as well as conditions imposing work conservation and non-preemptive priority:

(5.6) $\displaystyle
I:=1-\sum_{i=1}^{J}B_{i}=\Big{(}1-\sum_{i=1}^{J}X_{i}\Big{)}^{+},$ (5.7)
$\displaystyle
K_{i}(t)=\int_{[0,t]}1_{\\{\sum_{j=1}^{i-1}Q_{j,s}=0\\}}dK_{i}(s),\qquad i\geq
2,\,t\geq 0.$

Under the assumption of bounded reneging hazard rates, which is indeed
fulfilled when the reneging distribution is exponential, it was shown in [9,
Theorem 3.1] that uniqueness holds for solutions of the fluid equations for
any given data and initial conditions. Existence of solutions was also
established there by showing that the scaling limit of the underlying queueing
system is a solution.

By the same argument given in the proof of Theorem 2.6, it follows from the
results in Theorem 4.1 of [25] that the measure-valued age equation (5.1)
implies that for every $\psi\in{\mathcal{C}}_{b}([0,\infty))$ or $\psi=h$,

(5.8)
$\langle\psi,\nu_{i,t}\rangle=\int_{[0,\infty)}\frac{\bar{G}(x+t)}{\bar{G}(x)}\psi(x+t)\nu_{i,0}(dx)+\int_{[0,t]}\bar{G}(s-x)\psi(s-x)dK_{i}(s),$

where recall $\bar{G}=1-G$. In what follows, given a vector or vector-valued
process $Y$, we use $\tilde{Y}$ to be generic notation for the sum
$\sum_{i=1}^{J}Y_{i}$. By (5.8), $\tilde{\nu}$ and $\tilde{K}$ satisfy, for
every $\psi\in{\mathcal{C}}_{b}([0,\infty))$ or $\psi=h$,

(5.9)
$\langle\psi,\tilde{\nu}_{t}\rangle=\int_{[0,\infty)}\frac{\bar{G}(x+t)}{\bar{G}(x)}\psi(x+t)\tilde{\nu}_{0}(dx)+\int_{[0,t]}\bar{G}(t-s)\psi(t-s)d\tilde{K}(s).$

In other words, (2.19) holds with $(\nu,K)$ and $G^{s}$ replaced with
$(\tilde{\nu},\tilde{K})$ and $G$. We now argue that, $\tilde{K}$ and
$\tilde{B}$ satisfy the analog of (2.20). First, note that by (5.2),
$\tilde{B}=\langle 1,\tilde{\nu}\rangle$, and if $\tilde{B}_{t}<1$ then, on an
open interval containing $t$ we have $\tilde{X}<1$ due to (5.6). Hence,
$\tilde{Q}=0$ by (5.5) and $\tilde{R}=0$ by (5.2). Hence, subtracting (5.4)
from (5.3), $\tilde{K}=\tilde{E}+c$ on this interval (where $c$ does not
depend on time), and so $\tilde{K}^{\prime}(s)=\tilde{\lambda}$ holds on the
interval. Combining this with

$\tilde{K}(t)=\tilde{B}(t)-\tilde{B}(0)+\int_{0}^{t}\langle
h,\tilde{\nu}_{s}\rangle ds,$

which follows from (5.3) and (5.2), we obtain, exactly as in [9, Theorem 3.2],
that for a.e. $t$, $\tilde{K}^{\prime}(t)=\tilde{k}(t)$ where

(5.10) $\tilde{k}(t)=\begin{cases}\langle
h,\tilde{\nu}_{t}\rangle,&\tilde{B}(t)=1,\\\
\tilde{\lambda},&\tilde{B}(t)<1.\end{cases}$

###  5.2. Results for the Multiclass System

We will be interested in the supercritical case where $\sum_{i}\lambda_{i}>1$
and $\theta_{\min}=\min_{i}\theta_{i}>0$. Let $\rho_{i}$, $i=1,\ldots,J$ be
characterized by

$\sum_{i=1}^{j}\rho_{i}=\Big{(}\sum_{i=1}^{j}\lambda_{i}\Big{)}\wedge 1,\qquad
j=1,\ldots,J,$

and let

$q_{i}=\frac{\lambda_{i}-\rho_{i}}{\theta_{i}},\qquad i=1,\ldots,J.$

We now state the main result.

######  Theorem 5.2.

Suppose that $h$ satisfies Assumption 3.1, and
$\lambda,\theta\in(0,\infty)^{J}$ are such that
$\tilde{\lambda}=\sum_{i=1}^{J}\lambda_{i}>1$, and
$(X_{0},\nu_{0})\in{\mathbb{R}}_{+}^{J}\times({\mathcal{M}}_{F}[0,\infty))^{J}$
satisfies $1-\langle{\bf 1},\tilde{\nu}\rangle=(1-\tilde{X})^{+}$. Then any
solution $(B,X,Q,D,K,R,\nu)$ to the multiclass fluid equations with initial
condition $(X_{0},\nu_{0})$ and arrival and reneging rate vectors $\lambda$
and $\theta$ satisfies $\nu_{i,t}\Rightarrow\rho_{i}\nu^{*}$ and $Q_{i}(t)\to
q_{i}$ as $t\to\infty$ for $i=1,\ldots,J$.

######  Remark 5.3.

This validates Theorem 5.1 of [9] in the special case where for all $i$,
$h^{s}_{i}=h$, with $h$ satisfying Assumption 3.1.

######  Remark 5.4.

The characterizations in (5.8) and (5.10) show that the aggregate processes
$(\tilde{X},\tilde{\nu})$ and
$(\tilde{D},\tilde{K},\tilde{R},\tilde{S},\tilde{Q},\tilde{B})$ satisfy the
fluid equations of the single class case (see Definition 2.3), subject to the
simplification described at the beginning of Section 5.1, where in particular
reneging is given directly by (5.2) and the process $\eta$ is not used. Hence,
in the supercritical setting $\tilde{\lambda}>1$, we may conclude from Theorem
3.2(2) that, with $\nu^{*}(dx)=\bar{G}(x)dx$, one has
$\tilde{\nu}_{t}\Rightarrow\nu^{*}$ and that there exists $T<\infty$ such that
$\tilde{B}_{t}=1$ for all $t\geq T$. Moreover, by Proposition 4.8, $\langle
h,\tilde{\nu}_{t}\rangle\to 1$. As a result, by (2.20), one has
$\tilde{k}_{t}=\langle h,\tilde{\nu}_{t}\rangle$ for all large $t$, and hence
also $\tilde{k}_{t}\to 1$.

###### Proof of Theorem 5.2.

In this proof, the special case in which there exists
$i_{0}\in\\{1,\ldots,J-1\\}$ such that $\sum_{i=1}^{i_{0}}\lambda_{i}=1$ is
called the borderline case, and the more typical case, where such $i_{0}$ does
not exist, is called the typical case.

