $v_{3}$, then $v_{3},v_{4},$ and $v_{5}$ can be filled.

If $u$ is adjacent to $v_{6}$, then $v_{3},v_{4},$ and $v_{6}$ can be filled.

If $u$ is adjacent to $v$, then $v,v_{2},$ and $v_{5}$ can be filled. Since
all sub-cases for this case have no true blue vertex, then this case is
resolved. But there remains one more case.

If $v$ is adjacent to $v_{3}$, then $v_{2},v_{5},$ and $v_{6}$ can be filled.
Notice that $v_{6}$ is true blue, so this case must be extended further.
Suppose we add a new vertex called $u$.

If $u$ is adjacent to $v$, then $v,v_{2},$ and $v_{5}$ can be filled.

If $u$ is adjacent to $v_{6}$, then $v_{3},v_{4},$ and $v_{6}$ can be filled.

If $u$ is adjacent to $v_{4}$, then $v_{3},v_{4},$ and $v_{5}$ can be filled.
Notice in each sub-case there are no true blue vertices. Thus this case is
resolved and can be extended. Since all cases can be extended indefinitely,
this implies that graph 54 can be extended indefinitely without being forced
with $F(G)\geq 3$.

$v_{6}$$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$ Figure 5: Graph 58

Graph 58:  Let $v_{1}$ be the vertex of degree $1$ and let $v_{2}$ be the
neighbor of $v_{1}$; $v_{4}$ is the vertex of degree $3$ that is adjacent to
$v_{2}$, $v_{3}$, and $v_{5}$. Let $v_{5}$ be the vertex of degree $3$ that is
adjacent to $v_{6}$; vertices $v_{6}$ and $v_{3}$ both have degree 2. With
graph 58, although we have one True blue vertex, we can extend the graph along
vertices $v_{1},v_{3}$ and $v_{6}$ and be left without any true blue vertices
thus being able to extend indefinitely with $F(G)\geq 3$. Extending along
vertices $v_{4}$ and $v_{5}$ will give us one true blue vertex. We can then
add an additional vertex or an edge and with reshuffling the filled in
vertices, that will give us $F(G)\geq 3$ without a true blue vertex. For the
sake of completeness we include the details. We consider adding a new vertex
$v$ with possible neighbors. We note that $v$ cannot have degree $1$ and be
adjacent to $v_{2}$ since this would create a cherry. In each case of the
remaining cases we can find a coloring with $3$ blue vertices, none of which
are true blue vertices, and all of the remaining vertices including $v$ will
be colored white. If $v$ is adjacent to $v_{1}$, then we could color $v_{1}$,
$v_{3}$, and $v_{5}$ blue. If $v$ is adjacent to $v_{3}$, then we could color
$v_{2}$, $v_{3}$, and $v_{5}$ blue. If $v$ is adjacent to $v_{6}$, then we
could color $v_{1}$, $v_{5}$, and $v_{6}$ blue. The only problem cases is
where $v$ is adjacent to $v_{4}$ or $v_{5}$. Here we can find a failed zero
forcing set $S$ of size $3$ by selecting $v_{1}$, $v_{2}$, and $v_{4}$,
however we may not be able to extend $S$ to larger graphs. Without loss of
generality assume that $v$ is adjacent to $v_{4}$. We consider adding a new
vertex $u$ with possible neighbors. We note that $u$ cannot have degree $1$
and be adjacent to $v_{2}$ or $v_{4}$ since this would create a cherry. If $u$
is adjacent to $v$ then we could color $v$, $v_{3}$, and $v_{5}$ blue. If $u$
is adjacent to $v_{1}$ then we could color $v_{1}$, $v_{3}$, and $v_{5}$ blue.
If $u$ is adjacent to $v_{3}$ then we could color $v_{2},$ $v_{4}$, and
$v_{5}$ blue. If $u$ is adjacent to $v_{5}$ then we could color $v_{2},$
$v_{4}$, and $v_{5}$ blue. If $u$ is adjacent to $v_{6}$ then we could color
$v_{3},$ $v_{5}$, and $v_{6}$ blue.

$v_{6}$$v_{5}$$v_{4}$$v_{3}$$v_{2}$$v_{1}$ Figure 6: Graph 59

Graph 59:  Let $v_{1}$ be the vertex of degree $1$, $v_{4}$ be the vertex of
degree $4$ that is adjacent to $v_{1}$, $v_{3}$ be the vertex of degree $2$
that is adjacent to $v_{4}$, $v_{2}$ be the other vertex of degree $4$,
$v_{5}$ be the vertex of degree $3$ that is adjacent to $v_{2}$ and $v_{4}$,
and $v_{6}$ be the vertex of degree $2$ that is adjacent to $v_{2}$ and
$v_{5}$. Graph 59 follows with the case of graph 58. We can extend along any
vertex without any True Blue vertices with the only exception being when a new
vertex is adjacent to vertex $v_{3}$. Similarly to graph 58 however, upon any
additional extension, there are no issues arising when all the vertices are
reshuffled.

For the sake of completeness we include the details. We consider adding a new
vertex $v$ and possible neighbors of $v$. The vertex $v$ cannot have degree
$1$ and be adjacent to $v_{4}$ since this would create a cherry. In each case
of the remaining cases we can find a coloring with $3$ blue vertices, none of
which are true blue vertices, and all of the remaining vertices including $v$
will be colored white. If $v$ is adjacent to $v_{1}$, then $v_{1},$ $v_{3}$,
and $v_{5}$ can be colored blue. If $v$ is adjacent to $v_{3}$, then $v_{3},$
$v_{4}$, and $v_{6}$ can be colored blue. If $v$ is adjacent to $v_{5}$ then
$v_{2},$ $v_{4},$ and $v_{5}$ can be colored blue. If $v$ is adjacent to
$v_{6}$, then $v_{2}$, $v_{4},$ and $v_{6}$ can be colored blue. The only
problem case is if $v$ is adjacent to $v_{2}$. Here we can find a failed zero-
forcing set $S$ of size $3$ with vertices $v_{1}$, $v_{4}$, and $v_{6}$, but
we many not be able to extend this graph. Here we consider adding another
vertex $u$. The vertex $v$ cannot have degree $1$ and be adjacent to either
$v_{2}$ or $v_{4}$ since this would create a cherry. If $u$ is adjacent to
$v$, then we color $v$,$v_{3}$, and $v_{5}$ blue. If $u$ is adjacent to
$v_{1}$, then we color $v_{1}$, $v_{3}$, and $v_{5}$ blue. If $u$ is adjacent
to $v_{3}$, then we color $v_{2}$, $v_{3}$, and $v_{5}$ blue. If $u$ is
adjacent to $v_{5}$, then we color $v_{2}$, $v_{4}$, and $v_{5}$ blue. If $u$
is adjacent to $v_{6}$, then we color $v_{2}$,$v_{4}$, and $v_{6}$ blue.

$v_{6}$$v_{5}$$v_{4}$$v_{3}$$v_{1}$$v_{2}$ Figure 7: Graph 60

Graph 60:  Graph 60 has at least one true blue vertex but all extensions along
graph 60 result in $F(G)\geq 3$ and no true blue vertices. Let $v_{1}$ be the
vertex of degree $1$, $v_{3}$ be the neighbor of $v_{1}$, $v_{2}$ be the
vertex of degree $4$, $v_{4}$ be the vertex of degree $3$ that is adjacent to
$v_{2}$ and $v_{3}$, $v_{5}$ be the vertex of degree $3$ adjacent to $v_{4}$,
and let $v_{6}$ be the vertex of degree $2$. We consider adding another vertex
$v$ and possible neighbors of $v$. First if $v$ has degree $1$ then it cannot
be adjacent to $v_{3}$ since this would create a cherry. In each of the
remaining cases we can find a coloring with $3$ blue vertices, none of which
are true blue vertices, and all of the remaining vertices including $v$ will
be colored white. If $v$ is adjacent to $v_{1}$ then we can color
$v_{1}$,$v_{4}$, and $v_{6}$ blue, if $v$ is adjacent to $v_{2}$ then $v_{2}$,
$v_{3}$, and $v_{5}$ can be colored blue. If $v$ is adjacent to $v_{4}$ then
$v_{3}$, $v_{4},$ and $v_{6}$ can be colored blue and if $v$ is adjacent to
$v_{5}$ then $v_{2}$, $v_{3},$ and $v_{5}$ can be colored blue, and if $v$ is
adjacent to $v_{6}$ then $v_{2}$, $v_{3},$ and $v_{6}$ can be colored blue.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 8: Graph 76

Graph 76: We will show that all but two extensions of $G$ to a graph with 7
vertices will have no true blue vertex of degree 1. Let $v$ denote the new
vertex. Note that extensions which would result in a cherry are neglected as
those cases are resolved by Lemma 2.6. We have the following cases

If $v$ is adjacent to $v_{1}$ only, we color $v_{1}$,$v_{3}$, and $v_{4}$ blue
and $v$, $v_{2}$,$v_{5}$ and $v_{6}$ white.

Note each case above is analogous to another extension of $G$ due to its
symmetry. Now we show that the two extensions of $G$ which do contain a true
blue vertex of degree 1 may be further extended to remove said true blue
vertex. These two extensions are the same by symmetry, and so we consider them
as one case. Let our vertices be named as before and include $v_{7}$ as a
vertex of degree 1 adjacent to $v_{3}$.

If $v$ is adjacent to $v_{1}$ only, we color $v_{1}$, $v_{3}$, and $v_{4}$
blue and $v$,$v_{2}$,$v_{5}$,$v_{6}$ and $v_{7}$ white.

If $v$ is adjacent to $v_{4}$ only, we color $v_{2}$,$v_{3}$, and $v_{4}$ blue
and $v$,$v_{1}$,$v_{5}$,$v_{6}$ and $v_{7}$ white.

If $v$ is adjacent to $v_{7}$, we color $v_{2}$,$v_{5}$, and $v_{7}$ blue and
$v$,$v_{1}$,$v_{3}$,$v_{4}$ and $v_{6}$ white.

