#  Exact Controllability for the Wave Equation on a Graph with Cycle and
Delta-Prime Vertex Conditions

Sergei Avdonin, Department of Mathematics and Statistics, University of Alaska
Fairbanks, Fairbanks, AK 99775, U.S.A<EMAIL_ADDRESS>
and Moscow Center for Fundamental and Applied Mathematics, Moscow 119991,
Russia

Julian Edward, Department of Mathematics and Statistics, Florida International
University, Miami, FL 33199, U.S.A<EMAIL_ADDRESS>
Günter Leugering, Departments of Mathematics and Data Science Friedrich-
Alexander-University Erlangen-Nürnberg, Martenstr. 3, 91058 Erlangen, Germany;
<EMAIL_ADDRESS>
Abstract. Exact controllability for the wave equation on a metric graph
consisting of a cycle and two attached edges is proven. One boundary and one
internal control are used. At the internal vertices, delta-prime conditions
are satisfied. As a second example, we examine a tripod controlled at the root
and the junction, while the leaves are fixed. These examples are key to
understanding controllability properties in general metric graphs.

##  1 Introduction

Control problems for linear wave and beam equations on metric, in particular,
tree-like graphs, have been heavily researched because of their many
applications in science and engineering, see [2], [10], [14],[16], [17], [19]
and references therein. The theory has been extended to quasilinear wave and
beam equations in recent years, see e.g. [18] and the references therein for
the wave equation, as well as [12] for nonlinear beam equations on networks.
However, until very recently, very few positive results were obtained for
graphs with cycles. In fact, suppose controls are exerted at simple nodes
only. Then linear wave equations on graphs containing cycles, or even on trees
where two or more Dirichlet nodes are uncontrolled, are not exactly
controllable, and not always even approximately controllable. Also, applying
controls at multiple nodes is not possible as long as the classical
transmission conditions, i.e. the continuity condition and Kichhoff’s
condition across the joints, stay intact. This observation suggests we relax
these conditions in one way or another in order to gain a stronger access to
the system dynamics by the way of controls. We can distinguish two types of
transmission conditions at multiple nodes which have come to be known as
_delta_ and _delta-prime_ conditions in the context of electrical engineering,
nanotechnology and quantum graphs. In principle, there are two ways to
approach this concept: one is to relax the continuity condition of states
(delta) or derivatives (delta-prime) by inserting a control between two or
more incident edges, and the second one aims at relaxing the balance of forces
or states, respectively, (typically referred to as Kirchhoff conditions).
Significant progress in this direction recently appeared in [8], where an
exact controllability result for general graphs was formulated and proven. In
that work, the authors assume a mix of interior and boundary controls, and at
the uncontrolled interior vertices, standard (Kirchhoff–Neumann) conditions
are assumed to hold. For a related result, see [5]. We also mention another
concept of controllability, the so-called nodal profile controllability, where
at a given node the state and the velocity profile is exactly matched by
boundary controls. This concept is weaker than that of exact controllability
as discussed in this article, but admits cycles. See [15] for further
information and references.

In the present paper, we focus on two key elementary graphs in order to
understand the controllability properties for problems on general metric
graphs. These are a simple cycle with attached edges and a tripod with two
Dirichlet nodes, the former being more complicated.

##  2 Problem statement

To be more specific, we consider the graph shown in Figure 1 with delta-prime
vertex conditions holding at the uncontrolled interior vertices. Delta-prime
conditions arise naturally in problems in electrical engineering [1] and
nanotechnology [11]. We are unaware of any control theory papers on graphs
with delta-prime conditions at internal vertices, although the paper [4]
discussed an inverse problem for a quantum tree with delta prime conditions.
If we release the junction conditions at node $v_{3}$, keeping only edges 3
and 4 connected and pose instead a Dirichlet condition there for edge 2, then
Figure 2 emerges. Without changing the analysis significantly, we may also
replace edge 2 by two serially connected edges such that the circle becomes a
triangle. In this way, we can view our two examples as building blocks for the
understanding of the controllability properties for wave equations on general
metric graphs. A theory on exact controllability covering the general case,
however, is still lacking and subject to further investigations.

$v_{2}\quad$$e_{1}$$v_{1}$$v_{3}$$e_{4}$$v_{4}$$e_{3}$$e_{2}$ Figure 1: A ring
with two attached edges $v_{0}$$v_{2}$$v_{1}$$v_{3}$ Figure 2: Star with
coordinate system: $e_{j}$ identified with $(0,\ell_{j})$

We begin with the graph in Figure 1, so our graph ${\Omega}=\\{V,E\\}$
consists four vertices, $V=\\{v_{j},\,j=1,\ldots,4\\}$, and four edges,
$E=\\{e_{j},\,j=1,\ldots,4\\}$. Each edge is identified with an interval of
the real axis with positive length, $e_{1}(v_{1},v_{2})$ with $(0,l_{1}),$
$e_{2}(v_{2},v_{3})$ with $(0,l_{2}),$ $e_{3}(v_{2},v_{3})$ with $(0,l_{3}),$
and $e_{4}(v_{3},v_{4})$ with $(0,l_{4}).$ In what follows, we denote by
$\phi_{j}$ the restriction of function $\phi$ to the edge $e_{j}$. We consider
the initial boundary value problem on $[{\Omega}\setminus V]\times[0,T]$ with
one boundary control and one internal control:

$\partial_{tt}u-\partial_{xx}u+q(x)u=0,$ (2.1) $u(\cdot,0)=u_{t}(\cdot,0)=0,$
(2.2) $u_{1}(v_{1},t)=f_{1}(t),$ (2.3) $\partial u_{2}(v_{2},t)-\partial
u_{1}(v_{2},t)=f_{2}(t),\ \partial u_{3}(v_{2},t)-\partial u_{1}(v_{2},t)=0,\
\ \sum_{j=1}^{3}u_{j}(v_{2},t)=0.$ (2.4) $\partial u_{2}(v_{3},t)=\partial
u_{3}(v_{3},t)=\partial u_{4}(v_{3},t),\ \ \sum_{j=2}^{4}u_{j}(v_{3},t)=0.$
(2.5) $u_{4}(v_{4},t)=0.$ (2.6)

Here $f_{1}\in H^{1}(0,T),\;f_{1}(0)=0,\;f_{2}\in L^{2}(0,T),$ $q_{j}$ is a
continuous function for all $j$, and $\partial u_{j}$ denotes the derivative
of $u$ at a vertex, along the incident edge $e_{j}$ in the direction outward
from the vertex.

