Proof (i) Define the function $j:\lambda \mapsto J(\lambda u)$ for $\lambda> 0$ . Then

$$ \begin{align*} j(\lambda) :=&J(\lambda u) =\frac{\lambda^{2}}{2}\|\nabla u\|_2^{2}+\frac{\lambda^{2}}{2}\|\Delta u\|_2^{2}+\lambda^{3}\displaystyle\int_{\Omega}|\nabla u|^{2}\Delta u\textrm{ d}x-\frac{\lambda^{p+1}}{p+1}\|u\|_{p+1}^{p+1}. \end{align*} $$

We obtain $\operatorname *{\lim }\limits _{\lambda \rightarrow 0^{+}} J(\lambda u)=0$ and $\operatorname *{\lim }\limits _{\lambda \rightarrow +\infty } J(\lambda u)=-\infty $ .

(ii) Elementary calculations imply that

$$ \begin{align*} j'(\lambda) =&\lambda\|\nabla u\|_2^{2}+\lambda\|\Delta u\|_2^{2}+3\lambda^{2}\displaystyle\int_{\Omega}|\nabla u|^{2}\Delta u\textrm{d}x-\lambda^{p}\|u\|_{p+1}^{p+1}. \end{align*} $$

Let $k(\lambda ):=\lambda ^{-2} j'(\lambda )$ . After direct calculation, we have

$$ \begin{align*} k'(\lambda) =&-\lambda^{-2}\|\nabla u\|_2^{2}-\lambda^{-2}\|\Delta u\|_2^{2}-(p-2)\lambda^{p-3}\| u\|_{p+1}^{p+1}<0. \end{align*} $$

Since $\operatorname *{\lim }\limits _{\lambda \rightarrow 0^{+}} k(\lambda )=+\infty $ , $\operatorname *{\lim }\limits _{\lambda \rightarrow +\infty } k(\lambda )=-\infty $ , there exists a unique constant $\lambda ^{*}>0$ such that $k(\lambda )>0$ for $0<\lambda <\lambda ^{*}$ , $k(\lambda )<0$ for $\lambda ^{*}<\lambda <+\infty $ , and $k(\lambda ^{*})=0$ . By $j'(\lambda )=\lambda ^{2}k(\lambda )$ , $I(\lambda u)=\lambda j'(\lambda )$ , cases (ii) and (iii) hold.

Proof Employing H $\ddot {\textrm {o}}$ lder’s inequality, we obtain

$$ \begin{align*} -\int_{\Omega}|\nabla u|^{2}\Delta u\textrm{d}x\leq \left(\int_{\Omega}|\nabla u|^{4}\textrm{d}x\right)^{\frac{1}{2}}\left(\int_{\Omega}|\Delta u|^{2}\textrm{ d}x\right)^{\frac{1}{2}}\leq B_{2}^{2}\|\Delta u\|_{2}^{3}. \end{align*} $$

Fix $u\in \mathcal {N}$ . Since $p>2$ and by (1.4),

$$ \begin{align*} \|\Delta u\|_2^{2}&\leq\|u\|_{p+1}^{p+1}+3B_{2}^{2}\|\Delta u\|_{2}^{3}\leq B_{1}^{p+1}\|\Delta u\|_{2}^{p+1}+3B_{2}^{2}\|\Delta u\|_{2}^{3}, \end{align*} $$

where $B_1$ is the optimal constant in the embedding $H_0^{2}(\Omega )\hookrightarrow L^{p+1}(\Omega )$ , and $B_2$ is the optimal constant in the embedding $W_0^{1,2}(\Omega )\hookrightarrow L^{4}(\Omega )$ . Let

(2.1) $$ \begin{align} B_{0}:=\inf\left\{x\in(0,+\infty) |\ 1\leq B_1^{p+1}x^{p-1}+3B_2^{2}x\right\}. \end{align} $$

Then $ J(u) =\frac {1}{6}\|\nabla u\|_2^{2}+\frac {1}{6}\|\Delta u\|_2^{2}+\frac {p-2}{3(p+1)}\| u\|_{p+1}^{p+1}+\frac {1}{3}I(u) \geq \frac {1}{6}\|\Delta u\|_2^{2}\geq \frac {1}{6}B_{0}^{2}>0$ . Therefore, $d=\operatorname *{\inf }\limits _{u\in \mathcal {N}}J(u)>0$ .

Proof If $I_{\delta }(u)<0$ and $ \delta \|\Delta u\|_2^{2}<\delta \|\Delta u\|_2^{2}+\|\nabla u\|_{2}^{2}\leq B_1^{p+1}\|\Delta u\|_{2}^{p+1}+3B_2^{2}\|\Delta u\|_{2}^{3}$ , then $\|\Delta u\|_2>r(\delta )$ , and hence, cases (i) and (ii) hold. If $I_{\delta }(u)=0$ , we get $\|\Delta u\|_2\geq r(\delta )$ . If $\|\Delta u\|_2=0$ , $I_{\delta }(u)=0$ .

Proof For any $u\in \mathcal {N}$ , by the definition of d, we obtain

$$ \begin{align*} d&=\operatorname*{\inf}\limits_{u\in \mathcal{N}}J(u)\leq\frac{1}{6}B_{3}^{2}\|\Delta u\|_2^{2}+\frac{1}{6}\|\Delta u\|_2^{2}+\frac{p-2}{3(p+1)}B_{1}^{p+1}\|\Delta u\|_{2}^{p+1}, \end{align*} $$

which indicates that $ \frac {1}{6}B_{3}^{2}\|\Delta u\|_2^{2}+\frac {1}{6}\|\Delta u\|_2^{2}\geq \frac {d}{2}$ or $\frac {p-2}{3(p+1)}B_{1}^{p+1}\|\Delta u\|_{2}^{p+1}\geq \frac {d}{2}$ . Then $\|\Delta u\|_2\geq \left (\frac {3d}{B_{3}^{2}+1}\right )^{\frac {1}{2}}$ , or $\|\Delta u\|_{2}\geq \left (\frac {3d(p+1)}{2(p-2) B_{1}^{p+1}}\right )^{\frac {1}{p+1}}$ , where $B_{3}$ is the optimal constant in $W_0^{1,2}(\Omega )\hookrightarrow L^{2}(\Omega )$ . Let

