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Existence of four solutions for the elliptic problem with nonlinearity crossing one eigenvalue
Journal of Inequalities and Applications volume 2013, Article number: 187 (2013)
Abstract
We investigate the multiplicity of the weak solutions for the nonlinear elliptic boundary value problem. We get a theorem which shows the existence of at least four weak solutions for the asymptotically linear elliptic problem with Dirichlet boundary condition. We obtain this result by using the LeraySchauder degree theory, the variational reduction method and critical point theory.
MSC:35J15, 35J25.
1 Introduction
Let Ω be a bounded open subset of ${R}^{n}$ with smooth boundary ∂ Ω. Let L be the selfadjoint strongly elliptic partial differential operator
where ${a}_{ij}(x)\in {L}^{\mathrm{\infty}}(\mathrm{\Omega})$. Let $f:R\to R$ be a ${C}^{1}$ function. Let ${\lambda}_{k}$ be the eigenvalues and ${\varphi}_{k}$, $k=1,2,\dots $ , be the associated normalized eigenfunctions of the eigenvalue problem $Lu+\lambda u=0$ in Ω, $u=0$ on ∂ Ω. We note that $0<{\lambda}_{1}<{\lambda}_{2}\le \cdots \le {\lambda}_{k}\to \mathrm{\infty}$ and ${\varphi}_{1}>0$.
In this paper, we consider the multiplicity of solutions of the following elliptic problem with Dirichlet boundary condition and jumping nonlinearity:
The physical model for this kind of the jumping nonlinearity problem can be furnished by traveling waves in suspension bridges. The nonlinear equations with jumping nonlinearity have been extensively studied by Lazer and McKenna [1], McKenna and Walter [2], Tarantello [3], Micheletti and Pistoia [4, 5], Jung and Choi [6–8] and many other authors. McKenna and Walter [2] proved that if $\mathrm{\infty}<{f}^{\mathrm{\prime}}(\mathrm{\infty})<{\lambda}_{1}$ and ${\lambda}_{n}<{f}^{\mathrm{\prime}}(+\mathrm{\infty})$, then there exists ${s}_{0}$ such that for all $s\ge {s}_{0}$, (1.1) has at least two solutions if n is odd, and three solutions if n is even.
Our main result is as follows.
Theorem 1.1 Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$, $s>0$. Then (1.1) has at least four solutions.
For the proof of Theorem 1.1 we use the LeraySchauder degree theory, the variational reduction method and the critical point theory. The outline of this paper is as follows: In Section 2, we show the existence of the third weak solution for (1.1) by the variational reduction method and critical point theory. In Section 3, we show the existence of the fourth weak solution for (1.1) by the LeraySchauder degree theory.
2 Existence of the third weak solution
We note that (1.1) has a positive solution ${v}_{0}=\frac{s{\varphi}_{1}}{b{\lambda}_{1}}$ and a negative solution ${v}_{1}=\frac{s{\varphi}_{1}}{a{\lambda}_{1}}$. To show the existence of the third solution of (1.1), we shall use the variational reduction method and the critical point theory. We assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ and $s>0$. Let $g(\xi )=b{\xi}^{+}a{\xi}^{}$ and ${H}_{0}^{1}(\mathrm{\Omega})$ be the Sobolev space with the norm
We consider the following functional associated with (1.1):
where
For simplicity, we shall write $I={I}_{a,b}$ when a and b are fixed. By the condition $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ and $s>0$, I is well defined. By the following Lemma 2.1, $I(u,s)\in C({H}_{0}^{1}(\mathrm{\Omega}),R)$ and Fréchet differentiable in ${H}_{0}^{1}(\mathrm{\Omega})$, so the solutions of (1.1) coincide with the critical points of $I(u,s)$.
Lemma 2.1 Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ and $s>0$. Then $I(u,s)$ is continuous and Fréchet differentiable in ${H}_{0}^{1}(\mathrm{\Omega})$ and
for $h\in {H}_{0}^{1}(\mathrm{\Omega})$. Moreover, if we set
then ${D}_{u}F(u,s)$ is continuous with respect to weak convergence, ${D}_{u}F(u,s)$ is compact, and
This implies that $I\in {C}^{1}({H}_{0}^{1}(\mathrm{\Omega}),R)$ and $F(u,s)$ is weakly continuous.