The remaining cases of connecting $v$ to this extension of $G$ are analogous
to one of the previous five. Notice we can always color $v$ white and add
edges between $v$ and any subset of the remaining vertices without affecting
the number of blue vertices by Lemma 2.4.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 9: Graph 77

Graph 77: Suppose we add a new vertex called $v$. Due to the symmetry of the
graph, it is not necessary to extend on $v_{5}$ or $v_{1}$. Our goal with
extending is to find no true blue vertex. If we do have a true blue, we must
then extend further.

If $v$ is adjacent to $v_{6}$, then $v_{2},v_{4},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{3}$, then $v_{2},v_{3},$ and $v_{5}$ can be filled.

If $v$ is adjacent to $v_{1}$, then $v_{1},v_{3},$ and $v_{5}$ can be filled.

If $v$ is adjacent to $v_{4}$, then $v_{2},v_{4},$ and $v$ can be filled. For
this case, we must extend further since there is a true blue vertex. Suppose
we add a second vertex called $u$.

If $u$ is adjacent to $v_{3}$, then $v_{2},v_{3},$ and $v_{5}$ can be filled.

If $u$ is adjacent to $v$, then $v,v_{3},$ and $v_{5}$ can be filled.

Thus we can extend all cases indefinitely without ever forcing G. Since each
case is extendable, then graph 77 is extendable.

Notice there is no case where $v$ is adjacent to $v_{2}$ or $v_{5}$. This is
because connecting a vertex to either one would create a cherry structure and
that case is covered elsewhere. It is also important to notice that in each
case, we are able to find a failed zero-forcing number of three.

Every blue vertex has two white neighbors, so we can add any number of edges
or vertices to any these graphs without forcing G. This is because the three
blue vertices will always be adjacent to two white neighbors and cannot force
any other vertex. Thus $F(G)\geq 3$ for graph 77.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 10: Graph 80

Graph 80: Let the vertex on the far left be $v_{1}$. Let the vertex adjacent
to $v_{1}$ be $v_{2}$. Let the top vertex adjacent to $v_{2}$ be $v_{3}$ and
the bottom be $v_{4}$. Let the vertex that is adjacent to both $v_{3}$ and
$v_{4}$ on the right be $v_{4}$. And finally, let the vertex on the far right
side be $v_{6}$. We begin by extending graph 80 by one vertex, $v$ and
connecting $v$ to every vertex in $G$. Notice there will be no case where $v$
is adjacent to $v_{2}$ or $v_{5}$. This is because connecting a vertex to
either would create a cherry structure and that case is covered elsewhere.

If $v$ is adjacent to $v_{6}$, then $v_{3},v_{4},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{4}$, then $v_{2},v_{4},$ and $v_{5}$ can be filled.

If $v$ is adjacent to $v_{3}$, then $v_{2},v_{3},$ and $v_{5}$ can be filled.

If $v$ is adjacent to $v_{1}$, then $v_{1},$ $v_{3}$, and $v_{4}$ can be
filled.

In all cases there are no true blue vertices. Thus, all cases are extendable
and this implies graph 80 can be extended indefinitely. It is also important
to notice that in each case, we are able to find a failed zero-forcing number
of three.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 11: Graph 85

Graph 85: We will show that all but two extensions of $G$ to a graph with 7
vertices have no true blue vertex of degree 1. Let $v$ denote the new vertex.
Note that extensions which would result in a cherry are neglected as those
cases are resolved by Lemma 2.6. We consider the following cases.

If $v$ is adjacent to $v_{6}$ only, we color $v_{2}$, $v_{6}$, and $v_{4}$
blue and $v$,$v_{1}$, $v_{3}$, and $v_{5}$ white.

If $v$ is adjacent to $v_{1}$ only, we color $v_{1}$,$v_{3}$, and $v_{6}$ blue
and $v$,$v_{2}$,$v_{4}$, and $v_{5}$ white.

If $v$ is adjacent to $v_{3}$ only, we color $v_{2}$,$v_{3}$, and $v_{5}$ blue
and $v$, $v_{1}$,$v_{4}$, and $v_{6}$ white.

Now we show the two extensions of $G$ which do contain a true blue vertex of
degree 1 may be further extended to remove said true blue vertex. For the
first extension, let our vertices be named as before, and include $v_{7}$ as a
vertex of degree 1 adjacent to $v_{5}$.

If $v$ is adjacent $v_{1}$ only, we color $v_{1}$,$v_{3}$, and $v_{5}$ blue
and $v$,$v_{2}$,$v_{4}$,$v_{6}$, and $v_{7}$ white.

If $v$ is adjacent $v_{3}$ only, we color $v_{2}$,$v_{3}$, and $v_{5}$ blue
and $v$, $v_{1}$,$v_{4}$,$v_{6}$ and $v_{7}$ white.

If $v$ is adjacent $v_{4}$ only, we color $v_{2}$, $v_{4}$, and $v_{5}$ blue,
and $v$,$v_{1}$,$v_{3}$,$v_{6}$, and $v_{7}$ white.

If $v$ is adjacent $v_{6}$ only, we color $v_{2}$,$v_{4}$, and $v_{6}$ blue
and $v$,$v_{1}$,$v_{3}$,$v_{5}$, and $v_{7}$ white.

If $v$ is adjacent $v_{7}$ only, we color $v_{2}$,$v_{4}$, and $v_{7}$ blue
and $v$,$v_{1}$,$v_{3}$,$v_{5}$, and $v_{6}$ white.

For the second extension, let our vertices be named as before, and include
$v_{7}$ as a degree 1 vertex adjacent to $v_{4}$.

If $v$ is adjacent to $v_{1}$ only, we color $v_{1}$,$v_{4}$, and $v_{6}$ blue
and $v$,$v_{2}$,$v_{3}$,$v_{5}$ and $v_{7}$ white.

If $v$ is adjacent to $v_{3}$ only, we color $v_{2}$,$v_{3}$, and $v_{5}$ blue
and $v$,$v_{1}$,$v_{4}$,$v_{6}$, and $v_{7}$ white.

If $v$ is adjacent to $v_{5}$ only, we color $v_{2}$,$v_{4}$, and $v_{5}$ blue
and $v$,$v_{1}$,$v_{3}$,$v_{6}$, and $v_{7}$ white.

If $v$ is adjacent to $v_{6}$ only, we color $v_{2}$,$v_{4}$, and $v_{6}$ blue
and $v$,$v_{1}$,$v_{3}$,$v_{5}$ and $v_{7}$ white

If $v$ is adjacent to $v_{7}$ only, we color $v_{3}$,$v_{6}$, and $v_{7}$ blue
and $v$,$v_{1}$,$v_{2}$,$v_{4}$ and $v_{5}$ white.

Notice we can always color $v$ white and add edges between $v$ and any subset
of the remaining vertices without affecting the number of blue vertices by
Lemma 2.4.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 12: Graph 86

Graph 86: We will show that all extensions of $G$ to a graph with 7 vertices
have no true blue vertices of degree 1. Let $v$ be the new vertex. Let $v_{1}$
be the vertex of degree 1, $v_{2}$ be the vertex of degree 2 that is adjacent
to $v_{1}$, $v_{3}$ be the vertex of degree 3 that is adjacent to and on the
left of $v2$, $v_{4}$ be the vertex of degree 2 that is adjacent to $v_{3}$,
$v_{5}$ be the vertex of degree 2 that is adjacent to $v_{4}$, and $v_{6}$ be
the vertex of degree 3 that is adjacent to $v_{5}$. We have the following
cases.

If $v$ is adjacent to $v_{1}$ only, we color $v_{1}$,$v_{3}$, and $v_{6}$ blue
and $v$,$v_{2}$,$v_{4}$,$v_{5}$ white.

If $v$ is adjacent to $v_{2}$ only, the resulting graph contains a cherry. So
we color $v$,$v_{1}$ white, and color the remaining vertices blue.

If $v$ is adjacent to $v_{3}$ only, we color $v_{2}$,$v_{3}$, and $v_{4}$ blue
and $v$,$v_{1}$,$v_{5}$, and $v_{6}$ white.

If $v$ is adjacent to $v_{4}$ only, we color $v_{2}$,$v_{4}$, and $v_{6}$ blue
and $v$,$v_{1}$,$v_{3}$, and $v_{5}$ white.

The remaining cases of connecting $v$ to $G$ are analogous to one of the
previous four. Also note that in each case where the resulting graph does not
contain a cherry, we can always color $v$ white and add edges between $v$ and
any subset of the remaining vertices without affecting the number of blue
vertices.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 13: Graph 87

Graph 87: All but one of the single vertex extensions of $G$ is either a
cherry or suitable for extension by lemma 1.1. Let $v_{1}$ be the only vertex
of degree one that is adjacent to $v_{2}$; $v_{2}$ is adjacent to $v_{3}$ and
$v_{6}$; $v_{3}$ will be adjacent to $v_{2}$ and $v_{4}$ and $v_{4}$ will be
adjacent to $v_{5}$ and $v_{6}$.

If $v$ is adjacent to $v_{1}$, we can fill in $v_{1}$, $v_{4}$, and $v_{6}$.

If $v$ is adjacent to $v_{2}$, it creates a cherry and this case can be
discarded.

If $v$ is adjacent to $v_{3}$, we can fill in $v_{2}$, $v_{3}$, and $v_{5}$.

If $v$ is adjacent to $v_{5}$, we can fill in $v_{2}$, $v_{4}$, and $v_{5}$.

If $v$ is adjacent to $v_{6}$, we can fill in $v_{2}$, $v_{4}$, and $v_{6}$.

When $v$ is adjacent to $v_{4}$, it is unavoidable to have a True Blue vertex
and must be extended further. Now we will consider a graph in which $v$ is
adjacent to $v_{4}$. We can then add in an additional vertex, $u$.

If $u$ is adjacent to $v_{1}$, we can fill in $v_{1}$, $v_{4}$, and $v_{6}$.

If $u$ is adjacent to $v_{2}$, it again creates a cherry and we can discard
this case.

