Appendix
In this section, we will rewrite m-harmonic almost complex structure equation in a good divergence form in the spirit of [2] to prove Lemma 5. First we have the following,
Lemma 6
The Euler-Lagrange equation \([\varDelta ^{m}J,J]=0\) is equivalent to
$$\begin{aligned} \varDelta ^m J= \frac{1}{4} T_m\left( J,\nabla J, \cdots , \nabla ^{2m-1}J\right) \end{aligned}$$
(61)
where \(T_m=JQ_m+Q_mJ\) and
$$\begin{aligned} Q_m = \varDelta ^m\left( J^2\right) -\varDelta ^m J \,J- J\, \varDelta ^m J. \end{aligned}$$
Proof
This is a direct computation using the fact \(\varDelta (J^2)=0\). \(\square \)
Lemma 5 can be stated as follows,
Proposition 6
For \(m=2, 3\), \(T_m\) in Lemma 6 can be rewritten as
$$\begin{aligned} T_m=T_{\lambda _0} - \left[ J-\lambda _0, \left[ \varDelta ^m J, J\right] \right] \end{aligned}$$
(62)
where \(T_{\lambda _0}\) is a linear combination of the following terms
$$\begin{aligned} \nabla ^{\alpha }\bigg ((J-\lambda _{0})*\nabla ^{\beta }J*\nabla ^{\gamma }J\bigg ) \quad or\quad \lambda _0*\nabla ^{\alpha }\bigg ((J-\lambda _{0})*\nabla ^{\delta }J\bigg ), \end{aligned}$$
where \(\alpha ,\beta ,\gamma ,\delta \) are multi-indices such that \(1\le |\alpha | \le 2m-1\), \(0\le |\beta |,|\gamma |,|\delta |\le m\), \(|\alpha |+|\beta |+|\gamma |=2m\) and \(|\alpha |+|\delta |=2m\).
In what follows, we always assume J is a square matrix valued function and satisfies \(J^2=-id\). In this situation, we know \(\nabla \lambda _0=0\) for every constant matrix \(\lambda _0\). The reason for emphasizing this point is that if we consider the constant matrix \(\lambda _0 \) as a (1,1) tensor field on \((B_1,g)\), then (1, 2) tensor field \(\nabla \lambda _0\) might not be zero.
1.1 The case m=2: biharmonic almost complex structure
By the definition of \(T_m\) in Theorem 6, we have
$$\begin{aligned} T_2= J Q_2 + Q_2 J, \end{aligned}$$
where \(Q_2=2 \nabla \varDelta J \nabla J + 2\nabla J \nabla \varDelta J+ 2\varDelta J \varDelta J + 2 \varDelta ( \nabla J )^2.\) Set
$$\begin{aligned} {\mathbf {I}}&= J \big ( \nabla \varDelta J \nabla J + \nabla J \nabla \varDelta J \big ) + \big ( \nabla \varDelta J \nabla J + \nabla J \nabla \varDelta J \big ) J \\ \mathbf {II}&= J (\varDelta J)^2 + (\varDelta J)^2 J\\ \mathbf {III}&= J \varDelta ( \nabla J )^2 + \varDelta ( \nabla J )^2 J. \end{aligned}$$
Thus, we obtain \(T_2=2{\mathbf {I}}+2\mathbf {II}+2\mathbf {III}.\) Firstly, we compute the term \({\mathbf {I}}\):
$$\begin{aligned} {\mathbf {I}}&= J \nabla \varDelta J \nabla J + J \nabla J \nabla \varDelta J + \nabla \varDelta J \nabla J J+ \nabla J \nabla \varDelta J J \\&= J \nabla \varDelta J \nabla J - \nabla J J \nabla \varDelta J - \nabla \varDelta J J \nabla J + \nabla J \nabla \varDelta J J \\&= [J, \nabla \varDelta J] \nabla J + \nabla J [\nabla \varDelta J, J] \\&= [\nabla J, [\nabla \varDelta J, J]]. \end{aligned}$$
Since \(\nabla \bigg ( [\nabla \varDelta J, J] -[\varDelta J, \nabla J] \bigg )= [\varDelta ^2 J, J],\) we have
$$\begin{aligned}&\nabla [J -\lambda _0, [\nabla \varDelta J, J] -[\varDelta J, \nabla J]] \nonumber \\&\quad = [\nabla J, [\nabla \varDelta J, J] -[\varDelta J, \nabla J]]+ [J-\lambda _0, [\varDelta ^2 J, J]]. \end{aligned}$$
