Proof of Theorem 1.1 Because the proof will involve several rather technical arguments, we have preferred to start giving all the geometrical details in the special case that $k=3$ . We believe that this should help the reader to follow the various steps that will be necessary to handle the general case.

Case $\mathbf{k} {\boldsymbol =} \textbf{3}$ .

Using (1.8), we write

(2.5) $$ \begin{align} \tau_3(\varphi)=\bar\Delta^2\tau(\varphi)-{\operatorname{Tr}} R^N(\bar{\Delta}\tau(\varphi),d\varphi(\cdot))d\varphi(\cdot)-{\operatorname{Tr}} R^N(\bar{\nabla}_{(\cdot )}\tau(\varphi),\tau(\varphi))d\varphi(\cdot) . \end{align} $$

We fix the notation as follows:

$$ \begin{align*} \begin{aligned} \tau(\varphi)&=u_0^\alpha \frac{\partial}{\partial y^\alpha}, \\ \bar{\Delta}\tau(\varphi)&= \left (\Delta u_0^\alpha+A_1^\alpha \right )\displaystyle\frac{\partial}{\partial y^\alpha}= u_1^\alpha\displaystyle\frac{\partial}{\partial y^\alpha}, \end{aligned} \end{align*} $$

with

(2.6) $$ \begin{align} A_1^\alpha=A^\alpha \left ( u_0^\vartheta, \frac{\partial u_0^\vartheta}{\partial x^i}\right ), \end{align} $$

where

$$ \begin{align*} A=\left(A^{\alpha}\right):{\mathbb{R}}^n \times({\mathbb{R}}^m\times {\mathbb{R}}^n)\to {\mathbb{R}}^n \end{align*} $$

is defined, for $(\eta ^{\vartheta },\xi ^{\vartheta }_i)\in {\mathbb {R}}^n \times ({\mathbb {R}}^m\times {\mathbb {R}}^n)$ , according to (2.3), by

$$ \begin{align*} \begin{aligned} A^\alpha \left (\eta^{\vartheta},\xi^{\vartheta}_i\right ) &= \xi^{\vartheta}_i\, \left[-2\, g^{ij}\, \varphi_j^\beta\, \Gamma_{\beta \vartheta}^\alpha\right] \\ \nonumber &\quad +\eta^{\vartheta} \left [ (\Delta \varphi^\beta)\, \Gamma_{\beta \vartheta}^\alpha - g^{ij}\, \varphi_j^\beta\, \varphi_i^\omega \; S^{\alpha}_{\beta \omega \vartheta} \right]. \end{aligned} \end{align*} $$

It is important to point out that $A^\alpha $ is linear with respect to $\eta ^{\vartheta }$ and $\xi ^{\vartheta }_i$ .

All of this can be iterated once more and yields

$$ \begin{align*} \bar{\Delta}^2\tau(\varphi)= \left (\Delta u_1^\alpha+A_2^\alpha \right )\frac{\partial}{\partial y^\alpha}, \end{align*} $$

where now, of course,

$$ \begin{align*} A_2^\alpha=A^\alpha \left ( u_1^\vartheta, \frac{\partial u_1^\vartheta}{\partial x^i}\right ). \end{align*} $$

Now, let us assume that $\tau _3(\varphi )=0$ . Using (2.5) and performing a direct computation, we find that this is locally equivalent to the following system of equations:

(2.7) $$ \begin{align} \Delta u_1^\alpha = \left (F^3 \right )^\alpha , \end{align} $$

where

(2.8) $$ \begin{align} \begin{aligned} \left(F^3 \right)^\alpha &=-A^\alpha_2 -\Big [- {\operatorname{Tr}} R^N(\bar{\Delta}\tau(\varphi),d\varphi(\cdot))d\varphi(\cdot)-{\operatorname{Tr}} R^N(\bar{\nabla}_{(\cdot )}\tau(\varphi),\tau(\varphi))d\varphi(\cdot) \Big ]^\alpha\\ &=-A^\alpha_2-g^{ij}\, u_1^\vartheta\, \varphi_i^\beta\, \varphi_j^\omega\, R^{\alpha}_{\omega \beta \vartheta}+g^{ij}\,\displaystyle \frac{\partial u_0^\beta}{\partial x^i}\, u_0^\vartheta\, \varphi_j^\omega \, R^{\alpha}_{\omega \beta \vartheta}\\ &\quad +g^{ij}\, u_0^\sigma\, u_0^\vartheta\, \varphi_i^\gamma\, \varphi_j^\omega\, \Gamma_{\gamma \sigma}^\beta\, R^{\alpha}_{\omega \beta \vartheta}. \end{aligned} \end{align} $$

Now, we define the ${\mathbb {R}}^n$ -valued $1$ -form

$$ \begin{align*} v_0:= du_0=\left (du_0^1,\ldots,du_0^n \right )=\frac{\partial u_0^\alpha}{\partial x^i}\, dx^i \otimes e_\alpha , \end{align*} $$

where $\{e_\alpha \}$ is the canonical basis of ${\mathbb {R}}^n$ and $u_0$ is thought of as an ${\mathbb {R}}^n$ -valued function defined on U. The components of the $1$ -form $v_0$ give rise to an ${\mathbb {R}}^{mn}$ -valued function defined on U:

$$ \begin{align*} v_0= \left ( v_{0i}^\alpha \right )=\left ( \frac{\partial u_0^\alpha}{\partial x^i} \right ) , \qquad 1 \leq i \leq m,\quad 1 \leq \alpha \leq n . \end{align*} $$

For simplicity, we keep the same notation for the $1$ -form and the ${\mathbb {R}}^{mn}$ -valued function. Note that, here, the index i in $v_{0i}^\alpha $ does not mean the derivative with respect to the variable $x^i$ ( $v_{0}^\alpha $ does not even exist!). With this notation, (2.8) can be rewritten as follows:

(2.9) $$ \begin{align} \begin{aligned} \left (F^3 \right )^\alpha &= -A^\alpha_2-g^{ij}\, u_1^\vartheta\, \varphi_i^\beta\, \varphi_j^\omega\, R^{\alpha}_{\omega \beta \vartheta}+g^{ij}\, v_{0i}^\beta\, u_0^\vartheta\, \varphi_j^\omega\, R^{\alpha}_{\omega \beta \vartheta} \\ &\quad +g^{ij}\, u_0^\sigma\, u_0^\vartheta\, \varphi_i^\gamma\, \varphi_j^\omega\, \Gamma_{\gamma \sigma}^\beta\, R^{\alpha}_{\omega \beta \vartheta}. \end{aligned} \end{align} $$

Then, taking into account the definition of $A^\alpha _2$ , we conclude that $F^3=\left (\left (F^3 \right )^\alpha \right )$ depends on $u_0,v_0,u_1$ , and $\{\partial u_1^\alpha / \partial x^i\}_{i,\alpha }$ . Moreover, for future use, we deduce from (2.6) and inspection of (2.9) that there exists $C>0$ such that on D we have

(2.10) $$ \begin{align} \left | F^3 \right | \leq C \left [\sum_{\vartheta}\left ( |u_0^\vartheta | +|u_1^\vartheta |\right )+\sum_{i,\vartheta} |v_{0i}^\vartheta |+\sum_{i,\vartheta} \left |\frac{\partial u_{1}^\vartheta}{\partial x^i} \right | \right ], \end{align} $$

where D is an open subset of M such that its closure is compact and contained in U.

Indeed, in (2.6) and (2.9), it is possible to bound from above by means of a constant any of the functions which appear as a multiplicative coefficient of $u_0,v_0,u_1$ , and $\{\partial u_1^\alpha / \partial x^i\}_{i,\alpha }$ . From this, (2.10) can be obtained easily.

We shall also need to estimate the Laplacian of the ${\mathbb {R}}^{mn}$ -valued function $v_0$ . To this purpose, we perform a computation which gives the following output:

(2.11) $$ \begin{align} \begin{aligned} \left (\Delta v_0\right )_{i}^{\alpha} &= \Delta\displaystyle\frac{\partial u_0^\alpha}{\partial x^i}\\ &= \displaystyle\frac{\partial\left ( \Delta u_0^\alpha \right )}{\partial x^i}+\frac{\partial g^{kj}}{\partial x^i}\displaystyle\frac{\partial^2 u_0^\alpha}{\partial x^k \partial x^j} -\displaystyle\frac{\partial g^{\ell j}}{\partial x^i}\, \Gamma_{\ell j}^k\, \frac{\partial u_0^\alpha}{\partial x^k}-g^{\ell j}\, \displaystyle\frac{\partial \Gamma_{\ell j}^k}{\partial x^i}\displaystyle\frac{\partial u_0^\alpha}{\partial x^k}\\ &=\displaystyle\frac{\partial u_1^\alpha}{\partial x^i}+ \displaystyle\frac{\partial}{\partial x^i} \left\{ 2g^{\ell j}\, v_{0j}^\vartheta\, \varphi^\beta_{\ell}\, \Gamma^\alpha_{\beta \vartheta} -u_0^\vartheta \left [ (\Delta \varphi^\beta)\, \Gamma_{\beta \vartheta}^\alpha -g^{\ell j}\, \varphi_j^\beta\, \varphi_{\ell}^\omega\; S^{\alpha}_{\beta \omega \vartheta}\right] \right \}\\ &\quad{} + \displaystyle\frac{\partial g^{kj}}{\partial x^i} \displaystyle\frac{\partial v_{0k}^\alpha}{\partial x^j}-\displaystyle\frac{\partial g^{\ell j}}{\partial x^i}\, \Gamma_{\ell j}^k\, v_{0k}^\alpha - g^{\ell j}\, \displaystyle\frac{\partial \Gamma_{\ell j}^k}{\partial x^i}\, v_{0k}^\alpha. \end{aligned} \end{align} $$

Now, we call $F^2=\big( (F^2 )_i^\alpha \big)$ the right-hand side of (2.11). Thus, $F^2$ depends on $u_0,v_0,\{\partial v_{0i}^\alpha / \partial x^j\}_{i,j,\alpha }$ , and $\{\partial u_1^\alpha / \partial x^i\}_{i,\alpha }$ , and it is linear in each of them. Then, similarly to (2.10), it is easy to deduce from (2.11) the following estimate on D:

(2.12) $$ \begin{align} \left | F^2 \right | \leq C \left [\sum_{\vartheta} |u_0^\vartheta | + \sum_{i,\vartheta}|v_{0i}^\vartheta| + \sum_{i,\vartheta} \left |\frac{\partial u_{1}^\vartheta}{\partial x^i} \right | + \sum_{i,j,\vartheta} \left |\frac{\partial v_{0i}^\vartheta}{\partial x^j} \right |\right ]\, \end{align} $$

for a suitably large constant $C>0$ .

