1 Introduction
In this paper, we consider the following Schrödinger equation (see [18, 19]):
where \(n\geq 3\), \(0<\alpha <n\), \(2-\frac{\alpha }{n}< p\leq 2^{\ast } _{\alpha }=\frac{2n-\alpha }{n-2}\) and
In 2016, Qiao et al. [19] first considered the bound and ground states for the nonlinear Schrödinger equations under the condition
$$ - \Delta u +V(x)u= \biggl(\frac{1}{ \vert x \vert ^{\alpha }}\ast \vert u \vert ^{p} \biggr) \vert u \vert ^{p-2}u, \quad x \in \Re ^{n}, $$
(1.1)
(A1)
\(V\in C(\mathbb{R}^{n},\mathbb{R^{+}})\) is coercive, that is,
$$ \lim_{|x|\to +\infty }V(x)=+\infty . $$
(A2)
\(\inf_{\mathbb{R}^{n}} V>0\), and there exists a positive constant r satisfying
as \(|y|\to \infty \), where \(M>0\) and meas stands for the Lebesgue measure.
$$ \operatorname{meas}\bigl\{ x\in \mathbb{R}^{n}, \vert x-y \vert \leq r, V(x)\leq M\bigr\} \to 0 $$
The nonlinear system (1.1) has been proved to possess wide application fields in many real world problems such as anomalous diffusion [2, 4, 15], disease models [6, 9, 21], ecological models [26], synchronization of chaotic systems [1, 27], etc.
Anzeige
Put
is the Nevanlinna norm (see [8]), problem (1.1) is equivalent to the following Schrödinger problem defined by
where \(x\in \Re ^{n}\).
$$ \nu (x):=\frac{1}{2} \bigl\Vert u(x) \bigr\Vert ^{2} $$
$$ \min \nu (x), $$
(1.2)
The Phragmén–Lindelöf inequalities (see [23]) have the main objective to solve the so-called Schrödinger Phragmén–Lindelöf subproblem model to get the trial step \(\tau _{k}\)
$$\begin{aligned} \operatorname{Min}\quad & \mathcal{TW}_{k}(\tau ) = \frac{1}{2} \bigl\Vert u(x_{k})+\nabla S (x_{k}) \tau \bigr\Vert ^{2}, \\ &\Vert \tau \Vert \leq \triangle . \end{aligned}$$
In 2015, a modified Phragmén–Lindelöf inequality was proved by Wan [23]:
where p, c, and γ are three positive numbers.
$$\begin{aligned} \operatorname{Min}\quad & \phi _{k}(\tau ) = \frac{1}{2} \bigl\Vert u(x_{k})+\nabla u(x_{k})\tau \bigr\Vert ^{2}, \\ &\Vert \tau \Vert \leq c^{p} \bigl\Vert u(x_{k}) \bigr\Vert ^{\gamma }, \end{aligned}$$
Recently, another adaptive Schrödinger Phragmén–Lindelöf inequality has been defined by Qiao et al. [17]:
where \(\mathcal{B}_{k}\) is defined by
\(y_{k}=u(x_{k+1})-u(x_{k})\) and \(s_{k}=x_{k+1}-x_{k}\). This Schrödinger Phragmén–Lindelöf method can possess the global convergence without the nondegeneracy (see [1, 7, 11, 26] etc.), which shows that this paper made a further progress in theory. And there exist many applications about the Schrödinger Phragmén–Lindelöf inequalities (see [3, 25, 27, 28] etc.) for nonsmooth optimizations and other problems.