If $u$ is adjacent to $v_{3}$, we can fill in $v_{3}$, $v_{4}$, and $v_{6}$.

If $u$ is adjacent to $v_{4}$, we get another cherry and thus another
discarded case.

If $u$ is adjacent to $v_{5}$, we can fill in $v_{2}$, $v_{4}$, and $v_{5}$.

If $u$ is adjacent to $v_{6}$, we can fill in $v_{2}$, $v_{4}$, and $v_{6}$.

If $u$ is adjacent to $v$, then we can fill $v$, $v_{2}$, and $v_{5}$.

In every case, we see a suitable coloring for extension by Lemma 1.1. Thus we
may extend graph 87 to a super-graph $H$, and by Lemma 1.1, $F(H)\geq F(G)\geq
3$.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 14: Graph 96

Graph 96: We will show that any extension of $G$ to a graph with 7 vertices
will have at most one true blue vertex. Let the vertices of the triangle be
$v_{1},v_{2},$ and $v_{3}$ which are adjacent to vertices of degree $1$,
$v_{4},v_{5},$ and $v_{6}$ respectively.

Consider the addition of a new vertex $v$. We first consider when $v$ is
adjacent to one of the vertices $v_{4},v_{5},$or $v_{6}$. Without loss of
generality assume $v$ is adjacent to $v_{4}$. Then we color $v_{2},v_{3},$and
$v_{4}$ blue and $v,v_{1},v_{5},$ and $v_{6}$ white. Next we consider when $v$
is adjacent to one of the vertices $v_{1},v_{2},$or $v_{3}$. Without loss of
generality assume $v$ is adjacent to $v_{1}$. Then we color $v_{1},v_{3}$, and
$v_{4}$ blue and $v,v_{2},v_{5},$ and $v_{6}$ white.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 15: Graph 98

Graph 98: We begin by extending graph 98 by one vertex, $v$ and connecting $v$
to every vertex in $G$. Notice there will be no case where $v$ is adjacent to
$v_{5}$, $v_{4}$, or $v_{2}$. This is because connecting a vertex to either
one would create a cherry and that specific case is covered by the cherry
lemma.

If $v$ is adjacent to $v_{6}$, then $v_{2},v_{4},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{3}$, then $v_{2},v_{3},$ and $v_{5}$ can be filled.

If $v$ is adjacent to $v_{1}$, then $v_{1},v_{3},$ and $v_{5}$ can be filled.

We can see in all cases there are no true blue vertices. Thus, all cases are
extendable and this implies graph 98 can be extended indefinitely. It is also
important to notice that in each case, we are able to find a failed zero-
forcing set of size three.

$v_{1}$$v_{4}$$v_{6}$$v_{2}$$v_{3}$$v_{5}$ Figure 16: Graph 102

Graph 102: We begin by extending graph 102 by one vertex, $v$ and connecting
$v$ to every vertex in G. Due to the symmetry of the graph, we can ignore the
case where $v$ is adjacent to $v_{5}$. Notice there will be no case where $v$
is adjacent to $v_{3}$ or $v_{4}$. This is because connecting a vertex to
either one will create a cherry structure, and this is covered by the cherry
lemma.

If $v$ is adjacent to $v_{6}$, then $v_{1},v_{3},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{1}$, then $v_{1},v_{3},$ and $v_{5}$ can be filled.
Due to the symmetry of the graph, this case is the same as when $v$ is
adjacent to $v_{2}$. So it can be ignored. But, this case has a true blue and
so it must be extended. Suppose we add a new vertex $u$.

If $u$ is adjacent to $v_{6}$, then $v_{1},v_{3},$ and $v_{6}$ can be filled.

If $u$ is adjacent to $v_{2}$, then $u,v_{1},$ and $v_{3}$ can be filled. Here
vertex $u$ is true blue, so we must extend further. Suppose we add another
vertex called $w$.

If $w$ is adjacent to $v_{6}$, then $v_{2},v_{3},$ and $v_{6}$ can be filled.
Due to the symmetry of the graph, all remaining cases will have the same
results. Notice there are no true blue vertices and that $F(G)\geq 3$ in all
cases. Since there are no true blue vertices, then this case is now
extendable. Since each case is extendable, then graph 102 can be extended
indefinitely without being forced.

$v_{1}$$v_{2}$$v_{3}$$v_{5}$$v_{6}$$v_{4}$ Figure 17: Graph 103

Graph 103: We begin by extending graph 103 by one vertex, $v$ ,and connecting
$v$ to every vertex in G. Notice there will be no case where $v$ is adjacent
to $v_{5}$ or $v_{2}$. This is because connecting a vertex to either would
create a cherry structure and that case is covered by the cherry lemma. It is
also important to notice that in each case and sub-case, we are able to find a
failed zero-forcing number of three. Our goal with extending is to find no
true blue vertex within the graph. Due to the symmetry of the graph, we can
ignore the cases where $v$ is adjacent to $v_{4}$ and $v_{1}$.

If $v$ is adjacent to $v_{6}$, then $v_{3},v_{4},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{3}$, then $v_{2},v_{5},$ and $v_{6}$ can be filled.
Since $v_{6}$ is true blue, we must then extend further. Suppose we add a new
vertex called $u$.

If $u$ is adjacent to $v_{6}$, then $v_{3},v_{4},$ and $v_{6}$ can be filled.

If $u$ is adjacent to $v$, then $v,v_{2},$ and $v_{5}$ can be filled.

If $u$ is adjacent to $v_{4}$, then $v_{4},v_{5},$ and $v_{6}$ can be filled.
Here, $v_{6}$ is a true blue vertex, so we must extend to 9 vertices. Suppose
we add a new vertex called $w$.

If $w$ is adjacent to $v_{1}$, then $v_{1},v_{3},$ and $v_{4}$ can be filled.

If $w$ is adjacent to $v_{6}$, then $v_{3},v_{4},$ and $v_{6}$ can be filled.

If $w$ is adjacent to $v$, then $v,v_{2},$ and $v_{5}$ can be filled.

If $w$ is adjacent to $u$, then $v,v_{2},$ and $v_{5}$ can be filled.

Since there are no more true blue vertices, then this case is resolved and can
be extended indefinitely. Since each case is extendable, this implies that
graph 103 can also be extended indefinitely without being forced..

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 18: Graph 105

Graph 105: We begin by extending graph 105 by one vertex, $v$ and connecting
$v$ to every vertex in G. Notice there will be no case where $v$ is adjacent
to $v_{2}$. This is because adding an extension to $v_{2}$ would create a
cherry structure and this case is covered by the cherry lemma. It is also
important to notice that in each case, we are able to find a failed zero-
forcing number of three. Our goal with extending is to find no true blue
vertex within the graph.

If $v$ is adjacent to $v_{1}$, then $v_{1},v_{3},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{6}$, then $v_{2},v_{4},$ and $v_{6}$ can be filled.

If $v$ is adjacent to $v_{3}$, then $v_{2},v_{3},$ and $v_{5}$ can be filled.

If $v$ is adjacent to $v_{5}$, then $v_{1},v_{2},$ and $v_{5}$ can be filled.

If $v$ is adjacent to $v_{4}$, then $v_{1},v_{2},$ and $v_{4}$ can be filled.

Notice the last two cases have a true blue vertex. Due to the symmetry of the
graph, we can focus on one and ignore the other. Assume the case where $v$ is
adjacent to $v_{5}$. Suppose we add a new vertex called $u$.

If $u$ is adjacent to $v_{4}$, then $v_{2},v_{4},$ and $v_{5}$ can be filled.

If $u$ is adjacent to $v_{6}$, then $v_{2},v_{4},$ and $v_{6}$ can be filled.

If $u$ is adjacent to $v_{3}$, then $v_{2},v_{3},$ and $v_{5}$ can be filled.

If $u$ is adjacent to $v$, then $v,v_{2},$ and $v_{4}$ can be filled.

If $u$ is adjacent to $v_{1}$, then $v_{1},v_{3},$ and $v_{6}$ can be filled.

Since there are no more true blue vertices in these sub-cases, then this case
is resolved and can be extended indefinitely. Since each case is extendable,
this implies that graph 105 can be extended indefinitely without being forced
and $F(G)\geq 3$.

Graph 112:

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$ Figure 19: Graph 112

$P_{6}$: Any extension of $P_{6}$ to a graph with 7 vertices will have at most
one true blue vertex.

Let the vertices of $v_{1}$ or $v_{6}$ then we can color vertex $v$ white and
color vertices $v_{1},v_{3},$and $v_{5}$ blue to create a failed zero-forcing
set of size $3$. If $v$ is not adjacent to either $v_{1}$ or $v_{6}$ then $v$
must be adjacent to one or more of the vertices $v_{2},v_{3},v_{4},$ or
$v_{5}$. We consider different cases, each showing that we can create a failed
zero-forcing set of size $3$ with at most one true blue vertex. We consider a
series of different cases.

If $v$ is adjacent to $v_{2}$ and $v_{3}$ only then we color $v_{2},v_{3},$
and $v_{5}$ blue and $v,v_{1},v_{4},$ and $v_{6}$ white.

If $v$ is adjacent to $v_{2}$ and $v_{4}$ only then we color $v_{2},v_{3},$
and $v_{4}$ blue and $v,v_{1},v_{5},$ and $v_{6}$ white.

If $v$ is adjacent to $v_{2}$ and $v_{5}$ only then we color $v,v_{2}$ and
$v_{5}$ blue and $v_{1},v_{3},v_{4},$ and $v_{6}$ white.

If $v$ is adjacent to $v_{3}$ and $v_{4}$ only then we color $v,v_{2},$ and
$v_{5}$ blue and $v_{1},v_{3},v_{4},$ and $v_{6}$ white.

If $v$ is adjacent to $v_{2},v_{3},$ and $v_{4}$ only then we color
$v_{2},v_{3}$, and $v_{5}$ blue and $v_{1},v_{3},v_{4},$ and $v_{6}$ white.