Finally, we also define the function $F^1=\left (\left (F^1 \right )^\alpha \right )$ as follows:

$$ \begin{align*} \left (F^1\right )^\alpha =u_1^\alpha-A_1^\alpha. \end{align*} $$

We note that $F^1$ depends on $u_0,v_0$ , and $u_1$ , and it is linear in each of them. It follows that there exists $C>0$ such that

(2.13) $$ \begin{align} \left | F^1 \right | \leq C \left [\sum_{\vartheta}\left ( |u_0^\vartheta | + |u_1^\vartheta |\right ) + \sum_{i,\vartheta}|v_{0i}^\vartheta | \right ]. \end{align} $$

The next step is to consider $u_0,v_0,\{\partial v_{0i}^\alpha / \partial x^j\}_{i,j,\alpha },u_1$ , and $\{\partial u_1^\alpha / \partial x^i\}_{i,\alpha }$ not as vector-valued functions defined on U, but as a set of independent variables. More specifically, we define

(2.14) $$ \begin{align} u=\left (u_0,v_0,u_1 \right )\in {\mathbb{R}}^{r_2} \quad \mathrm{and}\quad F=\left ( F^1,F^2,F^3\right ) \colon \mathbb{R}^{r_1}\to \mathbb{R}^{r_2}, \end{align} $$

where $r_1=n+nm + n m^2 + n + nm$ and $r_2=n + n m + n$ . Thus, formally,

$$ \begin{align*} F=F \left ( u, \left (\frac{\partial u^a}{\partial x^i} \right )\right ). \end{align*} $$

Next, if we think of u as an $\mathbb {R}^{n(m+2)}$ -valued function defined on U, by construction, we have $\Delta u = F$ . Now, because, by assumption, $\varphi $ is harmonic on an open subset of D, it is clear that u vanishes on that open subset. Moreover, because we have proved the estimates (2.10), (2.12), and (2.13) on D, we can apply Theorem 2.1. Then, the statement follows precisely by the same globalization argument which was detailed in [Reference Branding and Oniciuc7, Proof of Theorem 1.3]. Thus, the proof of Theorem 1.1 is complete in the special case $k=3$ .

Remark 2.3 In the previous proof, we have worked with the partial derivatives $\{\partial v_{0i}^\alpha / \partial x^j\}_{i,j,\alpha }$ . This choice is the most suitable for the purposes of this paper and, particularly, for the extensions to the case $k \geq 4$ , which will be illustrated below. However, we point out that it is also possible to work with covariant derivatives

$$ \begin{align*} \left \{\nabla_j v_{0i}^\alpha\right \}_{i,j,\alpha}= \left \{ \frac{\partial v_{0i}^\alpha}{\partial x^j}-\Gamma_{ji}^k\, v_{0k}^\alpha \right \}_{i,j,\alpha}. \end{align*} $$

Indeed, the difference

$$ \begin{align*} \nabla_j v_{0i}^\alpha- \frac{\partial v_{0i}^\alpha}{\partial x^j} \end{align*} $$

is linear in $\{ v_{0i}^\alpha \}_{i,\alpha }$ . We also mention that $\nabla _iu^{\alpha }_1=\partial u^{\alpha }_1/\partial x^i$ . Consequently, the basic estimates (2.10), (2.12), and (2.13) still hold, conceptually with the same proofs. In particular, the choice of working with covariant derivatives was adopted in Reference Branding and Oniciuc[7].

Remark 2.4 We point out that, in Reference Branding and Oniciuc[7], the definition of $F^2=\big((F^2)_i^\alpha\big)$ should have been given as

(2.15) $$ \begin{align} F^2= dw - \left (\frac{\partial g^{kj}}{\partial x^i}\frac{\partial v_{0k}^{\alpha}}{\partial x^j}-\left (\frac{\partial g^{\ell j}}{\partial x^i}\, \Gamma_{\ell j}^k+g^{\ell j}\, \frac{\partial \Gamma_{\ell j}^k}{\partial x^i} \right ) v_{0k}^\alpha \right ). \end{align} $$

Note that a different sign convention was used in Reference Branding and Oniciuc[7]. Because the additional term is linear in $\{ v_{0i}^\alpha \}_{i,\alpha }$ and $\{\partial v_{0i}^\alpha / \partial x^j\}_{i,j,\alpha }$ , or $\{\nabla _j v_{0i}^\alpha \}_{i,j,\alpha }$ , the proofs in Reference Branding and Oniciuc[7] can be completed with minor changes.

Case $ \mathbf{k} \boldsymbol {\geq } \textbf {4}$ . First, we provide a short illustration of the case $k=4$ , because this step may help to understand the idea behind the introduction of a suitable set of recursively defined new variables, a fact which is a key point.

We recall from (1.7) that, when $k=4$ , the $4$ -tension field is

(2.16) $$ \begin{align} \begin{aligned} \tau_4(\varphi)&=\bar\Delta^3\tau(\varphi)-{\operatorname{Tr}} R^N(\bar{\Delta}^2\tau(\varphi),d\varphi(\cdot))d\varphi(\cdot)\\ &\quad -{\operatorname{Tr}} R^N(\bar{\nabla}_{(\cdot)}\bar{\Delta}\tau(\varphi),\tau(\varphi))d\varphi(\cdot)+{\operatorname{Tr}} R^N(\bar{\Delta}\tau(\varphi),\bar{\nabla}_{(\cdot)}\tau(\varphi))d\varphi(\cdot) . \end{aligned} \end{align} $$

Now, let us assume that $\tau _4(\varphi )=0$ . First, we set again

$$ \begin{align*} \begin{aligned} \tau(\varphi)&=u_0^\alpha \displaystyle\frac{\partial}{\partial y^\alpha}, \\ \bar{\Delta}\tau(\varphi)&= \left (\Delta u_0^\alpha+A_1^\alpha \right )\displaystyle\frac{\partial}{\partial y^\alpha}= u_1^\alpha\displaystyle\frac{\partial}{\partial y^\alpha}, \end{aligned} \end{align*} $$

where

$$ \begin{align*} A_1^\alpha=A^\alpha \left ( u_0^\vartheta, \frac{\partial u_0^\vartheta}{\partial x^i}\right ) \end{align*} $$

is defined in (2.6). Analogously, we continue with

$$ \begin{align*} \begin{aligned} \bar{\Delta}^2\tau(\varphi)&=\left (\Delta u_1^\alpha+A_2^\alpha \right )\displaystyle\frac{\partial}{\partial y^\alpha}= u_2^\alpha\frac{\partial}{\partial y^\alpha}, \\ \bar{\Delta}^3\tau(\varphi)&= \left (\Delta u_2^\alpha+A_3^\alpha \right )\displaystyle\frac{\partial}{\partial y^\alpha}, \end{aligned} \end{align*} $$

where

$$ \begin{align*} \begin{aligned} A_j^\alpha&=A^\alpha \left ( u_{j-1}^\vartheta, \displaystyle\frac{\partial u_{j-1}^\vartheta}{\partial x^i}\right ), \quad j=1,2,3 . \end{aligned} \end{align*} $$

Now, using (2.16), we find that, in our notation, the assumption $\tau _4(\varphi )=0$ is equivalent to

$$ \begin{align*} \left (\Delta u_2^\alpha \right )\frac{\partial}{\partial y^\alpha}= \left (F^4 \right )^\alpha \frac{\partial}{\partial y^\alpha}, \end{align*} $$

where we have set:

(2.17) $$ \begin{align} \begin{aligned} \left(F^4 \right )^\alpha&=-A_3^\alpha- \Big [-{\operatorname{Tr}} R^N(\bar{\Delta}^2\tau(\varphi),d\varphi(\cdot))d\varphi(\cdot) \\ &\quad -{\operatorname{Tr}} R^N(\bar{\nabla}_{(\cdot)}\bar{\Delta}\tau(\varphi),\tau(\varphi))d\varphi(\cdot)+{\operatorname{Tr}} R^N(\bar{\Delta}\tau(\varphi),\bar{\nabla}_{(\cdot)}\tau(\varphi))d\varphi(\cdot) \Big ]^\alpha . \end{aligned} \end{align} $$

