It is well known that critical groups and Morse theory are the main tools in solving elliptic partial differential equation. Let us recall some results which will be used later. We refer the readers to the book [16] for more information on Morse theory.

Let

*H* be a Hilbert space and

$I\in {C}^{1}(H,\mathbb{R})$ be a functional satisfying the (PS) condition or (C) condition, and

${H}_{q}(X,Y)$ be the

*q* th singular relative homology group with integer coefficients. Let

${u}_{0}$ be an isolated critical point of

*I* with

$I({u}_{0})=c$,

$c\in \mathbb{R}$, and

*U* be a neighborhood of

${u}_{0}$. The group

${C}_{q}(I,{u}_{0}):={H}_{q}({I}^{c}\cap U,{I}^{c}\cap U\setminus \{{u}_{0}\}),\phantom{\rule{1em}{0ex}}q\in Z$

is said to be the *q* th critical group of *I* at ${u}_{0}$, where ${I}^{c}=\{u\in H:I(u)\le c\}$.

Let

$K:=\{u\in H:{I}^{\prime}(u)=0\}$ be the set of critical points of

*I* and

$a<infI(K)$, the critical groups of

*I* at infinity are formally defined by [

17]

${C}_{q}(I,\mathrm{\infty}):={H}_{q}(H,{I}^{a}),\phantom{\rule{1em}{0ex}}q\in Z.$

From the deformation theorem, we see that the above definition is independent of the particular choice of

$c<infI(K)$. If

$c<infI(K)$ then

${C}_{q}(I,\mathrm{\infty}):={H}_{q}(H,{\dot{I}}^{a}),\phantom{\rule{1em}{0ex}}q\in Z.$

(20)

For the convenience of our proof, we first recall two interesting results and prove two important propositions.

**Proposition 3.1** [18]

*Under* (H_{2}), *if* $u\in E:={H}^{2}(\mathrm{\Omega})\cap {H}_{0}^{1}(\mathrm{\Omega})$ *is an isolated critical point of* *I*, *then* ${C}_{\ast}(I,u)\cong {C}_{\ast}(I{|}_{{C}_{0}^{3}(\mathrm{\Omega})},u)$.

**Proposition 3.2** [19]

*If* ${D}_{1}\subset D\subset {E}_{0}\subset {E}_{1}\subset X$ *and for some integer* $k\ge 0$ *we have* ${H}_{k}(E,D)\ne 0$ *and* ${H}_{k}({E}_{1},{D}_{1})=0$, *then either* ${H}_{k+1}({E}_{1},E)\ne 0$ *or* ${H}_{k-1}(D,{D}_{1})\ne 0$.

**Proposition 3.3** *If the assumptions of Theorem * 2.1

*hold*,

*then* ${C}_{k}(I,\mathrm{\infty})=0\phantom{\rule{1em}{0ex}}\mathit{\text{for all integers}}k\ge 0.$

*Proof* Under the guidance of [

8] and [

18], we begin to prove this result. Let

${I}_{1}=I{|}_{{C}_{0}^{3}(\overline{\mathrm{\Omega}})}$. Indeed, it follows from above Proposition 3.1 that

*I* and

${I}_{1}$ have same critical set. Since

${C}_{0}^{3}(\overline{\mathrm{\Omega}})$ is dense in

*E*, invoking Proposition 16 of Palais [

20], we have

${H}_{k}(E,{\dot{I}}^{a})={H}_{k}({C}_{0}^{3}(\overline{\mathrm{\Omega}}),{\dot{I}}_{1}^{a})\phantom{\rule{1em}{0ex}}\text{for all}a\in R\text{and all integers}k\ge 0.$

(21)

From equations (

20) and (

21), we see that in order to prove the proposition, it suffices to show that

${H}_{k}({C}_{0}^{3}(\overline{\mathrm{\Omega}}),{I}_{1}^{a})=0\phantom{\rule{1em}{0ex}}\text{for all}a0\text{with}|a|\text{large and all integers}k\ge 0.$

(22)

