On the Curvature Energy of Cartesian Surfaces

Abstract

We analyze the lower semicontinuous envelope of the curvature functional of Cartesian surfaces in codimension one. To this aim, following the approach by Anzellotti–Serapioni–Tamanini, we study the class of currents that naturally arise as weak limits of Gauss graphs of smooth functions. The curvature measures are then studied in the non-parametric case. Concerning homogeneous functions, some model examples are studied in detail. Finally, a new gap phenomenon is observed.

Appendices

Appendix A: Homogeneous Functions

Assume $u:B^2\rightarrow {\mathbb {R}}$ satisfies $u(x)=u(x/|x|)$ for each $x\in B^2\setminus \{0_{{\mathbb {R}}^2}\}$, whence we can write $u(x)=f(\theta )$ for some function $f:[0,2\pi ]\rightarrow {\mathbb {R}}$, where $x=(\rho \,\cos \theta ,\rho \,\sin \theta )$. Denoting for simplicity $s:=\sin \theta$ and $c:=\cos \theta$, we formally compute on $B^2\setminus \{0_{{\mathbb {R}}^2}\}$

$$\begin{aligned} \nabla u=\Bigl (-{{\dot{f}} s\over \rho },{{\dot{f}} c\over \rho } \Bigr ),\quad {{{\nu }}_u}={1\over (\rho ^2+{\dot{f}} ^2)^{1/2}}\,({\dot{f}} s,-{\dot{f}} c,\rho ) \end{aligned}$$

and hence, setting $F:=\rho ^2+{\dot{f}} ^2$, we get

$$\begin{aligned} \begin{array}{l} \displaystyle \partial _1{{{\nu }}_u}^1={-1\over \rho \,F^{3/2}}\bigl [ \rho ^2(2{\dot{f}} s\,c+\ddot{f} s^2)+{\dot{f}} ^3s\,c \bigr ] \\ \displaystyle \partial _2{{{\nu }}_u}^1={1\over \rho \,F^{3/2}}\bigl [ \rho ^2({\dot{f}} (c^2-s^2)+\ddot{f} s\,c)+{\dot{f}} ^3 c^2 \bigr ] \\ \displaystyle \partial _1{{{\nu }}_u}^2={1\over \rho \,F^{3/2}}\bigl [ \rho ^2({\dot{f}} (c^2-s^2)+\ddot{f} s\,c)-{\dot{f}} ^3 s^2 \bigr ] \\ \displaystyle \partial _2{{{\nu }}_u}^2={1\over \rho \,F^{3/2}}\bigl [ \rho ^2(2{\dot{f}} s\,c-\ddot{f} c^2)+{\dot{f}} ^3s\,c \bigr ] \\ \displaystyle \partial _1{{{\nu }}_u}^3={1\over F^{3/2}}\, {\dot{f}} ({\dot{f}} c+\ddot{f} s) \\ \displaystyle \partial _2{{{\nu }}_u}^3={1\over F^{3/2}}\, {\dot{f}} ({\dot{f}} s-\ddot{f} c). \end{array} \end{aligned}$$

This yields to

$$\begin{aligned} \begin{array}{rl} |\nabla {{{\nu }}_u}^1|^2= &{} \displaystyle {1\over \rho ^2F^3}\,\bigl [\rho ^4({\dot{f}} ^2+\ddot{f} ^2s^2+2s\,c\,{\dot{f}} \ddot{f} )+2\rho ^2{\dot{f}} ^3({\dot{f}} c^2+\ddot{f} s\,c)+{\dot{f}} ^6c^2 \bigr ]\\ |\nabla {{{\nu }}_u}^2|^2= &{} \displaystyle {1\over \rho ^2F^3}\,\bigl [\rho ^4({\dot{f}} ^2+\ddot{f} ^2c^2-2s\,c\,{\dot{f}} \ddot{f} )+2\rho ^2{\dot{f}} ^3({\dot{f}} s^2-\ddot{f} s\,c)+{\dot{f}} ^6s^2 \bigr ] \\ |\nabla {{{\nu }}_u}^3|^2= &{} \displaystyle {1\over F^3}\,{\dot{f}} ^2({\dot{f}} ^2+\ddot{f} ^2), \end{array} \end{aligned}$$

and hence

$$\begin{aligned} |\nabla {{{\nu }}_u}|^2= \displaystyle {1\over \rho ^2(\rho ^2+{\dot{f}} ^2)^3}\,\bigl [\rho ^4(\ddot{f} ^2+2{\dot{f}} ^2)+\rho ^2(3{\dot{f}} ^4+{\dot{f}} ^2\ddot{f} ^2)+{\dot{f}} ^6\bigr ]. \end{aligned}$$