We use the following set of variables:

$$ \begin{align*} \begin{aligned} u_0&= \left (u_0 ^\alpha\right ), \quad v_0:=du_0=\nabla u_0 \,;\\ u_1&= \left (u_1 ^\alpha\right ), \quad v_1:=\nabla u_1 \,;\\ u_2&= \left (u_2 ^\alpha\right ) . \end{aligned} \end{align*} $$

We define

$$ \begin{align*} u=\left ( \begin{array}{l} u_0\\ v_0\\ u_1\\ v_1\\ u_2 \end{array} \right ) \end{align*} $$

and $F=F\left (u_0,v_0,\nabla v_0, u_1,v_1,\nabla v_1,u_2,\nabla u_2\right )$ as follows:

$$ \begin{align*} F=\left ( \begin{array}{l} u_1-A_1\\ d\left (u_1-A_1\right )\\ u_2-A_2\\ d\left (u_2-A_2\right )\\ F^4 \end{array} \right ). \end{align*} $$

Now, $\Delta u = F\, +$ terms linear in $v_0$ and its first derivatives for the second component of F and $+$ terms linear in $v_1$ and its first derivatives for the fourth component as in (2.15). Using the same technique that we employed for the first two vector components of F in the case $k=3$ , now the first two vector components of F and, analogously, the third and fourth vector components, can be estimated to ensure the validity of (2.1). As for $F^4$ , the explicit analysis of (2.17) yields

(2.18) $$ \begin{align} \begin{aligned} \left (F^4 \right )^\alpha=&-A_3^\alpha-g^{ij}u_2^\vartheta \varphi_i^\beta \varphi_j^\omega\, R^\alpha_{\omega \beta \vartheta}\\ &+g^{ij}\,v_{1i}^\beta\, u_0^\vartheta \varphi_j^\omega \, R^\alpha_{\omega \beta \vartheta}\\ &+g^{ij}u_1^\sigma u_0^\vartheta \varphi_i^\gamma\varphi_j^\omega\, \Gamma^\beta_{\gamma \sigma} \,R^\alpha_{\omega \beta \vartheta}\\ &+g^{ij}\,v_{0i}^\beta \,u_1^\vartheta \varphi_j^\omega \, R^\alpha_{\omega \beta \vartheta}\\ &+g^{ij}u_0^\sigma u_1^\vartheta \varphi_i^\gamma\, \varphi_j^\omega\, \Gamma^\beta_{\gamma \sigma} \, R^\alpha_{\omega \beta \vartheta}\end{aligned} \end{align} $$

from which (2.1) follows easily. This ends the case $k=4$ . The general case can be handled similarly. Indeed, for any fixed value $k \geq 4$ , we have recursively defined functions $u_i=\left ( u_i^\alpha \right )$ by means of

(2.19) $$ \begin{align} \bar{\Delta}^{i+1}\tau(\varphi)= \left (\Delta u_{i}^\alpha+A_{i+1}^\alpha \right )\frac{\partial}{\partial y^\alpha}= u_{i+1}^\alpha\frac{\partial}{\partial y^\alpha}, \end{align} $$

for $0 \leq i \leq k-2$ , and also $A_{k-1}^\alpha $ is defined. Then, we introduce the vector-valued function

(2.20) $$ \begin{align} u=\left ( \begin{array}{l} u_0\\ v_0\\ u_1\\ v_1\\ \vdots \\ v_{k-3}\\ u_{k-2} \end{array} \right ), \end{align} $$

where $v_i= \nabla u_i$ , $0 \leq i \leq k-3$ . Note that we do not introduce $v_{k-2}$ in (2.20). Then, we define

$$ \begin{align*}F=F\left (u_0,v_0,\nabla v_0,\ldots, u_{k-3},v_{k-3},\nabla v_{k-3},u_{k-2},\nabla u_{k-2}\right )\end{align*} $$

as follows:

$$ \begin{align*} F=\left ( \begin{array}{l} u_1-A_1\\ d\left (u_1-A_1\right )\\ u_2-A_2\\ d\left (u_2-A_2\right )\\ \vdots\\ d\left (u_{k-2}-A_{k-2}\right )\\ F^k \end{array} \right ) , \end{align*} $$

with, similarly to (2.17), $\left (F^k\right )^\alpha =-A_{k-1}^\alpha -[\,\ldots \,]^\alpha $ . More precisely, here $[\,\ldots \,]$ is the right-hand side of (1.7) or (1.8) without the first term. Again, by construction,

$$ \begin{align*}\Delta u =F+\text{terms linear in~} v_j,\nabla v_j,\quad j=0,\ldots,\, k-3, \end{align*} $$

and it is easy to see that the first $2(k-2)$ vector components of F can be estimated in such a way that (2.1) holds. Finally, direct inspection of (1.7) or (1.8) (compare with (2.18)) shows that also $F^k$ can be estimated, so that (2.1) is verified. So the proof ends by application of Theorem 2.1.▪

Proof of Theorem 1.2 Let $\varphi , \tilde {\varphi }$ be two k-harmonic maps which coincide on an open subset. To simplify the notation, we shall also denote by $\varphi , \tilde {\varphi }$ the vector-valued functions which represent $\varphi , \tilde {\varphi }$ in local charts:

$$ \begin{align*} \varphi= \left(\varphi^1,\ldots,\varphi^n \right ) \,; \quad \tilde \varphi= \left(\tilde\varphi^1,\ldots,\tilde\varphi^n \right ) . \end{align*} $$

We define a vector-valued function u for the map $\varphi $ as follows:

(2.21) $$ \begin{align} u=\left ( \begin{array}{l} \varphi\\ d \varphi \\ u_0\\ v_0\\ u_1\\ v_1\\ \vdots \\ v_{k-3}\\ u_{k-2} \end{array} \right ), \end{align} $$

where $u_0,\ldots ,u_{k-2}$ and $v_0,\ldots ,v_{k-3}$ are defined as in the proof of Theorem 1.1. We also need to introduce the analogous vector-valued function associated with $\tilde {\varphi }$ , i.e.,

(2.22) $$ \begin{align} \tilde u=\left ( \begin{array}{l} \tilde \varphi\\ d \tilde\varphi \\ \tilde u_0\\ \tilde v_0\\ \tilde u_1\\ \tilde v_1\\ \vdots \\ \tilde v_{k-3}\\ \tilde u_{k-2} \end{array} \right ). \end{align} $$

Note that, in contrast to the proof of Theorem 1.1, the functions (2.21) and (2.22) also contain $\varphi $ , $\tilde {\varphi }$ and their first derivatives.

The proof of Theorem 1.2 amounts to showing that we can apply Aronszajn’s theorem, that is, Theorem 2.1, to the vector-valued function

$$ \begin{align*}z=u-\tilde u,\end{align*} $$

with $u, \,\tilde u$ defined in (2.21) and (2.22), respectively. As in the proof of Theorem 1.1, the following functions are also defined:

$$ \begin{align*} \begin{array}{l} A_i= \left (A_i ^\alpha\right ), \quad \tilde A_i= \left (\tilde A_i ^\alpha\right )\,\quad (1 \leq i \leq k-2), \\ \left (F^k\right )^\alpha=-A_{k-1}^\alpha-[\,\ldots\,]^\alpha, \quad \left (\tilde F^k\right )^\alpha=-\tilde A_{k-1}^\alpha-[\,\tilde{\ldots}\,]^\alpha. \end{array} \end{align*} $$

Next, we define G as follows:

(2.23) $$ \begin{align} G=\left ( \begin{array}{l} \Delta \left ( \varphi- \tilde \varphi \right )\\ \Delta \left ( d\varphi- d\tilde \varphi \right )\\ \left (u_1-A_1\right )-\left (\tilde u_1- \tilde A_1\right )\\ d\left (u_1-A_1\right )- d\left (\tilde u_1- \tilde A_1\right )\\ \left (u_2-A_2\right )-\left (\tilde u_2- \tilde A_2\right )\\ d\left (u_2-A_2\right )- d\left (\tilde u_2- \tilde A_2\right )\\ \vdots\\ d\left (u_{k-2}-A_{k-2}\right )-d\left (\tilde u_{k-2}- \tilde A_{k-2}\right )\\ F^k -\tilde F ^k \end{array} \right ) . \end{align} $$

Now, by construction, we have

$$ \begin{align*} \Delta z = G+\text{~terms linear in~} (v_j-\tilde v_j) ,(\nabla v_j-\nabla \tilde v_j), \quad j=0,\ldots, k-3. \end{align*} $$

Note that there are three different blocks in the definition of G. The first two rows only contain the Laplacian applied to $\varphi $ and its first partial derivatives, and we will explain in more detail below how one should think of $d\varphi $ in the definition of G. After that, we always have pairs of $u_j-A_j$ and its first derivatives, with $1\leq j\leq k-2$ . In the last row, we have the right-hand side of the polyharmonic map equations (1.7) and (1.8), denoted by $F^k$ . Note that this is the only place in (2.23) where the Euler–Lagrange equation for polyharmonic maps enters.

Our aim is now to apply the theorem of Aronszajn, that is, Theorem 2.1, to z. Hence, we have to estimate $\Delta z$ (equivalently, G) in terms of z and its first partial derivatives.

In the following, C will always represent a positive constant whose value may change from line to line.

To estimate the first row in (2.23), we use the following.