(2.2) $$ \begin{align} C_{0}:=\min\left\{\left(\frac{3d}{B_{3}^{2}+1}\right)^{\frac{1}{2}},\left[\frac{3d(p+1)}{2(p-2) B_{1}^{p+1}}\right]^{\frac{1}{p+1}}\right\}. \end{align} $$

Then $\textrm {dist}(0,\mathcal {N})=\operatorname *{\inf }\limits _{u\in \mathcal {N}}\|\Delta u\|_2\geq C_{0}>0$ . For any $u\in \mathcal {N}_{-}$ , we have $ \|\Delta u\|_2^{2}\leq B_1^{p+1}\|\Delta u\|_{2}^{p+1}+3B_2^{2}\|\Delta u\|_{2}^{3}$ , which implies $ \|\Delta u\|_2\geq B_{0}$ . Here, $B_{0}$ is given in (2.1). Then $\textrm {dist}(0,\mathcal {N}_{-})=\operatorname *{\inf }\limits _{u\in \mathcal {N}^{-}}\|\Delta u\|_2\geq B_{0}>0$ .

Proof For any $s>d$ and $p>2$ , $u\in J^{s}\cap \mathcal {N}_{+}$ , we have

$$ \begin{align*} s>J(u)=\frac{1}{6}\|\nabla u\|_2^{2}+\frac{1}{6}\|\Delta u\|_2^{2}+\frac{p-2}{3(p+1)}\| u\|_{p+1}^{p+1}+\frac{1}{3}I(u)>\frac{1}{6}\|\Delta u\|_2^{2}, \end{align*} $$

which yields $ \|\Delta u\|_2<C_{3}$ .

Proof For $u\in \mathcal {N}^{s}$ , we have $ \frac {1}{6}B_{4}^{-2}\|u\|_2^{2}\leq \frac {1}{6}\|\Delta u\|_2^{2}\leq J(u)<s$ , where $B_{4}$ is the optimal constant in the embedding $H_0^{2}(\Omega )\hookrightarrow L^{2}(\Omega )$ . Then $ \|u\|_{2}\leq K_{1}$ . By (2.3), we have

$$ \begin{align*} \|\Delta u\|_2^{2}&\leq\|u\|_{p+1}^{p+1}+3\|\nabla u\|_{4}^{2}\|\Delta u\|_{2}\\ &\leq C_{2}^{p+1}\|\Delta u\|_{2}^{b(p+1)}\| u\|_{2}^{(1-b)(p+1)}+3C_{1}^{2}\|\Delta u\|_{2}^{2a+1}\| u\|_{2}^{2(1-a)}, \end{align*} $$

and hence,

(2.4) $$ \begin{align} C_{2}^{p+1}\|\Delta u\|_{2}^{b(p+1)}\| u\|_{2}^{(1-b)(p+1)}\geq\frac{1}{2}\|\Delta u\|_2^{2},\quad \textrm{or} \quad 3C_{1}^{2}\|\Delta u\|_{2}^{2a+1}\| u\|_{2}^{2(1-a)}\geq\frac{1}{2}\|\Delta u\|_2^{2}. \end{align} $$

By the similar proof of Lemma 2.8, we get $\|\Delta u\|_{2}<C_{3}$ . Combining with (2.4), we have $ \| u\|_{2}\geq K_{2}$ or $\| u\|_{2}\geq K_{3}$ . Then $\|u\|_{2}\geq K_{4}>0$ ; hence, $\lambda _{s}>0$ .

Proof The proof is divided into two steps.

Step 1. Global existence. We would use the Galerkin’s approximation with some priori estimates. Let $\left \{\omega _{i}(x)\right \}$ be the orthogonal basis of $H_{0}^{2}(\Omega )$ . Construct the approximate solutions $u_{m}(x,t)$ of (1.1), $ u_{m}(x,t):=\operatorname *{\sum }\limits _{i=1}^{m} a_{mi}(t)\omega _{i}(x)$ , $m=1,2,\cdots $ , $i=1,2,\cdots ,m$ , which satisfy

(3.1) $$ \begin{align} &(u_{m}^{\prime},\omega_{i})+(\nabla u_{m},\nabla \omega_{i})+(\Delta u_{m},\Delta \omega_{i})-(2|\Delta u_{m}|^{2},\omega_{i})+(2|D^{2}u_{m}|^{2},\omega_{i})\nonumber\\&\quad=(|u_{m}|^{p-1}u_{m},\omega_{i}), \end{align} $$

and $u_{0m}:=\operatorname *{\sum }\limits _{i=1}^{m} b_{mi}(t)\omega _{i}(x)\rightarrow u_{0}(x)$ in $ H_{0}^{2}(\Omega )$ as $m\rightarrow +\infty $ .

By the standard theory of ODEs (e.g., the Peanos theorem), we deduce that the existence of a local solution to (3.1). Multiplying (3.1) by $a_{mi}'(t)$ , summing over i from 1 to m and integrating with respect to t, we have

(3.2) $$ \begin{align} \displaystyle\int_{0}^{t}\|u_{m}^{\prime}\|_{2}^{2}\textrm{d}\tau+J(u_{m}(x,t))=J(u_{m}(x,0)),\quad 0\leq t\leq T. \end{align} $$

Due to the convergence of $u_{0m}\rightarrow u_{0}(x)$ in $H_{0}^{2}(\Omega )$ , one has $ J(u_{m}(x,0))\rightarrow J(u_{0}(x))<d$ , $I(u_{m}(x,0))\rightarrow I(u_{0}(x))>0$ . Therefore, for sufficiently large m and any $0\leq t <+\infty $ , we obtain

(3.3) $$ \begin{align} \displaystyle\int_{0}^{t}\|u_{m}^{\prime}\|_{2}^{2}\textrm{d}\tau+J(u_{m})=J(u_{m}(x,0))<d, \quad I(u_{m}(x,0))>0, \end{align} $$

which implies that $u_{m}(x,0)\in \mathcal {W}$ for sufficiently large m.

