3.2.1 Motivation for our choice of recovery sequence

If $ (\boldsymbol {u}_n)_n \subset \mathcal {B}_s$ is a sequence satisfing INV and $ \boldsymbol {u}_n \rightharpoonup \boldsymbol {u}$ in $H^1(\Omega ,\mathbb {R}^3)$ , it might be that $\boldsymbol {u}$ does not satisfy INV because it is not true that $H^1$ embeds in $C^0$ for two-dimensional domains and hence the degree is not continuous for the weak $H^1$ convergence (for condition INV, we look at the degree of $ \boldsymbol {u}$ restricted to $2D$ spheres). The classical example of a sequence showing the noncontinuity of the degree for the weak convergence in $H^1$ is known as the bubbling-off of spheres and is given by

(3.2) $$ \begin{align} \mathfrak{S}_n: \mathbb{S}^2 \rightarrow \mathbb{S}^2, \quad \mathfrak{S}_n(\boldsymbol{x})=\pi_S^{-1}(n\pi_S(\boldsymbol{x})), \end{align} $$

where we have denoted by $\pi _S$ the stereographic projection from the south pole to the plane orthogonal to $\boldsymbol {e}_3$ and passing through the north pole $ \{ x_3=1\}$ . Note that $ \mathfrak {S}_n$ is conformal and $\mathfrak {S}_n$ does not converge uniformly to $ (0,0,-1)$ or strongly in $H^1$ , and that the Dirichlet energy $\int _{\mathbb {S}^2} |D \mathfrak {S}_n|^2 \mathrm {d} \boldsymbol {x}$ concentrates near the north pole.

Another way to see conformality appearing in our problem is to observe that a key point to obtain the lower bound on the relaxed energy in the axisymmetric case ([Reference Barchiesi, Henao, Mora-Corral and Rodiac9, Th. 1.1]) is the following ‘area-energy’ inequality

(3.3) $$ \begin{align} |(\operatorname{cof} D \boldsymbol{u}) \boldsymbol{e}_3|\leq \frac12 |D \boldsymbol{u}|^2. \end{align} $$

Let us recall briefly the proof of (3.3). Since $ \boldsymbol {e}_1 \wedge \boldsymbol {e}_2= \boldsymbol {e}_3$ , by using the properties of the cofactor matrix and Cauchy’s inequality, we find that

$$ \begin{align*} |(\operatorname{cof} D \boldsymbol{u}) \boldsymbol{e}_3| &=|(\operatorname{cof} D \boldsymbol{u}) \boldsymbol{e}_1 \wedge \boldsymbol{e}_2|= |(D \boldsymbol{u})\boldsymbol{e}_1 \wedge (D \boldsymbol{u}) \boldsymbol{e}_2| \\ & \leq |\partial_{x_1}\boldsymbol{u}| |\partial_{x_2}\boldsymbol{u}| \leq \frac12 (|\partial_{x_1}\boldsymbol{u}|^2+|\partial_{x_2} \boldsymbol{u}|^2) \leq \frac12 |D \boldsymbol{u}|^2, \end{align*} $$

where we recall that we are using the Frobenius norm of a matrix. From this proof, we see that there is equality in (3.3) if and only if

$$\begin{align*}\partial_{x_3} \boldsymbol{u}=0, \quad \partial_{x_1} \boldsymbol{u} \cdot \partial_{x_2} \boldsymbol{u} =0, \quad |\partial_{x_1} \boldsymbol{u}|=|\partial_{x_2} \boldsymbol{u}| .\end{align*}$$

The two last conditions mean that $\boldsymbol {u}$ restricted to the plane $(\boldsymbol {e}_1,\boldsymbol {e}_2)$ is locally conformal at $ \boldsymbol {x} = (x_1,x_2,x_3)$ .

The two previous paragraphs indicate that to construct a sequence showing the lack of compactness in $\mathcal {A}^r_s$ with optimal loss of energy for E, we must construct a sequence such that both $ \operatorname {cof} D \boldsymbol {u}_n$ and $D \boldsymbol {u}_n$ concentrate, and that inequality (3.3) becomes asymptotically an equality, thus involving conformality. Note that, because of the axisymmetry, the only place where the sequence can concentrate is the symmetry axis, as shown in [Reference Henao and Rodiac34]. To construct our recovery sequence, we will use the maps $\mathfrak {S}_n$ in (3.2). We can use spherical coordinates and see that $ \mathfrak {S}_n$ is given by

$$ \begin{align*} (\mathfrak{S}_n)_{\rho} =1, \quad (\mathfrak{S}_n)_{\theta} = \theta, \quad (\mathfrak{S}_n)_{\varphi}=2\arctan\left(n \tan \frac{\varphi}{2} \right). \end{align*} $$

The important information is carried by the zenith angle $( \mathfrak {S}_n)_{\varphi }$ . If we see the bubble as a map from $\mathbb {R}^2$ to $\mathbb {S}^2$ , elementary geometric relations show that $( \mathfrak {S}_n)_{\varphi }=2\arctan (nr)$ , where $(r,\theta )$ are the polar coordinates. However, we want to construct a bubble with values onto the sphere $ S((0,0,\frac 12),\frac 12)$ since it is what is done in [Reference Conti and De Lellis16]. In that case, by using the central angle theorem, since the origin of the sphere is at $(0,0,\frac 12)$ , we can see that we must take a modified sequence

$$ \begin{align*} (\tilde{\mathfrak{S}}_n)_{\rho}=2\cdot \frac12 \cos((\tilde{\mathfrak{S}}_n)_{\varphi}), \quad (\tilde{\mathfrak{S}}_n)_{\theta}=\theta, \quad (\tilde{\mathfrak{S}}_n)_{\varphi} =\arctan(n r). \end{align*} $$

We also observe that the image of a disk of radius $1/n$ in the plane $ x_3=0$ by $ \pi _S^{-1}(n\cdot )$ is the upper hemisphere of $\mathbb {S}^2$ . However, the image of a disk of radius $ \frac 1n$ in the plane $x_3=0$ by $ \pi _S^{-1}(n^2\cdot )$ is almost the all sphere $\mathbb {S}^2$ . We arrive at the conclusion that the map near our set of concentration should look like

$$\begin{align*}(\boldsymbol{u}_n)_{\rho}\approx \cos (\boldsymbol{u}_n)_{\varphi}, \quad (\boldsymbol{u}_n)_{\theta}= \theta, \quad (\boldsymbol{u}_n )_{\varphi} \approx \arctan(n^2 r).\end{align*}$$

This is to compare with the construction of Conti–De Lellis, where $ (\boldsymbol {u}_n )_{\varphi }=2\arctan (n r)$ . This latter map is not conformal because the sphere we want to use for the bubbling is not centred at the origin.

Another difference in our construction is that we will require that $\boldsymbol {u}_n$ is incompressible near the set of concentration. This is because we want to emphasize that the lack of compactness of the problem is due to the Dirichlet part of the neo-Hookean energy and not to the determinant part. Hence, our construction is valid for quite a general choice of the convex function H.

For notational simplicity, we set $ \varepsilon =1/n$ for $n \in \mathbb {N}$ with $n\geq 1$ . From what precedes, we can understand that the following function plays the main role in the construction:

(3.4) $$ \begin{align} f_{\kern-1.2pt\varepsilon} (r):=\arctan \Big ( \frac{r}{\varepsilon^2} \Big ) + \alpha_{\kern-1.2pt\varepsilon} \frac{r}{\varepsilon}, \quad 0\leq r \leq \varepsilon, \qquad \alpha_{\kern-1.2pt\varepsilon}:=\arctan(\varepsilon). \end{align} $$

Note that

$$ \begin{align*} f_{\kern-1.2pt\varepsilon}(0)=0, \quad f_{\kern-1.2pt\varepsilon}(\varepsilon)=\frac{\pi}{2}, \quad \text{and}\quad f_{\kern-1.2pt\varepsilon}^{\prime}(r)> 0\quad \text{for all }r. \end{align*} $$

The first term in $f_{\kern-1.2pt\varepsilon }(r)$ may be interpreted as a function that stretches the disk $\{\boldsymbol {x}\in \mathbb {R}^3:\ x_1^2+x_2^2 < \varepsilon ^2,\ x_3=0\}$ onto the disk $\{\boldsymbol {y}\in \mathbb {R}^3:\ y_1^2 +y_2^2 < \varepsilon ^{-2},\ y_3=1\},$ so as to subsequently wrap that disk (conformally, via the stereographic projection, see Figure 6) onto (a very large part of) the sphere $S((0,0,\frac 12),\frac 12)$ .

Figure 6 Conformal transformation of an $\varepsilon $ -disk onto the sphere, via the stereographic projection.

The function $f_{\kern-1.2pt\varepsilon }(r)$ will correspond to the angle formed between the positive $y_3$ axis and the segment joining the origin with a point on that bubble. The correction term $\alpha _{\kern-1.2pt\varepsilon } \frac {r}{\varepsilon }$ in $f_{\kern-1.2pt\varepsilon }(r)$ , which is chosen to be linear for simplicity in the calculations, has the effect of wrapping the disk onto the whole bubble (all the way up to $f_{\kern-1.2pt\varepsilon }(r)=\frac {\pi }{2}$ ) and not only to the large part consisting of all points with zenith angle between $0$ and $\frac {\pi }{2} - \alpha _{\kern-1.2pt\varepsilon }$ . From now on, a choice is made of

$$ \begin{align*} 0<\gamma \leq \frac{1}{3} \quad \text{a fixed positive exponent.} \end{align*} $$

3.2.2 Region $c_{\kern-1.2pt\varepsilon }$

We start by describing our recovery sequence in region

$$\begin{align*}c_{\kern-1.2pt\varepsilon}:=\{ x_1^2+x_2^2 < \varepsilon^2, 0 <x_3<1 \}.\end{align*}$$

This is where concentration of energy will occur. In this region, we use cylindrical coordinates in the domain; that is, we write $ \boldsymbol {x} (r, \theta , x_3) = r\boldsymbol {e}_r + x_3 \boldsymbol {e}_3, \ 0\leq r < \varepsilon ,\ 0\leq \theta \leq \pi , \ 0<x_3<1. $ In most of the regions, we describe $\boldsymbol {u}$ via its spherical coordinates; that is, we set

(3.5) $$ \begin{align} \boldsymbol{u}_{\kern-1.2pt\varepsilon }\big (\boldsymbol{x}(r,\theta,x_3)\big ) = u^\varepsilon_{\rho} \sin u^\varepsilon_{\varphi}\, \boldsymbol{e}_r(\theta) + u^\varepsilon_{\rho} \cos u^\varepsilon_{\varphi}\,\boldsymbol{e}_3. \end{align} $$

In region $c_{\kern-1.2pt\varepsilon }$ , we take

(3.6) $$ \begin{align} u^\varepsilon_{\rho}(r, x_3) = \Big ( \big (\cos \big (f(r)\big ) + 2\varepsilon^\gamma\big )^3 + x_3\cdot \frac{3r}{\partial_r \big ( - \cos f(r) \big )} \Big )^{1/3}, \qquad u^\varepsilon_{\varphi}(r) = f_{\kern-1.2pt\varepsilon}(r). \end{align} $$

These equations may be regarded as a perturbation of $u_{\rho } = \cos u_{\varphi }$ and $u_{\varphi }=\arctan (\frac {r}{\varepsilon ^2})$ – namely, the equations of a bubbling sequence (see Figure 7) from a disk of size $\varepsilon $ perpendicular to the symmetry axis to the sphere $S((0,0,\frac 12),\frac 12)$ .

Figure 7 Illustration of the deformation in the key region $c_{\kern-1.2pt\varepsilon }$ , where the singular energy originates. Even for the exaggeratedly large value of $\varepsilon =0.7$ used for these plots, the images of the disks $B^2(\boldsymbol {0}, \varepsilon )\times \{x_3\}$ , taken at different heights $x_3$ between 0 and 1, are almost indistinguishable. As $\varepsilon $ becomes smaller, the polar coordinates $\varepsilon ^\gamma $ and $2\varepsilon ^\gamma $ of the deformed points A and B are increasingly small, and the image of each of the disks resembles more and more the sphere $S\big ( (0,0,\frac {1}{2}), \frac {1}{2}\big )$ . The bubbling effect can also begin to be appreciated, since the angular sector $|u_{\varphi }|<\frac {\pi }{50}(1+ \varepsilon \arctan (\varepsilon ))$ that is zoomed out in Figure c) comes from the much smaller disks $B^2(\boldsymbol {0}, \frac {\pi }{50}\varepsilon ^2)\times \{x_3\}$ . Correspondingly, when $0<x_1^2+x_2^2<\varepsilon ^{2}$ , the huge tangential stretch $\frac {\partial \boldsymbol {u}_{\kern-1.2pt\varepsilon }}{\partial r}$ is of order $\varepsilon ^{-2}$ , and the normal compression $\frac {\partial \boldsymbol {u}_{\kern-1.2pt\varepsilon }}{\partial x_3}$ is of order $\varepsilon ^4$ .

The complicated expression for $u_{\rho }^\varepsilon $ arises as a solution of the incompressibility equation. Indeed, since in this construction $u_{\varphi }^\varepsilon $ is independent of $x_3$ , from (B.3) it follows that

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }\big ( \boldsymbol{x} (r,\theta,x_3)\big ) = \frac{1}{3} \frac{\sin \big (f_{\kern-1.2pt\varepsilon}(r)\big)\,f_{\kern-1.2pt\varepsilon}^{\prime}(r)}{r} \partial_{x_3} \big ( (u_{\rho}^\varepsilon)^3\big ) \equiv 1. \end{align*} $$

The ‘initial condition’

$$ \begin{align*} u_{\rho}^\varepsilon(r,x_3) = \cos\big ( f_{\kern-1.2pt\varepsilon}(r)\big) + 2\varepsilon^\gamma \quad\text{at} \quad x_3=0 \end{align*} $$

departs from the bubble $u_{\rho }=\cos u_{\varphi }$ only because of the small term $2\varepsilon ^{\gamma }$ , which is added for reasons of technical convenience that will become evident in the sequel (see, for example, (3.21)).

3.2.3 Region $a_{\kern-1.2pt\varepsilon }^{\prime }$

We call $a_{\kern-1.2pt\varepsilon }^{\prime }$ the region

$$ \begin{align*}a_{\kern-1.2pt\varepsilon}^{\prime}:=\{(x_1,x_2,x_3): x_1^2+x_2^2+x_3^2<\varepsilon^2,\, x_3< 0\}.\end{align*} $$

We parametrize this region by

$$ \begin{align*} \boldsymbol{x}(s,\theta,\varphi) = (1-s) g(\varphi)\boldsymbol{e}_r(\theta) + s\big ( \varepsilon \sin\varphi \boldsymbol{e}_r(\theta) - \varepsilon \cos \varphi \boldsymbol{e}_3\big ),\quad 0\leq s\leq 1,\ 0\leq \theta <2\pi,\ 0\leq \varphi < \frac{\pi}{2}, \end{align*} $$

where

(3.7) $$ \begin{align} g_{\kern-1.2pt\varepsilon}:[0,\frac{\pi}{2}]\to [0,\varepsilon] \text{ is the inverse of the function } f_{\kern-1.2pt\varepsilon} \text{ defined in } (3.4). \end{align} $$

For each fixed $\theta , \varphi $ the parametrization consists of an affine interpolation (with parameter s) between the preimage on the disk $\{r\leq \varepsilon ,\, x_3=0\}$ of the point on the bubble with zenith angle $\varphi $ (which is determined by the function $f(r)$ that characterizes the conformal mapping from the stack of horizontal disks in region $c_{\kern-1.2pt\varepsilon }$ to the corresponding stack of ‘copies’ of the bubble) and the point on the sphere $\{r^2+x_3^2=\varepsilon ^2,\, x_3\leq 0\}$ that (according to the eversion-type map $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ to be defined in region $a_{\kern-1.2pt\varepsilon }$ ) will be assigned the same zenith polar angle $\varphi $ (see Figure 8).

Figure 8 Schematic representation of the image of regions $a_{\kern-1.2pt\varepsilon }$ , $a_{\kern-1.2pt\varepsilon }^{\prime }$ , $c_{\kern-1.2pt\varepsilon }$ , $e_{\kern-1.2pt\varepsilon }^{\prime }$ and f after the deformation $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ . The parameter $\varphi $ in the definition of $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ corresponds to the zenith angle in the deformed configuration.

We define $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in region $a_{\kern-1.2pt\varepsilon }^{\prime }$ through its spherical coordinates

$$ \begin{align*} u_{\rho}^\varepsilon(s,\varphi) := \Big ( \big ( \cos\varphi + 2\varepsilon^\gamma\big ) ^3 - 3\int_{\sigma=0}^s h_{\kern-1.2pt\varepsilon}(\sigma,\varphi)\mathrm{d} \sigma \Big ) ^{\frac{1}{3}}, \qquad u_{\varphi}^\varepsilon(s,\varphi) &:= \varphi, \end{align*} $$

with

(3.8) $$ \begin{align} h_{\kern-1.2pt\varepsilon}(s,\varphi):= \varepsilon \Big ( (1-s) \frac{g_{\kern-1.2pt\varepsilon}(\varphi)}{\sin\varphi} +s\varepsilon \Big ) \Big ( (1-s) g_{\kern-1.2pt\varepsilon}^{\prime}(\varphi)\cos\varphi + s (\varepsilon-g_{\kern-1.2pt\varepsilon}(\varphi)\sin\varphi)\Big ). \end{align} $$

$\underline {\text {Derivatives of } \boldsymbol {u}_{\kern-1.2pt\varepsilon }:}$ they are given by

$$ \begin{align*} \partial_s \boldsymbol{u}_{\kern-1.2pt\varepsilon } = \partial_s u_{\rho}^\varepsilon\, \boldsymbol{e}_{\rho}(\theta,\varphi), \quad \partial_{\theta} \boldsymbol{u}_{\kern-1.2pt\varepsilon } = u_{\rho}^\varepsilon \sin\varphi\, \boldsymbol{e}_{\theta}(\theta), \quad \partial_{\varphi} \boldsymbol{u}_{\kern-1.2pt\varepsilon } = \partial_{\varphi} u_{\rho}^\varepsilon\,\boldsymbol{e}_{\rho}(\theta,\varphi) + u_{\rho}^\varepsilon\,\boldsymbol{e}_{\varphi}(\theta,\varphi), \end{align*} $$

with

$$ \begin{align*} \boldsymbol{e}_{\rho}(\theta,\varphi)= \sin\varphi\, \boldsymbol{e}_r(\theta) +\cos\varphi \boldsymbol{e}_3, \quad \boldsymbol{e}_{\varphi}(\theta,\varphi)= \cos\varphi\,\boldsymbol{e}_r(\theta)- \sin\varphi \boldsymbol{e}_3. \end{align*} $$

We now prove that with this definition, $ \det D \boldsymbol {u}_{\kern-1.2pt\varepsilon }=1$ in $a_{\kern-1.2pt\varepsilon }^{\prime }$ .

