Citation: Paul W. Mayne, Ethan Cargill, Bruce Miller. Geotechnical characteristics of sensitive Leda clay at Canada test site in Gloucester, Ontario[J]. AIMS Geosciences, 2019, 5(3): 390-411. doi: 10.3934/geosci.2019.3.390
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In this paper, we study a class of Schrödinger-Poisson system with the following version
{−Δu+u+K(x)ϕu=|u|p−1u+μh(x)u in R3,−Δϕ=K(x)u2 in R3, | (1) |
where
{−Δu+u+ϕu=f(u) in R3,−Δϕ=u2 in R3, | (2) |
which has been derived from finding standing waves of the Schrödinger-Poisson system
{iψt−Δψ+ϕψ=f(ψ) in R3,−Δϕ=|ψ|2 in R3. |
A starting point of studying system (1) is the following fact. For any
ϕu(x)=14π∫R3K(y)|u(y)|2|x−y|dy |
such that
−Δu+u+K(x)ϕuu=|u|p−1u+μh(x)u,u∈H1(R3). | (3) |
Problem (3) can be also looked on as a usual semilinear elliptic equation with an additional nonlocal perturbation
From Lemma 2.1, we know that under the condition (A1), the following eigenvalue problem
−Δu+u=μh(x)u,u∈ H1(R3) |
has a first eigenvalue
F(u):=∫R3K(x)ϕu(x)|u(x)|2dx |
and introduce the energy functional
Iμ(u)=12‖u‖2+14F(u)−∫R3(1p+1|u|p+1+μ2h(x)u2)dx, |
where
⟨I′μ(u),v⟩=∫R3(∇u∇v+uv+K(x)ϕuuv−|u|p−1uv+μh(x)uv)dx. |
It is known that there is a one to one correspondence between solutions of (3) and critical points of
Theorem 1.1. Suppose that the assumptions of
The second result is about
Theorem 1.2. Under the assumptions of
The proofs of Theorem 1.1 and Theorem 1.2 are based on critical point theory. There are several difficulties in the road of getting critical points of
While for nonautonomous version of Schrödinger-Poisson system, only a few results are known in the literature. Jiang et.al.[21] have studied the following Schrödinger-Poisson system with non constant coefficient
{−Δu+(1+λg(x))u+θϕ(x)u=|u|p−2u in R3,−Δϕ=u2in R3,lim |
in which the authors prove the existence of ground state solution and its asymptotic behavior depending on
\left\{ \begin{array}{ll} -\Delta u +u + \phi u = V(x)|u|^4u + \mu P(x)|u|^{q-2}u \quad \hbox{in}\ \mathbb{R}^3,& \\ -\Delta \phi = u^2 \quad \hbox{in} \ \mathbb{R}^3, \quad 2 < q < 6, \mu > 0& \end{array} \right. |
has been studied by Zhao et al. [31]. Besides some other conditions, Zhao et. al. [31] assume that
\begin{equation} \left\{ \begin{array}{ll} -\Delta u +u+L(x)\phi u = g(x, u) \quad & \ \hbox{in}\ \mathbb{R}^3,\\ -\Delta \phi = L(x)u^2\quad & \ \hbox{in} \ \mathbb{R}^3. \end{array} \right. \end{equation} | (4) |
Besides some other conditions and the assumption
\begin{equation} -\Delta u = a(x)|u|^{p-2}u + \tilde{\mu}k(x)u, \quad \ \hbox{in}\ \mathbb{R}^N, \end{equation} | (5) |
Costa et.al.[14] have proven the mountain pass geometry for the related functional of (5) when
This paper is organized as follows. In Section 2, we give some preliminaries. Special attentions are focused on several lemmas analyzing the Palais-Smale conditions of the functional
Notations. Throughout this paper,
S_{p+1} = {\inf\limits_{u\in H^1(\mathbb{R}^3)\setminus\{0\}}}\frac{\int_{\mathbb{R}^3}\left(|\nabla u|^2 + |u|^2\right)dx}{\left(\int_{\mathbb{R}^3}| u|^{p+1}dx\right)^{\frac2{p+1}}}. |
For any
In this section, we give some preliminary lemmas, which will be helpful to analyze the (PS) conditions for the functional
L_u(v) = \int_{{\mathbb{R}}^3} K(x)u^2vdx,\qquad v\in D^{1, 2}(\mathbb{R}^3), |
one may deduce from the Hölder and the Sobolev inequalities that
\begin{equation} |L_u(v)|\leq C\|u\|_{L^{\frac{12}{5}}}^2\|v\|_{L^6}\leq C \|u\|_{L^{\frac{12}{5}}}^2\|v\|_{D^{1, 2}}. \end{equation} | (6) |
Hence, for any
\phi_u(x) = \frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{K(y)u^2(y)}{|x-y|}dy. |
Clearly
\begin{equation} \|\phi_u\|_{D^{1, 2}}^2 = \int_{\mathbb{R}^3}|\nabla \phi_u|^2dx = \int_{\mathbb{R}^3}K(x)\phi_uu^2dx. \end{equation} | (7) |
Using (6) and (7), we obtain that
\begin{equation} \|\phi_{u}\|_{L^6} \leq C\|\phi_u\|_{D^{1, 2}}\leq C\|u\|_{L^{\frac{12}{5}}}^2\leq C\|u\|^2. \end{equation} | (8) |
Then we deduce that
\begin{equation} \int_{\mathbb{R}^3} K(x)\phi_u(x)u^2(x)dx\leq C \|u\|^4. \end{equation} | (9) |
Hence on
\begin{equation} F(u) = \int_{\mathbb{R}^3} K(x)\phi_u(x)u^2(x)dx \end{equation} | (10) |
and
\begin{equation} I_\mu(u) = \frac{1}{2}\|u\|^2+\frac{1}{4}F(u)-\int_{\mathbb{R}^3}\left({\frac{1}{p+1}} |u|^{p+1} + \frac{\mu}{2} h(x)u^2\right)dx \end{equation} | (11) |
are well defined and
\langle I'_\mu(u), v\rangle = \int_{\mathbb{R}^3}\left( \nabla u\nabla v+uv + K(x) \phi_uuv - |u|^{p-1}uv - \mu h(x)uv\right)dx. |
The following Lemma 2.1 is a direct consequence of [28,Lemma 2.13].