By applying the similar discussion of Theorem 8 in [Reference Xu and Su15], one could show from (3.3) that $u_{m}(x,t)\in \mathcal {W}$ for large m and $0\leq t <+\infty $ . Thus, $I(u_{m}(x,t))>0$ , $J(u_{m}(x,t))<d$ for all $t\in [0,T]$ . Then

(3.4) $$ \begin{align} \frac{1}{6}\|\nabla u_{m}\|_2^{2}+\frac{1}{6}\|\Delta u_{m}\|_2^{2}+\frac{p-2}{3(p+1)}\|u_{m}\|_{p+1}^{p+1}<J(u_{m}(t))<d. \end{align} $$

In addition, by using (3.2–3.4), we get for some positive constant C,

(3.5) $$ \begin{align} \|u_{m}^{\prime}\|_{L^{2}(0,T;L^{2}(\Omega))}\leq C, \end{align} $$

(3.6) $$ \begin{align} \|u_{m}\|_{L^{\infty}(0,T;H_{0}^{2}(\Omega))}\leq C, \end{align} $$

(3.7) $$ \begin{align} \|u_{m}\|_{L^{\infty}(0,T;W_{0}^{1,2}(\Omega))}\leq C, \end{align} $$

(3.8) $$ \begin{align} \|u_{m}\|_{L^{\infty}(0,T;L^{p+1}(\Omega))}\leq C. \end{align} $$

By the uniform estimates (3.5–3.8), it was seen that the local solutions can be extended globally. Thus, by the standard diagonal method and the Aubin-Lions-Simon theorem, we know there exists a function u and a sequence of $\left \{u_{m}\right \}$ (still by $\left \{u_{m}\right \}$ ) such that for each $T>0$ , one could obtain

(3.9) $$ \begin{align} u_{m}^{\prime}\rightharpoonup u', \quad &\textrm{weakly in}\ L^{2}(0,T;L^{2}(\Omega)), \end{align} $$

(3.10) $$ \begin{align} u_{m}\rightharpoonup u, \quad &\textrm{weakly in} \ L^{\infty}(0,T;W_{0}^{1,2}(\Omega)), \end{align} $$

(3.11) $$ \begin{align} u_{m}\rightharpoonup u, \quad &\textrm{weakly in} \ L^{\infty}(0,T;H_{0}^{2}(\Omega)), \end{align} $$

(3.12) $$ \begin{align} |u_{m}|^{p-1}u_{m}\rightarrow |u|^{p-1}u, \quad &\textrm{strongly in} \ L^{\frac{p+1}{p}}(\Omega\times(0,T)), \end{align} $$

(3.13) $$ \begin{align} |\Delta u_{m}|^{2}\rightharpoonup |\Delta u|^{2}, \quad &\textrm{weakly star in} \ L^{\infty}(0,T;L^{1}(\Omega)), \end{align} $$

(3.14) $$ \begin{align} |D^{2}u_{m}|^{2}\rightharpoonup |D^{2}u|^{2}, \quad &\textrm{weakly star in} \ L^{\infty}(0,T;L^{1}(\Omega)), \end{align} $$

as $m\rightarrow +\infty $ . Fix $k\in \mathcal {N}$ . In order to show the limit function u in (3.9–3.14) is a weak solution of (1.1), we choose a function $v\in C^1 ([0,T];H_{0}^{2}(\Omega ))$ defined as $ v(x,t):=\operatorname *{\sum }\limits _{i=1}^{k} l_{i}(t)\omega _{i}(x)$ , where $\left \{l_{i}(t)\right \}_{i=1}^{k}$ are arbitrarily given $C^1$ functions. Taking $m\geq k$ in (3.1), multiplying (3.1) by $l_{i}(t)$ , summing for i from 1 to k, and integrating with respect to t, we obtain

(3.15) $$ \begin{align} &\displaystyle\int_{0}^{T} (u_{m}^{\prime},v)+(\nabla u_{m},\nabla v)+(\Delta u_{m},\Delta v)-(2|\Delta u_{m}|^{2}, v)+(2|D^{2} u_{m}|^{2}, v)\textrm{ d}t\nonumber\\&\quad=\displaystyle\int_{0}^{T}(|u_{m}|^{p-1}u_{m},v)\textrm{d}t. \end{align} $$

Taking $m\rightarrow +\infty $ in (3.15) and recalling the convergence yield that

(3.16) $$ \begin{align} \displaystyle\int_{0}^{T} (u',v)+(\nabla u,\nabla v)+(\Delta u,\Delta v)-(2|\Delta u|^{2}, v)+(2|D^{2} u|^{2}, v)\textrm{ d}t=\displaystyle\int_{0}^{T}(|u|^{p-1}u,v)\textrm{d}t. \end{align} $$

Since functions of the form in (3.15) are dense in $L^{2}(0,T;H_{0}^{2}(\Omega ))$ , (3.16) also holds for all $v\in L^{2}(0,T;H_{0}^{2}(\Omega ))$ . By the arbitrariness of $T>0$ , we know

$$ \begin{align*} (u',v)+(\nabla u,\nabla v)+(\Delta u,\Delta v)-(2|\Delta u|^{2},v)+(2|D^{2}u|^{2},v)=(|u|^{p-1}u,v),\quad \text{a.e.} \ t>0. \end{align*} $$

Then u is a global weak solution to problem (1.1). To prove (1.6), we first assume that u was smooth enough that $u_{t}\in L^{2}(0,T;H_{0}^{2}(\Omega ))$ . Taking $v=u_{t}$ in (3.16), it is seen that (1.6) is true. By the density of $L^{2}(0,T;H_{0}^{2}(\Omega ))$ in $L^{2}(\Omega \times (0,T))$ , we know (1.6) also holds for weak solutions to (1.1). The existence of global solutions to (1.1) is obtained.