Derivatives of the parametrization of the reference domain:

$$ \begin{align*} \partial_s\boldsymbol{x} &= (\varepsilon \sin\varphi - g_{\kern-1.2pt\varepsilon}(\varphi))\boldsymbol{e}_r -\varepsilon \cos\varphi\,\boldsymbol{e}_3 \\ \partial_{\theta} \boldsymbol{x} &= \Big ( (1-s) g_{\kern-1.2pt\varepsilon}(\varphi) + s\varepsilon \sin\varphi \Big ) \boldsymbol{e}_{\theta} \\ \partial_{\varphi} \boldsymbol{x} &= \big ( (1-s) g^{\prime}_{\kern-1.2pt\varepsilon}(\varphi) + s\varepsilon \cos\varphi \big ) \boldsymbol{e}_r +s\varepsilon \sin\varphi\,\boldsymbol{e}_3. \end{align*} $$

Using the formulae

$$ \begin{align*} A_{1i} \boldsymbol{a}_i \wedge A_{2j} \boldsymbol{a}_j \wedge A_{3k} \boldsymbol{a}_k = (\det A) \boldsymbol{a}_1\wedge \boldsymbol{a}_2 \wedge \boldsymbol{a}_3, \qquad \boldsymbol{e}_r\wedge \boldsymbol{e}_{\theta} \wedge \boldsymbol{e}_3=1, \end{align*} $$

it can be seen that

$$ \begin{align*} \partial_s \boldsymbol{x} \wedge \partial_{\theta} \boldsymbol{x} \wedge \partial_{\varphi} \boldsymbol{x} =\Big ( (1-s) g_{\kern-1.2pt\varepsilon}(\varphi) + s\varepsilon \sin\varphi \Big ) \begin{vmatrix} \varepsilon\sin\varphi - g_{\kern-1.2pt\varepsilon}(\varphi) -\varepsilon\cos\varphi \\ (1-s)g^{\prime}_{\kern-1.2pt\varepsilon}(\varphi) + s\varepsilon\cos\varphi s\varepsilon\sin\varphi \end{vmatrix} = \sin \varphi\,h_{\kern-1.2pt\varepsilon}(s,\varphi). \end{align*} $$

Incompressibility: From the standard relation $\boldsymbol {e}_{\varphi }\wedge \boldsymbol {e}_{\theta } \wedge \boldsymbol {e}_{\rho }=1$ in spherical coordinates, it is clear that

$$ \begin{align*} \partial_s \boldsymbol{u}_{\kern-1.2pt\varepsilon } \wedge \partial_{\theta} \boldsymbol{u}_{\kern-1.2pt\varepsilon } \wedge \partial_{\varphi} \boldsymbol{u}_{\kern-1.2pt\varepsilon } &= - (u_{\rho}^\varepsilon)^2 \partial_s u_{\rho}^\varepsilon \sin \varphi = -\frac{1}{3}\partial_s \Big ( (u_{\rho}^\varepsilon)^3\Big ) \sin \varphi = h_{\kern-1.2pt\varepsilon}(s,\varphi) \sin \varphi. \end{align*} $$

However, using that the Jacobian of a composition is the product of the Jacobians, for $\boldsymbol {F}:= D\boldsymbol {u}_{\kern-1.2pt\varepsilon }(\boldsymbol {x})$ , we have that

$$ \begin{align*} \partial_s \boldsymbol{u}_{\kern-1.2pt\varepsilon } \wedge \partial_{\theta} \boldsymbol{u}_{\kern-1.2pt\varepsilon } \wedge \partial_{\varphi} \boldsymbol{u}_{\kern-1.2pt\varepsilon } = \boldsymbol{F} \partial_s \boldsymbol{x} \wedge \boldsymbol{F} \partial_{\theta} \boldsymbol{x} \wedge \boldsymbol{F} \partial_{\varphi} \boldsymbol{x} = \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \cdot \partial_s \boldsymbol{x} \wedge \partial_{\theta} \boldsymbol{x} \wedge \partial_{\varphi} \boldsymbol{x} = (\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }) h_{\kern-1.2pt\varepsilon}(s,\varphi) \sin \varphi. \end{align*} $$

Therefore, $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }(\boldsymbol {x})=1$ for all $\boldsymbol {x} \in a_{\kern-1.2pt\varepsilon }^{\prime }$ . For this, it is necessary to know that $h_{\kern-1.2pt\varepsilon }(s,\varphi )$ is nonzero; this is proved in Lemma A.2 below.

The formula

$$ \begin{align*} -\frac{1}{3}\partial_s \Big ((u_{\rho}^\varepsilon)^3\Big ) \sin \varphi = (\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }) h_{\kern-1.2pt\varepsilon}(s,\varphi) \sin \varphi \end{align*} $$

and the value of $u_{\rho }^\varepsilon $ at the disk $\{r\leq \varepsilon ,\, x_3=0\}$

$$\begin{align*}u_{\rho}^\varepsilon (r, \theta) = \cos f(r) + 2\varepsilon^\gamma \end{align*}$$

(which is prescribed by the construction of $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in the critical region $c_{\kern-1.2pt\varepsilon }$ ) explain the definition of $u_{\rho }^\varepsilon $ in the region $a_{\kern-1.2pt\varepsilon }^{\prime }$ . Although the construction here is more elaborate, it is based on the technique of using direction-preserving deformations developed in [Reference Henao and Serfaty35, Reference Henao, Mora-Corral and Xu33] to solve the incompressibility constraint. More precisely, this construction imposes that segments go to segments, which makes the incompressibility equation easily solvable. Indeed, the incompressibility equation when $\boldsymbol {u}^\varepsilon $ is direction-preserving only involves $\partial _s (u_{\rho }^\varepsilon )$ , so it can be integrated. Since we also impose conformality, the component $u_{\varphi }$ can be recovered and, hence, the equation can be solved explicitly for $\boldsymbol {u}$ .

3.2.4 Region $a_{\kern-1.2pt\varepsilon }$

We set

$$ \begin{align*}a_{\kern-1.2pt\varepsilon}:= \{ (x_1, x_2, x_3): \varepsilon^2 < x_1^2 + x_2^2 + x_3^2 <1,\ x_3< 0\}. \end{align*} $$

In this region, we use spherical coordinates $ (\rho ,\theta ,\varphi )$ , $ \varepsilon <\rho <1, \ 0\leq \theta <2 \pi , \ \pi /2\leq \varphi \leq \pi $ in the domain, and we define $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in region $a_{\kern-1.2pt\varepsilon }$ through its spherical coordinates

(3.9) $$ \begin{align} \begin{aligned} u_{\rho}^\varepsilon(\rho, \varphi) &:= \frac{1-\rho}{1-\varepsilon} \Big ( (\cos (\pi-\varphi) + 2\varepsilon^\gamma)^3 - 3\int_{\sigma=0}^1 h_{\kern-1.2pt\varepsilon}(\sigma, \varphi)\mathrm{d}\sigma\Big )^{\frac{1}{3}} + \varepsilon^\gamma \frac{\rho-\varepsilon}{1-\varepsilon}, \\ u_{\varphi}^\varepsilon (\rho,\varphi) &:= \pi-\varphi, \end{aligned} \end{align} $$

where $h_{\kern-1.2pt\varepsilon }(\sigma , \varphi )$ is the function defined in (3.8).

Regarding $u_{\rho }^\varepsilon $ , it is an affine interpolation connecting, on the one hand, the radial distance at the image of the interface between $a_{\kern-1.2pt\varepsilon }^{\prime }$ and $a_{\kern-1.2pt\varepsilon }$ (which is mapped to a ‘copy’ of the bubble lying slightly beneath it), and, on the other hand, the radial distance $\varepsilon ^\gamma $ at which all points on the reference lower unit hemisphere are mapped to. By Lemma A.5 below, the a.e. limit of $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in region $a_{\kern-1.2pt\varepsilon }$ is the Conti–De Lellis map.

3.2.5 Region b

We recall that this region is described by

$$\begin{align*}b=\{ \rho \sin \varphi \boldsymbol{e}_r(\theta) +\rho \cos \varphi \boldsymbol{e}_3 : 1\leq \rho <3, \ \frac{\pi}{2}\leq \varphi \leq \pi\}. \end{align*}$$

Here, a slight modification with respect to the original Conti–De Lellis maps is in order since in our construction we are now changing the scale of the spacings between the image of $A(\rho =1, x_3=0)$ , $B(\rho =\varepsilon , x_3=0)$ , $C(\rho =\varepsilon , x_3=1)$ and $D(\rho =1, x_3=1)$ to $\varepsilon ^\gamma $ with $\gamma \leq 1/3$ instead of just $\varepsilon $ .

Working with spherical coordinates $(\rho , \theta , \varphi )$ in the reference configuration and cylindrical coordinates in the deformed configuration, we first define the auxiliary function $\boldsymbol {\phi }_{\kern-1.2pt\varepsilon }$ by

$$ \begin{align*} \begin{aligned} \phi_r^\varepsilon (\rho, \varphi) &= (\rho - 1 +\sqrt{2}\varepsilon^\gamma ) \sin \left(\frac{\varphi + \pi}{2} \right) , \\ \phi_3^\varepsilon (\rho, \varphi) &= \varepsilon^\gamma + (\rho - 1 + \sqrt{2}\varepsilon^\gamma) \cos \left( \frac{\varphi + \pi}{2} \right) , \end{aligned} \qquad \rho \geq 1, \quad \frac{\pi}{2}\leq \varphi \leq \pi. \end{align*} $$

The factor $\sqrt {2}$ appears because, according to the definition of $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in region $a_{\kern-1.2pt\varepsilon }$ , the image of A is $(\varepsilon ^\gamma , 0)$ , whose distance to $(0,\varepsilon ^\gamma )$ is $\sqrt {2}\varepsilon ^\gamma $ . In region b, we define $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ to be

$$ \begin{align*}\boldsymbol{u}_{\kern-1.2pt\varepsilon } (\boldsymbol{x}) = \varepsilon^\gamma \boldsymbol{\psi} \Big (\varepsilon^{-\gamma} \boldsymbol{\phi}_{\kern-1.2pt\varepsilon}(\boldsymbol{x}) \Big ),\end{align*} $$

where $\boldsymbol {\psi }$ is any axisymmetric bi-Lipschitz bijection from

$$ \begin{align*} \Big \{(x_1, x_2, x_3):\ r^2 + (x_3-1)^2 \geq 2,\ x_3\leq 0,\ r\leq 1 + |x_3|\Big \} \end{align*} $$

onto

$$ \begin{align*} \Big \{(x_1, x_2, x_3):\ x_3\leq 0,\ r\leq 1 + |x_3|\Big \} \cup B\big ( (0,0,0),\, 1\big ) \end{align*} $$

such that

1. i) $\boldsymbol {\psi }(r, x_3)=(r,x_3)$ on the half-line $r=1+|x_3|$ , $x_3\leq 0$ ,

2. ii) $\boldsymbol {\psi } \big ( \sqrt {2}\sin (\bar \varphi ), 1 + \sqrt {2}\cos (\bar \varphi ) \big ) = \Big ( \sin \big ( 2(\pi -\bar \varphi )\big ), \cos \big ( 2(\pi - \bar \varphi )\big ) \Big ) $ , $\frac {3\pi }{4}\leq \bar \varphi \leq \pi $ ,

3. iii) $\boldsymbol {\psi }\equiv \mathbf {id}$ in $\{\boldsymbol {x}:\ r^2 +(x_3-1)^2 \geq 8,\ x_3\leq 0,\ r\leq 1+|x_3|\}.$

The first two properties ensure the continuity of $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ when passing from b to $d_{\kern-1.2pt\varepsilon }$ and to $a_{\kern-1.2pt\varepsilon }$ , respectively. When $\varepsilon \to 0$ , we can check that the resulting map $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ converges to the correct limit map.

3.2.6 Region $e_{\kern-1.2pt\varepsilon }^{\prime }$

The region $e_{\kern-1.2pt\varepsilon }^{\prime }$ is described by

$$ \begin{align*} e_{\kern-1.2pt\varepsilon}^{\prime} :=\{(x_1,x_2,x_3): x_1^2 +x_2^2 + (x_3-1)^2 < \varepsilon^2,\ x_3>1\}. \end{align*} $$

We parametrize this region by

$$ \begin{align*} \boldsymbol{x} (s,\theta,\varphi)= \boldsymbol{e}_3 + (1-s) g_{\kern-1.2pt\varepsilon}(\varphi) \boldsymbol{e}_r(\theta) &+ s \big ( \varepsilon \sin\varphi \, \boldsymbol{e}_r(\theta) +\varepsilon\cos\varphi\,\boldsymbol{e}_3\big ), \\ &\qquad\qquad\qquad \ \ 0\leq s\leq 1,\quad 0\leq \theta < 2\pi,\quad 0\leq \varphi < \frac{\pi}{2}, \end{align*} $$

where, as in region $a_{\kern-1.2pt\varepsilon }^{\prime }$ , the function $g_{\kern-1.2pt\varepsilon }:[0,\frac {\pi }{2}]\to [0,\varepsilon ]$ is defined by (3.7).

For each $\theta $ , $\varphi $ the parametrization consists of an affine interpolation (with parameter s) between the pre-image on the disk $\{r\leq \varepsilon , x_3=1\}$ of the point on the (outer wall of the) bubble with zenith angle $\varphi $ (which is determined by the function $f(r)$ , according to the definition of $u_{\varphi }^\varepsilon $ in region $c_{\kern-1.2pt\varepsilon }$ ) and the point on the hemisphere $\{r^2 + (x_3-1)^2 = \varepsilon ^2,\ x_3\geq 1\}$ that (according to the definition of the map in region $e_{\kern-1.2pt\varepsilon }$ ) will be assigned the same zenith polar angle $\varphi $ .

Define $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in region $e_{\kern-1.2pt\varepsilon }^{\prime }$ through its spherical coordinates

$$ \begin{align*} u_{\varphi}^\varepsilon (s, \varphi) &:= \varphi, \\ u_{\rho}^\varepsilon (s,\varphi)&:= \Bigg ( ( \cos \varphi + 2\varepsilon^\gamma )^3 + \left [ \frac{3r}{\partial_r \big ( -\cos f_{\kern-1.2pt\varepsilon}(r) \big ) }\right]_{r=g(\varphi)} + 3\int_{\sigma=0}^s h_{\kern-1.2pt\varepsilon}(\sigma, \varphi)\mathrm{d}\sigma \Bigg )^{\frac{1}{3}} \end{align*} $$

with $h_{\kern-1.2pt\varepsilon }(s,\varphi )$ defined as in (3.8). Note that, with this definition, $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is continuous at the interface $x_3=1$ , $0\leq r\leq \varepsilon $ , between this region and region $c_{\kern-1.2pt\varepsilon }$ .

3.2.7 Region $e_{\kern-1.2pt\varepsilon }$

Region $e_{\kern-1.2pt\varepsilon }$ is given by

$$\begin{align*}e_{\kern-1.2pt\varepsilon}= \{ \boldsymbol{e}_3+ \rho \sin \varphi \boldsymbol{e}_r(\theta) + \rho \cos \varphi \boldsymbol{e}_3 : \ \varepsilon < \rho< 1, \ 0\leq \varphi \leq \frac{\pi}{2}\}. \end{align*}$$

We define $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in region $e_{\kern-1.2pt\varepsilon }$ through its spherical coordinates

$$ \begin{align*} & u_{\varphi}^\varepsilon (\rho, \varphi) := \varphi, \qquad u_{\rho}^\varepsilon (\rho ,\varphi):= \frac{1-\rho}{1-\varepsilon} u_{\rho}^\varepsilon(\varepsilon,\varphi) + \frac{\rho-\varepsilon}{1-\varepsilon} (2\cos\varphi + 6\varepsilon^{\gamma}), \\ & u_{\rho}^\varepsilon(\varepsilon,\varphi)= \Bigg ( ( \cos \varphi + 2\varepsilon^\gamma )^3 + \left [ \frac{3r}{\partial_r \big ( -\cos f(r) \big ) }\right]_{r=g(\varphi)} + 3\int_{\sigma=0}^1 h_{\kern-1.2pt\varepsilon}(\sigma, \varphi)\mathrm{d}\sigma \Bigg )^{\frac{1}{3}}, \end{align*} $$

the function $h(\sigma ,\varphi )$ being defined in (3.8). Note that, with this definition, $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is continuous at the interface $x_1^2 +x_2^2 + (x_3-1)^2=\varepsilon ^2$ , $x_3\geq 1$ with region $e_{\kern-1.2pt\varepsilon }^{\prime }$ . Also, from the relation $(a^3 +b^3)^{\frac {1}{3}}\leq a+b$ , valid for $ a,b\geq 0$ , and Lemma 3.13, it can be seen that

(3.10) $$ \begin{align} u_{\rho}^\varepsilon(\varepsilon,\varphi) \leq (\cos\varphi +2\varepsilon^\gamma) + 4\varepsilon^{\frac{1}{3}} < \cos\varphi + 6\varepsilon^{\gamma}. \end{align} $$

Hence, $u_{\rho }^\varepsilon (\cdot , \varphi )$ is an affine interpolation between $u_{\rho }^\varepsilon (\varepsilon , \varphi )$ and a value (at $\rho =1$ ) that is strictly larger (even when $\cos \varphi =0$ ).

