The anisotropic polyharmonic curve flow for closed plane curves

Abstract

We study the curve diffusion flow for closed curves immersed in the Minkowski plane ${\mathcal {M}}$, which is equivalent to the Euclidean plane endowed with a closed, symmetric, convex curve called an indicatrix that scales the length of a vector in ${\mathcal {M}}$ depending on its length. The indiactrix $\partial {\mathcal {U}}$ (where ${\mathcal {U}}\subset {\mathbb {R}}^{2}$ is a convex, centrally symmetric domain) induces a second convex body, the isoperimetrix $\tilde{{\mathcal {I}}}$. This set is the unique convex set that miniminises the isoperimetric ratio (modulo homothetic rescaling) in the Minkowski plane. We prove that under the flow, closed curves that are initially close to a homothetic rescaling of the isoperimetrix in an averaged $L^{2}$ sense exists for all time and converge exponentially fast to a homothetic rescaling of the isoperimetrix that has enclosed area equal to the enclosed area of the initial immersion.

Appendix A: Appendix

Derivation of the tangent and normal vectors Let us now derive the equations (4) for the unit tangent and normal vectors associated to an immersed curve in the Minkowski setting. These allow us to develop a Minkowski analogue of the Frenet–Serret equations in Sect. 2.

Given an indicatrix as defined in the previous section, along with a curve $\Gamma$ with Euclidean tangent vector $\tau \left( {\theta }\right)$, it is most logical to define the Minkowski unit tangent vector T in the same direction as $\tau \left( {\theta }\right)$ by

$$\begin{aligned} T\left( {\theta }\right) =r\left( {\theta }\right) \tau \left( {\theta }\right) . \end{aligned}$$

The Euclidean tangent has been multiplied by the radius of the indicatrix at the corresponding angle to insure T is of Minkowski unit length. We wish to arrive at an analogue of the Frenet equations, and so wish to derive a set of equations in the form of

$$\begin{aligned} \left( { \begin{array}{c} T\\ N \end{array} }\right) _{\theta } = \left( { \begin{array}{cc} 0 &{} \alpha \left( {\theta }\right) \\ \beta \left( {\theta }\right) &{} 0 \end{array} }\right) \left( { \begin{array}{c} T\\ N \end{array} }\right) \end{aligned}$$

(58)

Presently, $\alpha ,\beta$ are unknown functions, but we do know that because T and N need to be $2\pi -$periodic that $\alpha ,\beta$ will also need to be $2\pi -$periodic. Now we already know from the Euclidean Frenet–Serret equations, (58) and from our definition of T that

$$\begin{aligned} T_\theta =r_{\theta }\tau +r\tau _{\theta }=r_{\theta }\tau +rn=\alpha N. \end{aligned}$$

Differentiating this identity yields

$$\begin{aligned} N_{\theta }=\frac{1}{r}\left[ {\left( {\frac{r_{\theta }}{\alpha }}\right) _{\theta }-\left( {\frac{r}{\alpha }}\right) }\right] T+\left[ {\left( {\frac{r_{\theta }}{\alpha }}\right) +\left( {\frac{r}{\alpha }}\right) _{\theta }}\right] n. \end{aligned}$$

We want $N_{\theta }$ to be solely in the direction of $\tau$ (and not n). Hence the last equation forces

$$\begin{aligned} 0=\left( {\frac{r_{\theta }}{\alpha }}\right) +\left( {\frac{r}{\alpha }}\right) _{\theta }=\frac{2r_{\theta }}{\alpha }-\frac{r\alpha _{\theta }}{\alpha ^{2}}\Leftrightarrow \frac{2r_{\theta }}{r}=\frac{\alpha _{\theta }}{\alpha }. \end{aligned}$$