If $\lambda_{1}<1$, set $\ell:=\max\\{j:\sum_{i=1}^{j}\lambda_{i}<1\\}$,
otherwise let $\ell=0$. Also, set $m=\ell+1$. Then, since by assumption
$\tilde{\lambda}>1$, by the definition of $m$, we have
$\sum_{i=1}^{m}\lambda_{i}=1$ [respectively, $>1$] in the borderline case
[respectively, in the typical case]. Also, in what follows, we use the hat
(when $\ell\geq 1$) and $\\#$ notation for summation up to $l$ and,
respectively, $m$, as in

(5.11) $\hat{Y}=\sum_{i=1}^{\ell}Y_{i},\quad\mbox{ and }\quad
Y^{\\#}=\sum_{i=1}^{m}Y_{i},\qquad Y=\lambda,X,\nu,D,K,R,B.$

(in addition to the notation already introduced,
$\tilde{Y}=\sum_{i=1}^{J}Y_{i}$).

The structure of the proof is as follows. In Step 1 we prove the assertions
for $i\leq\ell$. Steps 2 and 3 address the remaining classes $i\geq m$ in the
typical and borderline cases, respectively. First, note that since
$\tilde{\lambda}>1$, by Remark 5.4, there exists $T<\infty$ such that

(5.12) $\tilde{k}(t)\to 1,\langle h,\tilde{\nu}_{t}\rangle\to 1\mbox{ as
}t\rightarrow\infty\quad\mbox{ and }\quad\tilde{B}(t)=1\mbox{ for all }t\geq
T.$

Step 1. Consider the case $\ell\geq 1$ (that is, $\lambda_{1}<1$). In this
step we consider classes $1\leq i\leq\ell$ and establish the claim that there
exists $t_{1}<\infty$ such that $Q_{i}(t)=0$ for all $t\geq t_{1}$, and
moreover, that $\nu_{i}(t)\to\rho_{i}\nu^{*}$ as $t\rightarrow\infty$. (Note
that for $i\leq\ell$, $\lambda_{i}=\rho_{i}$, hence the asserted convergence
$Q_{i}(t)\to q_{i}=0$ would then follow).

Recalling the notational convention (5.11), by the definition of $\ell$,
$\hat{\lambda}=\sum_{i=1}^{\ell}\lambda_{i}<1$, and so there exist
$\varepsilon_{0}>0$ and $0<t_{0}<\infty$ such that $\langle
h,\tilde{\nu}_{t}\rangle>\hat{\lambda}+\varepsilon_{0}$ for all $t\geq t_{0}$.
If $\hat{Q}(t)=0$ for all $t\geq t_{0}$ then the claim follows trivially. So,
we now consider the converse case, when
${\mathcal{O}}:=\\{t>t_{0}:\hat{Q}(t)>0\\}$ is non-empty. Since $\hat{Q}$ is
continuous, ${\mathcal{O}}$ is open and is a union of countable open
intervals. For a.e. $s$ in each such interval, by (5.7), for all $i>\ell$,
$K_{i}^{\prime}(s)=0$. Moreover, since (5.5) and (5.6) together show that,
$\tilde{Q}(t)>0$ implies $\tilde{B}(t)=1$ for any $t>0$, we conclude in
particular that $\tilde{B}(s)=1$. In turn, (5.3), (5.5), (5.2) and the fact
that $\tilde{R}$ is non-decreasing together imply that for a.e.
$s\in{\mathcal{O}}$, $\tilde{D}^{\prime}(s)=\tilde{K}^{\prime}(s)=\langle
h,\tilde{\nu}_{s}\rangle$. Thus, we have for a.e. $s>t_{0}$,

$\displaystyle\hat{Q}(s)>0\quad\Rightarrow\quad\hat{Q}^{\prime}(s)=\hat{\lambda}-\hat{R}^{\prime}(s)-\hat{K}^{\prime}(s)\leq\hat{\lambda}-\langle
h,\tilde{\nu}_{s}\rangle\leq-\varepsilon_{0}.$

Thus, there must exist a finite time, $t_{1}\geq t_{0}$, when
$\hat{Q}(t_{1})=0$. Since the last display continues to hold for all $s\geq
t_{1}$, applying Lemma B.1 with $f=-Q$, $T=t_{1}$, $S=\infty$ and $c=0$, it
follows that for all $t\geq t_{1}$, $\hat{Q}(t)=0$, or equivalently,
$Q_{i}(t)=0$ for all $i\leq\ell$.

To finish proving the claim in Step 1, it only remains to show that
$\nu_{i}(t)\to\rho_{i}\nu^{*}=\lambda_{i}\nu^{*}$ for $i\leq\ell$. For $t\geq
t_{1}$, it follows from (5.5) and (5.2), respectively, that for $i\leq\ell$,
$X_{i}(t)=B_{i}(t)$ and $R_{i}^{\prime}(t)=0$. Hence by (5.3)–(5.4),
$K_{i}^{\prime}(t)=\lambda_{i}$ for such $i$ and $t$. Substituting these
relations in (5.8) and taking the large $t$ limit yields (exactly as in the
proof of Lemma 4.1), the convergence of $\nu_{i}(dx)$ to
$\lambda_{i}\bar{G}(x)dx$ as asserted.

Step 2. In this step we treat the typical case, proving the claim for all the
remaining classes $i\geq m=\ell+1$ (where possibly $\ell=0$, $m=1$). Recall
the notation in (5.11) and note that in this case one has
$\lambda^{\\#}=\sum_{i=1}^{m}\lambda_{i}>1$. By the definition of $m$ and
$\rho_{i}$, this implies $\rho_{m}<\lambda_{m}$.

Let $t_{1}<\infty$ be as in Step 1, and assume without loss of generality that
$t_{1}\geq T$, where $T$ is as in (5.12). Then given
$\varepsilon\in(0,1-\lambda^{\\#})$, there exists
$t_{2}=t_{2}(\varepsilon)\geq t_{1}$ such that for all $t\geq t_{2}$,
$|\langle h,\tilde{\nu}_{t}\rangle-1|<\varepsilon,$
$|\tilde{k}(t)-1|<\varepsilon$ and $\tilde{B}(t)=1$. Then (5.2) implies that
$\tilde{D}_{t}^{\prime}(t)=\langle
h,\tilde{\nu}_{t}\rangle\leq(1+\varepsilon)$, and since clearly,
$(D^{\\#})^{\prime}\leq\tilde{D}^{\prime}$, on $[t_{2},\infty)$, we have for
all $\varepsilon_{1}\leq\lambda^{\\#}-1-\varepsilon$,

$\frac{dX^{\\#}}{dt}=\lambda^{\\#}-\frac{dD^{\\#}}{dt}-\frac{dR^{\\#}}{dt}\geq\lambda^{\\#}-(1+\varepsilon)-\theta_{m}Q_{m}\geq\varepsilon_{1}-\theta_{m}Q_{m}.$

We now argue by contradiction to prove the claim that there exists $t_{3}\geq
t_{2}$ such that $Q_{m}(t_{3})>0$. Indeed, assume $Q_{m}$ vanishes on the
whole interval $[t_{2},\infty)$. Then the last display shows that
$X^{\\#}(t)\to\infty$, and hence by (5.5) and the fact that (5.6) implies
$\tilde{B}$ lies in $[0,1]$, $Q^{\\#}(t)\to\infty$. But since $\hat{Q}$
vanishes on $(t_{1},\infty)\supset(t_{2},\infty)$ by Step 1, this implies
$Q_{m}(t)=Q^{\\#}(t)-\hat{Q}(t)=Q^{\\#}(t)\to\infty$, which contradicts the
assumption that $Q_{m}$ is identically zero on $[t_{2},\infty)$. This proves
the claim.