Lemma 2.1. Assume that the hypothesis
Using the spectral theory of compact symmetric operators on Hilbert space, the above lemma implies the existence of a sequence of eigenvalues
-\Delta u+u = \mu h(x)u, \; \hbox{in}\; H^1(\mathbb{R}^3) |
with
\mu_1 = {\inf\limits_{u\in {H^1(\mathbb{R}^3)}\setminus \{0\}}}\frac{\|u\|^2}{ \int_{\mathbb{R}^3} h(x)u^2dx}, \quad \mu_n = {\inf\limits_{u\in {S_{n-1}^\perp}\setminus \{0\}}}\frac{\|u\|^2}{ \int_{\mathbb{R}^3} h(x)u^2dx}, |
where
Next we analyze the
Definition 2.2. For
Lemma 2.3. Let
Proof. For
\begin{eqnarray} &d& + 1 + o(1)\|u_n\| = I_\mu(u_n) -\frac14 \langle I_\mu'(u_n),u_n \rangle \\ & = & \frac{1}{4}\|u_n\|^2 - \frac{\mu}{4}\int_{\mathbb{R}^3}h(x)u_n^2dx + \frac{p-3}{4(p+1)}\int_{\mathbb{R}^3}|u_n|^{p+1} dx. \end{eqnarray} | (12) |
Note that
\begin{array}{rl} \int_{\mathbb{R}^3}h(x)u_n^2dx &\leq \left(\int_{\mathbb{R}^3}|u_n|^{p+1}dx\right)^{\frac2{p+1}} \left(\int_{\mathbb{R}^3}|h(x)|^{\frac{p+1}{p-1}}dx\right)^{\frac{p-1}{p+1}} \\ & \leq \frac{2\vartheta}{p+1} \int_{\mathbb{R}^3}|u_n|^{p+1}dx + \frac{p-1}{p+1} \vartheta^{-\frac2{p-1}}\int_{\mathbb{R}^3}|h(x)|^{\frac{p+1}{p-1}}dx. \end{array} |
Choosing
\begin{equation} d + 1 + o(1)\|u_n\| \geq \frac{1}{4}\|u_n\|^2 - D(p, h) \mu^{\frac{p+1}{p-1}}, \end{equation} | (13) |
where
The following lemma is a variant of Brezis-Lieb lemma. One may find the proof in [20].
Lemma 2.4. [20] If a sequence
\lim\limits_{n\to\infty}F(u_n) = F(u_0) + \lim\limits_{n\to\infty}F(u_n-u_0). |
Lemma 2.5. There is a
I_\mu(u) > - \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}}. |
Proof. Since
\begin{array}{rl} I_\mu(u) & = \frac12 \left(\|u\|^2 - \mu\int_{\mathbb{R}^3} h(x)u^2 dx\right) + \frac14 F(u) - \frac1{p+1}\int_{\mathbb{R}^3} |u|^{p+1} dx\\ & = \frac{p-1}{2(p+1)} \left(\|u\|^2 - \mu\int_{\mathbb{R}^3} h(x)u^2 dx\right) +\frac{p-3}{4(p+1)}F(u). \end{array} |
Noticing that
I_{\mu_1}(u) \geq \frac{p-3}{4(p+1)}F(u) > 0. |
Next, we claim: there is a
I_\mu(u) > - \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}}. |
Suppose this claim is not true, then there is a sequence
I_{\mu^{(n)}}(u_{\mu^{(n)}}) \leq - \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}}. |
Note that
\begin{eqnarray} I_{\mu^{(n)}}(u_{\mu^{(n)}}) + o(1)\|u_{\mu^{(n)}}\|&\geq & I_{\mu^{(n)}}(u_{\mu^{(n)}}) -\frac14 \langle I'_{\mu^{(n)}}(u_{\mu^{(n)}}), u_{\mu^{(n)}} \rangle \\ &\geq & \frac{1}{4}\|u_{\mu^{(n)}}\|^2 - D(p,h) \left(\mu^{(n)}\right)^{\frac{p+1}{p-1}}. \end{eqnarray} |
This implies that
\|u_{\mu^{(n)}}\|^2 - \mu^{(n)} \int_{\mathbb{R}^3} h(x)(u_{\mu^{(n)}})^2 dx \geq \left(1-\frac{\mu^{(n)}}{\mu_1}\right)\|u_{\mu^{(n)}}\|^2\to 0 |
because
\begin{array}{rl} I_{\mu^{(n)}}(u_{\mu^{(n)}}) & = \frac{p-1}{2(p+1)} \left(\|u_{\mu^{(n)}}\|^2 - \mu^{(n)} \int_{\mathbb{R}^3} h(x)(u_{\mu^{(n)}})^2 dx\right) \\ &\qquad + \frac{p-3}{4(p+1)}F(u_{\mu^{(n)}}), \end{array} |
we deduce that
\liminf\limits_{n\to\infty}I_{\mu^{(n)}}(u_{\mu^{(n)}}) \geq \frac{p-3}{4(p+1)}\liminf\limits_{n\to\infty} F(u_{\mu^{(n)}}) \geq 0, |
which contradicts to the
I_{\mu^{(n)}}(u_{\mu^{(n)}}) \leq - \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}}. |
This proves the claim and the proof of Lemma 2.5 is complete.
