On a generalization of the Cahn-Hilliard type equation with logarithmic nonlinearities in island formation

We denote by (.) the usual L²-scalar product with associated norm ||.||. We also set ||.||₋₁ = ||(−Δ)⁻¹||, where (−Δ)⁻¹ denotes the inverse of the minus Laplace operator associated with Neumann boundary conditions and acting on zero-mean functions.

More generally, we denote by ||.||_X the norm on the Banach space X.

We set 〈.〉=1 Vol (Ω)∫Ω.dx \langle {\kern 1pt} .{\kern 1pt} \rangle = \frac{1}{{{\kern 1pt} {\rm{Vol}}{\kern 1pt} (\Omega )}}\int_\Omega .dx , being understood that if ζ ∈ H⁻¹(Ω) = H¹(Ω)′ then 〈ζ〉=1 Vol (Ω)〈ζ,1〉H−1(Ω),H1(Ω). \langle \zeta \rangle = \frac{1}{{{\kern 1pt} {\rm{Vol}}{\kern 1pt} (\Omega )}}{\langle \zeta ,1\rangle _{{H^{ - 1}}(\Omega ),{H^1}(\Omega )}}. We also set, whenever this makes sense ζ¯=ζ−〈ζ〉. \overline \zeta = \zeta - \langle \zeta \rangle . We note that: ζ↦||ζ¯||−12+〈ζ2〉12,ζ↦||ζ¯||2+〈ζ2〉12,ζ↦||∇ζ||2+〈ζ2〉12,ζ↦||Δζ||2+〈ζ2〉12 \begin{array}{*{20}{c}}{\zeta \mapsto {{\left( {||\overline \zeta ||_{ - 1}^2 + \langle {\zeta ^2}\rangle } \right)}^{\frac{1}{2}}},}\\{\zeta \mapsto {{\left( {||\overline \zeta |{|^2} + \langle {\zeta ^2}\rangle } \right)}^{\frac{1}{2}}},}\\{\zeta \mapsto {{\left( {||\nabla \zeta |{|^2} + \langle {\zeta ^2}\rangle } \right)}^{\frac{1}{2}}},}\\{\zeta \mapsto {{\left( {||\Delta \zeta |{|^2} + \langle {\zeta ^2}\rangle } \right)}^{\frac{1}{2}}}}\end{array} are all norms on H⁻¹(Ω), L²(Ω), H¹(Ω) or H²(Ω) that are equivalent to the usual norms on these spaces. Furthermore, ||.||₋₁ is a norm on {ζ ∈ H⁻¹(Ω),〈ζ〉 = 0} which is equivalent to the usual H⁻¹ norm.

Note that the same letter c (and sometimes also c′ or c″) in this paper stands for (generally positive) constants that are independent of N and may change from line to line. The same applies to constants such as c_δ, cδ′ c_\delta ^\prime and cδ″ c_\delta ^{''} which depend on a parameter δ.

In this section, our objective is to establish estimates for u_N that are independent of N. These estimates can be rigorously justified based on the approximated problems. A crucial aspect involves obtaining a uniform (with respect to N) estimate for f_N (ς_N) in L²(Ω × (0, T)), where T > 0 is independent of N. This is essential for passing to the limit in the nonlinear term and obtaining a solution to the singular initial problem. Notably, achieving this goal necessitates a uniform (with respect to N) strict separation property for 〈ς_N〉 from the singular points −1 and 1. It is worth mentioning that such a strict separation property is straightforward for the original Cahn-Hilliard equation, given the conservation of the spatial average of the order parameter, provided that the same property holds for the initial datum.

We assume that ς₀ ∈ H¹(Ω) ∩ L^∞(Ω), with |ς₀(x)| < 1 almost everywhere in Ω, and (24) |〈ς0〉|≤1−2δ, δ∈(0,12], |\langle {\varsigma _0}\rangle | \le 1 - 2\delta ,{\kern 1pt} {\kern 1pt} \delta \in (0,\frac{1}{2}], given. If we first integrate (21) over Ω, we find due to (22), (25) d〈ςN〉dt+〈η(ςN)〉=0. \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \langle \eta ({\varsigma _N})\rangle = 0. In fact, we have that dςNdt+Δ2ςN−ΔfN(ςN)+η(ςN)=0. \frac{{d{\varsigma _N}}}{{dt}} + {\Delta ^2}{\varsigma _N} - \Delta {f_N}({\varsigma _N}) + \eta ({\varsigma _N}) = 0. Noting that ∫ΩΔ2ς dx=−∫Ω∇Δς.∇1dx+∫Γ∂Δς∂ν.ς ds=0. \int_\Omega {\Delta ^2}\varsigma {\kern 1pt} dx = - \int_\Omega \nabla \Delta \varsigma .\nabla 1dx + \int_\Gamma \frac{{\partial \Delta \varsigma }}{{\partial \nu }}.\varsigma {\kern 1pt} ds = 0. Furthermore, ∫ΩΔfN(ςN) dx=−∫Ω∇fN(ςN).