Let $t_{3}\geq t_{2}$ be such that $Q_{m}(t_{3})>0$, and let
${\mathcal{O}}_{m}:=\\{s\in[t_{3},\infty):Q_{m}(s)>0\\}$. We show below that
${\mathcal{O}}_{m}=[t_{3},\infty)$. Towards this goal, we will find it more
convenient to work with the balance equation for $Q^{\\#}$ than with
$X^{\\#}$. That is, using (5.3)–(5.5) and (5.2), note that

(5.13) $Q_{i}^{\prime}=\lambda_{i}-K_{i}^{\prime}-\theta_{i}Q_{i},\quad
i=1,\ldots,J,\qquad\mbox{ and
}\qquad\frac{dQ^{\\#}}{dt}=\lambda^{\\#}-\frac{dK^{\\#}}{dt}-\sum_{i=1}^{m}\theta_{i}Q_{i}^{\\#}.$

On any open interval in ${\mathcal{O}}_{m}$, $Q_{m}>0$ and $\hat{Q}=0$, and
hence the priority rule (5.7) implies $dK^{\\#}/dt=d\tilde{K}/dt=\tilde{k}$,
where recall $|\tilde{k}(t)-1|<\varepsilon$. Thus, for all $t\geq t_{3}$, we
have

$Q_{m}(t)>0\quad\Rightarrow\quad
Q_{m}^{\prime}(t)=\frac{dQ^{\\#}}{dt}(t)\geq\lambda^{\\#}-(1+\varepsilon)-\frac{dR^{\\#}}{dt}(t)\geq\varepsilon_{1}-\theta_{m}Q_{m}(t).$

Since this is strictly greater than $\varepsilon_{1}/2$ whenever
$Q_{m}(t)<\varepsilon_{1}/2\theta_{m}$, this clearly implies $Q_{m}(t)>0$ for
all $t\in[t_{3},\infty)$, as claimed. In turn, by the priority rule (5.7),
this implies that on $[t_{3},\infty)$, $K_{i}^{\prime}=0$ for all $i>m$, and
therefore by (5.3) and (5.2), $B_{i}^{\prime}=-\langle
h,\nu_{i}\rangle\leq-\varepsilon_{h}B_{i}$, where recall that
$\varepsilon_{h}$ is the strictly positive lower bound on $h$. This shows that
for $i>m$, $B_{i}(t)\rightarrow 0$ as $t\rightarrow\infty$ and hence,
$\nu_{i}\Rightarrow 0$. As $t\to\infty$, since we already have convergence of
the aggregate $\tilde{\nu}_{t}\Rightarrow\nu^{*}$ (see Remark 5.4) and
$\nu_{i,t}\Rightarrow\lambda_{i}\nu^{*}$ for all $i<m$ (by Step 1), we
conclude that $\nu_{m,t}\Rightarrow\rho_{m}\nu^{*}$.

To complete Step 2, it only remains to address the convergence of $Q_{i}$,
$i\geq m$. Since, as argued above, for $t\in[t_{3},\infty)$,
$K^{\prime}_{i}(t)=0$ for $i>m$, (5.13) shows that
$Q_{i}(t)\to\lambda_{i}/\theta_{i}=q_{i}$ as $t\to\infty$. As for $Q_{m}$,
note that since on $[t_{3},\infty)$, for $1\leq i\leq\ell=m-1$, $Q_{i}=0$ by
Step 1, (5.13) shows that $K_{i}^{\prime}=\lambda_{i}$, or equivalently,
$\hat{K}^{\prime}=\hat{\lambda}$. Thus, denoting $e(t):=\tilde{k}(t)-1$, we
have $e(t)\to 0$, and recalling that $\rho_{m}=\hat{\lambda}-1$,

$K_{m}^{\prime}(t)=\tilde{K}^{\prime}(t)-\hat{K}^{\prime}_{i}(t)=\tilde{k}(t)-\hat{\lambda}=(\rho_{m}+e(t)).$

Thus, we obtain

$Q_{m}^{\prime}(t)=(\lambda_{m}-\rho_{m}-e(t))-\theta_{m}Q_{m}(t).$

This implies that as $t\rightarrow\infty$, $Q_{m}(t)$ converges to
$q_{m}=(\lambda_{m}-\rho_{m})/\theta_{m}$. Here, we used the elementary fact
that for a differentiable function $u$ on $[0,\infty)$,

(5.14) $u^{\prime}(t)=w(t)-\theta u(t)\mbox{ and
}u(0)=u_{0}\qquad\Rightarrow\qquad
u(t)=\int_{0}^{t}e^{-\theta(t-s)}w(s)ds+u_{0}e^{-\theta t},$

which converges to $c/\theta$ whenever $w(t)\to c$ as $t\to\infty$.

Step 3. Lastly, we consider the borderline case, and establish the assertions
regarding the remaining classes $i\in\\{m,\ldots,J\\}$. In this case
$\lambda^{\\#}=\sum_{i=1}^{m}\lambda_{i}=1$.

As in (5.10), the priority structure specified by (5.7) dictates that
$dK^{\\#}/dt=k^{\\#}$, where $k^{\\#}(t)$ is given by $\langle
h,\tilde{\nu}_{t}\rangle$ when $B^{\\#}(t)=1$ and equal to $\lambda^{\\#}$
when $B^{\\#}(t)<1$. Since $\lambda^{\\#}=1$ and by (5.12), $\langle
h,\tilde{\nu}_{t}\rangle\to 1$ we infer that $k^{\\#}(t)\to 1$ as
$t\to\infty$. Summing (5.8) over $i\leq m$, using
$\int_{0}^{\infty}\bar{G}(x)dx=1$ and applying the test function $\psi={\bf
1}$ shows that $B^{\\#}_{t}\to 1$ as $t\to\infty$ (where the application of
bounded continuous $\psi$ can be justified in the usual manner). Applying
general compactly supported test functions gives
$\nu^{\\#}_{t}\Rightarrow\nu^{*}$, where $\nu^{*}(dx)=\bar{G}(x)dx$. Given the
convergence already established for $\nu_{i,t}$, $i\leq\ell=m-1$, the
convergence of $\nu^{\\#}_{t}$ yields that of $\nu_{m,t}\to\lambda_{m}\nu^{*}$
(note that in the borderline case currently considered,
$\lambda_{m}=\rho_{m}$). Moreover, the fact that $B^{\\#}(t)\to 1$ implies
that $\sum_{i=m+1}^{J}B_{i}(t)=\tilde{B}(t)-B^{\\#}(t)\to 0$, and hence, for
all $i>m$, $B_{i}(t)\to 0$ and consequently $\nu_{i,t}\Rightarrow 0$.