(63)
Now we compute the left-hand side of above equality:
$$\begin{aligned}&\nabla [J -\lambda _0, [\nabla \varDelta J, J] -[\varDelta J, \nabla J]] \\&\quad = \nabla [J -\lambda _0, \nabla [\varDelta J, J]] -2 \nabla [J -\lambda _0, [\varDelta J, \nabla J]] \\&\quad = \nabla [J -\lambda _0, \nabla [\varDelta J, J]]+ T_{\lambda _0}\\&\quad = \varDelta [J -\lambda _0, [\varDelta J, J]] - \nabla [\nabla J, [\varDelta J, J]] + T_{\lambda _0}\\&\quad = - \nabla [\nabla J, [\varDelta J, J]] + T_{\lambda _0}\\&\quad = - [\varDelta J, [\varDelta J, J]] -[\nabla J, [\nabla \varDelta J, J]] -[\nabla J, [\varDelta J, \nabla J]] + T_{\lambda _0} \end{aligned}$$
Substituting above equality into (63) yields
$$\begin{aligned} 2 {\mathbf {I}}= 2 [\nabla J, [\nabla \varDelta J, J]] = - [\varDelta J, [\varDelta J, J]] - [J-\lambda _0, [\varDelta ^2 J, J]] + T_{\lambda _0} \end{aligned}$$
(64)
We now turn to compute the term \(\mathbf {III}\). Since
$$\begin{aligned} J \varDelta (\nabla J)^2&= (J -\lambda _0 ) \varDelta (\nabla J)^2 + \lambda _0 \varDelta (\nabla J)^2 \\&= (J -\lambda _0 ) \varDelta (\nabla J)^2 + \lambda _0 \varDelta \nabla \big ( ( J -\lambda _0) \nabla J \big ) - \lambda _0 \varDelta \big ( (J-\lambda _0) \varDelta J \big ) \\&=(J -\lambda _0 ) \varDelta (\nabla J)^2 + T_{\lambda _0} \\&= \nabla _p \big ( (J -\lambda _0 ) \nabla _p (\nabla J)^2 \big ) -\nabla _p J \nabla _p (\nabla J)^2 + T_{\lambda _0} \\&= -\nabla _p J \nabla _p (\nabla J)^2 + T_{\lambda _0} \\&= -\nabla _p \big ( \nabla _p J (\nabla J)^2 \big ) + \varDelta J (\nabla J)^2 + T_{\lambda _0} \\&= \varDelta J (\nabla J)^2 + T_{\lambda _0} \end{aligned}$$
and similarly \(\varDelta (\nabla J)^2 J = (\nabla J)^2 \varDelta J + T_{\lambda _0},\) we have
$$\begin{aligned} \mathbf {III}= \varDelta J (\nabla J)^2 + (\nabla J)^2 \varDelta J + T_{\lambda _0} \end{aligned}$$
(65)
Now let us proceed to compute \(\mathbf {II}\):
$$\begin{aligned} \mathbf {II}&= J (\varDelta J)^2 + (\varDelta J)^2 J = - \big ( \varDelta J J + 2( \nabla J )^2 \big ) \varDelta J + \varDelta J \varDelta J J \\&= \varDelta J [\varDelta J, J] - 2 (\nabla J )^2 \varDelta J \end{aligned}$$
where we used the fact \(\varDelta (J^2)=0\) which implies
$$\begin{aligned} \varDelta J J=-J \varDelta J- 2 \nabla J \nabla J. \end{aligned}$$
(66)
On the other hand, we also have
$$\begin{aligned} \mathbf {II}&= J (\varDelta J)^2 + (\varDelta J)^2 J = J \varDelta J \varDelta J - \varDelta J \big ( J \varDelta J + 2 (\nabla J)^2 \big ) \\&= [J, \varDelta J] \varDelta J - 2 \varDelta J (\nabla J)^2 \end{aligned}$$
Hence, we obtain
$$\begin{aligned} 2 \mathbf {II}&= \varDelta J [\varDelta J, J]+ [J, \varDelta J] \varDelta J - 2\big ( (\nabla J )^2 \varDelta J+ \varDelta J (\nabla J)^2\big ) \nonumber \\&= [\varDelta J, [\varDelta J, J]] - 2(\nabla J )^2 \varDelta J-2 \varDelta J (\nabla J)^2 \nonumber \\&= [\varDelta J, [\varDelta J, J]] - 2 \mathbf {III} \end{aligned}$$
(67)
where in the last equality we used (65). Substituting (67) into (64), we get
$$\begin{aligned} T_2=2{\mathbf {I}}+ 2\mathbf {II} +2\mathbf {III} = T_{\lambda _0}-\left[ J-\lambda _0, \left[ \varDelta ^2 J, J\right] \right] \end{aligned}$$
which is the desired conclusion.