∇1dx+∫Γ∂fN(ςN)∂ν1 ds=0, \int_\Omega \Delta {f_N}({\varsigma _N}){\kern 1pt} dx = - \int_\Omega \nabla {f_N}({\varsigma _N}).\nabla 1dx + \int_\Gamma \frac{{\partial {f_N}({\varsigma _N})}}{{\partial \nu }}1{\kern 1pt} ds = 0, hence ddt∫ΩςNdx+∫Ωη(ςN)dx=0, \frac{d}{{dt}}\int_\Omega {\varsigma _N}dx + \int_\Omega \eta ({\varsigma _N})dx = 0, which yields d〈ςN〉dt+〈η(ςN)〉=0. \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \langle \eta ({\varsigma _N})\rangle = 0. On the other hand, setting ς_N = 〈ς_N〉 + ζ_N, (ζ_N) = 0 yields d〈ςN〉dt+〈ςN2−p(1−ςN)〉=0 \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \langle \varsigma _N^2 - p(1 - {\varsigma _N})\rangle = 0 and d〈ςN〉dt+〈〈ςN〉+ζN2−p1−ςN〉=0. \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \langle {\left( {\langle {\varsigma _N}\rangle + {\zeta _N}} \right)^2} - p\left( {1 - {\varsigma _N}} \right)\rangle = 0. Therefore, d〈ςN〉dt+〈〈ςN〉2+2〈ςN〉〈ζN〉+ζN2−p1−ςN〉=0, \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \langle {\langle {\varsigma _N}\rangle ^2} + 2\langle {\varsigma _N}\rangle \langle {\zeta _N}\rangle + \zeta _N^2 - p\left( {1 - {\varsigma _N}} \right)\rangle = 0, so that d〈ςN〉dt+〈ςN〉2+2〈ςN〉〈ζN〉+〈ζN2〉−p1−〈ςN〉=0, \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + {\langle {\varsigma _N}\rangle ^2} + 2\langle {\varsigma _N}\rangle \langle {\zeta _N}\rangle + \langle \zeta _N^2\rangle - p\left( {1 - \langle {\varsigma _N}\rangle } \right) = 0, it then follows d〈ςN〉dt+〈ςN〉2−p1−〈ςN〉=−〈ζN2〉. \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + {\langle {\varsigma _N}\rangle ^2} - p\left( {1 - \langle {\varsigma _N}\rangle } \right) = - \langle \zeta _N^2\rangle . Hence, we get (26) d〈ςN〉dt+η〈ςN〉=−〈ζN2〉. \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \eta \langle {\varsigma _N}\rangle = - \langle \zeta _N^2\rangle . Furthermore, ζ_N is a solution to: (27) ∂tζN+Δ2ζN−ΔfN(ςN)+η(ςN)¯=0, {\partial _t}{\zeta _N} + {\Delta ^2}{\zeta _N} - \Delta {f_N}({\varsigma _N}) + \overline {\eta ({\varsigma _N})} = 0, (28) ∂νζN=∂νΔζN=0, on Γ, {\partial _\nu }{\zeta _N} = {\partial _\nu }\Delta {\zeta _N} = 0,\;{\rm{on}}\;\Gamma , (29) ζN|t=0=ζ0(x), in Ω. {\zeta _N}{|_{t = 0}} = {\zeta _0}(x),\;{\rm{in}}\;\Omega . Since (21) gives, ∂ζN+〈ςN〉∂t+Δ2ςN−ΔfN(ςN)+η(ςN)=0∂ζN∂t+∂〈ςN〉∂t+Δ2ςN−ΔfN(ςN)+η(ςN)=0∂ζN∂t−〈g(ςN)〉+Δ2ζN−ΔfN(ςN)+η(ςN)=0∂ζN∂t+Δ2ζN−ΔfN(ςN)+η(ςN)¯=0. \begin{array}{*{20}{r}}{\frac{{\partial \left( {{\zeta _N} + \langle {\varsigma _N}\rangle } \right)}}{{\partial t}} + {\Delta ^2}{\varsigma _N} - \Delta {f_N}({\varsigma _N}) + \eta ({\varsigma _N}) = 0}\\{\frac{{\partial {\zeta _N}}}{{\partial t}} + \frac{{\partial \langle {\varsigma _N}\rangle }}{{\partial t}} + {\Delta ^2}{\varsigma _N} - \Delta {f_N}({\varsigma _N}) + \eta ({\varsigma _N}) = 0}\\{\frac{{\partial {\zeta _N}}}{{\partial t}} - \langle g({\varsigma _N})\rangle + {\Delta ^2}{\zeta _N} - \Delta {f_N}({\varsigma _N}) + \eta ({\varsigma _N}) = 0}\\{\frac{{\partial {\zeta _N}}}{{\partial t}} + {\Delta ^2}{\zeta _N} - \Delta {f_N}({\varsigma _N}) + \overline {\eta ({\varsigma _N})} = 0.