Proof Let $u\in {H}_{0}^{1}(\mathrm{\Omega})$. First, we will prove that $I(u,s)$ is continuous with respect to the variable u. We consider
Let $u=\sum {h}_{m}{\varphi}_{m}$, $v=\sum {\tilde{h}}_{m}{\varphi}_{m}$. Then we have
On the other hand,
and hence
With the above results, we see that $I(u,s)$ is continuous at u. To prove $I(u,s)$ is Fréchet differentiable at $u\in {H}_{0}^{1}(\mathrm{\Omega})$, it is enough to compute the following:
□
Let V be the onedimensional space spanned by the eigenfunction ${\varphi}_{1}$ whose eigenvalue is ${\lambda}_{1}$. Let W be the orthogonal complement of V in ${H}_{0}^{1}(\mathrm{\Omega})$. Let $P:{H}_{0}^{1}(\mathrm{\Omega})\to V$ be the orthogonal projection of ${H}_{0}^{1}(\mathrm{\Omega})$ onto V and $IP:{H}_{0}^{1}(\mathrm{\Omega})\to W$ denote that of ${H}_{0}^{1}(\mathrm{\Omega})$ onto W. Then every element $u\in {H}_{0}^{1}(\mathrm{\Omega})$ is expressed by $u=v+z$, $v\in Pu$, $z=(IP)u$. Then (2.1) is equivalent to the two systems in the two unknowns v and z:
The subspace W is spanned by eigenfunctions corresponding to the eigenvalues ${\lambda}_{k}$, $k\ge 2$. Let $v\in V$ be fixed and consider the function $h:W\to R$ defined by
The function h has continuous Fréchet derivatives Dh given by
for $y\in W$. By Lemma 2.1, I is a function of class ${C}^{1}$.
By the following Lemma 2.2, we can get the critical points of the functional $I(u,s)$ on the infinite dimensional space ${H}_{0}^{1}(\mathrm{\Omega})$ from that of the reduced functional $\tilde{I}(v,s)$ on the finite dimensional subspace V.
Lemma 2.2 (Reduction Method)
Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ and $s>0$. Then

(i)
there exists a unique solution $z\in W$ of the equation
If we put $z=\theta (v,s)$, then θ is continuous on V and satisfies a uniform Lipschitz condition in v with respect to ${L}^{2}$ norm (also norm $\parallel \cdot \parallel $). Moreover,

(ii)
There exists $m>0$ such that if w and y are in W, then
$$(Dh(w)Dh(y))(wy)\ge m{\parallel wy\parallel}^{2}.$$ 
(iii)
If $\tilde{I}:V\to R$ is defined by $\tilde{I}(v,s)=I(v+\theta (v,s),s)$, then $\tilde{I}$ has a continuous Fréchet derivative ${D}_{v}\tilde{I}$ with respect to v, and
$${D}_{v}\tilde{I}(v,s)(h)={D}_{u}I(v+\theta (v,s),s)(h)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}v,h\in V.$$(2.5) 
(iv)
If ${v}_{0}\in V$ is a critical point of $\tilde{I}$ if and only if ${v}_{0}+\theta ({v}_{0},s)$ is a critical point of I.

(v)
Let $S\subset V$ and $\mathrm{\Sigma}\subset {H}_{0}^{1}(\mathrm{\Omega})$ be open bounded regions such that
$$\{v+\theta (v,s);v\in S\}=\mathrm{\Sigma}\cap \{v+\theta (v,s);v\in V\}.$$
If $D\tilde{I}(v,s)\ne 0$ for $v\in \partial S$, then
where d denote the LeraySchauder degree.

(vi)
If ${u}_{0}={v}_{0}+\theta ({v}_{0})$ is a critical point of mountain pass type of I, then ${v}_{0}$ is a critical point of mountain pass type of $\tilde{I}$.