1.2 The Case m=3: 3-Harmonic Almost Complex Structure
By the definition of \(T_m\) in Theorem 6, we have
$$\begin{aligned} T_3= J Q_3 + Q_3 J, \end{aligned}$$
where
$$\begin{aligned} Q_3&= 2\nabla \varDelta ^{2}J\nabla J+2\nabla J\nabla \varDelta ^{2}J +\varDelta ^{2}J\varDelta J+\varDelta J\varDelta ^{2}J \nonumber \\&\quad +2\varDelta \big (\nabla \varDelta J\nabla J+\nabla J\nabla \varDelta J \big ) +2\varDelta \left( \varDelta J\right) ^{2}+2\varDelta ^{2}\left( \nabla J\right) ^{2}. \end{aligned}$$
For simplicity, we collect some terms which are \(T_{\lambda _0}\) type and appear frequently in the following proof.
Lemma 7
The following terms are \(T_{\lambda _0}\) type terms for any given constant matrix \(\lambda _0\):
$$\begin{aligned} \nabla \left( \nabla J*\nabla ^{2}J*\nabla ^{2}J\right) , \nabla ^{2}\left( \nabla J*\nabla J*\nabla ^{2}J\right) , \nabla \left( \nabla J*\nabla J*\nabla ^{3}J\right) , \nabla ^4 \left( \nabla J\right) ^2. \end{aligned}$$
Proof
For simplicity, we only show how to rewrite the first term and the third term. Other terms can be handled in much the same way. The first term:
$$\begin{aligned} \nabla \big (\nabla J*\nabla ^{2}J*\nabla ^{2}J \big )&=\nabla \left( \nabla \left( J-\lambda _{0}\right) *\nabla ^{2}J *\nabla ^{2}J \right) \\&=\nabla ^{2} \left( \left( J-\lambda _{0}\right) *\nabla ^{2}J *\nabla ^{2}J\right) -\nabla \left( \left( J-\lambda _{0}\right) *\nabla ^{3}J *\nabla ^{2}J \right) \\&\quad -\nabla \left( \left( J-\lambda _{0}\right) *\nabla ^{2}J *\nabla ^{3}J \right) \\&= T_{\lambda _0}. \end{aligned}$$
The third term:
$$\begin{aligned}&\nabla \big ( \nabla J *\nabla J *\nabla ^{3}J \big ) \\&\quad = \nabla ^2 \big ( \nabla J *\nabla J *\nabla ^2 J \big ) - \nabla \big ( \nabla ^2 J *\nabla J *\nabla ^{2}J \big ) -\nabla \big ( \nabla J *\nabla ^2 J *\nabla ^{2}J \big ) \\&\quad = T_{\lambda _0}. \end{aligned}$$
\(\square \)
Note that we will emphasize the terms of \(T_{\lambda _0}\) type by underlining it in the following proof. Set
$$\begin{aligned} {\mathbf {I}}&= \,J \nabla \varDelta ^{2} J \nabla J + J\nabla J \nabla \varDelta ^{2}J +\nabla \varDelta ^{2}J\nabla J\, J+\nabla J \nabla \varDelta ^{2}J \,J, \\ \mathbf {II}&= \, J \big (\varDelta ^{2}J\varDelta J+\varDelta J\varDelta ^{2}J \big ) +\big ( \varDelta ^{2}J\varDelta J+\varDelta J\varDelta ^{2}J \big )J, \\ \mathbf {III}&= \, J\varDelta \big ( \nabla \varDelta J\nabla J+\nabla J\nabla \varDelta J \big ) +\varDelta \big (\nabla \varDelta J\nabla J+\nabla J\nabla \varDelta J \big )J, \\ \mathbf {IV}&= \, J\varDelta \left( \varDelta J\right) ^{2} +\varDelta \left( \varDelta J\right) ^{2}J, \\ {\mathbf {V}}&= \, J\varDelta ^{2}\left( \nabla J\right) ^{2} +\varDelta ^{2}\left( \nabla J\right) ^{2}J. \end{aligned}$$