Lemma 2.5 Let $\varphi ,\tilde \varphi $ be two maps with corresponding variables (2.21) and (2.22). Then, the following estimate holds:

(2.24) $$ \begin{align} |\Delta (\varphi-\tilde\varphi)|\leq C(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|u_0-\tilde u_0|). \end{align} $$

Proof Recall that

$$ \begin{align*} \Delta\varphi^\alpha=-u_0^\alpha+\langle d\varphi^\beta,d\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}, \end{align*} $$

where, here and below, we have shortened the notation denoting

$$ \begin{align*} \langle d\varphi^\beta,d\varphi^\gamma\rangle=g^{ij}\, \frac{\partial\varphi^\beta}{\partial x^i} \frac{\partial\varphi^\gamma}{\partial x^j}. \end{align*} $$

We can rewrite

$$ \begin{align*} \Delta(\varphi^\alpha-\tilde\varphi^\alpha)=&-u_0^\alpha+\tilde u_0^\alpha+\langle d\varphi^\beta,d\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi) -\langle d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\tilde\varphi) \\ =&-u_0^\alpha+\tilde u_0^\alpha \\ &+\langle d\varphi^\beta-d\tilde\varphi^\beta,d\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi) +\langle d\tilde\varphi^\beta,d\varphi^\gamma-d\tilde\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi) \\ &+\langle d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\big(\Gamma^\alpha_{\beta\gamma}(\varphi)-\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)\big). \end{align*} $$

The first three terms on the right-hand side can be estimated directly. To estimate the difference of the Christoffel symbols, we make use of the mean-value inequality; for more details, we refer to the discussion before Lemma 2.6 in Reference Branding and Oniciuc[7]. The proof is now complete.▪

In the following, we will often apply the mean-value inequality without explicitly mentioning it.

As a second step, we estimate the second line of (2.23).

Lemma 2.6 Let $\varphi ,\tilde \varphi $ be two maps with corresponding variables (2.21) and (2.22). Then, the following estimate holds:

(2.25) $$ \begin{align} |\Delta (d\varphi-\tilde d\varphi)|\leq C\big(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|\nabla d\varphi-\nabla d\tilde\varphi|+|v_0-\tilde v_0|\big). \end{align} $$

Proof Recall that we use $d\varphi ,d\tilde \varphi $ in (2.23) to represent the partial derivatives of $\varphi $ and $\tilde \varphi $ . Hence, when we apply the Laplacian to the second line in (2.23), we will get some correction terms as already computed in (2.11); see also Remark 2.3.

Now, let us consider our two maps $\varphi ,\tilde \varphi $ : by combining (2.11) and using that $\Delta \varphi ^\alpha =-u_0^\alpha +\langle d\varphi ^\beta ,d\varphi ^\gamma \rangle \Gamma ^\alpha _{\beta \gamma }$ , we find

$$ \begin{align*} \Delta(\varphi^\alpha_i-\tilde\varphi^\alpha_i)=& -v^\alpha_{0i}+\tilde v^\alpha_{0i} +2\langle\nabla_i d\varphi^\beta,d\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi) -2\langle\nabla_i d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)\\ &+\langle d\varphi^\beta,d\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\varphi)}{\partial y^\delta}\varphi^\delta_i -\langle d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)}{\partial y^\delta} \tilde\varphi^\delta_i\\ &+\frac{\partial g^{k\ell}}{\partial x^i}\left(\frac{\partial\varphi_k^\alpha}{\partial x^\ell}-\frac{\partial\tilde\varphi_k^\alpha}{\partial x^\ell}\right) -\left(\frac{\partial g^{k\ell}}{\partial x^i}\Gamma^j_{k\ell} +g^{k\ell}\frac{\partial\Gamma^j_{k\ell}}{\partial x^i}\right)\big(\varphi^\alpha_j-\tilde\varphi^\alpha_j\big), \end{align*} $$

where, in the last line, for clarity, we have also added the linear terms. In order to estimate the second term on the right-hand side, we rewrite

$$ \begin{align*} \langle\nabla_i d\varphi^\beta,d\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi)& -\langle\nabla_i d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\tilde\varphi) \\ =&\langle\nabla_i d\varphi^\beta-\nabla_i d\tilde\varphi^\beta,d\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi) +\langle\nabla_i d\tilde\varphi^\beta,d\varphi^\gamma-d\tilde\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi) \\ &+\langle\nabla_i d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle \big(\Gamma^\alpha_{\beta\gamma}(\varphi)-\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)\big). \end{align*} $$

Then, it is easy to estimate

$$ \begin{align*} &|\langle\nabla_i d\varphi^\beta,d\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\varphi) -\langle\nabla_i d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)|\\ &\quad\leq C(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|\nabla d\varphi-\nabla d\tilde\varphi|). \end{align*} $$

Again, we rewrite

$$ \begin{align*} \langle d\varphi^\beta,d\varphi^\gamma\rangle&\frac{\partial\Gamma^\alpha_{\beta\gamma}(\varphi)}{\partial y^\delta}\varphi^\delta_i -\langle d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)}{\partial y^\delta}\tilde\varphi^\delta_i \\ =&\langle d\varphi^\beta-d\tilde\varphi^\beta,d\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\varphi)}{\partial y^\delta}\varphi^\delta_i +\langle d\tilde\varphi^\beta,d\varphi^\gamma-d\tilde\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\varphi)}{\partial y^\delta}\varphi^\delta_i\\ &+\langle d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle \left(\frac{\partial\Gamma^\alpha_{\beta\gamma}(\varphi)}{\partial y^\delta}-\frac{\partial\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)}{\partial y^\delta}\right)\varphi^\delta_i +\langle d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)}{\partial y^\delta} (\varphi^\delta_i-\tilde\varphi^\delta_i). \end{align*} $$

We deduce the estimate

$$ \begin{align*} |\langle d\varphi^\beta,d\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\varphi)}{\partial y^\delta}\varphi^\delta_i -\langle d\tilde\varphi^\beta,d\tilde\varphi^\gamma\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}(\tilde\varphi)}{\partial y^\delta}\tilde\varphi^\delta_i| \leq C(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|). \end{align*} $$

The claim now follows by combining the equations.▪

In the following two lemmata, we will estimate the pairs $u_{j+1}-A_{j+1}$ and their derivatives which are in the middle block of (2.23).

Lemma 2.7 Let $\varphi ,\tilde \varphi $ be two maps with corresponding variables (2.21) and (2.22). Assume that $0 \leq j\leq k-3$ . Then, the following estimate holds:

(2.26) $$ \begin{align} \big|\big(u_{j+1}&-A_{j+1}\big)-\big(\tilde u_{j+1}-\tilde A_{j+1}\big)\big| \\ \nonumber&\leq C(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|u_0-\tilde u_0|+|u_j-\tilde u_j|+|u_{j+1}-\tilde u_{j+1}|+|v_j-\tilde v_j|). \end{align} $$

Proof From (2.6), using $\Delta \varphi ^\alpha =-u_0^\alpha +\langle d\varphi ^\beta ,d\varphi ^\gamma \rangle \Gamma ^\alpha _{\beta \gamma }$ and recalling

$$ \begin{align*} A_{j+1}^\alpha=A^\alpha \left ( u_{j}^\vartheta, \frac{\partial u_{j}^\vartheta}{\partial x^i}\right ), \end{align*} $$

we can write

$$ \begin{align*} A_{j+1}^\alpha=& -2\langle du_j^\nu,d\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu} -u_j^\nu u_0^\beta\Gamma_{\beta\nu}^\alpha +u_j^\nu\langle d\varphi^\vartheta,d\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha, \end{align*} $$

where

$$ \begin{align*} C_{\vartheta\sigma\nu}^\alpha:= \Gamma^\mu_{\vartheta\sigma}\Gamma^\alpha_{\mu\nu} -S^\alpha_{\vartheta\sigma\nu}, \end{align*} $$

with S defined in (2.4). Hence, for two maps $\varphi ,\tilde \varphi $ , we get

$$ \begin{align*} A_{j+1}^\alpha-\tilde A_{j+1}^\alpha=& -2\langle du_j^\nu,d\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\varphi) +2\langle d\tilde u_j^\nu,d\tilde\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\tilde\varphi) \\ &-u_j^\nu u_0^\beta\Gamma_{\beta\nu}^\alpha(\varphi) +\tilde u_j^\nu \tilde u_0^\beta\Gamma_{\beta\nu}^\alpha(\tilde\varphi)\\ &+u_j^\nu\langle d\varphi^\vartheta,d\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\varphi) -\tilde u_j^\nu\langle d\tilde\varphi^\vartheta,d\tilde\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\tilde\varphi). \end{align*} $$

Now, we rewrite

$$ \begin{align*} &-\langle du_j^\nu,d\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\varphi) +\langle d\tilde u_j^\nu,d\tilde\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\tilde\varphi) \\ &\quad=-\langle du_j^\nu-d\tilde u_j^\nu,d\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\varphi) -\langle d\tilde u_j^\nu,d\varphi^\beta-d\tilde\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\varphi)\\ &\qquad-\langle d\tilde u_j^\nu,d\tilde\varphi^\beta\rangle \big(\Gamma^\alpha_{\beta\nu}(\varphi)-\Gamma^\alpha_{\beta\nu}(\tilde\varphi)\big). \end{align*} $$

This gives the estimate

$$ \begin{align*} |\langle du_j^\nu,d\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\varphi) -\langle d\tilde u_j^\nu,d\tilde\varphi^\beta\rangle\Gamma^\alpha_{\beta\nu}(\tilde\varphi)| \leq C(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|du_j-d\tilde u_j|). \end{align*} $$

