1 Introduction

The general law of hyperelasticity tells us that there exists an energy integral \( E[h] = \int _\mathbb {X}E(x, h, Dh) dx\) where \(E: \mathbb {X} \times \mathbb {Y} \times \mathbb {R}^{n\times n}\rightarrow \mathbb {R}\) is a given stored-energy function characterizing mechanical properties of the material. Here \(\mathbb {X}\) and \(\mathbb {Y}\) are nonempty bounded domains in \(\mathbb {R}^n, n > 2.\) The mathematical models of nonlinear elasticity have been first studied by Antman [1], Ball [4, 5], and Ciarlet [8]. One of the interesting and important problems in nonlinear elasticity is whether the radially symmetric minimizers are indeed global minimizers of the given physically reasonable energy. This leads us to study energy minimal homeomorphisms \(h: \mathbb {A}\xrightarrow []{{}_{\!\!\text {onto\,\,}\!\!}}\mathbb {A}^*\) of Sobolev class \(\mathscr {W}^{1,2}\) between annuli \(\mathbb {A}= \mathbb {A}(r, R) = \{x \in \mathbb {R}^n: r< |x| < R\}\) and \(\mathbb {A}^*=\mathbb {A}(r_*, R_*) = \{x \in \mathbb {R}^n: r_*< |x| < R_*\}\). Here \(0 \leqslant r < R\) and \(0 \leqslant r_*< R_*\) are the inner and outer radii of \(\mathbb {A}\) and \(\mathbb {A}^*\). The variational approach to Geometric Function Theory [2, 3] makes this problem more important. Indeed, several papers are devoted to understanding the expected radial symmetric properties see [17] and the references therein. Many times experimentally known answers to practical problems have led us to the deeper study of such mathematically challenging problems. We seek to minimize the p-harmonic energy of mappings between two annuli in \(\mathbb {R}^2\). We consider the modified Dirichlet energy \(\mathscr {F}_p[f]=\int _{\mathbb {A}}\frac{\Vert Df\Vert ^p}{|f|^p}\), \(1\leqslant p\leqslant 2\) and minimize it.

2 p-harmonic equation and statement of the main results

For natural number n, let \(A=(a_{i,j})_{n\times n}\in \mathbb {R}^{n\times n}\). We use \(A^{T}\) to denote the transpose of A. The Hilbert-Schmit norm, also called the Frobenius norm, of A is denoted by \(\Vert A\Vert \), where

$$\begin{aligned}\Vert A\Vert ^{2}=\sum _{1\le i,j\le n} \left| a_{i,j} \right| ^2= \text {tr} [A^{T}A].\end{aligned}$$

For \(p\ge 1\), we say that a mapping h belongs to the class \(\mathcal {W}^{1, p}(\mathbb {A},\mathbb {A}^*)\), if h belongs to the Sobolev space \(\mathcal {W}^{1, p}(\mathbb {A})\) and maps \(\mathbb {A}\) onto \(\mathbb {A}^*\). Let \(h=(h^{1},\ldots ,h^{n})\) belong to \(\mathcal {W}^{1,p}(\mathbb {A},\mathbb {A}^*)\). We denote the Jacobian matrix of h at the point \(x=(x_{1},\ldots ,x_{n})\) by Dh(x), where \(Dh(x)=\left( \frac{\partial h^{i}}{\partial x_{j}} \right) _{n\times n}\in \mathbb {R}^{n\times n}\). Then

$$\begin{aligned} \Vert Dh\Vert ^{2}=\sum _{1\le i,j\le n} \left| \frac{\partial h^{i}}{\partial x_{j}}\right| ^2. \end{aligned}$$

Here \(\frac{\partial h^{i}}{\partial x_{j}}\) denotes the weak partial derivatives of \(h^{i}\) with respect to \(x_{j}\). If h is continuous and belongs to \(\mathcal {W}^{1, p}(\mathbb {A},\mathbb {A}^*)\) \((p\ge 1)\), then the weak and ordinary partial derivatives coincide a.e. in \(\mathbb {A}\) (cf. [19, Proposition 1.2]). Let \(h=\rho S\), where \(S=\frac{h}{|h|}\) and \(\rho =|h|\). By [14, Equality (3.2)], we obtain that

$$\begin{aligned} Dh(x)=\nabla \rho (x)\otimes S(x) +\rho \cdot DS(x) \end{aligned}$$

and

$$\begin{aligned} \Vert Dh(x)\Vert ^{2}=|\nabla \rho (x)|^{2} +\rho ^{2} \Vert DS(x)\Vert ^{2}, \end{aligned}$$
(2.1)

where \(\nabla \rho \) denotes the gradient of \(\rho \).

We say that \(h:\mathbb {A} \rightarrow \mathbb {A}^{*} \) is a radial mapping, if \(h(x)=\rho (|x|)\frac{x}{|x|}\) and if \(\rho \) is real and positive function. We use \(\mathcal {R}(\mathbb {A},\mathbb {A}^*)\) to denote the class of radial homeomorphisms in \(\mathcal {W}^{1,p}(\mathbb {A},\mathbb {A}^*)\) and use \(\mathcal {P}(\mathbb {A},\mathbb {A}^*)\) to denote the class of generalized radial homeomorphisms in \(\mathcal {W}^{1,p}(\mathbb {A},\mathbb {A}^*)\). We also use \(\mathcal {H}(\mathbb {A},\mathbb {A}^*)\) to denote the class of homeomorphisms in \(\mathcal {W}^{1,p}(\mathbb {A},\mathbb {A}^*)\).