Lemma 2.6. If
Proof. Let
d + o(1) = \frac{1}{2}\|u_n\|^2 - \frac{\mu}{2}\int_{\mathbb{R}^3}h(x)u_n^2dx + \frac{1}{4}F(u_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|u_n|^{p+1}dx |
and
\langle I'_\mu (u_n), u_n\rangle = \|u_n\|^2 - \mu\int_{\mathbb{R}^3}h(x)u_n^2dx + F(u_n) - \int_{\mathbb{R}^3}|u_n|^{p+1}dx. |
Then we can prove that
\|u_n\|^2 = \|u_0\|^2 + \|w_n\|^2 +o(1), |
F(u_n) = F(u_0) + F(w_n) +o(1) |
and
\|u_n\|^{p+1}_{L^{p+1}} = \|u_0\|^{p+1}_{L^{p+1}} + \|w_n\|^{p+1}_{L^{p+1}} +o(1). |
Using Lemma 2.1, we also have that
\begin{eqnarray} d + o(1)& = & I_\mu(u_n) = I_\mu(u_0) + \frac{1}{2}\| w_n\|^2\\ &+& \frac{1}{4}F(w_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|w_n|^{p+1}dx. \end{eqnarray} | (14) |
Noticing
\begin{equation} \|u_0\|^2 - \mu\int_{\mathbb{R}^3}h(x)u_0^2dx + F(u_0) = \int_{\mathbb{R}^3}|u_0|^{p+1}dx. \end{equation} | (15) |
Since
o(1) = \|u_n\|^2 - \mu\int_{\mathbb{R}^3}h(x)u_n^2dx + F(u_n) - \int_{\mathbb{R}^3}|u_n|^{p+1}dx. |
Combining this with (15) as well as Lemma 2.1, we obtain that
\begin{equation} o(1) = \|w_n\|^2 + F(w_n) -\int_{\mathbb{R}^3}|w_n|^{p+1}dx. \end{equation} | (16) |
Recalling the definition of
Suppose that the case (ⅰ) occurs. We may obtain from (16) that
\|w_n\|^2 \geq S_{p+1} \left(\|w_n\|^2 + F(w_n) - o(1)\right)^{\frac2{p+1}}. |
Hence we get that for
\begin{equation} \|w_n\|^2 \geq S_{p+1}^{\frac{p+1}{p-1}} + o(1). \end{equation} | (17) |
Therefore using (14), (16) and (17), we deduce that for
\begin{eqnarray} & d& + o(1) = I_\mu(u_n) \\ & = & I_\mu(u_0) + \frac{1}{2}\|w_n\|^2 + \frac{1}{4}F(w_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|w_n|^{p+1}dx. \\ & = & I_\mu(u_0) + \frac{p-1}{2(p+1)}\|w_n\|^2 + \frac{p-3}{4(p+1)}F(w_n) \\ & > & - \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}} + \frac{p-1}{2(p+1)}\|w_n\|^2 + \frac{p-3}{4(p+1)}F(w_n) \\ & > & 0, \end{eqnarray} | (18) |
which contradicts to the condition
Next we give a mountain pass geometry for the functional
Lemma 2.7. There exist
Proof. For any
\begin{equation} u = te_1+v, \; \hbox{where}\; \int_{\mathbb{R}^3} \left(\nabla v \nabla e_1 + ve_1\right)dx = 0. \end{equation} | (19) |
Hence we deduce that
\begin{equation} \|u\|^2 = \|\nabla (te_1+v)\|_{L^2}^2 + \|te_1+v\|_{L^2}^2 = t^2+\|v\|^2, \end{equation} | (20) |
\begin{equation} \mu_2\int_{\mathbb{R}^3} h(x)v^2dx\leq \|v\|^2, \; \; \mu_1\int_{\mathbb{R}^3} h(x)e_1^2dx = \|e_1\|^2 = 1 \end{equation} | (21) |
and
\begin{equation} \mu_1\int_{\mathbb{R}^3} h(x)e_1 v dx = \int_{\mathbb{R}^3} \left(\nabla v \nabla e_1+ve_1\right)dx = 0. \end{equation} | (22) |
We first consider the case of
\begin{array}{rl} I_{\mu_1}(u)& = \frac{1}{2} \|u\|^2+\frac{1}{4}F(u)- \frac{\mu_1}{2}\int_{\mathbb{R}^3}h(x)u^2dx - \frac{1}{p+1}\int_{\mathbb{R}^3} |u|^{p+1}dx\\ & = \frac{1}{2} \|t e_1 + v\|^2 +\frac{1}{4} F(te_1+v)\\ & \quad -\frac{\mu_1}{2}\int_{\mathbb{R}^3}h(x)(te_1+v)^2dx-\frac{1}{p+1}\int_{\mathbb{R}^3} |te_1 + v|^{p+1}dx\\ & \geq\frac{1}{2}\left(1-\frac{\mu_1}{\mu_2}\right) \|v\|^2 +\frac{1}{4} F(te_1+v)-\frac{1}{p+1}\int_{\mathbb{R}^3} |te_1 + v|^{p+1}dx\\ & \geq \theta_1\|v\|^2 +\frac{1}{4} F(te_1+v) - C_1|t|^{p+1} - C_2\|v\|^{p+1}. \end{array} |
Next we estimate the term
F(te_1+v) = \frac{1}{4\pi}\int_{\mathbb{R}^3\times \mathbb{R}^3}\frac{K(x)K(y)(te_1(y)+v(y))^2(te_1(x)+v(x))^2}{|x-y|}dydx. |
Since
