We start with the following result.
Theorem 2.1 Let (X,p) be a complete partial metric space endowed with a directed graph G. If f,g:X\to X form a power graphic contraction pair, then the following hold:

(i)
F(f)\ne \mathrm{\varnothing} or F(g)\ne \mathrm{\varnothing} if and only if F(f)\cap F(g)\ne \mathrm{\varnothing}.

(ii)
If u\in F(f)\cap F(g), then the weight assigned to the vertex u is 0.

(iii)
F(f)\cap F(g)\ne \mathrm{\varnothing} provided that G satisfies the property (A).

(iv)
F(f)\cap F(g) is complete if and only if F(f)\cap F(g) is a singleton.
Proof To prove (i), let u\in F(f). By the given assumption, (u,gu)\in E(G). Assume that we assign a nonzero weight to the edge (u,gu). As (u,u)\in E(G) and f and g form a power graphic contraction, we have
\begin{array}{rcl}{p}^{\delta}(u,gu)& =& {p}^{\delta}(fu,gu)\\ \le & \varphi ({p}^{\alpha}(u,u){p}^{\beta}(u,fu){p}^{\gamma}(u,gu))\\ =& \varphi ({p}^{\alpha +\beta}(u,u){p}^{\gamma}(u,gu))\\ \le & \varphi ({p}^{\alpha +\beta}(u,gu){p}^{\gamma}(u,gu))\\ =& \varphi ({p}^{\delta}(u,gu))\\ <& {p}^{\delta}(u,gu),\end{array}
a contradiction. Hence, the weight assigned to the edge (u,gu) is zero and so u=gu. Therefore, u\in F(f)\cap F(g)\ne \mathrm{\varnothing}. Similarly, if u\in F(g), then we have u\in F(f). The converse is straightforward.
Now, let u\in F(f)\cap F(g). Assume that the weight assigned to the vertex u is not zero, then from (1.1), we have
\begin{array}{rcl}{p}^{\delta}(u,u)& =& {p}^{\delta}(fu,gu)\\ \le & \varphi ({p}^{\alpha}(u,u){p}^{\beta}(u,fu){p}^{\gamma}(u,gu))\\ =& \varphi ({p}^{\alpha +\beta +\gamma}(u,u))\\ =& \varphi ({p}^{\delta}(u,u))\\ <& {p}^{\delta}(u,u),\end{array}
a contradiction. Hence, (ii) is proved.
To prove (iii), we will first show that there exists a sequence \{{x}_{n}\} in X with f{x}_{2n}={x}_{2n+1} and g{x}_{2n+1}={x}_{2n+2} for all n\in \mathbb{N} with ({x}_{n},{x}_{n+1})\in E(G), and {lim}_{n\to \mathrm{\infty}}p({x}_{n},{x}_{n+1})=0.
Let {x}_{0} be an arbitrary point of X. If f{x}_{0}={x}_{0}, then the proof is finished, so we assume that f{x}_{0}\ne {x}_{0}. As ({x}_{0},f{x}_{0})\in E(G), so ({x}_{0},{x}_{1})\in E(G). Also, ({x}_{1},g{x}_{1})\in E(G) gives ({x}_{1},{x}_{2})\in E(G). Continuing this way, we define a sequence \{{x}_{n}\} in X such that ({x}_{n},{x}_{n+1})\in E(G) with f{x}_{2n}={x}_{2n+1} and g{x}_{2n+1}={x}_{2n+2} for n\in \mathbb{N}.
We may assume that the weight assigned to each edge ({x}_{2n},{x}_{2n+1}) is nonzero for all n\in \mathbb{N}. If not, then {x}_{2k}={x}_{2k+1} for some k, so f{x}_{2k}={x}_{2k+1}={x}_{2k}, and thus {x}_{2k}\in F(f). Hence, {x}_{2k}\in F(f)\cap F(g) by (i). Now, since ({x}_{2n},{x}_{2n+1})\in E(G), so from (1.1), we have
\begin{array}{rcl}{p}^{\delta}({x}_{2n+1},{x}_{2n+2})& =& {p}^{\delta}(f{x}_{2n},g{x}_{2n+1})\\ \le & \varphi ({p}^{\alpha}({x}_{2n},{x}_{2n+1}){p}^{\beta}({x}_{2n},f{x}_{2n}){p}^{\gamma}({x}_{2n+1},g{x}_{2n+1}))\\ =& \varphi ({p}^{\alpha}({x}_{2n},{x}_{2n+1}){p}^{\beta}({x}_{2n},{x}_{2n+1}){p}^{\gamma}({x}_{2n+1},{x}_{2n+2}))\\ =& \varphi ({p}^{\alpha +\beta}({x}_{2n},{x}_{2n+1}){p}^{\gamma}({x}_{2n+1},{x}_{2n+2}))\\ <& {p}^{\alpha +\beta}({x}_{2n},{x}_{2n+1}){p}^{\gamma}({x}_{2n+1},{x}_{2n+2}),\end{array}