We denote the space $L^{2}(\Omega)$ by ${\mathcal{H}}$ and the space of
functions on $\Omega$ whose restriction to each edge $e$ is in $H^{1}(e)$ by
${\mathcal{H}}^{1},$ with norm
$\|\phi\|^{2}_{{\mathcal{H}}^{1}}=\sum_{j=1}^{3}\|\phi^{\prime}\|_{L^{2}(e_{j})}^{2}+\|\phi\|_{L^{2}(e_{j})}^{2}.$
Functions $y\in{\mathcal{H}}^{1}$ such that

$y_{1}(v_{1})=y_{4}(v_{4})=0,\ \ \sum_{j=1}^{3}y_{j}(v_{2})=0,\ \
\sum_{j=2}^{4}y_{j}(v_{3})=0$

we denote by ${{\mathcal{H}}}^{1}_{0}.$ For presentational simplicity, we will
assume

$l_{2}\geq l_{3},$

leaving the simple adaptation to the other case to the reader.

###### Theorem 1

Let $T_{*}=\max\,\\{\,l_{1}+l_{2},\,l_{3}+l_{4}\,\\}$ and let $T\geq 2T_{*}$.
Let $(\phi^{1},\phi^{2})\in{\mathcal{H}}_{0}^{1}\times{\mathcal{H}}$. There
exist controls ${\bf f}:=(f_{1},f_{2})\in H^{1}_{0}(0,T)\times L^{2}(0,T)$
such that the solution $u^{\bf f}$ to System (2.1)-(2.6) satisfies

$u^{{\bf f}}(x,T)=\phi^{1}(x),\ u_{t}^{{\bf f}}(\cdot,T)=\phi^{2}(\cdot),$
(2.7)

and there exists a constant $C$ that depends only on $q,l_{j},j=1,\ldots,4,$
such that

$\|f_{1}\|_{H^{1}(0,2T)}+\|f_{2}\|_{L^{2}(0,2T)}\leq
C(\|\phi^{1}\|_{{\mathcal{H}}^{1}}+\|\phi^{2}\|_{{\mathcal{H}}}).$ (2.8)

The existence of ${\bf f}$ satisfying (2.7), (2.8) is called exact
controllability.

Key ingredients in the proof of Theorem 1 are shape and velocity control
results that can be formulated as follows.

###### Theorem 2

a) (Shape control) Let $T\geq T_{*}$. Let $\phi^{1}\in{\mathcal{H}}^{1}_{0}$.
There exist controls ${\bf f}:=(f_{1},f_{2})\in H_{0}^{1}(0,T)\times
L^{2}(0,T)$ such that

$u^{{\bf f}}(x,T)=\phi_{1}(x),$

and there exists a constant $C$ that depends only on $q,l_{j},j=1,\ldots,4,$
such that

$\|f_{1}\|_{L^{2}(0,T)}+\|f_{2}\|_{H^{1}(0,T)}\leq
C(\|\phi_{1}\|_{{\mathcal{H}}^{1}}).$ (2.9)

b) (Velocity control). Let $T\geq T_{*}$. Let $\phi^{2}\in{\mathcal{H}}$.
There exist controls ${\bf f}:=(f_{1},f_{2})\in H^{1}(0,T)\times L^{2}(0,T)$
such that

$u_{t}^{{\bf f}}(x,T)=\phi^{2}(x),$

and there exists a constant $C$ that depends only on $q,l_{j},j=1,\ldots,4,$
such that

$\|f_{1}\|_{L^{2}(0,T)}+\|f_{2}\|_{H^{1}(0,T)}\leq
C(\|\phi^{2}\|_{{\mathcal{H}}}).$ (2.10)

Theorem 1 follows from Theorem 2 by a moment argument. This argument was
successfully used in proving exact controllability of hybrid systems and of
the wave equation on graphs with the Kirchhoff–Neumann matching conditions
[2], [5], [7], [8].

We remark that compared with [8], this paper has a slightly sharper control
time.

This paper is organized as follows. In Section 3.1, we give a solution
representation of the wave equation on the interval for various boundary
conditions. These representations are used in the proof of shape and velocity
controllability. In Section 3.2, we prove shape controllability for a related
initial boundary value problem on a 3-point star as in Figure 2. The main
purpose of this section is to present the ideas of our shape control argument
in a technically simpler setting, but the result might also be of independent
interest. In Section 3.3, we then prove the shape controllability result
announced in Theorem 2. In Section 3.4, we prove Theorem 1.

##  3 Proofs.

###  3.1 Forward problem for the interval.

We begin by recalling solution for the forward problem on the interval.

$\displaystyle\partial_{tt}{v}-\partial_{xx}{v}+q(x){v}$ $\displaystyle=$
$\displaystyle 0,\ 0<x<l,\ t\in(0,T),$ (3.11)
$\displaystyle{v}(x,0)={v}_{t}(x,0)$ $\displaystyle=$ $\displaystyle 0,\
0<x<l,$ (3.12) $\displaystyle{v}_{x}(0,t)$ $\displaystyle=$ $\displaystyle
g(t),$ (3.13) $\displaystyle{v}_{x}(l,t)$ $\displaystyle=$
$\displaystyle-\partial v(l,t)=-h(t),\ t>0.$ (3.14)

For any function $p(t)$, we denote by $P(t)$ by the formula:
$P(t)=-\int_{0}^{t}p(s)ds$. In this case, see [7], the solution to the system
(3.16)-(3.19) is given by