As it is said before, an important problems in nonlinear elasticity is whether the radially symmetric minimizers are indeed global minimizers. For example, Iwaniec, and Onninen [12] discussed the minimizers of the following two energy integrals:

$$\begin{aligned}\mathfrak {E}[h]=\int _{\mathbb {A}} \Vert Dh(x)\Vert ^{n} dx \quad \text {and}\quad \mathfrak {F}[h]=\int _{\mathbb {A}} \frac{\Vert Dh(x)\Vert ^{n}}{|h(x)|^{n}} dx \end{aligned}$$

among all homeomorphisms in \(\mathcal {W}^{1,n}(\mathbb {A},\mathbb {A}^{*})\), respectively. The energy integral \(\mathfrak {F}\) for \(n=2\), has been considered previously by Astala, Iwaniec, and Martin in [2]. Further such energy has been generalized in planar annuli by Kalaj in [15, 16] and spatial annuli in [13]. On the other hand, Koski and Onninen [17] investigated the minimizers of the p-harmonic energy

$$\begin{aligned}\mathcal {E}_{p}[h]=\int _{\mathbb {A}} \Vert Dh(x)\Vert ^{p} dx \end{aligned}$$

among all homeomorphisms in \(\mathcal {W}^{1,p}(\mathbb {A},\mathbb {A}^{*})\), where \(\mathbb {A}\) and \(\mathbb {A}^{*}\) are planar annuli and \(1\le p<2\), provided the homeomorphisms fix the outer boundary. Recently, Kalaj [14] studied the Dirichlet-type energy \(\mathscr {F}[h]\) among mappings in \(\mathcal {H}(\mathbb {A},\mathbb {A}^*)\), where

$$\begin{aligned} \mathscr {F}[h]=\int _{\mathbb {A}} \frac{\Vert Dh(x)\Vert ^{n-1}}{|h(x)|^{n-1}}dx. \end{aligned}$$
(2.2)

For \(n=3\), the author proved that the minimizers of \(\mathscr {F}[h]\) are certain generalized radial diffeomorphism (cf. [14, Theorem 1.1]). Motivated by the case \(n=3\), in [14] it was posed the following question.

Question 2.1

For \(n\not =3\), does the Dirichlet integral of \(h\in \mathcal {H}(\mathbb {A},\mathbb {A}^{*}) \), i.e. the integral

$$\begin{aligned} \mathscr {F}[h]=\int _{\mathbb {A} } \frac{\Vert Dh(x)\Vert ^{n-1}}{|h(x)|^{n-1}}dx, \end{aligned}$$

achieve its minimum for generalized radial diffeomorphisms between annuli?

Then in the subsequent paper by Kalaj and Chen [9] was given the following answer.

Theorem 2.1

For \(n \ge 4\), we have

$$\begin{aligned}{} & {} \inf _{h\in \mathcal {H}(\mathbb {A},\mathbb {A}^{*})} \mathscr {F}[h] =\inf _{h\in \mathcal {P}(\mathbb {A},\mathbb {A}^{*})} \mathscr {F}[h] \end{aligned}$$

The last infimum is never attained.

In this paper, we consider the case of the p-energy Sobolev \(\mathcal {W}^{1,p}\) homeomorphisms between annuli \(\mathbb {A}\) and \(\mathbb {A}^*\) in the complex plane. Let

$$\begin{aligned}\mathscr {F}_p[h]=\int _{\mathbb {A}(1,r)}\frac{\Vert Dh\Vert ^p}{|h|^p}, \ \ 1\leqslant p<2.\end{aligned}$$

Then we seek the homeomorphisms h of the class \(\mathcal {W}^{1,p}\) which are furthermore assumed to preserve the order of the boundary components \(|h(z)|\rightarrow \)r when \(|z|\rightarrow r^*\) and \(|h(z)|\rightarrow R^*\) when \(|z|\rightarrow R\). Such a class of Sobolev homeomorphisms with the above property is denoted by \(\mathcal {H}^{1,p}(\mathbb {A},\mathbb {A}^*)\) and we say that they are admissible homeomorphisms. Since we minimize the \(\mathscr {F}_p\) energy in the class of homeomorphisms, we can perform the inner variation of the independent variable \(z_\epsilon = z + \epsilon \tau (z)\), which leads to the system (see for example [14])

$$\begin{aligned} \textrm{div}\left( \frac{1}{|h|^p}\Vert Dh\Vert ^{p-2}(Dh)^* Dh-\frac{1}{p|h|^p}\Vert Dh\Vert ^p I\right) =0,\end{aligned}$$
(2.3)

where

$$\begin{aligned}\textrm{div}\left( \begin{array}{cc} a(x,y) &{} b(x,y) \\ c(x,y) &{} d(x,y) \\ \end{array} \right) :=\left( \begin{array}{c} a_x+b_y \\ c_x+d_y \\ \end{array} \right) .\end{aligned}$$

Here \(z=(x,y)\). Our argument does not make direct use of the inner variational equation (2.3). Some important facts that follow from (2.3) are as follows.

  1. (1)

    If we assume that h is radial, then (2.3) reduces to the Euler-Lagrange equation (3.1) below.

  2. (2)

    Further if f is a solution of (2.3) then so is \(\tilde{f}=\frac{1}{f}\).

  3. (3)

    Let \(f_1(z)=\frac{1}{r_*} f(r z)\). Then \(f_1:\mathbb {A}(1,r_1)\xrightarrow []{{}_{\!\!\text {onto\,\,}\!\!}}\mathbb {A}(1,R_1)\), provided that \(f:\mathbb {A}(r,R)\xrightarrow []{{}_{\!\!\text {onto\,\,}\!\!}}\mathbb {A}(r^*, R^*)\), where \(R_1=R_*/r_*\) and \(r_1=R/r\). Moreover, f satisfies (2.3) if and only if \(f_1\) satisfies the same equation.