which implies that
{p}^{\alpha +\beta}({x}_{2n+1},{x}_{2n+2})<{p}^{\alpha +\beta}({x}_{2n},{x}_{2n+1}),
a contradiction if \alpha +\beta =0. So, take \alpha +\beta >0, and we have
p({x}_{2n+1},{x}_{2n+2})<p({x}_{2n},{x}_{2n+1})
for all n\in \mathbb{N}. Again from (1.1), we have
\begin{array}{rcl}{p}^{\delta}({x}_{2n+2},{x}_{2n+3})& =& {p}^{\delta}(g{x}_{2n+1},f{x}_{2n+2})\\ =& {p}^{\delta}(f{x}_{2n+2},g{x}_{2n+1})\\ \le & \varphi ({p}^{\alpha}({x}_{2n+2},{x}_{2n+1}){p}^{\beta}({x}_{2n+2},f{x}_{2n+2}){p}^{\gamma}({x}_{2n+1},g{x}_{2n+1}))\\ =& \varphi ({p}^{\alpha}({x}_{2n+1},{x}_{2n+2}){p}^{\beta}({x}_{2n+2},{x}_{2n+3}){p}^{\gamma}({x}_{2n+1},{x}_{2n+2}))\\ =& \varphi ({p}^{\alpha +\gamma}({x}_{2n+1},{x}_{2n+2}){p}^{\beta}({x}_{2n+2},{x}_{2n+3}))\\ <& {p}^{\alpha +\gamma}({x}_{2n+1},{x}_{2n+2}){p}^{\beta}({x}_{2n+2},{x}_{2n+3}),\end{array}
which implies that
{p}^{\alpha +\gamma}({x}_{2n+2},{x}_{2n+3})<{p}^{\alpha +\gamma}({x}_{2n+1},{x}_{2n+2}).
We arrive at a contradiction in case \alpha +\gamma =0. Therefore, we must take \alpha +\gamma >0; consequently, we have
p({x}_{2n+2},{x}_{2n+3})<p({x}_{2n+1},{x}_{2n+2})
for all n\in \mathbb{N}. Hence,
{p}^{\delta}({x}_{n},{x}_{n+1})\le \varphi ({p}^{\delta}({x}_{n1},{x}_{n}))<{p}^{\delta}({x}_{n1},{x}_{n})
(2.1)
for all n\in \mathbb{N}. Therefore, the decreasing sequence of positive real numbers \{{p}^{\delta}({x}_{n},{x}_{n+1})\} converges to some c\ge 0. If we assume that c>0, then from (2.1) we deduce that
0<c\le \underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\varphi ({p}^{\delta}({x}_{n1},{x}_{n}))\le \varphi (c)<c,
a contradiction. So, c=0, that is, {lim}_{n\to \mathrm{\infty}}{p}^{\delta}({x}_{n},{x}_{n+1})=0 and so we have {lim}_{n\to \mathrm{\infty}}p({x}_{n},{x}_{n+1})=0. Also,
{p}^{\delta}({x}_{n},{x}_{n+1})\le \varphi ({p}^{\delta}({x}_{n1},{x}_{n}))\le \cdots \le {\varphi}^{n}({p}^{\delta}({x}_{0},{x}_{1})).
(2.2)
Now, for m,n\in \mathbb{N} with m>n,
\begin{array}{rcl}{p}^{\delta}({x}_{n},{x}_{m})& \le & {p}^{\delta}({x}_{n},{x}_{n+1})+{p}^{\delta}({x}_{n+1},{x}_{n+2})+\cdots +{p}^{\delta}({x}_{m1},{x}_{m})\\ {p}^{\delta}({x}_{n+1},{x}_{n+1}){p}^{\delta}({x}_{n+2},{x}_{n+2})\cdots {p}^{\delta}({x}_{m1},{x}_{m1})\\ \le & {\varphi}^{n}({p}^{\delta}({x}_{0},{x}_{1}))+{\varphi}^{n+1}({p}^{\delta}({x}_{0},{x}_{1}))+\cdots +{\varphi}^{m1}({p}^{\delta}({x}_{0},{x}_{1}))\end{array}
implies that {p}^{\delta}({x}_{n},{x}_{m}) converges to 0 as n,m\to \mathrm{\infty}. That is, {lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m})=0. Since (X,p) is complete, following similar arguments to those given in Theorem 2.1 of [9], there exists a u\in X such that {lim}_{n,m\to \mathrm{\infty}}p({x}_{n},{x}_{m})={lim}_{n\to \mathrm{\infty}}p({x}_{n},u)=p(u,u)=0. By the given hypothesis, ({x}_{2n},u)\in E(G) for all n\in \mathbb{N}. We claim that the weight assigned to the edge (u,gu) is zero. If not, then as f and g form a power graphic contraction, so we have
\begin{array}{rcl}{p}^{\delta}({x}_{2n+1},u)& =& {p}^{\delta}(f{x}_{2n},gu)\\ \le & \varphi ({p}^{\alpha}({x}_{2n},u){p}^{\beta}({x}_{2n},f{x}_{2n}){p}^{\gamma}(u,gu))\\ =& \varphi ({p}^{\alpha}({x}_{2n},u){p}^{\beta}({x}_{2n},{x}_{2n+1}){p}^{\gamma}(u,gu)).\end{array}
(2.3)