If $v$ is adjacent to $v_{2},v_{3},$ and $v_{5}$ only then we color $v,v_{2},$
and $v_{4}$ blue and $v_{1},v_{3},v_{5},$ and $v_{6}$ white.

If $v$ is adjacent to $v_{2},v_{3},v_{4}$ and $v_{5}$ only, then we color
$v_{2},v_{3}$, and $v_{5}$ blue and $v,v_{1},v_{4}$, and $v_{6}$ white.

###  2.2 Proof of the main theorems

We now prove Theorem 2.1. Let $G$ be a graph with at least $7$ vertices. Then
$F(G^{\prime})\geq 3$.

###### Proof.

We will show that all graphs with $8$ vertices have a failed zero-forcing set
of size $3$ and have at most one true blue vertex. Now start with a graph $G$
with $n=9$ or more vertices. Remove $n-8$ vertices along with their incident
edges to create the graph $G^{\prime}$. By our previous work,
$F(G^{\prime})\geq 3$ and $G^{\prime}$ has no true blue vertices. By the
Extension Lemma 2.4, we can conclude that $F(G)\geq 3$. ∎

We have now proved our main theorem that gives us a characterization of all
graphs with $F(G)=2$.

######  Theorem 2.3.

The graphs with $F(G)=2$ are precisely those shown in Figure 1.

##  3 Acknowledgements

This research was supported by the National Science Foundation Research for
Undergraduates Award 1950189.

## References

  * [1] D. Cvetković, M. Petrić, A table of connected graphs on six vertices, Acta Math. Acad. Sci. Hungar. (1984), 50, 37-49.
  * [2] S. M. Fallat and L. Hogben, The minimum rank of symmetric matrices described by a graph: a survey _Linear Algebra Appl._ (2007), 426: 2-3, 558-582.
  * [3] K. Fetcie, B. Jacob, and D. Saavedra, The Failed Zero Forcing Number of a Graph. _Involve_ (2015), 8, 1, 99-117.
  * [4] M. Beaudouin-Lafona, M. Crawford, S. Chen, N. Karst, L. Nielsen, and D. Sakai Troxell, On the zero blocking number of rectangular, cylindrical, and Möbius grids, _Discrete Applied Mathematics_ , (2020), 282, 35-47.
  * [5] N. Karst, X. Shen, and M. Vu, Blocking zero forcing processes in Cartesian products of graphs, _Discrete Applied Mathematics_ , (2020), 285, 380-396.


# Nonergodic Brownian oscillator: Low-frequency response

Alex V. Plyukhin<EMAIL_ADDRESS>Department of Mathematics, Saint Anselm
College, Manchester, New Hampshire 03102, USA

###### Abstract

An undisturbed Brownian oscillator may not reach thermal equilibrium with the
thermal bath due to the formation of a localized normal mode. The latter may
emerge when the spectrum of the thermal bath has a finite upper bound
$\omega_{0}$ and the oscillator natural frequency exceeds a critical value
$\omega_{c}$, which depends on the specific form of the bath spectrum. We
consider the response of the oscillator with and without a localized mode to
the external periodic force with frequency $\Omega$ lower than $\omega_{0}$.
The results complement those obtained earlier for the high-frequency response
at $\Omega\geq\omega_{0}$ and require a different mathematical approach. The
signature property of the high-frequency response is resonance when the
external force frequency $\Omega$ coincides with the frequency of the
localized mode $\omega_{*}$. In the low-frequency domain $\Omega<\omega_{0}$
the condition of resonance $\Omega=\omega_{*}$ cannot be met (since
$\omega_{*}>\omega_{0}$). Yet, in the limits $\omega\to\omega_{c}$ and
$\Omega\to\omega_{0}^{-}$, the oscillator shows a peculiar quasi-resonance
response with an amplitude increasing with time sublinearly.

##  I Introduction.

Consider a dissipative harmonic oscillator, with the mass $m$ and natural
frequency $\omega$, driven by an external ac force $F_{ex}(t)=F_{0}\sin(\Omega
t)$, and described by the generalized Langevin equation Zwanzig ; Weiss

$\displaystyle
m\,\ddot{x}=-m\,\omega^{2}x-m\int_{0}^{t}K(t-\tau)\,\dot{x}(\tau)\,d\tau+F_{0}\sin(\Omega
t)+\xi(t).$ (1)

The noise $\xi(t)$ is zero-centered, $\langle\xi(t)\rangle=0$, and connected
with the dissipation kernel $K(t)$ by the standard fluctuation-dissipation
relation $\langle\xi(t)\xi(t^{\prime})\rangle=mk_{B}TK(t-t^{\prime})$.
Comparing to its Markovian (time-local) counterpart

$\displaystyle m\,\ddot{x}=-m\,\omega^{2}x-m\,\gamma\,\dot{x}+F_{0}\sin(\Omega
t)+\xi(t),$ (2)

the generalized Langevin equation (1) may hold on the time scale comparable
with the relaxation time of the thermal bath. The latter is important for
systems with a broad hierarchy of relevant time scales, e.g. for biomolecules
in aqueous solution Xie1 ; Xie2 ; Goychuk .

Some types of dissipation kernels $K(t)$ in Eq. (1) are of particular
interest. One special type, relevant to a broad class of open systems Weiss ;
Plyukhin_frac , includes power-law kernels $K(t)\sim t^{-\alpha}$ with
$0<\alpha<1$. Generalized Langevin equations with such kernels, and also open
systems they describe, are often referred to as fractional. For fractional
Langevin equations the integral representing the dissipation coefficient
$\gamma=\int_{0}^{\infty}K(t)\,dt$ diverges and therefore the Markovian
approximation $K(t)\to 2\gamma\delta(t)$ does not exist. Relaxation and
response properties of the fractional oscillator were studied in Barkai1 ;
Barkai2 . Note, however, that there are other definitions of a fractional
oscillator in the literature Achar ; Stanislav ; Ryabov .

In this paper we consider a kernel $K(t)$ of another special type for which
the spectrum $\rho(\omega)=\int_{0}^{\infty}K(t)\cos(\omega t)dt$ has a finite
upper bound $\omega_{0}$. The latter can be viewed as a highest frequency of
fictitious oscillators the thermal bath is comprised of. Solutions of the
generalized Langevin equation with such kernels may be qualitatively different
from those of the corresponding Markovian equations. A finite value of the
cutoff frequency $\omega_{0}$ is known to be a condition for the formation of
a localized normal mode with frequency $\omega_{*}$ outside of the bath
spectrum, $\omega_{*}>\omega_{0}$ Montroll ; Takeno ; Kashiwamura ; Rubin .
The localized mode does not exchange energy with the thermal bath and behaves
as an isolated oscillator. As a result, an open system with a localized mode
does not reach thermal equilibrium (does not thermalize). Instead, it evolves
into a cyclostationary state where dynamical variables and correlations,
averaged over an ensemble, oscillates in time with frequency of the localized
mode $\omega_{*}$. Cyclostationary states are not stationary and therefore are
manifestly nonergodic (ensemble and time averages are not equal). Following
Refs. Plyukhin ; Plyukhin2 , we shall refer to an open oscillator with a
localized mode as nonergodic.

Relaxation properties of nonergodic Brownian oscillators in the absence of an
external driving force ($F_{0}=0$) were studied in Kemeny ; Dhar ; Plyukhin .
Response properties to the external harmonic force were addressed in Ref.
Plyukhin2  for the case when the external frequency is higher than or equal to
the spectrum cutoff frequency, $\Omega\geq\omega_{0}$. That domain is of
particular interest because it covers setups when the external frequency
equals the frequency of the localized mode $\Omega=\omega_{*}$ (recall that
$\omega_{*}>\omega_{0}$). In that case the system shows unbounded resonance as
for an isolated oscillator with the natural frequency $\omega_{*}$. Another
conspicuous regime occurs when the oscillator natural frequency $\omega$ takes
a critical value $\omega_{c}$ separating ergodic and nonergodic configurations
(without and with a localized mode, respectively). In that case the cutoff
frequency $\omega_{0}$ has the meaning of the frequency of the incipient
localized mode, $\omega_{0}=\lim_{\omega\to\omega_{c}}\omega_{*}$. For the
setup $(\omega=\omega_{c},\,\Omega=\omega_{0})$ the oscillator shows an
unusual resonance (“quasi-resonance”) when the amplitude of the response
$q(t)=\langle x(t)\rangle$ increases with time sublinearly.

The goal of this paper is to extend the analysis of Ref. Plyukhin2  for
$\Omega\geq\omega_{0}$ to the low-frequency domain $\Omega<\omega_{0}$. Since
the localized mode frequency $\omega_{*}$ is higher than $\omega_{0}$, the
domain $\Omega<\omega_{0}$ does not include resonance and quasi-resonance
setups mentioned above. Still, the low-frequency response may be of interest
from both theoretical and experimental points of view. It requires a separate
consideration because the expressions obtained in Ref. Plyukhin2  for
$\Omega\geq\omega_{0}$ involve integrals which diverge for
$\Omega<\omega_{0}$. We will show that the results of Ref. Plyukhin2  can be
extended for the low-frequency domain merely by defining relevant improper
integrals in the sense of Cauchy principal value. That little correction,
however, leads to the significant difference in the explicit expressions for
the low- and high-frequency responses.

In order to make the paper self-consistent, in sections II and III we
recapitulate the model and special solutions, including quasi-resonance. That
sections overlap with Ref. Plyukhin2 . In section IV we show that the results
obtained for the high-frequency response cannot be directly applied for the
low-frequency domain and suggest an ad hoc remedy for the extension. An
alternative technique, addressing the low-frequency response, is developed in
sections V and VI.