Again, we rewrite

$$ \begin{align*} -u_j^\nu u_0^\beta\Gamma_{\beta\nu}^\alpha(\varphi) +\tilde u_j^\nu \tilde u_0^\beta\Gamma_{\beta\nu}^\alpha(\tilde\varphi) =&-(u_j^\nu-\tilde u_j^\nu) u_0^\beta\Gamma_{\beta\nu}^\alpha(\varphi) -\tilde u_j^\nu (u_0^\beta-\tilde u_0^\beta)\Gamma_{\beta\nu}^\alpha(\varphi) \\ &-\tilde u_j^\nu\tilde u_0^\beta\big(\Gamma_{\beta\nu}^\alpha(\varphi)-\Gamma_{\beta\nu}^\alpha(\tilde\varphi)\big) \end{align*} $$

and estimate

$$ \begin{align*} |u_j^\nu u_0^\beta\Gamma_{\beta\nu}^\alpha(\varphi) -\tilde u_j^\nu \tilde u_0^\beta\Gamma_{\beta\nu}^\alpha(\tilde\varphi)| \leq C(|\varphi-\tilde\varphi|+|u_0-\tilde u_0|+|u_j-\tilde u_j|). \end{align*} $$

Finally, we rewrite

$$ \begin{align*} u_j^\nu\langle d\varphi^\vartheta,&d\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\varphi) -\tilde u_j^\nu\langle d\tilde\varphi^\vartheta,d\tilde\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\tilde\varphi)\\ =&(u_j^\nu-\tilde u_j^\nu)\langle d\varphi^\vartheta,d\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\varphi) +\tilde u_j^\nu\langle d\varphi^\vartheta-d\tilde\varphi^\vartheta,d\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\varphi) \\ &+\tilde u_j^\nu\langle d\tilde\varphi^\vartheta,d\varphi^\sigma-d\tilde\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\varphi) +\tilde u_j^\nu\langle d\tilde\varphi^\vartheta,d\tilde\varphi^\sigma\rangle \big(C_{\vartheta\sigma\nu}^\alpha(\varphi)- C_{\vartheta\sigma\nu}^\alpha(\tilde\varphi)\big). \end{align*} $$

Hence, we find the estimate

$$ \begin{align*} |u_j^\nu\langle d\varphi^\vartheta,&d\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\varphi) \!-\!\tilde u_j^\nu\langle d\tilde\varphi^\vartheta,d\tilde\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^\alpha(\tilde\varphi)| \leq C(|\varphi-\tilde\varphi|\!+\!|d\varphi- d\tilde\varphi|+|u_j-\tilde u_j|). \end{align*} $$

This completes the proof.▪

Lemma 2.8 Let $\varphi ,\tilde \varphi $ be two maps with corresponding variables (2.21) and (2.22). Assume that $0 \leq j\leq k-3$ . Then, the following estimate holds:

(2.27) $$ \begin{align} \big|d\big(u_{j+1}&-A_{j+1}\big)-d\big(\tilde u_{j+1}-\tilde A_{j+1}\big)\big| \\ \nonumber\leq & C\big(|\varphi-\tilde\varphi|+|d\varphi-\tilde d\varphi|+|\nabla d\varphi-\nabla d\tilde\varphi| +|u_0-\tilde u_0|+|v_0-\tilde v_0|+|u_j-\tilde u_j| \\ \nonumber&+|v_j-\tilde v_j|+|\nabla v_j-\nabla\tilde v_j|+|u_{j+1}-\tilde u_{j+1}|\big). \end{align} $$

Proof By a direct calculation, we find

$$ \begin{align*} \nabla_i A^\alpha_{j+1}=&-2\langle\nabla_iv_j^\gamma,d\varphi^\beta\rangle\Gamma^\alpha_{\beta\gamma} -2\langle v_j^\gamma,\nabla_i d\varphi^\beta\rangle\Gamma^\alpha_{\beta\gamma} -2\langle v_j^\gamma,d\varphi^\beta\rangle\frac{\partial\Gamma^\alpha_{\beta\gamma}}{\partial y^\delta}\varphi^\delta_i \\ &-\nabla_iu_j^\gamma\, u_0^\beta\,\Gamma^\alpha_{\beta\gamma}-u_j^\gamma \,\nabla_iu_0^\beta\,\Gamma^\alpha_{\beta\gamma} -u_j^\gamma\, u_0^\beta\,\frac{\partial\Gamma^\alpha_{\beta\gamma}}{\partial y^\delta}\varphi^\delta_i \\ &+\nabla_i u^\nu_j\langle d\varphi^\vartheta,d\varphi^\sigma\rangle C^\alpha_{\vartheta\sigma\nu} + u^\nu_j\langle\nabla_i d\varphi^\vartheta,d\varphi^\sigma\rangle C^\alpha_{\vartheta\sigma\nu} + u^\nu_j\langle d\varphi^\vartheta,\nabla_id\varphi^\sigma\rangle C^\alpha_{\vartheta\sigma\nu} \\ &+ u^\nu_j\langle d\varphi^\vartheta,d\varphi^\sigma\rangle \frac{\partial C^\alpha_{\vartheta\sigma\nu}}{\partial y^\delta}\varphi^\delta_i. \end{align*} $$

Again, we have to be careful when applying the Laplacian to $v_j=\nabla u_j$ , as the Laplacian does not commute with partial derivatives, and we get several extra terms as demonstrated in (2.11). However, all these terms on the right-hand side can be easily estimated in terms of $v_r,\nabla v_r$ .

The statement of the lemma can now be derived as in the previous lemmata.▪

Finally, we estimate the contribution in (2.23) originating from the polyharmonic map equation. We only consider the case of polyharmonic maps of even order with $k=2s\geq 4$ , as the odd case follows by exactly the same arguments.

Lemma 2.9 Let $\varphi \colon M\to N$ be a polyharmonic map of even order $k=2s\geq 4$ with corresponding variables (2.21). Then,

(2.28) $$ \begin{align} (F^{2s})^\alpha=&-A^\alpha_{2s-1} -u^\delta_{2s-2}\langle d\varphi^\gamma,d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta} \\ \nonumber&+\sum_{\ell=1}^{s-1}\big(u^\delta_{s-\ell-1}\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta} +u^\vartheta_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta} \\ \nonumber &+u^\delta_{s+\ell-2}\langle v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}\big), \end{align} $$

where $E^\alpha _{\beta \delta \vartheta \eta }:=R^\alpha _{\beta \gamma \delta }\Gamma ^\gamma _{\vartheta \eta }{+R^\alpha _{\beta \gamma \vartheta }\Gamma ^\gamma _{\delta \eta }}$ .

Proof This follows directly from the Euler–Lagrange equation (1.7) using

$$ \begin{align*} \big(R^N(\bar\nabla_{e_j}\bar\Delta^{s+\ell-2}\tau(\varphi),\Delta^{s-\ell-1}\tau(\varphi))d\varphi(e_i)\big)^\alpha =&u^\delta_{s-\ell-1}\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta} \\ &+u^\vartheta_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}\Gamma^\gamma_{\vartheta\eta}. \end{align*} $$

▪

Lemma 2.10 Suppose $\varphi ,\tilde \varphi $ are two polyharmonic maps with corresponding variables (2.21) and (2.22). Then, the following estimate holds:

(2.29) $$ \begin{align} &|(F^k)^\alpha-(\tilde F^k)^\alpha|\nonumber\\ &\leq C\left(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+\sum_{\ell=0}^{k-2}|u_\ell-\tilde u_\ell|+\sum_{\ell=0}^{k-3}|v_\ell-\tilde v_\ell| +|\nabla u_{k-2}-\nabla\tilde u_{k-2}| \right). \end{align} $$

Proof Suppose we have two polyharmonic maps $\varphi ,\tilde \varphi $ of order $k=2s$ . Then, from (2.28), we get

$$ \begin{align*} &(F^{2s})^\alpha-(\tilde F^{2s})^\alpha\\ &\quad =-A^\alpha_{2s-1}+\tilde A^\alpha_{2s-1} \\ &\qquad-u^\delta_{2s-2}\langle d\varphi^\gamma,d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) +\tilde u^\delta_{2s-2}\langle d\tilde\varphi^\gamma,d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi) \\ &\qquad-\sum_{\ell=1}^{s-1}\big(u^\delta_{s-\ell-1}\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) -\tilde u^\delta_{s-\ell-1}\langle \tilde v^\gamma_{s+\ell-2},d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi) \big) \\ &\qquad-\sum_{\ell=1}^{s-1}\big(u^\vartheta_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\varphi) -\tilde u^\vartheta_{s+\ell-2}\tilde u^\delta_{s-\ell-1}\langle d\tilde\varphi^\eta,d\tilde\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\tilde\varphi) \big) \\ &\qquad-\sum_{\ell=1}^{s-1}\big(u^\delta_{s+\ell-2}\langle v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) -\tilde u^\delta_{s+\ell-2}\langle \tilde v^\gamma_{s-\ell-1},d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi)\big). \end{align*} $$