3.2.8 Region f

This region can be described by

$$\begin{align*}f=\{ \boldsymbol{e}_3 + \rho \sin \varphi \boldsymbol{e}_r(\theta)+ \rho \cos \varphi \boldsymbol{e}_3 : 1\leq \rho, \ 0 \leq \varphi < \frac{\pi}{2}\}\cap B(\boldsymbol{0},3). \end{align*}$$

Let $u_{\rho }(\rho , \varphi )$ , $u_{\varphi }(\rho ,\varphi )$ be the spherical coordinates for the limit Conti–De Lellis map. Recall that in region f, this map is not uniquely determined; nothing is imposed on the sequence producing $\boldsymbol {u}$ apart from the image region, the continuity across the interface with region e, and the bi-Lipschitz regularity. Our aim is to prove the upper bound of Theorem 1.2 regardless of the specific definition chosen for the limit map $\boldsymbol {u}$ in region f.

Define $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ through its spherical coordinates

$$ \begin{align*} u_{\varphi}^\varepsilon (\rho, \varphi) := u_{\varphi}(\rho, \varphi), \qquad u_{\rho}^\varepsilon (\rho ,\varphi):= u_{\rho}(\rho,\varphi) + 6\varepsilon{\gamma}. \end{align*} $$

Note that when $\rho =1$ , the radial coordinate

$$ \begin{align*}u_{\rho}^\varepsilon(1,\varphi)=u_{\rho}(1,\varphi) + 6\varepsilon^{\gamma} = 2\cos\varphi + 6\varepsilon^{\gamma}\end{align*} $$

coincides with the definition given in region $e_{\kern-1.2pt\varepsilon }$ .

3.2.9 Region $d_{\kern-1.2pt\varepsilon }$

Region $d_{\kern-1.2pt\varepsilon }$ is given by

$$ \begin{align*}d_{\kern-1.2pt\varepsilon} =\{ x_1^2 +x_2^2>\varepsilon^2,\quad 0<x_3<1\}. \end{align*} $$

We use cylindrical coordinates in the domain in this region. The definition of $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ must match the definition already provided in regions $a_{\kern-1.2pt\varepsilon }$ , b, $c_{\kern-1.2pt\varepsilon }$ , $e_{\kern-1.2pt\varepsilon }$ and f.

• • On the interface between $d_{\kern-1.2pt\varepsilon }$ and b, the deformation $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is already prescribed as

  $$ \begin{align*}r\geq 1\quad \Rightarrow\quad \boldsymbol{u}_{\kern-1.2pt\varepsilon }\big ( r\,\boldsymbol{e}_r(\theta) \big ) = \varepsilon^\gamma\boldsymbol{e}_r(\theta)+ (r-1) \frac{\boldsymbol{e}_r(\theta) - \boldsymbol{e}_3}{\sqrt{2}}. \end{align*} $$

• • On the interface between $a_{\kern-1.2pt\varepsilon }$ and $d_{\kern-1.2pt\varepsilon }$ , the deformation is prescribed as

  $$ \begin{align*}\varepsilon \leq r\leq 1\quad \Rightarrow\quad \boldsymbol{u}_{\kern-1.2pt\varepsilon } \big (r\,\boldsymbol{e}_r(\theta)\big ) = \left( \frac{1-r}{1-\varepsilon} (2\varepsilon^\gamma) + \varepsilon^\gamma \frac{r-\varepsilon}{1-\varepsilon}\right) \boldsymbol{e}_r(\theta). \end{align*} $$

• • On the interface between $c_{\kern-1.2pt\varepsilon }$ and $d_{\kern-1.2pt\varepsilon }$ , the deformation is prescribed as

  $$ \begin{align*}0\leq x_3\leq 1 \quad \Rightarrow\quad \boldsymbol{u}_{\kern-1.2pt\varepsilon }\big ( \varepsilon \,\boldsymbol{e}_r(\theta) + x_3\;\boldsymbol{e}_3\big) = \left( (2\varepsilon^\gamma)^3 + x_3\cdot \frac{3\varepsilon}{f'(\varepsilon)}\right)^{1/3} \boldsymbol{e}_r(\theta), \end{align*} $$
  where
  $$ \begin{align*}f'(\varepsilon)= 2 - \frac{\varepsilon^4}{\varepsilon^4 +\varepsilon^2} - \frac{\varepsilon -\alpha_{\kern-1.2pt\varepsilon}}{\varepsilon} = 2+O(\varepsilon). \end{align*} $$

• • On the interface between $e_{\kern-1.2pt\varepsilon }$ and $d_{\kern-1.2pt\varepsilon }$ , the deformation is

  $$ \begin{align*} \varepsilon\leq r\leq 1\ \Rightarrow\ \boldsymbol{u}_{\kern-1.2pt\varepsilon }\big (\boldsymbol{e}_3 + r\,\boldsymbol{e}_r(\theta)\big ) = \frac{1-r}{1-\varepsilon} \eta_{\kern-1.2pt\varepsilon} + \frac{r-\varepsilon}{1-\varepsilon} \cdot 6\varepsilon^{\gamma}, \end{align*} $$
  with
  $$ \begin{align*} \eta_{\kern-1.2pt\varepsilon}:=u_{\rho}^\varepsilon(\varepsilon,\frac{\pi}{2})= \Bigg ( (2\varepsilon^\gamma)^3 + \frac{3r}{\sin f_{\kern-1.2pt\varepsilon}(r)f^{\prime}_{\kern-1.2pt\varepsilon}(r)}\Bigg |_{r=\varepsilon}\Bigg )^{\frac{1}{3}} = \Bigg ( (2\varepsilon^\gamma)^3 + \frac{3\varepsilon}{f^{\prime}_{\kern-1.2pt\varepsilon}(\varepsilon)} \Bigg )^{\frac{1}{3}}. \end{align*} $$
  This relevant quantity $\eta _{\kern-1.2pt\varepsilon }$ is of order $\varepsilon ^{\gamma }$ , and we recall that we chose $\gamma \leq 1/3$ .

• • On the interface with f, the deformation is prescribed to be

  $$ \begin{align*} r\geq 1 \quad \Rightarrow \quad \boldsymbol{u}_{\kern-1.2pt\varepsilon }\big (\boldsymbol{e}_3 + r \,\boldsymbol{e}_r(\theta)\big ) = \big (u_{\rho}(r, \frac{\pi}{2}) + 6\varepsilon^{\gamma}\big )\,\boldsymbol{e}_r(\theta). \end{align*} $$

The map shall be defined by composing three different auxiliary transformations. The first one is given by

(3.11) $$ \begin{align} \hat r=\begin{cases} \displaystyle \frac{(r-\varepsilon)r}{\varepsilon^{2\gamma}-\varepsilon}, & \varepsilon < r\leq \varepsilon^{2\gamma},\\ r, & r\geq \varepsilon^{2\gamma}. \end{cases} \end{align} $$

After this transformation, the radial distance lies in the same interval $(0,+\infty )$ for all $\varepsilon $ . Note also that $\hat r=\varepsilon ^{2\gamma }$ when $r=\varepsilon ^{2\gamma }$ , so that $\hat r$ is continuous as a function of r. The decisions to take it quadratic in r, and to separately study the range $\varepsilon \leq r\leq \varepsilon ^{2\gamma }$ , are to control the determinants, as will be seen in Section 3.3.8. The second transformation is the fixed axisymmetric bi-Lipschitz map $\boldsymbol {g}$ used in section d; see Section 3.1. The third transformation is

(3.12) $$ \begin{align} \boldsymbol{w}_{\kern-1.2pt\varepsilon}\big ( s\,\boldsymbol{e}_r(\theta) + z\,\boldsymbol{e}_3\big ) = w_r^\varepsilon(s,z)\,\boldsymbol{e}_r(\theta) + w_3^\varepsilon(s,z)\, \boldsymbol{e}_3 \end{align} $$

given by

(3.13) $$ \begin{align} \begin{aligned} w_r^\varepsilon(s,z) &= \omega_{\kern-1.2pt\varepsilon}(z) + s\,\sin \big ( \varphi(z)\big ), \\ w_3^\varepsilon(s, z) &= -s\,\cos \big ( \varphi(z)\big ), \end{aligned} \qquad \text{with}\quad \varphi(z):= \frac{\pi}{4} (1+\frac{z}{3}). \end{align} $$

Set

$$ \begin{align*} \boldsymbol{u}_{\kern-1.2pt\varepsilon } \big ( r\boldsymbol{e}_r(\theta) + x_3 \boldsymbol{e}_3\big ) = \boldsymbol{w}_{\kern-1.2pt\varepsilon} \Big ( \boldsymbol{g}(\hat r\boldsymbol{e}_r(\theta) + x_3\,\boldsymbol{e}_3)\Big ) =w_r^\varepsilon(s,z) \boldsymbol{e}_r(\theta) + w_3^\varepsilon(s,z)\, \boldsymbol{e}_3, \end{align*} $$

where $s=s(\hat r,x_3)$ and $z=z(\hat r,x_3)$ are the cylindrical coordinates of $\boldsymbol {g}(\hat r\boldsymbol {e}_r(\theta ) + x_3\boldsymbol {e}_3)$ and $\hat r=\hat r(r)$ is that of (3.11). The function $\omega _{\kern-1.2pt\varepsilon }$ in (3.13) is chosen such that $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ takes the prescribed values on the interfaces with b, $a_{\kern-1.2pt\varepsilon }$ , $c_{\kern-1.2pt\varepsilon }$ , $e_{\kern-1.2pt\varepsilon }$ and f. To be precise, since $\boldsymbol {g}$ is affine in the segments joining the points $A'(\hat r=1, x_3=0)$ , $B'(\hat r=0, x_3=0)$ , $C'(\hat r=0, x_3=1)$ , $D'(\hat r=1, x_3=1)$ , then

$$ \begin{align*}\omega_{\kern-1.2pt\varepsilon}(0)=\varepsilon^\gamma,\quad \omega_{\kern-1.2pt\varepsilon}(1)=2\varepsilon^\gamma, \quad \omega_{\kern-1.2pt\varepsilon}(2)=\eta_{\kern-1.2pt\varepsilon}, \quad \omega_{\kern-1.2pt\varepsilon}(3)=6\varepsilon^{\gamma},\end{align*} $$

and

$$ \begin{align*}\omega_{\kern-1.2pt\varepsilon} (\xi) = \begin{cases} \xi (2\varepsilon^\gamma) + (1-\xi)\varepsilon^\gamma, & 0\leq \xi\leq 1, \\ \left( (2\varepsilon^\gamma)^3 + (\xi-1)\cdot \displaystyle \frac{3\varepsilon}{f'(\varepsilon)} \right)^{1/3}, & 1\leq \xi\leq 2, \\ (\xi-2)\eta_{\kern-1.2pt\varepsilon} + (3-\varepsilon)\cdot (6\varepsilon^{\gamma}),& 2\leq \xi\leq 3. \end{cases} \end{align*} $$

Note also that, since $s(\hat r, 1)=u_{\rho }(\hat r,\frac {\pi }{2})$ by (3.1) and $\hat r=r$ when $r\geq 1$ by (3.11), then

$$ \begin{align*} \boldsymbol{u}_{\kern-1.2pt\varepsilon }(r\,\boldsymbol{e}_r(\theta)+x_3\boldsymbol{e}_3)= (6\varepsilon^{\gamma} + u_{\rho}(r,\frac{\pi}{2}))\boldsymbol{e}_r(\theta), \quad r\geq 1, \quad x_3=1 , \end{align*} $$

as desired (so that $\boldsymbol {u}$ is continuous at the interface with f). Analogously,

$$ \begin{align*} \boldsymbol{u}_{\kern-1.2pt\varepsilon }(r\,\boldsymbol{e}_r(\theta)+x_3\boldsymbol{e}_3)= (\varepsilon^{\gamma} + \frac{r-1}{\sqrt{2}})\boldsymbol{e}_r(\theta) - \frac{r-1}{\sqrt{2}}\boldsymbol{e}_3, \quad r\geq 1, \quad x_3=0, \end{align*} $$

as desired for continuity across the interface with b.

The deformation $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ depends on $\varepsilon $ through $\omega _{\kern-1.2pt\varepsilon }(\xi )$ and through $\hat r=\frac {(r-\varepsilon )r}{\varepsilon ^{2\gamma }-\varepsilon }$ for $\varepsilon \leq r\leq \varepsilon ^{2\gamma }$ .

3.3.1 Extra energy in $c_{\kern-1.2pt\varepsilon }$ is $2\pi $ .

The key estimate is that of the energy in this region, since it is where the singular term of the energy completely originates. Since $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ maps all discs $B^2(\boldsymbol {0},\varepsilon )\times \{x_3\}$ onto essentially (see Figure 7) the sphere $S\big ( (0,0,\frac {1}{2}), \frac {1}{2}\big )$ , it follows that for every fixed $0<x_3<1$ ,

$$\begin{align*}\int_{B^2(\boldsymbol{0},\varepsilon)\times \{x_3\}} |\operatorname{cof} D\boldsymbol{u}_{\kern-1.2pt\varepsilon }| \mathrm{d}\mathcal H^2 \approx \mathcal H^2\Big ( S\big ( (0,0,\frac{1}{2}), \frac{1}{2}\big ) \Big ) = 4\pi \cdot (\frac{1}{2})^2 = 2\pi. \end{align*}$$

By integrating with respect to $x_3$ , recalling that the normal contraction $\frac {\partial \boldsymbol {u}_{\kern-1.2pt\varepsilon }}{x_3}$ is negligible (see Figure 7), and using that

$$\begin{align*}|D\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 = \Big |\frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_1}\Big |^2 + \Big |\frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_2}\Big |^2 + \Big |\frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_3}\Big |^2 \approx \Big |\frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_1}\Big |^2 + \Big |\frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_2}\Big |^2 \geq 2\Big | \frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_1} \wedge \frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_2} \Big |, \end{align*}$$

it follows that the Dirichlet energy $\int |D\boldsymbol {u}_{\kern-1.2pt\varepsilon }|^2 \mathrm {d}\boldsymbol {x}$ cannot be small in $c_{\kern-1.2pt\varepsilon }$ .

Proposition 3.1. Let $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ be as in (3.5). Then

$$\begin{align*}\lim_{\kern-1.2pt\varepsilon\to 0} \int_{c_{\kern-1.2pt\varepsilon}} [| D \boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2+H(\det D \boldsymbol{u}_{\kern-1.2pt\varepsilon }) ] \mathrm{d} \boldsymbol{x} = 2\pi. \end{align*}$$

Proof. First we notice that in the region $c_{\kern-1.2pt\varepsilon }$ , we have that $ \det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }=1$ . Hence, $ \int _{c_{\kern-1.2pt\varepsilon }} H(\det \boldsymbol {u}_{\kern-1.2pt\varepsilon } )=H(1)|c_{\kern-1.2pt\varepsilon }| \rightarrow 0$ as $\varepsilon $ tends to zero. Then, using (B.4), we observe that the above is equivalent to proving that

(3.14) $$ \begin{align} \lim_{\kern-1.2pt\varepsilon \rightarrow 0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} \Big [ \underbrace{r|\partial_r u^\varepsilon_{\rho}|^2 + r|u^\varepsilon_{\rho}\partial_r u^\varepsilon_{\varphi}|^2}_{:=I} + \underbrace{\frac{1}{r}|u^\varepsilon_{\rho} \sin u^\varepsilon_{\varphi}|^2}_{:=II} + \underbrace{r|\partial_{x_3} u^\varepsilon_{\rho}|^2}_{:=III} \Big ] \mathrm{d} r \mathrm{d} x_3 =1. \end{align} $$

We remark that our map satisfies $\partial _{x_3} \boldsymbol {u}^\varepsilon \approx 0$ and $|\partial _r u_{\rho }^\varepsilon |^2+|u^\varepsilon _{\rho }\partial _ru^\varepsilon _{\varphi }|^2\approx |u^\varepsilon _{\rho } \sin u^\varepsilon _{\varphi }|^2/r$ , which is due to the conformality of $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ in the planes orthogonal to $\boldsymbol {e}_3$ . Half of the energy will come from I and half from $II$ . The contribution of $III$ , as will be shown, is negligible. More precisely, the claim follows by combining (3.14) with Lemmas 3.2, 3.4 and 3.5.

Lemma 3.2. Let $u^{\varepsilon }_{\rho }$ and $u^{\varepsilon }_{\varphi }$ be defined as in (3.6). Then

$$ \begin{align*} \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|\partial_r u^{\varepsilon}_{\rho}|^2 + r|u^{\varepsilon}_{\rho}\partial_r u^{\varepsilon}_{\varphi}|^2\mathrm{d} r \mathrm{d} x_3 =\frac{1}{2}. \end{align*} $$

The proof consists in establishing $ \partial _r u^{\varepsilon }_{\rho } \approx \partial _r \big (\cos f_{\kern-1.2pt\varepsilon }(r)\big ), \ u_{\rho } \partial _r u_{\varphi }^\varepsilon \approx \big (\cos f_{\kern-1.2pt\varepsilon }(r) \big ) f^{\prime }_{\kern-1.2pt\varepsilon }(r), $ where $f_{\kern-1.2pt\varepsilon } $ is defined in (3.4), and in using the following result.