This last equation is equivalent to

$$\begin{aligned} 2\left( {\ln {r}}\right) _{\theta }=\left( {\ln {\alpha }}\right) _{\theta }, \end{aligned}$$

or

$$\begin{aligned} \alpha =Cr^{2} \end{aligned}$$

for some constant C. Noting that $\det \left( {T,N}\right) =C^{-1}$, we choose $C=1$ so that the Minkowski area element is identical to its Euclidean counterpart. Accordingly, the enclosed area ${\mathscr {A}}$ of a closed curve $\Gamma :{\mathbb {S}}^{1}\rightarrow {\mathcal {M}}^{2}$ is simply equal to

$$\begin{aligned} {\mathscr {A}}\left( {\Gamma }\right) =-\frac{1}{2}\int _{\Gamma }{{(\Gamma ,n)}\,ds}. \end{aligned}$$

Note that measure in the integral ds could have been swapped for $d\sigma$. We arrive at the following expression for the Minkowski tangent and normal vectors T and N:

$$\begin{aligned} \left( { \begin{array}{c} T\\ N \end{array} }\right) = \left( { \begin{array}{cc} r &{} 0\\ -h_{\theta } &{} h \end{array} }\right) \left( { \begin{array}{c} \tau \\ n \end{array}}\right) \end{aligned}$$

$\square$

Proposition A.1

Let $\Gamma :{\mathbb {S}}^{1}\rightarrow {\mathcal {M}}^{2}$ be a simple closed immersion in the Minkowski plane ${\mathcal {M}}$ with associated indicatrix $\partial {\mathcal {U}}$ and isoperimetrix $\tilde{{\mathcal {I}}}$. Then

$$\begin{aligned} \int _{\Gamma }{{\kappa }\,d\sigma }=2\,{\mathscr {A}}(\tilde{{\mathcal {I}}}), \end{aligned}$$

where ${\mathscr {A}}(\tilde{{\mathcal {I}}})$ denotes the enclosed area of the isoperimetrix.

Proof

Using the identities $d\sigma =h\,ds$ and $d\theta =k\,ds$, we have

$$\begin{aligned} \int _{\Gamma }{{\kappa }\,d\sigma }=\int _{\Gamma }{{k(h+h_{\theta \theta })}\,d\sigma }=\int _{0}^{2\pi }{h(h+h_{\theta \theta })\,d\theta }. \end{aligned}$$

Next, using the notation $\tau =(\cos \theta ,\sin \theta ),\,n=(-\sin \theta ,\cos \theta )$, we can write the isoperimetrix ${\tilde{I}}(\theta )$ as

$$\begin{aligned} \tilde{{\mathcal {I}}}(\theta )=\left\{ -h_{\theta }\,\tau +h\,n\,:\,\theta \in [0,2\pi )\right\} . \end{aligned}$$

A quick calculation gives $\tilde{{\mathcal {I}}}_{\theta }=-(h+h_{\theta \theta })\,\tau$, which imples the induced Euclidean arc length and normal to $\tilde{{\mathcal {I}}}$ are given by

$$\begin{aligned} d{\tilde{s}}=\sqrt{h+h_{\theta \theta }}\,d\theta \,\,\text {and}\,\,{\tilde{n}}=-\sqrt{h+h_{\theta \theta }}\,n, \end{aligned}$$

respectively. This implies that the signed enclosed area of $\tilde{{\mathcal {I}}}$ is given by

$$\begin{aligned} {\mathcal {A}}(\tilde{{\mathcal {I}}})=-\frac{1}{2}\int _{\Gamma }{{(\tilde{{\mathcal {I}}},{\tilde{n}})\,d{\tilde{s}}}\,d\sigma }=\frac{1}{2}\int _{0}^{2\pi }{h(h+h_{\theta \theta })\,d\theta }. \end{aligned}$$

Comparing to

$$\begin{aligned} \int _{\Gamma }{{\kappa }\,d\sigma }=\int _{\Gamma }{{k(h+h_{\theta \theta })}\,d\sigma }=\int _{0}^{2\pi }{h(h+h_{\theta \theta })\,d\theta }, \end{aligned}$$

the results of the proposition then follow. $\square$

Lemma A.2

Let $\Gamma :{\mathbb {S}}^{1}\rightarrow {\mathcal {M}}^{2}$ be a smooth closed curve with Minkowski curvature $\kappa$ and Minkowski arc length element $d\sigma$. Then for any $m\in {\mathbb {N}}$ we have