##  II Model

As in Ref. Plyukhin2 , we consider a Brownian oscillator described by the
generalized Langevin equation (1) with a specific dissipation kernel

$\displaystyle
K(t)=\frac{\omega_{0}^{2}}{4}\,[J_{0}(\omega_{0}t)+J_{2}(\omega_{0}t)]=\frac{\omega_{0}}{2}\,\frac{J_{1}(\omega_{0}t)}{t},$
(3)

where $J_{n}(x)$’s are Bessel functions of the first kind. The kernel has the
absolute maximum at $t=0$ and for $t>0$ it oscillates with an amplitude
decaying with time as $t^{-3/2}$. The kernel’s spectral density $\rho(\nu)$
has the upper bound $\omega_{0}$,

$\displaystyle\rho(\nu)=\int_{0}^{\infty}K(t)\cos(\nu
t)\,dt=\frac{1}{2}\,\sqrt{\omega_{0}^{2}-\nu^{2}}\,\,\theta(\omega_{0}-\nu).$
(4)

Here $\theta(x)$ is the step function. As was mentioned in the Introduction, a
finite upper bound of the bath’s spectrum is a condition of the formation of a
localized mode. The frequency of the latter $\omega_{*}$ is not explicitly
present in Eqs. (1) and (3), and can be viewed as a hidden parameter of the
model.

The generalized Langevin equation with kernel (3) describes dynamics of the
terminal atom of a semi-infinite harmonic chain, when the atom is subjected,
in addition to the force from the neighbour atom, to an external harmonic
potential and a time-periodic force. The rest of the chain plays the role of a
thermal bath. The setting can be considered as an extension of archetypal
Rubin’s model Weiss ; Rubin . Although our choice of the dissipation kernel
(3) is somewhat accidental, it appears to be relevant in a broad context of
the dynamics of low-dimensional systems with localized modes Chalopin . One
may expect that our model reflects some general features of systems and
processes whose spectral densities have finite upper bounds.

Taking average of Eq. (1), one gets for the average coordinate

$\displaystyle q(t)=\langle x(t)\rangle,$ (5)

which we shall refer below as the response, the deterministic equation

$\displaystyle\ddot{q}=-\omega^{2}q-\int_{0}^{t}K(t-\tau)\,\dot{q}(\tau)d\tau+\frac{F_{0}}{m}\sin(\Omega
t).$ (6)

We assume that the external force starts to act at $t=0$ and that at earlier
times the oscillator is in thermal equilibrium. The relevant initial
conditions are zero, $q(0)=\dot{q}(0)=0$. Then the solution of Eq. (6) in the
Laplace domain is

$\displaystyle\tilde{q}(s)=\frac{1}{m}\,\tilde{G}(s)\,\tilde{F}_{ex}(s)$ (7)

where the Laplace transforms of the Green’s function $G(t)$ and the external
force $F_{ex}(t)=F_{0}\,\sin(\Omega t)$ are

$\displaystyle\tilde{G}(s)=\frac{1}{s^{2}+s\,\tilde{K}(s)+\omega^{2}},\quad\tilde{F}_{ex}(s)=F_{0}\,\frac{\Omega}{s^{2}+\Omega^{2}}.$
(8)

In the time domain the solution has the form of the convolution

$\displaystyle
q(t)=\frac{F_{0}}{m}\,\int_{0}^{t}G(t-\tau)\,\sin(\Omega\,\tau)\,d\tau.$ (9)

One can show, see e.g. Plyukhin , that the Green’s function $G(t)$ defined by
Eq. (8) also has the meaning of a normalized correlation function $\langle
x(0)\dot{x}(t)\rangle$. Then Eq. (9) can be viewed as a Kubo relation, which
expresses a response in terms of an equilibrium correlation function.

There are two ways to proceed. In the first way, which we shall follow in Sec.
IV, the response is evaluated in the time domain using Eq. (9). This method
has the advantage that for the given model the Green’s function in the time
domain $G(t)$ is already known Plyukhin . In the second way, one finds the
response by evaluating the inverse Laplace transform of the right-hand side of
Eq. (7). Although the two methods are clearly equivalent, we shall see that
for the low-frequency response at $\Omega<\omega_{0}$ there are technical
reasons to prefer one method over the other.

It was shown in Plyukhin  that for the given model the Green’s function in the
time domain $G(t)$ takes the form

$\displaystyle G(t)=\begin{cases}G_{e}(t),&\text{if
$\omega\leq\omega_{c}$},\\\ G_{e}(t)+G_{0}\,\sin(\omega_{*}t),&\text{ if
$\omega>\omega_{c}$}.\end{cases}$ (10)

Here the critical oscillator frequency

$\displaystyle\omega_{c}=\omega_{0}/\sqrt{2}\approx 0.707\,\omega_{0}$ (11)

separates ergodic (thermalizing) configurations with $\omega\leq\omega_{c}$
and nonergodic (non-thermalizing) configurations with $\omega>\omega_{c}$. The
ergodic component $G_{e}(t)$ has the same form for $\omega\leq\omega_{c}$ and
for $\omega>\omega_{c}$,

$\displaystyle
G_{e}(t)=\frac{4}{\pi\omega_{0}}\int_{0}^{1}\frac{\sin(x\,\omega_{0}\,t)\,x\,\sqrt{1-x^{2}}\,dx}{(1-4\,\lambda^{2})\,x^{2}+4\,\lambda^{4}},$
(12)

where $\lambda$ denotes the dimentionless oscillator frequency in units of
$\omega_{0}$,

$\displaystyle\lambda=\omega/\omega_{0}.$ (13)

Note that in Ref. Plyukhin  $\lambda$ stands for $(\omega/\omega_{0})^{2}$.
For two values of the oscillator frequency, $\omega=\omega_{0}/2$ and
$\omega=\omega_{c}$, Eq. (12) reduces to more explicit forms in terms Bessel
functions, see Eqs. (15)-(16) below.

One can show that $G_{e}(t)$, given by Eq. (12), vanishes at long times for
all values of $\omega$ (or $\lambda$). Then for $\omega\leq\omega_{c}$ the
Green’s function $G(t)=G_{e}(t)$ goes at long times to zero too. Since $G(t)$
also has the meaning of the normalized correlation $\langle
x(0)\dot{x}(t)\rangle$ Plyukhin , the asymptotic vanishing of $G(t)$ means
that the initial correlations die out and the system approaches thermal
equilibrium.

Eq. (10) shows that the localized mode with frequency $\omega_{*}$ develops
for $\omega>\omega_{c}$. The localized mode amplitude $G_{0}$ and frequency
$\omega_{*}$ of the nonergodic component are given by the following
expressions Plyukhin

$\displaystyle\omega_{0}\,G_{0}=\frac{8\,\lambda^{2}-4}{(4\,\lambda^{2}-1)^{3/2}},\qquad\omega_{*}=\frac{2\,\lambda^{2}}{\sqrt{4\lambda^{2}-1}}\,\,\omega_{0}.$
(14)

Note that $G_{0}=0$ and the nonergodic component vanishes for
$\omega=\omega_{c}$ (for
$\lambda=\lambda_{c}=\omega_{c}/\omega_{0}=1/\sqrt{2}$). Therefore $q(t)$ as a
function of $\omega$ is continuous including the critical point
$\omega=\omega_{c}$. For configurations with $\omega>\omega_{c}$, with the
localized mode developed, the Green’s function and initial correlations do not
vanish at long times. Instead of evolving toward thermal equilibrium, the
oscillator in the course of time reaches the cyclostationary state where the
means and correlations oscillate with the frequency of the localized mode
$\omega_{*}$.

##  III Special cases and quasi-resonance

For the two values of the oscillator frequency, $\omega=\omega_{0}/2$ and
$\omega=\omega_{c}$, the inverse transform of Eq. (35) can be compactly
expressed in terms of Bessel functions,

$\displaystyle G(t)$ $\displaystyle=$
$\displaystyle\frac{8}{\omega_{0}^{2}t}\,J_{2}(\omega_{0}t)=\frac{2}{\omega_{0}}\,[J_{1}(\omega_{0}t)+J_{3}(\omega_{0}t)],\qquad\omega=\omega_{0}/2,$
(15) $\displaystyle G(t)$ $\displaystyle=$
$\displaystyle\frac{2}{\omega_{0}}\,J_{1}(\omega_{0}t),\qquad\omega=\omega_{c}=\omega_{0}/\sqrt{2}.$
(16)

These expressions can also be obtained from Eq. (12) using the integral forms
of the Bessel functions. Both special frequencies $\omega_{0}/2$ and
$\omega_{c}$ correspond to ergodic configurations: The Green’s function
vanishes at long times, and the oscillator evolves to thermal equilibrium.
However, one observes that for $\omega=\omega_{c}$ the Green’s function decays
slower, which can be viewed as a precursor of the formation of the localized
mode for $\omega>\omega_{c}$.

Substituting Eqs. (15) and (16) into Eq. (9) yields

$\displaystyle q(t)$ $\displaystyle=$
$\displaystyle\frac{2F_{0}}{m\omega_{0}}\,\int_{0}^{t}[J_{1}(\omega_{0}\tau)+J_{3}(\omega_{0}t)]\,\sin[\Omega\,(t-\tau)]\,dt,\quad\omega=\omega_{0}/2,$
(17) $\displaystyle q(t)$ $\displaystyle=$
$\displaystyle\frac{2F_{0}}{m\omega_{0}}\,\int_{0}^{t}J_{1}(\omega_{0}\tau)\,\sin[\Omega\,(t-\tau)]\,dt,\quad\omega=\omega_{c}.$
(18)

We can evaluate these convolution integrals analytically only for
$\Omega=\omega_{0}$. Of particular interest is the setting
$(\omega=\omega_{c},\,\Omega=\omega_{0})$, when Eq. (18) takes the form

$\displaystyle
q(t)=\frac{2F_{0}}{m\omega_{0}}\,\int_{0}^{t}J_{1}(\omega_{0}\tau)\,\sin[\omega_{0}\,(t-\tau)]\,dt,\qquad(\omega=\omega_{c},\,\Omega=\omega_{0}).$
(19)

Taking into account the convolution integral

$\displaystyle\int_{0}^{x}J_{1}(y)\,\sin(x-y)\,dy=\sin(x)-xJ_{0}(x),$ (20)

see Ref. Watson  and the Appendix in Ref. Plyukhin2 , one finds Plyukhin2

$\displaystyle
q(t)=\frac{2\,F_{0}}{m\,\omega_{0}^{2}}\,\Big{\\{}\sin(\omega_{0}t)-\omega_{0}t\,J_{0}(\omega_{0}t)\Big{\\}},\qquad(\omega=\omega_{c},\,\Omega=\omega_{0}).$
(21)

Here the first term is the anticipated steady-state solution oscillating with
the frequency of the driving force $\Omega=\omega_{0}$. The second term,
however, is neither stationary nor transient. It oscillates with an amplitude
increasing indefinitely in time as $\sqrt{t}$. Following Ref. Plyukhin2  we
refer to such response as quasi-resonance. It can be viewed as a resonance
between the external harmonic force and the incipient localized mode in the
critical configuration with $\omega=\omega_{c}$. Note that according Eq. (14)
the cutoff frequency $\omega_{0}$ also has the meaning of the frequency of the
incipient localized mode,
$\lim_{\omega\to\omega_{c}}\omega_{*}=\lim_{\lambda\to\lambda_{c}}\omega_{*}=\omega_{0}$.