\begin{array}{rl} & \left(te_1(y)+v(y)\right)^2\left(te_1(x)+v(x)\right)^2 = t^4 (e_1(y))^2(e_1(x))^2 + (v(y))^2(v(x))^2\\ & \qquad + 2t^3\left(e_1(y)(e_1(x))^2v(y) + e_1(x)(e_1(y))^2v(x)\right)\\ & \qquad + 2t\left(e_1(x)v(x)(v(y))^2 + e_1(y)v(y)(v(x))^2\right)\\ & \qquad +t^2\left((e_1(x))^2(v(y))^2 + 4e_1(y)e_1(x)v(y)v(x) + (e_1(y))^2(v(x))^2\right), \end{array} |
we know that
\begin{equation} \left|\int_{\mathbb{R}^3\times \mathbb{R}^3}\frac{K(x)K(y)\left(e_1(y)(e_1(x))^2v(y) + e_1(x)(e_1(y))^2v(x)\right)}{|x-y|}dydx\right|\leq C\|v\|; \end{equation} | (23) |
\begin{equation} \left|\int_{\mathbb{R}^3\times \mathbb{R}^3}\frac{K(x)K(y)\left(2(e_1(x))^2(v(y))^2 + 4e_1(y)e_1(x)v(y)v(x)\right)}{|x-y|}dydx\right|\leq C\|v\|^2 \end{equation} | (24) |
and
\begin{equation} \left|\int_{\mathbb{R}^3\times \mathbb{R}^3}\frac{K(x)K(y)\left(e_1(x)v(x)(v(y))^2 + e_1(y)v(y)(v(x))^2\right)}{|x-y|}dydx\right|\leq C\|v\|^3. \end{equation} | (25) |
Hence
\begin{array}{rl} I_{\mu_1}(u)& \geq \theta_1\|v\|^2 + \theta_2|t|^4 - C_1|t|^{p+1} - C_2\|v\|^{p+1} \\ & \qquad -C_3|t|^3\|v\| - C_4|t|^2\|v\|^2 - C_5|t|\|v\|^3+\frac{1}{4}F(v), \end{array} |
where
t^2\|v\|^2 \leq \frac2{p+1} |t|^{p+1} + \frac{p-1}{p+1} \|v\|^{\frac{2(p+1)}{p-1}}, |
|t|\|v\|^3 \leq \frac1{p+1} |t|^{p+1} + \frac{p}{p+1}\|v\|^{\frac{3(p+1)}{p}} |
and for some
|t|^3\|v\| \leq \frac{1}{q_0} \|v\|^{q_0} + \frac{q_0 - 1}{q_0} |t|^{\frac{3q_0}{q_0 - 1}}. |
Therefore we deduce that
\begin{equation} \begin{array}{rl} & I_{\mu_1}(u) \geq \theta_1 \|v\|^2 + \theta_2 |t|^4 - \frac{C_3}{q_0} \|v\|^{q_0} - \frac{C_3(q_0 - 1)}{q_0} |t|^{\frac{3q_0}{q_0 - 1}}\\ & \quad -\frac{2C_4}{p+1} |t|^{p+1} - \frac{(p-1)C_4}{p+1}\|v\|^{\frac{2(p+1)}{p-1}} - \frac{C_5}{p+1} |t|^{p+1} \\ & \quad - \frac{pC_5}{p+1}\|v\|^{\frac{3(p+1)}{p}} - C|t|^{p+1} - C\|v\|^{p+1}. \end{array} \end{equation} | (26) |
From
I_{\mu_1}(u) \geq \theta_3 \|v\|^2 + \theta_4 |t|^4 |
provided that
\begin{equation} I_{\mu_1}(u) \geq \theta_5 \|u\|^4\qquad \hbox{for}\quad \|u\|^2 \leq \tilde{\theta}_5^2. \end{equation} | (27) |
Set
\begin{array}{rl} I_\mu (u) & = I_{\mu_1}(u) + \frac{1}{2}(\mu_1 - \mu)\int_{\mathbb{R}}h(x)u^2dx\\ & \geq \theta_5 \|u\|^4 - \frac{\mu - \mu_1}{2\mu_1}\|u\|^2\\ & = \|u\|^2 \left(\theta_5 \|u\|^2 - \frac{\mu - \mu_1}{2\mu_1}\right)\\ & \geq \|u\|^2\left(\frac{1}{2} \theta_5 \tilde{\theta}^2_5 - \frac{1}{4}\theta_5\tilde{\theta}^2_5\right) = \frac{1}{4}\theta_5\tilde{\theta}^2_5\|u\|^2 \end{array} |
for
In this section, our aim is to prove Theorem 1.1. For
Proposition 3.1. Let the assumptions
d_{\mu_1} = \inf\limits_{\gamma \in \Gamma_1}\sup\limits_{t\in [0,1]}I_{\mu_1}(\gamma(t)) |
with
\Gamma_1 = \left\{ \gamma \in C([0,1], H^1(\mathbb{R}^3))\ : \ \gamma(0) = 0,\ I_{\mu_1}(\gamma(1)) < 0 \right\}. |
Then
Before proving Proposition 3.1, we analyze the
Lemma 3.2. If the assumptions
Proof. It suffices to find a path
\sup\limits_{t\in [0,1]}I_{\mu_1}(\gamma(t)) < \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}}. |
Define
\|U_R\|^2 -\mu_1\int h(x)U_R^2 dx+ T_R^2F(U_R) - T_R^{p-1}\int U_R^{p+1} dx = 0. |
If
\frac{1}{T_R^2}\left(\|U_R\|^2 -\mu_1\int h(x)U_R^2dx\right)+F(U_R) = T_R^{p-3}\int U_R^{p+1}dx \to \infty, |
which is impossible either. Hence we only need to estimate
I_{\mu_1}(tU_R) \leq g(t) + CF(U_R), |
where
g(t) = \frac{t^2}{2}\left(\|U_R\|^2 - \mu_1 \int_{\mathbb{R}^3} h(x) U_R^2 dx\right) - \frac{|t|^{p+1}}{p+1}\int_{\mathbb{R}^3} U_R^{p+1} dx. |
Noting that under the assumptions