}\end{array} ∂ςN∂ν=∂ζN∂ν and ∂Δςn∂ν=∂Δζn∂ν. \frac{{\partial {\varsigma _N}}}{{\partial \nu }} = \frac{{\partial {\zeta _N}}}{{\partial \nu }}{\kern 1pt} \;{\kern 1pt} {\rm{and}}\;{\kern 1pt} {\kern 1pt} \frac{{\partial \Delta {\varsigma _n}}}{{\partial \nu }} = \frac{{\partial \Delta {\zeta _n}}}{{\partial \nu }}. These equations can be written equivalently, if we multiply by −(Δ)⁻¹, as: (30) (−Δ)−1∂tζN−ΔζN+fN(ςN)¯+(−Δ)−1η(ςN)¯=0, in Ω×(0,T), {( - \Delta )^{ - 1}}{\partial _t}{\zeta _N} - \Delta {\zeta _N} + \overline {{f_N}({\varsigma _N})} + {( - \Delta )^{ - 1}}\overline {\eta ({\varsigma _N})} = 0,\;{\rm{in}}\;\Omega \times (0,T), (31) ∂νζN=0, on Γ, {\partial _\nu }{\zeta _N} = 0,\;{\rm{on}}\;\Gamma , (32) ζN|t=0=ζ0, in Ω. {\zeta _N}{|_{t = 0}} = {\zeta _0},\;{\rm{in}}\;\Omega . Multiplying (30) by ζ_N and integrating over Ω and by parts, we have: (33) 12ddt||ζN||−12+∥∇ζN∥2+[fN(ςN)¯,ζN]+[η(ςN),(−Δ−1)−1ζN]=0. \frac{1}{2}\frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + \parallel \nabla {\zeta _N}{\parallel ^2} + [\overline {{f_N}({\varsigma _N})} ,{\zeta _N}] + [\eta ({\varsigma _N}),{( - {\Delta ^{ - 1}})^{ - 1}}{\zeta _N}] = 0. Since ∫Ω(−Δ)−1∂ζN∂tζN dx=12ddt∫Ω(−Δ)−1ζN2 dx=12ddt||ζN||−12 \begin{array}{*{20}{l}}{\int_\Omega {{( - \Delta )}^{ - 1}}\frac{{\partial {\zeta _N}}}{{\partial t}}{\zeta _N}{\kern 1pt} dx}&{ = \frac{1}{2}\frac{d}{{dt}}\int_\Omega {{( - \Delta )}^{ - 1}}\zeta _N^2{\kern 1pt} dx}\\{}&{ = \frac{1}{2}\frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2}\end{array} and ∫Ω−ΔζN ζN dx=∫Ω∇ζN⋅∇ζN dx−∫Γ∂ζN∂νζN ds= ||∇ζN||2. \begin{array}{*{20}{l}}{\int_\Omega - \Delta {\zeta _N}{\kern 1pt} {\zeta _N}{\kern 1pt} dx}&{ = \int_\Omega \nabla {\zeta _N} \cdot \nabla {\zeta _N}{\kern 1pt} dx - \int_\Gamma \frac{{\partial {\zeta _N}}}{{\partial \nu }}{\zeta _N}{\kern 1pt} ds}\\{}&{ = \;||\nabla {\zeta _N}|{|^2}.}\end{array} Note that [fN(ςN)¯,ζN]=[fN(ςN)−fN(〈ςN〉),ζN] [\overline {{f_N}({\varsigma _N})} ,{\zeta _N}] = [{f_N}({\varsigma _N}) - {f_N}(\langle {\varsigma _N}\rangle ),{\zeta _N}] and taking s = ζ_N, m = 〈ς_N〉 in (17), we get (34) [fN(ςN)¯,ζN]≥c4∫ΩζN4+〈ςN〉2ζN2 dx−c. [\overline {{f_N}({\varsigma _N})} ,{\zeta _N}] \ge {c_4}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx - c. Indeed, from (17), we have fN(s+m)−fN(m)s≥c4s4+m2s2−c5fN(ςN)−fN〈ςN〉ζN≥c4ζN4+〈ςN〉2ζN2−c5, \begin{array}{*{20}{l}}{\left( {{f_N}(s + m) - {f_N}(m)} \right)s \ge {c_4}\left( {{s^4} + {m^2}{s^2}} \right) - {c_5}}\\{\left( {{f_N}({\varsigma _N}) - {f_N}\langle {\varsigma _N}\rangle } \right){\zeta _N} \ge {c_4}\left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right) - {c_5},}\end{array} so [fN(ςN)−fN〈ςN〉,ζN]=∫ΩfN(ςN)−fN〈ςN〉ζN dx≥c4∫ΩζN4+〈ςN〉2ζN2 dx−c5. \begin{array}{*{20}{l}}{[{f_N}({\varsigma _N}) - {f_N}\langle {\varsigma _N}\rangle ,{\zeta _N}]}&{ = \int_\Omega \left( {{f_N}({\varsigma _N}) - {f_N}\langle {\varsigma _N}\rangle } \right){\zeta _N}{\kern 1pt} dx}\\{}&{ \ge {c_4}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx - {c_5}.}\end{array} Furthermore, |[η(ςN)¯,(−Δ−1)ζN]|= |[η(ςN)−η(〈ςN〉),(−Δ−1)ζN]]|≤c||ζN|| ||η(ςN)−g(〈ςN〉)||, \begin{array}{*{20}{l}}{|[\overline {\eta ({\varsigma _N})} ,( - {\Delta ^{ - 1}}){\zeta _N}]|}&{ = \;|[\eta ({\varsigma _N}) - \eta (\langle {\varsigma _N}\rangle ),( - {\Delta ^{ - 1}}){\zeta _N}]]|}\\{}&{ \le c||{\zeta _N}||{\kern 1pt} ||\eta ({\varsigma _N}) - g(\langle {\varsigma _N}\rangle )||,}\end{array} by Cauchy Schwartz and using ||ζ_N||₋₁ ≤ c||ζ_N|| which results after reapplying Lemma (1) with s = ζ_N and m = 〈ς_N〉 and Young’s inequality: (35) |[η(ςN)¯,(−Δ−1)ζN]|≤c44∫ΩζN4+〈ςN〉2ζN2 dx+c||ζN||2≤c44∫ΩζN4+〈ςN〉2ζN2 dx+c. \begin{array}{*{20}{l}}{|[\overline {\eta ({\varsigma _N})} ,( - {\Delta ^{ - 1}}){\zeta _N}]|}&{ \le \frac{{{c_4}}}{4}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx + c||{\zeta _N}|{|^2}}\\{}&{ \le \frac{{{c_4}}}{4}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx + c.