Existence of nontrivial solution for a nonlocal problem with subcritical nonlinearity

Abstract

In this paper, we consider the following new nonlocal Dirichlet boundary value problem:

$$ \textstyle\begin{cases} -(a-b\int_{\Omega} \vert \nabla u \vert ^{2}\,dx)\Delta u=\lambda u+g(x,u),& x\in \Omega, \\ u=0,& x\in\partial\Omega, \end{cases} $$

(0.1)

where a and b are positive, λ is a positive parameter, $0\leq\lambda< a\lambda_{1}$, $\lambda_{1}$ is the first eigenvalue of operator −Δ. Under appropriate assumptions on the function g which is of subcritical growth, we obtain a nontrivial solution.

1 Introduction and main result

In this paper, we consider the following new nonlocal Dirichlet boundary value problem:

$$ \textstyle\begin{cases} -(a-b\int_{\Omega} \vert \nabla u \vert ^{2}\,dx)\Delta u=\lambda u+g(x,u),& x\in \Omega, \\ u=0,& x\in\partial\Omega, \end{cases} $$

(1.1)

where a and b are positive, λ is a positive parameter.

The search for a nontrivial solution of problem (1.1) is a new subject and of great significance. We put forward a new nonlocal term $a-b\int _{\Omega}|\nabla u|^{2}\,dx$, which is different from the well known nonlocal term $a+b\int_{\Omega}|\nabla u|^{2}\,dx$ and presents a lot of interesting difficulties.

Recently, mathematical studies have focused on the existence of solutions of the Kirchhoff type problem

$$ \textstyle\begin{cases} -(a+b\int_{\Omega} \vert \nabla u \vert ^{2}\,dx)\Delta u=g(x,u),& x\in\Omega, \\ u=0,& x\in\partial\Omega, \end{cases} $$

where $a>0$, $b>0$ and Ω is either a smooth bounded domain in $\mathbb{R}^{N}$ or $\Omega=\mathbb{R}^{N}$. The results about problem with subcritical nonlinearity can be seen in [1–5] and the critical cases in [6–13]. Here we do not present the results in detail, someone who is interested in them can consult the references therein.

However, there are only few results about problem (1.1). When $\lambda =0$ and $g(x,u)=|u|^{p-2}u$ was of subcritical growth, Yin and Liu [14] considered

$$ \textstyle\begin{cases} -(a-b\int_{\Omega} \vert \nabla u \vert ^{2}\,dx)\Delta u= \vert u \vert ^{p-2}u,& x\in\Omega, \\ u=0,& x\in\partial\Omega, \end{cases} $$

and obtained existence and multiplicity of nontrivial solutions. When $\lambda=0$ and $g(x,u)=f_{\lambda}(x)|u|^{p-2}u$, Lei [15] considered

$$ \textstyle\begin{cases} -(a-b\int_{\Omega} \vert \nabla u \vert ^{2}\,dx)\Delta u=f_{\lambda}(x) \vert u \vert ^{p-2}u,& x\in\Omega, \\ u=0,& x\in\partial\Omega. \end{cases} $$

Under some special conditions and for $1< p<2$, the author obtained two solutions. Lei [16] also investigated

$$ \textstyle\begin{cases} -(a-b\int_{\Omega} \vert \nabla u \vert ^{2}\,dx)\Delta u=\frac{\lambda}{u^{\gamma }},& x\in\Omega, \\ u=0,& x\in\partial\Omega, \end{cases} $$

and, when $0<\gamma<1$ and $0<\lambda<\lambda_{*}$, at least two positive solutions were obtained. Wang [17] studied a nonlocal problem involving critical exponent, namely

$$ \textstyle\begin{cases} -(a-b\int_{\Omega} \vert \nabla u \vert ^{2}\,dx)\Delta u= \vert u \vert ^{2}u+\mu f(x),& x\in \mathbb{R}^{4}, \\ u\in D^{1, 2}(\mathbb{R}^{4}), \end{cases} $$

for which infinitely many positive solutions and at least two positive solutions were found for $\mu=0$ and $\mu\in(0, \mu_{*}]$. For some other important results the interested reader is also referred to [18–21].