$$ \begin{aligned} \operatorname{Min}\quad & \mathcal{TM}_{k}(\tau ) = \frac{1}{2} \bigl\Vert u(x_{k})+\mathcal{B}_{k} \tau \bigr\Vert ^{2}, \\ &\Vert \tau \Vert \leq c^{p} \bigl\Vert u(x_{k}) \bigr\Vert , \end{aligned} $$
(1.3)
$$ \mathcal{B}_{k+1}=\mathcal{B}_{k} - \frac{\mathcal{B}_{k} s_{k} s_{k} ^{\mathcal{T}} \mathcal{B}_{k}}{s_{k}^{\mathcal{T}} \mathcal{B}_{k} s _{k}} + \frac{y_{k} {y_{k}}^{\mathcal{T}}}{{y_{k}}^{\mathcal{T}}s_{k}}, $$
(1.4)
Anzeige
We further consider the Schrödinger Phragmén–Lindelöf model for the nonlinear system \(u(x)\) at \(x_{k}\) (see [17])
where \(\nabla u(x_{k})\) is the Jacobian matrix of \(u(x)\) at \(x_{k}\).
$$ \vartheta (x_{k}+\tau )=u(x_{k})+\nabla u(x_{k})^{\mathcal{T}}d+ \frac{1}{2}\mathcal{T}_{k}d^{2}, $$
(1.5)
It is well known that the above model (1.5) can be written as the following extension (see [20, 23, 24]):
$$ \vartheta (x_{k}+\tau )=u(x_{k})+\nabla u(x_{k})^{\mathcal{T}}d+ \frac{3}{2}\bigl(s_{k-1}^{\mathcal{T}} \tau \bigr)^{2}s_{k-1}. $$
(1.6)
If we set the Schrödinger Phragmén–Lindelöf matrix \(\nabla u(x_{k})\), then we can use the Schrödinger Phragmén–Lindelöf matrix \(\mathcal{B}_{k}\) instead of it. Thus, our Schrödinger Phragmén–Lindelöf model can be defined as follows:
where \(\mathcal{B}_{k}=\mathcal{H}_{k}^{-1}\) and \(\mathcal{H}_{k}\) is generated by
where \(\rho _{k}=\frac{1}{s_{k}^{\mathcal{T}}y_{k}}\), \(\mathcal{V} _{k}=I-\rho _{k}y_{k}s_{k}^{\mathcal{T}}\) (see [23] etc.).
$$ \begin{aligned} \operatorname{Min}\quad& \mathcal{N}_{k}(\tau ) = \frac{1}{2} \biggl\Vert u(x_{k})+\mathcal{B}_{k}d+ \frac{3}{2}\bigl(s_{k-1}^{\mathcal{T}}\tau \bigr)^{2}s_{k-1} \biggr\Vert ^{2}, \\ &\Vert \tau \Vert \leq c^{p} \bigl\Vert u(x_{k}) \bigr\Vert ^{\gamma }, \end{aligned} $$
(1.7)
$$\begin{aligned} \mathcal{H}_{k+1} =&\mathcal{V}_{k}^{\mathcal{T}} \mathcal{H}_{k} \mathcal{V}_{k}+\rho _{k}s_{k}s_{k}^{\mathcal{T}} \\ =& \mathcal{V}_{k}^{\mathcal{T}}\bigl[\mathcal{V}_{k-1}^{\mathcal{T}} \mathcal{H}_{k-1}\mathcal{V}_{k-1}+\rho _{k-1}s_{k-1}s_{k-1}^{ \mathcal{T}} \bigr]\mathcal{V}_{k}+\rho _{k}s_{k}s_{k}^{\mathcal{T}} \\ =& \cdots \\ =& \bigl[\mathcal{V}_{k}^{\mathcal{T}}\cdots \mathcal{V}_{k-{m}+1}^{ \mathcal{T}} \bigr]\mathcal{H}_{k-{m}+1}[\mathcal{V}_{k-{m}+1} \cdots \mathcal{V}_{k}] \\ &{}+\rho _{k-{m}+1}\bigl[\mathcal{V}_{k-1}^{\mathcal{T}}\cdots \mathcal{V} _{k-{m}+2}^{\mathcal{T}}\bigr]s_{k-{m}+1}s^{\mathcal{T}}_{k-{m}+1}[ \mathcal{V}_{k-{m}+2}\cdots \mathcal{V}_{k-1}] \\ &{}+ \cdots \\ &{}+\rho _{k}s_{k}s_{k}^{\mathcal{T}}, \end{aligned}$$
(1.8)
Let \(\tau _{k}^{p}\) be the solution of (1.7). Define
and predict reduction by
$$ A\tau _{k}\bigl(\tau _{k}^{p} \bigr)=\nu \bigl(x_{k}+\tau _{k}^{p}\bigr)-\nu (x_{k}), $$
(1.9)
$$ \mathcal{P}\tau _{k}\bigl(\tau _{k}^{p} \bigr)=\mathcal{N}_{k}\bigl(\tau _{k}^{p}\bigr)- \mathcal{N}_{k}(0). $$
(1.10)
Based on definitions of \(A\tau _{k}(\tau _{k}^{p})\) and \(P\tau _{k}(\tau _{k}^{p})\), we design their ratio by
$$ r_{k}^{p} = \frac{A\tau _{k}(\tau _{k}^{p})}{\mathcal{P}\tau _{k}(\tau _{k}^{p})}. $$
(1.11)
Therefore, the Schrödinger-type algorithm for solving (1.1) is stated as follows.