Lemma 3.3. Let $f_{\kern-1.2pt\varepsilon }$ be defined as in (3.4). Then

$$\begin{align*}\lim_{\kern-1.2pt\varepsilon\to0} \int_{r=0}^{\varepsilon} rf_{\kern-1.2pt\varepsilon}^{\prime}(r)^2 \mathrm{d} r = \frac{1}{2}. \end{align*}$$

Proof. By using that $\alpha _{\kern-1.2pt\varepsilon }=\arctan (\varepsilon ) \leq \varepsilon $ , we can write that

$$ \begin{align*} \int_{r=0}^{\varepsilon} rf^{\prime}_{\kern-1.2pt\varepsilon}(r)^2 \mathrm{d} r &= \int_{r=0}^{\varepsilon} \left[\frac{1}{1+\frac{r^2}{\varepsilon^4}} \varepsilon^{-2} + \frac{\alpha_{\kern-1.2pt\varepsilon}}{\varepsilon} \right]^2 r\mathrm{d} r = \int_{u=1}^{1+\varepsilon^{-2}} \left[ \frac{1}{u}\varepsilon^{-2} + \frac{\alpha_{\kern-1.2pt\varepsilon}}{\varepsilon}\right]^2 \frac{\varepsilon^4}{2} \mathrm{d} u \\ &= \frac{1}{2}\left[ \int_{u=1}^{1+\varepsilon^{-2}} \frac{1}{u^2} + 2\frac{\varepsilon \alpha_{\kern-1.2pt\varepsilon}}{u} + \alpha_{\kern-1.2pt\varepsilon}^2\varepsilon^2 \mathrm{d} u \right] = \frac{1}{2} + O(\varepsilon^2 \ln|\varepsilon|).\\[-34pt] \end{align*} $$

For notational simplicity, we will drop the subscript and superscript $\varepsilon $ in the proofs of the following results. In the proof of Lemma 3.2, use shall be made of the following expressions. First,

$$ \begin{align*} \partial_r (u_{\rho})^3 = 3(\cos f(r) + 2\varepsilon^\gamma)^2 \partial_r \big ( \cos f(r)\big ) + x_3 \partial_r \left[ \frac{3r}{\partial_r (-\cos f(r))}\right]. \end{align*} $$

Second, since $ \partial _r u_{\rho } = \frac {\partial _r (u_{\rho }^3)}{3u_{\rho }^2}, $ it follows that

(3.15) $$ \begin{align} \partial_r u_{\rho} = \frac{(\cos f(r) + 2\varepsilon^\gamma)^2}{u_{\rho}^2} \partial_r \big ( \cos f(r)\big) + x_3 \frac{ \displaystyle \partial_r \left[ \frac{r}{\partial_r (-\cos f(r))}\right]}{u_{\rho}^2}. \end{align} $$

Proof of Lemma 3.2.

Claim 1:

$$ \begin{align*} \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|\partial_r u_{\rho}|^2 \,\mathrm{d} r \mathrm{d} x_3 = \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r(\sin^2 f(r))(f'(r))^2\mathrm{d} r \mathrm{d} x_3. \end{align*} $$

In order to prove this claim, we begin by applying the relation $a^2-b^2 = 2b(a-b) + (a-b)^2$ :

(3.16) $$ \begin{align} \int_{x_3=0}^1 &\int_{r=0}^{\varepsilon} r|\partial_r u_{\rho}|^2 - r(\sin^2 f(r))(f'(r))^2\,\mathrm{d} r \mathrm{d} x_3\notag \\ &\qquad\qquad = 2\int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r\partial_r\big (\cos f(r)\big ) \Big( \partial_r u_{\rho} -\partial_r\big ( \cos f(r) \big ) \Big ) \mathrm{d} r \mathrm{d} x_3 \notag \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad + \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r\Big ( \partial_r u_{\rho} -\partial_r\big ( \cos f(r)\big ) \Big )^2 \mathrm{d} r \mathrm{d} x_3. \end{align} $$

From (3.15), it follows that

(3.17) $$ \begin{align} \partial_ru_{\rho} -\partial_r\big ( \cos f(r)\big ) = \partial_r \big ( - \cos f(r)\big ) \left[ 1- \frac{(\cos f(r) + 2\varepsilon^\gamma)^2}{u_{\rho}^2} \right] + x_3 u_{\rho}^{-2} \partial_r \left( \frac{r}{\partial_r (-\cos f(r))}\right). \end{align} $$

The first term can be estimated via the relation

$$ \begin{align*}a^2-b^2 = (a+b)(a-b) = (a+b)\frac{a^3-b^3}{a^2+ab+b^2}, \end{align*} $$

yielding

(3.18) $$ \begin{align} u_{\rho}^2 - (\cos f(r) + 2\varepsilon^\gamma)^2 =\Big (u_{\rho} + (\cos f(r) + 2\varepsilon^\gamma)\Big ) \frac{ x_3 \cdot \frac{3r}{\partial_r (-\cos f(r))}}{ u_{\rho}^2 + u_{\rho} (\cos f(r)+2\varepsilon^\gamma)+ (\cos f(r)+2\varepsilon^\gamma)^2}. \end{align} $$

A first conclusion is that the expression on the left-hand side of (3.18) is positive since $0\leq f(r)<\frac {\pi }{2}$ and $-\cos f(r)$ is increasing. Second, since $ \partial _r(-\cos f(r))>0$ , it holds that

(3.19) $$ \begin{align} \cos f(r) + 2\varepsilon^\gamma \leq u_{\rho}. \end{align} $$

Therefore, bounding $u_{\rho }^2 + u_{\rho } (\cos f(r)+2\varepsilon ^\gamma ) +(\cos f(r)+2\varepsilon ^\gamma )^2$ from below by $u_{\rho }^2$ , it can be seen that

(3.20) $$ \begin{align} |\partial_r(-\cos f(r))| &\left |1- \frac{(\cos f(r) + 2\varepsilon^\gamma)^2}{u_{\rho}^2}\right| \notag \\ &\qquad\quad\leq \partial_r(-\cos f(r))\cdot u_{\rho}^{-2}\cdot (2u_{\rho})\cdot u_{\rho}^{-2}\cdot \frac{3r}{\partial_r (-\cos f(r))} \leq 6ru_{\rho}^{-3}. \end{align} $$

Putting together (3.17), (3.20) and Part c) of Lemma A.1, the following bound is obtained (for small enough $\varepsilon $ ):

$$ \begin{align*} | \partial_ru_{\rho} -\partial_r\big ( \cos f(r)\big )| \leq 6ru_{\rho}^{-3} + 64\sqrt{2} u_{\rho}^{-2} = u_{\rho}^{-2} (6 r u_{\rho}^{-1} + 64\sqrt{2}). \end{align*} $$

Thanks to (3.19) and by observing that

(3.21) $$ \begin{align} u_{\rho} \geq 2\varepsilon^\gamma \geq 2\varepsilon> 2r, \end{align} $$

we find $ | \partial _ru_{\rho } -\partial _r\big ( \cos f(r)\big )| \leq 25 \varepsilon ^{-2\gamma }. $ Plugging that into (3.16), and using Part b) of Lemma A.1, the conclusion is that

$$ \begin{align*} \left| \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|\partial_r u^{\varepsilon}_{\rho}|^2 - r(\sin^2 f(r))(f'(r))^2\,\mathrm{d} r \mathrm{d} x_3 \right| \\ \leq 25 \varepsilon^{-2\gamma} \cdot 12 \varepsilon^2 |\ln \varepsilon| + 25^2 \varepsilon^{-4\gamma} \int_0^\varepsilon r\mathrm{d} r, \end{align*} $$

which vanishes as $\varepsilon \to 0$ since $0<\gamma \leq \frac {1}{3} <\frac {1}{2}$ .

Claim 2:

$$ \begin{align*} \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|u_{\rho}^{\varepsilon}\partial_r u^{\varepsilon}_{\varphi}|^2 \,\mathrm{d} r \mathrm{d} x_3 = \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r(\cos^2 f(r))(f'(r))^2\mathrm{d} r \mathrm{d} x_3. \end{align*} $$

In order to prove this claim, we first observe that

$$ \begin{align*} & \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|u_{\rho}\partial_r u_{\varphi}|^2 - r(\cos^2 f(r))(f'(r))^2\mathrm{d} r \mathrm{d} x_3 \\ &\qquad \qquad \qquad = \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} rf'(r)^2(|u_{\rho}|^2 - \cos^2 f(r))\mathrm{d} r\mathrm{d} x_3 \\ &\qquad \qquad \qquad = \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} rf'(r)^2(|u_{\rho}|^2 - (\cos f(r)+2\varepsilon^\gamma)^2)\mathrm{d} r\mathrm{d} x_3 \\ &\qquad \qquad \qquad \qquad \qquad \qquad + 4\varepsilon^\gamma \int_{r=0}^{\varepsilon} rf'(r)^2 \cos f(r)\mathrm{d} r +4 \varepsilon^{2\gamma} \int_{r=0}^{\varepsilon} rf'(r)^2 \mathrm{d} r. \end{align*} $$

By Lemma 3.3, and considering that $0\leq \cos f(r)\leq 1$ , the last two terms are bounded by $4\varepsilon ^\gamma +4\varepsilon ^{2\gamma }$ and vanish as $\varepsilon $ goes to zero. As for the first term, by (3.18), (3.19) and Part a) of Lemma A.1,

$$ \begin{align*} u_{\rho}^2 -(\cos f(r) + 2\varepsilon^\gamma)^2 &\leq \frac{2u_{\rho} \cdot x_3\cdot \displaystyle \frac{3r}{\partial_r\big(\cos f(r)\big)}}{u_{\rho}^2+0+0} \leq 6 u_{\rho}^{-1}\cdot 2\varepsilon^{-2}(\varepsilon^4+r^2)^{3/2}\\ &\leq 6\varepsilon^{-\gamma}\cdot \varepsilon^{-2} \cdot (2\varepsilon^2)^{3/2} = 12\sqrt{2}\varepsilon^{1-\gamma}. \end{align*} $$

The claim follows by combining the above with Lemma 3.3.

Conclusion: By Claims 1 and 2, together with Lemma 3.3,

$$ \begin{align*} \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|\partial_r u^{\varepsilon}_{\rho}|^2 + r & |u^{\varepsilon}_{\rho}\partial_r u^{\varepsilon}_{\varphi}|^2\mathrm{d} r \mathrm{d} x_3 \\ & \ = \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r\sin^2 f(r)\,f'(r)^2 + r\cos^2 f(r)\,f'(r)^2 \mathrm{d} r = \frac{1}{2}. \end{align*} $$

Lemma 3.4. Let $u^{\varepsilon }_{\rho }$ and $u^{\varepsilon }_{\varphi }$ be defined as in (3.6). Then

$$ \begin{align*} \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} \frac{1}{r}|u^{\varepsilon}_{\rho} \sin u^{\varepsilon}_{\varphi}|^2 \mathrm{d} r \mathrm{d} x_3 =\frac{1}{2}. \end{align*} $$

Proof. Writing $u_{\rho }$ as $\cos f(r) + (u_{\rho }-\cos f(r))$ , the integral expression becomes

$$ \begin{align*} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} \frac{1}{r}&|u_{\rho} \sin u_{\varphi}|^2 \mathrm{d} r \mathrm{d} x_3 = \int_{r=0}^{\varepsilon} \frac{1}{r}\cos^2 f(r) \sin^2 f(r) \mathrm{d} r\\ & + \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} \frac{1}{r} \Big [2\cos f(r)\,(u_{\rho}-\cos f(r)) + (u_{\rho}-\cos f(r))^2 \Big ]\sin ^2 f(r) \mathrm{d} r \mathrm{d} x_3. \end{align*} $$

By (A.1), the first integral is given by

$$ \begin{align*} \int_{r=0}^{\varepsilon} \frac{1}{r}\cos^2 f(r) \sin^2 f(r) \mathrm{d} r &= \int_0^\varepsilon \frac{\varepsilon^4 r\mathrm{d} r }{(\varepsilon^4+r^2)^2} = \frac{\varepsilon^4}{2(\varepsilon^4+r^2)}\Big |_{r=\varepsilon}^0 \overset{\varepsilon\to0}{\longrightarrow}\frac{1}{2}. \end{align*} $$

Regarding the second integral, we use the relation $(a-b)(a^2+ab+b^2)=a^3-b^3$ :

$$\begin{align*}u_{\rho}-\cos f(r) = 2\varepsilon^\gamma + u_{\rho} - (\cos f(r) + 2\varepsilon^\gamma) =2\varepsilon^\gamma + \frac{x_3\cdot \frac{3r}{\partial_r(-\cos f(r))}}{u_{\rho}^2 + u_{\rho} (\cos f(r) + 2\varepsilon^\gamma) + (\cos f(r) + 2\varepsilon^\gamma)^2}. \end{align*}$$

Part a) of Lemma A.1 and (3.19) yield

$$ \begin{align*} u_{\rho}-\cos f(r) \leq 2\varepsilon^\gamma + (2\varepsilon^\gamma)^{-2}\varepsilon^{-2}(\varepsilon^4+r^2)^{3/2} = 2\varepsilon^\gamma + \frac{\sqrt{2}}{2}\varepsilon^{1-2\gamma}. \end{align*} $$

Hence, by using that $ \gamma \leq 1/3 <1/2$ , we arrive at

$$ \begin{align*} &\left| \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} \frac{1}{r} \Big [2\cos f(r)\,(u_{\rho}-\cos f(r)) + (u_{\rho}-\cos f(r))^2 \Big ]\sin ^2 f(r) \mathrm{d} r \mathrm{d} x_3 \right| \\ &\qquad\qquad \ \ \leq \Big [ 2\cdot \big ( 2\varepsilon^\gamma + \frac{\sqrt{2}}{2}\varepsilon^{1-2\gamma} \big ) + \big ( 2\varepsilon^\gamma + \frac{\sqrt{2}}{2}\varepsilon^{1-2\gamma} \big )^2 \Big ] \int_{r=0}^{\varepsilon} \frac{1}{r}\sin ^2 f(r) \mathrm{d} r\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad = \big [ O(\varepsilon^\gamma) + O(\varepsilon^{1-2\gamma})\big ] \underbrace{\int_{r=0}^{\varepsilon} \frac{2r}{\varepsilon^4+r^2}\mathrm{d} r }_{=\ln (1+\varepsilon^{-2})} \overset{\varepsilon\to0}{\longrightarrow} 0.\\[-24pt] \end{align*} $$

Lemma 3.5. Let $u^{(\varepsilon )}_{\rho }$ and $u^{(\varepsilon )}_{\varphi }$ be defined as in (3.6). Then

$$ \begin{align*} \lim_{\kern-1.2pt\varepsilon\to0} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|\partial_{x_3} u^{\varepsilon}_{\rho}|^2 \mathrm{d} r \mathrm{d} x_3 =0. \end{align*} $$

Proof. We start by observing that $ \partial _{x_3}u_{\rho } = \frac {1}{3} u_{\rho }^{-2} \partial _{x_3} \big ( u_{\rho }^3\big ) =u_{\rho }^{-2} \frac {r}{\partial _r\big (-\cos f(r)\big )}. $ By (3.19) and Part a) of Lemma A.1,

$$ \begin{align*} \int_{x_3=0}^1 \int_{r=0}^{\varepsilon} r|\partial_{x_3} u^{\varepsilon}_{\rho}|^2 \mathrm{d} r \mathrm{d} x_3 \leq \int_{r=0}^{\varepsilon} r |(2\varepsilon^\gamma)^{-2}\cdot 2\varepsilon^{-2}( \underbrace{\varepsilon^4 +r^2}_{\leq 2\varepsilon^2})^{3/2}|^2 \mathrm{d} r \leq \varepsilon^{4(1-\gamma)} \overset{\varepsilon\to 0}{\longrightarrow}0.\\[-34pt] \end{align*} $$

In the rest of this section, we will prove the following

Proposition 3.6. Let $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ be the recovery sequence defined in Section 3.2. Then

(3.22) $$ \begin{align} \lim_{\kern-1.2pt\varepsilon \rightarrow 0} \int_{a_{\kern-1.2pt\varepsilon}^{\prime}\cup e_{\kern-1.2pt\varepsilon}^{\prime}} \left[ |D \boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2+H(\det D \boldsymbol{u}_{\kern-1.2pt\varepsilon }) \right] \mathrm{d} \boldsymbol{x} =0 \end{align} $$

and

(3.23) $$ \begin{align} \lim_{\kern-1.2pt\varepsilon \rightarrow 0} \int_{a_{\kern-1.2pt\varepsilon}\cup b_{\kern-1.2pt\varepsilon} \cup d_{\kern-1.2pt\varepsilon} \cup f} \left[ |D \boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2+H(\det D \boldsymbol{u}_{\kern-1.2pt\varepsilon }) \right] \mathrm{d} \boldsymbol{x} =\int_{a \cup b \cup d \cup f }\left[ |D \boldsymbol{u}|^2+H(\det D \boldsymbol{u}) \right] \mathrm{d} \boldsymbol{x}. \end{align} $$

We remark that all the regions involved in the previous proposition are disjoint. Thus, we will work separately in each of these regions.

