1 Correction to: The Journal of Geometric Analysis (2019) 29:3055–3097 https://doi.org/10.1007/s12220-018-00104-z

In the sketch of the proof of Theorem 4 we claimed that a weak solution \((\xi , \kappa , \sigma )\) can be obtained by following the same lines as in the proof of [1, Proposition 3.4 and Theorem 3]. However, an appropriate argument should be slightly different due to different boundary conditions. The aim of this addendum is to fill that gap.

More precisely, in [1] at the free end one has \(\sigma ^\epsilon (t,0)=\kappa ^\epsilon (t,0)=0\) for all \(t\in [0,T]\). Therefore, the uniform (w.r.t. to \(\epsilon \)) \(L^2\) bound for \((\sigma ^\epsilon , \kappa ^\epsilon )\) follows directly from the Poincaré inequality and the uniform \(L^2\) bound for \((\partial _s\sigma ^\epsilon ,\partial _s\kappa ^\epsilon )\). In our case, the vanishing boundary conditions for \(\sigma ^\epsilon \) and \(\kappa ^\epsilon \) are no longer satisfied. We thus need some additional estimates to obtain the uniform \(L^2\) bound.

We first estimate the spatial average \(\overline{\sigma ^\epsilon }(t):=\int _{\mathbb {S}^1} \sigma ^{\epsilon }(s,t)\ \mathrm{{d}}s\). Indeed, an integration by parts and Cauchy–Schwarz yield

$$\begin{aligned}&\big |\overline{ \sigma ^\epsilon }(t)\big |=\big |\int _{\mathbb {S}^1}\kappa ^\epsilon \cdot \partial _s \xi ^\epsilon \,\mathrm{{d}}s\big | = \big |-\int _{\mathbb {S}^1} \partial _s \kappa ^\epsilon \cdot \xi ^\epsilon \,\mathrm{{d}}s \big |\le \Vert \partial _s\kappa ^\epsilon (t,\cdot )\Vert _{L^2(\mathbb {S}^1)} \Vert \xi ^\epsilon \Vert _{L^2(\mathbb {S}^1)}, \forall t\in [0,T]. \end{aligned}$$

Thus

$$\begin{aligned} \Vert \overline{\sigma ^\epsilon }\Vert _{L^2([0,T])} \le \sup _{t\in [0,T]} \Vert \xi ^{\epsilon }(t,\cdot )\Vert _{L^2(\mathbb {S}^1)} \Vert \partial _s\kappa ^\epsilon \Vert _{L^2(Q_T)}, \end{aligned}$$

where the right-hand side is uniformly bounded (cf. [1, Proposition 3.1]). Thus from Poincaré inequality \(\Vert \sigma ^\epsilon - \overline{\sigma ^\epsilon }\Vert _{L^2(Q_T)}\le C(T) \Vert \partial _s\sigma ^\epsilon \Vert _{L^2(Q_T)}\) we obtain \(\Vert \sigma ^\epsilon \Vert _{L^2(Q_T)}\le C\) with C independent of \(\epsilon \).

It remains to show that \(\kappa ^\epsilon \) is uniformly bounded in \(L^2(Q_T)\). To this end, we note that from the definition of \(\kappa ^\epsilon \) one has \(\partial _s\xi ^\epsilon =\epsilon \kappa ^\epsilon + \frac{\kappa ^\epsilon }{\sqrt{\epsilon + |\kappa ^\epsilon |^2}}\), whence

$$\begin{aligned} \sigma ^{\epsilon }= \kappa ^\epsilon \cdot \partial _s\xi ^\epsilon = |\kappa ^\epsilon | \left( \epsilon |\kappa ^\epsilon |+ \frac{|\kappa ^\epsilon |}{\sqrt{\epsilon +|\kappa ^\epsilon |^2}}\right) \ge |\kappa ^\epsilon ||\partial _s\xi ^\epsilon |. \end{aligned}$$

Observing that

$$\begin{aligned} |\partial _s\xi ^\epsilon |=\epsilon |\kappa ^\epsilon |+ \frac{|\kappa ^\epsilon |}{\sqrt{\epsilon +|\kappa ^\epsilon |^2}}\ge \epsilon + \frac{1}{\sqrt{1+\epsilon }} > 1 \end{aligned}$$

provided \(|\kappa ^\epsilon |\ge 1\), we infer that \(|\kappa ^\epsilon |\le \sigma ^\epsilon \) when \(|\kappa ^\epsilon |\ge 1\). Thus,

$$\begin{aligned} \int _{Q_T} |\kappa ^\epsilon |^2 \, \mathrm{{d}}s \mathrm{{d}}t&= \int _{\{(s,t)\in Q_T: |\kappa ^\epsilon |< 1\}} |\kappa ^\epsilon |^2 \, \mathrm{{d}}s\mathrm{{d}}t + \int _{\{(s,t)\in Q_T: |\kappa ^\epsilon |\ge 1\}} |\kappa ^\epsilon |^2 \, \mathrm{{d}}s\mathrm{{d}}t \\&\le |Q_T|+ \int _{Q_T}|\sigma ^\epsilon |^2 \ \mathrm{{d}}s\mathrm{{d}}t \le C, \end{aligned}$$

where C does not depend on \(\epsilon \).