Algorithm
Initial:
Let \(\mathfrak{B}_{0}=\mathfrak{H}_{0}^{-1}\in \Re ^{n}\times \Re ^{n}\) be a symmetric and positive definite matrix. \(x_{0}\in \Re ^{n}\) and \(\varrho =0\). ρ, c, and ϵ are three positive constants. Let \(l:=0\);
Step 1:
Stop if \(\|\chi (x_{l})\|<\epsilon \) holds;
Step 2:
Step 3:
Compute \(A\varsigma _{l}(\varsigma _{l}^{\varrho })\), \(\mathcal{P}\varsigma _{l}(\varsigma _{l}^{\varrho })\), and the ratio \(r_{l}^{\varrho }\). If \(r_{l}^{\varrho }<\rho \), let \(\varrho =\varrho +1\), go to Step 2. If \(r_{l}^{\varrho }\geq \rho \), go to the next step;
Step 4:
Set \(x_{l+1}=x_{l}+\varsigma _{l}^{\varrho }\), \(y_{l}=\chi (x_{l+1})-\chi (x_{l})\), update \(\mathfrak{B}_{l+1}= \mathfrak{H}_{l+1}^{-1}\) by (1.8) if \(y_{l}^{\mathfrak{T}} \varsigma _{l}^{p}>0\), otherwise set \(\mathfrak{B}_{l+1}=\mathfrak{B} _{l}\);
Step 5:
Let \(l:=l+1\) and \(\varrho =0\). Go to Step 1.
In this paper, we further focus on convergence results of the above algorithm under the following assumptions.
Assumptions
(A)
Define the set Ω by
It is easy to see that Ω is bounded.
$$ \varOmega =\bigl\{ x|\varphi (x)\leq \varphi (x_{0})\bigr\} . $$
(1.12)
(B)
The nonlinear system \(\chi (x)\) is twice continuously differentiable in \(\varOmega _{1}\), which is an open convex set containing Ω.
(C)
The following Phragmén–Lindelöf relation
holds.
$$ \bigl\Vert \bigl[\nabla \chi (x_{l})- \mathfrak{B}_{l}\bigr]\chi (x_{l}) \bigr\Vert =O\bigl( \bigl\Vert \varsigma _{l}^{p} \bigr\Vert \bigr) $$
(1.13)
(D)
The sequence matrices \(\{\mathfrak{B}_{l}\}\) are uniformly bounded in \(\varOmega _{1}\).
2 Convergence results
We first have the following new Phragmén–Lindelöf inequalities.
Lemma 2.1
Let
\(\tau _{k}^{p}\)
be the solution of (1.1). Then
holds.