Next we show that $Q_{i}(t)\to q_{i}$ for $i>m$, for which we again use
(5.13). Combining the convergence $k^{\\#}(t)\to 1$ that we just showed with
$\tilde{k}(t)\to 1$ from (5.12), it follows that $k_{i}(t)\to 0$ for all
$i>m$. Recalling that $K_{i}^{\prime}=k_{i}$ and using the first equation in
(5.13) and (5.14) yields $Q_{i}(t)\to\lambda_{i}/\theta_{i}=q_{i}$, for $i>m$.

We finally show that $Q_{m}(t)\to 0$. To this end, note that by the aggregate
equation in (5.13) and the property that for sufficiently large $t$,
$Q_{i}(t)=0$, for $i\leq\ell$, (from Step 1) giving
$dR^{\\#}(t)/dt=\theta_{m}Q_{m}(t)=\theta_{m}Q^{\\#}(t)$. Since
$\lambda^{\\#}=1$, (5.13) shows that the following is valid for all large $t$,

$\frac{dQ^{\\#}}{dt}=1-k^{\\#}(t)-\theta_{m}Q^{\\#}(t).$

Recalling that $k^{\\#}(t)\to 1$, and again using (5.14), it follows that
$Q^{\\#}(t)\to 0$. Consequently, $Q_{m}(t)\to 0$. This completes the proof. ∎

##  Appendix A Proof of Lemma 4.5.

###### Proof of Lemma 4.5.

Fix a measurable function $f:[0,\infty)\mapsto{\mathbb{R}}_{+}$ with
$\int_{0}^{\infty}fdx\leq 1$. For notational conciseness, define

$Af:=\int_{0}^{\infty}h^{s}(x)f(x)\log\frac{f(x)}{f^{*}(x)}dx-z_{f}\log
z_{f},$

where recall $z_{f}:=\int_{0}^{\infty}h^{s}fdx<\infty$. Let $U(x):=x\log x$,
$x>0$, $U(0)=0$. Fix a non-negative measurable function $\psi$ on $[0,\infty)$
with $c_{\psi}=\int_{0}^{\infty}\psi dx\leq 1$. Then, note from the definition
of $A$ in (4.8) that

(A.1)
$A(\psi)=\int_{0}^{\infty}U\left(\frac{\psi}{f^{*}}\right)h^{s}f^{*}dx-U\left(z_{\psi}\right).$

Since $\int_{0}^{\infty}h^{s}f^{*}dx=1$ and
$\int_{0}^{\infty}\left(\frac{\psi}{f^{*}}\right)h^{s}f^{*}dx=z_{\psi}$, the
convexity of $U$ and Jensen’s inequality imply the nonnegativity of $A(\psi)$.
To obtain the more refined estimate (4.8), define

$V(x):=U(x)-[U^{\prime}(z_{f})(x-z_{f})+U(z_{f})].$

Then, the strict convexity of $U$ implies $V(x)\geq 0$ and $V(x)=0$ if and
only if $x=z_{f}$. Using (A.1) we have

$A(\psi)=\int_{0}^{\infty}V\Big{(}\frac{\psi}{f^{*}}\Big{)}h^{s}f^{*}dx+\int_{0}^{\infty}U^{\prime}(z_{f})\Big{(}\frac{\psi}{f^{*}}-z_{f}\Big{)}\Big{]}h^{s}f^{*}dx=\int_{0}^{\infty}V\Big{(}\frac{\psi}{f^{*}}\Big{)}h^{s}f^{*}dx,$

where the last equality uses the definition of $z_{\psi}$. Since $V\geq 0$,
denoting $c_{\psi}:=\int_{0}^{\infty}\psi dx\leq 1$, and recalling the
functional $R$ from (4.5), we have

$\displaystyle A(\psi)$
$\displaystyle\geq\varepsilon_{h}\int_{0}^{\infty}V\Big{(}\frac{\psi}{f^{*}}\Big{)}f^{*}dx=\varepsilon_{h}\Big{[}\int_{0}^{\infty}\psi\log\frac{\psi}{f^{*}}dx-\int_{0}^{\infty}U^{\prime}(z_{\psi})\Big{(}\frac{\psi}{f^{*}}-z_{\psi}\Big{)}f^{*}dx-U(z_{\psi})\Big{]}$
$\displaystyle=\varepsilon_{h}[R(\mu^{\psi}\|\nu_{*})-c_{\psi}U^{\prime}(z_{\psi})+U^{\prime}(z_{\psi})z_{\psi}-U(z_{\psi})]$
$\displaystyle=\varepsilon_{h}[R(\mu^{\psi}\|\nu_{*})-c_{\psi}\log
z_{\psi}-c_{\psi}+z_{\psi}]$
$\displaystyle=\varepsilon_{h}\\{R(\mu^{\psi}\|\nu_{*})+c_{\psi}[-\log
z_{\psi}-1+z_{\psi}]+z_{\psi}(1-c_{\psi})\\}$
$\displaystyle\geq\varepsilon_{h}R(\mu^{\psi}\|\nu_{*}),$

where the third equality used the fact that $U^{\prime}(x)=\log x+1$ and
$U^{\prime}(x)x-U(x)=x$, and the last inequality uses the elementary
inequality $x-\log x\geq 1$ for all $x>0$. This proves (4.8). ∎

##  Appendix B An Elementary Property of Absolutely Continuous Functions

The following simple property is used in Section 4.3.

######  Lemma B.1.

Let $f$ be an absolutely continuous function defined on $[0,S)$ for some
$0<S\leq\infty$. Suppose that there exist a time $T\in(0,S)$ and a constant
$c>0$ such that $f(T)\geq c$ and for a.e. $t\in(T,S)$, $f^{\prime}(t)\geq 0$
if $f(t)<c$. Then $f(t)\geq c$ for all $t\in[T,S)$.

###### Proof.

Suppose the conclusion of the lemma does not hold. Then there must exist
$T<t_{1}<t_{2}$ for which $f(t_{1})\geq c$ and $f(t_{2})<c$. Since $f$ is
absolutely continuous, there must exist some interval
$(s_{1},s_{2})\subset(t_{1},t_{2})$ such that $f(s)<c$ for $s\in(s_{1},s_{2})$
and $f^{\prime}(s)<0$ for $s$ in a subset ${\mathcal{S}}\subset(s_{1},s_{2})$
of positive Lebesgue measure. However, this contradicts the assumption of the
lemma, that is, for a.e. $t\in[T,\infty)$, $f^{\prime}(t)\geq 0$ if $f(t)<c$.
Hence the lemma is proved. ∎

Acknowledgement. We would like to thank Amber Puha for raising the question of
a full justification of the convergence results in Theorem 3.3 of [24].