Setting moreover for simplicity $A_j:=\left| \begin{array}{cc} \partial _1u &{} \partial _2u \\ \partial _1{{{\nu }}_u}^j &{} \partial _2{{{\nu }}_u}^j \end{array}\right|$, we also compute

$$\begin{aligned} A_1= {{\dot{f}} ^2 s\over F^{3/2}},\quad A_2= -{{\dot{f}} ^2 c\over F^{3/2}},\quad A_3=- {{\dot{f}} ^3 \over \rho \,F^{3/2}} \end{aligned}$$

so that

$$\begin{aligned} {A_1^2+A_2^2+A_3^2}={{\dot{f}} ^4 \over \rho ^2\,(\rho ^2+{\dot{f}} ^2)^2}. \end{aligned}$$

Also, setting $B_{jk}:=\left| \begin{array}{cc} \partial _1{{{\nu }}_u}^j &{} \partial _2{{{\nu }}_u}^j \\ \partial _1{{{\nu }}_u}^k &{} \partial _2{{{\nu }}_u}^k \end{array}\right|$, we compute

$$\begin{aligned} B_{12}=-{{\dot{f}} ^2 \over F^2},\quad B_{13}=-{{\dot{f}} ^3\,c\over \rho \,F^2},\quad \quad B_{23}=-{{\dot{f}} ^3\,s\over \rho \,F^2}, \end{aligned}$$

so that

$$\begin{aligned} {B_{12}^2+B_{13}^2+B_{23}^2}={{\dot{f}} ^4 \over \rho ^2\,(\rho ^2+{\dot{f}} ^2)^{3}}. \end{aligned}$$

On account of (2.15) and (2.18), we, thus, obtain

$$\begin{aligned} \begin{array}{rl} |\xi _u^{(0)}|^2= &{} g_u = \displaystyle 1+|\nabla u|^2={\rho ^2+{\dot{f}} ^2 \over \rho ^2}\\ |\xi _u^{(1)}|^2= &{} \displaystyle |\nabla {{{\nu }}_u}|^2+A_1^2+A_2^2+A_3^2= \displaystyle {1\over \rho ^2(\rho ^2+{\dot{f}} ^2)^2}\,\bigl [\rho ^2\ddot{f} ^2+2\,(\rho ^2+{\dot{f}} ^2){\dot{f}} ^2\bigr ] \\ |\xi _u^{(2)}|^2= &{} \displaystyle {B_{12}^2+B_{13}^2+B_{23}^2}={{\dot{f}} ^4 \over \rho ^2\,(\rho ^2+{\dot{f}} ^2)^{3}}. \end{array} \end{aligned}$$

Now, by the formulas (2.11) and (2.12) we infer the explicit expressions of the Gauss and mean curvatures

$$\begin{aligned} \mathbf{K}_u=-{{\dot{f}} ^2 \over (\rho ^2+{\dot{f}} ^2)^2},\qquad \mathbf{H}_u={1\over 2}\,{\rho \,\ddot{f} \over (\rho ^2+{\dot{f}} ^2)^{3/2}}, \end{aligned}$$

and hence, we readily recover the expressions of the three terms $|\xi _u^{(i)}|$ given by the formulas (2.18). In particular, we get

$$\begin{aligned} \sqrt{g_u}\,\mathbf{K}_u = -{{\dot{f}} ^2 \over \rho \,(\rho ^2+{\dot{f}} ^2)^{3/2}},\qquad \sqrt{g_u}\,\mathbf{H}_u={1\over 2}\,{\ddot{f} \over \rho ^2+{\dot{f}} ^2}. \end{aligned}$$

We also compute

$$\begin{aligned} \begin{array}{rl} |\xi _u|^2= |\xi _u^{(0)}|^2+|\xi _u^{(1)}|^2+|\xi _u^{(2)}|^2= &{}\displaystyle {g\over (\rho ^2+{\dot{f}} ^2)^4}\,\bigl [ (\rho ^2+2{\dot{f}} ^2)^2+\rho ^2(\rho ^2+{\dot{f}} ^2)\,\ddot{f} ^2\bigr ] \\ = &{}\displaystyle {1\over \rho ^2(\rho ^2+{\dot{f}} ^2)^3}\,\bigl [ (\rho ^2+2{\dot{f}} ^2)^2+\rho ^2(\rho ^2+{\dot{f}} ^2)\,\ddot{f} ^2\bigr ]. \end{array} \end{aligned}$$