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma ^{m}}^{2}}\,d\sigma }\le \varepsilon {\mathscr {L}}^{2}\int _{\Gamma }{{\kappa _{\sigma ^{m+1}}^{2}}\,d\sigma }+\frac{1}{4\varepsilon ^{m}}{\mathscr {L}}^{-\left( {2m+1}\right) }K_{osc}, \end{aligned}$$

for any $\varepsilon >0$.

Proof

We will prove the lemma inductively. The case $m=1$ can be checked quite easily, by applying integration by parts and the Cauchy-Schwarz inequality:

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma }^{2}}\,d\sigma }&=\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) _{\sigma }^{2}}\,d\sigma }\\&=-\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) \left( {\kappa -{\bar{\kappa }}}\right) _{\sigma ^{2}}}\,d\sigma } \\&\quad \le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{2}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{2}}^{2}}\,d\sigma }}\right) ^{\frac{1}{2}} \\&\quad \le \varepsilon {\mathscr {L}}^{2}\int _{\Gamma }{{\kappa _{\sigma ^{2}}^{2}}\,d\sigma }+\frac{1}{4\varepsilon ^{1}}{\mathscr {L}}^{-2}\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }. \end{aligned}$$

Next assume inductively that the statement is true for $j=m$. That is, assume that

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma ^{j}}^{2}}\,d\sigma }\le \varepsilon {\mathscr {L}}^{2}\int _{\Gamma }{{\kappa _{\sigma ^{j+1}}^{2}}\,d\sigma }+\frac{1}{4\varepsilon ^{j}}{\mathscr {L}}^{-\left( {2j+1}\right) }K_{osc} \end{aligned}$$

(59)

for any $\varepsilon >0$. Again performing integration by parts and the Cauchy-Schwarz inequality, we have for any $\varepsilon >0$:

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma ^{j+1}}^{2}}\,d\sigma }&=-\int _{\Gamma }{{\kappa _{\sigma ^{j}}\cdot \kappa _{\sigma ^{j+2}}}\,d\sigma }\nonumber \\&\quad \le \left( {\int _{\Gamma }{{\kappa _{\sigma ^{j}}^{2}}\,d\sigma }}\right) ^{\frac{1}{2}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^j+2}^{2}}\,d\sigma }}\right) ^{\frac{1}{2}}\nonumber \\&\quad \le \frac{\varepsilon }{2} {\mathscr {L}}^{2}\int _{\Gamma }{{\kappa _{\sigma ^{j+2}}^{2}}\,d\sigma }+\frac{1}{2\varepsilon }{\mathscr {L}}^{-2}\int _{\Gamma }{{\kappa _{\sigma ^{j}}^{2}}\,d\sigma }. \end{aligned}$$

(60)

Substituting the inductive assumption (59) into (60) then gives

$$\begin{aligned}&\int _{\Gamma }{{\kappa _{\sigma ^{j+1}}^{2}}\,d\sigma }\\&\quad \le \frac{\varepsilon }{2} {\mathscr {L}}^{2}\int _{\Gamma }{{\kappa _{\sigma ^{j+2}}^{2}}\,d\sigma }+\frac{1}{2\varepsilon }{\mathscr {L}}^{-2}\left( {\varepsilon {\mathscr {L}}^{2}\int _{\Gamma }{{\kappa _{\sigma ^{j+1}}^{2}}\,d\sigma }+\frac{1}{4\varepsilon ^{j}}\left( {\varepsilon }\right) {\mathscr {L}}^{-\left( {2j+1}\right) }K_{osc}}\right) , \end{aligned}$$

meaning that

$$\begin{aligned} \frac{1}{2}\int _{\Gamma }{{\kappa _{\sigma ^{j+1}}^{2}}\,d\sigma }\le \frac{\varepsilon }{2} {\mathscr {L}}^{2}\int _{\Gamma }{{\kappa _{\sigma ^{j+2}}^{2}}\,d\sigma }+\frac{1}{2}\cdot \frac{1}{4\varepsilon ^{j+1}}{\mathscr {L}}^{-\left( {2\left( {j+1}\right) +1}\right) }K_{osc}. \end{aligned}$$