In order to estimate the first term on the right-hand side, we use (2.26). It is straightforward to estimate

$$ \begin{align*} &|u^\delta_{2s-2}\langle d\varphi^\gamma,d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) -\tilde u^\delta_{2s-2}\langle d\tilde\varphi^\gamma,d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi)|\\ &\quad\leq C(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|u_{2s-2}-\tilde u_{2s-2}|), \end{align*} $$

which controls the second term on the right-hand side. Concerning the first term inside the sum, we rewrite

$$ \begin{align*} &u^\delta_{s-\ell-1}\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) -\tilde u^\delta_{s-\ell-1}\langle \tilde v^\gamma_{s+\ell-2},d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi) \\ &\kern6pt =(u^\delta_{s-\ell-1} -\tilde u^\delta_{s-\ell-1})\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) +\tilde u^\delta_{s-\ell-1}\langle v^\gamma_{s+\ell-2}-\tilde v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) \\ &\quad+\tilde u^\delta_{s-\ell-1}\langle\tilde v^\gamma_{s+\ell-2},d\varphi^\beta-\tilde d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) +\tilde u^\delta_{s-\ell-1}\langle\tilde v^\gamma_{s+\ell-2},\tilde d\varphi^\beta\rangle \big(R^\alpha_{\beta\gamma\delta}(\varphi)-R^\alpha_{\beta\gamma\delta}(\tilde\varphi)\big). \end{align*} $$

Hence, we deduce the estimate

$$ \begin{align*} |u^\delta_{s-\ell-1}&\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) -\tilde u^\delta_{s-\ell-1}\langle \tilde v^\gamma_{s+\ell-2},d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi)| \\ &\leq C(|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|u_{s-\ell-1}-\tilde u_{s-\ell-1}| +|v_{s+\ell-2}-\tilde v_{s+\ell-2}|). \end{align*} $$

Regarding the second term in the sum, we get

$$ \begin{align*} &u^\vartheta_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\varphi) -\tilde u^\vartheta_{s+\ell-2}\tilde u^\delta_{s-\ell-1}\langle d\tilde\varphi^\eta,d\tilde\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\tilde\varphi) \\ &\quad=(u^\vartheta_{s+\ell-2}-\tilde u^\vartheta_{s+\ell-2})u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\varphi) \\ &\qquad+\tilde u^\vartheta_{s+\ell-2}(u^\delta_{s-\ell-1}-\tilde u^\delta_{s-\ell-1})\langle d\varphi^\eta,d\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\varphi)\\ &\qquad+\tilde u^\vartheta_{s+\ell-2}\tilde u^\delta_{s-\ell-1}\langle d\varphi^\eta-d\tilde\varphi^\eta,d\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\varphi)\\ &\qquad+\tilde u^\vartheta_{s+\ell-2}\tilde u^\delta_{s-\ell-1}\langle d\tilde\varphi^\eta,d\varphi^\beta-d\tilde\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\varphi)\\ &\qquad+\tilde u^\vartheta_{s+\ell-2}\tilde u^\delta_{s-\ell-1}\langle d\tilde\varphi^\eta,d\tilde\varphi^\beta\rangle \big(E^\alpha_{\beta\delta\vartheta\eta}(\varphi) - E^\alpha_{\beta\delta\vartheta\eta}(\tilde\varphi)\big) \end{align*} $$

and obtain the estimate

$$ \begin{align*} |u^\vartheta_{s+\ell-2}&u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\varphi) -\tilde u^\vartheta_{s+\ell-2}\tilde u^\delta_{s-\ell-1}\langle d\tilde\varphi^\eta,d\tilde\varphi^\beta\rangle E^\alpha_{\beta\delta\vartheta\eta}(\tilde\varphi)| \\ &\leq C\big(|u_{s+\ell-2}-\tilde u_{s+\ell-2}|+|u_{s-\ell-1}-\tilde u_{s-\ell-1}|+|d\varphi-d\tilde\varphi|+|\varphi-\tilde\varphi|\big). \end{align*} $$

The last term in the sum may be rewritten as

$$ \begin{align*} &u^\delta_{s+\ell-2}\langle v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) -\tilde u^\delta_{s+\ell-2}\langle \tilde v^\gamma_{s-\ell-1},d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi) \\ &\kern6pt =(u^\delta_{s+\ell-2} -\tilde u^\delta_{s+\ell-2})\langle v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) +\tilde u^\delta_{s+\ell-2}\langle v^\gamma_{s-\ell-1} -\tilde v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) \\ &\quad+\tilde u^\delta_{s+\ell-2}\langle \tilde v^\gamma_{s-\ell-1},d\varphi^\beta -d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) +\tilde u^\delta_{s+\ell-2}\langle \tilde v^\gamma_{s-\ell-1},d\tilde\varphi^\beta\rangle \big(R^\alpha_{\beta\gamma\delta}(\varphi)-R^\alpha_{\beta\gamma\delta}(\tilde\varphi)\big), \end{align*} $$

which leads us to the estimate

$$ \begin{align*} |u^\delta_{s+\ell-2}&\langle v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\varphi) -\tilde u^\delta_{s+\ell-2}\langle \tilde v^\gamma_{s-\ell-1},d\tilde\varphi^\beta\rangle R^\alpha_{\beta\gamma\delta}(\tilde\varphi)| \\ &\leq C\big(|u_{s+\ell-2}-\tilde u_{s+\ell-2}|+|v_{s-\ell-1}-\tilde v_{s-\ell-1}| +|d\varphi-d\tilde\varphi|+|\varphi-\tilde\varphi|\big). \end{align*} $$

The claim now follows by also using (2.26), where now we have to write $|\nabla u_{2s-2}- \nabla \tilde u_{2s-2}|$ instead of $|v_{2s-2}- \tilde v_{2s-2}|$ , because these last functions do not appear in the definition of z.▪

Combining the inequalities (2.24)–(2.27) and (2.29), we arrive at

(2.30) $$ \begin{align} |\Delta z|\leq C\big(&|\varphi-\tilde\varphi|+|d\varphi-d\tilde\varphi|+|\nabla d\varphi-\nabla d\tilde\varphi| \\ \nonumber&+\sum_{\ell=0}^{k-2}|u_\ell-\tilde u_\ell| +\sum_{\ell=0}^{k-3}(|v_\ell-\tilde v_\ell|+|\nabla v_\ell-\nabla\tilde v_\ell|) +|\nabla u_{k-2}-\nabla\tilde u_{k-2}| \big). \end{align} $$

Due to the estimate (2.30), the assumptions of Theorem 2.1 are satisfied, and we can conclude that $z=0$ , which, in particular, implies that $\varphi =\tilde \varphi $ . To complete the proof, we make use of the same globalization argument which was employed in [Reference Branding and Oniciuc7, Proof of Theorem 1.3].▪

Proof of Theorem 1.3 The proof is again based on Aronszajn’s Theorem 2.1 and the explicit expressions of the Christoffel symbols on $\mathbb {S}^n$ .

Let $\mathbb {S}^n$ be the Euclidean unit sphere and denote by $\mathcal N$ and $\mathcal S$ the north and south poles, respectively. It is well known that

$$ \begin{align*} \mathbb{S}^n\setminus\{\mathcal N, \mathcal S\}=\left(\mathbb{S}^{n-1}\times (0,\pi),\sin^2 s \cdot g_{\mathbb{S}^{n-1}}+ds^2\right). \end{align*} $$

Let $(y^a)$ be local coordinates on $\mathbb {S}^{n-1},a=1,\ldots ,n-1$ . Then, $y=(y^1,\ldots ,y^{n-1},{y^n=s})=(\tilde {y},s)$ are local coordinates on $\mathbb {S}^n\setminus \{\mathcal N,\mathcal S\}$ .

In this geometric setup, the Christoffel symbols on $\mathbb {S}^n$ are given by

(2.31) $$ \begin{align} \begin{aligned} \Gamma^a_{bc}(y)=&\tilde\Gamma^a_{bc}(\tilde{y}),\quad a,b,c=1,\ldots,n-1,\\ \Gamma^n_{bc}(y)=&-\sin s\cos s \ \tilde g_{bc}(\tilde{y})=-\displaystyle\frac{1}{2}\sin(2s) \ \tilde g_{bc}(\tilde{y}),\\ \Gamma^a_{nn}(y)=&\Gamma^n_{nn}(y)=\Gamma^n_{bn}(y)=0,\\ \Gamma^a_{bn}(y)=&\displaystyle\frac{\cos s}{\sin s}\delta^a_b,\\ \end{aligned} \end{align} $$

where we use a “ $\,\tilde{}\,$ ” to indicate objects on $\mathbb {S}^{n-1}$ . With respect to the local coordinates $y=(\tilde {y},s)$ on $\mathbb {S}^{n}$ , the equator $\mathbb {S}^{n-1}$ is given by the equation $s=\pi /2$ .

In the special case when the target manifold is $N={\mathbb {S}}^n$ , we know that

(2.32) $$ \begin{align} R(X,Y)Z= -\langle X,Z \rangle Y+\langle Y,Z \rangle X, \quad \forall \,X,Y,Z \in C(T{\mathbb{S}}^n) . \end{align} $$

Next, using (2.32), we express $R^\alpha _{\delta \beta \gamma }$ with respect to our local coordinates $(\tilde {y},s)$ . For our purposes, we only need to compute explicitly $R^n_{\delta \beta \gamma }$ . Using (1.10) and (2.32), we find that the only nonzero terms of this type are

(2.33) $$ \begin{align} - R^n_{ a n b}= R^n_{ a b n}=\langle R\left (\frac{\partial}{\partial y^b},\frac{\partial}{\partial s} \right ) \frac{\partial}{\partial y^a},\frac{\partial}{\partial s} \rangle=-(\sin s)^2\,\tilde{g}_{b a}(\tilde{y}). \end{align} $$

Now, let $(U,x^i)$ be a local chart on M and denote the domain of the above local coordinates on $\mathbb {S}^n\setminus \{\mathcal N,\mathcal S\}$ by V. In addition, we assume that $\varphi (U)\subset V$ .