Therefore, for f smooth, on account of (2.8) by the area formula we obtain

$$\begin{aligned} \mathcal{H}^2(\mathcal{G}\mathcal{G}_u)= \displaystyle \int _{B^2}|\xi _u|\,d\mathcal{L}^2 = \displaystyle \int _0^1\int _0^{2\pi }{[ (\rho ^2+2{\dot{f}} ^2)^2+\rho ^2(\rho ^2+{\dot{f}} ^2)\,\ddot{f} ^2 ]^{1/2} \over (\rho ^2+{\dot{f}} ^2)^{3/2}}\,d\theta \,d\rho . \end{aligned}$$

Remark A.1

Notice that $|\xi _u|\in L^1(B^2)$ if and only if $|\xi _u^{(i)}|\in L^1(B^2)$ for $i=0,1,2$, and we get

$$\begin{aligned} \begin{array}{rl} \displaystyle \int _{B^2}|\xi _u^{(0)}|\,d\mathcal{L}^2 = &{}\displaystyle \int _0^1\int _0^{2\pi }\sqrt{\rho ^2+{\dot{f}} ^2}\,d\theta \,d\rho \\ \displaystyle \int _{B^2}|\xi _u^{(1)}|\,d\mathcal{L}^2 = &{} \displaystyle \int _0^1\int _0^{2\pi } {\bigl [\rho ^2\ddot{f} ^2+2\,(\rho ^2+{\dot{f}} ^2){\dot{f}} ^2\bigr ]^{1/2} \over \rho ^2+{\dot{f}} ^2} \,\,d\theta \,d\rho \\ \displaystyle \int _{B^2}|\xi _u^{(2)}|\,d\mathcal{L}^2 = &{} \displaystyle \int _0^1\int _0^{2\pi } {{\dot{f}} ^2\over (\rho ^2+{\dot{f}} ^2)^{3/2}}\,\,d\theta \,d\rho .\end{array} \end{aligned}$$

Moreover, since $|\xi _u|=\sqrt{g_u}\,\sqrt{(1-\mathbf{K}_u)^2+(2\,\mathbf{H}_u)^2}$ we have

$$\begin{aligned} {1\over \sqrt{2}}\,\bigl (\sqrt{g_u}\,|1-\mathbf{K}_u|+\sqrt{g_u}\, |2\mathbf{H}_u| \bigr )\le |\xi _u| \le \sqrt{g_u}\,|1-\mathbf{K}_u|+\sqrt{g_u}\, |2\,\mathbf{H}_u|, \end{aligned}$$

where, we recall,

$$\begin{aligned} \sqrt{g_u}={\sqrt{\rho ^2+{\dot{f}} ^2}\over \rho },\quad |1-\mathbf{K}_u| = 1+ {{\dot{f}} ^2 \over (\rho ^2+{\dot{f}} ^2)^2}, \quad |2\,\mathbf{H}_u|= {\rho \,|\ddot{f} | \over (\rho ^2+{\dot{f}} ^2)^{3/2}}, \end{aligned}$$

and hence, we deduce that $|\xi _u|\in L^1(B^2)$ if and only if

$$\begin{aligned}&\int _{B^2}{(\rho ^2+{\dot{f}} ^2)^{1/2}\over \rho }\,d\mathcal{L}^2<\infty , \,\, \int _{B^2}{1\over \rho }\,{{\dot{f}} ^2 \over ({\rho ^2+{\dot{f}} ^2})^{3/2}}\,d\mathcal{L}^2<\infty ,\,\,{\text {and}}\,\, \\&\quad \int _{B^2}{|\ddot{f} | \over \rho ^2+{\dot{f}} ^2}\,d\mathcal{L}^2<\infty . \end{aligned}$$

Appendix B: An Energy Computation

We compute the energy of the various components of the current $\widetilde{\varSigma }_u$ in (8.1), referring to Example 8.1.

The Absolutely continuous term has energy $\mathrm{E}(GG_u^a)={{\mathcal {E}}}(u)$, and it is given by the sum of the three integrals in Remark A.1, with f given by (8.4). The Cantor component $GG_u^C$ is equal to zero.