Step 2. Decay rate. Taking $\varphi :=u$ in (1.5), we get $\frac {\textrm {d}}{\textrm {d}t} {\|u\|_2^{2}}=2(u_{t},u)=-2I(u)$ . From Lemma 2.6, we know that $u(x,t)\in \mathcal {W}_{\delta }$ for $\delta _{1} < \delta <\delta _{2}$ and $0 < t < \infty $ under the condition $J(u_0) < d$ and $I(u_0)> 0$ . Thus, $I_{\delta _{1}}(u)\geq 0$ for $0 < t < \infty $ . Therefore,

$$ \begin{align*} \frac{\textrm{d}}{\textrm{d}t}\frac{\|u\|_2^{2}}{2}=-I(u)&=(\delta_{1}-1)\|\Delta u\|_{2}^{2}-I_{\delta_{1}}(u) \leq (\delta_{1}-1)B_{4}^{-2}\| u\|_{2}^{2}, \end{align*} $$

where $B_{4}>0$ is the best embedding constant from $H_{0}^{2}(\Omega )$ to $L^{2}(\Omega )$ (i.e., $\|u\|_{2}\leq B_{4}\|\Delta u\|_{2}$ for $\forall u \in H_{0}^{2}(\Omega )$ ). Consequently, $ \|u\|_2^{2}\leq \|u_{0}\|_2^{2}\textrm {e}^{-\hat C t}$ with $\hat {C}:=2B_{4}^{-2}(1-\delta _{1})>0$ .

Proof We employ the concavity method. Assume on the contrary that u was a global weak solution to (1.2) with $J(u_0) < d$ , $J(u_0)\not =0$ , $I(u_0) < 0$ and define $ F(t):= \int _{0}^{t}\|u\|_{2}^{2}\textrm {d}\tau $ , $t\geq 0$ . Then $F'(t) =\|u\|_{2}^{2}$ ,

(3.17) $$ \begin{align} F"(t) = 2(u,u_{t})=-2I(u). \end{align} $$

By (1.6), (1.4), and (3.17), we have

$$ \begin{align*} F"(t)&\geq-6J(u)+\|\Delta u\|_{2}^{2}=-6J(u_{0})+6\displaystyle\int_{0}^{t}\|u'\|_{2}^{2}\textrm{d}\tau+\|\Delta u\|_{2}^{2}. \end{align*} $$

Noticing that $ \frac {1}{4}(F'(t)-F'(0))^{2}= ( \int _{0}^{t} \int _{\Omega }uu'\textrm {d}x\textrm {d}\tau )^{2}\leq \int _{0}^{t}\|u\|_{2}^{2}\textrm {d}\tau \int _{0}^{t}\|u'\|_{2}^{2}\textrm {d}\tau $ , we have

(3.18) $$ \begin{align} &F"(t)F(t)-\frac{3}{2}\left(F'(t)\right)^{2}\geq B_{4}^{-2}F(t)F'(t)-6J(u_{0})F(t)-3F'(0)F'(t). \end{align} $$

In the forthcoming proof, the cases that $J(u_0)< 0$ and $ 0< J(u_{0}) < d$ will be discussed separately.

(i) If $J(u_0)< 0$ , then (3.18) implies $ F"(t)F(t)-\frac {3}{2}\left (F'(t)\right )^{2} \geq B_{4}^{-2}F(t)F'(t)-3F'(0)F'(t)$ . Now we will prove that $I(u) < 0$ for all $t> 0$ . Otherwise, there must be a constant $t_{0}> 0$ such that $I(u(t_{0})) = 0$ and $I(u) < 0$ for $0 \leq t < t_{0}$ . From Lemma 2.3 (i, iii), $\|\Delta u\|_{2}> r(1)$ for $0 \leq t < t_{0}$ , and $\|\Delta u(t_{0})\|_{2} \geq r(1)$ , where $\|\Delta u(t_{0})\|_{2} \not = 0$ . In fact, using (1.6) and $J(u_0)< 0$ , we have $J(u(t_0))< 0$ . Since $0=F"(t_0)\geq -6J(u(t_0))+\|\Delta u(t_0)\|_{2}^{2}$ , we have $\|\Delta u(t_{0})\|_{2} \not = 0$ . Hence, we have $J(u(t_{0})) \geq d$ , which contradicts to (1.6). By (3.17), we get $F"(t)>0$ for $t\geq 0$ . Since $F'(0) =\|u_{0}\|_{2}^{2}\geq 0$ , $F'(t)>0$ , for large t, $F(t)>3B_{4}^{2}\|u_{0}\|_{2}^{2}$ , we have

(3.19) $$ \begin{align} F"(t)F(t)-\frac{3}{2}\left(F'(t)\right)^{2}>0. \end{align} $$

(ii) If $0<J(u_{0}) < d$ , then by Lemma 2.6 (ii), we know that $u(x,t)\in \mathcal {V}_{\delta }$ for $t\geq 0$ and $\delta _1 < \delta < \delta _2$ , where $\delta _1<1<\delta _2$ are the two roots of $d(\delta ) = J(u_0)$ . Hence, ${I_{\delta _2}(u)\leq 0}$ and $\|\Delta u\|_{2}\geq r(\delta _2)$ for $t\geq 0$ . It follows from (3.18) that for $t\geq 0$ , $ F"(t) \geq 2(\delta _2-1)r^{2}(\delta _2)$ , which shows for all $t \geq 0$ that $ F'(t)\geq 2(\delta _2-1)r^{2}(\delta _2)t$ and $ F(t)\geq (\delta _2-1)r^{2}(\delta _2)t^{2}$ . Therefore, for sufficiently large t, we have

$$ \begin{align*} B_{4}^{-2}F'(t)>12J(u_{0}),\quad B_{4}^{-2}F(t)>6F'(0). \end{align*} $$

Then we obtain (3.19). Due to $ (F^{-\frac {1}{2}}(t) )"=-\frac {1}{2}F^{-\frac {5}{2}}(t) [F(t)F"(t)-\frac {3}{2}\left (F'(t)\right )^{2} ]<0$ , $F^{-\frac {1}{2}}(t)$ is concave in $(0,+\infty )$ . So there exists a positive constant $t_{0}$ such that $F(t_0)>0$ , $F'(t_{0})>0$ and $\left (F^{-\frac {1}{2}}(t_{0})\right )'<0$ . Since $F(t)>0,F"(t)>0$ for $t\ge t_0$ and $F'(t_0)>0$ , one can find $(F^{-1/2})'(t)<0$ for $t\ge t_0$ , and hence, there is a constant $T>t_{0}$ such that $\operatorname *{\lim }\limits _{t\rightarrow T}F^{-\frac {1}{2}}(t)=0$ ; that is, $ \operatorname *{\lim }\limits _{t\rightarrow T} \int _{0}^{t}\|u(x,\tau )\|_{2}^{2}\textrm {d}\tau =+\infty $ .