For simplicity, we again denote the expression of $\varphi $ in local coordinates also by $\varphi $ , i.e.,

$$ \begin{align*} \varphi=(\varphi^1,\ldots,\varphi^n). \end{align*} $$

Assume that W is an open subset of U and $\varphi (W)\subset \mathbb {S}^{n-1}$ , i.e., $\varphi (W)\subset \mathbb {S}^{n-1}\cap V$ . Hence, in W, we have $\varphi ^n=\frac {\pi }{2}$ . Now, define $f\colon U\to {\mathbb {R}}, f:=\varphi ^n-\frac {\pi }{2}$ . Clearly, the function f vanishes when restricted to W.

Let D be an open subset of U such that its closure in M is compact and included in U, and $W\subset D\subset U$ .

We define a vector-valued function y for the nth component of the map $\varphi $ as follows:

(2.34) $$ \begin{align} y=\left ( \begin{array}{l} f\\ df \\ u_0^n\\ v_0^n\\ u_1^n\\ v_1^n\\ \vdots \\ v^n_{k-3}\\ u^n_{k-2} \end{array} \right ), \end{align} $$

where $u_0^n,\ldots ,u_{k-2}^n$ and $v_0^n,\ldots ,v_{k-3}^n$ are the n-components of the variables $u_0,\ldots ,u_{k-2}$ and $v_0,\ldots ,v_{k-3}$ defined as in the proof of Theorem 1.1. In addition, we define

(2.35) $$ \begin{align} F=\left ( \begin{array}{l} \Delta f\\ \Delta (df)\\ u_1^n-A_1^n\\ d(u_1^n-A_1^n)\\ u_2^n-A_2^n\\ d(u_2^n-A_2^n)\\ \vdots\\ d(u^n_{k-2}-A^n_{k-2})\\ (F^k)^n \end{array} \right ) . \end{align} $$

Again, by construction,

$$ \begin{align*} \Delta y =F+\text{\ terms linear in~} v^n_j,\ \ \nabla v^n_j, \qquad j=0,\ldots, k-3. \end{align*} $$

In the following, we will give the proof of Theorem 1.3. We will only consider the case of a polyharmonic map of even order, as the odd case can be treated by exactly the same methods.

Lemma 2.11 Let $\varphi \colon M\to {\mathbb {S}}^n$ be a map with corresponding variables (2.34). Assume that $0 \leq j\leq k-3$ . Then, the following estimates hold on D:

(2.36) $$ \begin{align} \begin{aligned} |\Delta f|\leq & C(|f|+|u_0^n|), \\ |\Delta (df)|\leq & C(|f|+|df|+|\nabla df|+|v_0^n|), \\ |A^n_{j+1}|\leq & C(|f|+|df|). \end{aligned} \end{align} $$

Proof Recall that

$$ \begin{align*} \Delta f=\Delta\varphi^n=-u_0^n+\langle d\varphi^\beta,d\varphi^\nu\rangle\Gamma^n_{\beta\nu}. \end{align*} $$

Using the explicit form of the Christoffel symbols (2.31), we obtain the following expansion:

$$ \begin{align*} \langle d\varphi^\beta,d\varphi^\nu\rangle\Gamma^n_{\beta\nu} &= \langle d\varphi^n,d\varphi^\nu\rangle\underbrace{\Gamma^n_{n\nu}}_{=0} +\langle d\varphi^b,d\varphi^c\rangle\Gamma^n_{bc} +\langle d\varphi^b,d\varphi^n\rangle\underbrace{\Gamma^n_{bn}}_{=0}\\ &=\langle d\varphi^b,d\varphi^c\rangle\Gamma^n_{bc}. \end{align*} $$

As we have

$$ \begin{align*} \left|\Gamma^n_{bc}\right| &= \left|-\frac{1}{2}\sin(2\varphi^n)\tilde{g}_{bc} \right| = \left|\frac{1}{2}\sin(2f)\tilde{g}_{bc} \right| \\ \nonumber &\leq C|f|, \end{align*} $$

we obtain

$$ \begin{align*} |\langle d\varphi^\beta,d\varphi^\nu\rangle\Gamma^n_{\beta\nu}|\leq C|f|, \end{align*} $$

establishing the first estimate.

Now, remember that

(2.37) $$ \begin{align} A_{j+1}^n=& -2\langle du_{j}^\nu,d\varphi^\beta\rangle\Gamma^n_{\beta\nu} -u_{j}^\nu u_0^\beta\Gamma_{\beta\nu}^n +u_{j}^\nu\langle d\varphi^\vartheta,d\varphi^\sigma\rangle C_{\vartheta\sigma\nu}^n, \end{align} $$

where

$$ \begin{align*} C_{\vartheta\sigma\nu}^n:=& \Gamma^\mu_{\vartheta\sigma}\Gamma^n_{\mu\nu} -\frac{\partial \Gamma_{\vartheta \nu}^n}{\partial y^\sigma} - \Gamma_{\vartheta \nu}^\gamma\, \Gamma_{\sigma \gamma}^n. \end{align*} $$

Using again the explicit form of the Christoffel symbols (2.31), we find the estimates

$$ \begin{align*} |\langle du_{j}^\beta,d\varphi^\nu\rangle\Gamma^n_{\beta\nu}|&\leq C|f|,\\ |u_{j}^\nu u_0^\beta\Gamma_{\beta\nu}^n|&\leq C|f|. \end{align*} $$

As for the terms proportional to $C_{\vartheta \sigma \nu }^\alpha $ , we note that

$$ \begin{align*} u_{j}^\nu\langle d\varphi^\vartheta,d\varphi^\sigma\rangle\Gamma^\mu_{\vartheta\sigma}\Gamma^n_{\mu\nu} =u_{j}^c\langle d\varphi^\vartheta,d\varphi^\sigma\rangle\Gamma^b_{\vartheta\sigma}\Gamma^n_{bc}. \end{align*} $$

This allows us to derive the estimate

$$ \begin{align*} |u_{j}^\nu\langle d\varphi^\vartheta,d\varphi^\sigma\rangle\Gamma^\mu_{\vartheta\sigma}\Gamma^n_{\mu\nu}| \leq C|f|. \end{align*} $$

In order to estimate $u_{j}^\nu \langle d\varphi ^\vartheta ,d\varphi ^\sigma \rangle \frac {\partial \Gamma _{\vartheta \nu }^n}{\partial y^\sigma }$ , we make use of the same strategy as before taking into account that

$$ \begin{align*} \frac{\partial \Gamma_{ab}^n}{\partial y^c}=-\frac{1}{2}\sin(2f)\frac{\partial\tilde g_{ab}}{\partial y^c}, \end{align*} $$

which follows from (2.31).

Because the last two terms are proportional to $C_{\vartheta \sigma \nu }^\alpha $ , they can then be estimated as

$$ \begin{align*} |u_{j}^\nu\langle d\varphi^\vartheta,d\varphi^\sigma\rangle\frac{\partial \Gamma_{\vartheta \nu}^n}{\partial y^\sigma}| &\leq C(|f|+|df|), \\ |u_{j}^\nu\langle d\varphi^\vartheta,d\varphi^\sigma\rangle\Gamma_{\vartheta \nu}^\gamma\, \Gamma_{\sigma \gamma}^n| &\leq C|f|. \end{align*} $$

This proves the third estimate of the lemma. The estimate on $\Delta (df)$ can be achieved by exactly the same methods.▪

Lemma 2.12 Let $\varphi \colon M\to {\mathbb {S}}^n$ be a map with corresponding variables (2.34). Assume that $0 \leq j\leq k-3$ . Then, the following estimate holds:

(2.38) $$ \begin{align} |dA^n_{j+1}|\leq C(|f|+|df|+|\nabla df|). \end{align} $$

Proof By a direct calculation, we find

$$ \begin{align*} \nabla_i A^n_{j+1}=&-2\langle\nabla_iv_j^\gamma,d\varphi^\beta\rangle\Gamma^n_{\beta\gamma} -2\langle v_j^\gamma,\nabla_i d\varphi^\beta\rangle\Gamma^n_{\beta\gamma} -2\langle v_j^\gamma,d\varphi^\beta\rangle\frac{\partial\Gamma^n_{\beta\gamma}}{\partial y^\delta}\varphi^\delta_i \\ &-\nabla_iu_j^\gamma\, u_0^\beta\,\Gamma^n_{\beta\gamma}-u_j^\gamma \,\nabla_iu_0^\beta\,\Gamma^n_{\beta\gamma} -u_j^\gamma\, u_0^\beta\,\frac{\partial\Gamma^n_{\beta\gamma}}{\partial y^\delta}\varphi^\delta_i \\ &+\nabla_i u^\nu_j\langle d\varphi^\vartheta,d\varphi^\sigma\rangle C^n_{\vartheta\sigma\nu} + u^\nu_j\langle\nabla_i d\varphi^\vartheta,d\varphi^\sigma\rangle C^n_{\vartheta\sigma\nu} + u^\nu_j\langle d\varphi^\vartheta,\nabla_id\varphi^\sigma\rangle C^n_{\vartheta\sigma\nu} \\ &+ u^\nu_j\langle d\varphi^\vartheta,d\varphi^\sigma\rangle \frac{\partial C^n_{\vartheta\sigma\nu}}{\partial y^\delta}\varphi^\delta_i. \end{align*} $$

All the terms in this expression can be estimated by expanding the Christoffel symbols and using the same strategy as in the proof of Lemma 2.11 except the last term which requires a more careful inspection.