Multiplying out by 2 then gives us the inductive step, completing the lemma. $\square$

Lemma A.3

Let $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be an absolutely continuous and periodic function of period P. Then, if $\int _{0}^{P}{f\,dx}=0$ we have

$$\begin{aligned} \int _{0}^{P}{f^{2}\,dx}\le \frac{P^{2}}{4\pi ^{2}}\int _{0}^{P}{f_{x}^{2}\,dx}, \end{aligned}$$

with equality if and only if

$$\begin{aligned} f\left( {x}\right) =A\cos \left( {\frac{2\pi }{P}x}\right) +B\sin \left( {\frac{2\pi }{P}x}\right) \end{aligned}$$

for some constants A, B.

Proof

The proof is relatively straightforward, and follows from finding a function f that extremises the integral $\int _{0}^{P}{f^{2}\,dx}$, given a constrained value of $\int _{0}^{P}{f_{x}^{2}\,dx}$, using the calculus of variations. $\square$

Lemma A.4

Let $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be an absolutely continuous and periodic function of period P. Then, if $\int _{0}^{P}{f\,dx}=0$ we have

$$\begin{aligned} \left\| {f}\right\| _{\infty }^{2}\le \frac{P}{2\pi }\int _{0}^{P}{f_{x}^{2}\,dx}. \end{aligned}$$

Proof

Since $\int _{0}^{P}{f\,dx}=0$ and f is $P-$periodic we conclude that there exists distinct $0\le p<q<P$ such that

$$\begin{aligned} f\left( {p}\right) =f\left( {q}\right) =0. \end{aligned}$$

Next, the fundamental theorem of calculus tells us that for any $x\in \left( {0,P}\right)$,

$$\begin{aligned} \frac{1}{2}\left[ {f\left( {x}\right) }\right] ^{2}=\int _{p}^{x}{ff_{u}\,du}=\int _{q}^{x}{ff_{u}\,du}. \end{aligned}$$

Hence

$$\begin{aligned} \left( {f\left( {x}\right) }\right) ^{2}&=\int _{p}^{u}{ff_{u}\,du}-\int _{x}^{q}{ff_{u}\,du}\le \int _{p}^{q}{\left| {ff_{x}}\right| \,dx}\le \int _{0}^{P}{\left| {ff_{x}}\right| \,dx}\\&\quad \le \left( {\int _{0}^{P}{f^{2}\,dx}\cdot \int _{0}^{P}{f_{x}^{2}\,dx}}\right) ^{\frac{1}{2}}\le \frac{P}{2\pi }\int _{0}^{P}{f_{x}^{2}\,dx}, \end{aligned}$$

where the last step follows from Lemma A.3. We have also utilised Hölder’s inequality with $p=q=2$. $\square$

Lemma A.5

(Wirtinger’s inequality [8]) Let $f:{\mathbb {R}}\rightarrow {\mathbb {R}}$ be an absolutely continuous and periodic function of period P. If there exists a point $p\in \left[ {0,P}\right]$ with $f(p)=0$, then

$$\begin{aligned} \int _{0}^{P}{f^{2}\,ds}\le \left( {\frac{P}{\pi }}\right) ^{2}\int _{0}^{P}{f_{x}^{2}\,dx}. \end{aligned}$$