Proof Let $\lambda _{k}=1-\frac {1}{k}$ , $k=2,3,\cdots $ . Consider the approximation problems

(4.1) $$ \begin{align} \left\{ \begin{array}{lll} u_{t}-\Delta u+\Delta^{2}u-\textrm{div}(2\nabla u\Delta u)+\Delta|\nabla u|^{2}=|u|^{p-1}u, & (x,t) \in \Omega\times (0,T),\\ u=0,\quad \frac{\partial u}{\partial \eta}=0, & (x,t) \in \partial\Omega\times (0,T),\\ u(x, 0)=u_{0}^{k}(x):=\lambda_{k}u_{0}(x),& x \in \Omega. \end{array}\right. \end{align} $$

Noticing that $I(u_0) \geq 0$ , by Lemma 2.1 (iii), we could deduce that there exists a unique constant $\lambda ^{*}=\lambda ^{*}(u_{0})\geq 1$ such that $I(\lambda ^{*}u_{0})=0$ . By $\lambda _{k}<1\leq \lambda ^{*}$ , we get $I(u_{0}^{k}) = I(\lambda _{k}u_{0})> 0$ and $J(u_{0}^{k})= J(\lambda _{k}u_{0}) <J(u_{0}) = d$ . In view of Theorem 3.1, for each k, problem (4.1) admits a global weak solution $u^{k}(x,t)\in L^{\infty }(0,\infty ;H_{0}^{2}(\Omega ))\cap \mathcal {W}$ with $u_{t}^{k}\in L^{2}(0,\infty ;L^{2}(\Omega ))$ for $0 \leq t < \infty $ satisfying $ \int _{0}^{t}\|u_{\tau }^{k}\|_{2}^{2}\textrm {d}\tau +J(u^{k})=J(u_{0}^{k})<d$ . Applying the similar discussion in Theorem 3.1, there exist a subsequence of $ \{u^{k} \}$ and a function u such that u is a weak solution of (1.2) with $I(u) \geq 0$ and $J(u) \leq d$ for $0 \leq t < \infty $ .

Let us discuss the decay rate of $\|u\|_{2}^{2}$ . First, suppose that $I(u)> 0$ for $0 < t < \infty $ ; then $u $ does not vanish in finite time. Combining with (3.17), we have $u_{t} \not \equiv 0$ . By (1.6), for small $t_0> 0$ , we have $ 0<J(u(t_{0}))<J(u_{0})=d$ . Taking $t = t_0$ as the initial time and by Lemma 2.6, we get $u\in \mathcal {W}_{\delta }$ for $\delta _1 < \delta < \delta _2$ and $t_0<t<\infty $ . Hence, $I_{\delta _1}(u) \geq 0$ for $t_0<t<\infty $ and

$$ \begin{align*} \frac{1}{2}\frac{\textrm{d}}{\textrm{d}t}\|u\|_{2}^{2}=-I(u)=(\delta_1-1)\|\Delta u\|_{2}^{2}-I_{\delta_1}(u)\leq B_{4}^{-2}(\delta_1-1)\|u\|_{2}^{2}, \end{align*} $$

which implies $ \|u\|_{2}^{2}\leq \|u(t_{0})\|_{2}^{2}\textrm {e}^{-2B_{4}^{-2}(1-\delta _1)(t-t_{0})}$ . The decay rate holds with $\hat C_{2} := 2B_{4}^{-2}(1-\delta _1)$ and $\hat C_{1}:=\|u(t_{0})\|_2^{2}\textrm {e}^{2 B_{4}^{-2}(1-\delta _{1})t_{0}}$ .

Next, suppose there is a positive constant $t_1$ such that $I(u)> 0$ for $0 < t < t_1$ and $I(u(x, t_1)) = 0$ . Obviously, $u_t \not \equiv 0$ for $0 < t < t_1$ and $ \int _{0}^{t_{1}}\|u_{\tau }\|_{2}^{2}\textrm {d}\tau>0$ . Applying (1.6) again, we have $ J(u(t_{1}))=J(u_{0})- \int _{0}^{t_{1}}\|u_{\tau }\|_{2}^{2}\textrm {d}\tau <J(u_{0})=d$ . By the definition of d, we know $\|u(t_1)\|_{2} = 0$ and $\|u\|_{2} \equiv 0$ for all $t \geq t_1$ . It is seen that such a function $u $ is a weak solution of (1.2) which vanishes in finite time.

Proof Similarly to the proof of Theorem 4.1, we could get

$$ \begin{align*} F"(t)F(t)-\frac{3}{2}\left(F'(t)\right)^{2} \geq B_{4}^{-2}F'(t)F(t)-6J(u_{0})F(t)-3F'(t)F'(0). \end{align*} $$

Since $J(u_0) = d$ , $I(u_0) < 0$ , by the continuity of $J(u)$ and $I(u)$ with respect to t, there exists a constant $t_0> 0$ such that $J(u(x, t))> 0$ and $I(u(x, t)) < 0$ for $0 < t \leq t_0$ . From $(u_t, u) =-I(u)$ , we have $u_t \not \equiv 0$ for $0 < t \leq t_0$ . Furthermore, we have $ J(u(t_{0}))=d- \int _{0}^{t_{0}}\|u_{\tau }\|_{2}^{2}\textrm {d}\tau <d$ . Taking $t = t_0$ as the initial time and by Lemma 2.6, we know that $u(x, t)\in \mathcal {V}_{\delta }$ for $\delta _1 < \delta < \delta _2$ and $t> t_0$ , where $\delta _1 < 1 < \delta _2$ are the two roots of the equation $d(\delta ) = J(u_{0})$ . Thus, $I_{\delta _2}(u)\leq 0$ for $t> t_0$ . The rest of the proof is similar to that of Theorem 3.2.

Proof Let $T_{\max }$ be the maximal existence time of (1.1). If $T_{\max } =\infty $ , we denote the $\omega $ -limit set of $u_0$ as $ \omega (u_{0})=\operatorname *{\bigcap }\limits _{t\geq 0}\overline {\left \{u(\tau ):\tau \geq t\right \}}^{H_{0}^{2}(\Omega )}$ .