$$ \mathcal{P}\tau _{k}\bigl(\tau _{k}^{p} \bigr)\leq -\frac{1}{2} \bigl\Vert \mathcal{B}_{k}u(x _{k}) \bigr\Vert \min \biggl\{ \triangle _{k}, \frac{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert }{M_{l} ^{2}}\biggr\} +O\bigl(\triangle _{k}^{2}\bigr) $$
(2.1)
Proof
Define
$$ \mathcal{J}(u)=\frac{1}{2} \int _{\Re ^{n}} \vert \nabla u \vert ^{2} + u^{2}\,dx- \frac{1}{2p} \int _{\Re ^{n}} \int _{\Re ^{n}} \frac{ \vert u(x) \vert ^{p} \vert u(y) \vert ^{p}}{ \vert x-y \vert ^{ \alpha }} \,dx \,dy. $$
It follows from (1.7) that
$$ j_{0}< j_{1}=\inf_{\mathcal{N}^{-}} \mathcal{J}(u)< j_{0}+\frac{p-1}{2p}S _{\alpha , p}^{\frac{p}{p-1}}. $$
Consider \(V(x)\) is a minimizer for both \(S_{\alpha , p}\). By the continuity of \(\mathcal{J}\), we know that
where \(0\leq t<\gamma \).
$$ \mathcal{J}(u_{0}+tV)< j_{0}+\frac{p-1}{2p}S_{\alpha , p}^{ \frac{p}{p-1}}, $$
So
$$\begin{aligned} \mathcal{J}(u_{0}+tV) &=\frac{1}{2} \Vert u_{0}+tV \Vert ^{2}-\frac{1}{2p} \tilde{B}(u_{0}+tV) - \int _{\Re ^{n}}h(u_{0}+tV)\,dx \\ &=\mathcal{J}(u_{0})+\frac{t^{2}}{2}\biggl[ \Vert V \Vert ^{2}-\frac{t^{p-2}}{p} \tilde{B}(V)\biggr] +\tilde{B}(u_{0})+ \tilde{B}(tV)-\tilde{B}(u_{0}+tV) \\ &< j_{0}+\frac{p-1}{2p}S_{\alpha , p}^{\frac{p}{p-1}}. \end{aligned}$$
It follows from \(t\geq \gamma \) that
$$\begin{aligned} \mathcal{J}(u_{0}+tV) &=\frac{1}{2} \Vert u_{0}+tV \Vert ^{2}-\frac{1}{2p} \tilde{B}(u_{0}+tV) - \int _{\Re ^{n}}h(u_{0}+tV)\,dx \\ &=\frac{1}{2} \Vert u_{0} \Vert ^{2}+t \int _{\Re ^{n}}\nabla u_{0}\nabla V+u_{0}V \,dx +\frac{t^{2}}{2} \Vert V \Vert ^{2}-\frac{1}{2p} \tilde{B}(u_{0}) \\ &\quad{}+\frac{1}{2p}\bigl[\tilde{B}(u_{0})+\tilde{B}(tV)- \tilde{B}(u_{0}+tV)\bigr] - \frac{1}{2p}\tilde{B}(tV) \\ &\quad{}- \int _{\Re ^{n}}hu_{0}\,dx- \int _{\Re ^{n}}htV \,dx \\ &=\mathcal{J}(u_{0})+\frac{t^{2}}{2}\biggl[ \Vert V \Vert ^{2}-\frac{t^{2(p-1)}}{2p} \tilde{B}(V)\biggr] +\frac{1}{2p}\biggl[ \tilde{B}(u_{0})+\tilde{B}(tV) \\ &\quad{}-\tilde{B}(u_{0}+tV)+2p \int _{\Re ^{n}} \int _{\Re ^{n}}\frac{ \vert u _{0}(x) \vert ^{p} \vert u_{0}(y) \vert ^{p-2}u_{0}(y)}{ \vert x-y \vert ^{\alpha }} \,dx \,dy\biggr] \\ &< j_{0}+\frac{p-1}{2p}S_{\alpha , p}^{\frac{p}{p-1}}. \end{aligned}$$
Here, we use that \(\langle J'(u_{0}), tV\rangle =0\) and \(V(x)\) is a solution of (1.1). By the definition of \(\tau _{k}^{p}\) [14, 16] and it being the solution of (1.7), we get
for any \(\alpha \in [0,1]\).