}\end{array} Indeed, we have that |[η(ςN)¯,(−Δ−1)ζN]|≤c||ζN|| ||η(ςN)−η(〈ςN〉)||≤c||ζN||22+||η(ςN)−η(〈ςN〉)||22≤c2||ζN||2+c9∫ΩζN4+〈ςN〉2ζN2+ζN2 dx+c10≤c42∫ΩζN4+〈ςN〉2ζN2 dx+c||ζN||2+c′≤c42∫ΩζN4+〈ςN〉2ζN2 dx+c. \begin{array}{*{20}{l}}{|[\overline {\eta ({\varsigma _N})} ,( - {\Delta ^{ - 1}}){\zeta _N}]|}&{ \le c||{\zeta _N}||{\kern 1pt} ||\eta ({\varsigma _N}) - \eta (\langle {\varsigma _N}\rangle )||}\\{}&{ \le c\left( {\frac{{||{\zeta _N}|{|^2}}}{2} + \frac{{||\eta ({\varsigma _N}) - \eta (\langle {\varsigma _N}\rangle )|{|^2}}}{2}} \right)}\\{}&{ \le \frac{c}{2}\left( {||{\zeta _N}|{|^2} + {c_9}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2 + \zeta _N^2} \right){\kern 1pt} dx + {c_{10}}} \right)}\\{}&{ \le \frac{{{c_4}}}{2}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx + c||{\zeta _N}|{|^2} + {c^\prime}}\\{}&{ \le \frac{{{c_4}}}{2}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx + c.}\end{array} It follows from (33)–(35), that (36) ddt||ζN||−12+c||ζN||H1(Ω)2+∫ΩζN4+〈ςN〉2ζN2dx≤c′, c>0. \frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + c\left( {||{\zeta _N}||_{{H^1}(\Omega )}^2 + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx} \right) \le {c^\prime},{\kern 1pt} {\kern 1pt} c > 0. Since 12ddt||ζN||−12+||∇ζN||2=−[fN(ςN)¯,ζN]−[η(ςN)¯,−Δ−1ζN]≤−c4∫ΩζN4+〈ςN〉2ζN2dx+c+c42∫ΩζN4+〈ςN〉2ζN2dx+c. \begin{array}{*{20}{l}}{\frac{1}{2}\frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + ||\nabla {\zeta _N}|{|^2}}&{ = - [\overline {{f_N}({\varsigma _N})} ,{\zeta _N}] - [\overline {\eta ({\varsigma _N})} , - {\Delta ^{ - 1}}{\zeta _N}]}\\{}&{ \le - {c_4}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + c + \frac{{{c_4}}}{2}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + c.}\end{array} Next, multiplying (30) by −Δζ_N and integrating over Ω, we obtain (37) 12ddt||ζN||2+||ΔζN||2+[fN(ςN)¯,−ΔζN]+[η(ςN)¯,ζN]=0. \frac{1}{2}\frac{d}{{dt}}||{\zeta _N}|{|^2} + ||\Delta {\zeta _N}|{|^2} + [\overline {{f_N}({\varsigma _N})} , - \Delta {\zeta _N}] + [\overline {\eta ({\varsigma _N})} ,{\zeta _N}] = 0. In fact, we have ∫Ω−Δ−1∂ζN∂t (−ΔζN) dx=∫Ω∂ζN∂t ζN dx=12ddt∫ΩζN2 dx=12ddt||ζN||2. \begin{array}{*{20}{l}}{\int_\Omega - {\Delta ^{ - 1}}\frac{{\partial {\zeta _N}}}{{\partial t}}{\kern 1pt} ( - \Delta {\zeta _N}){\kern 1pt} dx}&{ = \int_\Omega \frac{{\partial {\zeta _N}}}{{\partial t}}{\kern 1pt} {\zeta _N}{\kern 1pt} dx}\\{}&{ = \frac{1}{2}\frac{d}{{dt}}\int_\Omega \zeta _N^2{\kern 1pt} dx}\\{}&{ = \frac{1}{2}\frac{d}{{dt}}||{\zeta _N}|{|^2}.}\end{array} Furthermore, owing to (14) (38) [fN(ςN)¯,−ΔζN]=[fN(ςN),−ΔζN]=−∫ΩfN(ςN) ΔζN dx=∫Ω∇fN(ςN).∇ζN dx=∫Ωf′(ςN)∇ςN∇ζN dx=[f′(ςN)∇ςN,∇ζN]≥−λ1||∇ζN||2 \begin{array}{*{20}{l}}{[\overline {{f_N}({\varsigma _N})} , - \Delta {\zeta _N}]}&{ = [{f_N}({\varsigma _N}), - \Delta {\zeta _N}]}\\{}&{ = - \int_\Omega {f_N}({\varsigma _N}){\kern 1pt} \Delta {\zeta _N}{\kern 1pt} dx}\\{}&{ = \int_\Omega \nabla {f_N}({\varsigma _N}).\nabla {\zeta _N}{\kern 1pt} dx}\\{}&{ = \int_\Omega {f^\prime}({\varsigma _N})\nabla {\varsigma _N}\nabla {\zeta _N}{\kern 1pt} dx}\\{}&{ = [{f^\prime}({\varsigma _N})\nabla {\varsigma _N},\nabla {\zeta _N}]}\\{}&{ \ge - {\lambda _1}||\nabla {\zeta _N}|{|^2}}\end{array} and owing once more to Lemma (1), we have (39) |[η(ςN)¯,ζN]|= |[η(ςN)−η(〈ςN〉),ζN]|≤12||ΔζN||2+c∫ΩζN4+〈ςN〉2ζN2 dx+c′. \begin{array}{*{20}{l}}{|[\overline {\eta ({\varsigma _N})} ,{\zeta _N}]|}&{ = \;|[\eta ({\varsigma _N}) - \eta (\langle {\varsigma _N}\rangle ),{\zeta _N}]|}\\{}&{ \le \frac{1}{2}||\Delta {\zeta _N}|{|^2} + c\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx + {c^\prime}.