(i) Assume that $u_0\in \mathcal {N}_{+}$ with $\|u_{0}\|_{2}\leq \lambda _J(u_{0})$ . We first claim that $u(t)\in \mathcal {N}_{+}$ for all $t\in [0, T_{\max })$ . If not, there would exist a constant $t_0\in (0, T_{\max })$ such that $u(t)\in \mathcal {N}_{+}$ for $0 \leq t < t_{0}$ and $I(u(t_0))=0$ . It follows from $I(u(t))=- \int _{\Omega }u_tudx$ that $u_t\not \equiv 0$ for $(x,t)\in \Omega \times [0, t_0)$ . Using (1.6), we have $J(u(t_0)) < J(u_0)$ , which deduces $u(t_0)\in J^{J(u_0)}$ . Therefore, $u(t_0)\in J^{J(u_{0})}\cap \mathcal {N}=\mathcal {N}^{J(u_0)}$ . By the definition of $\lambda _{J(u_0)}$ , we have

(5.1) $$ \begin{align} \lambda_{J(u_{0})}\leq\|u(t_{0})\|_{2}. \end{align} $$

Noticing $I(u(t))> 0$ for $t\in [0, t_0)$ , $ \|u(t_{0})\|_{2}<\|u_{0}\|_{2}\leq \lambda _{J(u_{0})}$ , a contradiction to (5.1). So $u(t)\in \mathcal {N}_{+}$ and $u(t)\in J^{J(u_0)}\cap \mathcal {N}_{+}$ for all $t\in [0, T_{\max })$ . Lemma 2.8 shows $ \|\Delta u\|_{2}<C_{3}$ , $t\in [0,T_{\max })$ , so that $T_{\max } =\infty $ . Let $\omega $ be an arbitrary element in $\omega (u_{0})$ . We have $ J(\omega )<J(u_{0})$ , $\|\omega \|_{2}<\lambda _{J(u_{0})}$ , which implies $\omega \notin \mathcal {N}^{J(u_{0})}$ . Recalling the definition of $\lambda _{J(u_0)}$ , $\omega (u_0)\cap \mathcal {N} = \emptyset $ . Hence, $\omega (u_0)=\left \{0\right \}$ . Therefore, the weak solution u of problem (1.1) exists globally and $u\rightarrow 0$ as $t\rightarrow +\infty $ .

(ii) Assume that $u_0\in \mathcal {N}_{-}$ with $\|u_{0}\|_{2}\geq \Lambda _{J(u_{0})}$ . We claim that $u(t)\in \mathcal {N}_{-}$ for all $t\in [0, T_{\max })$ . If not, there would be a constant $t_{0}\in (0, T_{\max })$ such that $u(t)\in \mathcal {N}_{-}$ for $0 \leq t < t_0$ and $I(u(t_{0}))=0$ . Similarly to case (i), one has $J(u(t_0)) < J(u_0)$ , which implies $u(t_0)\in J^{J(u_0)}$ , and $u(t_0)\in \mathcal {N}^{J(u_0)}$ . By the definition of $\Lambda _{J(u_0)}$ , we have

(5.2) $$ \begin{align} \Lambda_{J(u_{0})}\geq\|u(t_{0})\|_{2}. \end{align} $$

However, $I(u(t)) < 0$ for $t\in [0, t_0)$ , and we get $ \|u(t_{0})\|_{2}>\|u_{0}\|_{2}\geq \Lambda _{J(u_{0})}$ , a contradiction with (5.2). So $u(t)\in \mathcal {N}_{-}$ , $t\in [0,T_{\max })$ . Suppose $T_{\max } = \infty $ ; for every $\omega \in \omega (u_0)$ , we get $\|\omega \|_{2}>\Lambda _{J(u_{0})}$ , $J(\omega )<J(u_{0})$ , and hence, $\omega \in J^{J(u_{0})}$ and $\omega \not \in \mathcal {N}^{J(u_{0})}$ . Recalling the definition of $\Lambda _{J(u_0)}$ , we obtain $\omega (u_0)\cap \mathcal {N}=\emptyset $ . Thus, $\omega (u_0) = \left \{0\right \}$ , a contradiction to Lemma 2.7. Hence, $T_{\max } <+\infty $ .

Proof Let $K(t):=-J(u(t))$ . Then $L(0)>0$ and $K(0)>0$ . By (1.6), we get $ K'(t)=\|u'\|_{2}^{2}\geq 0$ , which implies $K(t)\geq K(0)>0$ for all $t\in [0,T)$ . By (1.4), we have

(6.2) $$ \begin{align} L'(t) =-I(u) \geq-3J(u) =3K(t). \end{align} $$

Combining (6.2) with the Cauchy inequality, we obtain

(6.3) $$ \begin{align} L(t)K'(t)=\frac{1}{2}\|u\|_{2}^{2}\|u'\|_{2}^{2}\geq\frac{1}{2}(u,u')^{2}=\frac{1}{2}(L'(t))^{2}\geq\frac{3}{2}L'(t)K(t). \end{align} $$

According to (6.3), $ (K(t)L^{-\frac {3}{2}}(t))'=L^{-\frac {5}{2}}(t)(K'(t)L(t)-\frac {3}{2}K(t)L'(t))\geq 0$ . Therefore,

(6.4) $$ \begin{align} 0<K(0)L^{-\frac{3}{2}}(0)\leq K(t)L^{-\frac{3}{2}}(t)\leq\frac{1}{3}L'(t)L^{-\frac{3}{2}}(t)=-\frac{2}{3}(L^{-\frac{1}{2}}(t))'. \end{align} $$

Integrating (6.4) with respect to t, we have $ \eta t\leq -\frac {2}{3} (L^{-\frac {1}{2}}(t)-L^{-\frac {1}{2}}(0) )$ , where ${\eta :=K(0)L^{-\frac {3}{2}}(0)}$ , and hence, we obtain that there exists a constant $T_{*}<+\infty $ such that $\operatorname *{\lim }\limits _{t\rightarrow T_{*}}L(t)=\infty $ (i.e., the weak solution blows up). Hence, (6.1) holds. Similarly, integrating (6.4) from t to $T_{*}$ , we have $ L(t)\leq \left [3\cdot 2^{-1}\eta \right ]^{-2}(T_{*}-t)^{-2}$ .