Proof (i) Let $\delta =\frac{a+b}{2}$. If ${g}_{1}(\xi )=g(\xi )\delta \xi $, then equation (2.3) is equivalent to
The operator ${(L\delta )}^{1}(IP)$ is self adjoint, compact and linear map from $(IP){L}^{2}(\mathrm{\Omega})$ into itself and its ${L}_{2}$ norm is ${{\lambda}_{2}\delta }^{1}$. Since ${g}_{1}({\xi}_{2}){g}_{1}({\xi}_{1})\le max\{a\delta ,b\delta \}{\xi}_{2}{\xi}_{1}=\delta {\xi}_{2}{\xi}_{1}$, it follows that the righthand side of (2.6) defines, for fixed $v\in V$, a Lipschitz mapping of $(IP){L}^{2}(\mathrm{\Omega})$ into itself with Lipschitz constant $r=\frac{\delta}{{\lambda}_{2}\delta }<1$. Therefore, by the contraction mapping principle, for given $v\in V$, there exists a unique $z=(IP){L}^{2}(\mathrm{\Omega})$ which satisfies (2.6). If $\theta (v,s)$ denote the unique $z\in (IP){L}^{2}(\mathrm{\Omega})$ which solves (2.3), then θ is continuous and satisfies a uniform Lipschitz condition in v with respect to ${L}^{2}$ norm (also norm $\parallel \cdot \parallel $). In fact, if ${z}_{1}=\theta ({v}_{1},s)$ and ${z}_{2}=\theta ({v}_{2},s)$, then
Hence,
Let $u=v+z$, $v\in V$ and $z=\theta (v,s)$. If $w\in (IP){L}^{2}(\mathrm{\Omega})\cap {H}_{0}^{1}(\mathrm{\Omega})$,
From (2.3), we see that
Since
we have

(ii)
If w and y are in W, then
$$(Dh(w)Dh(y))(wy)={\int}_{\mathrm{\Omega}}[L(wy)(wy)(g(v+w)g(v+y))(wy)]\phantom{\rule{0.2em}{0ex}}dx.$$
Since ${\int}_{\mathrm{\Omega}}[L(wy)(wy)]\phantom{\rule{0.2em}{0ex}}dx={\parallel wy\parallel}^{2}$ and $(g({\xi}_{2})g({\xi}_{1}))({\xi}_{2}{\xi}_{1})<b{({\xi}_{2}{\xi}_{1})}^{2}$, we see that if w and y are in W, then ${\parallel wy\parallel}_{{L}^{2}(\mathrm{\Omega})}\le \frac{1}{{\lambda}_{2}}{\parallel wy\parallel}^{2}$ and
and
where $(1+b\frac{b}{{\lambda}_{2}})>0$.

(iii)
Since the functional I has a continuous Fréchet derivative ${D}_{u}I$, $\tilde{I}$ has a continuous Fréchet derivative ${D}_{v}\tilde{I}$ with respect to v.

(iv)
Suppose that there exists ${v}_{0}\in V$ such that ${D}_{v}\tilde{I}({v}_{0},s)=0$. From ${D}_{v}\tilde{I}(v,s)(h)={D}_{u}I(v+\theta (v,s),s)(h)$ for all $v,h\in V$, ${D}_{u}I({v}_{0}+\theta ({v}_{0},s),s)(h)={D}_{v}\tilde{I}({v}_{0},s)(h)=0$ for all $h\in V$. Since ${D}_{u}I(v+\theta (v,s),s)(w)=0$ for all $w\in W$ and E is the direct sum of V and W, it follows that ${D}_{u}I({v}_{0}+\theta ({v}_{0},s),s)=0$. Thus, ${v}_{0}+\theta ({v}_{0},s)$ is a solution of (1.1). Conversely, if u is a solution of (1.1) and $v=Pu$, then ${D}_{v}\tilde{I}(v)=0$.

(v)
The proof of part (v) follows by arguing as in Lemma 2.6 of [1].