We are inspired by the above articles and consider a new problem which is different from the mentioned above. Assume that nonlinearity g satisfies the following assumptions:

$(g_{1})$ :

g is continuous, $1\leq i\leq N$, $|g(x, u)|\leq C(1+|u|^{p-1})$ for some $C>0$ and $2< p<2^{*}$, where $2^{*}=\frac {2N}{N-2}$ if $N\geq3$, $2^{*}=\infty$ if $N=1\text{ or }2$;

$(g_{2})$ :

$g(x, u)=o(u)$ uniformly in x as $u\rightarrow0$;

$(g_{3})$ :

$u\mapsto\frac{g(x, u)}{u}$ is positive for $u\neq 0$, nonincreasing on $(-\infty, 0)$ and nondecreasing on $(0, +\infty)$.

Now, we state our main result.

Theorem 1.1

Suppose that conditions $(g_{1})$–$(g_{3})$ and $0\leq\lambda< a\lambda _{1}$ hold, then problem (1.1) has a nontrivial solution.

2 Preliminary results

In this section, we present the variational results which will be used in the proof of Theorem 1.1. Let $E:=H_{0}^{1}(\Omega)$ be endowed with the usual norm

$$\Vert u \Vert =\langle u, u\rangle^{1/2}= \biggl( \int_{\Omega} \vert \nabla u \vert ^{2} \biggr)^{1/2}. $$

The usual norm in the Lebesgue space $L^{p}(\Omega)$ is denoted by $|u|_{p}$.

A function $u\in E$ is called a weak solution of problem (1.1) if

$$a \int_{\Omega}\nabla u \nabla v\,dx-b \Vert u \Vert ^{2} \int_{\Omega}\nabla u \nabla v\,dx=\lambda \int_{\Omega}uv\,dx- \int_{\Omega}g(x, u)v\,dx,\quad \forall v\in E. $$

Moreover, our assumptions imply that the solutions of (1.1) are the critical points of the functional defined in E by

$$I(u)=\frac{a}{2} \Vert u \Vert ^{2}-\frac{b}{4} \Vert u \Vert ^{4}-\frac{\lambda}{2} \int _{\Omega} \vert u \vert ^{2}\,dx- \int_{\Omega}G(x, u)\,dx. $$

It is easy to see for $\forall u, v\in E$,

$$\bigl\langle I'(u), v\bigr\rangle =a \int_{\Omega}\nabla u \nabla v\,dx-b \Vert u \Vert ^{2} \int_{\Omega}\nabla u \nabla v\,dx-\lambda \int_{\Omega}uv\,dx- \int _{\Omega}g(x, u)v\,dx. $$

Let $\lambda_{i}$ ($i=1,2,\dots$) be the eigenvalues of operator −Δ with zero Dirichlet boundary condition. It is well known that each eigenvalue $\lambda_{i}$ is positive, isolated and has finite multiplicity, the smallest eigenvalue $\lambda_{1}$ being simple and $\lambda_{i}\rightarrow\infty$ as $i\rightarrow\infty$. Here we only need the first eigenvalue of −Δ, where $\lambda_{1}=\inf_{u\neq 0}\frac{\int_{\Omega}|\nabla u|^{2}}{\int_{\Omega}|u|^{2}}$ and assume that $0\leq\lambda< a\lambda_{1}$.

3 Proof of Theorem 1.1

In this section, we will prove Theorem 1.1, so from now on we always suppose that $(g_{1})$–$(g_{3})$ hold. First, $(g_{1})$ and $(g_{2})$ imply that for each $\varepsilon>0$ there is a $C_{\varepsilon}>0$ such that

$$ \bigl\vert g(x, u) \bigr\vert \leq\varepsilon \vert u \vert +C_{\varepsilon} \vert u \vert ^{p-1} \quad\text{for all } u \in\mathbb{R}. $$

(3.1)

And using $(g_{2})$ and $(g_{3})$, one can easily check that

$$ G(x, u)\geq0 \quad\text{and} \quad g(x, u)u\geq2G(x, u)>0 \quad \text{if } u\neq0. $$

(3.2)

Lemma 3.1

If $0\leq\lambda< a\lambda_{1}$, then there exists a sequence $\{u_{n}\} \subset E$ satisfying $I(u_{n})\rightarrow c$, $I'(u_{n})\rightarrow 0$, where $0< c<\frac{a^{2}}{4b}$.