This is why we reduce the problem to the annuli \(\mathbb {A}=\mathbb {A}(1,r)\) and \(\mathbb {A}^*=\mathbb {A}(1,R)\). Now we formulate the main results.

Theorem 2.2

Let \(\mathbb {A}\) and \(\mathbb {A}^*\) be planar annuli and \(1< p\leqslant 2.\) Then there exists a radially symmetric mapping \(h_\circ :\mathbb {A}\rightarrow \mathbb {A}^*\) such that

$$\begin{aligned} \min _{{\mathcal {H}}^{1,p}(\mathbb {A},\mathbb {A}^*)} \mathscr {F}_p[h] =\mathscr {F}_p[h_\circ ] .\end{aligned}$$
(2.4)

The map \(h_\circ \) is the unique minimizer, up to a rotation, in the class \(\mathcal {H}^{1,p}(\mathbb {A},\mathbb {A}^*)\). Furthermore, the minimizer \(h_\circ \) is a homeomorphism.

Theorem 2.3

Let \(\mathbb {A}\) and \(\mathbb {A}^*\) be planar annuli. Then there exists a radially symmetric mapping \(h_\circ :\mathbb {A}\rightarrow \mathbb {A}^*\) which is a homeomorphism such that

$$\begin{aligned} \min _{{\mathcal {H}}^{1,1}(\mathbb {A},\mathbb {A}^*)}\mathscr {F}_1[h] =\mathscr {F}_p[h_\circ ],\end{aligned}$$
(2.5)

if and only if

$$\begin{aligned} {\frac{\pi }{2}-\tan ^{-1}\left[ \frac{1}{\sqrt{r^2-1}}\right] }\geqslant \log R.\end{aligned}$$
(2.6)

The map \(h_\circ \) is the unique minimizer, up to a rotation, in the class \(\mathcal {H}^{1,1}(\mathbb {A},\mathbb {A}^*)\).

Remark 2.4

Note that the case \(p=2\) of Theorem 2.2 has been already considered by Astala, Iwaniec, and Martin in [2].

On the other hand side our result can be seen as a variation of minimization property of radial mappings of p-Dirichlet energy throughout Sobolev mappings from the unit ball \(\mathbb {B}\subset \mathbb {R}^n\) onto the unit sphere \(\mathbb {S}^{n-1}\), fixing the boundary. This is an old problem solved by several authors (see for example [6, 7, 18]).

Furthermore, as was remarked before, Koski and Onninen [17] have considered \(\mathcal {E}_{p}\) energy and proved the minimization property, under a certain constrain. Indeed, if we denote the outer boundary of \(\mathbb {A}\) by \(\partial _\circ \mathbb {A}\) and consider the subfamily of homomorphisms \(\mathcal {H}_\circ =\{f\in \mathcal {H}^{1,p}(\mathbb {A},\mathbb {A}^*): f(x)=\frac{R_*}{R} x, \ \text { for } x\in \partial _\circ \mathbb {A}\}\), then the minimizer of \(\mathcal {E}_{p}\) energy is a radial mapping \(h(x)=\rho (x)\frac{x}{|x|}\) provided that R and r satisfies some inequality that depends on p ([17, Theorem 1.5]). In the same paper they proved that this constraint is crucial and there exists annuli, where the minimizer of \(\mathcal {E}_{p}\) is not a radial mapping.

Remark 2.5

By virtue of the density of diffeomorphisms in \(\mathcal {H}^{1,p}(\mathbb {A},\mathbb {A}^*)\), see [10, 11], we can equivalently replace the admissible homeomorphisms by sense preserving diffeomorphims. Indeed, for \(p \geqslant 1\), we have

$$\begin{aligned} \inf _{f\in \mathcal {H}^{1,p}(\mathbb {A},\mathbb {A}^*)}\mathcal {E}_p[h] = \inf _{f\in \textrm{Diff}(\mathbb {A},\mathbb {A}^*)}\mathcal {E}_p[h].\end{aligned}$$
(2.7)

Here by \(\textrm{Diff}(\mathbb {A},\mathbb {A}^*)\) we denote the class of orientation preserving diffeomorphisms from \(\mathbb {A}\) onto \(\mathbb {A}^*\) which also preserve the order of the boundary components. A similar result hold for the \(\mathscr {F}_p\) energy. Indeed

$$\begin{aligned} \inf _{f\in \mathcal {H}^{1,p}(\mathbb {A},\mathbb {A}^*)}\mathscr {F}_p[h] = \inf _{f\in \textrm{Diff}(\mathbb {A},\mathbb {A}^*)}\mathscr {F}_p[h].\end{aligned}$$
(2.8)

3 Radial minimizer of the energy \(\mathscr {F}_p[h]\), \(1<p<2\)

This section aims is to find the radial minimizer \(h_\circ \) of \(\mathscr {F}_p\) energy that maps annuli \(\mathbb {A}(1,r)\) onto \(\mathbb {A}(1,R)\) keeping the boundary order. Moreover, we will use that solution to prove the minimization property of \(h_\circ \) in the class of all Sobolev homeomorphisms. Contrary to the case \(p=1\), which will be considered later, we will not have any restriction on r and R. Assume that \(h(z) = H(t) e^{i \theta }\), where \(z=t e^{i\theta }\), where H is a differentiable function and that \(t\in [1,r]\), \(\theta \in [0,2\pi ]\). Then

$$\begin{aligned}\Vert D h\Vert ^2 = |h_t|^2 + \frac{|h_{\theta }|^2}{t^2}=\dot{H}(t)^2+\frac{H(t)^2}{t^2}.\end{aligned}$$

