We multiply (3.4) by $\frac{\partial u}{\partial t}$ and integrate over Ω and by parts.
This gives

$\frac{d}{dt}\left({\parallel {A}_{k}^{1/2}u\parallel}^{2}+{B}_{k}^{1/2}[u]+2{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\right)+2{\parallel \frac{\partial u}{\partial t}\parallel}^{2}=0,$(3.9)

where

${B}_{k}^{1/2}[u]=\sum _{i=1}^{k-1}\sum _{|\alpha |=i}{a}_{\alpha}{\parallel {\mathcal{\mathcal{D}}}^{\alpha}u\parallel}^{2}$

(note that ${B}_{k}^{1/2}[u]$ is not necessarily nonnegative).
We can note that, owing to the interpolation inequality

${\parallel v\parallel}_{{H}^{i}(\mathrm{\Omega})}\le c(i){\parallel v\parallel}_{{H}^{m}(\mathrm{\Omega})}^{\frac{i}{m}}{\parallel v\parallel}^{1-\frac{i}{m}},v\in {H}^{m}(\mathrm{\Omega}),i\in \{1,\mathrm{\dots},m-1\},m\in \mathbb{N},m\ge 2,$(3.10)

there holds

$|{B}_{k}^{1/2}[u]|\le \frac{1}{2}{\parallel {A}_{k}^{1/2}u\parallel}^{2}+c{\parallel u\parallel}^{2}.$

This yields, employing (3.8),

${\parallel {A}_{k}^{1/2}u\parallel}^{2}+{B}_{k}^{1/2}[u]+2{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\ge \frac{1}{2}{\parallel {A}_{k}^{1/2}u\parallel}^{2}+{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x+c{\parallel u\parallel}_{{L}^{4}(\mathrm{\Omega})}^{4}-{c}^{\prime}{\parallel u\parallel}^{2}-{c}^{\prime \prime},$

whence

${\parallel {A}_{k}^{1/2}u\parallel}^{2}+{B}_{k}^{1/2}[u]+2{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\ge c\left({\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\right)-{c}^{\prime},c>0,$(3.11)

noting that, owing to Young’s inequality,

${\parallel u\parallel}^{2}\le \u03f5{\parallel u\parallel}_{{L}^{4}(\mathrm{\Omega})}^{4}+c(\u03f5)\mathit{\hspace{1em}}\text{for all}\u03f50.$(3.12)

We then multiply (3.4) by *u* and have, owing to (3.7) and the interpolation
inequality (3.10),

$\frac{d}{dt}{\parallel u\parallel}^{2}+c\left({\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\right)\le {c}^{\prime}{\parallel u\parallel}^{2}+{c}^{\prime \prime},$

hence, proceeding as above and employing, in particular, (3.8),

$\frac{d}{dt}{\parallel u\parallel}^{2}+c\left({\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\right)\le {c}^{\prime},c>0.$(3.13)

Summing (3.9) and (3.13),
we obtain a differential inequality of the form

$\frac{d{E}_{1}}{dt}+c\left({E}_{1}+{\parallel \frac{\partial u}{\partial t}\parallel}^{2}\right)\le {c}^{\prime},c>0,$(3.14)

where

${E}_{1}={\parallel {A}_{k}^{1/2}u\parallel}^{2}+{B}_{k}^{1/2}[u]+2{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x+{\parallel u\parallel}^{2}$

satisfies, owing to (3.11),

${E}_{1}\ge c\left({\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\right)-{c}^{\prime},c>0.$(3.15)

Note indeed that

${E}_{1}\le c{\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+2{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\le c\left({\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\right)-{c}^{\prime},c>0,{c}^{\prime}\ge 0.$

It follows from (3.14)–(3.15) and Gronwall’s lemma that

${\parallel u(t)\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}\le c{e}^{-{c}^{\prime}t}\left({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F({u}_{0})\mathit{d}x\right)+{c}^{\prime \prime},{c}^{\prime}>0,t\ge 0,$(3.16)

and

${\int}_{t}^{t+r}{\parallel \frac{\partial u}{\partial t}\parallel}^{2}\mathit{d}s\le c{e}^{-{c}^{\prime}t}\left({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F({u}_{0})\mathit{d}x\right)+{c}^{\prime \prime},{c}^{\prime}>0,t\ge 0,r>0\text{given}.$(3.17)

Next, we multiply (3.4) by ${A}_{k}u$ and find, owing to the interpolation inequality (3.10),

$\frac{d}{dt}{\parallel {A}_{k}^{1/2}u\parallel}^{2}+c{\parallel u\parallel}_{{H}^{2k}(\mathrm{\Omega})}^{2}\le c({\parallel u\parallel}^{2}+{\parallel f(u)\parallel}^{2}).$

It follows from the continuity of *f* and *F*, the continuous embedding
${H}^{k}(\mathrm{\Omega})\subset \mathcal{\mathcal{C}}(\overline{\mathrm{\Omega}})$ (recall that $k\ge 2$)
and (3.16) that