Proof We assert that for any $t\in [0,T)$ , $I(u)<0$ . From (1.4), $ I(u_{0}) \leq 3J(u_{0})-\frac {1}{2}\|\Delta u_{0}\|_{2}^{2} <0$ . If the assertion was not true, there would be a constant $t_{0}\in (0,T)$ such that $I(u)<0$ and $I(u(t_{0}))=0$ . From (6.2), we know $\|u\|_{2}^{2}$ is strictly increasing with $t\in [0,t_{0})$ . Therefore,

(6.5) $$ \begin{align} {J(u(t_{0}))\leq J(u_{0})<\frac{1}{6}\|\Delta u_{0}\|_{2}^{2}\leq \frac{1}{6}\|\Delta u(t_{0})\|_{2}^{2}.} \end{align} $$

From the definition of $J(u)$ and $I(u(t_{0}))=0$ , we have $J(u(t_{0}))>\frac {1}{6}\|\Delta u(t_{0})\|_{2}^{2}$ . It contradicts to (6.5). Therefore, we have $I(u)<0$ for any $t\in [0,T)$ , and $\|u\|_{2}^{2}$ is strictly increasing with respect to t. For any $T\in (0,T^{*})$ , $\rho>0$ , $\xi>0$ , we define an auxiliary function

$$ \begin{align*} G(t):=\displaystyle\int_{0}^{t}\|u\|_{2}^{2}\textrm{d}\tau+(T-t)\|u_{0}\|_{2}^{2}+\rho(t+\xi)^{2},\quad t\in[0,T^{*}]. \end{align*} $$

By direct calculation, we have $ G'(t) =\|u\|_{2}^{2}-\|u_0\|_{2}^{2}+2\rho (t+\xi )= \int _{0}^{t} \frac {\textrm {d}}{\textrm {d}\tau }\|u\|_{2}^{2}\textrm {d}\tau + 2\rho (t+\xi )=2 \int _{0}^{t}(u,u_{\tau })\textrm {d}\tau +2\rho (t+\xi )$ , and

$$ \begin{align*} G"(t)\geq-6J(u)+\|\Delta u\|_{2}^{2}+2\rho\geq-6J(u_{0})+6\displaystyle\int_{0}^{t}\|u_{\tau}\|_{2}^{2}\textrm{d}\tau+\|\Delta u\|_{2}^{2}+2\rho. \end{align*} $$

It is obvious that $G"(t)>0$ . Then $G'(t)$ is increasing on $[0,T]$ and $ G'(t)\geq G'(0)>0$ . This indicates that $G(t)$ is increasing on $[0,T]$ . Define

$$ \begin{align*} f(t):=\left[\displaystyle\int_{0}^{t}\|u\|_{2}^{2}\textrm{d}\tau+\rho(t+\xi)^{2} \kern-1pt\right]\left(\displaystyle\int_{0}^{t}\|u_{\tau}\|_{2}^{2}\textrm{ d}\tau+\rho\right)-\left[\displaystyle\int_{0}^{t}\left(u,u_{\tau}\right)\textrm{d}\tau+\rho(t+\xi) \kern-1pt\right]^{2}\!. \end{align*} $$

By Cauchy-Schwarz inequality, we have $f(t)\geq 0$ . For any $t\in [0,T]$ ,

$$ \begin{align*} G(t)G"(t)-\frac{3}{2}(G'(t))^{2}\geq G(t)\left[-6J(u_{0})+B_{4}^{-2}\|u_{0}\|_{2}^{2}-4\rho\right]\geq 0, \end{align*} $$

where $0<\rho \leq -\frac {3}{2}J(u_{0})+\frac {1}{4}B_{4}^{-2}\|u_{0}\|_{2}^{2}:=\hat \rho $ . From Lemma 2.10, $ T\leq \frac {T\|u_{0}\|_{2}^{2}+\rho \xi ^{2}}{\rho \xi }$ . Fixing $\rho _{0}\in (0,\hat \rho ]$ , we have $0<\frac {\|u_{0}\|_{2}^{2}}{\rho \xi }<1$ for any $\xi \in (\frac {\|u_{0}\|_{2}^{2}}{\rho _{0}},+\infty )$ . This indicates $ T\leq \frac {\rho _{0}\xi ^{2}}{\rho _{0}\xi -\|u_{0}\|_{2}^{2}}$ . The right side of the above formula takes the minimum value at $\xi :=\frac {2\|u_{0}\|_{2}^{2}}{\rho _{0}}$ . Then $ T\leq \frac {4\|u_{0}\|_{2}^{2}}{\rho _{0}}$ . For $\rho _{0}\in (0,\hat \rho ]$ , $T<T^{*} \leq \frac {16\|u_{0}\|_{2}^{2}}{B_{4}^{-2}\|u_{0}\|_{2}^{2}-6J(u_{0})}$ .

Proof Let $M(t):=\|\Delta u\|_{2}^{2}$ and $M(0)>0$ . When $N=1$ , problem (1.1) can be simplified to

(6.7) $$ \begin{align} \left\{ \begin{array}{lll} u_{t}-\Delta u+\Delta^{2}u=|u|^{p-1}u, & (x,t) \in \Omega\times (0,T),\\ u=0,\quad \frac{\partial u}{\partial \eta}=0, & (x,t) \in \partial\Omega\times (0,T),\\ u(x, 0)=u_{0}(x),& x \in \Omega. \end{array}\right. \end{align} $$

By computation, we have

(6.8) $$ \begin{align} M'(t)=2\displaystyle\int_{\Omega}(\Delta u-\Delta^{2}u+|u|^{p-1}u)\Delta^{2}u\textrm{d}x. \end{align} $$

Let us estimate the integrals in (6.8). We have

(6.9) $$ \begin{align} 2\displaystyle\int_{\Omega}\Delta^{2}u \Delta u\textrm{d}x &\leq\frac{1}{\lambda}\displaystyle\int_{\Omega}|\Delta u|^{2}\textrm{ d}x+\lambda\displaystyle\int_{\Omega}(\Delta^{2} u)^{2}\textrm{d}x, \end{align} $$

(6.10) $$ \begin{align} 2\displaystyle\int_{\Omega} |u|^{p-1}u\Delta^{2}u\textrm{d}x &\leq\frac{1}{\mu}\displaystyle\int_{\Omega}|u|^{2p}\textrm{ d}x+\mu\displaystyle\int_{\Omega}(\Delta^{2} u)^{2}\textrm{d}x, \end{align} $$

with two arbitrary positive constants $\lambda $ , $\mu $ . By replacing (6.9) and (6.10) in (6.8), we obtain

(6.11) $$ \begin{align} M'(t)\leq\frac{1}{\mu}\displaystyle\int_{\Omega}|u|^{2p}\textrm{d}x+\frac{1}{\lambda}\displaystyle\int_{\Omega}|\Delta u|^{2}\textrm{ d}x+(\mu+\lambda-2)\displaystyle\int_{\Omega}(\Delta^{2}u)^{2}\textrm{d}x. \end{align} $$

Choosing $\lambda +\mu \leq 2$ , inequality (6.11) reduces to $ M'(t)\leq \tilde \alpha M^{p}(t)+\tilde \beta M(t)$ . Then we get (6.6).