The Jump component is $GG_u^J=\varPhi ^J_\#{[\![\, I\times I\, ]\!]}$, see (8.7). Consider the 2-vector $\overrightarrow{\xi }:=\partial _s \varPhi ^J\wedge \partial _\lambda \varPhi ^J$. The mass of $GG_u^J$ agrees with the area of the Gauss graph surface $\varPhi ^J({I\times I})$. Moreover, the stratification $\overrightarrow{\xi }=\xi ^{(0)}+\xi ^{(1)}+\xi ^{(2)}$ gives $\xi ^{(2)}=0$ and

$$\begin{aligned}&\xi ^{(0)}=[u(\gamma _0)]^\pm \bigl ( \mathbf{{t}}_1\,e_1\wedge e_3+ \mathbf{{t}}_2\,e_2\wedge e_3),\qquad \\&\xi ^{(1)}=-[u(\gamma _0)]^\pm \,\sigma _u\,\mathbf{{k}}\,\bigl (\mathbf{{t}}_1\,e_3\wedge \varepsilon _1+\mathbf{{t}}_2\,e_3\wedge \varepsilon _2 \bigr ), \end{aligned}$$

where $\gamma _0(s):=(s,0)$, $\mathbf{{t}}(s)=(\mathbf{{t}}_1,\mathbf{{t}}_2)$ is the unit tangent vector to the jump set, $\sigma _u(s):=\mathrm{sgn}[u(\gamma _0(s))]^\pm$ is the sign of the Jump, and $\mathbf{{k}}(s)$ the signed curvature, whence $\mathbf{{t}}(s)\equiv (1,0)$), $\sigma _u(s)\equiv 1$, $\mathbf{{k}}(s)\equiv 0$. We, thus, have $\mathrm{E}_2(GG_u^J)=0$, whereas by the area formula, we get

$$\begin{aligned} \begin{array}{rl} \mathrm{E}_0(GG_u^J)= &{} \displaystyle \int _{I\times I}|[u(\gamma _0(s))]^\pm |\,ds\,d\lambda =|Du^J|(B^2)=1, \\ \mathrm{E}_1(GG_u^J)= &{} \displaystyle \int _{I}|[u(\gamma _0(s))]^\pm |\,\mathbf{{k}}(s)\,ds=0. \end{array} \end{aligned}$$

The Jump-edge component is $S_u^{Je}= \varPhi ^{Je}_{+\,\#}{[\![\, I\times I\, ]\!]} - \varPhi ^{Je}_{-\,\#}{[\![\, I\times I\, ]\!]}$, whence the mass decomposition

$$\begin{aligned} \mathbf{M}(S_u^{Je})= \mathbf{M}(\varPhi ^{Je}_{+\,\#}{[\![\, I\times I\, ]\!]})+\mathbf{M}(\varPhi ^{Je}_{-\,\#}{[\![\, I\times I\, ]\!]}) \end{aligned}$$

and a similar energy splitting hold. Furthermore, using the formulas after equation (8.10), we get

$$\begin{aligned} \partial _s{\varPhi ^{Je}_\pm }&= {\dot{\gamma }}^1_0\,e_1+{\dot{\gamma }}^2_0\,e_2+ \nabla u^\pm (\gamma _0)\bullet {\dot{\gamma }}_0\,e_3+ \sum _{j=1}^3\partial _s{(\phi ^{Je}_\pm )}^j\,\varepsilon _j,\quad \\&\quad \partial _\lambda {\varPhi ^{Je}_\pm } = \sum _{j=1}^3\partial _\lambda {(\phi ^{Je}_\pm )}^j\,\varepsilon _j, \end{aligned}$$

and hence, we may decompose $\overrightarrow{\xi }:= \partial _s{\varPhi ^{Je}_\pm }\wedge \partial _\lambda {\varPhi ^{Je}_\pm }=\xi ^{(0)}+\xi ^{(1)}+\xi ^{(2)}$, where this time the first stratum $\xi ^{(0)}=0$ and

$$\begin{aligned} \xi ^{(1)}=\sum _{j=1}^3\,\partial _\lambda {(\phi ^{Je}_\pm )}^j\Bigl (\sum _{i=1}^2 {\dot{\gamma }}^i_0\,e_i\wedge \varepsilon _j + \nabla u^\pm (\gamma _0)\bullet {\dot{\gamma }}_0\,e_3\wedge \varepsilon _j \Bigr ),\\ \xi ^{(2)}=\sum _{1\le j_1<j_2\le 3}\bigl (\partial _s{(\phi ^{Je}_\pm )}^{j_1} \partial _\lambda {(\phi ^{Je}_\pm )}^{j_2} -\partial _\lambda {(\phi ^{Je}_\pm )}^{j_1} \partial _s{(\phi ^{Je}_\pm )}^{j_2}\bigr )\,\varepsilon _{j_1}\wedge \varepsilon _{j_2}. \end{aligned}$$