\begin{equation} \begin{array}{rl} F(U_R) & \leq \left(\int_{\mathbb{R}^3}K(x)^{\frac65}U_R^{\frac{12}{5}}dx\right)^{\frac56} \left(\int_{\mathbb{R}^3} \phi_{U_R}^6dx\right)^{\frac16}\\ & \leq C\left(\int_{\mathbb{R}^3} e^{-\frac65 a|x+R\theta|}(U(x))^{\frac{12}{5}} dx\right)^{\frac56}\\ & \leq C\left(\int_{\mathbb{R}^3} e^{-{\frac65}aR}e^{({\frac65}a-{\frac{12}5}(1-\varepsilon))|x|} dx\right)^{\frac56} \leq C e^{-aR} \end{array} \end{equation} | (28) |
since
\begin{equation} \begin{array}{rl} & \int_{\mathbb{R}^3}h(x)U_R^2dx = \int_{\mathbb{R}^3}h(x+R\theta)U^2(x)dx\\ & \geq C\int_{\mathbb{R}^3} e^{-b|x+R\theta|}U^2(x) dx \geq C\int_{\mathbb{R}^3} e^{-b|x| -bR } U^2(x) dx\\ & \geq C e^{-bR} \int_{\mathbb{R}^3} e^{-b|x|} U^2(x)dx \geq C e^{-bR}. \end{array} \end{equation} | (29) |
It is now deduced from (28) and (29) that
\begin{array}{rl} & \sup\limits_{t > 0} I_{\mu_1}(tU_R) \leq \sup\limits_{t > 0} g(t) + C e^{-aR} \\ &\leq \frac{p-1}{2(p+1)} \left(\|U_R\|^2 - \mu_1 \int_{\mathbb{R}^3} h(x) U_R^2 dx\right)^{\frac{p+1}{p-1}}(\|U_R\|_{L^{p+1}}^{-2})^{\frac{p+1}{p-1}} + C e^{-aR} \\ &\leq \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}} - Ce^{-bR} + o(e^{-bR}) + C e^{-aR} \\ & < \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}} \end{array} |
for
Lemma 3.3. Under the assumptions
Proof. Let
d + o(1) = \frac{1}{2}\|u_n\|^2 - \frac{\mu_1}{2}\int_{\mathbb{R}^3}h(x)u_n^2dx + \frac{1}{4}F(u_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|u_n|^{p+1}dx |
and
\langle I'_{\mu_1} (u_n), u_n\rangle = \|u_n\|^2 - \mu_1\int_{\mathbb{R}^3}h(x)u_n^2dx + F(u_n) - \int_{\mathbb{R}^3}|u_n|^{p+1}dx. |
Hence we can deduce that
\|u_n\|^2 = \|u_0\|^2 + \|w_n\|^2 +o(1),\quad F(u_n) = F(u_0) + F(w_n) +o(1) |
and
\|u_n\|^{p+1}_{L^{p+1}} = \|u_0\|^{p+1}_{L^{p+1}} + \|w_n\|^{p+1}_{L^{p+1}} +o(1). |
Since
\begin{eqnarray} & & d + o(1) = I_{\mu_1}(u_n) = I_{\mu_1}(u_0) + \frac{1}{2}\| w_n\|^2\\ & & \qquad \quad + \frac{1}{4}F(w_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|w_n|^{p+1}dx. \end{eqnarray} | (30) |
From
\|u_0\|^2 - \mu_1\int_{\mathbb{R}^3}h(x)u_0^2dx + F(u_0) = \int_{\mathbb{R}^3}|u_0|^{p+1}dx |
and then
I_{\mu_1}(u_0) \geq \frac{p-1}{2(p+1)}\left(\|u_0\|^2 - \mu_1 \int_{\mathbb{R}^3}h(x)u_0^2dx\right) + \frac{p-3}{4(p+1)}F(u_0) \geq 0. |
Now using an argument similar to the proof of (16), we obtain that
\begin{equation} o(1) = \|w_n\|^2 + F(w_n) - \int_{\mathbb{R}^3}|w_n|^{P+1}dx. \end{equation} | (31) |
By the relation
Suppose that the case (I) occurs. Then up to a sbusequence, we may obtain from (31) that
\|w_n\|^2 \geq S_{p+1} \left(\| w_n\|^2 + F(w_n) - o(1)\right)^{\frac2{p+1}}, |
which implies that for
\|w_n\|^2 \geq S_{p+1}^{\frac{p+1}{p-1}} + o(1). |
It is deduced from this and (30) that
Proof of Proposition 3.1. Since
Proof of Theorem 1.1. By Proposition 3.1, the
\begin{equation} d_{\mu_1} \leq \max\limits_{t\in [0, 1]} I_{\mu_1}(\gamma_n(t)) < d_{\mu_1} + \frac{1}{n}. \end{equation} | (32) |
By Ekeland's variational principle [5], there exists
\begin{equation} \left\{ \begin{array}{ll} d_{\mu_1} \leq \max\limits_{t\in [0, 1]} I_{\mu_1} (\gamma_n^*(t))\leq \max\limits_{t\in [0, 1]} I_{\mu_1} (\gamma_n(t)) < d_{\mu_1} +\frac{1}{n};\\ \max\limits_{t\in [0, 1]}\|\gamma_n(t)-\gamma_n^*(t)\| < \frac{1}{\sqrt{n}}; \\ \hbox{ there exists} \; t_n\in [0, 1]\; \hbox{such that}\; z_n = \gamma_n^*(t_n) \hbox{ satisfies}: \\ I_{\mu_1}(z_n) = \max\limits_{t\in [0, 1]} I_{\mu_1}(\gamma_n^*(t)), \;\hbox{and}\; \|I'_{\mu_1}(z_n)\|\leq\frac{1}{\sqrt{n}}. \end{array} \right. \end{equation} | (33) |
