L>0$, and thus $N^{(k_{\ell})}_{1}(u)>N^{(k_{\ell})}_{2}(u)>0$ for all
$u\in[t,t+\epsilon]$, for all $\ell$ large enough. It follows that $X(j)=0$
for all $j=\lfloor k_{\ell}t\rfloor+1,\dots,\lfloor
k_{\ell}(t+\epsilon)\rfloor$, and that the minima in equations (27) and (28)
are never attained in the third case (otherwise we would have
$N^{(k_{\ell})}_{2}(u)=0$ for some $u\in[t,t+\epsilon]$). Therefore, we have

$\displaystyle M_{1}(j)$
$\displaystyle=\min\Big{\\{}Q_{2}(j)-D_{2}(j)+S_{2}(j),\,\,Q_{1}(j)-D_{1}(j)+S_{1}(j)\Big{\\}},$
$\displaystyle M_{2}(j)$
$\displaystyle=\min\Big{\\{}Q_{3}(j)-D_{3}(j)+S_{3}(j)-M_{1}(j),\,\,Q_{2}(j)-D_{2}(j)+S_{2}(j)\Big{\\}},$

where $Q(\cdot)$ is defined recursively as

$\displaystyle Q_{1}(u+1)$ $\displaystyle=Q_{1}(u)-D_{1}(u)+S_{1}(u)-M_{1}(u)$
$\displaystyle Q_{2}(u+1)$
$\displaystyle=Q_{2}(u)-D_{2}(u)+S_{2}(u)-M_{1}(u)-M_{2}(u)$ $\displaystyle
Q_{3}(u+1)$ $\displaystyle=Q_{3}(u)-D_{3}(u)+S_{3}(u)-M_{2}(u).$

Note that this corresponds to Case 1 in Lemma 8. Therefore, since
$\omega\in\mathcal{C}$, Lemma 8 implies

$\displaystyle r_{1}(t+\epsilon)-r_{1}(t)$
$\displaystyle=\lim\limits_{\ell\to\infty}R^{(k_{\ell})}_{1}(t+\epsilon)-R^{(k_{\ell})}_{1}(t)$
$\displaystyle=\epsilon.\,\overline{C}_{1},$

and

$\displaystyle r_{2}(t+\epsilon)-r_{2}(t)$
$\displaystyle=\lim\limits_{\ell\to\infty}R^{(k_{\ell})}_{2}(t+\epsilon)-R^{(k_{\ell})}_{2}(t)$
$\displaystyle=\epsilon.\,\underline{C}_{2}.$

Dividing by $\epsilon$ and taking the limit as $\epsilon$ goes to zero we
conclude that, when $n_{1}(t)>n_{2}(t)>0$, we have

$\displaystyle\frac{dn_{1}(t)}{dt}$
$\displaystyle=\lambda_{1}-\overline{C}_{1}\qquad\text{and}\qquad\frac{dn_{2}(t)}{dt}=\lambda_{2}-\underline{C}_{2}.$

Case (b): $n_{1}(t)=n_{2}(t)>0$

First note that, if $n_{1}(t)>0$ and $n_{2}(t)>0$, the same argument as in the
previous case gives us

$r_{i}(t+\epsilon)-r_{i}(t)\in\big{[}\epsilon\underline{C}_{i},\,\epsilon\overline{C}_{i}\big{]},$
(30)

and

$r_{1}(t+\epsilon)+r_{2}(t+\epsilon)-r_{1}(t)-r_{2}(t)=\epsilon C_{1,2},$ (31)

for all sufficiently small $\epsilon$.