Lemma A.6

(Dziuk et al. [9], Lemma 2.4) Let $\Gamma :{\mathbb {S}}^{1}\rightarrow {\mathbb {R}}^{2}$ be a smooth closed curve. Let $\phi :{\mathbb {S}}^{1}\rightarrow {\mathbb {R}}$ be a sufficiently smooth function. Then for any $l\ge 2,K\in {\mathbb {N}}$ and $0\le i<K$ we have

$$\begin{aligned} {\mathscr {L}}^{i+1-\frac{1}{l}}\left( {\int _{\Gamma }{{\left( {\phi }\right) _{\sigma ^{i}}^{2}}\,d\sigma }}\right) ^{\frac{1}{l}}\le c\left( {K}\right) {\mathscr {L}}^{\frac{1-\alpha }{2}}\left( {\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) ^{\frac{1-\alpha }{2}}\left\| {\phi }\right\| _{K,2}^{\alpha }. \end{aligned}$$

Here $\alpha =\frac{i+\frac{1}{2}-\frac{1}{l}}{K}$, and

$$\begin{aligned} \left\| {\phi }\right\| _{K,2}:=\sum _{j=0}^{K}{\mathscr {L}}^{j+\frac{1}{2}}\left( {\int _{\Gamma }{{\left( {\phi }\right) _{\sigma ^{j}}^{2}}\,d\sigma }}\right) ^{\frac{1}{2}}. \end{aligned}$$

In particular, if $\phi =\kappa -{\bar{\phi }}$, then

$$\begin{aligned} {\mathscr {L}}^{i+1-\frac{1}{l}}\left( {\int _{\Gamma }{{\left( {k-{\bar{k}}}\right) _{\sigma ^{i}}^{2}}\,d\sigma }}\right) ^{\frac{1}{l}}\le c\left( {K}\right) \left( {K_{osc}}\right) ^{\frac{1-\alpha }{2}}\left\| {k-{\bar{k}}}\right\| _{K,2}^{\alpha }. \end{aligned}$$

Proof

The proof is identical to that of Lemma 2.4 from [9] and is of a standard interpolative nature. Note that although we use $k-{\bar{k}}$ in the identity (as opposed to Kuwert et all who use k). $\square$

Lemma A.7

(Dziuk et al. [9], Proposition 2.5) Let $\Gamma :{\mathbb {S}}^{1}\rightarrow {\mathbb {R}}^{2}$ be a smooth closed curve. Let $\phi :{\mathbb {S}}^{1}\rightarrow {\mathbb {R}}$ be a sufficiently smooth function. Then for any term $P_{\nu }^{\mu }\left( {\phi }\right)$ (where $P_{\nu }^{\mu }\left( {\cdot }\right)$ denotes the same P-style notation introduced in Sect. 1) with $\nu \ge 2$ which contains only derivatives of $\kappa$ of order at most $K-1$, we have

$$\begin{aligned} \int _{\Gamma }{{\left| {P_{\nu }^{\mu }\left( {\phi }\right) }\right| }\,d\sigma }\le c\left( {K,\mu ,\nu }\right) {\mathscr {L}}^{1-\mu -\nu }\left( {{\mathscr {L}}\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) ^{\frac{\nu -\eta }{2}}\left\| {\phi }\right\| _{K,2}^{\eta }. \end{aligned}$$

(61)

In particular, for $\phi =\kappa -{\bar{\kappa }}$ we have the estimate

$$\begin{aligned} \int _{\Gamma }{{\left| {P_{\nu }^{\mu }\left( {\kappa -{\bar{\kappa }}}\right) }\right| }\,d\sigma }\le c\left( {K,\mu ,\nu }\right) {\mathscr {L}}^{1-\mu -\nu }\left( {K_{osc}}\right) ^{\frac{\nu -\eta }{2}}\left\| {\kappa -{\bar{\kappa }}}\right\| _{K,2}^{\eta } \end{aligned}$$

(62)

where $\eta =\frac{\mu +\frac{\nu }{2}-1}{K}$.