(vi)
Suppose ${v}_{0}$ is not of mountain pass type of $\tilde{I}$. Let S be an open neighborhood of ${v}_{0}$ in V such that ${\tilde{I}}^{1}(\mathrm{\infty},\tilde{I}({v}_{0}))\cap S$ is empty or path connected. If ${\tilde{I}}^{1}(\mathrm{\infty},\tilde{I}({v}_{0}))\cap S$ is empty, by part (i) we see that $\{v+w:v\in V,w\in W\}\cap {I}^{1}(\mathrm{\infty},I({u}_{0}))$ is also empty. Thus ${u}_{0}$ is not of mountain pass type for I. If ${\tilde{I}}^{1}(\mathrm{\infty},\tilde{I}({v}_{0}))\cap S$ is path connected, Letting $T=\{v+w:v\in V,\parallel w\theta (v)\parallel <1\}$ and using again (i) it is seen that $T\cap {I}^{1}(\mathrm{\infty},I({u}_{0}))$ is also path connected. □
We define the functional on ${H}_{0}^{1}(\mathrm{\Omega})$
The critical points of ${I}_{a,b}^{\ast}(u)$ coincide with the solutions of the equation
Under the assumption $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$, (2.9) has only the trivial solution and hence ${I}_{a,b}^{\ast}(u)$ has only one critical point $u=0$. In fact, from (2.9), we obtain
under the assumption $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ the lefthand side of (2.10) is nonnegative, so the only possibility to hold (2.10) is that $u=0$. Given $v\in V$, let ${\theta}^{\ast}(v)=\theta (v,0)$ be the unique solution of the equation
Let us define ${\tilde{I}}_{a,b}^{\ast}(v)={I}_{a,b}^{\ast}(v+{\theta}^{\ast}(v))$. We note that ${\tilde{I}}_{a,b}^{\ast}(v)$ has only the trivial critical point $v=0$.
Lemma 2.3 Let $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$. Then we have
Proof To prove the conclusion, it suffices to show that ${\tilde{I}}_{a,b}^{\ast}(v)$ does not satisfies the following cases:

(i)
${\tilde{I}}_{a,b}^{\ast}(v)\le 0$ and ${\tilde{I}}_{a,b}^{\ast}({v}_{0})=0$ for some ${v}_{0}\in V$ with ${v}_{0}\ne 0$.

(ii)
${\tilde{I}}_{a,b}^{\ast}(v)\ge 0$ and ${\tilde{I}}_{a,b}^{\ast}({v}_{1})=0$ for some ${v}_{1}\in V$ with ${v}_{1}\ne 0$.

(iii)
${\tilde{I}}_{a,b}^{\ast}(v)<0$ for all $v\in V$ with $v\ne 0$.

(iv)
There exist ${v}_{1}$ and ${v}_{2}$ in V such that ${\tilde{I}}_{a,b}^{\ast}({v}_{1})<0$ and ${\tilde{I}}_{a,b}^{\ast}({v}_{2})>0$.
Suppose that (i) holds. It follows that ${\tilde{I}}_{a,b}^{\ast}(v)$ has an absolute maximum at ${v}_{0}$, and hence $D{\tilde{I}}_{a,b}^{\ast}({v}_{0})=0$. Therefore, by Lemma 2.2, ${u}_{0}={v}_{0}+{\theta}^{\ast}({v}_{0})$ is a nontrivial solution of (2.9), which is a contradiction. A similar argument show that it is impossible that (ii) holds. Suppose that (iii) holds. Then there exists ${t}_{0}\in (0,1)$ such that for all t with $0<t\le {t}_{0}$,
We note that there exist ${v}_{0}\ne 0$ and t with $0<t\le {t}_{0}$ such that $t{\tilde{I}}_{a,b}^{\ast}({v}_{0})+(1t){\tilde{I}}_{0,b}^{\ast}({v}_{0})=0$ for some ${v}_{0}\ne 0$ and t. Let ${t}_{1}$ be the greatest number such that
Then $0<{t}_{1}\le {t}_{0}$. Since ${t}_{1}{\tilde{I}}_{a,b}^{\ast}(v)+(1{t}_{1}){\tilde{I}}_{0,b}^{\ast}(v)\le 0$ for all $v\ne 0$, and hence ${v}_{0}$ is a point of maximum of ${t}_{1}{\tilde{I}}_{a,b}^{\ast}(v)+(1{t}_{1}){\tilde{I}}_{0,b}^{\ast}(v)$, we have
Let $v\in V$ be given and $0<{t}_{1}<1$. Let ${\theta}_{{t}_{1}}^{\ast}(v)$ be the unique solution of the equation