By Lemma 3.2 and Lemma 3.3 we get a convergent subsequence (still denoted by
In this section, we always assume the conditions
\inf\{ I_\mu(u)\ : \ u\in \mathcal{M}\},\quad \mathcal{M} = \{u\in H^1(\mathbb{R}^3)\ :\ \langle I'_\mu(u), u\rangle = 0\} |
to get a ground state solution. But for
\mathcal{N} = \{u\in H^1(\mathbb{R}^3)\backslash\{0\}:I_\mu'(u) = 0\}. |
And then we consider the following minimization problem
\begin{equation} c_{0,\mu} = \inf\{I_\mu(u):u\in\mathcal{N}\}. \end{equation} | (34) |
Lemma 4.1. Let
d_{0, \mu} = \inf\limits_{\|u\| < \rho} I_\mu (u). |
Then the
Proof. Firstly, we prove that
\begin{eqnarray} I_\mu(u) & = &\frac12 \|u\|^2-\frac{\mu}{2}\int_{\mathbb{R}^3} h(x)u^2dx + \frac14 F(u) -\frac{1}{p+1}\int_{\mathbb{R}^3}|u|^{p+1}dx \\ &\geq& \frac12 \|u\|^2-\frac{\mu}{2\mu_1}\|u\|^2 - C\|u\|^{p+1} > -\infty \end{eqnarray} |
as
I_\mu(te_1) = \frac{t^2}2 \|e_1\|^2-\frac{\mu t^2}{2}\int_{\mathbb{R}^3} h(x)e_1^2dx + \frac{t^4}4 F(e_1) - \frac{t^{p+1}}{p+1}\int_{\mathbb{R}^3}|e_1|^{p+1}dx. |
It is now deduced from
I_\mu(te_1) = \frac{t^2}2\left(1-\frac{\mu}{\mu_1}\right)\|e_1\|^2 + \frac{t^4}4 F(e_1) -\frac{t^{p+1}}{p+1}\int_{\mathbb{R}^3}|e_1|^{p+1}dx. |
Since
Secondly, let
I_\mu(u_n) \to d_{0, \mu}\qquad \hbox{and}\qquad I'_\mu(u_n) \to 0. |
Then we can prove that
We emphasize that the above lemma does NOT mean that
Lemma 4.2. For
Proof. By Lemma 4.1, we know that
For any
I_\mu(u) = I_\mu(u) - \frac{1}{4}\langle I'_\mu(u), u\rangle \geq \frac{1}{4}\|u\|^2 - D(p,h) \mu^{\frac{p+1}{p-1}}. |
Therefore the
Now let
I_\mu(u_n) \to c_{0,\mu}\qquad \hbox{and}\qquad I'_\mu(u_n) = 0. |
Since
Next, to analyze further the
Lemma 4.3. There exists
Proof. The proof is divided into two steps. In the first place, for
\|u\|^2 - \mu_1\int_{\mathbb{R}^3} h(x)u^2dx + F(u) = \int_{\mathbb{R}^3} |u|^{p+1}dx |
and hence
I_{\mu_1}(u) = \frac{p-1}{2(p+1)} \left(\|u\|^2 - \mu_1\int_{\mathbb{R}^3} h(x)u^2dx\right) + \frac{p-3}{4(p+1)}F(u). |
Since
I_{\mu_1}(u) \geq \frac{p-3}{4(p+1)}F(u) > 0. |
In the second place, denoted by
I'_{\mu^{(n)}}(u_{0,\mu^{(n)}}) = 0 |
and we also have that
c_{0, \mu^{(n)}} = I_{\mu^{(n)}}(u_{0,\mu^{(n)}}) < 0. |
Hence we deduce that
\begin{array}{rl} I_{\mu^{(n)}}(u_{0,\mu^{(n)}}) & = \frac{p-1}{2(p+1)} \left(\|u_{0,\mu^{(n)}}\|^2 - \mu^{(n)}\int_{\mathbb{R}^3} h(x)(u_{0,\mu^{(n)}})^2dx\right)\\ & \qquad \quad + \frac{p-3}{4(p+1)}F(u_{0,\mu^{(n)}}). \end{array} |
Using the definition of
\|u_{0,\mu^{(n)}}\|^2 - \mu^{(n)}\int_{\mathbb{R}^3} h(x)(u_{0,\mu^{(n)}})^2dx \geq \left(1-\frac{\mu^{(n)}}{\mu_1}\right)\|u_{0,\mu^{(n)}}\|^2\to 0 |
because
Claim. As
Proof of the Claim. From
\begin{eqnarray} o(1) + I_{\mu^{(n)}}(u_{0,\mu^{(n)}}) & = & I_{\mu^{(n)}}(\tilde{u}_0) + \frac{1}{2}\|\tilde{w}_n\|^2 \\ & & + \frac{1}{4}F(\tilde{w}_n) - \frac{1}{p+1}\int_{\mathbb{R}^3}|\tilde{w}_n|^{p+1}dx, \end{eqnarray} | (35) |
where
Now we distinguish two cases:
Suppose that the case (ⅰ) occurs. We may deduce from a proof similar to Lemma 2.6 that
I_{\mu^{(n)}}(u_{0,\mu^{(n)}}) + o(1) \geq I_{\mu_1}(\tilde{u}_0) + \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}}, |
which is a contradiction because
Next we prove that
\liminf\limits_{n\to\infty} I_{\mu^{(n)}}(u_{0,\mu^{(n)}}) \geq \frac{p-3}{4(p+1)} F(\tilde{u}_0) > 0, |
which is also a contradiction since
Hence there is
Remark 4.4. The proof of Lemma 4.3 implies that (1) of Theorem 1.2 holds.