Furthermore

$$\begin{aligned}t\frac{\Vert Dh\Vert ^p}{|h|^p} = t\left( \frac{1}{t^2}+\frac{ \dot{H}(t)^2}{H(t)^2}\right) ^{p/2}.\end{aligned}$$

Let

$$\begin{aligned}L(t, H, \dot{H})=t\left( \frac{1}{t^2}+\frac{ \dot{H}(t)^2}{H(t)^2}\right) ^{p/2}.\end{aligned}$$

Then Euler-Lagrange equation

$$\begin{aligned}L_H=\partial _t L_{\dot{H}},\end{aligned}$$

can be written in the following form

$$\begin{aligned} \ddot{H}= \frac{\dot{H} \left( (p-3) H^3+t H^2 \dot{H}-t^2 H \dot{H}^2+(p-1) t^3 \dot{H}^3\right) }{t H^3+(p-1) t^3 H \dot{H}^2},\end{aligned}$$
(3.1)

where \(H=H(t)\), \(\dot{H}= H'(t)\) and \(\ddot{H}=H''(t)\). Then by straightforward calculation (3.1) can be reduced to the following differential equation

$$\begin{aligned} \frac{t \dot{H}(t)}{H(t)}=\frac{\sqrt{g(t)}}{\sqrt{1-g(t)}},\end{aligned}$$
(3.2)

where g is a solution to the following differential equation

$$\begin{aligned} \dot{g}(t)= F[t, g(t)]:=\frac{2 (2-p) (g(t)-1) g(t)}{t+(p-2) t g(t)}.\end{aligned}$$
(3.3)

Show that \(F<0\) provided that \(t\geqslant 1\) and \(g(t)\in (0,1)\). Namely

$$\begin{aligned}t+(-2+p) t g(t)\geqslant t +(p-2)t = (p-1)t>0.\end{aligned}$$

Since \(2 (2-p) (g(t)-1) g(t)<0\) we infer that g is a decreasing function.

The general solution of (3.3) is given by \(g=k^{-1}\), where the function k is defined by

$$\begin{aligned} k(s)=b\exp \left( {\frac{(p-1) \log (1-s)-\log s}{2 (2-p)}}\right) ,\end{aligned}$$
(3.4)

where b is a positive constant and \(s\in (0,1)\).

By (3.2) we infer that H is given by

$$\begin{aligned} H(t)= C\exp \left[ \int _1^t \frac{\sqrt{g(x)}}{\sqrt{1-g(x)} x} \, dx\right] .\end{aligned}$$
(3.5)

By using the change \(t=k(s)\) in (3.5) we obtain

$$\begin{aligned} H(t) = C \exp \left[ \int _{g(t)}^{g(1)} \frac{\left( \frac{p-1}{1-s}+\frac{1}{s}\right) \sqrt{s}}{2 (2-p) \sqrt{1-s}} ds\right] .\end{aligned}$$
(3.6)

Since we seek increasing homeomorphic mappings \(H:[1,r]\xrightarrow []{{}_{\!\!\text {onto\,\,}\!\!}}[1,R]\), we have the initial conditions \(H(1)=1\) and \(H(r)=R\). Then \(C=1\). Let \(0<\tau <1\) and chose \(b=b(\tau )\) so that

$$\begin{aligned}b=\exp \left( \frac{(p-1) \log (1-\tau )-\log \tau }{2 (p-2)}\right) .\end{aligned}$$

Denote the corresponding g by \(g_\tau \). Then we have \(g_\tau (1)=\tau \).

Moreover by (3.4)

$$\begin{aligned}g_\tau \left[ \exp \left( {\frac{(p-1) \log (\frac{1-t}{1-\tau })-\log \frac{t}{\tau }}{2 (2-p)}}\right) \right] =t.\end{aligned}$$

Define the function

$$\begin{aligned}\mathcal {R}(\tau )=\exp \left[ \int _{g_\tau (r)}^{\tau } \frac{\left( \frac{p-1}{1-s}+\frac{1}{s}\right) \sqrt{s}}{2 (2-p) \sqrt{1-s}} dx\right] .\end{aligned}$$

Then we also define

$$\begin{aligned}H_\tau (t) = \exp \left[ \int _{g_\tau (t)}^{\tau } \frac{\left( \frac{p-1}{1-s}+\frac{1}{s}\right) \sqrt{s}}{2 (2-p) \sqrt{1-s}} ds\right] .\end{aligned}$$

Then

$$\begin{aligned}H_\tau (1)=1\end{aligned}$$

and

$$\begin{aligned} H_\tau (r)=\mathcal {R}(\tau ).\end{aligned}$$
(3.7)

Let us show that there is a unique \(s_\circ =s(r,\tau )\in (0,\tau )\) such that \(B(s_\circ )=0\), where

$$\begin{aligned}B(s):={\frac{(p-1) \log (\frac{1-s}{1-\tau })-\log \frac{s}{\tau }}{2 (2-p)}}-\log r.\end{aligned}$$

Note that B is continuous, \(B(\tau )=\log \frac{1}{r}<0\) and \(B(0)=+\infty \). Moreover

$$\begin{aligned}B'(s)=\frac{1+(-2+p) s}{2 (2-p) (-1+s) s}<0.\end{aligned}$$

Thus there is a unique \(s_\circ \) so that \(B(s_\circ )=0\). Then \(g_\tau (r)=s_\circ \). Since for \(0<s<\tau \) and \(p\in (1,2]\), we have

$$\begin{aligned}{\frac{ \log (\frac{1-s}{1-\tau })-\log \frac{s}{\tau }}{2 (2-p)}}\geqslant {\frac{(p-1) \log (\frac{1-s}{1-\tau })-\log \frac{s}{\tau }}{2 (2-p)}},\end{aligned}$$

it follows that

$$\begin{aligned}{\frac{ \log (\frac{1-s_\circ }{1-\tau })-\log \frac{s_\circ }{\tau }}{2 (2-p)}}-\log r\geqslant B(s_\circ )={\frac{(p-1) \log (\frac{1-s_\circ }{1-\tau })-\log \frac{s_\circ }{\tau }}{2 (2-p)}}-\log r=0.\end{aligned}$$