Then, we obtain \(T_3= 2{\mathbf {I}}+ \mathbf {II}+ 2\mathbf {III}+ 2\mathbf {IV}+ 2{\mathbf {V}}.\)
Step One: dealing with I. Now Let us compute the first term I:
$$\begin{aligned} {\mathbf {I}}&=J\nabla \varDelta ^{2}J\nabla J + J\nabla J\nabla \varDelta ^{2}J + \nabla \varDelta ^{2}J\nabla J J + \nabla J\nabla \varDelta ^{2}J J \\&=J\nabla \varDelta ^{2}J\nabla J-\nabla JJ\nabla \varDelta ^{2}J -\nabla \varDelta ^{2}JJ\nabla J + \nabla J\nabla \varDelta ^{2}J J \\&=\left[ J,\nabla \varDelta ^{2}J\right] \nabla J + \nabla J\left[ \nabla \varDelta ^{2}J,J\right] \\&=\left[ \nabla J,\left[ \nabla \varDelta ^{2}J,J\right] \right] , \end{aligned}$$
Since \(\nabla \bigg ([\nabla \varDelta ^{2}J,J]-[\varDelta ^{2}J,\nabla J] +[\nabla \varDelta J,\varDelta J] \bigg )=[\varDelta ^3 J, J],\) we have
$$\begin{aligned}&\nabla \left[ J-\lambda _{0},\left[ \nabla \varDelta ^{2}J,J\right] -\left[ \varDelta ^{2}J,\nabla J\right] +[\nabla \varDelta J,\varDelta J] \right] \nonumber \\&\quad = \big [\nabla J,\left[ \nabla \varDelta ^{2}J,J\right] -\left[ \varDelta ^{2}J,\nabla J\right] +[\nabla \varDelta J,\varDelta J]\big ] + \left[ J-\lambda _{0}, \left[ \varDelta ^3 J, J\right] \right] . \end{aligned}$$
(68)
Now we compute the left-hand side of above equality.
$$\begin{aligned}&\nabla \left[ J-\lambda _{0},\left[ \nabla \varDelta ^{2}J,J\right] -\left[ \varDelta ^{2}J,\nabla J\right] +\underline{[\nabla \varDelta J,\varDelta J]} \right] \\&\quad = \nabla \left[ J-\lambda _{0},\left[ \nabla \varDelta ^{2}J,J\right] -\left[ \varDelta ^{2}J,\nabla J\right] \right] +{T_{\lambda _0}}\\&\quad = \nabla [J-\lambda _{0},\nabla \left[ \varDelta ^{2}J,J\right] -2\left[ \varDelta ^{2}J,\nabla J\right] ] +{T_{\lambda _0}}\\&\quad = \varDelta \left[ J-\lambda _{0},[\varDelta ^{2}J,J]\right] - \nabla \big [\nabla J,[\varDelta ^{2}J,J]\big ] - 2\nabla [J-\lambda _{0},\left[ \varDelta ^{2}J,\nabla J\right] ] + T_{\lambda _0}\\&\quad = \varDelta \big [J-\lambda _{0},\nabla [\nabla \varDelta J,J] -\underline{[\nabla \varDelta J,\nabla J]}\big ]\\&\qquad -2\nabla _p \big [J-\lambda _{0},\nabla _q[\nabla _q\varDelta J,\nabla _p J] -\underline{\left[ \nabla _q\varDelta J,\nabla _{qp}^2 J\right] } \big ]\\&\qquad {-\nabla \left[ \nabla J,\left[ \varDelta ^{2}J,J\right] \right] }+ T_{\lambda _0}\\&\quad =\underline{\varDelta \nabla \left[ J-\lambda _{0},\left[ \nabla \varDelta J,J\right] \right] } -\varDelta [\nabla J,[\nabla \varDelta J,J]]\\&\qquad -2 \underline{\nabla _{pq}^{2} \big [ J-\lambda _{0},[\nabla _q \varDelta J,\nabla J] \big ]} +2 \underline{\nabla _{p}[\nabla _{q}J,[\nabla _{q}\varDelta J,\nabla _{p}J]]}\\&\qquad {-\nabla [\nabla J,[\varDelta ^{2}J,J]]}+T_{\lambda _0}\\&\quad = -\varDelta [\nabla J,[\nabla \varDelta J,J]] -\nabla [\nabla J,[\varDelta ^{2}J,J]]+ T_{\lambda _0}, \end{aligned}$$