${\parallel u\parallel}^{2}+{\parallel f(u)\parallel}^{2}\le Q({\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})})\le {e}^{-ct}Q({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})})+{c}^{\prime},c>0,t\ge 0,$(3.18)

so that

$\frac{d}{dt}}{\parallel {A}_{k}^{1/2}u\parallel}^{2}+c{\parallel u\parallel}_{{H}^{2k}(\mathrm{\Omega})}^{2}\le {e}^{-{c}^{\prime}t}Q({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})})+{c}^{\prime \prime},c,{c}^{\prime}>0,t\ge 0.$(3.19)

Summing (3.14) and (3.19), we have a differential inequality of the form

$\frac{d{E}_{2}}{dt}+c\left({E}_{2}+{\parallel u\parallel}_{{H}^{2k}(\mathrm{\Omega})}^{2}+{\parallel \frac{\partial u}{\partial t}\parallel}^{2}\right)\le {e}^{-{c}^{\prime}t}Q({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})})+{c}^{\prime \prime},c,{c}^{\prime}>0,t\ge 0,$

where

${E}_{2}={E}_{1}+{\parallel {A}_{k}^{1/2}u\parallel}^{2}$

satisfies

${E}_{2}\ge c\left({\parallel u\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}+{\int}_{\mathrm{\Omega}}F(u)\mathit{d}x\right)-{c}^{\prime},c>0.$

We then rewrite (3.4) as an elliptic equation, for $t>0$ fixed,

${A}_{k}u=-\frac{\partial u}{\partial t}-{B}_{k}u-f(u),{\mathcal{\mathcal{D}}}^{\alpha}u=0\mathit{\hspace{1em}}\text{on}\mathrm{\Gamma},|\alpha |\le k-1.$(3.20)

Multiplying (3.20) by ${A}_{k}u$, we obtain, owing to the interpolation inequality (3.10),

${\parallel {A}_{k}u\parallel}^{2}\le c\left({\parallel u\parallel}^{2}+{\parallel f(u)\parallel}^{2}+{\parallel \frac{\partial u}{\partial t}\parallel}^{2}\right),$

hence, owing to (3.18),

${\parallel u\parallel}_{{H}^{2k}(\mathrm{\Omega})}^{2}\le c\left({e}^{-{c}^{\prime}t}Q({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})})+{\parallel \frac{\partial u}{\partial t}\parallel}^{2}\right)+{c}^{\prime \prime},{c}^{\prime}>0.$(3.21)

Next, we differentiate (3.4) with respect to time and find

$\frac{\partial}{\partial t}\frac{\partial u}{\partial t}+{A}_{k}\frac{\partial u}{\partial t}+{B}_{k}\frac{\partial u}{\partial t}+{f}^{\prime}(u)\frac{\partial u}{\partial t}=0,$(3.22)${\mathcal{\mathcal{D}}}^{\alpha}\frac{\partial u}{\partial t}=0\mathit{\hspace{1em}}\text{on}\mathrm{\Gamma},|\alpha |\le k-1,$(3.23)${\frac{\partial u}{\partial t}|}_{t=0}=-{A}_{k}{u}_{0}-{B}_{k}{u}_{0}-f({u}_{0}).$(3.24)

We can note that, if ${u}_{0}\in {H}^{2k}(\mathrm{\Omega})\cap {H}_{0}^{k}(\mathrm{\Omega})\phantom{\rule{veryverythickmathspace}{0ex}}(=D({A}_{k}))$,
then $\frac{\partial u}{\partial t}(0)\in {L}^{2}(\mathrm{\Omega})$ and

$\parallel \frac{\partial u}{\partial t}(0)\parallel \le Q({\parallel {u}_{0}\parallel}_{{H}^{2k}(\mathrm{\Omega})}).$(3.25)

We multiply (3.22) by $\frac{\partial u}{\partial t}$ and have, owing to (3.6) and
the interpolation
inequality (3.10),

$\frac{d}{dt}{\parallel \frac{\partial u}{\partial t}\parallel}^{2}+c{\parallel \frac{\partial u}{\partial t}\parallel}_{{H}^{k}(\mathrm{\Omega})}^{2}\le {c}^{\prime}{\parallel \frac{\partial u}{\partial t}\parallel}^{2},c>0.$(3.26)

It follows from (3.17) (for $r=1$), (3.26) and the uniform
Gronwall’s lemma that

${\parallel \frac{\partial u}{\partial t}(t)\parallel}^{2}\le {e}^{-ct}Q({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})})+{c}^{\prime},c>0,t\ge 1,$(3.27)

and from (3.25)–(3.26) and Gronwall’s lemma that

${\parallel \frac{\partial u}{\partial t}(t)\parallel}^{2}\le {e}^{ct}Q({\parallel {u}_{0}\parallel}_{{H}^{2k}(\mathrm{\Omega})}),t\ge 0.$(3.28)

We finally deduce from (3.21) and (3.27)–(3.28) that

${\parallel u(t)\parallel}_{{H}^{2k}(\mathrm{\Omega})}\le {e}^{-ct}Q({\parallel {u}_{0}\parallel}_{{H}^{k}(\mathrm{\Omega})})+{c}^{\prime},c>0,t\ge 1,$(3.29)

and

${\parallel u(t)\parallel}_{{H}^{2k}(\mathrm{\Omega})}\le {e}^{-ct}Q({\parallel {u}_{0}\parallel}_{{H}^{2k}(\mathrm{\Omega})})+{c}^{\prime},c>0,t\ge 0.$(3.30)

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