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$\langle\textrm{out}|H_{I}(\infty)|m_{j}^{2},\vec{p},J,J_{z};\infty\rangle=d^{(J)}_{hh^{\prime}}(\theta)(|k_{34}|)^{J}\mathcal{C}_{J,h_{34}}~{}.$
(B.26)

Then in $3+1$ dimensions, and when all the external particles are massless,
the scattering amplitude in (B.12) takes the following form

$\widetilde{\mathcal{N}}_{J;h,h^{\prime}}\frac{1}{s-m_{J}^{2}+{\bf
i}\varepsilon}\frac{s^{J}}{2}\,d^{(J)}_{hh^{\prime}}(\theta)$ (B.27)

where $\widetilde{\mathcal{N}}_{J;h,h^{\prime}}$ is

$\widetilde{\mathcal{N}}_{J;h,h^{\prime}}=\mathcal{C}_{J,s}\,\mathcal{C}_{J,h^{\prime}}\qquad\text{and}\qquad\mathcal{C}_{J,h}=2^{\frac{J}{2}-h}\sqrt{\frac{\Gamma(J-h+1)\,\Gamma(J+h+1)}{\Gamma(2J+1)}}~{}.$
(B.28)

##  Appendix C Orthogonal polynomials

Orthogonal polynomials denote a family of polynomials such that any two
distinct members of the family are orthogonal to each other with respect to
the inner product. We discuss some key features of Gegenbauer polynomials,
Jacobi polynomials, and Wigner matrices for our purposes.

###  C.1 Short summary on Gegenbauer polynomials

Gegenbauer polynomial is the higher dimensional analogue of Legendre
polynomial. Gegenbauer polynomial appears in the multipole expansion of
newtonian potential in higher dimensional spacetime

$\displaystyle\frac{1}{(1-2\,z\,t+t^{2})^{\beta}}=\sum_{n=0}^{\infty}\mathcal{G}^{(\beta)}_{n}(z)\,t^{n}~{}.$
(C.1)

In standard literature, the variable in the superscript is usually denoted by
the letter $\alpha$. We choose to use $\beta$ to avoid confusion with string
length squared $\alpha^{\prime}$. This reduces to Legendre polynomial for
$\beta=1/2$. For $D$ spacetime dimension131313In 3 space-dimension, the
multipole expansion of coulomb potential gives Legendre polynomial. In $D-1$
space dimensions, if we do analogous expansion of Columb potential then we
obtain Gegenbauer polynomial.

$\displaystyle\beta=\frac{(D-3)}{2}$ (C.2)

An explicit expression of Gegenbauer polynomial is

$\mathcal{G}^{(\beta)}_{n}(z)=\sum_{a=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^{a}\frac{\Gamma(n-a+\beta)}{a!\Gamma(\beta)(n-2a)!}(2z)^{n-2a}~{}.$
(C.3)

Another useful formula for Gegenbauer polynomial is the Rodrigues formula

$\mathcal{G}^{(\beta)}_{n}(z)=\frac{1}{2^{n}n!}\frac{\Gamma(\beta+\frac{1}{2})\Gamma(n+2\beta)}{\Gamma(2\beta)\Gamma(\beta+n+\frac{1}{2})}(z^{2}-1)^{-\beta+\frac{1}{2}}\frac{d^{n}}{dz^{n}}\Big{[}(z^{2}-1)^{n+\beta-\frac{1}{2}}\Big{]}~{}.$
(C.4)

Gegenbauer polynomial (and hence Legendre polynomial) satisfies the following
property

$\mathcal{G}^{(\beta)}_{n}(-z)=(-1)^{n}\,\mathcal{G}^{(\beta)}_{n}(z)~{}.$
(C.5)

##### Connection to the four scalar amplitude

The numerator of the four scalar amplitude is given by

$\sum_{a=0}^{\lfloor\frac{J}{2}\rfloor}~{}A(J,a,D)\Big{[}(k_{12}\cdot
k_{12})(k_{34}\cdot k_{34})\Big{]}^{a}(k_{12}\cdot k_{34})^{J-2a}$ (C.6)

here $A(J,a,D)$ is given by

$A(J,a,D)=\Bigg{[}\frac{(-1)^{a}J!(2J+D-2a-5)!!}{2^{a}a!(J-2a)!(2J+D-5)!!}\Bigg{]}~{}.$
(C.7)

Comparing this with eqn (C.3) we can see that it is same as

$s^{J}\frac{\Gamma(J+1)\,\Gamma\left(\beta\right)}{2^{J}\Gamma\left(\beta+J\right)}\mathcal{G}^{(\beta)}_{J}(\hat{k}_{12}\cdot\hat{k}_{34})\qquad\text{where}\qquad\beta=\frac{D-3}{2}~{}.$
(C.8)

In order to find normalization consider $a=0$

$A(J,0,D)=\Bigg{[}\frac{J!(2J+D-5)!!}{J!(2J+D-5)!!}\Bigg{]}=1$ (C.9)

whereas the coefficient of the $a=0$ term in eqn (C.3) is

$2^{J}\frac{\Gamma(J+\beta)}{\Gamma(\beta)J!}=2^{J}\frac{\Gamma(J+\beta)}{\Gamma(\beta)\,\Gamma(J+1)}~{}.$
(C.10)

This determines the normalization; the factor of $s^{J}$ can be fixed by
dimensional analysis.

##### Vector derivative of Gegenbauer polynomial

Let’s consider two space-like unit vector $\widehat{n}$ and $\widehat{q}$ such
that

$\widehat{n}\cdot\widehat{q}=\cos\theta=z~{}.$ (C.11)

Consider the Gegenbauer polynomial of the form

$\mathcal{G}^{(\beta)}_{\ell}(\widehat{n}\cdot\widehat{q})\equiv\mathcal{G}^{(\beta)}_{\ell}(z)~{}.$
(C.12)

Then we can show that

$\widehat{r}\cdot\frac{\partial}{\partial{\widehat{n}}}\Big{[}\mathcal{G}^{(\beta)}_{\ell}(z)\Big{]}=(\hat{r}\cdot\hat{q})\mathcal{G}^{(\beta)\prime}_{\ell}(z)-(\hat{r}\cdot\hat{n})\mathcal{G}^{(\beta)\prime}_{\ell-1}(z)$
(C.13)

where $\hat{r}$ is some other unit vector and ′ denotes derivative of
Gegenbauer polynomial w.r.t $z$.