where in the second equality from bottom we employ lemma 7. By substituting above equality into (68), we obtain
$$\begin{aligned} {\mathbf {I}}&= [\nabla J,[\nabla \varDelta ^{2}J,J]]\\&= [\nabla J,[\varDelta ^{2}J,\nabla J]]-[\nabla J,[\nabla \varDelta J,\varDelta J]]-\varDelta [\nabla J,[\nabla \varDelta J,J]]-\nabla \left[ \nabla J,\left[ \varDelta ^{2}J,J\right] \right] \\&\quad +{T_{\lambda _0}} - \left[ J-\lambda _0, \left[ \varDelta ^3 J, J\right] \right] . \end{aligned}$$
Since \(\nabla [\nabla J,[\varDelta ^{2}J,J]]=[\varDelta J,[\varDelta ^{2}J,J]]+[\nabla J,[\nabla \varDelta ^{2}J,J]]+[\nabla J,[\varDelta ^{2}J,\nabla J]],\) we deduce
$$\begin{aligned} 2{\mathbf {I}}&= -[\nabla J,[\nabla \varDelta J,\varDelta J]]-\varDelta [\nabla J,[\nabla \varDelta J,J]]-\left[ \varDelta J,\left[ \varDelta ^{2}J,J\right] \right] \nonumber \\&\quad +{T_{\lambda _0}}- \left[ J-\lambda _0, \left[ \varDelta ^3 J, J\right] \right] . \end{aligned}$$
(69)
By lemma 7, we can derive
$$\begin{aligned}&[\nabla J,[\nabla \varDelta J,\varDelta J]] \nonumber \\&= \nabla J\left( \nabla \varDelta J\varDelta J-\varDelta J\nabla \varDelta J\right) -\left( \nabla \varDelta J\varDelta J-\varDelta J\nabla \varDelta J\right) \nabla J \nonumber \\&= \nabla J\nabla \varDelta J\varDelta J+\varDelta J\nabla \varDelta J\nabla J -\nabla J\varDelta J\nabla \varDelta J-\nabla \varDelta J\varDelta J\nabla J \nonumber \\&= \underline{\nabla \left( \nabla J\varDelta J\varDelta J\right) } +\underline{\nabla \left( \varDelta J\varDelta J\nabla J\right) } -2\left( \varDelta J\right) ^{3} -2\nabla J\varDelta J\nabla \varDelta J -2\nabla \varDelta J\varDelta J\nabla J \nonumber \\&= -2\left( \varDelta J\right) ^{3} -2\nabla J\varDelta J\nabla \varDelta J -2\nabla \varDelta J\varDelta J\nabla J +{T_{\lambda _0}} \end{aligned}$$
(70)
and
$$\begin{aligned}&\varDelta [\nabla J,[\nabla \varDelta J,J]] \nonumber \\&\quad = \varDelta \big [\nabla J,\nabla [\varDelta J,J]-[\varDelta J,\nabla J] \big ] \nonumber \\&\quad = \varDelta [\nabla J,\nabla [\varDelta J,J]] -\underline{\varDelta [\nabla J,[\varDelta J,\nabla J]]} \nonumber \\&\quad = \underline{\varDelta \nabla [\nabla J,[\varDelta J,J]]} -\varDelta [\varDelta J,[\varDelta J,J]]+{T_{\lambda _0}}\nonumber \\&\quad = -\varDelta \left( \left( \varDelta J\right) ^{2}J+J\left( \varDelta J\right) ^{2}-2\varDelta JJ\varDelta J\right) +{T_{\lambda _0}} \nonumber \\&\quad = -\varDelta \left( 2\left( \varDelta J\right) ^{2}J +2J\left( \varDelta J\right) ^{2} +2\underline{\left( \nabla J\right) ^{2}\varDelta J} +2\underline{\varDelta J\left( \nabla J\right) ^{2}} \right) +T_{\lambda _0}\nonumber \\&\quad =-2\varDelta \left( \left( \varDelta J\right) ^{2}J+J\left( \varDelta J\right) ^{2}\right) +{T_{\lambda _0}}. \end{aligned}$$