3.3.2 Extra energy in $a_{\kern-1.2pt\varepsilon }^{\prime }$ is negligible

Inverse of the parametrization of the reference configuration

Using Lemma A.2 and, for example, Ball’s global invertibility theorem [Reference Ball4] (considering that $(s, \cdot , \varphi )\mapsto \boldsymbol {x}(s, \cdot , \varphi )$ , seen as a map to $\mathbb {R}^2$ , is one-to-one on the boundary of $[0,1]\times [0, \frac {\pi }{2}-\delta ]$ for every small $\delta $ ), we obtain that the parametrization of the reference domain (excluding $\varphi =\frac {\pi }{2}$ , which collapses to the circle $\{r=\varepsilon ,\, x_3=0\}$ ) is a diffeomorphism and that s, $\theta $ and $\varphi $ can be obtained as functions of $\boldsymbol {x}$ in the interior of region $a_{\kern-1.2pt\varepsilon }^{\prime }$ . Inverting the coefficient matrix for $\partial _s\boldsymbol {x}$ , $\partial _{\theta } \boldsymbol {x}$ , $\partial _{\varphi } \boldsymbol {x}$ in the basis $(\boldsymbol {e}_r, \boldsymbol {e}_{\theta }, \boldsymbol {e}_3)$ , we find that

(3.24) $$ \begin{align} \nabla s &= \frac{ s\sin \varphi\,\boldsymbol{e}_r - \big ( (1-s)\varepsilon^{-1} g'(\varphi) + s\cos \varphi\big ) \boldsymbol{e}_3}{ (1-s) g'(\varphi)\cos\varphi + s (\varepsilon-g(\varphi)\sin\varphi)}, \nonumber \\ \nabla \theta &= \frac{1}{(1-s) g(\varphi) + s \varepsilon\sin\varphi} \boldsymbol{e}_{\theta}, \nonumber \\ \nabla \varphi &= \frac{\cos \varphi \,\boldsymbol{e}_r + (\sin \varphi - \varepsilon^{-1}g(\varphi))\boldsymbol{e}_3}{ (1-s) g'(\varphi)\cos\varphi + s (\varepsilon-g(\varphi)\sin\varphi)}. \end{align} $$

The Dirichlet energy: Based on the representation

$$ \begin{align*} D\boldsymbol{u}_{\kern-1.2pt\varepsilon } = \partial_s\boldsymbol{u}_{\kern-1.2pt\varepsilon } \otimes \nabla s + \partial_{\theta} \boldsymbol{u}_{\kern-1.2pt\varepsilon } \otimes \nabla \theta + \partial_{\varphi} \boldsymbol{u}_{\kern-1.2pt\varepsilon } \otimes \nabla \varphi, \qquad |D\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 = \operatorname{tr} D\boldsymbol{u}_{\kern-1.2pt\varepsilon }^T D\boldsymbol{u}_{\kern-1.2pt\varepsilon }, \end{align*} $$

it follows that

$$ \begin{align*} |D\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 &= |\nabla s|^2 |\partial_s\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 + 2(\nabla s\cdot \nabla \varphi)(\partial_s\boldsymbol{u}_{\kern-1.2pt\varepsilon } \cdot \partial_{\varphi} \boldsymbol{u}_{\kern-1.2pt\varepsilon }) + |\nabla \theta|^2 |\partial_{\theta} \boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 + |\nabla \varphi|^2 |\partial_{\varphi} \boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 \\ & \! \! \! = |\nabla s|^2|\partial_s u_{\rho}^\varepsilon|^2 + 2(\nabla s\cdot \nabla \varphi) \partial_s u_{\rho}^\varepsilon \partial_{\varphi} u_{\rho}^\varepsilon + |\nabla \theta|^2 |u_{\rho}^\varepsilon|^2 \sin^2\varphi + |\nabla \varphi|^2 (|u_{\rho}^\varepsilon|^2 + |\partial_{\varphi} u_{\rho}^\varepsilon|^2) \\ & \! \! \! \leq |\nabla s|^2|\partial_s u_{\rho}^\varepsilon|^2 + 2(|\nabla s||\partial_s u_{\rho}^\varepsilon|)(|\nabla \varphi| |\partial_{\varphi} u_{\rho}^\varepsilon|) + |\nabla \theta|^2 |u_{\rho}^\varepsilon|^2 \sin^2\varphi + |\nabla \varphi|^2 (|u_{\rho}^\varepsilon|^2 + |\partial_{\varphi} u_{\rho}^\varepsilon|^2). \end{align*} $$

Cauchy’s inequality then yields that

(3.25) $$ \begin{align} |D\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 &\leq 2|\nabla s|^2|\partial_s u_{\rho}^\varepsilon|^2 + |\nabla \theta|^2 |u_{\rho}^\varepsilon|^2 \sin^2\varphi + |\nabla \varphi|^2 (|u_{\rho}^\varepsilon|^2 + 2|\partial_{\varphi} u_{\rho}^\varepsilon|^2). \end{align} $$

Note also that

$$ \begin{align*} \int_{a_{\kern-1.2pt\varepsilon}^{\prime}} |D\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 \mathrm{d}\boldsymbol{x} = 2\pi \int_{s=0}^1 \int_{\varphi=0}^{\frac{\pi}{2}} |D\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 h_{\kern-1.2pt\varepsilon}(s,\varphi) \sin\varphi \mathrm{d}\varphi \mathrm{d} s. \end{align*} $$

$\underline {\text {Estimates for } g_{\kern-1.2pt\varepsilon } \text { and } h_{\kern-1.2pt\varepsilon }}$

In order to estimate the partial derivatives of $ u_{\rho }^\varepsilon $ , it is important to control first the derivatives of the functions $g_{\kern-1.2pt\varepsilon }$ and $h_{\kern-1.2pt\varepsilon }$ that appear in its definition. This is the object of Lemma A.2 in the appendix.

$\underline {\text {Estimates for } u_{\rho }^\varepsilon }$

Lemma 3.7. For all $\varphi \in [0, \frac {\pi }{2}]$ , all $s\in [0,1]$ and all positive $\varepsilon $ such that $\varepsilon ^{2-2\gamma } < \frac {7}{9\pi \sqrt {2}}$ ,

$$ \begin{align*} \frac{1}{4}(\cos \varphi + 2\varepsilon^\gamma)\leq u_{\rho}^\varepsilon(s,\varphi) \leq \cos \varphi + 2\varepsilon^\gamma\leq 2 , \quad |\partial_s u_{\rho}^\varepsilon|\leq C \varepsilon^{2-2\gamma}\cos\varphi, \quad |\partial_{\varphi} u_{\rho}^\varepsilon|=O(1). \end{align*} $$

Proof. Since, by Lemma A.5,

$$ \begin{align*} 3\left| \int_{\sigma=0}^1 h(\sigma,\varphi)\mathrm{d}\sigma \right| \leq \frac{9\pi\sqrt{2}}{2} \varepsilon^2 \cos\varphi, \end{align*} $$

then

$$ \begin{align*} (\cos\varphi +2\varepsilon^\gamma )^3\geq u_{\rho}^\varepsilon(s,\varphi)^3 \geq (\cos\varphi +2\varepsilon^\gamma )^3 - \frac{9\pi\sqrt{2}}{2}\varepsilon^2\cos\varphi. \end{align*} $$

$\underline {\text {If } 2\varepsilon ^\gamma \leq \cos \varphi }$ , then

$$ \begin{align*}\frac{7}{8}\cos^2\varphi \geq \frac{7}{2}\varepsilon^{2\gamma} \geq \frac{9\pi\sqrt{2}}{2}\varepsilon^2 \end{align*} $$

and

$$ \begin{align*} u_{\rho}^\varepsilon(s,\varphi)^3 \geq \cos^3\varphi - \frac{7}{8}\cos^2 \varphi \cdot \cos\varphi \geq \left( \frac{1}{2}\cos\varphi\right)^3. \end{align*} $$

Consequently,

$$ \begin{align*} u_{\rho}^\varepsilon(s,\varphi)\geq \frac{1}{2}\cos\varphi = \frac{1}{2} (\cos\varphi + 2\varepsilon^\gamma) \frac{1}{1 + \frac{2\varepsilon^\gamma}{\cos\varphi}} \geq \frac{1}{4}(\cos\varphi + 2\varepsilon^\gamma). \end{align*} $$

$\underline {\text {If } 2\varepsilon ^\gamma> \cos \varphi ,}$

$$ \begin{align*} u_{\rho}^\varepsilon(s,\varphi)^3 \geq 8\varepsilon^{3\gamma} - \frac{9\pi\sqrt{2}}{2}\varepsilon^2 \cdot 2\varepsilon^\gamma = 8\varepsilon^{3/2} \left( 1- \frac{9\pi\sqrt{2}}{8}\varepsilon^{2-2\gamma} \right) \geq \left((2\sqrt{\varepsilon})\cdot \frac{1}{2}\right)^3. \end{align*} $$

Therefore,

$$ \begin{align*} u_{\rho}^\varepsilon(s,\varphi) \geq \frac{1}{2} (\cos\varphi + 2\varepsilon^\gamma) \frac{1}{\frac{\cos\varphi}{2\varepsilon^\gamma}+1} \geq \frac{1}{4}(\cos\varphi + 2\varepsilon^\gamma). \end{align*} $$

Regarding $\partial _{\varphi } u_{\rho }^\varepsilon $ ,

$$ \begin{align*} 3\partial_{\varphi} u_{\rho}^\varepsilon = (u_{\rho}^\varepsilon)^{-2}\partial_{\varphi} \Big ( (u_{\rho}^\varepsilon)^3\Big ) =-3 \underbrace{\left(\frac{\cos\varphi + 2\varepsilon^\gamma}{u_{\rho}^\varepsilon}\right)^2}_{\leq 4^2}\sin\varphi -3(u_{\rho}^\varepsilon)^{-2} \underbrace{\int_{\sigma=0}^s \partial_{\varphi} h(\sigma, \varphi) \mathrm{d}\sigma}_{=O(\varepsilon)}. \end{align*} $$

Since

$$ \begin{align*} (u_{\rho}^\varepsilon)^2 \geq \frac{1}{4^2}(\cos \varphi + 2\varepsilon^\gamma)^2 \geq \frac{1}{4}\varepsilon^{2\gamma}, \end{align*} $$

and $\gamma \leq \frac {1}{3} < \frac {1}{2}$ , the derivative $\partial _{\varphi } u_{\rho }^\varepsilon $ is bounded uniformly with respect to $\varepsilon $ .

The estimate for the derivative with respect to s is now straightforward:

$$ \begin{align*} 3|\partial_s u_{\rho}^\varepsilon| = \left| (u_{\rho}^\varepsilon)^{-2}\partial_s \Big ( (u_{\rho}^\varepsilon)^3\Big ) \right| \leq 3\cdot \frac{ \frac{3\pi\sqrt{2}}{2} \varepsilon^2\cos \varphi}{\frac{1}{4}\varepsilon^{2\gamma}} \leq C\varepsilon^{2-2\gamma}\cos\varphi. \end{align*} $$

$\underline {\text {Integral estimates for } \nabla s, \nabla \theta \text { and } \nabla \varphi }$

Lemma 3.8.

$$ \begin{align*} & \int_{a_{\kern-1.2pt\varepsilon}^{\prime}} |\nabla \varphi|^2\, \mathrm{d}\boldsymbol{x} = O (\varepsilon^2|\ln \varepsilon|), \qquad \int_{a_{\kern-1.2pt\varepsilon}^{\prime}}(\varepsilon \cos\varphi)^2|\nabla s|^2\,\mathrm{d}\boldsymbol{x} = O(\varepsilon|\ln \varepsilon|),\\ & \text{and}\quad \int_{a_{\kern-1.2pt\varepsilon}^{\prime}} |\nabla \theta|^2 \sin^2\! \varphi\,\mathrm{d}\boldsymbol{x} = O (\varepsilon|\ln \varepsilon|^2). \end{align*} $$

Proof. By (A.5) and Lemma A.6, the modulus of the numerator in (3.24) can be estimated by

$$ \begin{align*} \left| \cos\varphi \right| + \left| \sin \varphi -\varepsilon^{-1}g(\varphi) \right| &= \left| \cos\varphi \right| + \left| \varepsilon^{-1} (\varepsilon-g(\varphi)) - (1-\sin\varphi) \right| \leq \varepsilon^{-1} (\varepsilon-g(\varphi)) +2\cos \varphi \\ &\leq \varepsilon^{-1} (\varepsilon-g(\varphi)) + 6\frac{\varepsilon-g(\varphi)}{\max\{\varepsilon, \frac{g(\varphi)}{\varepsilon}\}} \leq 7\varepsilon^{-1}(\varepsilon-g(\varphi)). \end{align*} $$

Therefore, by Lemma A.5, for all $\varphi \in [0,\frac {\pi }{2}]$ and $s\in [0,1]$ ,

$$ \begin{align*} |\nabla \varphi|^2 h(s,\varphi) &\leq \frac{49\varepsilon^{-2}(\varepsilon-g(\varphi))^2 }{ |(1-s)g'(\varphi)\cos \varphi + s(\varepsilon - g(\varphi)\sin\varphi) |^2} \cdot \\ & \quad \quad \cdot \varepsilon \Big ( \underbrace{(1-s) \frac{g(\varphi)}{\sin \varphi}+ s\varepsilon}_{\leq \sqrt{2} \varepsilon} \Big ) \Big ( (1-s)g'(\varphi)\cos \varphi + s(\varepsilon - g(\varphi)\sin\varphi) \Big ) \\ & \leq \frac{C(\varepsilon-g(\varphi))^2}{(1-s)g'(\varphi)\cos \varphi + s(\varepsilon - g(\varphi)\sin\varphi)}. \end{align*} $$

Hence, by Lemma A.9,

$$ \begin{align*} \int_{\varphi=0}^{\frac{\pi}{2}}\int_{s=0}^1 |\nabla \varphi|^2 h(s,\varphi) \sin\varphi\,\mathrm{d} s \mathrm{d}\varphi & \leq \int_{\varphi=0}^{\frac{\pi}{2}} C\underbrace{(\varepsilon-g(\varphi))^2}_{\leq \varepsilon^2}|\ln \varepsilon| \mathrm{d}\varphi =O(\varepsilon^2|\ln \varepsilon|). \end{align*} $$

As for $\nabla \theta $ , observe first that

$$ \begin{align*} |\nabla \theta|^2 \sin^2 \varphi \, h(s,\varphi) \sin \varphi &= \frac{\varepsilon \Big ( (1-s)g'(\varphi)\cos \varphi + s(\varepsilon - g(\varphi)\sin\varphi) \Big )}{(1-s) \frac{g(\varphi)}{\sin\varphi}+ s\varepsilon } \cdot\sin\varphi. \end{align*} $$

By Lemma A.6,

$$ \begin{align*} |\nabla \theta|^2 \sin^2 \varphi \, h(s,\varphi) \sin \varphi \leq \frac{ C\varepsilon \min \{\frac{1}{\varepsilon}, \frac{\varepsilon}{g(\varphi)}\}g'(\varphi)\cos\varphi}{(1-s) \frac{g(\varphi)}{\sin\varphi}+ s\varepsilon }. \end{align*} $$

$\underline {\text {Case when } g(\varphi )\leq \varepsilon ^2:}$ calling $r:=g(\varphi )$ , it is easy to see that

$$ \begin{align*} g'(\varphi) = \frac{1}{ \frac{\varepsilon^2}{r^2 +\varepsilon^4} + \frac{\alpha_{\kern-1.2pt\varepsilon}}{\varepsilon} } \leq 2\varepsilon^2 \quad \text{and}\quad g(\varphi)= \int_{t=0}^\varphi g'(t)\mathrm{d} t \geq \int_{t=0}^\varphi \frac{\varepsilon^2}{2} \mathrm{d} t \geq \frac{\varepsilon^2}{2} \varphi \geq \frac{\varepsilon^2}{2}\sin\varphi. \end{align*} $$

Hence,

$$ \begin{align*} \int_{s=0}^1 |\nabla \theta|^2 \sin^2 \varphi \, h(s,\varphi) \sin \varphi&\leq \int_{s=0}^1\frac{ C\varepsilon^2}{(1-s) \varepsilon^2+ s\varepsilon } \leq C\frac{\varepsilon^2}{\varepsilon-\varepsilon^2}|\ln \varepsilon| = O(\varepsilon|\ln \varepsilon|). \end{align*} $$

$\underline {\text {Case when } g(\varphi )\geq \varepsilon ^2:}$ here, $g'(\varphi )\leq 2\frac {r^2}{\varepsilon ^2}$ . Also, applying Lemma A.7 with

$$ \begin{align*} a= \frac{g(\varphi)}{\sin \varphi},\quad b= \varepsilon,\quad \lambda= \varepsilon , \end{align*} $$

it can be seen that

$$ \begin{align*} \int_{s=0}^1 |\nabla \theta|^2 \sin^2 \varphi \, h(s,\varphi) \sin \varphi \,\mathrm{d} s &\leq C\varepsilon \frac{\varepsilon}{r}\cdot 2 \frac{r^2}{\varepsilon^2}\cos\varphi \cdot \varepsilon^{-1}\frac{1}{1-\varepsilon} |\ln \varepsilon| = C\frac{r}{\varepsilon}|\ln \varepsilon|. \end{align*} $$

Integrating now over $\varphi $ , changing variables to $t= \frac {g(\varphi )}{\varepsilon }$ , $\varphi = f \Big ( \varepsilon t\Big )$ :

$$ \begin{align*} &\int_{\varphi=f(\varepsilon^2)}^{\frac{\pi}{2}} \int_{s=0}^1 |\nabla \theta|^2 \sin^2 \varphi \, h(s,\varphi)\sin \varphi \,\mathrm{d} s \mathrm{d}\varphi \leq C|\ln \varepsilon| \int_{t=\varepsilon}^1 t \underbrace{f'\Big ( \varepsilon t\Big )}_{\leq 2t^{-2}} \varepsilon \mathrm{d} t \leq C\varepsilon |\ln \varepsilon|^2. & \end{align*} $$

Finally, for the result for $\nabla s$ , note that

$$ \begin{align*} (\varepsilon\cos\varphi)^2 |\nabla s|^2 h(s,\varphi)\sin\varphi & \leq \frac{ C\varepsilon^2 \Big (\sin^2 \varphi + \cos^2\varphi + \varepsilon^{-2}|g'(\varphi)|^2\Big )\cdot \varepsilon \Big ( (1-s) \frac{g(\varphi)}{\sin \varphi}+ s\varepsilon \Big )}{(1-s)g'(\varphi)\cos \varphi + s(\varepsilon - g(\varphi)\sin\varphi)}\cos^2 \varphi \\ & \leq \frac{C\varepsilon^2\cos^2 \varphi}{(1-s)g'(\varphi)\cos \varphi + s(\varepsilon - g(\varphi)\sin\varphi)}. \end{align*} $$

By Lemma A.9,

$$ \begin{align*} & \int_{\varphi=0}^{\frac{\pi}{2}}\int_{s=0}^1 (\varepsilon\cos\varphi)^2|\nabla s|^2 h(s,\varphi) \sin\varphi\,\mathrm{d} s \mathrm{d}\varphi \leq C\varepsilon^2 \int_{\varphi=0}^{\frac{\pi}{2}} \frac{\cos^2\varphi}{\varepsilon-g(\varphi)}|\ln \varepsilon| \mathrm{d}\varphi \\ &\leq C\varepsilon^2|\ln \varepsilon| \int_{\varphi=0}^{\frac{\pi}{2}} \frac{\varepsilon - g(\varphi)}{\max\{\varepsilon, \frac{g(\varphi)}{\varepsilon}\}^2 } \mathrm{d}\varphi \leq C\varepsilon^3|\ln \varepsilon| \left( \int_0^{f(\varepsilon^2)}\varepsilon^{-2}\mathrm{d}\varphi + \int_{f(\varepsilon^2)}^{\frac{\pi}{2}} \frac{\varepsilon^2}{g(\varphi)^2} \mathrm{d}\varphi \right). \end{align*} $$

For the last integral, change variables to $t=\frac {g(\varphi )}{\varepsilon }$ , $\varphi = f(\varepsilon t)$ :

$$ \begin{align*} \int_{f(\varepsilon^2)}^{\frac{\pi}{2}} \frac{\varepsilon^2}{g(\varphi)^2} \mathrm{d}\varphi &= \int_{\kern-1.2pt\varepsilon}^1 t^{-2} f'(\varepsilon t) \varepsilon \mathrm{d} t \leq 2\varepsilon \int_{\kern-1.2pt\varepsilon}^1 t^{-4}\mathrm{d} t \leq \frac{2}{3}\varepsilon^{-2}, \end{align*} $$

and the conclusion follows.