###  C.2 Jacobi polynomial

Derivatives of the Jacobi polynomial $\mathcal{J}^{(\alpha,\beta)}_{n}(z)$
satisfy the following relations

$\frac{d^{k}}{dz^{k}}\mathcal{J}^{(\alpha,\beta)}_{n}(z)=\frac{1}{2^{k}}\frac{\Gamma(\alpha+\beta+n+k+1)}{\Gamma(\alpha+\beta+n+1)}\mathcal{J}^{(\alpha+k,\beta+k)}_{n-k}(z)~{}.$
(C.14)

Jacobi polynomial also obeys the following recursion relation

$(z-1)\frac{d}{dz}\mathcal{J}^{(\alpha,\beta)}_{n}(z)=(\alpha+n)\mathcal{J}^{(\alpha-1,\beta+1)}_{n}(z)-\alpha\,\mathcal{J}^{(\alpha,\beta)}_{n}(z)~{}.$
(C.15)

Legendre polynomial is simply one particular Jacobi polynomial

$P_{J}(z)\equiv\mathcal{J}^{(0,0)}_{J}(z)~{}.$ (C.16)

Then from this equation and eqn (C.17) it follows

$\frac{d^{k}}{dz^{k}}P_{J}(z)=\frac{1}{2^{k}}\frac{\Gamma(J+k+1)}{\Gamma(J+1)}\mathcal{J}^{(k,\,k)}_{J-k}(z)~{}.$
(C.17)

We use this equation many times to show the angular distributions. More
generally Gegenbauer polynomial is a special case of Jacobi polynomial

$\mathcal{G}^{(\beta)}_{n}(z)=\frac{(2\alpha)_{n}}{(\alpha+1/2)_{n}}\mathcal{J}^{(\beta-1/2,\,\beta-1/2)}_{n}(z)$
(C.18)

where $(a)_{n}$ is Pochhammer function. Jacobi polynomials satisfy the
following symmetry property

$\displaystyle\mathcal{J}^{(\alpha,\beta)}_{J}(-z)=(-1)^{J}\mathcal{J}^{(\beta,\alpha)}_{J}(z)~{}.$
(C.19)

###  C.3 Wigner matrix

Wigner (small-)$d$ matrix is defined as

$\displaystyle d_{h^{\prime}h}^{(j)}(\beta)$ $\displaystyle=$
$\displaystyle\left\langle j,h^{\prime}\left|\exp\left[-\frac{{\bf
i}J_{y}}{\hbar}\beta\right]\right|j,h\right\rangle~{}.$ (C.20)

It is quantum rotation matrix in $x$-$z$ plane; it rotates the states
belonging to $2j+1$ dimensional irreducible representation by angle $\theta$.
A few important property of the Wigner matrices are the following

$\displaystyle d_{h^{\prime},h}^{(j)}(\pi+\theta)$
$\displaystyle=(-1)^{j-h}d_{h^{\prime},-h}^{(j)}(\theta)\quad\text{and}$
(C.21) $\displaystyle d_{h^{\prime},h}^{(j)}(\theta)$
$\displaystyle=(-1)^{h-h^{\prime}}d_{h,h^{\prime}}^{(j)}(\theta)=d_{-h,-h^{\prime}}^{(j)}(\theta)~{}.$
(C.22)

For the purpose of this paper, the following mathematical identities are
useful

  1. 1.

Wigner matrix with one argument being zero can be written as derivative of
Legendre polynomial

$d^{(j)}_{h0}(\theta)=\sqrt{\frac{(j-h)!}{(j+h)!}}(-1)^{h}(1-z^{2})^{h/2}\frac{d^{h}}{dx^{h}}P_{j}(z)~{}.$
(C.23)

  2. 2.

Wigner matrices can written in terms of Jacobi polynomial in the following way

$d^{(j)}_{h,h^{\prime}}(\theta)=\sqrt{\frac{(j-h)!(j+h)!}{(j-h^{\prime})!(j+h^{\prime})!}}\,\Big{(}\cos\frac{\theta}{2}\,\Big{)}^{h+h^{\prime}}\Big{(}\sin\frac{\theta}{2}\,\Big{)}^{h-h^{\prime}}\mathcal{J}^{(h-h^{\prime},h+h^{\prime})}_{j-h}(\cos\theta)~{}.$
(C.24)

We list a few cases which will be relevant for us (here $z=\cos\theta$)

    1. (a)

$d^{(j)}_{1,1}(\theta)=\Big{(}\cos\frac{\theta}{2}\,\Big{)}^{2}\mathcal{J}^{(0,2)}_{j-1}(\cos\theta)=\bigg{[}\frac{z+1}{2}\bigg{]}\mathcal{J}^{(0,2)}_{j-1}(z)\quad,$
(C.25)

    2. (b)

$d^{(j)}_{2,2}(\theta)=\Big{(}\cos\frac{\theta}{2}\,\Big{)}^{4}\mathcal{J}^{(0,4)}_{j-2}(\cos\theta)=\bigg{[}\frac{z+1}{2}\bigg{]}^{2}\mathcal{J}^{(0,4)}_{j-2}(z)\quad\text{and}$
(C.26)

    3. (c)

$d^{(j)}_{4,4}(\theta)=\Big{(}\cos\frac{\theta}{2}\,\Big{)}^{8}\mathcal{J}^{(0,8)}_{j-4}(\cos\theta)=\bigg{[}\frac{z+1}{2}\bigg{]}^{4}\mathcal{J}^{(0,8)}_{j-4}(z)~{}.$
(C.27)

###  C.4 Mathematical identities

Now we derive various mathematical identities which have been used in the
paper.

##### Mathematical identity I

In this section we show that

$\Bigg{[}\frac{\partial P_{j}(z)}{\partial
z}+(z-1)\frac{\partial^{2}P_{j}(z)}{\partial
z^{2}}\Bigg{]}=\frac{j(j+1)}{2}\mathcal{J}^{(0,2)}_{j-1}(z)~{}.$ (C.28)

We start from the LHS

$\bigg{[}\frac{d}{dz}\mathcal{J}^{(0,0)}_{j}(z)+(z-1)\frac{d^{2}}{dz^{2}}\mathcal{J}^{(0,0)}_{j}(z)\bigg{]}~{}.$
(C.29)

Using the expression for derivatives of Jacobi polynomial in eqn (C.17) we get

$\frac{(j+1)}{2}\bigg{(}\mathcal{J}^{(1,1)}_{j-1}(z)+(z-1)\frac{d}{dz}\mathcal{J}^{(1,1)}_{j-1}(z)\bigg{)}~{}.$
(C.30)

Now using the recursion relation (C.15) we get

$\frac{(j+1)}{2}\bigg{(}\cancel{\mathcal{J}^{(1,1)}_{j-1}(z)}-\cancel{\mathcal{J}^{(1,1)}_{j-1}(z)}+j\mathcal{J}^{(0,2)}_{j-1}(z)\bigg{)}=\frac{j(j+1)}{2}\mathcal{J}^{(0,2)}_{j-1}(z)~{}.$
(C.31)

As a consequence of (C.28) and (C.25) we obtain

$d_{11}^{(j)}(\theta)=\bigg{[}\frac{z+1}{2}\bigg{]}\mathcal{J}^{(0,2)}_{j-1}(z)=\frac{(z+1)}{j(j+1)}\Bigg{[}\frac{\partial
P_{j}(z)}{\partial z}+(z-1)\frac{\partial^{2}P_{j}(z)}{\partial
z^{2}}\Bigg{]}~{}.$ (C.32)

We use this identity for psps amplitude.