(71)
where in the second equality from bottom we used (66). Substituting equalities (70) and (71) into equality (69) yields
$$\begin{aligned} 2{\mathbf {I}}&= 2\left( \varDelta J\right) ^{3}+2\nabla J\varDelta J\nabla \varDelta J+2\nabla \varDelta J\varDelta J\nabla J+2\varDelta \left( \left( \varDelta J\right) ^{2}J+J\left( \varDelta J\right) ^{2}\right) \nonumber \\&\quad -\left[ \varDelta J,\left[ \varDelta ^{2}J,J\right] \right] +{T_{\lambda _0}} - \left[ J-\lambda _0, \left[ \varDelta ^3 J, J\right] \right] . \end{aligned}$$
(72)
Step Two: dealing with V and II. Firstly, we deal with fifth term V. It follows from Lemma 7 that
$$\begin{aligned} {\mathbf {V}}&=J\varDelta ^{2}\left( \nabla J\right) ^{2}+\varDelta ^{2}\left( \nabla J\right) ^{2}J\\&=\left( J-\lambda _{0}\right) \varDelta ^{2}\left( \nabla J\right) ^{2} +\varDelta ^{2}\left( \nabla J\right) ^{2}\left( J-\lambda _{0}\right) +\underline{\lambda _{0}\varDelta ^{2}\left( \nabla J\right) ^{2}} +\underline{\varDelta ^{2}\left( \nabla J\right) ^{2}\lambda _{0}}\\&= \nabla \left( \left( J-\lambda _{0}\right) \nabla \varDelta \left( \nabla J\right) ^{2}\right) -\nabla J\nabla \varDelta \left( \nabla J\right) ^{2} \\&\quad +\nabla \left( \nabla \varDelta \left( \nabla J\right) ^{2}\left( J-\lambda _{0}\right) \right) -\nabla \varDelta \left( \nabla J\right) ^{2}\nabla J +{T_{\lambda _0}}\\&= \underline{\varDelta \bigg ((J-\lambda _{0})\varDelta (\nabla J)^{2}\bigg )} -\nabla \left( \nabla J\varDelta \left( \nabla J\right) ^{2}\right) -\nabla J\nabla \varDelta \left( \nabla J\right) ^{2} \\&\quad + \underline{\varDelta \bigg (\varDelta (\nabla J)^{2}(J-\lambda _{0})\bigg )} -\nabla \left( \varDelta \left( \nabla J\right) ^{2}\nabla J\right) -\nabla \varDelta \left( \nabla J\right) ^{2}\nabla J +{T_{\lambda _0}}\\&=-\nabla \left( \nabla J\varDelta \left( \nabla J\right) ^{2}\right) -\nabla J\nabla \varDelta \left( \nabla J\right) ^{2} -\nabla \left( \varDelta \left( \nabla J\right) ^{2}\nabla J\right) \\&\quad -\nabla \varDelta \left( \nabla J\right) ^{2}\nabla J +{T_{\lambda _0}}. \end{aligned}$$
Since
$$\begin{aligned} \nabla _{p}\bigg (\nabla _{p}J\varDelta \left( \nabla J\right) ^{2}\bigg ) =\nabla _{pq}^2\bigg (\nabla _{p}J\nabla _{q}\left( \nabla J\right) ^{2}\bigg ) -\nabla _{p}\bigg ( \nabla _{qp}^{2}J\nabla _{q}\left( \nabla J\right) ^{2}\bigg ) ={T_{\lambda _0}} \end{aligned}$$
and