In order to prove equation (

22), we proceed as follows. We define the sets

$\partial {B}_{1}^{c}=\{u\in {C}_{0}^{3}(\overline{\mathrm{\Omega}}):{\parallel u\parallel}_{{C}_{0}^{3}(\overline{\mathrm{\Omega}})}=1\}$

and

$\partial {B}_{1,+}^{c}=\{u\in \partial {B}_{1}^{c}:u(x)>0\text{for some}x\in \mathrm{\Omega}\}.$

Consider the map

${h}_{+}:[0,1]\times \partial {B}_{1,+}^{c}\to \partial {B}_{1,+}^{c}$ defined by

${h}_{+}(t,u)=\frac{(1-t)u+t{\varphi}_{1}}{{\parallel (1-t)u+t{\varphi}_{1}\parallel}_{{C}_{0}^{3}(\overline{\mathrm{\Omega}})}}\phantom{\rule{1em}{0ex}}\text{for all}(t,u)\in [0,1]\times \partial {B}_{1,+}^{c}.$

Clearly, ${h}_{+}$ is a continuous homotopy and $h(1,u)={\varphi}_{1}$ for all $x\in \partial {B}_{1,+}^{c}$. Therefore, $\partial {B}_{1,+}^{c}$ is contractible in itself.

By equation (

3) in (H

_{4}), given any

$\gamma >0$, we can find

$C=C(\gamma )>0$ such that

$F(x,t)\ge \frac{\gamma}{2}{t}^{2}\phantom{\rule{1em}{0ex}}\text{for all}x\in \mathrm{\Omega}\text{and all}t\ge C.$

(23)

Similarly, from condition (H

_{3}), and by choosing

$C>0$ even bigger if necessary, we observe that there is a number

${\gamma}_{0}>0$ such that

$F(x,t)\ge -\frac{{\gamma}_{0}}{2}{t}^{2}\phantom{\rule{1em}{0ex}}\text{for all}x\in \mathrm{\Omega}\text{and all}t\le -C(\gamma ).$

(24)

Moreover, by condition (H

_{2}), we have

$|F(x,t)|\le {C}_{3}\phantom{\rule{1em}{0ex}}\text{for all}x\in \mathrm{\Omega}\text{and all}|t|\le C(\gamma )$

(25)

for some ${C}_{3}>0$.

Let

$u\in \partial {B}_{1,+}^{c}$. By inequalities (23), (24), and (25), for all

$t>0$ we have

$\begin{array}{rl}I(tu)& =\frac{{t}^{2}}{2}{\parallel u\parallel}^{2}-{\int}_{\mathrm{\Omega}}F(x,tu)\phantom{\rule{0.2em}{0ex}}dx\\ =\frac{{t}^{2}}{2}{\parallel u\parallel}^{2}-{\int}_{tu\ge C}F(x,tu)\phantom{\rule{0.2em}{0ex}}dx-{\int}_{tu\le -C}F(x,tu)\phantom{\rule{0.2em}{0ex}}dx-{\int}_{|tu|\le C}F(x,tu)\phantom{\rule{0.2em}{0ex}}dx\\ \le \frac{{t}^{2}}{2}[(1+\frac{{\gamma}_{0}}{{\lambda}_{1}}){\parallel u\parallel}^{2}-\gamma {\int}_{tu\ge C}{u}^{2}\phantom{\rule{0.2em}{0ex}}dx]+{C}_{3}|\mathrm{\Omega}|.\end{array}$

(26)

Recalling that

$\gamma >0$ is arbitrary, from (26), we have

$I(tu)\to -\mathrm{\infty}\phantom{\rule{1em}{0ex}}\text{as}t\to -\mathrm{\infty}.$

(27)

Using formula (4) in condition (H

_{4}), we see that there exist constants

${\xi}_{0}$ and

$M>0$ such that

$f(x,t)t-2F(x,t)\ge {\xi}_{0}\phantom{\rule{1em}{0ex}}\text{for all}x\in \mathrm{\Omega}\text{and all}t\ge M.$

(28)

By (H

_{2}) and formula (2) in condition (H

_{4}), we have

$2F(x,t)-f(x,t)t\le C\phantom{\rule{1em}{0ex}}\text{for all}x\in \mathrm{\Omega}\text{and all}tM$