Suppose that

$\lambda_{1}-\overline{C}_{1}>\lambda_{2}-\underline{C}_{2}.$

Combining this with Equation (30), it follows that

$n_{1}(u)>n_{2}(u)$

for all $u\in(t,t+\epsilon]$. Therefore

$\displaystyle\frac{dn_{1}(u)}{du}$
$\displaystyle=\lambda_{1}-\overline{C}_{1}\qquad\text{and}\qquad\frac{dn_{2}(u)}{du}=\lambda_{2}-\underline{C}_{2},$

for all $u\in(t,t+\epsilon]$. Since $t$ is a regular time, we also have

$\displaystyle\frac{dn_{1}(t)}{dt}$
$\displaystyle=\lambda_{1}-\overline{C}_{1}\qquad\text{and}\qquad\frac{dn_{2}(t)}{dt}=\lambda_{2}-\underline{C}_{2}.$

Analogously, if

$\lambda_{2}-\overline{C}_{2}>\lambda_{1}-\underline{C}_{1},$

we have

$\displaystyle\frac{dn_{1}(t)}{dt}$
$\displaystyle=\lambda_{1}-\underline{C}_{1}\qquad\text{and}\qquad\frac{dn_{2}(t)}{dt}=\lambda_{2}-\overline{C}_{2}.$

Finally, suppose that

$\frac{\lambda_{1}+\lambda_{2}-C_{1,2}}{2}\geq\max\left\\{\lambda_{1}-\overline{C}_{1},\,\,\lambda_{2}-\overline{C}_{2}\right\\}.$

In particular, this means that

$\lambda_{1}-\overline{C}_{1}\leq\lambda_{2}-\underline{C}_{2}\qquad\text{and}\qquad\lambda_{2}-\overline{C}_{2}\leq\lambda_{1}-\underline{C}_{1}.$

Therefore, we have

$\frac{dn_{1}(u)}{du}-\frac{dn_{2}(u)}{du}\leq 0$

when $n_{1}(u)>n_{2}(u)$, and

$\frac{dn_{1}(u)}{du}-\frac{dn_{2}(u)}{du}\geq 0$

when $n_{1}(u)<n_{2}(u)$.

######  Lemma 10.

We have

$\frac{dn_{1}(t)}{dt}-\frac{dn_{2}(t)}{dt}=0,$

and

$\frac{dn_{1}(t)}{dt}=\frac{dn_{2}(t)}{dt}=\frac{\lambda_{1}+\lambda_{2}-C_{1,2}}{2}.$

###### Proof.

Proof: We prove this by contradiction. Suppose that

$\frac{dn_{1}(t)}{dt}-\frac{dn_{2}(t)}{dt}>0.$

Since $n_{1}(t)=n_{2}(t)$ and $t$ is a regular point, we must have
$n_{1}(u)>n_{2}(u)$ for all sufficiently small $u>t$. However, we have

$\frac{dn_{1}(u)}{du}-\frac{dn_{2}(u)}{du}\leq 0$

for all sufficiently small $u>t$, which contradicts the regularity of $t$.
Assuming

$\frac{dn_{1}(t)}{dt}-\frac{dn_{2}(t)}{dt}<0$

yields the same contradiction. Therefore, we must have

$\frac{dn_{1}(t)}{dt}-\frac{dn_{2}(t)}{dt}=0.$

Combining this with Equation (31), we obtain

$\frac{dn_{1}(t)}{dt}=\frac{dn_{2}(t)}{dt}=\frac{\lambda_{1}+\lambda_{2}-C_{1,2}}{2}.$

$\square$
∎

Case (c): $n_{1}(t)>n_{2}(t)=0$

Suppose that $\lambda_{2}<\underline{C}_{2}$. Then, using the same argument as
in Case (a) but now with only $N_{1}(\cdot)$ being infinitely backlogged and
$N_{2}(\cdot)$ being stable (Lemma 3), we conclude that

$\displaystyle\frac{dn_{1}(t)}{dt}$
$\displaystyle=\lambda_{1}-C_{1}(\lambda_{2})\qquad\text{and}\qquad\frac{dn_{2}(t)}{dt}=0.$

Suppose that $\lambda_{2}>\underline{C}_{2}$. Since $n_{2}(u)<n_{1}(u)$ for
all sufficiently small $u>t$, we have