Thus

$$\begin{aligned} 0<s_\circ <\tau _\circ =\frac{1}{1+r^{4-2 p} \left( -1+\frac{1}{\tau }\right) }.\end{aligned}$$
(3.8)

Then

$$\begin{aligned}\mathcal {R}(\tau )=\exp \left[ \int _{s_\circ }^{\tau } \frac{\left( \frac{p-1}{1-s}+\frac{1}{s}\right) \sqrt{s}}{2 (2-p) \sqrt{1-s}} ds\right] .\end{aligned}$$

Let us show now that, if \(p>1\), then for every \(R\in (1,+\infty )\), there is \(\tau \in (0,1)\) so that \(\mathcal {R}(\tau )=R\). It is clear that \(\mathcal {R}\) is continuous and also it is clear that \(\lim _{\tau \rightarrow 0} \mathcal {R}(\tau )=1\). Let us show that \(\lim _{\tau \rightarrow 1} \mathcal {R}(\tau )=+\infty \). Observe that \(0\leqslant s\leqslant \sqrt{s}\leqslant 1\). Then from (3.8) we have that

$$\begin{aligned}\mathcal {R}(\tau )\geqslant K(\tau ),\end{aligned}$$

where

$$\begin{aligned}K(\tau )=\exp \left[ \int _{\tau _\circ }^{\tau } \frac{\left( \frac{p-1}{s-1}+\frac{1}{s}\right) s}{2 (2-p) \sqrt{1-s}} ds\right] .\end{aligned}$$

Then \(K(\tau )=\exp (k(\tau )-k(\tau _0))\), where

$$\begin{aligned}k(s)=\frac{3+p (s-2)-2 s}{(2-p) \sqrt{1-s}}.\end{aligned}$$

Then

$$\begin{aligned}\lim _{\tau \rightarrow 1^-}\sqrt{1-\tau } \log K(\tau )=\frac{(p-1) \left( r^2+r^p\right) }{(2-p) r^2}.\end{aligned}$$

We notice that here is the moment where \(p\in (1,2)\) is an important assumption. In particular \(\lim _{\tau \rightarrow 1} R(\tau )=\infty \). So there is \(\tau =\tau (r, R)\) so that \(\mathcal {R}(\tau )=R\). In view of (3.7), we have constructed a smooth increasing mapping \(H=H_\circ =H_{r,R}:[1,r]\rightarrow [1,R]\) so that \(H(1)=1\) and \(H(r)=R\), See Fig. 1 below. Let us show that

$$\begin{aligned} h_\circ (z)= H(t)e^{i\theta }, \ \ z=t e^{i\theta },\end{aligned}$$
(3.9)

is the minimizer in the class of radial homeomorphisms between \(\mathbb {A}\) and \(\mathbb {A}^*\).

Fig. 1
figure 1

The graphic of \(H_\circ \) satisfying initial conditions \(H(1)=1\), \(H(2)=2\) is far from being identity

Full size image

Assume now that \(H: [1,r]\rightarrow [1,R]\) is any smooth homeomorphism and assume that \(h(z)=H(t)e^{i\theta }\). Prove that

$$\begin{aligned} \mathscr {F}_p[h]\geqslant \mathscr {F}_p[h_\circ ].\end{aligned}$$
(3.10)

We start from a simple inequality from [17]

$$\begin{aligned} (a+b)^{q/2}\geqslant s^{1-q/2}a^{q/2}+(1-s)^{1-q/2}b^{q/2}, \ \ q\in [1,2], \ \ s\in [0,1]. \end{aligned}$$
(3.11)

By inserting \(q=p\), \(s=g(t)\),

$$\begin{aligned}a=t^{\frac{2}{p}-2}, \ \ b=t^{2/p} \frac{\dot{H}^2}{H^2}\end{aligned}$$

in (3.11) we have

$$\begin{aligned} \begin{aligned}t \left( \frac{1}{t^2}+ \frac{\dot{H}^2}{H^2}\right) ^{p/2}&= \left( t^{2/p-2}+ t^{2/p}\frac{\dot{H}^2}{H^2}\right) ^{p/2}\\&\geqslant (1-g(t))^{1-p/2}t^{1-p}+g(t)^{1-p/2}t\frac{|\dot{H}|^p}{|H|^p}.\end{aligned}\end{aligned}$$
(3.12)

The equality in (3.11) is attained precisely when

$$\begin{aligned}\frac{b}{a}= \frac{s}{1-s}\end{aligned}$$

and thus the equality is attained in (3.12) precisely when

$$\begin{aligned} \frac{t\dot{H}}{H}=\frac{\sqrt{g(t)}}{\sqrt{1-g(t)}}.\end{aligned}$$
(3.13)

Then by

$$\begin{aligned} a^p\geqslant p x^{p-1} a -(p-1) x^p,\end{aligned}$$
(3.14)

where \(a=\frac{\dot{H}(t)}{H(t)}\) and \(x=\frac{\sqrt{g(t)}}{\sqrt{1-g(t)} t}\) we get

$$\begin{aligned} t\left( \frac{1}{t^2}+ \frac{\dot{H}(t)^2}{H(t)^2}\right) ^{p/2}\geqslant t^{1-p}\frac{1-pg(t)}{(1-g(t))^{p/2}}+\frac{s^{2-p}\sqrt{g(t)}}{(1-g(t))^{(p-1)/2}}p \frac{\dot{H}}{H}.\end{aligned}$$
(3.15)

Notice that, the condition (3.13) is precisely satisfied when we have equality in (3.15).