$\displaystyle{v}_{NN}^{(g,h)}(x,t)$ $\displaystyle=$
$\displaystyle\sum_{n\geq
0}\left(G(t-2nl-x)+\int_{2nl+x}^{t}w_{+}(2nl+x,s)G(t-s)ds\right)$ (3.15)
$\displaystyle+\sum_{n\geq
1}\left(G(t-2nl+x)+\int_{2nl-x}^{t}w_{-}(2nl-x,s)G(t-s)ds\right)$
$\displaystyle+\sum_{n\geq
1}\left(H(t-(2n-1)l+x)+\int_{(2n-l)\ell-x}^{t}w_{-}((2n-1)l-x,s)H(t-s)\,ds\right)$
$\displaystyle+\sum_{n\geq
1}\left(H(t-(2n-1)l-x)+\int_{(2n-l)\ell+x}^{t}w_{+}((2n-1)l+x,s)H(t-s)\,ds\right),$

where $w_{\pm}$ are continuous functions with $w_{x}\in L^{2},$ and we set
$G(t)=H(t)=0$ for $t<0.$

Similarly, the solution to

$\displaystyle\partial_{tt}{v}-\partial_{xx}{v}+q(x){v}$ $\displaystyle=$
$\displaystyle 0,\ 0<x<l,\ t\in(0,T),$ (3.16)
$\displaystyle{v}(x,0)={v}_{t}(x,0)$ $\displaystyle=$ $\displaystyle 0,\
0<x<l,$ (3.17) $\displaystyle{v}_{x}(0,t)$ $\displaystyle=$ $\displaystyle
g(t),$ (3.18) $\displaystyle{v}(l,t)$ $\displaystyle=$ $\displaystyle f(t),\
t>0,$ (3.19)

is given by

$\displaystyle{v}_{ND}^{(g,f)}(x,t)$ $\displaystyle=$
$\displaystyle\sum_{n\geq 0}\mbox{
{$(-1)^{n}$}}\left(G(t-2nl-x)+\int_{2nl+x}^{t}w_{+}(2nl+x,s)G(t-s)ds\right)$
$\displaystyle+\sum_{n\geq 1}\mbox{
{$(-1)^{n}$}}\left(G(t-2nl+x)+\int_{2nl-x}^{t}w_{-}(2nl-x,s)G(t-s)ds\right),$
$\displaystyle+\sum_{n\geq
1}(-1)^{n-1}\left(f(t-(2n-1)l+x)+\int_{(2n-1)l-x}^{t}k_{-}((2n-1)l-x,s)f_{1}(t-s)ds\right)$
$\displaystyle+\sum_{n\geq
1}(-1)^{n-1}\left(f(t-(2n-1)l-x)+\int_{(2n-1)l+x}^{t}k_{+}((2n-1)l+x,s)f_{1}(t-s)ds\right),$

where the properties of $k_{\pm}$ are analogous to those of $w_{\pm}$, and we
set $f(t)=0$ for $t<0.$

###  3.2 Shape control for three-point star

In this section, we demonstrate our shape control argument on a simpler
example: a three edge star. In what follows, we denote by $v_{0}$ the central
vertex, and by $v_{j}$ the boundary vertex for $e_{j}$, with $j=1,2,3,$ see
Figure 2.

We consider the initial boundary value problem with one boundary control and
one internal control:

$\partial_{tt}u-\partial_{xx}u+q(x)u=0,\ x\in e_{j},t>0$ (3.21)
$u(\cdot,0)=u_{t}(\cdot,0)=0,$ (3.22) $u_{1}(v_{1},t)=f_{1}(t),\
u_{2}(v_{2},t)=u_{3}(v_{3},t)=0.$ (3.23) $\partial u_{2}(v_{0},t)-\partial
u_{1}(v_{0},t)=f_{2}(t),\ \partial u_{3}(v_{0},t)-\partial u_{1}(v_{0},t)=0,\
\ \sum_{j=1}^{3}u_{j}(v_{0},t)=0.$ (3.24)

For simplicity of exposition, we assume $l_{2}\geq l_{3}$ in this section. Let
${\mathcal{H}}^{1}_{0}$ be a space of functions $y$ on this graph whose
restriction $y_{j}$ to each $e_{j}$ belongs to $H^{1}(e_{j})$ and

$y_{1}(v_{1})=y_{1}(v_{1})=y_{1}(v_{1})=0,\ \ \sum_{j=1}^{3}y_{j}(v_{0})=0.$

###### Theorem 3

(Shape control) Let $T_{*}=\max(l_{2},l_{1}+l_{3}).$ Let $T\geq T_{*}$. Let
$\phi\in{\mathcal{H}}^{1}_{0}$. There exist controls ${\bf
f}:=(f_{1},f_{2})\in H_{0}^{1}(0,T)\times L^{2}(0,T)$ such that

$u^{{\bf f}}(x,T)=\phi(x),$

and there exists a constant $C$ that depends only on $q,l_{j},j=1,\ldots,3,$
such that

$\|f_{1}\|_{L^{2}(0,T)}+\|f_{2}\|_{H^{1}(0,T)}\leq
C(\|\phi\|_{{\mathcal{H}}^{1}}).$ (3.25)

Proof:

Set $\partial g_{j}(t)=u_{j}(v_{0},t).$ Applying (3.24), we get the equations:

$\displaystyle g_{1}(t)+f_{2}(t)$ $\displaystyle=$ $\displaystyle g_{2}(t),$
(3.26) $\displaystyle g_{1}(t)$ $\displaystyle=$ $\displaystyle g_{3}(t).$
(3.27)

For readability of what follows, we now set $q=0$, so that the integral terms
in (LABEL:nd) vanish. The necessary modifications in the general case will be
given at the end of section 3.3. Using $\sum_{1}^{3}\partial_{t}u_{j}(0,t)=0$,
we get the equation:

$\displaystyle 0$ $\displaystyle=$ $\displaystyle g_{1}(t)+2\sum_{n\geq
1}\mbox{ {$(-1)^{n}$}}g_{1}(t-2nl_{1})+2\sum_{n\geq
1}(-1)^{n}f_{1}^{\prime}(t-(2n-1)l_{1})$ (3.28)
$\displaystyle+g_{2}(t)+2\sum_{n\geq 1}\mbox{ {$(-1)^{n}$}}g_{2}(t-2nl_{2})$
$\displaystyle+g_{3}(t)+2\sum_{n\geq 1}\mbox{ {$(-1)^{n}$}}g_{3}(t-2nl_{3}).$

Combining these with (3.26), (3.27), we get

$0=3g_{1}(t)+f_{2}(t)+2\sum_{n\geq 1}\mbox{
{$(-1)^{n}$}}g_{1}(t-2nl_{1})+2\sum_{n\geq 1}\mbox{
{$(-1)^{n}$}}(g_{1}(t-2nl_{2})+f_{2}(t-2nl_{2}))+2\sum_{n\geq 1}\mbox{
{$(-1)^{n}$}}g_{1}(t-2nl_{3})$ $+2\sum_{n\geq
1}(-1)^{n}f_{1}^{\prime}(t-(2n-1)l_{1}).$ (3.29)

The construction of the controls is a two step process using the equations
above.