}\end{array} Indeed, |[η(ςN)−η(〈ςN〉),ζN]|≤ ||η(ςN)−η(〈ςN〉)|| ||ζN||≤||ζN||22+||η(ςN)−η(〈ςN〉)||22≤12||ΔζN||2+c∫ΩζN4+〈ςN〉2ζN2 dx+c′. \begin{array}{*{20}{l}}{|[\eta ({\varsigma _N}) - \eta (\langle {\varsigma _N}\rangle ),{\zeta _N}]|}&{ \le \;||\eta ({\varsigma _N}) - \eta (\langle {\varsigma _N}\rangle )||{\kern 1pt} ||{\zeta _N}||}&{}&{}\\{}&{ \le \frac{{||{\zeta _N}|{|^2}}}{2} + \frac{{||\eta ({\varsigma _N}) - \eta (\langle {\varsigma _N}\rangle )|{|^2}}}{2}}&{}&{}\\{}&{ \le \frac{1}{2}||\Delta {\zeta _N}|{|^2} + c\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx + {c^\prime}.}&{}&{}\end{array} From (37)–(39) it follows that (40) ddt||ζN||2+||ΔζN||2≤c||ζN||H1(Ω)2+∫ΩζN4+〈ςN〉2ζN2 dx+c′. \frac{d}{{dt}}||{\zeta _N}|{|^2} + ||\Delta {\zeta _N}|{|^2} \le c\left( {||{\zeta _N}||_{{H^1}(\Omega )}^2 + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx} \right) + {c^\prime}. By (37), we have 12ddt||ζN||2+||ΔζN||2= [fN(ςN)¯,−ΔζN]−[η(ςN)¯,ζN]≤λ1||∇ζN||2+12||ΔζN||2+c∫ΩζN4+〈ςN〉2ζN2 dx+c′, \begin{array}{*{20}{l}}{\frac{1}{2}\frac{d}{{dt}}||{\zeta _N}|{|^2} + ||\Delta {\zeta _N}|{|^2}}&{ = \;[\overline {{f_N}({\varsigma _N})} , - \Delta {\zeta _N}] - [\overline {\eta ({\varsigma _N})} ,{\zeta _N}]}&{}&{}\\{}&{ \le {\lambda _1}||\nabla {\zeta _N}|{|^2} + \frac{1}{2}||\Delta {\zeta _N}|{|^2} + c\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx + {c^\prime},}&{}&{}\end{array} so, ddt||ζN||2+||ΔζN||2≤c||∇ζN||2+∫ΩζN4+〈ςN〉2ζN2 dx+c′. \frac{d}{{dt}}||{\zeta _N}|{|^2} + ||\Delta {\zeta _N}|{|^2} \le c\left( {||\nabla {\zeta _N}|{|^2} + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx} \right) + {c^\prime}. Finally, summing (36) and γ₁ times (40), where γ₁ > 0 is small enough and independent of N, we find (41) ddt||ζN||−12+γ1||ζN||2+c||ζN||H2(Ω)2+∫ΩζN4+〈ςN〉2ζN2 dx≤c′, c>0. \frac{d}{{dt}}\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) + c\left( {||{\zeta _N}||_{{H^2}(\Omega )}^2 + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx} \right) \le {c^\prime},\;\;\;c > 0. Indeed, we have (36)+γ₁(40) gives ddt||ζN||2+γ1||ζN||2+c||ζN||H1(Ω)+γ1||ΔζN||2+c∫ΩζN4+〈ςN〉2ζN2 dx≤c′, \frac{d}{{dt}}\left( {||{\zeta _N}|{|^2} + {\gamma _1}||{\zeta _N}|{|^2}} \right) + c||{\zeta _N}|{|_{{H^1}(\Omega )}} + {\gamma _1}||\Delta {\zeta _N}|{|^2} + c\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx \le {c^\prime}, then ddt||ζN||2+γ1||ζN||2+c||ζN||H2(Ω)+∫ΩζN4+〈ςN〉2ζN2 dx≤c′. \frac{d}{{dt}}\left( {||{\zeta _N}|{|^2} + {\gamma _1}||{\zeta _N}|{|^2}} \right) + c\left( {||{\zeta _N}|{|_{{H^2}(\Omega )}} + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx} \right) \le {c^\prime}. We now come back to (25)–(26). Noting that g(s) ≥ ps − p, we have d〈ςN〉dt+η(〈ςN〉)=−〈ζN2〉≤0d〈ςN〉dt≤−η(〈ςN〉)≤−p〈ςN〉+pd〈ςN〉dt+p〈ςN〉≤p. \begin{array}{*{20}{c}}{\frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \eta (\langle {\varsigma _N}\rangle ) = - \langle \zeta _N^2\rangle \le 0}\\{\frac{{d\langle {\varsigma _N}\rangle }}{{dt}} \le - \eta (\langle {\varsigma _N}\rangle ) \le - p\langle {\varsigma _N}\rangle + p}\\{\frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + p\langle {\varsigma _N}\rangle \le p.}\end{array} Consider the ODE: d〈ςN〉dt+p〈ςN〉=p, \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + p\langle {\varsigma _N}\rangle = p, then d〈ςN〉dt=−p〈ςN〉−1, \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} = - p\left( {\langle {\varsigma _N}\rangle - 1} \right), hence d〈ςN〉〈ςN〉−1=−p dt, \frac{{d\langle {\varsigma _N}\rangle }}{{\left( {\langle {\varsigma _N}\rangle - 1} \right)}} = - p{\kern 1pt} dt, we infer ln|〈ςN〉−1|=−p t+k, \ln |\langle {\varsigma _N}\rangle - 1| = - p{\kern 1pt} t + k, so 〈ςN〉=ce−p t+1, \langle {\varsigma _N}\rangle = c{e^{ - p{\kern 1pt} t}} + 1, but at t = 0, 〈ς_N〉 = 〈ς₀〉, hence c = 〈ς₀〉 − 1.