Define

$$\begin{aligned}P(t) = t^{2-p} \left( {1-g(t)}\right) ^{\frac{1}{2} (1-p)} \sqrt{g(t)},\end{aligned}$$

and show that it is a constant. This fact is crucial for our approach.

By (3.3) we obtain that

$$\begin{aligned}\frac{P'(t)}{P(t)}=\frac{2-p}{t}+\frac{(1+(p-2) g(t)) \dot{g}(t)}{2 (1-g(t)) g(t)}=0.\end{aligned}$$

Thus

$$\begin{aligned} P(t)\equiv c=P(r)=r^{2-p} \left( {1-g(r)}\right) ^{\frac{1}{2} (1-p)} \sqrt{g(r)}.\end{aligned}$$
(3.16)

Observe that

$$\begin{aligned}g(r)=g_\tau (r)=c_\circ (r,\tau )=c_\circ (r, \tau (r,R)).\end{aligned}$$

Thus \(c=c(r,R)\). Now we have

$$\begin{aligned}\begin{aligned}\mathscr {F}_p[h]&=2\pi \int _1^rt\left( \frac{1}{t^2}+\frac{\dot{H}(t)^2}{H^2(t)}\right) ^{p/2}dt \\&\geqslant 2\pi \int _1^r\left( t^{1-p}\frac{1-pg(t)}{(1-g(t))^{p/2}}+c(r,R) \frac{\dot{H}}{H}\right) dt\\&=2\pi \int _1^r\left( t^{1-p}\frac{1-pg(t)}{(1-g(t))^{p/2}}\right) dt+2\pi \int _1^r c(r,R)\frac{\dot{H}(t)}{H(t)}dt\\&=2\pi \int _1^r\left( t^{1-p}\frac{1-pg(t)}{(1-g(t))^{p/2}}\right) dt+2\pi c(r, R)\log R\\&=\mathscr {F}_p[h^\circ ]. \end{aligned} \end{aligned}$$

4 Radial minimizers for the case \(p=1\)

The corresponding subintegral expression for the functional \(\mathscr {F}_1[h]=\int _{\mathbb {A}(1,r)} \frac{|Df(z)|}{|f(z)|}\), for radial function \(h(z)=H(t)e^{i\theta }\), \(z=te^{i\theta }\) is given by

$$\begin{aligned}L(t, H, \dot{H})=\left( 1+\frac{t^2 \dot{H}(t)^2}{H(t)^2}\right) ^{1/2}.\end{aligned}$$

The corresponding differential equation (3.1) for \(p=1\) reduces to

$$\begin{aligned} \left( -t H(t) \dot{H}(t)^2+t^2 \dot{H}(t)^3+H(t)^2 \left( 2 \dot{H}(t)+t \ddot{H}(t)\right) \right) =0\end{aligned}$$
(4.1)

which can be written in the following form

$$\begin{aligned}\frac{t\dot{H}(t)}{H(t)} =\frac{\sqrt{g(t)}}{\sqrt{1-g(t)}}\end{aligned}$$

where g is a solution of the differential equation (see (3.3) for \(p=1\)):

$$\begin{aligned} 2 g(t)+t \dot{g}(t)=0.\end{aligned}$$
(4.2)

Then the general solution of (4.2) is given by \(g(t) = {b t^{-2}}.\) Then the solution of (4.1) is the solution of the equation

$$\begin{aligned}\frac{t\dot{H}(t)}{H(t)}=\frac{1}{\sqrt{b^2 t^2-1}}\end{aligned}$$

and it is given by

$$\begin{aligned}H(t)=c \exp \left( {-\cot ^{-1}\left[ \sqrt{b^2t^2-1 }\right] }\right) .\end{aligned}$$

If we let that \(H(1)=1\) then

$$\begin{aligned} H(t)=\exp \left( \cot ^{-1}\left[ \sqrt{b^2 -1}\right] -\cot ^{-1}\left[ \sqrt{ b^2 t^2-1}\right] \right) .\end{aligned}$$
(4.3)

Here \(b\geqslant 1\). Moreover, if we assume that \(H(r)=R\), then after straightforward computations we get

$$\begin{aligned}b=\frac{\sqrt{\left( 1+r^2-2 r \cos \log R\right) } \csc [\log R]}{r}.\end{aligned}$$

The corresponding minimizer is denoted by \(h_\circ (z)=H(r)e^{i\theta },\) \(z=re^{i\theta }\). Hence

$$\begin{aligned}\mathscr {F}[h]=2\pi \int _1 ^r \left( 1+\frac{t^2(\dot{H}(t))^2}{H^2(t)}\right) ^{1/2}dt \geqslant 2\pi \int _1^r \sqrt{1-\frac{1}{b^2 t^2}}+\frac{ \dot{H}(t)}{bH(t)}dt\end{aligned}$$

Thus

$$\begin{aligned}\mathscr {F}[h]\geqslant \mathscr {F}[h_\circ ]\end{aligned}$$

where

$$\begin{aligned}\mathscr {F}[h_\circ ]=2\pi \frac{-\sqrt{b^2-1}+\sqrt{b^2r^2-1}-\csc ^{-1}\left[ b\right] +\csc ^{-1}\left[ b r\right] }{b}+ \frac{2\pi \log R}{b}.\end{aligned}$$

Lemma 4.1

It exists a radial homeomorphism \(h: \mathbb {A}(1,r)\rightarrow \mathbb {A}(1,R)\) if and only if

$$\begin{aligned}{\frac{\pi }{2}-\tan ^{-1}\left[ \frac{1}{\sqrt{r^2-1}}\right] }>\log R.\end{aligned}$$