(29)

for some

$C>0$. By inequalities (28) and (29), for any

$u\in E$ we have

${\int}_{\mathrm{\Omega}}(2F(x,t)-f(x,t)t)\phantom{\rule{0.2em}{0ex}}dx\le C,$

(30)

where

*C* is a positive constant. Let

$i:{C}_{0}^{3}(\overline{\mathrm{\Omega}})\to E$ be the continuous embedding map. Let

${\u3008\cdot ,\cdot \u3009}_{0}$ denote the duality brackets for the pair

$({C}_{0}^{3}{(\overline{\mathrm{\Omega}})}^{\ast},{C}_{0}^{3}(\overline{\mathrm{\Omega}}))$. We let

${I}_{2}=I\circ i$, and so

$\begin{array}{r}{I}_{2}^{\prime}(u)={i}^{\ast}{I}^{\prime}(i(u))\phantom{\rule{1em}{0ex}}\text{for all}u\in {C}_{0}^{3}(\overline{\mathrm{\Omega}}),\\ \frac{d}{dt}{I}_{2}(tu)={\u3008{I}_{2}^{\prime}(tu),u\u3009}_{0}=t{\parallel u\parallel}^{2}-{\int}_{\mathrm{\Omega}}f(x,tu)u\phantom{\rule{0.2em}{0ex}}dx\le \frac{1}{t}(2I(tu)+{C}^{\ast}).\end{array}$

(31)

Then, from equation (

27), we obtain

$\frac{d}{dt}{I}_{2}(tu)<0\phantom{\rule{1em}{0ex}}\text{for all}t0\text{large such that}I(tu)-\frac{{C}^{\ast}}{2}.$

(32)

From conditions (H

_{2}) and (H

_{3}), we see that given

$\u03f5>0$, we can find

$M>0$ such that

$F(x,t)\le \frac{1}{2}(l+\u03f5){t}^{2}+M\phantom{\rule{1em}{0ex}}\text{for all}x\in \mathrm{\Omega}\text{and all}t\le 0.$

(33)

Using inequality (33), we have

$\begin{array}{rl}I(u)& \ge \frac{1}{2}({\parallel u\parallel}^{2}-l{|u|}_{2}^{2}-\u03f5{|u|}_{2}^{2})-M\\ \ge C{\parallel u\parallel}^{2}-M\end{array}$

for

$u\in -{C}_{+}$, where

${C}_{+}$ is defined as

$\{u\in {C}_{0}^{3}(\mathrm{\Omega}):u(x)\ge 0\text{for all}x\in \mathrm{\Omega}\}$

and

$C>0$ is a positive constant. So

$I{|}_{-{C}^{+}}$ is coercive, thus we find

${C}^{\ast \ast}>0$ such that

$I{|}_{-{C}^{+}}\ge -{C}^{\ast \ast}$. We pick

$a<min\{-\frac{{C}^{\ast}}{2},-{C}^{\ast \ast},\underset{\partial {B}_{1}^{c}}{inf}{I}_{2}\}.$

Then inequality (32) implies that we can find

$k(u)>1$ such that

$\{\begin{array}{cc}{I}_{2}(tu)>a\hfill & \text{if}t\in [0,k(u)),\hfill \\ {I}_{2}(tu)=a\hfill & \text{if}t=k(u),\hfill \\ {I}_{2}(tu)a\hfill & \text{if}tk(u).\hfill \end{array}$

Moreover, the implicit function theorem implies that $k\in C(\partial {B}_{1,+}^{c},[1,+\mathrm{\infty}))$.