Proof

Using Hölder’s inequality and Lemma A.6 with $K=\nu$, if $\sum _{j=1}^{\nu }i_{j}=\mu$ we have

$$\begin{aligned} \int _{\Gamma }{{\left| {\phi _{\sigma ^{i_{1}}}*\cdots *\phi _{\sigma ^{i_{\nu }}}}\right| }\,d\sigma }&\le \prod _{j=1}^{\nu }\left( {\int _{\Gamma }{{\phi _{\sigma ^{i_{j}}}^{\nu }}\,d\sigma }}\right) ^{\frac{1}{\nu }}\nonumber \\&={\mathscr {L}}^{1-\mu -\nu }\prod _{j=1}^{\nu }{\mathscr {L}}^{i_{j}+1-\frac{1}{\nu }}\left( {\int _{\Gamma }{{\phi _{\sigma ^{i_{j}}}^{\nu }}\,d\sigma }}\right) ^{\frac{1}{\nu }}\nonumber \\&\le c\left( {K,\mu ,\nu }\right) {\mathscr {L}}^{1-\mu -\nu }\prod _{j=1}^{\nu }\left( {{\mathscr {L}}\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) ^{\frac{1-\alpha _{j}}{2}}\left\| {\phi }\right\| _{K,2}^{\alpha _{j}} \end{aligned}$$

(63)

where $\alpha _{j}=\frac{i_{j}+\frac{1}{2}-\frac{1}{\nu }}{K}$. Now

$$\begin{aligned} \sum _{j=1}^{\nu }\alpha _{j}=\frac{1}{K}\sum _{j=1}^{\nu }\left( {i_{j}+\frac{1}{2}-\frac{1}{\nu }}\right) =\frac{\mu +\frac{\nu }{2}-1}{K}=\eta , \end{aligned}$$

and so substituting this into (63) gives the first inequality of the lemma. It is then a simple matter of substituting $\phi =\kappa -{\bar{\kappa }}$ into this result to prove statement (67). $\square$

Lemma A.8

(Dziuk et al. [9]) Let $\Gamma :{\mathbb {S}}^{1}\rightarrow {\mathbb {R}}^{2}$ be a smooth closed curve and $\phi :{\mathbb {S}}^{1}\rightarrow {\mathbb {R}}$ a sufficiently smooth function. Then for any term $P_{\nu }^{\mu }\left( {\phi }\right)$ with $\nu \ge 2$ which contains only derivatives of $\kappa$ of order at most $K-1$, we have for any $\varepsilon >0$

$$\begin{aligned}&\int _{\Gamma }{{\left| {P_{\nu }^{\mu ,K-1}\left( {\phi }\right) }\right| }\,d\sigma }\nonumber \\&\quad \le c\left( {K,\mu ,\nu }\right) {\mathscr {L}}^{1-\mu -\nu }\left( {{\mathscr {L}}\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) ^{\frac{\nu -\eta }{2}}\left( {{\mathscr {L}}^{2K+1}\int _{\Gamma }{{\phi _{\sigma ^{K}}^{2}}\,d\sigma }+{\mathscr {L}}\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) ^{\frac{\eta }{2}}.\nonumber \\ \end{aligned}$$

(64)

Moreover if $\mu +\frac{1}{2}\nu <2K+1$ then $\eta <2$ and we have for any $\varepsilon >0$

$$\begin{aligned} \int _{\Gamma }{{\left| {P_{\nu }^{\mu ,K-1}\left( {\phi }\right) }\right| }\,d\sigma }\le \varepsilon \int _{\Gamma }{{\phi _{\sigma ^{K}}^{2}}\,d\sigma }+c\cdot \varepsilon ^{-\frac{\eta }{2-\eta }}\left( {\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) ^{\frac{\nu -\eta }{2-\eta }}+c\left( {\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) ^{\mu +\nu -1}. \end{aligned}$$

(65)

In particular, for $\phi =\kappa -{\bar{\kappa }}$, we have the estimate

$$\begin{aligned} \int _{\Gamma }{{\left| {P_{\nu }^{\mu }\left( {\kappa -{\bar{\kappa }}}\right) }\right| }\,d\sigma }\le c\left( {K,\mu ,\nu }\right) {\mathscr {L}}^{1-\mu -\nu }\left( {{\mathscr {K}}_{osc}}\right) ^{\frac{\nu -\eta }{2}}\left( {{\mathscr {L}}^{2K+1}\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) _{\sigma ^{K}}^{2}}\,d\sigma }}\right) ^{\frac{\eta }{2}}. \end{aligned}$$

Here, as before, $\eta =\frac{\mu +\frac{\nu }{2}-1}{K}$.