Proof

By differentiating (4.3) w.r.t. b we get

$$\begin{aligned}{} & {} \partial _b H(t)=\frac{\exp \left( {\cot ^{-1}\left[ \sqrt{-1+b^2}\right] -\cot ^{-1}\left[ \sqrt{-1+b^2 t^2}\right] }\right) \left( -\frac{1}{\sqrt{-1+b^2}}+\frac{1}{\sqrt{-1+b^2 t^2}}\right) }{b}.\end{aligned}$$

Hence H is decreasing in b. The largest value is for \(b=1\) and it is equal to

$$\begin{aligned}R_\circ (r):=\exp \left( \frac{\pi }{2}-\tan ^{-1}\left[ \frac{1}{\sqrt{r^2-1}}\right] \right) \end{aligned}$$

for \(t=r\). In other words, there is a increasing diffeomorphism of [1, r] onto [1, R] if and only if \(R\leqslant R_\circ (r).\) \(\square \)

Remark 4.2

Observe that \(\lim _{r\rightarrow \infty } \mathcal {R}(r)=e^{\pi /2}\), so there is not any homeomorphic minimizer of the \(\mathscr {F}\) energy between annuli \(\mathbb {A}(1,r)\) and \(\mathbb {A}(1,e^{\pi /2})\). Note that the conformal modulus of \(\mod \mathbb {A}(1,e^{\pi /2})\) is \(\log e^{\pi /2}=\pi /2\). So the case \(p=1\) differs from the case \(p>1\). Moreover, this case is also opposite to the Nitsche type phenomenon for Dirichlet energy \(\mathcal {E}\). Namely Nitsche type phenomenon asserts that the modulus of image domain could be arbitrarily large, but not small enough.

5 Proof of Theorem 2.2 and Theorem 2.3

We begin with the following proposition

Proposition 5.1

Assume that \(h=\rho (z)e^{i\Theta (z)}\) is a diffeomorphism between annuli \(\mathbb {A}(1,r)\) and \(\mathbb {A}(1,R)\). Then for every \(t\in [1,r]\) and \(\theta \in [0,2\pi ]\) we have

$$\begin{aligned} \int _{t\mathbb {T}}|\nabla \Theta (z)||dz|\geqslant 2\pi . \end{aligned}$$
(5.1)

If the equality hold in (5.1) for every \(\theta \in [0,2\pi ]\), then \(\Theta (z) =e^{i\varphi (\theta )},\) \(z=te^{i\theta }\), for a diffeomorphism \(\varphi :[0,2\pi ]\xrightarrow []{{}_{\!\!\text {onto\,\,}\!\!}}[\alpha , 2\pi +\alpha ]\). Further, we have

$$\begin{aligned} \int _{1}^{R}\frac{|\nabla \rho (te^{i\theta }) |}{\rho (te^{i\theta })}dt\geqslant \log R. \end{aligned}$$
(5.2)

If the equality hold in (5.2) for every \(t\in [1,R]\), then \(\rho (te^{i\theta }) =\rho (t)\).

Proof of Proposition 5.1

First of all, for fixed t, \(\gamma (\theta )=e^{i\Theta (t e^{i\theta })}\) is a surjection of \([0,2\pi ]\) onto \(\mathbb {T}=\{z:|z|=1\}\). Further

$$\begin{aligned}|\nabla \Theta (t e^{i\theta })|^2=|\Theta _t|^2+\frac{|\Theta _\theta |^2}{t^2}. \end{aligned}$$

So

$$\begin{aligned} |\gamma '(\theta )|=|\Theta _\theta |\leqslant t |\nabla \Theta (t e^{i\theta })|.\end{aligned}$$
(5.3)

The equality is attained in (5.3) if and only if \(\Theta _t\equiv 0\). In this case \(\gamma (\theta )=e^{i\varphi (\theta )}\), for a smooth function of \(\varphi :[0,2\pi ] \xrightarrow []{{}_{\!\!\text {onto\,\,}\!\!}}[\alpha ,2\pi +\alpha ]\).

We obtain that

$$\begin{aligned}|\mathbb {T}|=2\pi \leqslant \int _0^{2\pi }|\gamma '(\theta )|d\theta \leqslant \int _{t\mathbb {T}}|\nabla \Theta (z)| |dz|,\end{aligned}$$

with an equality if and only if \(\Theta (se^{i\theta })\) does not depend on t. Thus the first statement of the proposition is proved.

Similarly the function \(\alpha (t)=\log \rho (t e^{i\theta })\) is a surjection of [1, r] onto \([0,\log R]\) and hence

$$\begin{aligned}\log R= \int _1^r \alpha '(t)dt\leqslant \int _1^r \frac{|\nabla \rho (te^{i\theta })|}{\rho (te^{i\theta })}dt.\end{aligned}$$

The equality statement can be proved in the same way as the former part. We only need to use the formula

$$\begin{aligned}|\nabla \rho (t e^{i\theta })|^2=|\rho _t|^2+\frac{|\rho _\theta |^2}{t^2}\geqslant |\rho _t|^2. \end{aligned}$$

\(\square \)