Proof

For $\lambda_{1}=\inf_{u\neq0}\frac{\int_{\Omega}|\nabla u|^{2}}{\int_{\Omega}|u|^{2}}$, then

$$\biggl(a-\frac{\lambda}{\lambda_{1}} \biggr) \int_{\Omega} \vert \nabla u \vert ^{2}\leq a \int_{\Omega} \vert \nabla u \vert ^{2}-\lambda \int_{\Omega} \vert u \vert ^{2}\leq a \int_{\Omega} \vert \nabla u \vert ^{2}. $$

Also by (3.1), we can choose a sufficiently small $\varepsilon=\frac {\lambda_{1}}{2} (a-\frac{\lambda}{\lambda_{1}} )$, and then

$$ \begin{aligned} I(u)&=\frac{a}{2} \Vert u \Vert ^{2}-\frac{b}{4} \Vert u \Vert ^{4}- \frac{\lambda}{2} \int _{\Omega} \vert u \vert ^{2}- \int_{\Omega}G(x, u) \\ &\geq\frac{1}{2} \biggl(a-\frac{\lambda}{\lambda_{1}} \biggr) \int_{\Omega } \vert \nabla u \vert ^{2}- \frac{b}{4}\biggl( \int_{\Omega} \vert \nabla u \vert ^{2} \biggr)^{2} -\frac{\varepsilon}{2} \int_{\Omega} \vert u \vert ^{2}-\frac{C_{\varepsilon}}{p} \int _{\Omega} \vert u \vert ^{p} \\ &\geq\frac{1}{2} \biggl(a-\frac{\lambda}{\lambda_{1}} \biggr) \int_{\Omega } \vert \nabla u \vert ^{2}- \frac{b}{4} \Vert u \Vert ^{4}-\frac{\varepsilon}{2\lambda _{1}} \int_{\Omega} \vert \nabla u \vert ^{2}- \frac{C_{1}C_{\varepsilon}}{p} \Vert u \Vert ^{p} \\ &\geq\frac{1}{4} \biggl(a-\frac{\lambda}{\lambda_{1}} \biggr) \Vert u \Vert ^{2}-\frac{b}{4} \Vert u \Vert ^{4}- \frac{C_{1}C_{\varepsilon}}{p} \Vert u \Vert ^{p}, \end{aligned} $$

Since $4< p<2^{*}$, for small enough $\rho>0$, for all $u\in E$ and $\| u\|=\rho$, it holds that $I(u)=\gamma>0$. On the other hand, for $u\neq 0$ and $t\in\mathbb{R}$,

$$ I(tu)=\frac{at^{2}}{2} \Vert u \Vert ^{2}-\frac{bt^{4}}{4} \Vert u \Vert ^{4}-\frac{\lambda t^{2}}{2} \int_{\Omega} \vert u \vert ^{2}- \int_{\Omega}G(x, tu), $$

so that when $t\rightarrow\infty$, we have $I(tu)\rightarrow-\infty$. This means that there is a $t_{1}$ such that $u_{1}=t_{1}u\in E$, $\| u_{1}\|>\rho$ and $I(u_{1})<0$. As a consequence, by the mountain pass lemma without (PS) condition [22], there exists a sequence $\{ u_{n}\}\subset E$ such that $I(u_{n})\rightarrow c$, $I'(u_{n})\rightarrow0$ for

$$c=\inf_{h\in\Gamma}\max_{u\in h([0, 1])}I(u)\geq\gamma>0, $$

where

$$\Gamma=\bigl\{ h\in C\bigl([0, 1], E\bigr): h(0)=0, h(1)=u_{1} \bigr\} . $$

Because

$$ \begin{aligned} \max_{t\in[0, 1]}I(tu_{1})&= \max_{t\in[0, 1]} \biggl\{ \frac{at^{2}}{2} \Vert u_{1} \Vert ^{2}-\frac{bt^{4}}{4} \Vert u_{1} \Vert ^{4}-\frac{\lambda t^{2}}{2} \int _{\Omega} \vert u_{1} \vert ^{2}- \int_{\Omega}G(x, tu_{1}) \biggr\} \\ &< \max_{t\in[0, 1]} \biggl\{ \frac{at^{2}}{2} \Vert u_{1} \Vert ^{2}-\frac {bt^{4}}{4} \Vert u_{1} \Vert ^{4} \biggr\} \\ &\leq\frac{a^{2}}{4b}, \end{aligned} $$

it is easy to obtain that $0< c<\frac{a^{2}}{4b}$ according to the definition of c. □

Lemma 3.2

Under the condition $c<\frac{a^{2}}{4b}$, I satisfies the $(PS)_{c}$ condition, i.e., any $(PS)_{c}$ sequence of I has a convergent subsequence.