3.3.3 Extra energy in $a_{\kern-1.2pt\varepsilon }$ is negligible

In this section, we prove the part of Proposition 3.6 relative to the region $a_{\kern-1.2pt\varepsilon }$ .

In order to prove this proposition, we first deduce an integrability property of the function H implied by the assumption that $ E(\boldsymbol {u})<+\infty $ . Indeed, using (B.1), the following expression is obtained for the Jacobian of the limit map:

(3.26) $$ \begin{align} \det D\boldsymbol{u} = \frac{u_{\rho}^2 \sin u_{\varphi} \cdot \begin{vmatrix} \partial_{\rho} u_{\rho} & \partial_{\varphi} u_{\rho} \\ \partial_{\rho} u_{\varphi} & \partial_{\varphi} u_{\varphi} \end{vmatrix} }{ \rho^2 \sin\varphi } = (1-\rho)^2\rho^{-2}\cos^3\varphi. \end{align} $$

Lemma 3.9. Let H be as in the statement of Theorem 1.2. Then

$$ \begin{align*} \int_{s=1}^\infty H(s) s^{-5/2} \mathrm{d} s < \infty. \end{align*} $$

Proof. First note that the convexity assumption of H together with the growth (1.1) implies that there exists $\delta>0$ such that

(3.27) $$ \begin{align} H \text{ is decreasing in } (0,\delta) \text{ and increasing in } (\frac{1}{\delta}, +\infty). \end{align} $$

Let $\boldsymbol {u}$ be the Conti–De Lellis map as in Theorem 1.2. By assumption (1.8),

$$ \begin{align*} \infty &> \int_{a} H(\det D\boldsymbol{u}) \mathrm{d} \boldsymbol{x} > \int\limits_{\{\cos \varphi > \frac{1}{2}\ \wedge\ \rho < \frac{1}{2} \}} H(\det D\boldsymbol{u})\mathrm{d}\boldsymbol{x} \\ & = 2\pi \int_{\rho=0}^{\frac{1}{2}}\int_{\cos\varphi=\frac{1}{2}}^{\cos\varphi=1} H \big ( (1-\rho)^2 \rho^{-2}\cos^3\varphi \big ) \rho^2 \mathrm{d} \big ( \cos\varphi\big ) \mathrm{d}\rho \\ & = 2\pi \int_{\rho=0}^{\frac{1}{2}} \int_{t=\frac{1}{2}}^1 H \big ( (1-\rho)^2 \rho^{-2} t^3 \big ) \rho^2 \mathrm{d} t \mathrm{d} \rho. \end{align*} $$

At this point, we observe that $ (1-\rho )^2 \rho ^{-2} t^3 \geq 2^{-5} \rho ^{-2}$ . In the above integral, we keep only those values of $\rho $ such that $2^{-5}\rho ^{-2} \geq \frac {1}{\delta }$ (i.e., $\rho \leq \sqrt {\delta /32}$ ). Since, by (3.27), H is increasing in $(\frac {1}{\delta }, +\infty )$ , it follows that

$$ \begin{align*} \infty &> 2\pi \int_{\rho=0}^{\sqrt{\delta/32}} \int_{t=\frac{1}{2}}^1 H(2^{-5}\rho^{-2}) \rho^2 \mathrm{d} t \mathrm{d} \rho = \pi \int_{\rho=0}^{\sqrt{\delta/32}} H(2^{-5}\rho^{-2})\rho^2 \mathrm{d} \rho. \end{align*} $$

Changing the integration variable to $s=2^{-5}\rho ^{-2}$ yields

$$ \begin{align*} \infty> \int_{s=\frac{1}{\delta}}^\infty H(s) \frac{2^{-5/2}}{s} s^{-3/2} \mathrm{d} s. \end{align*} $$

This finishes the proof since in $[1, \frac {1}{\delta }]$ the function H is continuous (hence bounded).

Lemma 3.10. Let us suppose that $\varepsilon $ is sufficiently small so that $\varepsilon ^{2(1-\gamma )} \leq \frac {\sqrt {2}}{\pi }. $ Then, for all $0\leq \varphi < \frac {\pi }{2}$ ,

$$ \begin{align*} \left( (\cos \varphi + 2\varepsilon^\gamma)^3 - 3\int_{\sigma=0}^1 h(\sigma, \varphi)\mathrm{d}\sigma \right)^{\frac{1}{3}} \geq \cos\varphi + \varepsilon^{\gamma}. \end{align*} $$

Proof. By Lemma A.5,

$$ \begin{align*} (\cos \varphi + 2\varepsilon^\gamma)^3 & - 3\int_{\sigma=0}^1 h(\sigma, \varphi)\mathrm{d}\sigma \geq (\cos \varphi + 2\varepsilon^\gamma)^3 - \frac{9\pi\sqrt{2}}{2}\varepsilon^2\cos\varphi \\ &\geq \big (\cos^3\varphi + 6\varepsilon^{\gamma}\cos^2\varphi+12\varepsilon^{2\gamma}\cos\varphi +8\varepsilon^{3\gamma}\big ) - \frac{9\pi\sqrt{2}}{2}\varepsilon^2\cos\varphi \\ &\qquad = (\cos \varphi + \varepsilon^{\gamma})^3 + \underbrace{3\varepsilon^{\gamma}\cos^2\varphi}_{\geq 0} + \frac{9\pi\sqrt{2}}{2}\varepsilon^{2\gamma} \underbrace{\big ( \frac{\sqrt{2}}{\pi} - \varepsilon^{2(1-\gamma)}\big )}_{\geq 0}\cos\varphi + \underbrace{7\varepsilon^{3\gamma}}_{\geq 0}. \end{align*} $$

Lemma 3.11. Let $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ be the recovery sequence of Section 3.2 and $\boldsymbol {u}$ the Conti–De Lellis map. Suppose that $\varepsilon $ is sufficiently small so that $ \varepsilon ^{2(1-\gamma )} \leq \frac {\sqrt {2}}{\pi }. $ Then

$$ \begin{align*}\det D\boldsymbol{u}^\varepsilon (\boldsymbol{x}) \geq \det D\boldsymbol{u}(\boldsymbol{x}) \quad \text{for all } \boldsymbol{x} \text{ in region } a_{\kern-1.2pt\varepsilon}. \end{align*} $$

Proof. The same calculation as in (3.26) gives

(3.28) $$ \begin{align} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } = \rho^{-2}(u_{\rho}^\varepsilon)^2\partial_{\rho} u_{\rho}^\varepsilon. \end{align} $$

By Lemma 3.10,

(3.29) $$ \begin{align} \partial_{\rho} u_{\rho}^\varepsilon &= \underbrace{\frac{1}{1-\varepsilon}}_{\geq 1} \Bigg ( \Big ( (\cos (\pi-\varphi) + 2\varepsilon^\gamma)^3 - 3\int_{\sigma=0}^1 h(\sigma, \varphi)\mathrm{d}\sigma \Big )^{\frac{1}{3}} -\varepsilon^\gamma \Bigg ) \geq -\cos \varphi. \end{align} $$

Regarding $(u_{\rho }^\varepsilon )^2$ , using again Lemma 3.10,

(3.30) $$ \begin{align} u_{\rho}^\varepsilon \geq \underbrace{\frac{1-\rho}{1-\varepsilon}}_{\geq (1-\rho)} (\cos (\pi\varphi) + \varepsilon^\gamma) + \underbrace{\varepsilon^\gamma \frac{\rho-\varepsilon}{1-\varepsilon}}_{\geq 0} \geq -(1-\rho)\cos\varphi. \end{align} $$

Combining (3.28), (3.29) and (3.30), it can be seen that

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \geq -(1-\rho)^2 \rho^{-2}\cos^3 \varphi = \det D\boldsymbol{u}.\\[-38pt] \end{align*} $$

Now bounds from above are needed for $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ .

Lemma 3.12. Suppose that $\varepsilon $ is small enough so that $\frac {(1+2\varepsilon ^\gamma )^3}{1-\varepsilon } < 2.$ Then, for all $\rho \in (\varepsilon ,1)$ , $\theta \in [0,2\pi ]$ , and $\varphi \in [\frac {\pi }{2},\pi ]$ ,

$$ \begin{align*}\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big ( \boldsymbol{x} (\rho, \theta, \varphi)\big ) \leq 2\rho^{-2}.\end{align*} $$

Proof. From (3.9), it can be seen (since $h\geq 0$ ) that

$$ \begin{align*}|u_{\rho}^\varepsilon(\rho,\varphi)|\leq \frac{1-\rho}{1-\varepsilon} (1+2\varepsilon^\gamma) + \varepsilon^\gamma \frac{\rho-\varepsilon}{1-\varepsilon} \leq (1+2\varepsilon^\gamma). \end{align*} $$

However,

$$ \begin{align*} \partial_{\rho} u_{\rho}^\varepsilon &= \frac{1}{1-\varepsilon} \Bigg ( \Big ( (\underbrace{\cos (\pi-\varphi)}_{\leq 1} + 2\varepsilon^\gamma)^3 - 3\int_{\sigma=0}^1 h(\sigma, \varphi)\mathrm{d}\sigma \Big )^{\frac{1}{3}} -\varepsilon^\gamma \Bigg ) \leq \frac{1+2\varepsilon^\gamma}{1-\varepsilon}. \end{align*} $$

Plugging both into (3.28), the desired upper bound for $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is obtained.

We are now ready to prove the part of Proposition 3.6 relative to the region $a_{\kern-1.2pt\varepsilon }$ .

Proof of Proposition 3.6 in $a_{\kern-1.2pt\varepsilon }$ .

We start with the Dirichlet part of the energy. Since $\boldsymbol {u}$ is given in spherical coordinates in this region, we use (B.2). By direct computation, we can check that $ \chi _{a_{\kern-1.2pt\varepsilon }}D\boldsymbol {u}_{\kern-1.2pt\varepsilon } \rightarrow \chi _a D\boldsymbol {u}$ pointwise. Furthermore, using that $ h(s,\varphi )=O(\varepsilon ^2)$ from Lemma A.5, we can see that $ |\partial _{\rho } u_{\rho }^\varepsilon |, |u_{\rho }^\varepsilon \partial _{\varphi } u_{\varphi }^\varepsilon |, |u_{\rho }^\varepsilon \sin u_{\varphi }^\varepsilon |$ are uniformly bounded in the region $a_{\kern-1.2pt\varepsilon }$ . Now, using the same computation as in the last item of Lemma 3.7, we also obtain that $ | \partial _{\varphi } u_{\rho }^\varepsilon |$ is uniformly bounded. We can thus apply the dominated convergence theorem to obtain the convergence of the Dirichlet energy in this region.

We then deal with the determinant part of the energy. Combining the bounds for $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ from below and above, one obtains

$$ \begin{align*} H\big (\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }(\boldsymbol{x})\big ) \leq C + H\big (\det D\boldsymbol{u}(\boldsymbol{x})\big ) + H(2\rho^{-2}), \end{align*} $$

the constant C being $C:=\max \{ H(J): \delta \leq J\leq 1/\delta \}$ . In order to use the dominated convergence theorem, it suffices, then, to show that $H(2\rho ^{-2})$ is integrable. We just note that

$$ \begin{align*} \int_a H(2\rho^{-2})\mathrm{d} \boldsymbol{x} &= 2\pi \int_{\rho=0}^1\int_{\varphi=0}^\pi H(2\rho^{-2}) \rho^2 \mathrm{d} (-\cos \varphi)\mathrm{d}\rho = 2\pi \int_{\rho=0}^1 H(2\rho^{-2}) \rho^2\mathrm{d}\rho. \end{align*} $$

Changing the integration variable to $t:=2\rho ^{-2}$ and using Lemma 3.9,

$$ \begin{align*} \int_a H(2\rho^{-2})\mathrm{d} \boldsymbol{x} &=2\sqrt{2}\pi \int_{t=2}^\infty H(t)t^{-5/2} \mathrm{d} t < \infty.\\[-38pt] \end{align*} $$

3.3.4 Extra energy in b is negligible

Here, we prove the part of Proposition 3.6 concerning region b.

Proof of Proposition 3.6 in b.

We start with the Dirichlet energy. Since in this region $\boldsymbol {u}_{\kern-1.2pt\varepsilon }= \varepsilon ^\gamma \boldsymbol {\psi } \left (\varepsilon ^\gamma \boldsymbol {\phi }_{\kern-1.2pt\varepsilon } (\boldsymbol {x}) \right )$ , we find that $ D \boldsymbol {u}_{\kern-1.2pt\varepsilon } (\boldsymbol {x})= D \boldsymbol {\psi } (\varepsilon ^{-\gamma }\boldsymbol {\phi }_{\kern-1.2pt\varepsilon } (\boldsymbol {x}))D\boldsymbol {\phi }_{\kern-1.2pt\varepsilon }(\boldsymbol {x}))$ . From this and the expressions of $ \boldsymbol {\phi }_{\kern-1.2pt\varepsilon }$ and $ \boldsymbol {\psi }$ , we can see that $ D \boldsymbol {u}_{\kern-1.2pt\varepsilon } \rightarrow D \boldsymbol {u}$ almost everywhere in $ b$ . Since $ \boldsymbol {\psi } $ is bi-Lipschitz and $ |D \boldsymbol {\phi }_{\kern-1.2pt\varepsilon }|\leq C$ for $\varepsilon $ small enough, we can apply the dominated convergence theorem to get the convergence of the Dirichlet energy.

For the determinant part, as in Subsection 3.3.3, we start by deriving an integrability property of the function H in Theorem 1.2. We can compute that the Jacobian of the limit map is

$$ \begin{align*} \det D\boldsymbol{u} (\rho, \varphi) = \frac{ (\rho-1)^2 \sin \Big (\frac{\varphi+\pi}{2} \Big )}{2\rho^2 \sin\varphi} = \frac{(\rho-1)^2}{4\rho^2 \sin \frac{\varphi}{2}}. \end{align*} $$

We know from (3.27) that H is decreasing in some interval $(0,\delta )$ . Since

$$ \begin{align*} \rho\geq 1,\ \frac{\pi}{2}\leq \varphi\leq \pi\quad\Rightarrow\quad \det D\boldsymbol{u} \leq \frac{(\rho-1)^2}{2\sqrt{2}}, \end{align*} $$

it follows that

$$ \begin{align*} \frac{(\rho-1)^2}{2\sqrt{2}}\leq\delta\ \Rightarrow\ H(\det D\boldsymbol{u}) \geq H\Big ( \frac{(\rho-1)^2}{2\sqrt{2}} \Big ). \end{align*} $$

Hence,

$$ \begin{align*} \infty &> \int_b H(\det D\boldsymbol{u})\mathrm{d}\boldsymbol{x} \geq 2\pi \int_{\rho=1}^3 \int_{\varphi=\frac{\pi}{2}}^{\pi} H(\det D\boldsymbol{u}) \cdot 1^2 \cdot \sin\varphi\mathrm{d}\varphi \mathrm{d}\rho \\ & \geq 2\pi \int_{\{ \frac{(\rho-1)^2}{2\sqrt{2}}\leq\delta \}} \int_{\varphi=\frac{\pi}{2}}^\pi H\Big ( \frac{(\rho-1)^2}{2\sqrt{2}} \Big )\sin\varphi\mathrm{d}\varphi\mathrm{d}\rho = 2\pi \int_{t=0}^{\sqrt{\delta}} H(t^2)\cdot \sqrt{2\sqrt{2}} \mathrm{d} t, \end{align*} $$

which yields that the function $t\mapsto H(t^2)$ is $L^1\big ((0,1)\big )$ . This is helpful in what follows.