Finally, (42) 〈ςN(t)〉≤〈ς0〉−1e−p t+1, \langle {\varsigma _N}(t)\rangle \le \left( {\langle {\varsigma _0}\rangle - 1} \right){e^{ - p{\kern 1pt} t}} + 1, as long as it exists. In particular, (43) 〈ςN(t)〉<1. \langle {\varsigma _N}(t)\rangle < 1. Note that it follows from (41) that: ddt||ζN||−12+γ1||ζN||2+c||ζN||−12+γ1||ζN||2≤c′, c>0. \frac{d}{{dt}}\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) + c\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) \le {c^\prime},{\kern 1pt} {\kern 1pt} c > 0. Indeed, we have that c||ζN||H2(Ω)2+∫ΩζN4+〈ςN〉2ζN2 dx≥c||ζN||H(Ω)2−12||ΔζN||2−c′≥c||ΔζN||2−c′≥c||ζN||2−c′≥c2||ζN||2+c2||ζN||2−c′≥c||ζN||−12+γ1||ζN||2−c′, \begin{array}{*{20}{l}}{c\left( {||{\zeta _N}||_{{H^2}(\Omega )}^2 + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx} \right)}&{ \ge c||{\zeta _N}||_{{H^(}\Omega )}^2 - \frac{1}{2}||\Delta {\zeta _N}|{|^2} - {c^\prime}}&{}&{}&{}&{}&{}\\{}&{ \ge c||\Delta {\zeta _N}|{|^2} - {c^\prime}}&{}&{}&{}&{}&{}\\{}&{ \ge c||{\zeta _N}|{|^2} - {c^\prime}}&{}&{}&{}&{}&{}\\{}&{ \ge \frac{c}{2}||{\zeta _N}|{|^2} + \frac{c}{2}||{\zeta _N}|{|^2} - {c^\prime}}&{}&{}&{}&{}&{}\\{}&{ \ge c\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) - {c^\prime},}&{}&{}&{}&{}&{}\end{array} but by (41) ddt||ζN||−12+γ1||ζN||2+c||ζN||−12+γ1||ζN||2≤ddt||ζN||−12+γ1||ζN||2+c||ζN||H2(Ω)2+∫ΩζN4+〈ςN〉2ζN2 dx+c′≤2c′. \begin{array}{*{20}{c}}{\frac{d}{{dt}}\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) + c\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right)}\\{ \le \frac{d}{{dt}}\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) + c\left( {||{\zeta _N}||_{{H^2}(\Omega )}^2 + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right){\kern 1pt} dx} \right) + {c^\prime}}\\{ \le 2{c^\prime}.}\end{array} Which yields (44) ||ζN(t)||2≤ce−c′t||ζ0||2+c′, c″>0. ||{\zeta _N}(t)|{|^2} \le c{e^{ - {c^\prime}t}}||{\zeta _0}|{|^2} + {c^\prime},{\kern 1pt} {\kern 1pt} {c^{''}} > 0. In fact, ddt||ζN||−12+γ1||ζN||2+c||ζN||−12+γ1||ζN||2≤c′, c>0, \frac{d}{{dt}}\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) + c\left( {||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2}} \right) \le {c^\prime},{\kern 1pt} {\kern 1pt} c > 0, then by Gronwall lemma ||ζN||−12+γ1||ζN||2≤||ζ0||−12+γ1||ζ0||2+c″γ1||ζN||2≤e−ct||ζ0||−12+γ1||ζ0||2+c″||ζN||2≤1γ1e−ct||ζ0||−12+γ1||ζ0||2+c″γ1, \begin{array}{*{20}{c}}{||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2} \le \left( {||{\zeta _0}||_{ - 1}^2 + {\gamma _1}||{\zeta _0}|{|^2}} \right) + {c^{''}}}\\{{\gamma _1}||{\zeta _N}|{|^2} \le {e^{ - ct}}\left( {||{\zeta _0}||_{ - 1}^2 + {\gamma _1}||{\zeta _0}|{|^2}} \right) + {c^{''}}}\\{||{\zeta _N}|{|^2} \le \frac{1}{{{\gamma _1}}}{e^{ - ct}}\left( {||{\zeta _0}||_{ - 1}^2 + {\gamma _1}||{\zeta _0}|{|^2}} \right) + \frac{{{c^{''}}}}{{{\gamma _1}}},}\end{array} but ||ζ0||−12≤q||ζ0||2 ||{\zeta _0}||_{ - 1}^2 \le q||{\zeta _0}|{|^2} , then ||ζN||2≤1γ1e−ctq||ζ0||2+γ1||ζ0||2+c″γ1≤ce−c′t||ζ0||2+c″. \begin{array}{*{20}{l}}{||{\zeta _N}|{|^2}}&{ \le \frac{1}{{{\gamma _1}}}{e^{ - ct}}\left( {q||{\zeta _0}|{|^2} + {\gamma _1}||{\zeta _0}|{|^2}} \right) + \frac{{{c^{''}}}}{{{\gamma _1}}}}&{}&{}\\{}&{ \le c{e^{ - {c^\prime}t}}||{\zeta _0}|{|^2} + {c^{''}}.}&{}&{}\end{array} As long as it exists, in particular 〈ζN2〉=k||ζN2(t)||≤c||ζ0||2+c′, since e−c′t≤1, \langle \zeta _N^2\rangle = k||\zeta _N^2(t)|| \le c||{\zeta _0}|{|^2} + {c^\prime},{\kern 1pt} \;{\rm{since}}\;{\kern 1pt} {e^{ - {c^\prime}t}} \le 1, and ζ₀ = ς₀ − 〈ς₀〉 and |〈ς₀〉| ≤ 1 − 2δ, so (45) 〈ζN2(t)〉≤c(ς0,δ), \langle \zeta _N^2(t)\rangle \le c({\varsigma _0},\delta ), as long as it exists.

Let y_± be the solution of the Ricatti ODE’s (46) y+′+η(y+)=0, y+(0)=〈ς0〉 y_ + ^\prime + \eta ({y_ + }) = 0,{\kern 1pt} {\kern 1pt} {y_ + }(0) = \langle {\varsigma _0}\rangle (47) y−′+η(y−)=−c(ς0,δ), y−(0)=〈ς0〉, y_ - ^\prime + \eta ({y_ - }) = - c({\varsigma _0},\delta ),{\kern 1pt} {\kern 1pt} {y_ - }(0) = \langle {\varsigma _0}\rangle , where c(ς₀, δ) is the constant in (45). Then it follows from the comparison principle that, as long as this makes sense (48) y−(t)≤〈ςN(t)〉≤y+(t). {y_ - }(t) \le \langle {\varsigma _N}(t)\rangle \le {y_ + }(t). Indeed, we have that d〈ςN〉dt+g(〈ςN〉)=−〈ζN2〉≤0, \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + g(\langle {\varsigma _N}\rangle ) = - \langle \zeta _N^2\rangle \le 0, so, 〈ς_N (t)〉 ≤ y₊(t), also 〈ζN2〉≤c(ς0,δ), \langle \zeta _N^2\rangle \le c({\varsigma _0},\delta ), hence d〈ςN〉dt+η(〈ςN〉)≥−c(ς0,δ), \frac{{d\langle {\varsigma _N}\rangle }}{{dt}} + \eta (\langle {\varsigma _N}\rangle ) \ge - c({\varsigma _0},\delta ), so 〈ς_N (t)) ≥ y₋(t).