Proof

We drew on the experience of [14]. Let $\{u_{n}\} \subset E$ be such that $I(u_{n})\rightarrow c$, $I'(u_{n})\rightarrow 0$. Since by (3.2)

$$\begin{aligned} c+o(1)&=I(u_{n})-\frac{1}{2}\bigl\langle I'(u_{n}), u_{n}\bigr\rangle \\ &=\frac{a}{2} \Vert u_{n} \Vert ^{2}- \frac{b}{4} \Vert u_{n} \Vert ^{4}- \frac{\lambda }{2} \int_{\Omega} \vert u_{n} \vert ^{2}- \int_{\Omega}G(x, u_{n}) \\ &\quad{}- \biggl[\frac{a}{2} \Vert u_{n} \Vert ^{2}- \frac{b}{2} \Vert u_{n} \Vert ^{4}- \frac{\lambda }{2} \int_{\Omega} \vert u_{n} \vert ^{2}- \frac{1}{2}g(x, u_{n}) \biggr] \\ &\geq\frac{b}{4} \Vert u_{n} \Vert ^{4}, \end{aligned}$$

we know that $\{u_{n}\}$ is bounded in E. By passing to a subsequence, still denoted $\{u_{n}\}$, we may assume that there is a $u\in E$ such that

$$ \begin{aligned} &u_{n}\rightharpoonup u \quad \text{in } E, \\ &u_{n}\rightarrow u \quad \text{in } L^{s}(\Omega) \text{ for } s\in [1, 2^{*}), \\ &u_{n}(x)\rightarrow u(x) \quad \text{for a.e. } x\in\Omega. \end{aligned} $$

On account of

$$ \begin{aligned}o(1)&=\bigl\langle I'(u_{n}), u_{n}-u\bigr\rangle \\ &=\bigl(a-b \Vert u_{n} \Vert ^{2}\bigr) \int_{\Omega}\nabla u_{n} \nabla(u_{n}-u)- \lambda \int_{\Omega}u_{n}(u_{n}-u)- \int_{\Omega}g(x, u_{n}) (u_{n}-u) \end{aligned} $$

and

$$\biggl\vert \int_{\Omega}u_{n}(u_{n}-u) \biggr\vert \leq \biggl( \int_{\Omega } \vert u_{n} \vert ^{2} \biggr)^{\frac{1}{2}} \biggl( \int_{\Omega} \vert u_{n}-u \vert ^{2} \biggr)^{\frac{1}{2}}, $$

also by (3.1)

$$ \begin{aligned} & \biggl\vert \int_{\Omega}g(x, u_{n}) (u_{n}-u) \biggr\vert \\ &\quad \leq\varepsilon \biggl\vert \int_{\Omega}u_{n}(u_{n}-u) \biggr\vert +C_{\varepsilon } \biggl\vert \int_{\Omega} \vert u_{n} \vert ^{p-2}u_{n}(u_{n}-u) \biggr\vert \\ &\quad \leq\varepsilon \biggl( \int_{\Omega} \vert u_{n} \vert ^{2} \biggr)^{\frac{1}{2}} \biggl( \int_{\Omega} \vert u_{n}-u \vert ^{2} \biggr)^{\frac{1}{2}} +C_{\varepsilon} \biggl( \int_{\Omega}\bigl( \vert u_{n} \vert ^{p-1} \bigr)^{\frac{p}{p-1}} \biggr)^{\frac{p-1}{p}} \biggl( \int_{\Omega}\bigl( \vert u_{n}-u \vert ^{p}\bigr) \biggr)^{\frac{1}{p}}, \end{aligned} $$

because $u_{n}\rightarrow u$ in $L^{s}(\Omega)$, $s\in[1, 2^{*})$, the above two formulas show that when $n\rightarrow\infty$,

$$ \bigl(a-b \Vert u_{n} \Vert ^{2}\bigr) \int_{\Omega}\nabla u_{n} \nabla(u_{n}-u) \rightarrow0. $$

(3.3)