Now we estimate $H(\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon })$ . In the region $\rho \geq 1+\sqrt {2}\varepsilon ^\gamma $ , we have that

$$\begin{align*}\boldsymbol{u}_{\kern-1.2pt\varepsilon }(\boldsymbol{x}) =\varepsilon^\gamma \mathbf{id} \Big ( \varepsilon^{-\gamma} \boldsymbol{\phi}_{\kern-1.2pt\varepsilon} (\boldsymbol{x}) \Big ) = \varepsilon^\gamma \boldsymbol{e}_3 + (\rho -1 +\sqrt{2}\varepsilon^\gamma ) \Big ( \Big ( \sin \frac{\varphi+\pi}{2}\Big ) \boldsymbol{e}_r + \Big ( \cos \frac{\varphi+\pi}{2}\Big ) \,\boldsymbol{e}_{\theta}\Big ). \end{align*}$$

Hence,

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } = \frac{(\rho-1+\sqrt{2}\varepsilon^\gamma)^2}{4\rho^2 \sin \frac{\varphi}{2}}. \end{align*} $$

On the one hand, $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }(\boldsymbol {x})\geq \det D\boldsymbol {u}(\boldsymbol {x})$ for all $\boldsymbol {x}$ in that region. On the other hand,

$$ \begin{align*} 1\leq \rho\leq 3,\ \frac{\pi}{2}\leq \varphi\leq \pi\quad\Rightarrow\quad \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \leq \frac{(2+\sqrt{2}\varepsilon^\gamma)^2}{4\cdot 1^2 \cdot \frac{\sqrt{2}}{2}} \leq 2 \end{align*} $$

for small $\varepsilon $ . Therefore,

$$ \begin{align*} H(\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }(\boldsymbol{x})) \leq \max\Big \{ H(\det D\boldsymbol{u}(\boldsymbol{x})), \max_{\delta\leq J\leq 2} H(J)\Big \} \end{align*} $$

and, by dominated convergence,

$$ \begin{align*} \int_{b\cap \{\rho \geq 1+\sqrt{2}\varepsilon^\gamma\}} H (\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }) \mathrm{d}\boldsymbol{x} \overset{\varepsilon\to0}{\longrightarrow} \int_b H(\det D\boldsymbol{u}) \mathrm{d} \boldsymbol{x}. \end{align*} $$

In the region $1\leq \rho \leq 1+\sqrt {2}\varepsilon ^\gamma $ ,

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }(\boldsymbol{x}) = \Big ( \det (D\boldsymbol{\psi})\big (\varepsilon^{-\gamma}\boldsymbol{\phi}_{\kern-1.2pt\varepsilon}(\boldsymbol{x})\big ) \Big ) \cdot \frac{(\rho-1+\sqrt{2}\varepsilon^\gamma)^2}{4\rho^2 \sin \frac{\varphi}{2}}, \end{align*} $$

so

$$ \begin{align*} (\min \det D\boldsymbol{\psi})\cdot \frac{(\rho -1+\sqrt{2}\varepsilon^{\gamma})^2}{4(1+\sqrt{2}\varepsilon^\gamma)^2\cdot 1} \leq \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }(\boldsymbol{x}) \leq (\max \det D\boldsymbol{\psi})\frac{(\rho -1+\sqrt{2}\varepsilon^{\gamma})^2}{4\cdot 1^2 \cdot \frac{\sqrt{2}}{2}}. \end{align*} $$

Since $\rho -1+\sqrt {2}\varepsilon ^\gamma \leq 2\sqrt {2}\varepsilon ^\gamma $ , for $\varepsilon $ small enough, $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }(\boldsymbol {x})$ lies in the region where H is decreasing and

$$ \begin{align*} H(\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }(\boldsymbol{x})) \leq H\big ( C^2 (\rho-1 + \sqrt{2}\varepsilon^\gamma)^2 \big ),\quad \text{with} \quad C^2= \frac{\min \det D\boldsymbol{\psi}}{5}. \end{align*} $$

Consequently,

$$ \begin{align*} \int_{\{ 1\leq \rho \leq 1+\sqrt{2}\varepsilon^\gamma \}} H(\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }) \mathrm{d}\boldsymbol{x} &\leq 2\pi \int_{\{ 1\leq \rho \leq 1+\sqrt{2}\varepsilon^\gamma \}} H\big ( C^2 (\rho-1 + \sqrt{2}\varepsilon^\gamma)^2 \big ) \int_{\varphi=\frac{\pi}{2}}^\pi 3^2 \sin\varphi \mathrm{d}\varphi\mathrm{d}\rho \\ &= 18\pi \int_{t=0}^{2\sqrt{2}C\varepsilon^\gamma} H(t^2)\mathrm{d} t , \end{align*} $$

which vanishes as $\varepsilon \to 0$ because $t\mapsto H(t^2)$ is integrable in $(0,1)$ .

3.3.5 Extra energy in $e_{\kern-1.2pt\varepsilon }^{\prime }$ is negligible

We prove here the part of Proposition 3.6 concerning region $e_{\kern-1.2pt\varepsilon }^{\prime }$ and, in particular, (3.22).

We start with the following.

Lemma 3.13. For all $\varphi \in [0,\frac {\pi }{2}]$ , all $s\in [0,1]$ , and all $\varepsilon $ sufficiently small,

(3.31) $$ \begin{align} \cos \varphi + 2\varepsilon^\gamma \leq u_{\rho}^\varepsilon \leq \Big ( ( \cos \varphi + 2\varepsilon^\gamma )^3 + 12\sqrt{2}\varepsilon + \frac{9\pi \sqrt{2}}{2}\varepsilon^2 \cos\varphi \Big )^{\frac{1}{3}} \leq 2, \end{align} $$

(3.32) $$ \begin{align} |\partial_s u_{\rho}^\varepsilon|\leq C\varepsilon^{2-2\gamma}\cos\varphi \quad \text{and}\quad |\partial_{\varphi} u_{\rho}^\varepsilon|\leq C \varepsilon^{-2\gamma}. \end{align} $$

Proof. The bounds for $u_{\rho }^\varepsilon $ can be obtained using Lemmas A.1.a) and A.5. The bound for $\partial _s u_{\rho }^\varepsilon $ is proved exactly as in Lemma A.5. Regarding $\partial _{\varphi } u_{\rho }^\varepsilon $ ,

$$ \begin{align*} 3\partial_{\varphi} u_{\rho}^\varepsilon = (u_{\rho}^\varepsilon)^{-2} &\partial_{\varphi} \big ( (u_{\rho}^\varepsilon)^3\big ) = -3 \underbrace{\left( \frac{\cos \varphi +2\varepsilon^\gamma}{u_{\rho}^\varepsilon}\right)^2}_{\leq 1 \text{ by}\ (3.31)} \sin \varphi \\ &+ (u_{\rho}^\varepsilon)^{-2} \left [ \partial_r \left ( \frac{3r}{\partial_r \big ( -\cos f(r) \big ) }\right)\right]_{r=g(\varphi)}g'(\varphi) + 3(u_{\rho}^\varepsilon)^{-2} \underbrace{\int_{\sigma=0}^s \partial_{\varphi} h(\sigma, \varphi)\mathrm{d}\sigma}_{=O(\varepsilon) \text{ by Lemma}\ A.5}. \end{align*} $$

The claim (3.32) follows from $u_{\rho }^\varepsilon \geq 2\varepsilon ^\gamma $ (see (3.31)), Lemma A.1.c) and Lemma A.4.

Proof of Proposition 3.6 in $e_{\kern-1.2pt\varepsilon }^{\prime }$ .

We note that in this region, $ \det D \boldsymbol {u}_{\kern-1.2pt\varepsilon } =1$ and thus

$$\begin{align*}\int_{e_{\kern-1.2pt\varepsilon}^{\prime}} H(\det D \boldsymbol{u}_{\kern-1.2pt\varepsilon }) \mathrm{d} \boldsymbol{x}=O(\varepsilon^3) .\end{align*}$$

We deal with the Dirichlet energy. First, the parametrization of the region in the reference configuration is equivalent up to a reflection (changing the sign of $\nabla \varphi $ but not its magnitude) to the parametrization of the region $a_{\kern-1.2pt\varepsilon }^{\prime }$ . By (3.25) and Lemma 3.13,

$$ \begin{align*} \int_{e_{\kern-1.2pt\varepsilon}^{\prime}} |D\boldsymbol{u}_{\kern-1.2pt\varepsilon }|^2 \mathrm{d}\boldsymbol{x} \leq \underbrace{C\varepsilon^{4(\frac{1}{2}-\gamma)}}_{\leq 1}\int_{e_{\kern-1.2pt\varepsilon}^{\prime}} (\varepsilon\cos\varphi)^2|\nabla s|^2 + 4\int_{e_{\kern-1.2pt\varepsilon}^{\prime}} |\nabla \theta|^2 \sin^2\varphi\, \mathrm{d}\boldsymbol{x} + C\varepsilon^{-4\gamma}\int_{e_{\kern-1.2pt\varepsilon}^{\prime}} |\nabla \varphi|^2\,\mathrm{d}\boldsymbol{x}. \end{align*} $$

The estimate for $\int |D\boldsymbol {u}_{\kern-1.2pt\varepsilon }|^2$ in this region now follows from Lemma 3.8, which can be applied in region $e_{\kern-1.2pt\varepsilon }^{\prime }$ thanks to a reflection argument.

3.3.6 Extra energy in $e_{\kern-1.2pt\varepsilon }$ is negligible

We show here that the neo-Hookean energy of the recovery sequence in region $e_{\kern-1.2pt\varepsilon }$ converges to the neo-Hookean energy of the limit map in region e; cf. (3.23) in Proposition 3.6.

Before doing the proof of this fact, we start by stating some integrability properties of the function H in Theorem 1.2.

Lemma 3.14. Let H be as in the statement of Theorem 1.2. Then

$$ \begin{align*} \int_0^1 H(s^3)\mathrm{d} s <\infty \quad \text{and}\quad \int_1^\infty s^{-5/2}H(s)\mathrm{d} s < \infty. \end{align*} $$

Proof. We remark the expression of the Jacobian of the limit map:

$$ \begin{align*} \det D\boldsymbol{u} = \frac{u_{\rho}^2\partial_{\rho} u_{\rho}}{\rho^2} = (1+\rho)^2 \rho^{-2}\cos^3 \varphi. \end{align*} $$

It vanishes as $\varphi \to \frac {\pi }{2}^-$ and goes to infinity as $\rho \to 0^+$ (and $\cos \varphi $ remains away from zero). The neo-Hookean energy of the limit map is being assumed to be finite, so

$$ \begin{align*} \infty> \int_{e} H(\det D\boldsymbol{u})\, \mathrm{d}\boldsymbol{x} &= 2\pi \int_{\varphi=0}^1\int_{\varphi=0}^{\pi/2} H\Big ( (1+\rho)^2\rho^{-2}\cos^3 \varphi \Big ) \rho^2 \sin\varphi\,\mathrm{d}\varphi\, \mathrm{d}\rho \\ &\underset{t=\cos\varphi}{=}\int_{\rho=0}^1\int_{t=0}^1 H\Big ( (1+\rho)^2\rho^{-2}t^3\Big)\rho^2 \,\mathrm{d} t \,\mathrm{d}\rho. \end{align*} $$

Note that

$$ \begin{align*} \int_{\rho=0}^1\int_{t=0}^1 H\Big ( (1+\rho)^2\rho^{-2}t^3\Big)\rho^2 \,\mathrm{d} t \,\mathrm{d}\rho> \int_{\rho=\frac{1}{2}}^1 \int_{t=0}^{(\delta/16)^{1/3}} H(16t^3)\,\mathrm{d} t\,\mathrm{d} \rho \end{align*} $$

because if $\frac {1}{2}<\rho < 1$ and $t <(\delta /16)^{1/3}$ , then

$$ \begin{align*} (1+\rho)^2\rho^{-2}t^3 < (1+1)^2 (1/2)^{-2}t^3 < \delta, \end{align*} $$

and in $(0,\delta )$ , the function H decreases (see (3.27)). The integrand is now independent of $\rho $ , and changing variables to $s=16^{1/3}t$ yields the integrability of $H(s^3)$ near zero.

Regarding the integrability at infinity, it was already established in Lemma 3.9.

Proof of Proposition 3.6 in $e_{\kern-1.2pt\varepsilon }$ .

For the Dirichlet part of the energy, we use spherical coordinates and formula (B.2). We note that

$$ \begin{align*} \left[\frac{3r}{\partial_r(-\cos f(r))}\right]_{r=g(\varphi)}= \frac{3 g(\varphi)}{f'(g(\varphi))\sin \varphi}. \end{align*} $$

We can compute $D \boldsymbol {u}_{\kern-1.2pt\varepsilon }$ and use that $ h(s,\varphi )=O(\varepsilon ^2)$ and Lemmas A.4–A.5 to prove that $ D\boldsymbol {u}_{\kern-1.2pt\varepsilon } \rightarrow D \boldsymbol {u}$ pointwise and that $ | D\boldsymbol {u}_{\kern-1.2pt\varepsilon }| \leq C$ in $e_{\kern-1.2pt\varepsilon }$ . Hence, from the dominated convergence theorem, we obtain the convergence of the Dirichlet energy.

For the other part of the energy, using (B.1), one obtains the Jacobian determinant for $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ :

(3.33) $$ \begin{align} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } = \rho^{-2}(u_{\rho}^\varepsilon)^2\partial_{\rho} u_{\rho}^\varepsilon = \frac{1}{1-\varepsilon}\rho^{-2}(u_{\rho}^\varepsilon)^2 \big ( 2\cos\varphi + 6\varepsilon^{\gamma} - u_{\rho}^\varepsilon (\varepsilon,\varphi)\big ). \end{align} $$

Note that

$$ \begin{align*} (1+\varepsilon)\cos\varphi \leq \cos\varphi + \varepsilon <\cos \varphi + 2\varepsilon^\gamma \leq u_{\rho}^\varepsilon(\varepsilon, \varphi), \end{align*} $$

so

$$ \begin{align*} u_{\rho}(\rho,\varphi)= (1+\rho)\cos\varphi = \frac{1-\rho}{1-\varepsilon}\big ((1+\varepsilon)\cos\varphi\big) + \frac{\rho-\varepsilon}{1-\varepsilon} (2\cos\varphi) \leq u_{\rho}^\varepsilon(\rho,\varphi). \end{align*} $$

Using now (3.10),

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \geq \frac{1}{1-\varepsilon} \rho^{-2} (u_{\rho})^2 \cos\varphi = \frac{1}{1-\varepsilon}\det D\boldsymbol{u}\geq \det D\boldsymbol{u}. \end{align*} $$

Now a bound from above is needed for $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ . Plugging (3.10) along with the lower bound $u_{\rho }^\varepsilon (\varepsilon ,\varphi )\geq \cos \varphi $ into (3.33) yields

$$ \begin{align*} \det D\boldsymbol{u}\big (\boldsymbol{x}(\rho, \theta, \varphi)\big ) \leq \frac{1}{1-\varepsilon}\rho^{-2}\cdot \big (\cos\varphi + 2\varepsilon^\gamma +4\varepsilon^{\frac{1}{3}}\big )^2 \big ( \cos\varphi + 6\varepsilon^{\gamma}\big ) \leq 2\rho^{-2} \end{align*} $$

for every $\rho $ and $\varphi $ (provided $\varepsilon $ is sufficiently small).

These bounds, with exactly the same proof as the part of Proposition 3.6 relative to region $a_{\kern-1.2pt\varepsilon }$ , gives the desired conclusion:

$$ \begin{align*}\lim_{\kern-1.2pt\varepsilon\to 0} \int_{e_{\kern-1.2pt\varepsilon}} H(\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon }) \mathrm{d}\boldsymbol{x} = \int_{e} H(\det D\boldsymbol{u})\mathrm{d} \boldsymbol{x}.\\[-45pt] \end{align*} $$

3.3.7 Extra energy in f is negligible

We now deal with region f and prove Proposition 3.6 for this region.

Proof of Proposition 3.6 in f.

It is direct to see that, in region f, we have $ D \boldsymbol {u}_{\kern-1.2pt\varepsilon } \rightarrow D \boldsymbol {u}$ almost everywhere and $ |D\boldsymbol {u}_{\kern-1.2pt\varepsilon }|\leq C(|D \boldsymbol {u}|^2+1)$ . Then by dominated convergence theorem, we obtain the convergence of the Dirichlet energy. For the other part of the energy. First note (using (B.1)) that

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } = \frac{(u_{\rho} + 6\varepsilon^{\gamma})^2 \sin u_{\varphi}}{\rho^2\sin \varphi} \begin{vmatrix} \partial_{\rho} u_{\rho} & \partial_{\varphi} u_{\rho} \\ \partial_{\rho} u_{\varphi} & \partial_{\varphi} u_{\varphi} \end{vmatrix}, \quad \det D\boldsymbol{u} = \frac{u_{\rho}^2 \sin u_{\varphi}}{\rho^2\sin \varphi} \begin{vmatrix} \partial_{\rho} u_{\rho} & \partial_{\varphi} u_{\rho} \\ \partial_{\rho} u_{\varphi} & \partial_{\varphi} u_{\varphi} \end{vmatrix}. \end{align*} $$

On the one hand, this shows that $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon } \geq \det D\boldsymbol {u}$ . It remains to bound the Jacobian from above. Here, we decompose f in the part where $0\leq \varphi <\frac {\pi }{4}$ and where $\frac {\pi }{4}\leq \varphi \leq \frac {\pi }{2}$ . In the first part, use that

$$ \begin{align*} u_{\rho}(\rho,\varphi)\geq u_{\rho}(1,\varphi)=2\cos\varphi \geq \sqrt{2}\geq 6\varepsilon^{\gamma}, \end{align*} $$

which holds (at least close to $\rho =1$ , where the regularization of constructing the sequence $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is required) since $\boldsymbol {u}$ is orientation-preserving and injective, so increasing in $\rho $ . Then

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \leq \frac{(u_{\rho} + u_{\rho})^2 \sin u_{\varphi}}{\rho^2\sin \varphi} \begin{vmatrix} \partial_{\rho} u_{\rho} & \partial_{\varphi} u_{\rho} \\ \partial_{\rho} u_{\varphi} & \partial_{\varphi} u_{\varphi} \end{vmatrix} =4\det D\boldsymbol{u}. \end{align*} $$

In $\{\varphi \in [\frac {\pi }{4},\frac {\pi }{2}]\}$ , just consider that

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \leq \frac{(u_{\rho} + 6\varepsilon^{\gamma})^2 \sin u_{\varphi}}{1^2 \cdot \sin \frac{\pi}{4}} \begin{vmatrix} \partial_{\rho} u_{\rho} & \partial_{\varphi} u_{\rho} \\ \partial_{\rho} u_{\varphi} & \partial_{\varphi} u_{\varphi} \end{vmatrix} \end{align*} $$

and that the right-hand side is uniformly bounded since the limit map $\boldsymbol {u}$ is Lipschitz. With these bounds at hand and the hypothesis on H in Theorem 1.2, it is possible to obtain the conclusion invoking the dominated convergence theorem.

