# Correction to: Global Attractors of Sixth Order PDEs Describing the Faceting of Growing Surfaces

• M. D. Korzec
• P. Nayar
• P. Rybka
Correction

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

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.

## Reference

1. 1.
Korzec, M.D., Nayar, P., Rybka, P.: Global attractors of sixth order PDEs describing the faceting of growing surfaces. J. Dyn. Differ. Equ. 28, 49–67 (2016)