The second configuration, for which the response can be expressed in a compact
form, is $(\omega=\omega_{0}/2,\,\Omega=\omega_{0})$. In that case Eq. (17)
reads

$\displaystyle
q(t)=\frac{2F_{0}}{m\omega_{0}}\,\int_{0}^{t}[J_{1}(\omega_{0}\tau)+J_{3}(\omega_{0}t)]\,\sin[\omega_{0}\,(t-\tau)]\,dt.$
(22)

Taking into account integral (20) and its generalization Bailey

$\displaystyle\int_{0}^{t}J_{3}(y)\,\sin(x-y)\,dy=-x\,J_{2}(x)+6J_{1}(x)-3\sin(x)=x\,J_{0}(x)+4J_{1}(x)-3\sin(x)$
(23)

(here in the second equality we use the recurrence relation
$\frac{2n}{x}J_{n}=J_{n-1}+J_{n+1}$ with $n=1$), yields

$\displaystyle
q(t)=\frac{4\,F_{0}}{m\,\omega_{0}^{2}}\,\Big{\\{}\sin(\omega_{0}t-\pi)+2J_{1}(\omega_{0}t)\Big{\\}},\qquad(\omega=\omega_{0}/2,\,\Omega=\omega_{0}).$
(24)

The structure of this solution is similar to that for the normal damped and
fractional oscillators. The first term is the steady-state solution which
oscillates with the frequency of the external force $\Omega=\omega_{0}$. The
second term is a transient which vanishes at long times. One notable feature
is a slow decay of the transient term; its amplitude decays as $1/\sqrt{t}$.
Another remarkable feature is that the phase shift $\pi$ of the steady-state
response is the same as for an isolated (undamped) oscillator when
$\Omega>\omega$ French . Recall that for the normal damped oscillator,
described by Eq. (2), the phase shift reaches the value $\pi$ only in the
limit $\Omega\to\infty$ French .

For $\Omega\neq\omega_{0}$ the integral expressions (17) and (18) apparently
cannot be further simplified and may be used to evaluate the response
numerically. In that way, for the ergodic configuration with
$\omega=\omega_{0}/2$ we find that $q(t)$ evolves into a steady-state form

$\displaystyle q(t)\to A\,\sin(\Omega t-\delta)$ (25)

with the amplitude $A>0$ weakly decreasing with increasing $\Omega$, and the
phase shift $\delta$ monotonically increasing from $0$ for small $\Omega$ to
$\pi$ for $\Omega\gg\omega$. There is no resonance when $\Omega$ is close to
$\omega$, and on the whole, the response is qualitatively similar to that of
the overdamped normal damped oscillator, described by Eq. (2) with
$\gamma>\omega$ French .

For the critical configuration with $\omega=\omega_{c}$, the response,
described by Eq. (18), may show a variety of patterns. For $\Omega=\omega_{0}$
the response has the quasi-resonance form (21) and never reaches the steady
state (25). When $|\Omega/\omega_{0}-1|$ is finite but small, the response
evolves at long times to the steady state (25), but at shorter times shows
beats. The duration and amplitude of the first beat increase when $\Omega$ is
getting closer to $\omega_{0}$ and diverge in the limit $\Omega\to\omega_{0}$.
When $\Omega$ is not close to $\omega_{0}$ the response quickly reaches the
steady state form (25), where the amplitude weakly depends on $\Omega$ for
$\Omega<\omega_{0}$ and decreases with increasing $\Omega$ when
$\Omega>\omega_{0}$.

##  IV Ad hoc approach

Consider general ergodic configurations with $\omega\leq\omega_{c}$. The
localized mode is absent, the Green’s function has only ergodic component,
$G(t)=G_{e}(t)$, and the response, according to Eq. (9), is

$\displaystyle
q(t)=\frac{F_{0}}{m}\,\int_{0}^{t}G_{e}(t-\tau)\,\sin(\Omega\,\tau)\,d\tau.$
(26)

Substituting Eq. (12) for $G_{e}(t)$ into Eq. (26), changing the integration
order, and integrating over $\tau$ yields

$\displaystyle q(t)=\frac{F_{0}}{m\omega_{0}^{2}}\Big{\\{}B\,\sin(\Omega
t)+\varphi(t)\Big{\\}},$ (27)

where the dimensionless amplitude $B$ and the transient $\varphi(t)$ are

$\displaystyle B$ $\displaystyle=$
$\displaystyle-\frac{4}{\pi}\int_{0}^{1}\frac{x^{2}\sqrt{1-x^{2}}\,dx}{\Big{[}(1-4\lambda^{2})\,x^{2}+4\lambda^{4}\Big{]}\left(\Lambda^{2}-x^{2}\right)},$
(28) $\displaystyle\varphi(t)$ $\displaystyle=$
$\displaystyle\frac{4\Lambda}{\pi}\int_{0}^{1}\frac{x\sqrt{1-x^{2}}\,\sin(x\,\omega_{0}\,t)\,dx}{\Big{[}(1-4\lambda^{2})\,x^{2}+4\lambda^{4}\Big{]}\left(\Lambda^{2}-x^{2}\right)}.$
(29)

In these expressions $\lambda$ and $\Lambda$ denote the dimensionless
oscillator and external force frequencies in units of the cutoff frequency
$\omega_{0}$,

$\displaystyle\lambda=\omega/\omega_{0},\qquad\Lambda=\Omega/\omega_{0}.$ (30)

One observes that Eqs. (27)-(29) make sense, and have been used in Ref.
Plyukhin2 , only for the high-frequency response at $\Omega\geq\omega_{0}$, or
$\Lambda\geq 1$. In the low-frequency domain $\Omega<\omega_{0}$, or
$\Lambda<1$, the integrands in Eqs. (28) and (29) are singular at $x=\Lambda$
and the integrals diverge.

Recall that in deriving the above equations we substituted integral (12) over
$x$ into integral (26) over $\tau$ and then, in a rather cavalier manner,
changed the integration order $\int\\!d\tau\int\\!dx\to\int\\!dx\int\\!d\tau$.
The conditions of that last operation, regulated by Fubini’s theorem, are not
satisfied for $\Lambda<1$ since the integrand is not continuous in the
integration domain.

However, one notices that the integrals (28) and (29) converge in Cauchy’s
principal value sense:

$\displaystyle\text{p.v.}\int_{0}^{1}f(x)\,dx=\lim_{\epsilon\to
0^{+}}\left\\{\int_{0}^{\Lambda-\epsilon}f(x)\,dx+\int_{\Lambda+\epsilon}^{1}f(x)\,dx\right\\}.$
(31)

This suggests that Eqs. (27)-(29) are valid for the low-frequency response as
well, if one interpret the integral expressions for $B$ and $\varphi(t)$ in
the principal value sense,

$\displaystyle B$ $\displaystyle=$
$\displaystyle-\frac{4}{\pi}\,\text{p.v.}\int_{0}^{1}\frac{x^{2}\sqrt{1-x^{2}}\,dx}{\Big{[}(1-4\lambda^{2})\,x^{2}+4\lambda^{4}\Big{]}\left(\Lambda^{2}-x^{2}\right)},$
(32) $\displaystyle\varphi(t)$ $\displaystyle=$
$\displaystyle\frac{4\Lambda}{\pi}\,\text{p.v.}\int_{0}^{1}\frac{x\sqrt{1-x^{2}}\,\sin(x\,\omega_{0}\,t)\,dx}{\Big{[}(1-4\lambda^{2})\,x^{2}+4\lambda^{4}\Big{]}\left(\Lambda^{2}-x^{2}\right)}.$
(33)

Whereas this guess seems reasonable, it is rather ad hoc. In the next section
we will justify Eqs. (32) and (33) with another method, briefly mentioned at
the end of Sec. II. In that method the principal value of the integral for
$\varphi(t)$ emerges naturally, and the expression for $B$ has a more explicit
form, see Eqs. (58) and (59) below.

##  V Response of ergodic configurations

An alternative method to evaluate the response is to start with Eq. (7) for
its Laplace transform $\tilde{q}(s)$ and to perform the inversion
$q(t)=\frac{1}{2\pi i}\int_{\text{Br}}e^{st}\tilde{q}(s)\,ds$, which gives

$\displaystyle q(t)=\frac{F_{0}\Omega}{m}\frac{1}{2\pi
i}\,\int_{\text{Br}}\frac{e^{st}\,\tilde{G}(s)\,ds}{(s+i\Omega)(s-i\Omega)}.$
(34)

Here Br denotes the Bromwich path in the complex plane (a vertical line to the
right of all singularities of the integrand) and the transform of the Green’s
function is

$\displaystyle\tilde{G}(s)=\frac{2}{s^{2}+s\,\sqrt{s^{2}+\omega_{0}^{2}}+2\,\omega^{2}}.$
(35)

This expression follows from Eq. (8) and the Laplace transform of the kernel
(3)

$\displaystyle\tilde{K}(s)=\frac{1}{2}\,\left(\sqrt{s^{2}+\omega_{0}^{2}}-s\right).$
(36)

As shown in Ref. Plyukhin , for ergodic configurations with
$\omega\leq\omega_{c}$ the only singularities of the Green’s function
$\tilde{G}(s)$ are two branch points at $\pm i\omega_{0}$. Two other
singularities of the integrand in Eq. (34) are simple poles at $\pm i\Omega$.
Since all singular points are on the imaginary axis, contour Br in Eq. (34) is
an arbitrary vertical line to the right of the origin.