Step 1.

In this step, we determine $f_{1}^{1}$ supported in
$(T_{*}-l_{1}-l_{2},T_{*}-l_{1})$, together with $f_{2}(t)$, supported on
$(T_{*}-l_{2},T_{*}),$ so that

$u^{(f_{1}^{1},f_{2})}_{j}(x,T_{*})=\phi_{j},\ j=2,3.$

Since $q=0$, the integral kernels in (LABEL:nd) vanish, so we can easily find
$g_{j}\in L^{2}(T_{*}-l_{j},T_{*})$ such that for $j=2,3$,

$u_{j}(x,T_{*})=v_{ND}^{(g_{j},0)}(x,T_{*})=\phi_{j}(x).$ (3.30)

We then set $g_{3}(t)=0$ for $t\leq T_{*}-l_{3},$ and $g_{2}(t)=0$ for $t\leq
T_{*}-l_{2}.$ Then $f_{2}$ is found immediately for all $t$ by applying
(3.26),(3.27), and clearly $f_{2}$ has its support in $[T_{*}-l_{2},T_{*}]$.

We now solve for $(f_{1}^{1})^{\prime}$. Denote by $\alpha(t)$ various
functions that are now known for $t\in[T_{*}-l_{2},T_{*}]$, which include
$g_{1}$ and $f_{2}$. Thus (3.28) simplifies to

$\sum_{n\geq 1}(-1)^{n}(f_{1}^{1})^{\prime}(t-(2n-1)l_{1})=\alpha(t),\
t\in(T_{*}-l_{2},T_{*}),$

hence

$\sum_{n\geq 0}(-1)^{n}(f_{1}^{1})^{\prime}(t-2nl_{1})=\alpha(t),\
t\in(0,T_{*}-l_{1}).$ (3.31)

We solve for $f_{1}^{\prime}$ iteratively. For $t<2l_{1}$ and $n>0$,
$(f_{1}^{1})^{\prime}(t-2nl_{1})=0$ so (3.31) simplifies to

$(f_{1}^{1})^{\prime}=\alpha(t);$ (3.32)

thus we have solved for $(f_{1}^{1})^{\prime}(t)$ for $t<2l_{1}.$ Now suppose
we have solved for $(f_{1}^{1})^{\prime}(t)$ for $t<2n_{0}l_{1}.$ Then for
$t<2(n_{0}+1)l_{1}$ and $n>0$, $(f_{1}^{1})^{\prime}(t-2nl_{1})$ is known, so
again we have (3.32) holding, and we thus solve for $(f_{1}^{1})^{\prime}(t)$
for $t<2(n_{0}+1)l_{1}$. This process can be iterated until $t>T_{*}-l_{1}$,
where we set $(f_{1}^{1})^{\prime}(t)=0$. Setting $f_{1}^{1}(0)=0$, this
uniquely determines $f_{1}^{1}$ on $(0,T_{*}-l_{1}).$ We now use any extension
of $f_{1}^{1}$ to $H_{0}^{1}(0,T_{*}).$

Step 2. We have defined $f_{2}(t)$ for all $t$. It remains to find $f_{1}(t)$.
Let $\tilde{\phi}=\phi-u^{(f_{1}^{1},f_{2})}(*,T_{*}),$ so
$\tilde{\phi}_{j}=0$ for $j=2,3.$ By continuity, we have
$\tilde{\phi}(v_{0})=0$, and also $\tilde{\phi}(v_{1})=0$. Then by (LABEL:nd),
we find $f_{1}^{2}\in H^{1}_{0}(0,T_{*})$, with support in
$[T_{*}-l_{1},T_{*}]$, so that
$v_{ND}^{(0,f_{1}^{2})}(x,T_{*})=\tilde{\phi}_{1}(x),$ and hence
$u=u^{(f_{1}^{2},0)}$ will solve $u_{1}(x,T_{*})=\tilde{\phi}_{1}(x).$ We then
set $f_{1}=f_{1}^{1}+f_{1}^{2}$, and the shape control problem is solved.
$\Box$

One can similarly prove velocity controllability, in time $T_{*}$, and then by
using the arguments of the Section 3.4, we get

###### Theorem 4

Let $T_{*}=\max\,\\{\,l_{2},\,l_{1}+l_{3}\,\\}$ and let $T\geq 2T_{*}$. Let
$(\phi^{1},\phi^{2})\in{\mathcal{H}}_{0}^{1}\times{\mathcal{H}}$. There exist
controls ${\bf f}:=(f_{1},f_{2})\in H^{1}_{0}(0,T)\times L^{2}(0,T)$ such that
the solution $u^{\bf f}$ to System (2.1)-(2.6) satisfies

$u^{{\bf f}}(x,T)=\phi^{1}(x),\ u_{t}^{{\bf f}}(\cdot,T)=\phi^{2}(\cdot),$
(3.33)

and there exists a constant $C$ that depends only on $q,l_{j},j=1,\ldots,4,$
such that

$\|f_{1}\|_{H^{1}(0,2T)}+\|f_{2}\|_{L^{2}(0,2T)}\leq
C(\|\phi^{1}\|_{{\mathcal{H}}^{1}}+\|\phi^{2}\|_{{\mathcal{H}}}).$ (3.34)

###  3.3 Shape control for graph with cycle

In this section we prove part a) of Theorem 2. The proof for part b) is
velocity controllability is similar, and is left to the reader. We begin by
using (2.4),(2.5), to derive some equations, (3.37)-(3.41), that will be used
in constructing the shape control.