Again, a direct calculation yields

$$ \begin{align*} \frac{\partial C^n_{\vartheta\sigma\nu}}{\partial y^\delta}= \frac{\partial \Gamma^\mu_{\vartheta\sigma}}{\partial y^\delta} \Gamma^n_{\mu\nu}+ \Gamma^\mu_{\vartheta\sigma} \frac{\partial \Gamma^n_{\mu\nu}}{\partial y^\delta} -\frac{\partial^2\Gamma_{\vartheta \nu}^n}{\partial y^\sigma\partial y^\delta} -\frac{\partial\Gamma_{\vartheta \nu}^\gamma}{\partial y^\delta}\Gamma_{\sigma \gamma}^n -\Gamma_{\vartheta \nu}^\gamma\frac{\partial\Gamma_{\sigma \gamma}^n}{\partial y^\delta}. \end{align*} $$

We realize that all terms in $u^\nu _j\langle d\varphi ^\vartheta ,d\varphi ^\sigma \rangle \frac {\partial C^n_{\vartheta \sigma \nu }}{\partial y^\delta }\varphi ^\delta _i$ can be estimated by the same reasoning used before; only the contribution that is proportional to the second derivative of the Christoffel symbols needs to be treated in more detail. Hence, let us have a closer look at

$$ \begin{align*} \frac{\partial^2\Gamma_{\vartheta \nu}^n}{\partial y^\sigma\partial y^\delta} u^\nu_j\langle d\varphi^\vartheta,d\varphi^\sigma\rangle \varphi^\delta_i =& \frac{\partial^2\Gamma_{ab}^n}{\partial y^c\partial y^d} u^b_j\langle d\varphi^a,d\varphi^c\rangle \varphi^d_i + \frac{\partial^2\Gamma_{ab}^n}{\partial y^n\partial y^d} u^b_j\langle d\varphi^a,d\varphi^n\rangle \varphi^d_i \\ &+ \frac{\partial^2\Gamma_{ab}^n}{\partial y^c\partial y^n} u^b_j\langle d\varphi^a,d\varphi^c\rangle \varphi^n_i + \frac{\partial^2\Gamma_{ab}^n}{\partial y^n\partial y^n} u^b_j\langle d\varphi^a,d\varphi^n\rangle \varphi^n_i. \end{align*} $$

Using (2.31), we obtain

$$ \begin{align*} \frac{\partial^2\Gamma_{ab}^n}{\partial y^c\partial y^d}=-\frac{1}{2}\sin(2f)\frac{\partial^2\tilde g_{ab}}{\partial y^c\partial y^d}, \end{align*} $$

and we can conclude that all terms can be estimated in such a way that the statement of the lemma holds true.▪

Lemma 2.13 Let $\varphi \colon M\to {\mathbb {S}}^n$ be a polyharmonic map of even order $k=2s$ with corresponding variables (2.34). Then, the following estimate holds:

(2.39) $$ \begin{align} |(F^{2s})^n|\leq C\left(|f|+|df|+\sum_{\ell=0}^{2s-2}|u^n_\ell|+\sum_{\ell=0}^{2s-3}|v^n_\ell|+|\nabla u^n_{2s-2}|\right). \end{align} $$

Proof Recall that

$$ \begin{align*} (F^{2s})^n=&-A^n_{2s-1} -u^\delta_{2s-2}\langle d\varphi^\gamma,d\varphi^\beta\rangle R^n_{\beta\gamma\delta} \\ \nonumber&-\sum_{\ell=1}^{s-1}\big(u^\delta_{s-\ell-1}\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^n_{\beta\gamma\delta} +u^\vartheta_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^n_{\beta\delta\vartheta\eta} \\ \nonumber &\hspace{1.5cm} +u^\delta_{s+\ell-2}\langle v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^n_{\beta\gamma\delta}\big), \end{align*} $$

where

$$ \begin{align*} E^n_{\beta\delta\vartheta\eta}:=&R^n_{\beta\gamma\delta}\Gamma^\gamma_{\vartheta\eta}+R^n_{\beta\gamma\vartheta}\Gamma^\gamma_{\delta\eta}. \end{align*} $$

First, we use the third inequality in (2.36) to estimate $A^n_{2s-1}$ . It remains to estimate the terms involving the curvature tensor. Inserting the nonzero components of the curvature tensor given in (2.33), we find

$$ \begin{align*} u^\delta_{2s-2}\langle d\varphi^\gamma,d\varphi^\beta\rangle R^n_{\beta\gamma\delta} =u^\delta_{2s-2}\langle d\varphi^n,d\varphi^\beta\rangle R^n_{\beta n\delta} +u^n_{2s-2}\langle d\varphi^\gamma,d\varphi^\beta\rangle R^n_{\beta\gamma n}. \end{align*} $$

Hence, we may estimate

$$ \begin{align*} |u^\delta_{2s-2}\langle d\varphi^\gamma,d\varphi^\beta\rangle R^n_{\beta\gamma\delta}|\leq C\big (|df|+|u^n_{2s-2}| \big ). \end{align*} $$

By the same reasoning, we find

$$ \begin{align*} |u^\delta_{s-\ell-1}\langle v^\gamma_{s+\ell-2},d\varphi^\beta\rangle R^n_{\beta\gamma\delta}| &\leq C \big (|u^n_{s-\ell-1}|+|v^n_{s+\ell-2}| \big ),\\ |u^\delta_{s+\ell-2}\langle v^\gamma_{s-\ell-1},d\varphi^\beta\rangle R^n_{\beta\gamma\delta}| &\leq C \big (|u^n_{s+\ell-2}|+|v^n_{s-\ell-1}| \big ). \end{align*} $$

Regarding the term that is proportional to $E^n_{\beta \delta \vartheta \eta }$ , we expand

$$ \begin{align*} u^\vartheta_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^n_{\beta\delta\vartheta\eta} =& \, u^b_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^c,d\varphi^\beta\rangle R^n_{\beta n\delta}\Gamma^n_{bc}\\ &+u^\vartheta_{s+\ell-2}u^n_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle R^n_{\beta\gamma n}\Gamma^\gamma_{\vartheta\eta}\\ &+u^\vartheta_{s+\ell-2}u^b_{s-\ell-1}\langle d\varphi^c,d\varphi^\beta\rangle R^n_{\beta n\vartheta}\Gamma^n_{bc}\\ &+u^n_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle R^n_{\beta\gamma n}\Gamma^\gamma_{\delta\eta}, \end{align*} $$

where we used both (2.31) and (2.33). Hence, we can infer the estimate

$$ \begin{align*} |u^\vartheta_{s+\ell-2}u^\delta_{s-\ell-1}\langle d\varphi^\eta,d\varphi^\beta\rangle E^n_{\beta\delta\vartheta\eta}| \leq C\big (|f|+|u^n_{s+\ell-2}|+|u^n_{s-\ell-1}| \big ). \end{align*} $$

The claim now follows by combining all the single estimates.▪

Now, we show that the vector variable y defined in (2.34) satisfies

(2.40) $$ \begin{align} y \equiv 0 \quad \mathrm{on}\,\, W. \end{align} $$

To prove this claim, we observe that $f=\varphi ^n -\pi /2$ , and consequently $df$ , vanish on W, because $\varphi $ maps W into the equator. Next, using $\Delta f=0$ on W and the explicit expression (2.31) of the Christoffel symbols, we deduce that $u^n_0=0$ . The functions $A^n_{j}$ , $j=1, \ldots ,k-2$ , also vanish identically on W. This follows easily from the definition (2.37), using again the explicit expression (2.31) of the Christoffel symbols together with $d\varphi ^n=0$ . Finally, from these facts, it is easy to deduce that all the components of y vanish on W, and so the claim (2.40) holds.

Next, using the inequalities (2.36), (2.38), and (2.39), we find

(2.41) $$ \begin{align} |\Delta y|\leq C\big(|f|+|df|+|\nabla df|+\sum_{\ell=0}^{k-2}|u^n_\ell|+\sum_{\ell=0}^{k-3}(|v^n_\ell|+|\nabla v^n_\ell|) +|\nabla u^n_{k-2}|\big). \end{align} $$

Now, because of (2.40) and the estimate (2.41), the assumptions of Theorem 2.1 are satisfied, and we can conclude that $y=0$ on D, which, in particular, implies that $\varphi ^n=\frac {\pi }{2}$ , i.e., $\varphi $ maps the whole of D into $\mathbb {S}^{n-1}$ . We finish the proof of Theorem 1.3 by setting $A:=\{p\in M : \varphi (p)\in \mathbb {S}^{n-1}\}$ and using the same globalization argument as above.▪

Proof of Theorems 3.1 and 3.2 In order to prove Theorem 3.1, we define the vector-valued function

$$ \begin{align*} u=\left ( \begin{array}{l} u_0\\ v_0\\ u_1\\ v_1\\ u_2 \end{array} \right ), \end{align*} $$

and by the same analysis as in the proof of Theorem 1.1, we show using Aronszajn’s theorem that $u=0$ .

In order to prove Theorem 3.2, we define the vector-valued function

$$ \begin{align*} u=\left ( \begin{array}{l} \varphi\\ d\varphi\\ u_0\\ v_0\\ u_1\\ v_1\\ u_2 \end{array} \right ), \end{align*} $$

and $\tilde u$ will be defined accordingly. Employing the same strategy as in the proof of Theorem 1.2, now it is easy to complete the proof of Theorem 3.2.▪