##### Mathematical identity II

In this section we show that

$\Bigg{[}2\,\frac{\partial^{2}P_{J}(z)}{\partial
z^{2}}+(z-1)^{2}\,\frac{\partial^{4}P_{J}(z)}{\partial
z^{4}}+4(z-1)\frac{\partial^{3}P_{J}(z)}{\partial
z^{3}}\Bigg{]}=\frac{1}{4}(J+2)(J+1)J(J-1)\mathcal{J}^{(0,4)}_{J-2}(z)~{}.$
(C.33)

We start from the LHS and using (C.17) we can write it as

$\frac{1}{2}(J+2)(J+1)\Bigg{[}\mathcal{J}^{(2,2)}_{J-2}(z)+\frac{(z-1)^{2}\,(J+3)}{4}\Big{[}\frac{d}{dz}\mathcal{J}^{(3,3)}_{J-3}(z)\Big{]}+(z-1)(J+3)\mathcal{J}^{(3,3)}_{J-3}(z)\Bigg{]}~{}.$
(C.34)

We use the definition of derivative of Jacobi polynomial to get

$\frac{1}{2}(J+2)(J+1)\Bigg{[}\frac{(z-1)J(J+3)}{4}\mathcal{J}^{(2,4)}_{J-3}(z)+\frac{J}{2}\mathcal{J}^{(1,3)}_{J-2}(z)\Bigg{]}~{}.$
(C.35)

We rewrite the first term as a derivative and then use identity (C.15) to get
the answer,

$\frac{J}{2}\Bigg{[}(J-1)\mathcal{J}^{(0,4)}_{J-2}(z)-\cancel{\mathcal{J}^{(1,3)}_{J-2}(z)}+\cancel{\mathcal{J}^{(1,3)}_{J-2}(z)}\Bigg{]}$
(C.36)

From (C.26) and (C.33) we get

$(z+1)^{2}\Bigg{[}2\,\frac{\partial^{2}P_{J}(z)}{\partial
z^{2}}+(z-1)^{2}\,\frac{\partial^{4}P_{J}(z)}{\partial
z^{4}}+4(z-1)\frac{\partial^{3}P_{J}(z)}{\partial
z^{3}}\Bigg{]}=4(J+2)(J+1)J(J-1)\,d^{(j)}_{2,2}(\theta)~{}.$ (C.37)

We use this identity for pppp and gsgs amplitudes.

##### Mathematical identity III

In this section, we show that

$\begin{split}\frac{1}{256}\frac{\Gamma(J+5)}{\Gamma(J-3)}\mathcal{J}^{(0,8)}_{J-4}=&\Big{[}\frac{3}{2}\frac{d^{4}}{dz^{4}}P_{J}(z)+6(z-1)\frac{d^{5}}{dz^{5}}P_{J}(z)+\frac{9(z-1)^{2}}{2}\frac{d^{6}}{dz^{6}}P_{J}(z)\\\
&+(z-1)^{3}\frac{d^{7}}{dz^{7}}P_{J}(z)+\frac{(z-1)^{4}}{16}\frac{d^{8}}{dz^{8}}P_{J}(z)\Big{]}\end{split}$
(C.38)

We start from RHS. First we write

$\frac{d^{4+n}}{dz^{4+n}}P_{J}(z)=\frac{\Gamma(J+5)}{16\Gamma(J+1)}\frac{d^{n}}{dz^{n}}\mathcal{J}_{J-4}^{(4,4)}(z)$
(C.39)

Now we successively use (C.15) for the right-hand side of above equation

$\begin{split}\frac{d^{4}}{dz^{4}}P_{J}(z)&=\frac{\Gamma(J+5)}{16\Gamma(J+1)}\mathcal{J}_{J-4}^{(4,4)}(z)\\\
(z-1)\frac{d^{5}}{dz^{5}}P_{J}(z)&=\frac{\Gamma(J+5)}{16\Gamma(J+1)}\Big{[}J\mathcal{J}_{J-4}^{(3,5)}(z)-4\mathcal{J}_{J-4}^{(4,4)}(z)\Big{]}\\\
(z-1)^{2}\frac{d^{6}}{dz^{6}}P_{J}(z)&=\frac{\Gamma(J+5)}{16\Gamma(J+1)}\Bigg{[}\frac{\Gamma(J+1)}{\Gamma(J-1)}\mathcal{J}_{J-4}^{(2,6)}(z)-8J\mathcal{J}_{J-4}^{(3,5)}(z)+20\mathcal{J}_{J-4}^{(4,4)}(z)\Bigg{]}\\\
(z-1)^{3}\frac{d^{7}}{dz^{7}}P_{J}(z)&=\frac{\Gamma(J+5)}{16\Gamma(J+1)}\Bigg{[}\frac{\Gamma(J+1)}{\Gamma(J-2)}\mathcal{J}_{J-4}^{(1,7)}(z)-12\frac{\Gamma(J+1)}{\Gamma(J-1)}\mathcal{J}_{J-4}^{(2,6)}(z)\
\\\
&\qquad\qquad\qquad+60J\mathcal{J}_{J-4}^{(3,5)}(z)-120\mathcal{J}_{J-4}^{(4,4)}(z)\Bigg{]}\\\
(z-1)^{4}\frac{d^{8}}{dz^{8}}P_{J}(z)&=\frac{\Gamma(J+5)}{16\Gamma(J+1)}\Bigg{[}\frac{\Gamma(J+1)}{\Gamma(J-3)}\mathcal{J}_{J-4}^{(0,8)}(z)-16\frac{\Gamma(J+1)}{\Gamma(J-2)}\mathcal{J}_{J-4}^{(1,7)}(z)\\\
&\qquad\qquad\qquad+120\frac{\Gamma(J+1)}{\Gamma(J-1)}\mathcal{J}_{J-4}^{(2,6)}(z)-480J\mathcal{J}_{J-4}^{(3,5)}(z)+840\mathcal{J}_{J-4}^{(4,4)}(z)\Bigg{]}\end{split}$
(C.40)

Putting these in the RHS of (C.38) we get the LHS. We use this identity for
gggg amplitude.

##  Appendix D Form factors for the gggg amplitude

The expressions for 6 of the form factors
($\mathcal{F}_{1\text{({gggg})}},\mathcal{F}_{2\text{({gggg})}},\mathcal{F}_{3\text{({gggg})}},\mathcal{F}_{4\text{({gggg})}},\mathcal{F}_{7\text{({gggg})}},\mathcal{F}_{8\text{({gggg})}}$)
are very long in terms of Gegenbauer polynomial. We write those expressions
here.