Now let $u$ be the solution to the system (2.1)-(2.6). We adopt the following
notation:

$g_{j}(t)=\partial u_{j}(v_{2},t),\ j=1,2,3,\mbox{ and }h_{k}(t)=\partial
u_{k}(v_{3},t),\ k=2,3,4..$ (3.35)

Then for each $j$, we can use (3.15) or (LABEL:nd) to represent $u_{j}$. In
particular,

$u_{1}(x,t)=v^{(g_{1},f_{1})}_{ND}(x,t);\
u_{j}(x,t)=v_{NN}^{(g_{j},h_{j})}(x,t)\mbox{ for }j=2,3;\
u_{4}(x,t)=v_{ND}^{h_{4},0}(x,t).$ (3.36)

Applying (2.4), we get the equations:

$\displaystyle g_{1}(t)+f_{2}(t)$ $\displaystyle=$ $\displaystyle g_{2}(t),$
(3.37) $\displaystyle g_{1}(t)$ $\displaystyle=$ $\displaystyle g_{3}(t).$
(3.38)

For readability of what follows, we now set $q=0$, so that the integral terms
in (3.15) and (LABEL:nd) vanish. The necessary modifications in the general
case will be given at the end of section 3.3. Using
$\sum_{1}^{3}\partial_{t}u_{j}(0,t)=0$, we get the equation:

$\displaystyle 0$ $\displaystyle=$ $\displaystyle g_{1}(t)+2\sum_{n\geq
1}\mbox{ {$(-1)^{n}$}}g_{1}(t-2nl_{1})+2\sum_{n\geq
1}(-1)^{n-1}f_{1}^{\prime}(t-(2n-1)l_{1})$ (3.39)
$\displaystyle+g_{2}(t)+2\sum_{n\geq 1}g_{2}(t-2nl_{2})+2\sum_{n\geq
1}h_{2}(t-(2n-1)l_{2})$ $\displaystyle+g_{3}(t)+2\sum_{n\geq
1}g_{3}(t-2nl_{3})+2\sum_{n\geq 1}h_{3}(t-(2n-1)l_{3}).$

Combining these and using $h_{2}(t)=h_{3}(t)=h_{4}(t)$ , we get

$0=3g_{1}(t)+f_{2}(t)+2\sum_{n\geq 1}g_{1}(t-2nl_{1})+2\sum_{n\geq
1}(g_{1}(t-2nl_{2})+f_{2}(t-2nl_{2}))+2\sum_{n\geq 1}g_{1}(t-2nl_{3})$
$+2\sum_{n\geq 1}(-1)^{n-1}f_{1}^{\prime}(t-(2n-1)l_{1})+2\sum_{n\geq
1}h_{4}(t-(2n-1)l_{2})+2\sum_{n\geq 1}h_{4}(t-(2n-1)l_{3}),\forall t.$ (3.40)

A similar argument at $v_{3}$ gives

$0=3h_{4}(t)+2\sum_{n\geq 1}h_{4}(t-2nl_{4})+2\sum_{n\geq
1}h_{4}(t-2nl_{2})+2\sum_{n\geq 1}h_{4}(t-2nl_{3})$ $-2\sum_{n\geq
1}(g_{1}+f_{2})(t-(2n-1)l_{2})-2\sum_{n\geq 1}g_{1}(t-(2n-1)l_{3}),\forall t.$
(3.41)

By definition, we have $g_{j}(t)=0$ for $t<0$, and by unit speed of
propagation and (3.35), we have $h_{j}(t)=0$ for $t<\min(l_{2},l_{3}),$ so the
sums in (3.40),(3.41) are all finite.

To construct our shape control, we will use $f_{2}$ alone to attain the
desired shape on $e_{4}$, while $f_{1},f_{2}$ will be used jointly to attain
control on the other edges. Recall we have assumed $l_{3}\leq l_{2};$ for
presentational simplicity. Recall $T_{*}=\max(l_{1}+l_{2},l_{3}+l_{4})$. In
the construction that follows, $f_{2}$ will be supported in the interval
$(T_{*}-\max(l_{3}+l_{4},l_{2}),T_{*}))$ while $f_{1}$ will be supported in
$(T_{*}-l_{2}-l_{1},T_{*}).$

Let $\phi\in{\mathcal{H}}^{1}$. For the moment, we assume $q=0$, pointing out
the modifications necessary in the general case at the section’s end. In what
follows, we set $f_{j}(t)=g_{k}(t)=0$ for $t<0$ and all $j,k$, and
$h_{j}(t)=0$ for $t<l_{3}$. When convenient, we will denote $u$ as
$u^{(f_{1},f_{2})}.$

Step 1.

For this step, we construct $f_{2}^{1}(t)$ supported in
$(T_{*}-l_{4}-l_{3},T_{*}-l_{3})$ so that

$u_{4}^{(0,f_{2}^{1})}(x,T_{*})=\phi_{4}(x).$

Setting $v^{(h_{4},0)}(x,T_{*})=\phi_{4}(x)$ in (LABEL:nd), we get
$h_{4}(T_{*}-x)=-\phi_{4}^{\prime}(x)$ on $(T_{*}-l_{4},T_{*})$, and we then
extend $h_{4}$ trivially to $(0,T_{*}).$ We now compute the control
$f_{2}^{1}$, uniquely determined on $(T_{*}-l_{4}-l_{3},T_{*}-l_{3})$, that
generates $h_{4}$, i.e. such that

$u^{(0,f_{2}^{1})}(v_{3},t)=h_{4}(t),\ t\in(T_{*}-l_{4},T_{*}).$

Let $\Delta=\min(l_{1},l_{2}-l_{3},l_{2})$ if $l_{3}>l_{2}$, and otherwise
$\Delta=\min(l_{1},l_{2}).$ Define $t_{n}^{*}=n\Delta$ for $n\geq 0.$ We will
use (3.40), (3.41) to solve for $g_{1},f_{2}^{1}$ on successive intervals
$[t_{j}^{*},t_{j+1}^{*}]$ in terms of the known function $h_{4}$. As $j$
increases, the number of non-zero terms in (3.40) and/or (3.41) increases. In
what follows, we will denote by $\alpha(t)$ various functions, which change
from line to line, that are known for $t<t_{j}^{*}$. Also, $\beta(t)$ are
functions, that varying from line to line, that are determined by $h_{4}$, and
hence are known for all $t$ in the argument below.