Proof When $N=1$ , we get (6.7). First, we give the upper bound of extinction rate. Multiplying the equation in (6.7) by u and integrating it over $\Omega \times (t,t+h)$ with $h>0$ and then dividing the result by h yields that

(7.2) $$ \begin{align} &\frac{1}{h}\displaystyle\int_{t}^{t+h}\displaystyle\int_{\Omega}u_{\tau}u\textrm{d}x\textrm{d}\tau+\frac{1}{h}\displaystyle\int_{t}^{t+h}\|\nabla u\|_{2}^{2}\textrm{d}\tau +\frac{1}{h}\displaystyle\int_{t}^{t+h}\|\Delta u\|_{2}^{2}\textrm{d}\tau =\frac{1}{h}\displaystyle\int_{t}^{t+h}\|u\|_{p+1}^{p+1}\textrm{d}\tau. \end{align} $$

Let $h\rightarrow 0^{+}$ in (7.2). Using the Lebesgue differentiation theorem in [Reference Antontsev and Shmarev2], one could obtain

(7.3) $$ \begin{align} H'(t)+2\|\nabla u\|_{2}^{2}+2\|\Delta u\|_{2}^{2}=2\|u\|_{p+1}^{p+1}, \end{align} $$

where $H(t):=\|u\|_{2}^{2}$ . According to the embedding relationship $W_{0}^{1,2}(\Omega )\hookrightarrow L^{2}(\Omega )\hookrightarrow L^{p+1}(\Omega )$ and $H_{0}^{2}(\Omega )\hookrightarrow L^{2}(\Omega )$ , we have $\|u\|_{2}\leq B_{3}\|\nabla u\|_{2}$ , $\|u\|_{2}\leq B_{4}\|\Delta u\|_{2}$ , $\|u\|_{p+1}\leq B_{5}\|u\|_{2}$ . Then (7.3) can be written $ H'(t)+2D_{1}H(t)\leq 2D_{2}H^{\frac {p+1}{2}}(t)$ , where $D_{1}:=B_{3}^{-2}+B_{4}^{-2}$ , $D_{2}:=B_{5}^{p+1}$ . Defining $\varphi (t):=H^{\frac {1-p}{2}}(t)$ , we get the following differential inequality:

(7.4) $$ \begin{align} \varphi'(t)\leq(1-p)\left(D_{2}-D_{1}\varphi(t)\right):=\zeta(t). \end{align} $$

It is clear from (7.4) that $\zeta (0) < 0$ . Recalling the continuity of $\zeta (t)$ , there exists a sufficiently small $T_0> 0$ such that $ \zeta (t) <\frac {\zeta (0)}{2}<0$ for $0<t\leq T_{0}$ . Then $\varphi '(t) \leq {\zeta (0)}/{2}$ , which implies that

$$ \begin{align*} \left\{ \begin{array}{llll} \varphi(t)\leq\varphi(0)+\frac{\zeta(0)t}{2},\ &0<t<T_{1},\\ \varphi(t)=0,\ &t\geq T_{1}. \end{array} \right. \end{align*} $$

Obviously, by the definition of $\zeta (t)$ , (7.1) holds.

Proof The proof of a upper bound of extinction rate for u is the same as Theorem 7.1. We only give the proof of a lower bound of extinction rate. Let $H(t):=\|u\|_{2}^{2}$ . We have

(7.6) $$ \begin{align} H'(t)&\geq-6J(u_{0})+\|\nabla u\|_{2}^{2}+\|\Delta u\|_{2}^{2}+\frac{2(p-2)}{p+1}\|u\|_{p+1}^{p+1}\nonumber\\ &\geq (B_{3}^{-2}+B_{4}^{-2})H(t)+\frac{2(p-2)}{p+1}B_{5}^{p+1}H^{\frac{p+1}{2}}(t). \end{align} $$

By the definition of $\varphi (t)$ in Theorem 7.1, we have $ \varphi '(t)\geq D_{3}\varphi (t)-D_{4}$ ; hence, we get (7.5).

Proof Let $H(t):=\|u\|_{2}^{2}$ . In the forthcoming proof, the cases $p<1$ , $p=1$ , and $p\geq 2$ will be discussed separately.

(i) $p<1$ , $\|u_{0}\|_{2}^{1-p}>D_{4}D_{3}^{-1}$ . By the similar proof of Corollary 7.1, we have

$$ \begin{align*} \|u\|_{2}\geq\left[\left(\|u_{0}\|_{2}^{1-p}-D_{4}D_{3}^{-1}\right)\textrm{e}^{D_{3}t}+D_{4}D_{3}^{-1}\right]^{\frac{1}{1-p}}. \end{align*} $$

(ii) $p=1$ . By (7.6), we obtain $ H'(t)\geq D_{5}H(t)$ , where $D_{5}:=B_{3}^{-2}+B_{4}^{-2}{-B_5^2}$ . Hence, $\|u\|_{2}\geq \|u_{0}\|_{2}^{2}\textrm {e}^{\frac {D_{5}}{2}t}$ .

(iii) $p\geq 2$ . By (7.6), we obtain $ H'(t)\geq D_{6}H(t)$ , where $D_{6}:=B_{3}^{-2}+B_{4}^{-2}$ . Hence, $\|u\|_{2}\geq \|u_{0}\|_{2}^{2}\textrm {e}^{\frac {D_{6}}{2}t}$ .