By Lemma 2.2, ${v}_{0+{\theta}_{{t}_{1}}^{\ast}({v}_{0})}$ is a nontrivial solution of the equation
that is,
which contradicts the fact that the above equation has only the trivial solution because $\mathrm{\infty}<{t}_{1}a<{\lambda}_{1}<b<{\lambda}_{2}$. Suppose that (iv) holds. Since the equation (2.9) has only the trivial solution, ${I}_{a,b}^{\ast}(u)$ has only one critical point $u=0$. By the assumption (iv), the trivial solution $u=0$ is a point of inflexion, and hence by Lemma 2.2, $v=0$ is the critical point of ${\tilde{I}}_{a,b}^{\ast}(v)$, which is a point of inflexion. By the assumption (iv), there exist ${v}_{3}$, ${v}_{4}$ in V and ${t}_{0}\in (0,1)$ such that for all t with $0<t\le {t}_{0}$, $t{\tilde{I}}_{a,b}^{\ast}({v}_{3})+(1t){\tilde{I}}_{0,b}^{\ast}({v}_{3})<0$ and $t{\tilde{I}}_{a,b}^{\ast}({v}_{4})+(1t){\tilde{I}}_{0,b}^{\ast}({v}_{4})>0$. We note that there exists t with $0<t\le {t}_{0}$ such that $t{\tilde{I}}_{a,b}^{\ast}({v}_{3})+(1t){\tilde{I}}_{0,b}^{\ast}({v}_{3})<0$ and $t{\tilde{I}}_{a,b}^{\ast}({v}_{4})+(1t){\tilde{I}}_{0,b}^{\ast}({v}_{4})>0$. Let ${t}_{1}$ be the greatest number such that
Then $0<{t}_{1}\le {t}_{0}$. Since ${t}_{1}{\tilde{I}}_{a,b}^{\ast}({v}_{3})+(1{t}_{1}){\tilde{I}}_{0,b}^{\ast}({v}_{3})<0$ and ${t}_{1}{\tilde{I}}_{a,b}^{\ast}({v}_{4})+(1{t}_{1}){\tilde{I}}_{0,b}^{\ast}({v}_{4})>0$. Thus, there exists a point of inflexion ${v}_{0}\ne 0$ of ${t}_{1}{\tilde{I}}_{a,b}^{\ast}(v)+(1{t}_{1}){\tilde{I}}_{0,b}^{\ast}(v)$ and we have
Let $v\in V$ be given and $0<{t}_{1}<1$. Let ${\theta}_{{t}_{1}}^{\ast}(v)$ be the unique solution of the equation
By Lemma 2.2, ${v}_{0}+{\theta}_{{t}_{1}}^{\ast}({v}_{0})$ is a nontrivial solution of the equation
that is,
which contradicts the fact that the above equation has only the trivial solution because $\mathrm{\infty}<{t}_{1}a<{\lambda}_{1}<b<{\lambda}_{2}$. □
From now on, we shall denote that ${I}_{a,b}=I$.
Lemma 2.4 Let $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$. Then we have
Proof We shall prove the lemma by contradiction. We suppose that there exists a sequence ${\{{v}_{n}\}}_{1}^{\mathrm{\infty}}$ in V and a number $M>0$ such that $\parallel {v}_{n}\parallel \to \mathrm{\infty}$ as $n\to \mathrm{\infty}$ and $\tilde{I}({v}_{n},s)\le M$. For given ${v}_{n}\in V$, let ${w}_{n}=\theta ({v}_{n},s)$ be the unique solution of the equation
By Lemma 2.2, we have that for some constant k,
From this the sequence $\{\frac{\parallel {v}_{n}+{w}_{n}\parallel}{\parallel {v}_{n}\parallel}\}$ is bounded in H. Let ${u}_{n}={v}_{n}+{w}_{n}$, ${v}_{n}^{\ast}=\frac{{v}_{n}}{\parallel {v}_{n}\parallel}$, ${w}_{n}^{\ast}=\frac{{w}_{n}}{\parallel {v}_{n}\parallel}$ and ${u}_{n}^{\ast}={v}_{n}^{\ast}+{w}_{n}^{\ast}$ for $n\ge 1$. For ${w}_{n}=\theta ({v}_{n},s)$ and ${w}_{n}^{\ast}=\frac{{w}_{n}}{\parallel {v}_{n}\parallel}$, we have