In the following, we are going to prove the existence of another nonnegative bound state solution of (3). To obtain this goal, we have to analyze further the
Lemma 4.5. Under the assumptions of
Proof. Let
d + o(1) = \frac{1}{2}\|u_n\|^2 - \frac{\mu}{2}\int_{\mathbb{R}^3}h(x)u_n^2dx + \frac{1}{4}F(u_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|u_n|^{p+1}dx |
and
\langle I'_\mu (u_n), u_n\rangle = \|u_n\|^2 - \mu\int_{\mathbb{R}^3}h(x)u_n^2dx + F(u_n) - \int_{\mathbb{R}^3}|u_n|^{p+1}dx. |
Similar to the proof in Lemma 2.3, we can deduce that
\|u_n\|^2 = \|u_0\|^2 + \|w_n\|^2 +o(1), |
F(u_n) = F(u_0) + F(w_n) +o(1) |
and
\|u_n\|^{p+1}_{L^{p+1}} = \|u_0\|^{p+1}_{L^{p+1}} + \|w_n\|^{p+1}_{L^{p+1}} +o(1). |
Using Lemma 2.1, we also have that
\begin{eqnarray} & & d + o(1) = I_\mu(u_n) = I_\mu(u_0) + \frac{1}{2}\| w_n\|^2\\ & & \qquad \quad + \frac{1}{4}F(w_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|w_n|^{p+1}dx. \end{eqnarray} | (36) |
Since
I_\mu(u_0) \geq c_{0,\mu} |
and
\|u_0\|^2 - \mu\int_{\mathbb{R}^3}h(x)u_0^2dx + \int_{\mathbb{R}^3}\phi_{u_0}u_0^2 = \int_{\mathbb{R}^3}|u_0|^{p+1}dx. |
Note that
o(1) = \|u_n\|^2 - \mu\int_{\mathbb{R}^3}h(x)u_n^2dx + F(u_n) - \int_{\mathbb{R}^3}|u_n|^{p+1}dx |
imply that
\begin{equation} o(1) = \| w_n\|^2 + F(w_n) -\int_{\mathbb{R}^3}|w_n|^{p+1}dx. \end{equation} | (37) |
Using
Suppose (I) occurs. Up to a subsequence, we may obtain from (37) that
\|w_n\|^2 \geq S_{p+1} \left(\|w_n\|^2 + F(w_n) - o(1)\right)^{\frac2{p+1}}. |
Hence we get that for
\begin{equation} \|w_n\|^2 \geq S_{p+1}^{\frac{p+1}{p-1}} + o(1). \end{equation} | (38) |
Therefore using (36) and (38), we deduce that for
\begin{eqnarray} & d & + o(1) = I_\mu(u_n) \\ & = & I_\mu(u_0) + \frac{1}{2}\|w_n\|^2 + \frac{1}{4}F(w_n) -\frac{1}{p+1}\int_{\mathbb{R}^3}|w_n|^{p+1}dx \\ & = & I_\mu(u_0) + \frac{p-1}{2(p+1)}\|w_n\|^2 + \frac{p-3}{4(p+1)}F(w_n) \\ & \geq & c_{0,\mu} + \frac{p-1}{2(p+1)}\|w_n\|^2 + \frac{p-3}{4(p+1)}F(w_n) \\ & > & c_{0,\mu} + \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}}, \end{eqnarray} | (39) |
which contradicts to the assumption
Next, for the
d_{2, \mu} = \inf\limits_{\gamma \in \Gamma_2}\sup\limits_{t\in [0,1]}I_\mu(\gamma(t)) |
with
\Gamma_2 = \{ \gamma \in C([0,1], H^1(\mathbb{R}^3))\ : \ \gamma(0) = w_{0, \mu},\ I_\mu(\gamma(1)) < c_{0,\mu} \}. |
Lemma 4.6. Suppose that the conditions
d_{2, \mu} < c_{0,\mu} + \frac{p-1}{2(p+1)} S_{p+1}^{\frac{p+1}{p-1}}. |
Proof. It suffices to find a path starting from
\begin{array}{rl} I_\mu(w_0 + tU_R) & = \frac{1}{2}\left(\left\|w_0 + tU_R\right\|^2 - \mu \int_{\mathbb{R}^3} h(x)|w_0 + t U_R|^2dx\right) \\ & \quad + \frac{1}{4}F(w_0 + tU_R) - \frac1{p+1} \int_{\mathbb{R}^3}|w_0 + t U_R|^{p+1}dx \\ & = I_\mu(w_0) + A_1 + A_2 + A_3 +\frac{t^2}{2}\|U_R\|^2 - \frac{\mu}{2}\int_{\mathbb{R}^3}h(x)U_R^2dx, \end{array} |
where
A_1 = \langle w_0, tU_R\rangle - \mu t\int_{\mathbb{R}^3} h(x)w_0 U_R dx, |
A_2 = \frac{1}{4}\left(F(w_0 + tU_R) - F(w_0)\right) |
and
A_3 = \frac1{p+1} \int_{\mathbb{R}^3}\left(|w_0|^{p+1} - |w_0 + t U_R|^{p+1}\right)dx. |
Since
A_1 = \int_{\mathbb{R}^3}(w_0)^p t U_R dx - \int_{\mathbb{R}^3} K(x)\phi_{w_0}w_0 t U_R dx. |
From an elementary inequality:
(a+b)^q - a^q \geq b^q + q a^{q-1}b,\qquad q > 1,\ \ a > 0, b > 0, |
we deduce that
|A_3| \leq - \frac1{p+1}\int_{\mathbb{R}^3} |t U_R|^{p+1}dx - \int_{\mathbb{R}^3} |w_0|^p tU_Rdx. |