In particular, it follows that (at least) a local in times solution exists on some [0, T ], where T > 0 is independent of N. Note also that y+(t)=y2−cy1etc01−cetc0, {y_ + }(t) = \frac{{{y_2} - c{y_1}{e^{\frac{t}{{{c_0}}}}}}}{{1 - c{e^{\frac{t}{{{c_0}}}}}}}, with y1=p+p2+4p2,y2=p−p2+4p2 \begin{array}{*{20}{c}}{{y_1} = \frac{{p + \sqrt {{p^2} + 4p} }}{2},}\\{{y_2} = \frac{{p - \sqrt {{p^2} + 4p} }}{2}}\end{array} and c=y2−〈ς0〉y1−〈ς0〉. c = \frac{{{y_2} - \langle {\varsigma _0}\rangle }}{{{y_1} - \langle {\varsigma _0}\rangle }}. y₊(t) is the solution of the ODE: y′ + η(y) = 0, so y′ + y² − p(1 − y) = 0, so dydt=p+py−y2∫dyp+py−y2=∫dt. \begin{array}{*{20}{c}}{\frac{{dy}}{{dt}}}&{ = p + py - {y^2}}&{}&{}\\{\int \frac{{dy}}{{p + py - {y^2}}}}&{ = \int dt.}&{}&{}\end{array} We have p+py−y2=0:Δ=p2−4(p)(−1)=p2+4p, \begin{array}{*{20}{c}}{p + py - {y^2} = 0:}\\{\Delta = {p^2} - 4(p)( - 1) = {p^2} + 4p,}\end{array} hence the roots of the quadratic equation are y1=−p−p2+4p−2=y1=p+p2+4p2,y2=−p+p2+4p−2=y2=p−p2+4p2, \begin{array}{*{20}{c}}{{y_1} = \frac{{ - p - \sqrt {{p^2} + 4p} }}{{ - 2}} = {y_1} = \frac{{p + \sqrt {{p^2} + 4p} }}{2},}\\{{y_2} = \frac{{ - p + \sqrt {{p^2} + 4p} }}{{ - 2}} = {y_2} = \frac{{p - \sqrt {{p^2} + 4p} }}{2},}\end{array} so ∫−dy(y−y1)(y−y2)=t+c,1(y−y1)(y−y2)=Ay−y1+By−y2. \begin{array}{*{20}{c}}{\int \frac{{ - dy}}{{(y - {y_1})(y - {y_2})}} = t + c,}\\{\frac{1}{{(y - {y_1})(y - {y_2})}} = \frac{A}{{y - {y_1}}} + \frac{B}{{y - {y_2}}}.}\end{array} Where A=limy→y11y−y2=1y1−y2=1p2+4p=c0,B=limy→y21y−y1=1y2−y1=−1p2+4p=−c0. \begin{array}{*{20}{c}}{A = \mathop {\lim }\limits_{y \to {y_1}} \frac{1}{{y - {y_2}}} = \frac{1}{{{y_1} - {y_2}}} = \frac{1}{{\sqrt {{p^2} + 4p} }} = {c_0},}\\{B = \mathop {\lim }\limits_{y \to {y_2}} \frac{1}{{y - {y_1}}} = \frac{1}{{{y_2} - {y_1}}} = \frac{{ - 1}}{{\sqrt {{p^2} + 4p} }} = - {c_0}.}\end{array} We infer ∫−c0y−y1+c0y−y2dt=t+c,c0ln|y−y2|−ln|y−y1|=t+lnc,c0ln|y−y2y−y1|=t+lnc,ln|y−y2y−y1|c0=t+lnc,y−y2y−y1c0=cet,y−y2y−y1=cetc0,y−y2=cetc0(y−y1),y1−cetc0=y2−cy1etc0,y=y2−cy1etc01−cetc0. \begin{array}{*{20}{c}}{\int \left( {\frac{{ - {c_0}}}{{y - {y_1}}} + \frac{{{c_0}}}{{y - {y_2}}}} \right)dt = t + c,}\\{{c_0}\left( {\ln |y - {y_2}| - \ln |y - {y_1}|} \right) = t + \ln c,}\\{{c_0}\ln |\frac{{y - {y_2}}}{{y - {y_1}}}| = t + \ln c,}\\{\ln |\frac{{y - {y_2}}}{{y - {y_1}}}{|^{{c_0}}} = t + \ln c,}\\{{{\left( {\frac{{y - {y_2}}}{{y - {y_1}}}} \right)}^{{c_0}}} = c{e^t},}\\{\frac{{y - {y_2}}}{{y - {y_1}}} = c{e^{\frac{t}{{{c_0}}}}},}\\{y - {y_2} = c{e^{\frac{t}{{{c_0}}}}}(y - {y_1}),}\\{y\left( {1 - c{e^{\frac{t}{{{c_0}}}}}} \right) = {y_2} - c{y_1}{e^{\frac{t}{{{c_0}}}}},}\\{y = \frac{{{y_2} - c{y_1}{e^{\frac{t}{{{c_0}}}}}}}{{1 - c{e^{\frac{t}{{{c_0}}}}}}}.}\end{array} But y|_t=0 = 〈ς₀〉, so 〈ς0〉=y2−cy11−c,y2−cy1=〈ς0〉1−c,y2−〈ς0〉=cy1−〈ς0〉,c=y2−〈ς0〉y1−〈ς0〉. \begin{array}{*{20}{c}}{\langle {\varsigma _0}\rangle = \frac{{{y_2} - c{y_1}}}{{1 - c}},}\\{{y_2} - c{y_1} = \langle {\varsigma _0}\rangle \left( {1 - c} \right),}\\{{y_2} - \langle {\varsigma _0}\rangle = c\left( {{y_1} - \langle {\varsigma _0}\rangle } \right),}\\{c = \frac{{{y_2} - \langle {\varsigma _0}\rangle }}{{{y_1} - \langle {\varsigma _0}\rangle }}.