Since $\{\frac{{v}_{n}+{w}_{n}}{\parallel {v}_{n}\parallel}\}$ is bounded and $\frac{{\varphi}_{1}}{\parallel {v}_{n}\parallel}\to 0$, $b\frac{{({v}_{n}+{w}_{n})}^{+}}{\parallel {v}_{n}\parallel}+a\frac{{({v}_{n}+{w}_{n})}^{}}{\parallel {v}_{n}\parallel}s\frac{{\varphi}_{1}}{\parallel {v}_{n}\parallel}\frac{h(x)}{\parallel {v}_{n}\parallel}$ is bounded in H. Since ${L}^{1}$ is a compact operator, by passing to a subsequence, we get that ${\{{w}_{n}^{\ast}\}}_{1}^{\mathrm{\infty}}$ converges to ${w}^{\ast}$. Since V is 1dimensional subspace, we may assume that ${\{{v}_{n}^{\ast}\}}_{1}^{\mathrm{\infty}}$ converges to ${v}^{\ast}\in V$ with $\parallel {v}^{\ast}\parallel =1$. Therefore, ${\{{u}_{n}^{\ast}\}}_{1}^{\mathrm{\infty}}$ converges to ${u}^{\ast}$ in ${H}_{0}^{1}(\mathrm{\Omega})$. Since $\tilde{I}({v}_{n},s)\le M$ for all n, we have, for all n,
Dividing the above inequality by ${\parallel {v}_{n}\parallel}^{2}$, we obtain
From the definition of ${w}_{n}=\theta ({v}_{n},s)$, we have that for any $y\in W$, $n\ge 1$,
Let us set $y={w}_{n}$ in (2.12) and divide by ${\parallel {v}_{n}\parallel}^{2}$. Then we have
for all $n\ge 1$. Let $y\in W$. Dividing (2.12) by $\parallel {v}_{n}\parallel $ and letting $n\to \mathrm{\infty}$, we obtain
(2.14) can be rewritten in the form
By ${w}^{\ast}=\theta ({v}^{\ast})$, letting $n\to \mathrm{\infty}$ in (2.13), we have
Hence,
Letting $n\to \mathrm{\infty}$ in (2.11), we obtain
Since $\parallel {v}^{\ast}\parallel =1$, this contradicts the fact that ${\tilde{I}}^{\ast}(v)>0$ for all $v\ne 0$. Thus, $\tilde{I}(v,s)\to \mathrm{\infty}$ as $\parallel v\parallel \to \mathrm{\infty}$. □
Lemma 2.5 Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ and $s>0$. Then there exists a small open neighborhood B of ${v}_{0}$ in V such that ${v}_{0}$ is a strict local point maximum of $\tilde{I}(v,s)$ and there exists a small open neighborhood D of ${v}_{1}$ in V such that ${v}_{1}$ is a strict point of minimum of $\tilde{I}(v,s)$, where ${v}_{0}=\frac{s{\varphi}_{1}}{b{\lambda}_{1}}$ is a positive solution and ${v}_{1}=\frac{s{\varphi}_{1}}{a{\lambda}_{1}}$ is a negative solution of (1.1).
Proof Since the positive solution ${v}_{0}=\frac{s{\varphi}_{1}}{b{\lambda}_{1}}$ is in V, $\theta ({v}_{0},s)=0$ and $I+\theta $, where I is an identity map on V, is continuous on V, it follows that there exists a small open neighborhood B of ${v}_{0}$ in V such that if ${v}_{0}+v\in B$, then ${v}_{0}+v+\theta ({v}_{0}+v,s)>0$. Here, $\theta ({v}_{0}+v,s)=\theta ({v}_{0},s)=0$. Therefore, if ${v}_{0}+v\in B$, then for $z=\theta {v}_{0}+v,s=\theta ({v}_{0})=0$, we have
where
Each $v\in V$ has the form $v=c{\varphi}_{1}$ Therefore, we have, in B,
Since ${\lambda}_{1}<b<{\lambda}_{2}$, ${v}_{0}$ is a strict local point maximum of $\tilde{I}(v,s)$. Since the negative solution ${v}_{1}=\frac{s{\varphi}_{1}}{a{\lambda}_{1}}$ is in V, $\theta ({v}_{1},s)=0$ and $I+\theta $, where I is an identity map on V, is continuous on V, it follows that there exists a small open neighborhood D of ${v}_{1}$ in V such that if ${v}_{1}+v\in B$, then ${v}_{1}+v+\theta ({v}_{1}+v,s)<0$. Here, $\theta ({v}_{1}+v,s)=\theta ({v}_{1},s)=0$. Therefore, if ${v}_{1}+v\in B$, then for $z=\theta {v}_{1}+v,s=\theta ({v}_{1},s)=0$, we have
where
Each $v\in V$ has the form $v=c{\varphi}_{1}$ Therefore, we have, in B,
since $\mathrm{\infty}<a<{\lambda}_{1}$. Thus, ${v}_{1}$ is a strict local point minimum of $\tilde{I}(v,s)$. □
Lemma 2.6 (Existence of the Third Solution)
Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ and $s>0$. Then (1.1) has the third weak solution which is a strict point of minimum of ${I}_{a,b}(u,s)$.