By the choice of

*a*, we have

${I}_{2}^{a}=\{tu:u\in \partial {B}_{1,+}^{c},t\ge k(u)\}.$

(34)

We define the set

${E}_{+}=\{tu:u\in \partial {B}_{1,+}^{c},t\ge 1\}$. The map

${\stackrel{\u02c6}{h}}_{+}:[0,1]\times {E}_{+}\to {E}_{+}$ defined by

${\stackrel{\u02c6}{h}}_{+}(s,tu)=\{\begin{array}{cc}(1-s)tu+sk(u)u\hfill & \text{if}1\le tk(u),\hfill \\ tu\hfill & \text{if}t\ge k(u),\hfill \end{array}\phantom{\rule{1em}{0ex}}s\in [0,1],$

(35)

is a continuous deformation of

${E}_{+}$,

${\stackrel{\u02c6}{h}}_{+}(1,{E}_{+})\subset {I}_{2}^{a}$ and

${\stackrel{\u02c6}{h}}_{+}(s,\cdot ){|}_{{I}_{2}^{a}}=\mathit{id}{|}_{{I}_{2}^{a}}$ for all

$s\in [0,1]$ (see equations (

34) and (

35)). Therefore,

${I}_{2}^{a}$ is a strong deformation retract of

${E}_{+}$. Hence we have

${H}_{k}({C}_{0}^{3}(\overline{\mathrm{\Omega}}),{I}_{2}^{a})={H}_{k}({C}_{0}^{3}(\overline{\mathrm{\Omega}}),{E}_{+})={H}_{k}({C}_{0}^{3}(\overline{\mathrm{\Omega}}),\partial {B}_{1,+}^{c})\phantom{\rule{1em}{0ex}}\text{for all}k\ge 0.$

(36)

Recalling that in the first part of the proof, we established that

$\partial {B}_{1,+}^{c}$ is contractible. This yields

${H}_{k}({C}_{0}^{3}(\overline{\mathrm{\Omega}}),\partial {B}_{1,+}^{c})=0\phantom{\rule{1em}{0ex}}\text{for all integers}k\ge 0.$

Combining with equation (36) leads to equation (22), which completes the proof. □

**Proposition 3.4** *If the assumptions of Theorem * 2.1

*hold*,

*then* *where* $d=dimV$ (*V* *being defined in Lemma * 2.3).

*Proof* By condition (H

_{5}), given

$\u03f5>0$, we can find

${\delta}_{\ast}>0$ such that

$\frac{1}{2}({\vartheta}_{1}(x)-\u03f5){t}^{2}\le F(x,t)\phantom{\rule{1em}{0ex}}\text{for all}x\in \mathrm{\Omega}\text{and all}|t|\le {\delta}_{\ast}.$

(37)

Since

*V* is finite dimensional, all norms are equivalent. Thus we can find

$\rho >0$ small such that

$\parallel u\parallel \le \rho \phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}{\parallel u\parallel}_{\mathrm{\infty}}\le {\delta}_{\ast}$

(38)

for all

$u\in V$. Taking inequalities (37) and (38) into account, for all

$u\in V$ with

$\parallel u\parallel \le \rho $ we have

$I(u)\le \frac{1}{2}{\int}_{\mathrm{\Omega}}({\lambda}_{k}-{\vartheta}_{1}(x)){u}^{2}\phantom{\rule{0.2em}{0ex}}dx+\frac{\u03f5}{2}{|u|}_{2}^{2}.$

(39)

Similar to the proof of Lemma 2.3, there exists

$C>0$ such that

$I(u)\le (-C+\u03f5){\parallel u\parallel}^{2}\le 0$

(40)

for all $u\in V$ and $\parallel u\parallel \le \rho $.

On the other hand, for given

$\u03f5>0$, it follows from (H

_{2}) and (H

_{5}) that

$F(x,t)\le \frac{{\vartheta}_{2}(x)+\u03f5}{2}{t}^{2}+{C}_{\u03f5}{|t|}^{r}$

(41)

for all

$x\in \mathrm{\Omega}$ and

$t\in R$. By (41) and Lemma 2.3, we have

$I(u)\ge {C}_{4}{\parallel u\parallel}^{2}-{C}_{5}{\parallel u\parallel}^{r}$

(42)

for all

$u\in W$. From inequality (42), we infer that for

*ρ* small enough we have

$I(u)>0\phantom{\rule{1em}{0ex}}\text{for all}u\in W\text{with}0\parallel u\parallel \le \rho .$

(43)

From inequalities (40) and (43), we know that *I* has a local linking at 0. Then invoking Proposition 2.3 of Bartsch and Li [17], we obtain ${C}_{d}(I,0)\ne 0$. □