We deduce, by taking upper limit as n\to \mathrm{\infty} in (2.3), that
\begin{array}{rcl}{p}^{\delta}(u,gu)& \le & \underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\varphi ({p}^{\alpha}({x}_{2n},u){p}^{\beta}({x}_{2n},{x}_{2n+1}){p}^{\gamma}(u,gu))\\ \le & \varphi ({p}^{\alpha}(u,u){p}^{\beta}(u,u){p}^{\gamma}(u,gu))\\ \le & \varphi ({p}^{\alpha +\beta +\gamma}(u,gu))\\ <& {p}^{\delta}(u,gu),\end{array}
a contradiction. Hence, u=gu and u\in F(f)\cap F(g) by (i).
Finally, to prove (iv), suppose the set F(f)\cap F(g) is complete. We are to show that F(f)\cap F(g) is a singleton. Assume on the contrary that there exist u and v such that u,v\in F(f)\cap F(g) but u\ne v. As (u,v)\in E(G) and f and g form a power graphic contraction, so
\begin{array}{rcl}0& <& {p}^{\delta}(u,v)={p}^{\delta}(fu,fv)\\ \le & \varphi ({p}^{\alpha}(u,v){p}^{\beta}(u,fu){p}^{\gamma}(v,gv))\\ =& \varphi ({p}^{\alpha}(u,v){p}^{\beta}(u,u){p}^{\gamma}(v,v))\\ \le & \varphi ({p}^{\delta}(u,v)),\end{array}
a contradiction. Hence, u=v. Conversely, if F(f)\cap F(g) is a singleton, then it follows that F(f)\cap F(g) is complete. □
Corollary 2.2 Let (X,p) be a complete partial metric space endowed with a directed graph G. If we replace (1.1) by
{p}^{\delta}({f}^{s}x,{g}^{t}y)\le \varphi ({p}^{\alpha}(x,y){p}^{\beta}(x,{f}^{s}x){p}^{\gamma}(y,{g}^{t}y)),
(2.4)
where \alpha ,\beta ,\gamma \ge 0 with \delta =\alpha +\beta +\gamma \in (0,\mathrm{\infty}) and s,t\in \mathbb{N}, then the conclusions obtained in Theorem 2.1 remain true.
Proof It follows from Theorem 2.1, that F({f}^{s})\cap F({g}^{t}) is a singleton provided that F({f}^{s})\cap F({g}^{t}) is complete. Let F({f}^{s})\cap F({g}^{t})=\{w\}, then we have f(w)=f({f}^{s}(w))={f}^{s+1}(w)={f}^{s}(f(w)), and g(w)=g({g}^{t}(w))={g}^{t+1}(w)={g}^{t}(g(w)) implies that fw and gw are also in F({f}^{s})\cap F({g}^{t}). Since F({f}^{s})\cap F({g}^{t}) is a singleton, we deduce that w=fw=gw. Hence, F(f)\cap F(g) is a singleton. □
The following remark shows that different choices of α, β and γ give a variety of power graphic contraction pairs of two mappings.
Remarks 2.3 Let (X,p) be a complete partial metric space endowed with a directed graph G.
(R1) We may replace (1.1) with the following:
{p}^{3}(fx,gy)\le \varphi (p(x,y)p(x,fx)p(y,gy))
(2.5)
to obtain conclusions of Theorem 2.1. Indeed, taking \alpha =\beta =\gamma =1 in Theorem 2.1, one obtains (2.5).
(R2) If we replace (1.1) by one of the following condition:
then the conclusions obtained in Theorem 2.1 remain true. Note that

(i)
if we take \alpha =\beta =1 and \gamma =0 in (1.1), then we obtain (2.6),

(ii)
take \alpha =\gamma =1, \beta =0 in (1.1) to obtain (2.7),

(iii)
use \beta =\gamma =1, \alpha =0 in (1.1) and obtain (2.8).
(R3) Also, if we replace (1.1) by one of the following conditions:
then the conclusions obtained in Theorem 2.1 remain true. Note that

(iv)
take \alpha =1 and \beta =\gamma =0 in (1.1) to obtain (2.9),

(v)
to obtain (2.10), take \beta =1, \alpha =\gamma =0 in (1.1),

(vi)
if one takes \gamma =1, \alpha =\beta =0 in (1.1), then we obtain (2.11).
Remark 2.4 If we take f=g in a power graphic contraction pair, then we obtain fixed point results for a power graphic contraction.