We rewrite (3.40), resp. (3.41) as:

$\alpha(t)=3g_{1}(t)+f_{2}^{1}(t)+2\sum_{n\geq 1}\mbox{
{$(-1)^{n}$}}g_{1}(t-2nl_{1})+2\sum_{n\geq
1}\big{(}g_{1}(t-2nl_{2})+f_{2}^{1}(t-2nl_{2})\big{)}+2\sum_{n\geq
1}g_{1}(t-2nl_{3}),$ (3.42) $\beta(t)=\sum_{n\geq
1}(g_{1}(t-(2n-1)l_{2}+l_{3})+f_{2}(t-(2n-1)l_{2}+l_{3})+g_{1}(t)+\sum_{n\geq
1}g_{1}(t-2nl_{3}).$ (3.43)

For $t<t_{1}^{*}$ and $n\geq 1$, we have $t-(2n-1)l_{j}<0$ for $j=1,2,3$ by
the definition of $t_{j}^{*}$, and hence

$\alpha(t)=3g_{1}(t)+f_{2}^{1}(t),$ (3.44)

and similarly (3.43) becomes

$\beta(t)=\big{(}g_{1}(t-l_{2}+l_{3})+f_{2}^{1}(t-l_{2}+l_{3})\big{)}+g_{1}(t).$
(3.45)

We have two cases.

Case 1: $l_{2}>l_{3}.$ Then for $t<t_{1}^{*}$, we have $t-l_{2}+l_{3}<0$, so

$\beta(t)=g_{1}(t),\ t\in[0,t_{1}^{*}].$

Combining this with (3.44), we can clearly solve for $f_{2}^{1},g_{1}$ for
$t<t_{1}^{*}.$

Case 2: $l_{2}=l_{3}$. Then (3.45) becomes

$\alpha(t)=2g_{1}(t)+f_{2}^{1}(t),\ t\in[0,t_{1}^{*}],$

and thus we solve for $f_{2}^{1},g_{1}$ for $t<t_{1}^{*}.$

Now assume $f_{2}^{1}(t),g_{1}(t)$ have been determined for $t<t_{k}^{*}$. In
this case, $t<t_{k+1}^{*}$ implies that for $n>0$ and $j=1,2,3$, we have
$(t-2nl_{j})<t_{k}^{*}$, and so the functions
$g_{1}(t-2nl_{j}),f_{2}^{1}(t-2nl_{j})$ have been previously calculated. Thus
(3.42) becomes:

$\alpha(t)=3g_{1}(t)+f_{2}^{1}(t),\ t\in[t_{k}^{*},t_{k+1}^{*}].$ (3.46)

We now rewrite (3.43) by identifying known terms. First, if $t<t_{k+1}^{*}$
and $j=2,3$, then for $n>1$ we have $t-(2n-1)l_{j}+l_{3}<t_{k}^{*},$ so (3.45)
becomes

$\alpha(t)=g_{1}(t-l_{2}+l_{3})+f_{2}^{1}(t-l_{2}+l_{3})+2g_{1}(t),\
t<t_{k+1}^{*}.$ (3.47)

By considering the same cases as above, we solve for $f_{2}^{1}(t),g_{1}(t)$
for $t<t_{k+1}^{*}$.

We iterate this argument until $t_{k+1}^{*}\geq T_{*}-l_{3}$; this determines
$f_{2}^{1}(t)$ for $t\in(T_{*}-l_{4}-l_{3},T_{*}-l_{3})$. We then extend
$f_{2}^{1}$ trivially to the rest of $[0,T_{*}].$

Step 2. Let $\psi=\phi-u^{(0,f_{2}^{1})}(\cdot,T_{*}),$ so $\psi_{4}=0.$

In this step, we determine $f_{1}^{1}$ supported on
$(T_{*}-l_{1}-l_{2},T_{*}-l_{1})$, together with $f_{2}^{2}(t)$, supported on
$(T_{*}-l_{2},T_{*}),$ so that

$u^{(f_{1}^{1},f_{2}^{2})}_{j}(x,T_{*})=\psi_{j},\ j=2,3.$

Setting $f_{2}=f_{2}^{1}+f_{2}^{2}$, we will then have

$u^{(f_{1}^{1},f_{2})}_{j}(x,T_{*})=\phi_{j},\ j=2,3,4.$

Since $q=0$, the integral kernels in (LABEL:nd) vanish, so for $j=2,3$ we can
easily find $g_{j}\in L^{2}(T_{*}-l_{j},T_{*})$ such that

$u^{g_{j}}(x,T_{*})=v_{NN}^{(g_{j},0)}(x,T_{*})=\psi_{j}(x).$ (3.48)

We then set $g_{3}(t)=0$ for $t\leq T_{*}-l_{3},$ and $g_{2}(t)=0$ for $t\leq
T_{*}-l_{2}.$ Then $f_{2}^{2}$ is found immediately for all $t$ by applying
(3.37),(3.38), and clearly $f_{2}^{2}$ has its support in
$[T_{*}-l_{2},T_{*}]$. The support of $f_{2}=f_{2}^{1}+f_{2}^{2}$ will be

$[T_{*}-l_{2},T_{*}]\cup[T_{*}-l_{3}-l_{4},T_{*}-l_{3}]=[T_{*}-\max(l_{2},l_{3}+l_{4}),T_{*}].$

We now solve for $(f_{1}^{1})^{\prime}$. First note that by unit speed of wave
propagation, for $t<T_{*}$ the waves generated by $g_{2},g_{3}$ will not reach
beyond $v_{3},$ so $h_{j}(t)=0$. Thus, (3.40) simplifies to

$0=3g_{1}(t)+f_{2}^{1}(t)+2\sum_{n\geq 1}(-1)^{n}g_{1}(t-2nl_{1})+2\sum_{n\geq
1}(g_{1}(t-2nl_{2})+f_{2}^{1}(t-2nl_{2}))+2\sum_{n\geq 1}g_{1}(t-2nl_{3})$
$+2\sum_{n\geq 1}(-1)^{n}(f_{1}^{1})^{\prime}(t-(2n-1)l_{1}),\
t\in(T_{*}-l_{2},T_{*}).$ (3.49)

Since the functions $g_{1},f_{2}^{2}$ are known, we can solve for
$(f_{1}^{1})^{\prime}(t)$ iteratively for
$t\in[T_{*}-l_{2}-l_{1},T_{*}-l_{1}].$ We then set $f_{1}^{1}(t)=0$ for $t\leq
T_{*}-l_{2}-l_{1}$.