$\displaystyle\frac{dr_{2}(t)}{dt}$ $\displaystyle\leq\underline{C}_{2}.$ (32)

Therefore, we have $n_{2}(u)>0$ for all sufficiently small $u>t$. By Case (a),
we have

$\displaystyle\frac{dn_{1}(u)}{du}$
$\displaystyle=\lambda_{1}-\overline{C}_{1}\qquad\text{and}\qquad\frac{dn_{2}(u)}{du}=\lambda_{2}-\underline{C}_{2}$

for all sufficiently small $u>t$. Using once more that $t$ is a regular time,
it follows that

$\displaystyle\frac{dn_{1}(t)}{dt}$
$\displaystyle=\lambda_{1}-\overline{C}_{1}\qquad\text{and}\qquad\frac{dn_{2}(t)}{dt}=\lambda_{2}-\underline{C}_{2}.$

Finally, suppose that $\lambda_{2}=\underline{C}_{2}$. Using the same argument
as in Lemma 10, it can be checked that

$\displaystyle\frac{dn_{2}(t)}{dt}$ $\displaystyle=0.$

Case (d): $n_{1}(t)=n_{2}(t)=0$

Since $t$ is a regular point, $n_{i}(0)>0$, and cases (a), (b), and (c) imply
that $n_{i}(\cdot)$ can only become $0$ at a non-regular point, we must have
$n_{1}(u)=n_{2}(u)=0$ for all sufficiently large $u<t$. It follows that

$\frac{dn_{1}(u)}{du}=\frac{dn_{2}(u)}{du}=0$

for all sufficiently large $u<t$. Using once again that $t$ is a regular point
yields

$\frac{dn_{1}(t)}{dt}=\frac{dn_{2}(t)}{dt}=0.$

$\square$
∎

###  F.3 Completing the proof of Theorem 3

For every sample path in $\mathcal{C}$, we have established the following.
Proposition 2 implies the existence of limit points of the sequence of
processes $\left\\{N^{(k)}(\cdot)\right\\}_{k=1}^{\infty}$. Furthermore,
according to Proposition 3 these limit points verify the differential
equations of the fluid model. Combining this with the fact that the
trajectories are Lipschitz continuous, and thus they are differentiable almost
everywhere, the limit points are fluid solutions. In particular, this means
that fluid solutions exist.

Finally, since the trajectories are piece-wise linear and any component that
hits zero stays at zero forever, the uniqueness of solutions follows from the
uniqueness of the pieces that comprise them. In particular, this implies that
all limit points are the same, and therefore the limit converges.

##  Appendix G Proof of Theorem 2

  * (i)

When $\lambda_{1}+\lambda_{2}<C_{1,2}$, $\lambda_{1}<C_{1}(\lambda_{2})$, and
$\lambda_{2}<C_{2}(\lambda_{1})$, theorems 3 and 4 imply that there exists
$\delta>0$ such that, for any $n^{0}\in\mathbb{R}_{+}^{2}$ with
$\|n^{0}\|_{1}>0$, if $N(0)=kn^{0}$, we have

$\lim\limits_{k\to\infty}\frac{N_{i}(kt)}{k}=0,\qquad a.s.,$

for all $t\geq\delta\|n^{0}\|_{1}$. Moreover, Equation (13) implies that, for
any $q^{0}\in\mathbb{R}_{+}^{3}$ with $\|q^{0}\|_{1}>0$, if $Q(0)=kq^{0}$, we
have

$\lim\limits_{k\to\infty}\frac{Q_{j}(kt)}{k}=0,\qquad a.s.,$

for all $t>0$. Therefore, the positive recurrence of $\big{(}{\bf
N}(\cdot),{\bf Q}(\cdot)\big{)}$ when $\lambda_{1}+\lambda_{2}<C_{1,2}$,
$\lambda_{1}<C_{1}(\lambda_{2})$, and $\lambda_{2}<C_{2}(\lambda_{1})$ follows
from these finite-time convergences to $0$, and [11, Theorem 6.2].