$\displaystyle\mathcal{F}_{1\text{({gggg})}}$
$\displaystyle=-\left(\frac{1}{32}(y_{s})^{J-8}\
\frac{d^{4}\mathcal{G}^{(\beta)}_{J}(z_{s})}{dz_{s}^{4}}\right)-\frac{1}{32}(y_{s})^{J-8}\left(-50\
\frac{d^{3}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{3}}+73\frac{d^{4}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{4}}\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.-18\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{5}}+\frac{d^{6}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{6}}-4z_{s}\
\left(12\frac{d^{4}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{4}}-10\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{5}}\right.\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\left.+2z_{s}\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{5}}+\frac{d^{6}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{6}}-z_{s}\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}\right)\right)$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+\frac{1}{256}(y_{s})^{J-8}\
\left(1800\frac{d^{2}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{2}}-13800\
\frac{d^{3}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{3}}+16689\
\frac{d^{4}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{4}}\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.-6228\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-5}(z_{s})}{dz_{s}^{5}}+898\frac{d^{6}\mathcal{G}^{(\beta)}_{J-6}(z_{s})}{dz_{s}^{6}}-52\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-7}(z_{s})}{dz_{s}^{7}}+\frac{d^{8}\mathcal{G}^{(\beta)}_{J-8}(z_{s})}{dz_{s}^{8}}\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.+8z_{s}\
\left(1000\frac{d^{3}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{3}}-3616\
\frac{d^{4}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{4}}+2441\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{5}}-525\frac{d^{6}\mathcal{G}^{(\beta)}_{J-5}(z_{s})}{dz_{s}^{6}}\right.\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\left.+41\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-6}(z_{s})}{dz_{s}^{7}}-\frac{d^{8}\mathcal{G}^{(\beta)}_{J-7}(z_{s})}{dz_{s}^{8}}+z_{s}\
\left(872\frac{d^{4}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{4}}-1880\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{5}}\right.\right.\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\left.\left.+757\frac{d^{6}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{6}}-90\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-5}(z_{s})}{dz_{s}^{7}}+3\frac{d^{8}\mathcal{G}^{(\beta)}_{J-6}(z_{s})}{dz_{s}^{8}}+2z_{s}\
\left(112\frac{d^{5}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{5}}\right.\right.\right.\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\left.\left.\left.-160\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{6}}+38\frac{d^{7}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{7}}-2\
\frac{d^{8}\mathcal{G}^{(\beta)}_{J-5}(z_{s})}{dz_{s}^{8}}+z_{s}\left(8\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}\right.\right.\right.\right.\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\left.\left.\left.\left.-8\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{7}}+\frac{d^{8}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{8}}\right)\right)\right)\right)\right)~{}.$
(D.1)

$\displaystyle\mathcal{F}_{2\text{({gggg})}}$
$\displaystyle=\frac{1}{32}(y_{s})^{J-8}\left[1764\
\frac{d^{4}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{4}}-1791\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{5}}+457\frac{d^{6}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{6}}-39\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-5}(z_{s})}{dz_{s}^{7}}\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.+\frac{d^{8}\mathcal{G}^{(\beta)}_{J-6}(z_{s})}{dz_{s}^{8}}+2z_{s}\
\left\\{900\frac{d^{5}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{5}}-573\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{6}}+82\frac{d^{7}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{7}}\right.\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\left.-3\
\frac{d^{8}\mathcal{G}^{(\beta)}_{J-5}(z_{s})}{dz_{s}^{8}}+2z_{s}\left(116\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}-47\frac{d^{7}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{7}}+8z_{s}\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{7}}\right.\right.\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\left.+3\frac{d^{8}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{8}}-2z_{s}\
\frac{d^{8}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{8}}\right)\right\\}$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-\left.4\ \left(-9\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{5}}+\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}-2z_{s}\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{6}}\right)\right]~{}.$
(D.2) $\displaystyle\mathcal{F}_{3\text{({gggg})}}=$
$\displaystyle\frac{1}{4}(y_{s})^{J-8}\left[-162\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{5}}+161\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}-26\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{7}}+\frac{d^{8}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{8}}\right.$
$\displaystyle~{}~{}~{}\left.-4z_{s}\
\left(20\frac{d^{6}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{6}}-14\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{7}}+2z_{s}\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{7}}+\frac{d^{8}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{8}}\right.\right.$
$\displaystyle~{}~{}~{}\left.\left.-z_{s}\
\frac{d^{8}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{8}}\right)-2\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J}(z_{s})}{dz_{s}^{6}}~{}.\right]$ (D.3)
$\displaystyle\mathcal{F}_{4\text{({gggg})}}$
$\displaystyle=\frac{1}{16}(y_{s})^{J-8}\left[143\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}-26\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{7}}+\frac{d^{8}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{8}}\right.$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.+4z_{s}\ \left(13\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{7}}-\frac{d^{8}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{8}}+z_{s}\
\frac{d^{8}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{8}}\right)\right]~{}.$
(D.4) $\displaystyle\mathcal{F}_{7\text{({gggg})}}$
$\displaystyle=\frac{1}{4}(y_{s})^{J-6}\left[-50\
\frac{d^{3}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{3}}+73\frac{d^{4}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{4}}-18\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{5}}+\frac{d^{6}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{6}}\right.$
$\displaystyle~{}~{}~{}\left.-4z_{s}\
\left(12\frac{d^{4}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{4}}-10\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{5}}+2z_{s}\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{5}}+\frac{d^{6}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{6}}\right.\right.$
$\displaystyle~{}~{}~{}\left.\left.-z_{s}\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}\right)-2\frac{d^{4}\mathcal{G}^{(\beta)}_{J}(z_{s})}{dz_{s}^{4}}\right]~{}.$
(D.5) $\displaystyle\mathcal{F}_{8\text{({gggg})}}$
$\displaystyle=\frac{1}{4}(y_{s})^{J-7}\left[-98\
\frac{d^{4}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{4}}+113\frac{d^{5}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{5}}-22\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{6}}+\frac{d^{7}\mathcal{G}^{(\beta)}_{J-4}(z_{s})}{dz_{s}^{7}}\right.$
$\displaystyle~{}~{}~{}\left.-4z_{s}\
\left(16\frac{d^{5}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{5}}-12\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{6}}+2z_{s}\
\frac{d^{6}\mathcal{G}^{(\beta)}_{J-1}(z_{s})}{dz_{s}^{6}}+\frac{d^{7}\mathcal{G}^{(\beta)}_{J-3}(z_{s})}{dz_{s}^{7}}\right.\right.$
$\displaystyle~{}~{}~{}\left.\left.-z_{s}\
\frac{d^{7}\mathcal{G}^{(\beta)}_{J-2}(z_{s})}{dz_{s}^{7}}\right)-2\
\frac{d^{5}\mathcal{G}^{(\beta)}_{J}(z_{s})}{dz_{s}^{5}}~{}.\right]$ (D.6)

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