$$\begin{aligned} \mathcal{P}\tau _{k}\bigl(\tau _{k}^{p}\bigr) \leq & \mathcal{P}\tau _{k}\biggl(-\alpha \frac{\triangle _{k}}{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert } \mathcal{B}_{k}u(x _{k})\biggr) \\ =&\frac{1}{2}\biggl[\alpha ^{2} \triangle _{k}^{2} \frac{ \Vert \mathcal{B}_{k} \mathcal{B}_{k}u(x_{k}) \Vert ^{2}}{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert ^{2}}+\alpha ^{4} \triangle _{k}^{4} \frac{9}{4}\frac{(s_{k-1}^{\mathcal{T}} \mathcal{B}_{k}u(x_{k}))^{4}}{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert ^{4}} \\ &{}+ 3\alpha ^{2} \triangle _{k}^{2} \frac{(s_{k-1}^{\mathcal{T}} \mathcal{B}_{k}u(x_{k}))^{2}}{ \Vert \mathcal{B}_{k}s_{k-1} \Vert ^{2}}u(x_{k})^{ \mathcal{T}}s_{k-1}-2\alpha \triangle _{k} \frac{(u(x_{k})^{ \mathcal{T}}\mathcal{B}_{k}\mathcal{B}_{k}u(x_{k}))}{ \Vert \mathcal{B} _{k}u(x_{k}) \Vert } \\ &{}- 3\alpha ^{3}\triangle _{k}^{3} \frac{(s_{k-1}^{\mathcal{T}} \mathcal{B}_{k}u(x_{k}))^{2}s_{k-1}^{\mathcal{T}}\mathcal{B}_{k} \mathcal{B}_{k}u(x_{k})}{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert ^{3}}\biggr] \\ =& \frac{1}{2}\biggl[\alpha ^{2} \triangle _{k}^{2} \frac{ \Vert \mathcal{B}_{k} \mathcal{B}_{k}u(x_{k}) \Vert ^{2}}{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert ^{2}} -2 \alpha \triangle _{k} \frac{(u(x_{k})^{\mathcal{T}}\mathcal{B}_{k} \mathcal{B}_{k}u(x_{k}))}{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert }+O\bigl( \triangle _{k}^{2}\bigr)\biggr] \\ \leq & -\alpha \triangle _{k} \bigl\Vert \mathcal{B}_{k}u(x_{k}) \bigr\Vert +\frac{1}{2} \alpha ^{2} \triangle _{k}^{2} M_{l}^{2}+O\bigl(\triangle _{k}^{2}\bigr) \end{aligned}$$
Therefore
□
$$\begin{aligned} \mathcal{P}\tau _{k}\bigl(\tau _{k}^{p}\bigr) \leq & \min_{0\leq \alpha \leq 1}\biggl[- \alpha \triangle _{k} \bigl\Vert \mathcal{B}_{k}u(x_{k}) \bigr\Vert + \frac{1}{2} \alpha ^{2} \triangle _{k}^{2} M_{l}^{2}\biggr]+O\bigl(\triangle _{k}^{2} \bigr) \\ \leq &-\frac{1}{2} \bigl\Vert \mathcal{B}_{k}u(x_{k}) \bigr\Vert \min \biggl\{ \triangle _{k},\frac{ \Vert \mathcal{B}_{k}u(x_{k}) \Vert }{M_{l}^{2}}\biggr\} +O \bigl(\triangle _{k}^{2}\bigr). \end{aligned}$$
Lemma 2.2
Let Assumptions (A), (B), (C), and (D) hold. Then
where
\(\tau _{k}\)
is the solution of (1.7).