If there exists a subsequence of $\{u_{n}\}$, still denoted $\{u_{n}\}$, satisfying $\|u_{n}\|^{2}\rightarrow\frac{a}{b}$, define a functional

$$\varphi(u)=\frac{\lambda}{2} \int_{\Omega} \vert u \vert ^{2}+ \int_{\Omega}G(x, u), \quad u\in E. $$

Then

$$\bigl\langle \varphi'(u), v\bigr\rangle =\lambda \int_{\Omega}uv+ \int_{\Omega}g(x, u)v, \quad u, v\in E, $$

and

$$\bigl\langle \varphi'(u_{n})-\varphi'(u), v\bigr\rangle =\lambda \int_{\Omega }(u_{n}-u)v+ \int_{\Omega}\bigl[g(x, u_{n})-g(x, u)\bigr]v. $$

Claim. $\langle\varphi'(u_{n})-\varphi'(u), v\rangle \rightarrow0$, $\forall v\in E$.

Firstly,

$$\lambda \int_{\Omega}(u_{n}-u)v\leq\lambda \biggl( \int_{\Omega } \vert u_{n}-u \vert ^{2} \biggr)^{\frac{1}{2}} \biggl( \int_{\Omega} \vert v \vert ^{2} \biggr)^{\frac{1}{2}}, $$

since $u_{n}\rightarrow u$ in $L^{2}(\Omega)$, thus $\lambda\int_{\Omega }(u_{n}-u)v\rightarrow0$.

Secondly, to prove the claim, we only need to prove

$$ \lim_{n\rightarrow\infty} \int_{\Omega} \bigl\vert g(x, u_{n})-g(x, u) \bigr\vert \vert v \vert =0. $$

(3.4)

If (3.4) is not true, then there exist a constant $\varepsilon_{0}>0$ and a subsequence $u_{k_{i}}$ such that

$$ \int_{\Omega} \bigl\vert g(x, u_{k_{i}})-g(x, u) \bigr\vert \vert v \vert \geq\varepsilon_{0}, \quad \forall i\in \mathbb{N}, $$

(3.5)

Since $u_{n}\rightarrow u$ in $L^{p}(\Omega)$, passing to a subsequence if necessary, we can assume that $\sum_{i=1}^{\infty }|u_{k_{i}}-u|_{p}^{p}<+\infty$. Set

$$\omega(x)= \Biggl[\sum_{i=1}^{\infty} \bigl\vert u_{k_{i}}(x)-u(x) \bigr\vert ^{p} \Biggr]^{\frac {1}{p}},\quad \forall x\in\Omega. $$

Then $\omega\in L^{p}(\Omega)$. Note that for $\forall i\in\mathbb{N}$, $x\in\Omega$,

$$ \begin{aligned}[b] & \bigl\vert g(x, u_{k_{i}})-g(x, u) \bigr\vert \vert v \vert \\ &\quad \leq\bigl( \bigl\vert g(x, u_{k_{i}}) \bigr\vert + \bigl\vert g(x, u) \bigr\vert \bigr) \vert v \vert \\ &\quad \leq\bigl[\varepsilon\bigl( \vert u_{k_{i}} \vert + \vert u \vert \bigr)+C_{\varepsilon }\bigl( \vert u_{k_{i}} \vert ^{p-1}+ \vert u \vert ^{p-1}\bigr)\bigr] \vert v \vert \\ &\quad \leq\bigl[2^{2}\varepsilon\bigl( \vert u_{k_{i}}-u \vert + \vert u \vert \bigr)+2^{p}C_{\varepsilon }\bigl( \vert u_{k_{i}}-u \vert ^{p-1}+ \vert u \vert ^{p-1}\bigr) \bigr] \vert v \vert \\ &\quad \leq\bigl[2^{2}\varepsilon\bigl( \vert \omega \vert + \vert u \vert \bigr)+2^{p}C_{\varepsilon}\bigl( \vert \omega \vert ^{p-1}+ \vert u \vert ^{p-1}\bigr)\bigr] \vert v \vert \\ &\quad :=f(x), \end{aligned} $$