Step 3. We have defined $f_{2}(t)$ for all $t$. It remains to find $f_{1}(t)$.
Choose some extension of $f_{1}^{1}$ to $H^{1}_{0}(0,T_{*}).$ Let
$\tilde{\psi}=\psi-u^{(f_{1}^{1},f_{2})}(*,T_{*}),$ so $\tilde{\psi}_{j}=0$
for $j=2,3,4.$ By continuity, we have $\tilde{\psi}(v_{2})=0$, and also
$\tilde{\psi}(v_{1})=0$. Then by (LABEL:nd), we solve $f_{1}^{2}\in
H^{1}_{0}(T_{*}-l_{1},T_{*})$, with support in $(T_{*}-l_{1},T_{*})$, so that
$v_{ND}^{(f_{1}^{2},0)}(x,T_{*})=\tilde{\psi}_{1}(x),$ and hence
$u=u^{f_{1}^{2},0}$ will solve $u_{1}(x,T_{*})=\tilde{\psi}_{1}.$ We then set
$f_{1}=f_{1}^{1}+f_{1}^{2}$, and the shape control problem is solved.

We now indicate the changes necessary for the case $q\neq 0$. In this case,
the only two changes. First, solving (3.48) now requires a standard integral
equation argument. Second, (3.49) changes to

$0=3g_{1}(t)+f_{2}(t)+2\sum_{n\geq 1}(-1)^{n}g_{1}(t-2nl_{1})+2\sum_{n\geq
1}(g_{1}(t-2nl_{2})+f_{2}(t-2nl_{2}))+2\sum_{n\geq 1}g_{1}(t-2nl_{3})$
$+2\sum_{n\geq 1}(-1)^{n}(L_{n}(f_{1}^{1})^{\prime})(t-(2n-1)l_{1}),\
t<T_{*},$ (3.50)

with

$(L_{n}(f_{1}^{1})^{\prime})(t-(2n-1)l_{1})=2(f_{1}^{1})^{\prime}(t-(2n-1)l_{1})+2q((2n-1)l_{1})f_{1}^{1}(t-(2n-1)l_{1})$
$-\int_{s=0}^{t-(2n-1)l_{1}}\big{(}k_{+}((2n-1)l_{1},s+(2n-1)l_{1})+k_{-}((2n-1)l_{1},s+(2n-1)l_{1})\big{)}f_{1}^{1}(t-(2n-1)l_{1}-s)ds.$

A standard argument using the theory of Volterra integral equations shows that
$L_{n}$ is an isomorphism on $L^{2}(0,T_{*}-(2-1)l_{1})$. Thus, using (3.50),
we solve for $L_{n}(f_{1}^{1})^{\prime}$, then $(f_{1}^{1})^{\prime}$, then
$f_{1}^{1}$, using at iterative argument. The rest of the proof of shape
controllability is the same as in the case of $q=0$.$\Box$

Remark Recalling that $T_{*}=\max\,\\{\,l_{1}+l_{2},\,l_{3}+l_{4}\,\\}$, we
summarize the supports of the controls $f_{1},f_{2}$.

1) if $l_{1}+l_{2}+\mu=l_{3}+l_{4},\ \mu\geq 0,$ then $T_{*}=l_{3}+l_{4},$
supp $\;f_{1}\subset[\mu,T_{*}]$ and supp $\;f_{2}\subset[0,T_{*}]$;

2) if $l_{1}+l_{2}=l_{3}+l_{4}+\mu,\ \mu\geq 0,$ then $T_{*}=l_{1}+l_{2},$
supp $\;f_{1}\subset[0,T_{*}]$ and supp $\;f_{2}\subset[\tau,T_{*}],$ where
$\tau=\min\,\\{\mu,l_{1}\\}.$

###  3.4 Proof of Theorem 1

Now we prove the exact controllability in time $2T,$ for $T\geq T_{*},$ using
the shape and velocity controllability in time $T$. Let
$\\{{\omega}_{n}^{2}\\}_{1}^{\infty}$ and $\\{\varphi_{n}\\}_{1}^{\infty}$ be
the eigenvalues and corresponding normalized eigenfunctions of the spectral
problem associated to system (2.1)-(2.6):

$-\varphi_{n}^{\prime\prime}+q(x)\varphi_{n}={\omega}_{n}^{2}\varphi_{n},\
x\in e_{j},j=1,...,4,$ (3.51)
$(\varphi_{n})_{1}(v_{1})=(\varphi_{n})_{4}(v_{4})=0,$ (3.52)
$\partial(\varphi_{n})_{1}(v_{2})=\partial(\varphi_{n})_{2}(v_{2})=\partial(\varphi_{n})_{3}(v_{2}),\
\sum_{j=1}^{3}(\varphi_{n})_{j}(v_{2})=0,$ (3.53)
$\partial(\varphi_{n})_{2}(v_{3})=\partial(\varphi_{n})_{3}(v_{3})=\partial(\varphi_{n})_{4}(v_{3}),\
\sum_{j=2}^{4}(\varphi_{n})_{j}(v_{3})=0.$ (3.54)

Here and in what follows, we denote the restriction of $\varphi_{n}$ to edge
$e_{j}$ by $(\varphi_{n})_{j}.$

The following estimates can be extracted from [6], [9], and [13]:

$|\partial(\varphi_{n})_{1}(v_{1})|\prec n,\ |(\varphi_{n})_{2}(v_{2})|\prec
1.$

We write $u(x,t)=\sum_{1}^{\infty}a_{n}(t)\varphi_{n}(x).$ To find the
coefficients $a_{n}$ we integrate by parts in $x$ the identity

$0=\int_{0}^{T}\int_{\Omega}[\partial_{tt}u-\partial_{xx}u+q(x)u]\,\varphi_{n}(x)\mu(t)\,dxdt$

with arbitrary $\mu\in C^{2}_{0}[0,T].$ We obtain the initial value problem

$a_{n}^{\prime\prime}(t)+\omega_{n}^{2}a_{n}(t)=-f_{1}(t)(\varphi_{n})_{1}^{\prime}(v_{1})+f_{2}(t)(\varphi_{n})_{2}(v_{2}),\
a_{n}(0)=a_{n}^{\prime}(0)=0.$

By variation of parameters, we get

$a_{n}(T)=\int_{0}^{T}[-f_{1}(t)(\partial\varphi_{n})_{1}(v_{1})+f_{2}(t)(\varphi_{n})_{2}(v_{2})]\,\frac{\sin{\omega}_{n}(T-t)}{{\omega}_{n}}\,dt.$
(3.55)

In what follows, the controls solving the shape control problem will be
denoted by $f_{j}=f_{j0}$, and the controls solving the velocity control
problem will be denoted by $f_{j}=f_{j1},$ for $j=1,2.$ We apply integration
by parts for the terms with $f_{1},f_{2}$ in (3.55), using
$f_{1}(0)=f_{1}(T)=0$. Setting $a_{n}=a_{n}(T)$, the shape controllability
result can be formulated as solvability of the moment problem

${{\omega}_{n}}a_{n}=\int_{0}^{T}f_{10}^{\prime}(t){\kappa}_{n1}{\cos{\omega}_{n}(T-t)}+f_{20}(t){\kappa}_{n2}\,{\sin{\omega}_{n}(T-t)}\,dt,\
\forall n,$ (3.56)

for arbitrary sequence $\\{a_{n}{\omega}_{n}\\}\in\ell^{2}$ by the functions
$f_{10}^{\prime},f_{20}\in L^{2}(0,T).$ Here

${\kappa}_{n1}=\frac{\partial(\varphi_{n})_{1}(v_{1})}{{\omega_{n}}},\
{\kappa}_{n2}={(\varphi_{n})_{2}(v_{2})}.$

Differentiating (3.55) with respect to $T$, and then integrating by parts
while using $f_{1}(0)=f_{1}(T)=0$, we reformulate the the velocity
controllability result as solvability of the moment problem

$b_{n}=\int_{0}^{T}[f_{21}(t)\,{\kappa}_{n2}\,\cos{\omega}_{n}(T-t)+f^{\prime}_{11}(t)\,{\kappa}_{n1}\,\sin{\omega}_{n}(T-t)]\,dt,\
\forall n,$ (3.57)

for arbitrary sequences $\\{b_{n}\\}\in\ell^{2}$ by the functions
$f_{11}^{\prime},f_{21}\in L^{2}(0,T).$

Now we extend the functions $f^{\prime}_{11},f_{20}$ in the odd way with
respect to $t=T$ from the interval $[0,T]$ to $[0,2T],$ and the functions
$f^{\prime}_{10},f_{21}$ — in the even way. Define the functions

$\ f^{\prime}_{1}=\frac{f^{\prime}_{10}+f^{\prime}_{11}}{2},\
f_{2}=\frac{f_{20}+f_{21}}{2},$

and observe that these functions solve the moment problems

$a_{n}{\omega}_{n}=\int_{0}^{2T}[f^{\prime}_{1}(t)\,{\kappa}_{n1}\,\cos{\omega}_{n}(T-t)+f_{2}(t)\,{\kappa}_{n2}\,\sin{\omega}_{n}(T-t)]\,dt$
(3.58)
$b_{n}=\int_{0}^{2T}[f_{2}(t)\,{\kappa}_{n2}\,\cos{\omega}_{n}(T-t)+f^{\prime}_{1}(t)\,{\kappa}_{n1}\,\sin{\omega}_{n}(T-t)]\,dt$
(3.59)

for arbitrary sequences $\\{a_{n}{\omega}_{n}\\},\;\\{b_{n}\\}\in\ell^{2}$.
Also, defining $f_{1}(t)=\int_{0}^{t}f^{\prime}_{1}(s)\,ds,$ it’s not hard to
prove that $f_{1}\in H^{1}_{0}(0,2T)$. Thus these moment problems can be
rewritten in the form

$a_{n}{\omega}_{n}=\int_{0}^{2T}[-f_{2}(t)\,{\kappa}_{n2}\
-f_{1}(t)\,{\kappa}_{n1}\omega_{n}]\,\sin{\omega}_{n}(T-t)\,dt$ (3.60)
$b_{n}=\int_{0}^{2T}[-f_{2}(t)\,{\kappa}_{n2}-f_{1}(t)\,{\kappa}_{n1}\omega_{n}]\,\cos{\omega}_{n}(T-t)\,dt,$
(3.61)

It is an algebra exercise to check that solvability of the moment problem
(3.60), (3.61) is equivalent to solvability of the moment problem

$\alpha_{n}=\int_{0}^{2T}[-f_{2}(t)\,{\kappa}_{n2}\
+f_{1}(t)\,{\kappa}_{n1}\omega_{n}]\,\sin{\omega}_{n}(2T-t)\,dt$ (3.62)
$\beta_{n}=\int_{0}^{2T}[-f_{2}(t)\,{\kappa}_{n2}+f_{1}(t)\,{\kappa}_{n1}\omega_{n}]\,\cos{\omega}_{n}(2T-t)\,dt,$
(3.63)

for arbitrary sequences $\\{\alpha_{n}\\},\\{\beta_{n}\\}\in\ell^{2}.$ Having
solved these moment problems is equivalent to proving exact controllability in
time $2T$. The proof of Theorem 1 is complete.$\Box$

Acknowledgments
The research of Sergei Avdonin was supported in part by the National Science
Foundation, grant DMS 1909869.

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