Proof

Combining the previous lemma with the following standard interpolation inequality from that follows from repeated applications of Lemma A.2 (and is also found in [4])

$$\begin{aligned} \left\| {\phi }\right\| _{K,2}^{2}\le c\left( {K}\right) \left( {{\mathscr {L}}^{2K+1}\int _{\Gamma }{{\phi _{\sigma ^{K}}^{2}}\,d\sigma }+{\mathscr {L}}\int _{\Gamma }{{\phi ^{2}}\,d\sigma }}\right) \end{aligned}$$

yields the identity (64) immediately. To prove (65) we simply combine (64) with the Cauchy-Schwarz identity. The final identity of the Lemma follow by letting $\phi =\kappa -{\bar{\kappa }}$ in (64) and combining this with the identity

$$\begin{aligned} K_{osc}\le {\mathscr {L}}\left( {\frac{{\mathscr {L}}^{2}}{4\pi ^{2}}}\right) ^{K}\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) _{\sigma ^{K}}^{2}}\,d\sigma }= c\left( {K}\right) {\mathscr {L}}^{2K+1}\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) _{\sigma ^{K}}^{2}}\,d\sigma }, \end{aligned}$$

which is a direct consequence of applying Lemma A.3$\left( {p+1}\right)$ times repeatedly. $\square$

Lemma A.9

Let $\Gamma :{\mathbb {S}}^{1}\rightarrow {\mathcal {M}}^{2}$ be a closed Minkowski curve with Minkowski curvature $\kappa$. Then for any $m,L\in {\mathbb {N}}$ with $m<L$ we have the estimate

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma ^{m}}^{2}}\,d\sigma }\le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{1-\frac{m}{L}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{L}}^{2}}\,d\sigma }}\right) ^{\frac{m}{L}}. \end{aligned}$$

Proof

First we prove the intermediate identity

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma ^{m}}^{2}}\,d\sigma }\le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{m+1}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{m+1}}^{2}}\,d\sigma }}\right) ^{\frac{m}{m+1}},\,\,m\in {\mathbb {N}}. \end{aligned}$$

(66)

For $m=1$ we simply use integration by parts and Hölder’s inequality:

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma }^{2}}\,d\sigma }&=-\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) \kappa _{\sigma \sigma }}\,d\sigma }\le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{2}}\left( {\int _{\Gamma }{{\kappa _{\sigma \sigma }^{2}}\,d\sigma }}\right) ^{\frac{1}{2}}\\&=\left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{1+1}}\left( {\int _{\Gamma }{{\kappa _{\sigma \sigma }^{2}}\,d\sigma }}\right) ^{\frac{1}{1+1}}. \end{aligned}$$

Next we inductively assume that (66) is true for $m=1,2,\dots ,M$. Using integration by parts and Hölder’s inequality once more gives

$$\begin{aligned}&\int _{\Gamma }{{\kappa _{\sigma ^{M+1}}^{2}}\,d\sigma }=-\int _{\Gamma }{{\kappa _{\sigma ^{M}}\kappa _{\sigma ^{M+1}}}\,d\sigma }\le \left( {\int _{\Gamma }{{\kappa _{\sigma ^{M}}^{2}}\,d\sigma }}\right) ^{\frac{1}{2}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{M+1}}^{2}}\,d\sigma }}\right) ^{\frac{1}{2}} \\&\quad \le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{2\left( {M+1}\right) }}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{M+1}}^{2}}\,d\sigma }}\right) ^{\frac{M}{2\left( {M+1}\right) }}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{M+2}}^{2}}\,d\sigma }}\right) ^{\frac{1}{2}} \\&\quad \le \frac{M}{2\left( {M+1}\right) }\int _{\Gamma }{{\kappa _{\sigma ^{M+1}}^{2}}\,d\sigma }+\frac{M+2}{2\left( {M+1}\right) }\left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{M+2}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{M+2}}^{2}}\,d\sigma }}\right) ^{\frac{M+1}{M+2}}. \end{aligned}$$