  * (ii)

We first consider a coupled process $\left(\underline{N}(\cdot),\underline{\bf
Q}(\cdot)\right)$, where the coupling with the original processes is done in a
way so that $\underline{N}(0)={\bf N}_{1}(0)+{\bf N}_{2}(0)$, $\underline{\bf
Q}_{1}(0)={\bf Q}_{1}(0)+{\bf Q}_{3}(0)$, and $\underline{\bf Q}_{2}(0)={\bf
Q}_{2}(0)$ and so that they have the same arrival processes $A_{1}(\cdot)$,
$A_{2}(\cdot)$, $S_{1}(\cdot)$, $S_{2}(\cdot)$, and $S_{3}(\cdot)$, and the
same abandonment primitives $Z_{\ell}^{(i)}(\cdot)$. These new processes are
defined recursively as

$\displaystyle\underline{N}(t+1)$
$\displaystyle=\underline{N}(t)+A_{1}(t)+A_{2}(t)-\sum\limits_{\ell=1}^{\overline{\overline{M}}(t)}Y_{\ell}(t)$
$\displaystyle\underline{Q}_{1}(t+1)$
$\displaystyle=\underline{Q}_{1}(t)-\underline{D}_{1}(t)+S_{1}(t)+S_{3}(t)-\overline{\overline{M}}(t)$
$\displaystyle\underline{Q}_{2}(t+1)$
$\displaystyle=\underline{Q}_{2}(t)-\underline{D}_{2}(t)+S_{2}(t)-\overline{\overline{M}}(t),$

where

$\overline{\overline{M}}(t)=\min\left\\{\underline{Q}_{1}(t)-\underline{D}_{1}(t)+S_{1}(t)+S_{3}(t),\,\,\underline{Q}_{2}(t)-\underline{D}_{2}(t)+S_{2}(t)\right\\}.$

This corresponds to a system where three-way matchings are always attempted if
there are enough qubits, regardless of the requests available. Then,
$\underline{N}(\cdot)$ keeps track of the difference between the total number
of requests that arrived to the system, and the total number of successful
three-way matchings. This can be negative, and it can be checked that
$\underline{N}(t)\leq{\bf N}_{1}(t)+{\bf N}_{2}(t)$, for all $t\geq 0$, for
any non-idling policy. Moreover, since $\underline{Q}_{1}(\cdot)$ and
$\underline{Q}_{2}(\cdot)$ behave as two-sided queues with abandonments, we
have

$\lim\limits_{t\to\infty}\frac{1}{t}\sum\limits_{s=0}^{t}\sum\limits_{\ell=1}^{\overline{\overline{M}}(s)}Y_{\ell}(s)=C_{1,2}.$

Therefore, since $\lambda_{1}+\lambda_{2}>C_{1,2}$, we have that
$\underline{N}(t)$ diverges to $+\infty$ almost surely and, since
$\underline{N}(t)\leq{\bf N}_{1}(t)+{\bf N}_{2}(t)$ for all $t\geq 0$, the
process $\big{(}{\bf N}(\cdot),{\bf Q}(\cdot)\big{)}$ is transient.

On the other hand, suppose that $\lambda_{1}+\lambda_{2}\leq C_{1,2}$, and
that $\lambda_{j}<\underline{C}_{j}$ and $\lambda_{i}>C_{i}(\lambda_{j})$.
Analogously to the previous case, we can construct a coupled process
$\underline{N}_{i}(\cdot)$ such that $N_{i}(t)\geq\underline{N}_{i}(t)$ for
all $t\geq 0$, and such that its throughput is $C_{i}(\lambda_{j})$. Then,
since $\lambda_{i}>C_{i}(\lambda_{j})$, we have that
$\underline{N}_{i}(\cdot)$ diverges to $+\infty$ almost surely and, since
$N_{i}(t)\geq\underline{N}_{i}(t)$ for all $t\geq 0$, the process $\big{(}{\bf
N}(\cdot),{\bf Q}(\cdot)\big{)}$ is transient.