Proof of Theorem 2.2

Assume as before that \(h(z)=\rho (z)e^{i\Theta (z)}\) is a mapping from the annulus \(\mathbb {A}\) onto the annulus \(\mathbb {A}^{*}\). We start from the following inequality which follows from Hölder inequality

$$\begin{aligned}\mathscr {F}_p[h]=\int _{\mathbb {A}(1,r)}\frac{\Vert Dh\Vert ^p}{|h|^p}\geqslant \frac{\left( \int _{\mathbb {A}(1,r)}\frac{\Vert Dh\Vert }{|h|} \cdot \frac{\Vert Dh_\circ \Vert ^{p-1}}{|h_\circ |^{p-1} }\right) ^p}{\left( \int _{\mathbb {A}(1,r)}\frac{\Vert Dh_\circ \Vert ^p}{|h_\circ |^p}\right) ^{p-1}}.\end{aligned}$$

In view of (2.1)

$$\begin{aligned} \Vert Dh\Vert ^{2}=|\nabla \rho |^2+\rho ^2|\nabla \Theta |^2, \end{aligned}$$

where \(\rho (z)=|h(z)|\). And thus

$$\begin{aligned}\frac{\Vert Dh\Vert }{|h|}=\left( {|\nabla \Theta |^2}+\frac{|\nabla \rho |^2}{\rho ^2}\right) ^{1/2}.\end{aligned}$$

Then by (3.10), for \(q=1\) we have

$$\begin{aligned} \frac{\Vert Dh\Vert }{|h|}\geqslant \left( \sqrt{1-g(t)}|\nabla \Theta |+\sqrt{g(t)}\frac{|\nabla \rho |}{\rho }\right) . \end{aligned}$$
(5.4)

From (5.4) we get

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {A}(1,r)}\frac{\Vert Dh\Vert }{|h|} \cdot \frac{\Vert Dh_\circ \Vert ^{p-1}}{|h_\circ |^{p-1} }\\&\quad = \int _{\mathbb {A}(1,r)} \left( {|\nabla \Theta |^2}+\frac{|\nabla \rho |^2}{\rho ^2}\right) ^{1/2} \cdot \frac{\Vert Dh_\circ \Vert ^{p-1}}{|h_\circ |^{p-1} }\\&\quad \geqslant \int _0^{2\pi }\int _1^r t\frac{\Vert Dh_\circ \Vert ^{p-1}}{|h_\circ |^{p-1} }\left( \sqrt{1-g(t)}|\nabla \Theta |+\sqrt{g(t)}\frac{|\nabla \rho |}{\rho }\right) dtd\theta . \end{aligned}\end{aligned}$$

Let

$$\begin{aligned}K(t) = t\sqrt{g(t)}\frac{\Vert Dh_\circ \Vert ^{p-1}}{|h_\circ |^{p-1}}.\end{aligned}$$

Then

$$\begin{aligned}K(t)=t \sqrt{g(t)} \left( \frac{1}{t^2}+\frac{g(t)}{t^2 (1-g(t))}\right) ^{\frac{1}{2} (p-1)}=P(t).\end{aligned}$$

Thus we again use (3.16) to conclude that \(K(t)=c(r,R)\). Furthermore

$$\begin{aligned} \begin{aligned}t\frac{\Vert Dh\Vert }{|h|} \cdot \frac{\Vert Dh_\circ \Vert ^{p-1}}{|h_\circ |^{p-1}}&\geqslant t \left( {t^2 (1-g(t))}\right) ^{\frac{1}{2} (1-p)}\left[ \sqrt{1-g(t)}|\nabla \theta |+\sqrt{g(t)}\frac{|\nabla \rho |}{\rho }\right] \\&=t^{2-p} \left( 1-g(t))\right) ^{1-p/2}|\nabla \Theta |+c(r,R) \frac{|\nabla \rho |}{\rho }. \end{aligned} \end{aligned}$$

Now by Proposition 5.1 we have

$$\begin{aligned} \int _{\mathbb {A}} \frac{|\nabla \rho |}{|\rho |}\geqslant 2\pi \log R \end{aligned}$$

and

$$\begin{aligned}t\int _0^{2\pi }|\nabla \Theta (t e^{i\theta })|d\theta \geqslant 2\pi .\end{aligned}$$

So we have

$$\begin{aligned} \int _{\mathbb {A}(1,r)}\frac{\Vert Dh\Vert }{|h|} \cdot \frac{\Vert Dh_\circ \Vert ^{p-1}}{|h_\circ |^{p-1} }\geqslant 2\pi \left( c \log R +\int _1^r t^{2-p} (1-g(t))^{1-p/2}dt\right) =\mathscr {F}_p[h_\circ ]. \end{aligned}$$

Thus

$$\begin{aligned}\mathscr {F}_p[h]\geqslant \frac{\mathscr {F}^p_p[h_\circ ]}{\mathscr {F}_p^{p-1}[h_\circ ]}=\mathscr {F}_p[h_\circ ].\end{aligned}$$

The uniqueness part of this theorem follows from Proposition 5.1. The equation in (5.4) is satisfied if and only if

$$\begin{aligned}\frac{\rho (te^{i\theta })|\nabla \Theta (te^{i\theta })|}{|\nabla \rho (te^{i\theta })|}\end{aligned}$$

is a function that depends only on t. Since \(\Theta (\theta )=e^{i\varphi (\theta )}\), we get \(|\nabla \Theta (\theta )|=\varphi '(\theta )=\textrm{const}\). Because \(\varphi :[0,2\pi ]\xrightarrow []{{}_{\!\!\text {onto\,\,}\!\!}}[\alpha , 2\pi +\alpha ]\), it follows that \(\varphi (\theta )=\theta +\alpha \). In other words h(z) is a minimizer if and only if \(h(z) = H_\circ (t)e^{i(\theta +\alpha )}=e^{i\alpha } h_\circ (z)\). This finishes the proof. \(\square \)

Proof of Theorem 2.3

The proof of Theorem 2.3 is the same as the proof of Theorem 2.2 up to the part concerning the existence of the radial solutions given in Sect. 4 (See Lemma 4.1). \(\square \)