Proof We know that (1.1) has a positive solution ${v}_{0}=\frac{s{\varphi}_{1}}{b{\lambda}_{1}}$ and a negative solution ${v}_{1}=\frac{s{\varphi}_{1}}{a{\lambda}_{1}}$. By Lemma 2.2 (reduction method), we shall show the existence of the third weak solution of (1.1). By Lemma 2.1, $\tilde{I}(v,s)$ is continuous and Fréchet differentiable. By Lemma 2.4, $\tilde{I}(v,s)\to \mathrm{\infty}$ as $\parallel v\parallel \to \mathrm{\infty}$, so $\tilde{I}(v,s)$ is bounded from below, satisfies the (P.S.) condition. By Lemma 2.5, there exists a small open neighborhood B of ${v}_{0}$ in V such that ${v}_{0}$ is a strict local point of maximum of $\tilde{I}(v,s)$ and there exists a small open neighborhood D of ${v}_{1}$ in V such that ${v}_{1}$ is a strict local point of minimum of $\tilde{I}(v,s)$. By the shape of the graph of the functional $\tilde{I}$ on the 1dimensional subspace V, there exists the third critical point ${v}_{2}\ne 0$ which is a strict local point of minimum of $\tilde{I}(v,s)$. By Lemma 2.2, ${v}_{2}+\theta ({v}_{2},s)$ is a solution of (1.1) and a strict local point of minimum of ${I}_{a,b}(u,s)$. □
3 Existence of the fourth weak solution
In this section, we shall consider the LeraySchauder degree of the elliptic operator for the multiplicity of the solutions of (1.1).
Lemma 3.1 Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$. Then there exists ${s}_{1}>0$, $\u03f5>0$ such that the LeraySchauder degree
for $s\ge {s}_{1}$. Here, ${B}_{\tau}^{\ast}$ denotes a ball of radius τ in ${H}_{0}^{1}(\mathrm{\Omega})$ and ${v}_{0}=\frac{s{\varphi}_{1}}{b{\lambda}_{1}}$ is a positive solution of (1.1).
The proof of this lemma has the similar process to that of Theorem 1 in [2].
Lemma 3.2 If $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$, then there exist positive constants ${s}_{2}$, ϵ such that
for $s\ge {s}_{2}$, where ${v}_{1}=\frac{s{\varphi}_{1}}{a{\lambda}_{1}}$ is a negative solution of (1.1).
The proof of this lemma has the same process as that of Lemma 3.1.
Lemma 3.3 If $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$, then there exist positive constants ${s}_{3}$, $\gamma >0$ such that
for $s\ge {s}_{3}$, where ${v}_{2}$ is the third solution of (1.1).
Proof We know that the third solution ${v}_{2}$ of (1.1) is an isolated local minimum of ${I}_{a,b}(u,s)$. By [9], there exist positive constants ${s}_{3}$, $\gamma >0$ such that
for $s\ge {s}_{3}$, where ${B}_{\gamma}^{\ast}({v}_{2})\cap V$ is a ball centered at 0 containing no other critical point. By (v) of Lemma 2.2,
So we prove the lemma. □
Lemma 3.4 Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$ and s is bounded. Then there exist C and ${s}^{\ast}>0$ such that any solution to (1.1) satisfies $\parallel u\parallel \le C$ for any s with $s\le {s}^{\ast}$.