Absorbing gives the statement (66) for $M+1$, completing the induction. Here we have used our inductive assumption in the second line, and Hölder’s inequality with $p=\frac{2\left( {M+1}\right) }{M},q=\frac{2\left( {M+1}\right) }{M+2}$ in the last line. To arrive at the claim of the lemma, we simply employ the inequality (66) repeatedly:

$$\begin{aligned} \int _{\Gamma }{{\kappa _{\sigma ^{m}}^{2}}\,d\sigma }&\le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{m+1}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{m+1}}^{2}}\,d\sigma }}\right) ^{\frac{m}{m+1}}\nonumber \\&\le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{m+1}}\left( {\left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{1}{m+2}}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{m+2}}^{2}}\,d\sigma }}\right) ^{\frac{m+1}{m+2}}}\right) ^{\frac{m}{m+1}}\nonumber \\&=\left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{\frac{m}{m\left( {m+1}\right) }+\frac{m}{\left( {m+1}\right) \left( {m+2}\right) }}\left( {\int _{\Gamma }{{\kappa _{\sigma ^{m+2}}^{2}}\,d\sigma }}\right) ^{\frac{m}{m+2}}\nonumber \\&\vdots \nonumber \\&\le \left( {\int _{\Gamma }{{\left( {\kappa -{\bar{\kappa }}}\right) ^{2}}\,d\sigma }}\right) ^{m\sum _{i=0}^{L-m-1}\left\{ \frac{1}{\left( {m+i}\right) \left( {m+i+1}\right) }\right\} }\left( {\int _{\Gamma }{{\kappa _{\sigma ^{L}}^{2}}\,d\sigma }}\right) ^{\frac{m}{L}}. \end{aligned}$$

(67)

Now $\sum _{i=0}^{L-m-1}\left\{ \frac{1}{\left( {m+i}\right) \left( {m+i+1}\right) }\right\}$ is a telescoping series, which sums to $1/m-1/L$. Substituting this value into (67) then proves the lemma. $\square$

Theorem A.10

([5], Theorem 1.1) Let $q\in {\mathbb {R}}^{n}$, $m,p\in {\mathbb {N}}$ with $p>m$. Additionally, let ${\mathcal {A}},{\mathcal {V}}>0$ be some fixed constants. Let ${\mathfrak {T}}$ be the set of all mappings $f:\Sigma :\rightarrow {\mathbb {R}}^{n}$ with the following properties:

• $\Sigma$ is an m-dimensional, compact manifold (without boundary)

• f is an immersion in $W^{2,p}\left( {\Sigma ,{\mathbb {R}}^{n}}\right)$ satisfying

  $$\begin{aligned} \left\| {A\left( {f}\right) }\right\| _{p}&\le {\mathcal {A}},\\ \text {vol}\left( {\Sigma }\right)&\le {\mathcal {V}},\text { and }\\ q&\in f\left( {\Sigma }\right) . \end{aligned}$$

Then for every sequence $f^{i}:\Sigma ^{i}\rightarrow {\mathbb {R}}^{n}$ in ${\mathfrak {T}}$ there is a subsequence $f^{j}$, a mapping $f:\Sigma \rightarrow {\mathbb {R}}^{n}$ in ${\mathfrak {T}}$ and a sequence of diffeomorphisms $\phi ^{j}:\Sigma \rightarrow \Sigma ^{j}$ such that $f^{j}\circ \phi ^{j}$ converges in the $C^{1}$-topology to f.