(3.6)

and

$$ \begin{aligned} \int_{\Omega}f(x)\,dx&= \int_{\Omega}\bigl[2^{2}\varepsilon\bigl( \vert \omega \vert + \vert u \vert \bigr)+2^{p}C_{\varepsilon}\bigl( \vert \omega \vert ^{p-1}+ \vert u \vert ^{p-1}\bigr)\bigr] \vert v \vert \\ &\leq2^{2}\varepsilon\bigl( \vert \omega \vert _{2}+ \vert u \vert _{2}\bigr) \vert v \vert _{2}+2^{p}C_{\varepsilon } \bigl( \vert \omega \vert _{p}^{p-1}+ \vert u \vert _{p}^{p-1}\bigr) \vert v \vert _{p}< +\infty. \end{aligned} $$

(3.7)

Since $u_{k_{i}}\rightarrow u$ a.e. in Ω, then by (3.6), (3.7) and Lebesgue Dominated Convergence Theorem, we have

$$\lim_{i\rightarrow\infty} \int_{\Omega} \bigl\vert g\bigl(x, u_{k_{i}}(x)\bigr)-g \bigl(x, u(x)\bigr) \bigr\vert \vert v \vert =0, $$

which contradicts (3.5). Hence (3.4) holds. Then the claim follows. By arbitrariness of v, then

$$\bigl\Vert \varphi'(u_{n})-\varphi'(u) \bigr\Vert _{E'}\rightarrow0, $$

and $\varphi'(u_{n})\rightarrow\varphi'(u)$ in $E'$. While $\langle I'(u_{n}), v\rangle=(a-b\|u_{n}\|^{2})\langle u_{n}, v\rangle-\langle \varphi'(u_{n}), v\rangle$, $\langle I'(u_{n}), v\rangle\rightarrow0$, $a-b\|u_{n}\|^{2}\rightarrow 0$, hence $\varphi'(u_{n})\rightarrow0$, i.e.,

$$\bigl\langle \varphi'(u), v\bigr\rangle =\lambda \int_{\Omega}uv+ \int_{\Omega}g(x, u)v=0,\quad\forall v\in E, $$

and then we have

$$\lambda u(x)+g\bigl(x, u(x)\bigr)=0\quad\text{for a.e. }x\in\Omega, $$

by the fundamental lemma of the variational method (see [23]). It follows that $u=0$. So

$$\varphi(u_{n})=\frac{\lambda}{2} \int_{\Omega} \vert u_{n} \vert ^{2}+ \int_{\Omega }G(x, u_{n})\rightarrow\frac{\lambda}{2} \int_{\Omega} \vert u \vert ^{2}+ \int _{\Omega}G(x, u)=0. $$

Hence we see that $I(u_{n})=\frac{a}{2}\|u_{n}\|^{2}-\frac{b}{4}\| u_{n}\|^{4}-\frac{\lambda}{2}\int_{\Omega}|u_{n}|^{2}-\int_{\Omega }G(x, u_{n})\rightarrow\frac{a^{2}}{4b}$ from $\|u_{n}\|^{2}\rightarrow \frac{a}{b}$. This is a contradiction to $I(u_{n})\rightarrow c<\frac {a^{2}}{4b}$. Then $a-b\|u_{n}\|^{2}\rightarrow0$ is not true and any subsequence of $\{a-b\|u_{n}\|^{2}\rightarrow0\}$ does not converge to zero. Therefore there exists a $\delta>0$ such that $|a-b\|u_{n}\| ^{2}|>\delta>0$ when n is large enough. It is clear that $\{a-b\| u_{n}\|^{2}\rightarrow0\}$ is bounded. It follows from (3.3) that $\int_{\Omega}\nabla u_{n}\nabla (u_{n}-u)\rightarrow0$. So $\|u_{n}\|\rightarrow\|u\|$. Hence $u_{n}\rightarrow u$ in E due to the uniform convexity of E. □

Proof of Theorem 1.1

According to Lemma 3.1, there exists a sequence $\{u_{n}\}\in E$ satisfying $I(u_{n})\rightarrow c>0$, $I'(u_{n})\rightarrow0$. By Lemma 3.2, $\{u_{n}\}$, which is the sequence obtained by Lemma 3.1, possesses a convergent to u subsequence (still denoted by $\{u_{n}\}$). So it follows from the continuity that $I(u_{n})\rightarrow c>0$, $I'(u_{n})\rightarrow0$. But $I(0)=0$, therefore $u\neq0$, that is, u is a nontrivial solution of problem (1.1). □