We, thus, get

$$\begin{aligned} |\xi ^{(1)}|=\sqrt{1+(\partial _\mathbf{{t}}u^\pm (\gamma _0))^2}\,|\partial _\lambda {\phi ^{Je}_\pm }|,\quad |\xi ^{(2)}|=|J_2 \phi ^{Je}_\pm |, \end{aligned}$$

and hence,

$$\begin{aligned} \mathrm{E}(S_u^{Je})=\mathrm{E}(\varPhi ^{Je}_{+\,\#}{[\![\, I\times I\, ]\!]})+\mathrm{E}(\varPhi ^{Je}_{-\,\#}{[\![\, I\times I\, ]\!]}), \end{aligned}$$

where $\mathrm{E}_0(\varPhi ^{Je}_{\pm \,\#}{[\![\, I\times I\, ]\!]})=0$ and, again by the area formula,

$$\begin{aligned} \mathrm{E}_1(\varPhi ^{Je}_{\pm \,\#}{[\![\, I\times I\, ]\!]})= \int _{I}\sqrt{1+(\partial _\mathbf{{t}}u^\pm (\gamma _0(s)))^2}\Bigl \{\int _I|\partial _\lambda {\phi ^{Je}_\pm (s,\lambda ))}|\,d\lambda \Bigr \}\,ds ={\pi \over 2}, \end{aligned}$$

where we used that $\partial _\mathbf{{t}}u^\pm (\gamma _0(s))\equiv 0$, whereas

$$\begin{aligned} \mathrm{E}_2(\varPhi ^{Je}_{\pm \,\#}{[\![\, I\times I\, ]\!]})=\mathcal{H}^2(\phi ^{Je}_\pm (I\times I))={\pi \over 2} \end{aligned}$$

and correspondingly

$$\begin{aligned} \mathbf{M}(\varPhi ^{Je}_{\pm \,\#}{[\![\, I\times I\, ]\!]})=\int _{I\times I} \Bigl ( (1+(\partial _\mathbf{{t}}u^\pm (\gamma _0))^2)\,(\partial _\lambda {\phi ^{Je}_\pm })^2+ (J_2 \phi ^{Je}_\pm )^2\Bigr )^{1/2}\,ds\,d\lambda ={\sqrt{2}\over 2}\,\pi . \end{aligned}$$

Finally, the Edge component is $S_u^{e}= \sum _{k=1}^2 {\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]}$, see (8.13). We compute:

$$\begin{aligned} \mathbf{M}(S_u^{e})= \sum _{k=1}^2 \mathbf{M}({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]}) ,\qquad \mathrm{E}(S_u^{e})= \sum _{k=1}^2 \mathrm{E}({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]}), \end{aligned}$$

where for $k=1,2$ we have $\mathrm{E}_0({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]})=0$,

$$\begin{aligned} \mathrm{E}_1({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]})= \int _{I}\sqrt{1+(\partial _\mathbf{{t}}u(\gamma _k(s)))^2}\Bigl \{\int _I|\partial _\lambda {\phi ^{e}_k(s,\lambda ))}|\,d\lambda \Bigr \}\,ds \end{aligned}$$

with $\gamma _1(s):=(0,s)$ and $\gamma _2(s):=(0,s-1)$, and

$$\begin{aligned} \mathrm{E}_2({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]})=\mathcal{H}^2(\phi ^{e}_k(I\times I)) \end{aligned}$$

whereas concerning the mass, we get

$$\begin{aligned} \mathbf{M}({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]})=\int _{I\times I} \Bigl ( (1+(\partial _\mathbf{{t}}u(\gamma _k))^2)\,(\partial _\lambda {\phi ^{e}_k})^2+ (J_2 \phi ^{e}_k)^2\Bigr )^{1/2}\,ds\,d\lambda . \end{aligned}$$

By (8.14), we check $|\partial _\lambda {\phi ^{e}_1(s,\lambda ))}|=(\pi /2-\arctan s)$, $|\partial _\lambda {\phi ^{e}_2(s,\lambda ))}|=\pi /2-\arctan (1-s)$, and $J_2 \phi ^{e}_k\equiv 0$, whereas $\partial _\mathbf{{t}}u(\gamma _k(s))\equiv 0$. We, thus, conclude that for $k=1,2$

$$\begin{aligned} \begin{array}{c} \displaystyle \mathrm{E}_1({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]})=\int _I\Bigl ({\pi \over 2}-\arctan s \Bigr )\,ds={\pi \over 4}+{1\over 2}\,\log 2 \\ \displaystyle \mathrm{E}_2({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]})=0,\qquad \mathbf{M}({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]})=\mathrm{E}_1({\varPhi ^{e}_k}_\#{[\![\, I\times I\, ]\!]}). \end{array} \end{aligned}$$