3.3.8 Extra energy in $d_{\kern-1.2pt\varepsilon }$ is negligible

We conclude this section by proving the convergence of the neo-Hookean energy of the recovery sequence towards the energy of the limit map in region $d_{\kern-1.2pt\varepsilon }$ .

Proof of Proposition 3.6 in $d_{\kern-1.2pt\varepsilon }$ .

We recall that in region $ d_{\kern-1.2pt\varepsilon }$ , the map $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is the composition of three maps: $\boldsymbol {u}_{\kern-1.2pt\varepsilon }=\boldsymbol {w}_{\kern-1.2pt\varepsilon } \circ \boldsymbol {g} \circ \hat {\boldsymbol {R}}$ , where we denoted by $ \hat { \boldsymbol {R}}$ the map which is described in spherical coordinates by $ (r, \theta ,\varphi ) \mapsto (\hat {r},\theta ,\varphi )$ with $ \hat {r}$ defined in (3.11). We thus have $ D \boldsymbol {u}_{\kern-1.2pt\varepsilon }=D \boldsymbol {w}_{\kern-1.2pt\varepsilon } (\boldsymbol {g}\circ \hat {\boldsymbol {R}}) D \boldsymbol {g}( \hat {\boldsymbol {R}})D \hat {\boldsymbol {R}}$ . From this and the expressions of $ \boldsymbol {w}_{\kern-1.2pt\varepsilon } $ and $ \boldsymbol {R}$ , it is easily seen that $ \chi _{d_{\kern-1.2pt\varepsilon }}D \boldsymbol {u}_{\kern-1.2pt\varepsilon } \rightarrow \chi _d D \boldsymbol {u}$ almost everywhere and that $ |D \boldsymbol {u}_{\kern-1.2pt\varepsilon } |\leq C$ for some $C>0$ . Hence, by dominated convergence, we obtain the convergence of the Dirichlet energy.

We now examine the convergence of the determinant part of the energy. We first examine the determinant of the differential of the limit map to understand the integrability property of H. Let $\boldsymbol {g}$ be the bi-Lipschitz map from $\{\hat r\boldsymbol {e}_r+ x_3 \boldsymbol {e}_3: \hat r\geq 0, x_3\in [0,1]\}$ onto $\{s\boldsymbol {e}_r + z\boldsymbol {e}_3: s\geq 0, z\in [0,3]\}$ of Section 3.1. Since $\boldsymbol {g}$ is Lipschitz and $s(\hat r,x_3)=0$ when $\hat r=0$ (because that corresponds to the values of $\boldsymbol {g}$ on the segment $B'C'$ ),

(3.34)

Since $\boldsymbol {g}^{-1}$ is Lipschitz,

(3.35) $$ \begin{align} 0< c \leq \det D\boldsymbol{g} = \frac{s(\hat r,x_3)}{\hat r} \frac{\partial(s,z)}{\partial(\hat r,x_3)} \leq \|D\boldsymbol{g}\|_{\infty} \frac{\partial(s,z)}{\partial(\hat r,x_3)}. \end{align} $$

It can be seen that

(3.36) $$ \begin{align} \det D\boldsymbol{u} \big ( \hat r\boldsymbol{e}_r(\theta) + x_3\boldsymbol{e}_3\big ) = \frac{\pi}{12} \frac{s^2\sin\varphi}{\hat r} \frac{\partial (s,z)}{\partial (\hat r,x_3)}. \end{align} $$

In addition,

(3.37) $$ \begin{align} \frac{\partial (s,z)}{\partial (\hat r,x_3)} = \left | \frac{\partial \boldsymbol{g}}{\partial \hat r}\wedge \frac{\partial \boldsymbol{g}}{\partial x_3}\right |\leq \|D\boldsymbol{g}\|_{\infty}^2. \end{align} $$

From the above, it follows that

$$ \begin{align*} \frac{c}{\|D\boldsymbol{g}\|_{\infty}}\frac{\pi}{12} \frac{s^2\cdot \sin \frac{\pi}{4}}{\hat r} \leq \det D\boldsymbol{u} \big ( \hat r\boldsymbol{e}_r(\theta) + x_3\boldsymbol{e}_{\theta}\big ) \leq \frac{\pi}{12} \frac{s^2 \cdot 1}{\hat r}\|D\boldsymbol{g}\|_{\infty}^2. \end{align*} $$

This amounts to saying that $\det D\boldsymbol {u} \sim s^2/\hat r$ and that

$$ \begin{align*} \infty> \int_{ d} H(s^2/\hat r)\mathrm{d}\boldsymbol{x} \geq 2\pi \int_{\hat r=0}^{3 \sin(\arccos(1/3))} \int_{x_3=0}^1 H \Big ( \frac{s^2(\hat r,x_3)}{\hat r}\Big ) \hat r\mathrm{d} x_3 \mathrm{d} \hat r. \end{align*} $$

Since $s(\hat r,x_3)=0$ in the segments $A'B'$ and $C'D'$ , where $\hat r>0$ , the determinant vanishes as $\boldsymbol {x}$ tends to those segments from region d. In contrast, the determinant does not blow up to infinity since the singularity $\hat r=0$ in the denominator is removable (recall that $s/\hat r\leq \|D\boldsymbol {g}\|_{\infty }$ ).

Now we consider the determinant of $ D \boldsymbol {u}_{\kern-1.2pt\varepsilon }$ . Let $\boldsymbol {w}_{\kern-1.2pt\varepsilon }$ be the auxiliary transformation of (3.12). We observe that

(3.38) $$ \begin{align} \begin{aligned} &\frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial s} \wedge \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial \theta} \wedge \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial z} \\ &= (\sin\varphi\,\boldsymbol{e}_r -\cos \varphi\,\boldsymbol{e}_3) \wedge (w_r^\varepsilon \boldsymbol{e}_{\theta}) \wedge \Big ( \omega_{\kern-1.2pt\varepsilon}^{\prime}(z)\boldsymbol{e}_r+\frac{\pi}{12} s\,\cos \varphi\,\boldsymbol{e}_r + \frac{\pi}{12} s\,\sin\varphi\,\boldsymbol{e}_3\Big ) \\ &= \begin{vmatrix} \sin \varphi & 0 & \omega_{\kern-1.2pt\varepsilon}^{\prime} + \frac{\pi}{12}s\cos\varphi \\ 0 & w_r^\varepsilon & 0 \\ -\cos\varphi & 0 & \frac{\pi}{12}s\sin\varphi \end{vmatrix} = w_r^\varepsilon \cdot \Big (\frac{\pi}{12}s +\omega_{\kern-1.2pt\varepsilon}^{\prime}(z)\cos\varphi \Big ). \end{aligned} \end{align} $$

Since $\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is defined with cylindrical coordinates,

(3.39) $$ \begin{align} & \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big ( r\boldsymbol{e}_r(\theta) + x_3 \boldsymbol{e}_3 \big ) = \frac{1}{r} \frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial r}\wedge \frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial \theta} \wedge \frac{\partial \boldsymbol{u}_{\kern-1.2pt\varepsilon }}{\partial x_3} \nonumber \\ &= \frac{1}{r} \Big ( \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial s} \frac{\partial s}{\partial \hat r}\frac{\partial \hat r}{\partial r} + \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial z}\frac{\partial z}{\partial \hat r} \frac{\partial \hat r}{\partial r} \Big ) \wedge \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial \theta}\wedge \Big ( \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial s}\frac{\partial s}{\partial x_3} + \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial z}\frac{\partial z}{\partial x_3}\Big ) \nonumber \\ & =\frac{1}{r} \Big ( \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial s} \wedge \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial \theta} \wedge \frac{\partial \boldsymbol{w}_{\kern-1.2pt\varepsilon}}{\partial z}\Big ) \frac{\partial \hat r}{\partial r} \Big ( \frac{\partial s}{\partial \hat r}\frac{\partial z}{\partial x_3} - \frac{\partial s}{\partial x_3}\frac{\partial z}{\partial \hat r}\Big ). \end{align} $$

Consider first the range $\varepsilon \leq r\leq \varepsilon ^{2\gamma }$ :

(3.40) $$ \begin{align} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big ( r\boldsymbol{e}_r(\theta) + x_3 \boldsymbol{e}_3 \big ) = \frac{w_r^\varepsilon (2r-\varepsilon) }{r(\varepsilon^{2\gamma}-\varepsilon)} \Big (\frac{\pi}{12}s +\omega_{\kern-1.2pt\varepsilon}^{\prime}(z)\cos\varphi(z) \Big )\frac{\partial (s,z)}{\partial (\hat r,x_3)}. \end{align} $$

By (3.37), and since $\varepsilon < \frac {1}{2}\varepsilon ^{2\gamma }$ for small $\varepsilon $ ,

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big ( r\boldsymbol{e}_r(\theta) + x_3\boldsymbol{e}_3\big ) \leq \frac{w_r^\varepsilon \cdot (2r)}{r (\frac{1}{2}\varepsilon^{2\gamma})}\Big (\frac{\pi}{12}s +\omega_{\kern-1.2pt\varepsilon}^{\prime}(z)\cos\varphi(z) \Big ) \|D\boldsymbol{g}\|_{\infty}^2. \end{align*} $$

By (3.34),

$$\begin{align*}w_r^\varepsilon \big ( s(\hat r, x_3), z(\hat r, x_3) \big ) = \omega_{\kern-1.2pt\varepsilon}(z) + s(\hat r, x_3)\sin (\varphi(z)) \leq \omega_{\kern-1.2pt\varepsilon}(3) + \|D\boldsymbol{g}\|_{\infty} \hat r \leq 6\varepsilon^{\gamma} + \|D\boldsymbol{g}\|_{\infty} \frac{(r-\varepsilon)r}{\varepsilon^{2\gamma}-\varepsilon}. \end{align*}$$

Since $r\leq \varepsilon ^{2\gamma }$ , it follows that $w_r^\varepsilon \leq C\varepsilon ^{\gamma }$ . By direct computation, we can check that $ |\omega _{\kern-1.2pt\varepsilon }^{\prime }(z)|\leq C \varepsilon ^\gamma $ . Thus, we find

$$ \begin{align*} \frac{\pi}{12}s(\hat r, x_3) +\omega_{\kern-1.2pt\varepsilon}^{\prime}\big ( z(\hat r, x_3)\big ) \cos\varphi\big(z(r,x_3)\big ) \leq C\varepsilon^{\gamma}. \end{align*} $$

Therefore,

$$ \begin{align*} \sup \Big\{ \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big (r\boldsymbol{e}_r(\theta)+x_3\varepsilon\big ): \varepsilon\leq r\leq \varepsilon^{2\gamma}\Big \} < \infty. \end{align*} $$

Here, we recall that we chose $ \gamma \leq 1/3$ . The radial distance r is greater than $\varepsilon $ ; hence, $r>2r-\varepsilon $ . Also, $w_r^\varepsilon (r)\geq 2\varepsilon ^\gamma $ and $\omega _{\kern-1.2pt\varepsilon }^{\prime }(z)\geq C\varepsilon ^\gamma $ . Using these bounds in (3.40) yields

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big ( r\boldsymbol{e}_r(\theta) + x_3 \boldsymbol{e}_3 \big ) \geq \frac{2\varepsilon^\gamma}{\varepsilon^{2\gamma}-\varepsilon} \big ( 0 + C\varepsilon^\gamma\cos\varphi\big ) \frac{\partial(s,z)}{\partial(\hat r, x_3)} \geq C\cos\varphi, \end{align*} $$

where in the last inequality also (3.35) was used. At this point, let us observe that $\cos \varphi $ is bounded below away from zero when $\varepsilon \leq r\leq \varepsilon ^{2\gamma }$ . Indeed, when $\hat r=0$ , the point $(\hat r, x_3)$ lies on the segment $BC$ and $z(\hat r, x_3)$ lies between $z=1$ and $z=2$ . Therefore, when $\hat r$ is close to zero, $z(\hat r, x_3)$ must be near the interval $[1,2]$ . Hence, $\cos \varphi $ should be bounded below by a value close to $\cos \varphi (2)=\cos (5\pi /12)$ . This is made rigorous by noting that

$$ \begin{align*} z(\hat r,x_3)&= z(0,x_3) + \int_0^{\hat r} \frac{\partial z}{\partial r} (t,x_3)\mathrm{d} t \leq 2 + \int_0^{\hat r} \big ( ( D\boldsymbol{g})\boldsymbol{e}_r\big ) \cdot \boldsymbol{e}_3 \mathrm{d} t \leq 2 + \|D\boldsymbol{g}\|_{\infty} \hat r. \end{align*} $$

Since $\hat r \leq \varepsilon ^{2\gamma }$ when $r\leq \varepsilon ^{2\gamma }$ , then, for $\varepsilon $ sufficiently small, $\cos \varphi $ is greater than any value smaller than $\cos (5\pi /12)$ . All in all, in the region $\varepsilon \leq r\leq \varepsilon ^{2\gamma }$ , the determinants are controlled from below and above, and

$$ \begin{align*} \lim_{\kern-1.2pt\varepsilon\to0} 2\pi \int_0^1 \int_{\kern-1.2pt\varepsilon}^{\varepsilon^{2\gamma}} H(\det D\boldsymbol{u}_{\kern-1.2pt\varepsilon })\, r\,\mathrm{d} r\,\mathrm{d} x_3 = 0. \end{align*} $$

Consider now the range $\varepsilon ^{2\gamma }\leq r\leq 3$ . By the definition of $\omega _{\kern-1.2pt\varepsilon }$ and (3.34),

$$ \begin{align*} w_r^\varepsilon = \omega_{\kern-1.2pt\varepsilon}(z(r,x_3)) + s(r,x_3)\sin \varphi \leq 6\varepsilon^{\gamma} + \|D\boldsymbol{g}\|_{\infty} r \leq (6+\|D\boldsymbol{g}\|_{\infty})\max\{r, \varepsilon^{\gamma}\}. \end{align*} $$

Analogously,

$$ \begin{align*} \frac{\pi}{12}s(r,x_3) + \omega_{\kern-1.2pt\varepsilon}^{\prime}(z(r,x_3))\cos \varphi\big ( z (r,x_3)\big ) \leq C\max\{r,\varepsilon^{\gamma}\}. \end{align*} $$

Hence, using (3.38) and (3.39) (recall that $\hat r=r$ for $r\geq \varepsilon ^{2\gamma }$ ), it follows that

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } (r\,\boldsymbol{e}_r(\theta) + x_3\boldsymbol{e}_3\big ) \leq C\frac{\max\{r, \varepsilon^{\gamma}\}^2}{r} \leq \begin{cases} C\varepsilon^{2\gamma}/r, & \text{if } \varepsilon^{2\gamma}\leq r \leq \varepsilon^{\gamma} \\ Cr^2/r, & \text{if } \varepsilon^{\gamma}\leq r\leq 3. \end{cases} \end{align*} $$

In both cases, the expression at the right-hand side is clearly bounded. Therefore, $\det D\boldsymbol {u}_{\kern-1.2pt\varepsilon }$ is bounded from above in $d_{\kern-1.2pt\varepsilon }$ uniformly with respect to $\varepsilon $ .

Regarding the lower bound, it suffices to see, from (3.38) and (3.39), that when $r\geq \varepsilon ^{2\gamma }$ ,

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big (r\,\boldsymbol{e}_r(\theta)+ x_3\,\boldsymbol{e}_3\big ) \geq \frac{1}{r} \cdot \big (s\,\sin \varphi \big)\Big (\frac{\pi}{12}s \Big )\cdot 1 \cdot \frac{\partial (s,z)}{(r,x_3)}, \end{align*} $$

where $s=s(r,x_3)$ and $\varphi =\varphi \big (z(r,x_e)\big )$ (recall that $\hat r\equiv r$ in the range of values of r being considered in this part of the proof). Comparing with (3.36), the conclusion is obtained that

$$ \begin{align*} \det D\boldsymbol{u}_{\kern-1.2pt\varepsilon } \big (r\,\boldsymbol{e}_r(\theta)+ x_3\,\boldsymbol{e}_3\big ) \geq \det D\boldsymbol{u} \big (r\,\boldsymbol{e}_r(\theta)+ x_3\,\boldsymbol{e}_3\big ) \qquad \forall\, r\geq \varepsilon^{2\gamma}, \ \forall\, x_3\in[0,1]. \end{align*} $$

With this estimate at hand, it is possible to find a dominating function to prove the desired convergence of the energies in region $d_{\kern-1.2pt\varepsilon }$ .