Proof If not, then there exist $({u}_{n},{s}_{n})$ with $\parallel {u}_{n}\parallel \to \mathrm{\infty}$, ${s}_{n}\to {s}^{\ast}$, ${s}_{n}$ is bounded, which satisfy equation (1.1). Now let ${v}_{n}=\frac{{u}_{n}}{\parallel {u}_{n}\parallel}$, and ${v}_{n}$ satisfies
Taking the inner product of both sides of (3.2) with ${\varphi}_{1}$, we have
Thus,
Thus, if ${v}_{n}$ is a solution of (3.3), then
Since ${v}_{n}$’s are precompact in ${H}_{0}^{1}(\mathrm{\Omega})$, there exists v with $\parallel v\parallel =1$ such that ${v}_{n}\to v$. Taking the limit of both sides of (3.4), we have
Thus, we have
and hence $v=0$. This is impossible, since $\parallel v\parallel =1$. □
We have the following no solvability condition for (1.1).
Lemma 3.5 Assume that $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$. Then there exists a constant ${s}_{0}<0$ so small enough that if $s\le {s}_{0}$, the problem
has no solution.
Proof We can rewrite (1.1) as
Taking the inner product of both sides of (3.5) with ${\varphi}_{1}$, we have
Thus, there is no solution for $s\le {s}_{0}$ if ${s}_{0}<0$. This completes the proof. □
Lemma 3.6 Let $\mathrm{\infty}<a<{\lambda}_{1}<b<{\lambda}_{2}$. Then there exists $\beta >0$, depending on C and ${s}^{\ast}$ such that
for $s\le {s}^{\ast}$ and $\beta >C$.
Proof By Lemma 3.5, there exists a constant ${s}_{0}<0$ so small enough that if $s\le {s}_{0}$, (1.1) has no solution. By Lemma 3.4, there exist a constant C and ${s}^{\ast}>0$ such that if u is a solution of (1.1) with $s\le {s}^{\ast}$, then $\parallel u\parallel \le C$. Let us choose β so large that $\beta >C$. We note that
and
for $0\le \lambda \le 1$. By the homotopy invariance property, we have that the LeraySchauder degree
where $s\le {s}^{\ast}$ and $0\le \lambda \le 1$. Thus, we proved the lemma. □
Proof of Theorem 1.1 By Lemma 3.4 and Lemma 3.6, there exists a large number $\beta >0$ (depending on C and ${s}^{\ast}$) such that all solutions of (1.1) are contained in the ball ${B}_{\beta}(0)$ and the LeraySchauder degree
for $\beta >C$ and $0<s<{s}^{\ast}$. Let us choose a positive number ${s}^{\ast}$ such that ${s}_{\ast}=max\{{s}_{1},{s}_{2},{s}_{3}\}$ and ${s}_{\ast}<{s}^{\ast}$. Then by Lemma 3.1, Lemma 3.2 and Lemma 3.3, for all s with ${s}_{\ast}\le s\le {s}^{\ast}$, there exist three disjoint balls ${B}_{\u03f5s}^{\ast}({v}_{0})$, ${B}_{\u03f5s}^{\ast}({v}_{1})$ and ${B}_{\gamma}^{\ast}({v}_{2})$ such that
By the excision property of LeraySchauder degree,
Thus, (1.1) has at least four solutions. □
Abbreviations
 EFSEPNCOE:

Existence of Four Solutions for the Elliptic Problem with Nonlinearity Crossing One Eigenvalue.
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Acknowledgements
This work (Tacksun Jung) was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (KRF20100023985).
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TJ carried out (EFSEPNCOE) studies, participated in the sequence alignment and drafted the manuscript. QC participated in the sequence alignment. All authors read and approved the final manuscript.
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Jung, T., Choi, Q. Existence of four solutions for the elliptic problem with nonlinearity crossing one eigenvalue. J Inequal Appl 2013, 187 (2013). https://doi.org/10.1186/1029242X2013187
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Keywords
 elliptic boundary value problem
 jumping nonlinearity
 inverse compact operator
 variational reduction method
 LeraySchauder degree theory