$$ \bigl\vert A\tau _{k}\bigl(\tau _{k}^{p} \bigr)-\mathcal{P}\tau _{k}\bigl(\tau _{k}^{p} \bigr) \bigr\vert =O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{2}\bigr), $$
Proof
It follows from (1.9) and (1.10) that
□
$$\begin{aligned} \bigl\vert A\tau _{k}\bigl(\tau _{k}^{p} \bigr)-\mathcal{P}\tau _{k}\bigl(\tau _{k}^{p} \bigr) \bigr\vert =& \bigl\vert \nu \bigl(x_{k}+\tau _{k}^{p}\bigr)-\mathcal{N}_{k}\bigl(\tau _{k}^{p}\bigr) \bigr\vert \\ =& \frac{1}{2} \biggl\vert \bigl\Vert u(x_{k})+\nabla u(x_{k}) \tau _{k}^{p} +O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{2}\bigr) \bigr\Vert ^{2} \\ &{}- \biggl\Vert u(x_{k})+\mathcal{B}_{k}\tau _{k}^{p}+\frac{3}{2}\bigl(s_{k-1}^{ \mathcal{T}} \tau _{k}^{p}\bigr)^{2}s_{k-1} \biggr\Vert ^{2} \biggr\vert \\ =& \bigl\vert u(x_{k})^{\mathcal{T}}\nabla u(x_{k}) \tau _{k}^{p}-u(x_{k})^{ \mathcal{T}} \mathcal{B}_{k}\tau _{k}^{p}+O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{2}\bigr) \\ &{}+O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{3}\bigr)+O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{4}\bigr) \bigr\vert \\ \leq & \bigl\Vert \bigl[\nabla u(x_{k})-\mathcal{B}_{k} \bigr]u(x_{k}) \bigr\Vert \bigl\Vert \tau _{k}^{p} \bigr\Vert +O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{2}\bigr) \\ &{}+O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{3}\bigr)+O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{4}\bigr) \\ =&O\bigl( \bigl\Vert \tau _{k}^{p} \bigr\Vert ^{2}\bigr). \end{aligned}$$
Theorem 2.1
Let Assumptions (A), (B), (C), and (D) hold. Then Algorithm either finitely stops or generates an infinite sequence
\(\{x_{k}\}\)
satisfying
where
\(\{x_{k}\}\)
is defined as in Algorithm.
$$ \lim_{k\rightarrow \infty } \bigl\Vert u(x_{k}) \bigr\Vert =0, $$
(2.2)
Proof
We know that \(t^{-}(u)\) is a continuous function of u. Consequently, the manifold \(\varLambda ^{-}\) disconnects \(D^{1,2}( {\Re ^{n}})\) in exactly two connected components \(\mathcal{U}_{1}\) and \(\mathcal{U}_{2}\), where
$$\begin{aligned}& \mathcal{U}_{1}= \biggl\{ u\in D^{1,2}\bigl({\Re ^{n}}\bigr): u=0 \text{ or } \Vert u \Vert _{D}< t^{-} \biggl(\frac{u}{ \Vert u \Vert _{D}} \biggr) \biggr\} , \\& \mathcal{U}_{2}= \biggl\{ u\in D^{1,2}\bigl({\Re ^{n}}\bigr): \Vert u \Vert _{D}>t^{-} \biggl( \frac{u}{ \Vert u \Vert _{D}} \biggr) \biggr\} . \end{aligned}$$
So \(D^{1,2}({\Re ^{n}})=\varLambda ^{-}\cup \mathcal{U}_{1} \cup \mathcal{U}_{2}\). In particular, \(u_{0}\in \varLambda ^{+} \subset \mathcal{U}_{1}\). Since
we have
uniformly for \(t\in \mathbb{R}\).
$$ t^{-} \biggl(\frac{u_{0}+tW}{ \Vert u_{0}+tW \Vert _{D}} \biggr)\frac{u_{0}+tW}{ \Vert u _{0}+tW \Vert _{D}}\in \varLambda , $$
$$ 0< t^{-} \biggl(\frac{u_{0}+tW}{ \Vert u_{0}+tW \Vert _{D}} \biggr)< C_{0} $$
On the other hand,
where \(t\geq \tilde{t}\).
$$ \Vert u_{0}+tW \Vert _{D}\geq t \Vert W \Vert _{D}- \Vert u_{0} \Vert _{D}\geq C_{0}, $$
So that we can fix a positive number \(t_{0}\) such that
which yields that
$$ \Vert u_{0}+t_{0}W \Vert _{D}> t^{-}\biggl(\frac{u_{0}+t_{0}W}{ \Vert u_{0}+t_{0}W \Vert _{D}}\biggr), $$
$$ u_{0}+t_{0}W\in \mathcal{U}_{2}. $$
Combining this and the fact \(u_{0}\in \mathcal{U}_{1}\), we know that
for some \(0< t_{1}< t_{0}\).