}\end{array} We assume from now on that t ∈ [0, T ], where T is as above, we again multiply (30) by ζ_N and we have 12ddt||ζN||−12+||∇ζN||2+ [fN(ςN),ζN]≤c∫ΩζN4+〈ςN〉2ζN2dx+c′, \frac{1}{2}\frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + ||\nabla {\zeta _N}|{|^2} + \;[{f_N}({\varsigma _N}),{\zeta _N}] \le c\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + {c^\prime}, since by (33) (49) 12ddt||ζN||−12+||∇ζN||2+[fN(ςN)¯,ζN]=−[η(ςN)¯,−Δ−1ζN]≤c∫ΩζN4+〈ςN〉2ζN2dx+c′. \begin{array}{*{20}{c}}{\frac{1}{2}\frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + ||\nabla {\zeta _N}|{|^2} + [\overline {{f_N}({\varsigma _N})} ,{\zeta _N}]}&{ = - [\overline {\eta ({\varsigma _N})} , - {\Delta ^{ - 1}}{\zeta _N}]}&{}&{}\\{}&{ \le c\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + {c^\prime}.}&{}&{}\end{array} Which yields, employing (18) with s = ζ_N and m = 〈ς_N〉, (50) ddt||ζN||−12+cδ||fN(ςN)||L1(Ω)+∫ΩFN(ςN)dx≤c′∫ΩζN4+〈ςN〉2ζN2dx+cδ″, cδ>0. \frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + {c_\delta }\left( {||{f_N}({\varsigma _N})|{|_{{L^1}(\Omega )}} + \int_\Omega {F_N}({\varsigma _N})dx} \right) \le {c^\prime}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + c_\delta ^{''},{\kern 1pt} {\kern 1pt} {c_{\delta > 0}}. In fact, we know from (18) that fN(ςN)ζN≥cδFN(ςN)+|f1,N(ςN)|−cδ′∫ΩfN(ςN)ζNdx≥cδ∫ΩFN(ςN)dx+∫Ω|f1,N(ςN)|dx−cδ′, \begin{array}{*{20}{l}}{{f_N}({\varsigma _N}){\zeta _N}}&{ \ge {c_\delta }\left( {{F_N}({\varsigma _N}) + |{f_{1,N}}({\varsigma _N})|} \right) - c_\delta ^\prime}&{}&{}&{}\\{\int_\Omega {f_N}({\varsigma _N}){\zeta _N}dx}&{ \ge {c_\delta }\left( {\int_\Omega {F_N}({\varsigma _N})dx + \int_\Omega |{f_{1,N}}({\varsigma _N})|dx} \right) - c_\delta ^\prime,}&{}&{}&{}\end{array} 12ddt||ζN||−12+cδ||f1,1(ςN)||L1(Ω)+∫ΩFN(ςN)dx−cδ′ ≤c∫ΩζN4+〈ςN〉2ζN2dx+c′, \begin{array}{*{20}{c}}{\frac{1}{2}\frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + {c_\delta }\left( {||{f_{1,1}}({\varsigma _N})|{|_{{L^1}(\Omega )}} + \int_\Omega {F_N}({\varsigma _N})dx} \right) - c_\delta ^\prime}\\{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le c\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + {c^\prime},}\end{array} but f₁(ς_N) = f_N (ς_N) + λς_N. We then infer that ddt||ζN||−12+cδ||fN(ςN)||L1(Ω)+∫ΩFN(ςN)dx≤c′∫ΩζN4+〈ςN〉2ζN2dx+cδ″, cδ>0. \frac{d}{{dt}}||{\zeta _N}||_{ - 1}^2 + {c_\delta }\left( {||{f_N}({\varsigma _N})|{|_{{L^1}(\Omega )}} + \int_\Omega {F_N}({\varsigma _N})dx} \right) \le {c^\prime}\int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + c_\delta ^{''},{\kern 1pt} {\kern 1pt} {c_{\delta > 0}}. Summing (41) and γ₂× (50), where γ₂ > 0 is small enough and independent of N and δ, we obtain a differential inequality of the form (51) dE1,Ndt+cδE1,N+||ζN||H2(Ω)2+∫ΩζN4+〈ςN〉2ζN2dx+||fN(ςN)||L1(Ω)+∫ΩFN(ςN)dx≤cδ′, cδ>0, \begin{array}{*{20}{c}}{\frac{{d{E_{1,N}}}}{{dt}} + {c_\delta }\left( {{E_{1,N}} + ||{\zeta _N}||_{{H^2}(\Omega )}^2} \right) + \int_\Omega \left( {\zeta _N^4 + {{\langle {\varsigma _N}\rangle }^2}\zeta _N^2} \right)dx + ||{f_N}({\varsigma _N})|{|_{{L^1}(\Omega )}}}\\{ + \int_\Omega {F_N}({\varsigma _N})dx \le c_\delta ^\prime,{\kern 1pt} {\kern 1pt} {c_\delta } > 0,}\end{array} where E1,N=(1+γ2)||ζN||−12+γ1||ζN||2 {E_{1,N}} = (1 + {\gamma _2})||{\zeta _N}||_{ - 1}^2 + {\gamma _1}||{\zeta _N}|{|^2} .