The evaluation of the integral (34) can be performed with the standard
technique based on the Cauchy’s integral and residue theorems, but there are
two subtleties. The first is that the poles lie on the branch cut, the second
is that $\tilde{G}(s)$ has two branches. Below we outline the main steps of
the evaluation.

The first step is to close the Bromwich contour as shown in Fig. 1 (left) and
to consider the integral $I$ over the resulting closed contour $\Gamma$

$\displaystyle I=\frac{F_{0}\Omega}{m}\frac{1}{2\pi
i}\,\int_{\Gamma}\frac{e^{st}\,\tilde{G}(s)\,ds}{(s+i\Omega)(s-i\Omega)},$
(37)

assuming the limit when the radius $R$ of the large arc is infinitely large,
$R\to\infty$, and the radii $\epsilon$ of the small (semi-) circles about the
branch points $\pm i\omega_{0}$ and poles $\pm i\Omega$ are infinitely small,
$\epsilon\to 0$. The contribution to $I$ from the arc vanishes, the
contributions from the line above and below the real axis cancel each other,
and the contribution from the rightmost vertical line equals $q(t)$. Then we
get

$\displaystyle I=q(t)+I_{C},$ (38)

where $I_{C}$ is the contribution from the contour $C$ depicted in Fig. 1
(right). On the other hand, since there are no singular points in the interior
of $\Gamma$, Cauchy’s integral theorem asserts that $I=0$, and therefore

$\displaystyle q(t)=-I_{C}.$ (39)

Figure 1: Left: Integration contour $\Gamma$ for integral $I$ in Eq. (37).
Right: Integration contour $C$ for integral $I_{C}$ in Eqs. (38) and (45)

To proceed, one has to take into account that the Green’s function (35)
involves the two-branch function $f(s)=\sqrt{s^{2}+\omega^{2}}$. To choose the
branch which is physically meaningful, one may recall that the Green’s
function $G(t)$ also determines the equilibrium correlation $\langle
q(0)\dot{q}(t)\rangle$ Plyukhin

$\displaystyle\langle q(0)\dot{q}(t)\rangle=\frac{k_{B}T}{m}\,G(t).$ (40)

That implies the initial condition $G(0)=0$, which in the Laplace domain has
the form

$\displaystyle\lim_{s\to\infty}s\,\tilde{G}(s)=0.$ (41)

One can show that this condition is only satisfied when $\tilde{G}(s)$
involves the branch of $f(s)=\sqrt{s^{2}+\omega^{2}}$ which takes real
positive (negative) values when $s$ is real and positive (negative).
Explicitly, the physical branch of the function

$\displaystyle
f(s)=\sqrt{s^{2}+\omega_{0}^{2}}=\sqrt{(s-i\omega_{0})(s+i\omega_{0})}$ (42)

is convenient to describe using a pair of polar coordinates
$(r_{1},\theta_{1})$ and $(r_{2},\theta_{2})$ with the origins at
$s=i\omega_{0}$ and $s=-i\omega_{0}$, respectively:

$\displaystyle
f(s)=\sqrt{r_{1}r_{2}}\,\exp\left(i\,\frac{\theta_{1}+\theta_{2}}{2}\right).$
(43)

For the physical branch the polar angles may be chosen from the same interval

$\displaystyle-\frac{3\pi}{2}<\theta_{1}\leq\frac{\pi}{2},\quad-\frac{3\pi}{2}<\theta_{2}\leq\frac{\pi}{2}.$
(44)

For instance, for $s=x$ being real and positive we get
$\theta_{1}+\theta_{2}=0$, then $f(s)=\sqrt{r_{1}r_{2}}$ is real and positive.

To evaluate the integral

$\displaystyle I_{C}=\frac{F_{0}\Omega}{m}\frac{1}{2\pi
i}\,\int_{C}\frac{e^{st}\,\tilde{G}(s)\,ds}{(s+i\Omega)(s-i\Omega)}$ (45)

we split it into five contributions

$\displaystyle I_{C}=I_{1}+I_{2}+I_{3}+I_{4}+I_{5},$ (46)

corresponding to the regions $C_{i}$ of the integration path, see Fig. 1
(right). Contributions $I_{1}$ and $I_{2}$ come from the integration over
circles $C_{1}$ and $C_{2}$ of radius $\epsilon$ about the branch points $\pm
i\omega_{0}$. One finds that the integrals are of order of $\epsilon$ for
$\omega\neq\omega_{c}$, and of order of $\sqrt{\epsilon}$ for
$\omega=\omega_{c}$. Thus $I_{1}$ and $I_{2}$ for any $\omega$ vanish when
$\epsilon\to 0$,

$\displaystyle I_{1}=I_{2}=0.$ (47)

Next consider the integrals $I_{3},I_{4}$ over the semi-circles
$C_{3}^{\pm},C_{4}^{\pm}$ of radius $\epsilon$ about $s=\pm i\Omega$. Using
Eqs. (43) and (44), for $s\in C_{3}^{\pm},C_{4}^{\pm}$ in the limit
$\epsilon\to 0$ one finds

$\displaystyle f(s)=\pm\sqrt{\omega_{0}^{2}-\Omega^{2}},$ (48)

where the sign is positive for the right semi-circles and negative for left
semi-circles. The Green’s function (35) for $s\in C_{3}^{\pm},C_{4}^{\pm}$
takes the form

$\displaystyle\tilde{G}(s)=\frac{2}{2\omega^{2}-\Omega^{2}\pm
i\Omega\,\sqrt{\omega_{0}^{2}-\Omega^{2}}},$ (49)

where again the plus and minus in the denominator correspond to the right and
left semi-circles, respectively. Then the straightforward evaluation gives

$\displaystyle
I_{3}+I_{4}=-\frac{F_{0}}{m}\,\frac{2(2\omega^{2}-\Omega^{2})}{4\omega^{4}+\Omega^{2}(\omega_{0}^{2}-4\omega^{2})}\,\sin(\Omega
t).$ (50)

The last contribution $I_{5}=I_{5}^{+}+I_{5}^{-}$ comes from the vertical
segments $C_{5}^{+}$ and $C_{5}^{-}$ along the right and left shores of the
branch cut, excluding the $2\epsilon$-wide intervals about $\pm i\Omega$. In
that case $s=iy\pm\delta$, where $\delta$ is infinitesimal and

$\displaystyle
y\in(-\omega_{0}+\epsilon,-\Omega-\epsilon)\cup(-\Omega+\epsilon,\Omega-\epsilon)\cup(\Omega+\epsilon,\omega_{0}-\epsilon).$
(51)

For $s\in C_{5}^{\pm}$ one finds
$f(s)=\sqrt{s^{2}+\omega_{0}^{2}}=\pm\sqrt{\omega_{0}^{2}-y^{2}}$, and the
Green’s function (35) takes the form

$\displaystyle\tilde{G}(s)=\frac{2}{2\omega^{2}-y^{2}\pm
iy\sqrt{\omega_{0}^{2}-y^{2}}},$ (52)

where the plus and minus correspond to $C_{5}^{+}$ and $C_{5}^{-}$,
respectively. Taking that into account we obtain

$\displaystyle
I_{5}=-\frac{F_{0}\,\Omega}{m}\,\frac{4}{\pi}\,\text{p.v.}\int_{0}^{\omega_{0}}\frac{y\,\sqrt{\omega_{0}^{2}-y^{2}}\,\sin(yt)\,dy}{[4\omega^{4}+y^{2}(\omega_{0}^{2}-4\omega^{2})]\,(\Omega^{2}-y^{2})}.$
(53)

The principal value of the integral emerges here quite naturally since the
integration region (51) has symmetric gaps about $\pm\Omega$.

Summing up contributions yields

$\displaystyle I_{C}$ $\displaystyle=$
$\displaystyle-\frac{F_{0}}{m\omega_{0}^{2}}\left\\{B\,\sin(\Omega
t)+\varphi(t)\right\\},$ (54) $\displaystyle B$ $\displaystyle=$
$\displaystyle\frac{2\omega_{0}^{2}(2\omega^{2}-\Omega^{2})}{4\omega^{4}+(\omega_{0}^{2}-4\omega^{2})\Omega^{2}},$
(55) $\displaystyle\varphi(t)$ $\displaystyle=$
$\displaystyle\frac{4\Omega\,\omega_{0}^{2}}{\pi}\,\text{p.v.}\int_{0}^{\omega_{0}}\frac{y\,\sqrt{\omega_{0}^{2}-y^{2}}\,\sin(yt)\,dy}{[4\omega^{4}+y^{2}(\omega_{0}^{2}-4\omega^{2})]\,(\Omega^{2}-y^{2})}.$
(56)

Finally, using Eq. (39) and expressing $B$ and $\varphi(t)$ in terms of
dimensionless frequencies $\lambda=\omega/\omega_{0}$ and
$\Lambda=\Omega/\omega_{0}$ we get

$\displaystyle q(t)$ $\displaystyle=$
$\displaystyle\frac{F_{0}}{m\omega_{0}^{2}}\Big{\\{}B\,\sin(\Omega
t)+\varphi(t)\Big{\\}},$ (57) $\displaystyle B$ $\displaystyle=$
$\displaystyle\frac{2\,(2\lambda^{2}-\Lambda^{2})}{4\lambda^{4}+(1-4\lambda^{2})\,\Lambda^{2}},$
(58) $\displaystyle\varphi(t)$ $\displaystyle=$
$\displaystyle\frac{4\Lambda}{\pi}\,\text{p.v.}\int_{0}^{1}\frac{x\,\sqrt{1-x^{2}}\,\sin(x\omega_{0}t)\,dx}{[4\lambda^{4}+x^{2}(1-4\lambda^{2})]\,(\Lambda^{2}-x^{2})}.$
(59)

These results, which hold for $\omega<\omega_{c}$ and $\Omega<\omega_{0}$ (or,
for $\lambda<\lambda_{c}$ and $\Lambda<1$), are equivalent to Eqs. (32) and
(33) obtained in the previous section with method I, but Eq. (58) for $B$ is
more explicit, and the principle value in the integral expression for
$\varphi(t)$ is now totally justified, see the remark after Eq. (53).