$$ u_{0}+t_{1}W\in \varLambda ^{-} $$
So
$$ c_{1}=\inf_{\varLambda ^{-}}I(u)\leq \max_{0\leq t\leq t_{0}}I(u_{0}+tW) < c _{0}+\frac{N+2-\alpha }{4N-2\alpha }S^{\frac{2N-\alpha }{N+2-\alpha }} _{H,L}. $$
And there exists a minimizing sequence \(\{u_{n}\}\subset \varLambda ^{-}\) satisfying
where \(w\in \varLambda ^{-}\).
$$\begin{aligned}& I(u_{n})< c_{1}+\frac{1}{n}; \\& I(w)\geq I(u_{n})-\frac{1}{n} \Vert u-w \Vert _{D}, \end{aligned}$$
So that
which implies \(\|u_{n}\|\) has an upper bound.
$$\begin{aligned} c_{1}+1 &>I(u_{n})=\frac{1}{2} \Vert u_{n} \Vert ^{2}_{D}-\frac{1}{2\cdot 2^{ \ast }_{\alpha }}B(u_{n})- \int _{\Re ^{n}}h(x)u_{n} \,dx \\ &\geq \biggl(\frac{1}{2}-\frac{1}{2\cdot 2^{\ast }_{\alpha }} \biggr) \Vert u _{n} \Vert ^{2}_{D} - \biggl(1- \frac{1}{2\cdot 2^{\ast }_{\alpha }} \biggr) \Vert h \Vert _{H^{-1}} \Vert u_{n} \Vert _{D}, \end{aligned}$$
It follows from \(\{u_{n}\}\subset \varLambda ^{-}\) that
$$ \Vert u_{n} \Vert ^{2}_{D}\leq \bigl(2\cdot 2^{\ast }_{\alpha }-1\bigr) \frac{ \Vert u_{n} \Vert ^{2^{\ast }_{\alpha }}_{D}}{S^{2^{\ast }_{\alpha }}_{H,L}}. $$
Thus, \(\|u_{n}\|_{D}\) has a uniform positive lower bound.
Similarly,
$$ I(u_{n})\to c_{1}, \qquad I'(u_{n}) \to 0\quad \text{in } H^{-1}. $$
By Lemma 2.2 and
we obtain that
which leads to a contradiction.
$$ c_{1}< c_{0}+\frac{N+2-\alpha }{4N-2\alpha }S^{\frac{2N-\alpha }{N+2- \alpha }}_{H,L}, $$
$$ \int _{\Re ^{n}}h(x)u_{1}\,dx>0 \quad \text{and} \quad u_{1}\in \varLambda ^{+}, $$
In the case \(h>0\). Applying Lemma 2.1 to \(u_{1}\) and \(|u_{1}|\), we know that there exists \(t^{-}(|u_{1}|)\) such that
$$ t^{-}\bigl( \vert u_{1} \vert \bigr) \vert u_{1} \vert \in \varLambda ^{-}. $$
Moreover,
$$ t^{-}\bigl( \vert u_{1} \vert \bigr)\geq t_{0}\bigl( \vert u_{1} \vert \bigr)=t_{0}(u_{1}) = \biggl[\frac{A(u_{1})}{(2^{ \ast }_{\alpha }-1)B(u_{1})} \biggr]^{\frac{1}{2^{\ast }_{\alpha }-2}}. $$
So
which implies that \(u_{1}\geq 0\). According to the maximum principle, we get \(u_{1}>0\).
$$ \int _{\Re ^{n}}h(x)u_{1}\,dx= \int _{\Re ^{n}}h(x) \vert u_{1} \vert dx, $$
It is easy to see that \(\|u_{n}\|\) is bounded, which yields that
and
as \(n\to \infty \).