For the estimate of
\begin{array}{rl} |A_2| &\leq t\int_{\mathbb{R}^3}K(x)\phi_{w_0}w_0 U_R dx + \frac{t^2}{2}\int_{\mathbb{R}^3}K(x)\phi_{w_0} (U_R)^2 dx \\ & \qquad \quad + \frac{t^4}{4}\int_{\mathbb{R}^3} K(x)\phi_{U_R}(U_R)^2 dx + t^3\int_{\mathbb{R}^3} K(x)\phi_{U_R}w_0 U_R dx \\ & \qquad \qquad \quad + t^2\int_{\mathbb{R}^3\times\mathbb{R}^3}\frac{K(x)K(y)w_0(x)w_0(y)U_R(x)U_R(y)}{|x-y|}dxdy. \end{array} |
Since
\begin{array}{rl} & \int_{\mathbb{R}^3\times\mathbb{R}^3}\frac{K(x)K(y)w_0(x)w_0(y)U_R(x)U_R(y)}{|x-y|}dxdy \\ & = \int_{\mathbb{R}^3}K(x)\phi_{\sqrt{w_0 U_R}}w_0 U_R dx\\ & \leq \|\phi_{\sqrt{w_0 U_R}}\|_{L^6} \left(\int_{\mathbb{R}^3} K(x)^{\frac65}(w_0U_R)^{\frac65} dx\right)^{\frac56} \\ & \leq C \left(\int_{\mathbb{R}^3} e^{-\frac65 aR} e^{\left(\frac65 a - \frac65(1-\delta)\right)|x|} dx\right)^{\frac56}\\ & \leq C e^{-a R}\quad \hbox{since}\quad 0 < a < 1. \end{array} |
Similarly we can deduce that for
\int_{\mathbb{R}^3}K(x)\phi_{w_0}w_0 U_R dx \leq C e^{-a R}, \ \ \int_{\mathbb{R}^3}K(x)\phi_{w_0} (U_R)^2 dx \leq C e^{-a R}, |
\int_{\mathbb{R}^3} K(x)\phi_{U_R}(U_R)^2 dx \leq C e^{-a R}\ \ \hbox{and}\ \ \int_{\mathbb{R}^3} K(x)\phi_{U_R}w_0 U_R dx \leq C e^{-a R}. |
Since
\begin{array}{rl} & I_\mu(w_0 + tU_R) \leq I_\mu(w_0) +\frac{t^2}{2}\|U_R\|^2dx - \frac{\mu}{2}\int_{\mathbb{R}^3}h(x)U_R^2dx\\ & \qquad \qquad - \frac1{p+1}\int_{\mathbb{R}^3} |t U_R|^{p+1}dx+C e^{-a R}\\ & \leq I_\mu(w_0) + \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}} +C e^{-a R} - C e^{-b R} + o(e^{-b R})\\ & < c_{0,\mu} + \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}} \end{array} |
for
Proposition 4.7. Under the conditions (A1)-(A4), if
Proof. Since for
Proof of Theorem 1.2. The conclusion (1) of Theorem 1.2 follows from Lemma 4.3 and Remark 4.4. It remains to prove (2) of Theorem 1.2. By Proposition 4.7, the
\begin{equation} d_{2, \mu} \leq \max\limits_{t\in [0, 1]} I_{\mu} (\gamma_n(t)) < d_{2, \mu} + \frac{1}{n}. \end{equation} | (40) |
By Ekeland's variational principle, there exists
\begin{equation} \left\{ \begin{array}{ll} d_{2, \mu} \leq \max\limits_{t\in [0, 1]} I_{\mu} (\gamma_n^*(t))\leq \max\limits_{t\in [0, 1]} I_{\mu} (\gamma_n(t)) < d_{2, \mu} +\frac{1}{n};\\ \max\limits_{t\in [0, 1]}\|\gamma_n(t)-\gamma_n^*(t)\| < \frac{1}{\sqrt{n}}; \\ \hbox{ there exists} \; t_n\in [0, 1]\; \hbox{such that}\; z_n : = \gamma_n^*(t_n) \hbox{ satisfies}: \\ I_{\mu}(z_n) = \max\limits_{t\in [0, 1]} I_{\mu} (\gamma_n^*(t)), \;\hbox{and}\; \|I'_{\mu}(z_n)\|\leq\frac{1}{\sqrt{n}}. \end{array} \right. \end{equation} | (41) |
By Lemma 4.6 we get a convergent subsequence (still denoted by
Next, let
0 < \alpha \leq d_{2, \mu^{(n)}}\leq \max\limits_{s > 0}I_{\mu^{(n)}}(w_{0,\mu^{(n)}} + sU_R) |
and
I_{\mu^{(n)}}(w_{0,\mu^{(n)}} + sU_R) \leq \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}} +C e^{-a R} - C e^{-b R} + o(e^{-b R}), |
\begin{equation} \limsup\limits_{n\to\infty} d_{2, \mu^{(n)}} \leq \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}}. \end{equation} | (42) |
Next, similar to the proof in Lemma 2.3, we can deduce that
I_{\mu^{(n)}}(u_{2,\mu^{(n)}}) \geq I_{\mu_1}(\tilde{u}_2) + \frac{p-1}{2(p+1)}S_{p+1}^{\frac{p+1}{p-1}}, |
which contradicts to (42). Hence
The author thanks the unknown referee for helpful comments.
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