$$\begin{aligned} \nabla _{p}J\nabla _{p}\varDelta \left( \nabla J\right) ^{2}&=\underline{\nabla _{p}\bigg (\nabla _{p}J\varDelta \left( \nabla J\right) ^{2}\bigg )} -\varDelta J\varDelta \left( \nabla J\right) ^{2}\\&=-\underline{\nabla _{p}\left( \varDelta J\nabla _{p}\left( \nabla J\right) ^{2}\right) } +\nabla _{p}\varDelta J\nabla _{p}\left( \nabla J\right) ^{2}+{T_{\lambda _0}}\\&=\underline{\nabla _{p}\left( \nabla _{p}\varDelta J\left( \nabla J\right) ^{2}\right) } -\varDelta ^{2}J\left( \nabla J\right) ^{2} +{T_{\lambda _0}}\\&=-\varDelta ^{2}J\left( \nabla J\right) ^{2}+{T_{\lambda _0}}, \end{aligned}$$
we have
$$\begin{aligned} {\mathbf {V}}=\varDelta ^{2}J\left( \nabla J\right) ^{2}+\left( \nabla J\right) ^{2}\varDelta ^{2}J+{T_{\lambda _0}}. \end{aligned}$$
(73)
Next, we deal with the second term
$$\begin{aligned} \mathbf {II}&=J\left( \varDelta ^{2}J\varDelta J+\varDelta J\varDelta ^{2}J\right) +\left( \varDelta ^{2}J\varDelta J+\varDelta J\varDelta ^{2}J\right) J\nonumber \\&=\left[ \varDelta J,\left[ \varDelta ^{2}J,J\right] \right] -2\left( \nabla J\right) ^{2}\varDelta ^{2}J-2\varDelta ^{2}J\left( \nabla J\right) ^{2} \nonumber \\&=\left[ \varDelta J,\left[ \varDelta ^{2}J,J\right] \right] -2{\mathbf {V}}+T_{\lambda _0}, \end{aligned}$$
(74)
where we have used (66) and (73).
Step Three: dealing with III Here we begin to deal with the third term:
$$\begin{aligned} \mathbf {III}&=J\varDelta \bigg (\nabla \varDelta J\nabla J+\nabla J\nabla \varDelta J\bigg ) +\varDelta \bigg (\nabla \varDelta J\nabla J+\nabla J\nabla \varDelta J\bigg )J\\&=J\varDelta \left( \nabla \big (\varDelta J\nabla J+\nabla J\varDelta J\big ) -2\left( \varDelta J\right) ^{2}\right) \\&\quad \, +\varDelta \left( \nabla \big (\varDelta J\nabla J+\nabla J\varDelta J\big ) -2\left( \varDelta J\right) ^{2}\right) J\\&=J\varDelta \nabla \bigg (\varDelta J\nabla J+\nabla J\varDelta J\bigg ) +\varDelta \nabla \bigg (\varDelta J\nabla J+\nabla J\varDelta J\bigg )J\\&\quad \, -2\left( J\varDelta \left( \varDelta J\right) ^{2} +\varDelta \left( \varDelta J\right) ^{2}J \right) \\&=J\varDelta \nabla \bigg (\varDelta J\nabla J+\nabla J\varDelta J\bigg ) +\varDelta \nabla \bigg (\varDelta J\nabla J+\nabla J\varDelta J\bigg )J -2\mathrm {\mathbf {IV}}. \end{aligned}$$
Since
$$\begin{aligned}&J\varDelta \nabla \big (\varDelta J\nabla J\big ) \\&= \big (J-\lambda _{0}\big )\varDelta \nabla \left( \varDelta J\nabla J\right) +\lambda _{0}\varDelta \nabla \left( \varDelta J\nabla J\right) \\&= \left( J-\lambda _{0}\right) \varDelta \nabla \left( \varDelta J\nabla J\right) +\underline{\lambda _{0}\varDelta \nabla \bigg (\nabla \left( \varDelta J\big (J-\lambda _{0}\big )\right) -\nabla \varDelta J\big (J-\lambda _{0}\big )\bigg )}\\&= \left( J-\lambda _{0}\right) \varDelta \nabla \big (\varDelta J\nabla J\big ) +{T_{\lambda _0}}\\&= \nabla _p \bigg ( (J-\lambda _0) \nabla ^2_{pq} \big ( \varDelta J \nabla _q J\big )\bigg ) -\nabla _p J \nabla ^2_{pq} \big ( \varDelta J \nabla _q J\big ) +{T_{\lambda _0}}\\&= \underline{\varDelta \bigg ( (J-\lambda _0) \nabla _q \big (\varDelta J \nabla _q J \big )\bigg )} -\underline{\nabla _p \bigg ( \nabla _p J \nabla _q \big ( \varDelta J \nabla _q J\big )\bigg )} \\&\quad -\nabla _p J \nabla ^2_{pq} \big ( \varDelta J \nabla _q J\big ) +{T_{\lambda _0}}\\&= - \underline{\nabla _p \bigg ( \nabla _p J \nabla _q \big (\varDelta J \nabla _q J \big )\bigg )} + \varDelta J \nabla _q \big ( \varDelta J \nabla _q J \big ) +{T_{\lambda _0}}\\&= \underline{\nabla _q \bigg ( \varDelta J \varDelta J \nabla _q J \bigg )} -\nabla \varDelta J \varDelta J \nabla J +{T_{\lambda _0}}\\&= -\nabla \varDelta J\varDelta J\nabla J+{T_{\lambda _0}}, \end{aligned}$$
we have
$$\begin{aligned} \mathbf {III}&= -\nabla \varDelta J\bigg (\varDelta J\nabla J+\nabla J\varDelta J\bigg ) -\bigg (\varDelta J\nabla J+\nabla J\varDelta J\bigg )\nabla \varDelta J -2\mathbf {IV}+{T_{\lambda _0}}\nonumber \\&= -\nabla \varDelta J\varDelta J\nabla J-\nabla J\varDelta J\nabla \varDelta J -\bigg (\nabla \varDelta J\nabla J\varDelta J+\varDelta J\nabla J\nabla \varDelta J\bigg ) \nonumber \\&\quad -2\mathbf {IV}+{T_{\lambda _0}}\nonumber \\&= -\nabla \varDelta J\varDelta J\nabla J -\nabla J\varDelta J\nabla \varDelta J -\underline{\nabla \big (\varDelta J\nabla J\varDelta J \big )} +(\varDelta J)^{3} -2\mathbf {IV}+{T_{\lambda _0}}\nonumber \\&= -\nabla \varDelta J\varDelta J\nabla J-\nabla J\varDelta J\nabla \varDelta J+\left( \varDelta J\right) ^{3}-2\mathbf {IV}+{T_{\lambda _0}}. \end{aligned}$$
(75)
Step Four: dealing with IV Since
$$\begin{aligned} J\varDelta \left( \varDelta J\right) ^{2}&=\nabla _{p}\left( J\nabla _{p}\left( \varDelta J\right) ^{2}\right) -\nabla _{p}J\nabla _{p}\left( \varDelta J\right) ^{2}\\&=\varDelta \left( J\left( \varDelta J\right) ^{2}\right) -\underline{\nabla _{p}\left( \nabla _{p}J\left( \varDelta J\right) ^{2}\right) } -\nabla _{p}J\nabla _{p}\left( \varDelta J\right) ^{2}\\&=\varDelta \left( J\left( \varDelta J\right) ^{2}\right) -\underline{\nabla _{p}\left( \nabla _{p}J\left( \varDelta J\right) ^{2}\right) } +\left( \varDelta J\right) ^{3} +{T_{\lambda _0}}\\&=\varDelta \left( J\left( \varDelta J\right) ^{2}\right) +\left( \varDelta J\right) ^{3}+{T_{\lambda _0}}, \end{aligned}$$
we have
$$\begin{aligned} \mathbf {IV}&=J\varDelta \left( \varDelta J\right) ^{2}+\varDelta \left( \varDelta J\right) ^{2}J\nonumber \\&=\varDelta \bigg (J\left( \varDelta J\right) ^{2}+\left( \varDelta J\right) ^{2}J\bigg ) +2\left( \varDelta J\right) ^{3} +{T_{\lambda _0}}. \end{aligned}$$
(76)
Step Five: divergence forms of nonlinearity Combining the equalities (72), (74), (75) and (76), we derive that
$$\begin{aligned} 2{\mathbf {I}}+\mathbf {II}+2\mathbf {III}+2\mathbf {IV}+2{\mathbf {V}}=T_{\lambda _0}, \end{aligned}$$
which completes the proof.