Figure 2: The average coordinate $q(t)$, in units of
$q_{0}=F_{0}/(m\omega_{0}^{2})$, for ergodic configurations with the
dimensionless oscillator frequency $\lambda=\omega/\omega_{0}=0.7$ (just below
the critical value $\lambda_{c}=1/\sqrt{2}\approx 0.707$) and three values of
the dimensionless external frequency $\Lambda=\Omega/\omega_{0}$, evaluated
with Eqs. (57)-(59). At much longer times the beat pattern dies out and $q(t)$
takes the steady-state form Eq. (60). The plot at the bottom is close to that
for the quasi-resonance solution Eq. (21). The integral in Eq. (59) is
evaluated numerically with the NIntegration routine of Wolfram Mathematica
using a principal value option Mathematica .

One can verify numerically that the function $\varphi(t)$ defined by Eq. (59)
evolves at long times into a harmonic form oscillating with frequency
$\Omega$. Therefore, at long times both terms in Eq. (57) oscillate with
frequency $\Omega$, and so does the total solution,

$\displaystyle q(t)\to A\,\sin(\Omega t-\delta).$ (60)

Here the amplitude $A$ and the phase shift $\delta$ depend on $\omega$ and
$\Omega$ (or, on $\lambda$ and $\Lambda$) and are difficult to express in an
analytical forms. The response relaxes to the steady state form (60) on the
time scale which increases and, in fact, diverges, when $\omega$ and $\Omega$
are getting close to $\omega_{c}$ and $\omega_{0}$, respectively. For such
settings, the solution on a long initial time interval shows beat patterns,
see Fig. 2. For the critical configuration $\omega=\omega_{c}$ the duration
and amplitude of the first beat indefinitely increases as
$\Omega\to\omega_{0}^{-}$, and the solution converges to the quasi-resonance
form (21). Such behavior is similar (though not numerically identical) to that
for the high-frequency response of the critical configuration
($\omega=\omega_{c}$) when $\Omega$ approaches $\omega_{0}$ from the above,
$\Omega\to\omega_{0}^{+}$, see Fig. 1 in Ref. Plyukhin2 .

##  VI Response of nonergodic configurations

Consider nonergodic configurations with $\omega>\omega_{c}$. The Green’s
function has now both ergodic and nonergodic (periodic) components,
$G(t)=G_{e}(t)+G_{0}\,\sin(\omega_{*}t)$. We can follow the same approach as
in the previous section, but taking into account that the Green’s function
$\tilde{G}(s)$ in the Laplace domain now has two extra poles at $\pm
i\omega_{*}$. Since $\omega_{*}>\omega_{0}$, the integral $I$ given by Eq.
(37) is not zero but equals to the sum of residues at $s=\pm i\omega_{*}$,

$\displaystyle
I=Res[e^{st}\,\tilde{q}(s),i\omega_{*}]+Res[e^{st}\,\tilde{q}(s),-i\omega_{*}],$
(61)

where

$\displaystyle\tilde{q}(s)=\frac{1}{m}\,\tilde{G}(s)\,\tilde{F}_{ex}(s)=\frac{1}{m}\,\tilde{G}(s)\,\frac{\Omega}{s^{2}+\Omega^{2}}.$
(62)

On the other hand $I=q(t)+I_{C}$, see Eq. (38), and therefore

$\displaystyle
q(t)=-I_{C}+Res[e^{st}\,\tilde{q}(s),i\omega_{*}]+Res[e^{st}\,\tilde{q}(s),-i\omega_{*}].$
(63)

Here the integral $I_{C}$ is still given by Eq. (54) and the sum of residues
can be written as

$\displaystyle
Res[e^{st}\tilde{q}(s),i\omega_{*}]+Res[e^{st}\tilde{q}(s),-i\omega_{*}]=\frac{F_{0}\Omega}{m\,(\Omega^{2}-\omega_{*}^{2})}\,\Big{\\{}Res[e^{st}\tilde{G}(s),i\omega_{*}]+Res[e^{st}\tilde{G}(s),i\omega_{*}]\Big{\\}}.$
(64)

The sum of residues of the Green’s function can be readily evaluated (see
Appendix B in Plyukhin  for $\mu=1$), or one may just notice that it gives the
nonergodic part of the Green’s function $G_{0}\sin(\omega_{*}t)$, which gives

$\displaystyle
Res[e^{st}\tilde{q}(s),i\omega_{*}]+Res[e^{st}\tilde{q}(s),-i\omega_{*}]=\frac{F_{0}\Omega}{m\,(\Omega^{2}-\omega_{*}^{2})}\,G_{0}\,\sin(\omega_{*}t).$
(65)

Combining the results and expressing in terms of dimensionless frequencies, we
obtain

$\displaystyle q(t)$ $\displaystyle=$
$\displaystyle\frac{F_{0}}{m\omega_{0}^{2}}\Big{\\{}B\sin(\Omega
t)+C\,\sin(\omega_{*}t)+\varphi(t)\Big{\\}},$ (66) $\displaystyle B$
$\displaystyle=$
$\displaystyle\frac{2\,(2\lambda^{2}-\Lambda^{2})}{4\lambda^{4}+(1-4\lambda^{2})\,\Lambda^{2}},$
(67) $\displaystyle C$ $\displaystyle=$
$\displaystyle\omega_{0}G_{0}\,\frac{\Lambda}{\Lambda^{2}-\lambda_{*}^{2}}=-\frac{4\Lambda(2\,\lambda^{2}-1)}{\sqrt{4\,\lambda^{2}-1}\,(4\lambda^{4}+\Lambda^{2}-4\lambda^{2}\Lambda^{2})},$
(68) $\displaystyle\varphi(t)$ $\displaystyle=$
$\displaystyle\frac{4\Lambda}{\pi}\,\text{p.v.}\int_{0}^{1}\frac{x\,\sqrt{1-x^{2}}\,\sin(x\omega_{0}t)\,dx}{(\Lambda^{2}-x^{2})[4\lambda^{4}+x^{2}(1-4\lambda^{2})]}.$
(69)

Here, in the first equation for $C$, $\lambda_{*}$ stands for the
dimensionless localized frequency $\lambda_{*}=\omega_{*}/\omega_{0}$ and
$G_{0}$ is the amplitude of the nonergodic part of the Green’s function, both
are given by Eq. (14).

Figure 3: The average coordinate $q(t)$, in units of
$q_{0}=F_{0}/(m\omega_{0}^{2})$, for nonergodic configurations with the
dimensionless oscillator frequency $\lambda=\omega/\omega_{0}=0.71$ (just
above the critical value $\lambda_{c}=1/\sqrt{2}\approx 0.707$) and three
values of the dimensionless external frequency $\Lambda=\Omega/\omega_{0}$,
evaluated with Eqs. (66)-(69). At longer times $q(t)$ shows beats which do not
die out and continue indefinitely. The plot at the bottom is close to that for
the quasi-resonance solution Eq. (21).

Compared to the response of ergodic configurations, solution (66) involves the
additional term $\frac{F_{0}}{m\omega_{0}^{2}}\,C\,\sin(\omega_{*}t)$. The
term $\varphi(t)$ at long times still oscillates with frequency $\Omega$.
Therefore, the asymptotic long-time response of nonergodic configurations does
not oscillate with the frequency of the external force $\Omega$ but has the
form of the superposition of two harmonic terms with frequencies $\Omega$ and
$\omega_{*}$,

$\displaystyle q(t)\to A\,\sin(\Omega t-\delta)+C\,\sin(\omega_{*}t),$ (70)

In the considered domain $(\omega>\omega_{c},\,\Omega<\omega_{0})$ the
frequencies $\omega_{*}$ and $\Omega$ are separated, $\omega_{*}>\omega_{0}$
and $\Omega<\omega_{0}$, yet may be close. In that cases the asymptotic
solution (70) shows beats which, in contrast to ergodic configurations, do not
vanish at long times but continue indefinitely. When $\omega\to\omega_{c}^{+}$
and $\Omega\to\omega_{0}^{-}$ the response converges to the quasi-resonance
form (21), see Fig. 3.

##  VII Conclusion

The nonergodic Brownian oscillator considered in this paper is a quite unusual
open system. Its signature properties are the luck of thermalization (when
unperturbed) and unbounded resonance and quasi-resonance response when the
frequency of the external force $\Omega$ coincides with the frequency of the
localized mode $\omega_{*}$. In this paper we considered the low-frequency
response at $\Omega<\omega_{0}$, where $\omega_{0}$ is the upper bound of the
thermal bath’s spectrum. In that case the two frequencies $\Omega$ and
$\omega_{*}$ are separated, $\Omega<\omega_{0}$ and
$\omega_{*}\geq\omega_{0}$, and the condition of resonance cannot be met.
However, for the critical and near-to-critical configurations ($\omega$ is
equal or close to $\omega_{c}$) the localized frequency $\omega_{*}$ is close
to $\omega_{0}$, and the frequencies $\Omega$ and $\omega_{*}$ may be close
even for the low-frequency domain $\Omega<\omega_{0}$. In such case the low-
frequency response may show a quasi-resonance response, with the amplitude
increasing with time sublinearly, on a finite but long time interval.

Comparing the results with those obtained in Ref. Plyukhin , we found the
response to be a continuous function of the external frequency, including the
point $\Omega=\omega_{0}$. Although for $\Omega<\omega_{0}$ and for
$\Omega\geq\omega_{0}$ the response is given by different expressions, they
match at $\Omega=\omega_{0}$.

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