1 Correction to: J Dyn Diff Equat (2016) 28:49–67 https://doi.org/10.1007/s10884-015-9510-6

The original version of this article, unfortunately, contained an error.

In [1], we studied

$$\begin{aligned} \begin{array}{ll} h_t = \frac{\delta }{2}| \nabla h|^2 + \Delta (\Delta ^2 h - \Delta \mathrm{div}\,D_F W(\nabla h))&{} \hbox {in }\Omega \times \mathbb {R}_+,\\ h(x, 0) = h_0(x)&{} \hbox {for } x \in \Omega , \end{array} \end{aligned}$$
(1)

for \(\Omega = (0,L)^d\), \(d=1\) or \(d=2\) with periodic boundary conditions. The nonlinearity had the following form,

$$\begin{aligned} \begin{array}{ll} W(F) = \frac{1}{4}(F^2-1)^2, &{} d=1,\\ W(F_1,F_2)= \frac{\alpha }{12}(F_1^4 + F_2^4) + \frac{\beta }{2} F_1^2 F_2^2 - \frac{1}{2}(F_1^2 + F_2^2) + A, &{} d=2, \end{array} \end{aligned}$$

where \(\alpha , \beta >0\) are anisotropy coefficients.

The way to obtain long-time results was through the study of the differentiated system (1), \(u=\nabla h\), i.e. we differentiated (1) with respect to x. Here is the resulting problem,

$$\begin{aligned} \begin{array}{ll} u_t = \frac{\delta }{2} \nabla |u|^2 + \Delta ^3 u - \nabla \Delta \mathrm{div}\,D_u W(u) &{} \hbox {in } \Omega \times \mathbb {R}_+,\\ u(x,0) = u_0(x)&{} \hbox {for }x\in \Omega , \end{array} \end{aligned}$$
(2)

where \(u = (u_1, u_2) =(h_x, h_y)\) (resp. \(u= h_x\)), if \(d=2\), (resp. \(d=1\)).

We proved in [1] the following result about (2).

Theorem 1

([1, Theorem 4], [1, Theorem 5]) Let us consider \(\Omega =(0,L)^d\) with \(d=1,2\) and \(L>0\) arbitrary. The semigroup \(S(t):\dot{H}^2_{per}(\Omega ) \rightarrow \dot{H}^2_{per}(\Omega ), u_0 \mapsto S(t)u_0 = u(t) \) generated by equation (2) with periodic boundary conditions has a global attractor.

We also claimed that the following result holds true.

Theorem 2

([1, Theorem 6]) The semigroup generated by equation (1) has a global attractor in \(H^3_{per}\) for \(d=1\) and \(d=2\).

However, this claim is not valid, because if h is solution to (1), then due to [1, Lemma 13] we know that \(\nabla h\in L^2(0,T; \dot{H}^5_{per})\) and integration of (1) over \(\Omega \) yields,

$$\begin{aligned} \frac{d}{dt} \int _\Omega h (x,t)\,dx = \int _\Omega \frac{\partial h}{\partial t} (x,t)\,dx = \int _\Omega \delta |\nabla h|^2\,dx \ge 0. \end{aligned}$$

However, \(\frac{d}{dt} \int _\Omega h (x,t)\,dx = 0\) if and only if \(h\equiv const.\) Moreover, \(h = const.\) is a steady state of (1). As a result, if h is not a constant steady state, then

$$\begin{aligned} 0< \frac{d}{dt} \int _\Omega h (x,t)\,dx. \end{aligned}$$

This fact was overlooked in [1], making the claim in Theorem 2 invalid.