In this post, we would like to discuss some combinatorial aspects of the $p$Laplacian. Namely, let $\int_\Omega \nabla u^p \, dx$ be the $p$Dirichlet energy, where $u \in W^{1,p}(\Omega)$ and $p>1$. Its first variation is given by
\[D^1 \left(\int_\Omega \nabla u^p \, dx\right) (\xi_1) = p \int_\Omega \nabla u^{p2} (\nabla u, \nabla \xi_1) \, dx,\]where $\xi_1 \in W^{1,p}(\Omega)$.
The second variation (if exists) is also easy to compute:
\begin{align} \notag D^2 \left(\int_\Omega \nabla u^p \, dx\right) (\xi_1, \xi_2) &= p(p2) \int_\Omega \nabla u^{p4} (\nabla u, \nabla \xi_1) (\nabla u, \nabla \xi_2) \, dx \newline \notag &+ p \int_\Omega \nabla u^{p2} (\nabla \xi_1, \nabla \xi_2) \, dx, \end{align}
where $\xi_2 \in W^{1,p}(\Omega)$.
Let us make some effort to calculate the third variation:
\begin{align} \notag D^3 \left(\int_\Omega \nabla u^p \, dx\right) (\xi_1, \xi_2, \xi_3) &= p(p2)(p4) \int_\Omega \nabla u^{p6} (\nabla u, \nabla \xi_1) (\nabla u, \nabla \xi_2) (\nabla u, \nabla \xi_3) \, dx \newline \notag &+ p(p2) \int_\Omega \nabla u^{p4} (\nabla u, \nabla \xi_1) (\nabla \xi_2, \nabla \xi_3) \, dx \newline \notag &+ p(p2) \int_\Omega \nabla u^{p4} (\nabla u, \nabla \xi_2) (\nabla \xi_1, \nabla \xi_3) \, dx \newline \notag &+ p(p2) \int_\Omega \nabla u^{p4} (\nabla u, \nabla \xi_3) (\nabla \xi_1, \nabla \xi_2) \, dx, \end{align}
where $\xi_3 \in W^{1,p}(\Omega)$.
We already start seeing some structure. So, let us now try to derive a general formula for the $n$th variation of the energy functional. Our main result is the following one.
Proposition. Let $u \in W^{1,p}(\Omega)$. If for a natural $n \geq 1$ there exists $n$th variation of the $p$Direchlet energy of $u$ in direction $(\xi_1,\dots,\xi_n) \in (W^{1,p}(\Omega))^n$, then
\begin{align} \notag &D^n \left(\int_\Omega \nabla u^p \, dx\right) (\xi_1, \dots, \xi_n) \newline \notag &= \int_\Omega \left( \sum\limits_{i=0}^{\lfloor \frac{n}{2} \rfloor} \nabla u^{p2(ni)} \prod\limits_{j=0}^{ni1} (p2j) \left[\sum\limits_{\sigma \in B(n,n2i)} \prod\limits_{k=1}^{n2i} (\nabla u, \nabla \xi_{\sigma(k)}) \left(\sum\limits_{\omega \in P(n,\sigma)} \prod\limits_{l=1}^{i} (\nabla \xi_{\omega(l,1)}, \nabla \xi_{\omega(l,2)}) \right) \right] \right) dx, \end{align}
where

$B(n,n2i)$ is the set of all possible $(n2i)$combinations of $\{1,2,\dots,n\}$ such that the ordering inside each $\sigma \in B(n,n2i)$ is immaterial. Evidently, the cardinality of $B(n,n2i)$ is ${n \choose n2i}$. In particular, if $i=0$, then $card(B(n,n2i)) = 1$.

$P(n,\sigma)$ is the set of all possible partitions of the set $\{1,2,\dots,n\} \setminus \sigma$ into pairs such that the ordering of pairs and inside a pair is immaterial. Note that $card(\sigma)=n2i$, and hence the number of pairs in each $\omega \in P(\sigma)$ is $i$. It is not hard to see that the cardinality of $P(\sigma)$ is $\frac{(2i)!}{2^i i!}$. We represent $\omega$ as a $i \times 2$matrix $(\omega(s,t))_{s=1..i,~t=1,2}$. For instance, if $n=6$ and $\sigma = \{1,2\}$, then
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