$$ \Vert u_{n} \Vert ^{2}= \Vert w_{n} \Vert ^{2}+ \Vert v \Vert ^{2}+o(1),\quad n\to \infty , $$
$$\begin{aligned} & \int _{\Re ^{n}} \frac{ \vert w_{n}(x) \vert ^{p} \vert w_{n}(y) \vert ^{p}}{ \vert x-y \vert ^{\alpha }} \,dw \\ &\quad = \int _{\Re ^{n}} \frac{ \vert w_{n}(x) \vert ^{p} \vert w_{n}(y) \vert ^{p}}{ \vert x-y \vert ^{\alpha }} \,dw + \int _{\Re ^{n}} \frac{ \vert w(x) \vert ^{p} \vert w(y) \vert ^{p}}{ \vert x-y \vert ^{\alpha }} \,dw+o_{n}(1) \end{aligned}$$
So
and
$$\begin{aligned} c\leftarrow \mathcal{J}(w_{n}) &=\frac{1}{2} \Vert w_{n} \Vert ^{2}- \frac{1}{2p} \int _{\Re ^{n}} \frac{ \vert w_{n}(x) \vert ^{p} \vert w_{n}(y) \vert ^{p}}{ \vert x-y \vert ^{ \alpha }} \,dw - \int _{\Re ^{n}}h(x)w_{n} \,dx \\ &=\frac{1}{2} \Vert w_{n} \Vert ^{2}-\frac{1}{2p} \int _{\Re ^{n}} \frac{ \vert w_{n}(x) \vert ^{p} \vert w _{n}(y) \vert ^{p}}{ \vert x-y \vert ^{\alpha }} \,dw - \int _{\Re ^{n}}h(x)w_{n} \,dx \\ &\quad{}+\frac{1}{2} \Vert v \Vert ^{2}-\frac{1}{2p} \int _{\Re ^{n}} \frac{ \vert v(x) \vert ^{p} \vert v(y) \vert ^{p}}{ \vert x-y \vert ^{ \alpha }} \,dw - \int _{\Re ^{n}}h(x)v \,dx+o_{n}(1) \\ &=\mathcal{J}(v)+\frac{1}{2} \Vert w_{n} \Vert ^{2}-\frac{1}{2p}\tilde{B}(w _{n})+o_{n}(1) \end{aligned}$$
$$ \frac{1}{2} \Vert w_{n} \Vert ^{2}- \frac{1}{2p}\tilde{B}(w_{n})+o_{n}(1) < \frac{p-1}{2p}S_{\alpha , p}^{\frac{p}{p-1}}. $$
(2.3)
Notice that
which yields that
$$ o(1)=\bigl\langle J'(u_{n}),u_{n}\bigr\rangle = \bigl\langle J'(v),v\bigr\rangle + \Vert w_{n} \Vert ^{2}-\tilde{B}(w_{n})+o(1), $$
$$ \Vert w_{n} \Vert ^{2}- \tilde{B}(w_{n})=o(1). $$
(2.4)
It follows from (2.4) that
and
which also leads to a contradiction.
$$ \Vert w_{n} \Vert ^{2}=\tilde{B}(w_{n})\leq \frac{ \Vert w_{n} \Vert ^{2p}}{S_{\alpha , p} ^{p}} $$
$$\begin{aligned} \frac{1}{2}\frac{p-1}{p}S_{\alpha , p}^{\frac{p}{p-1}} &= \frac{1}{2}\biggl(1- \frac{1}{p}\biggr)S^{\frac{p}{p-1}} \\ &\leq \frac{1}{2}\biggl(1-\frac{1}{p}\biggr) \Vert w_{n} \Vert ^{2} \\ &=\frac{1}{2} \Vert w_{n} \Vert ^{2}- \frac{1}{2p}\tilde{B}(w_{n})+o_{n}(1) \\ &< \frac{p-1}{2p}S_{\alpha , p}^{\frac{p}{p-1}}, \end{aligned}$$
Suppose that
holds. Using Assumption (C) we get (2.2). It follows from (2.5) that the subsequence \(\{k_{j}\}\) satisfies
$$ \lim_{k\rightarrow \infty } \bigl\Vert \mathcal{B}_{k}u(x_{k}) \bigr\Vert =0 $$
(2.5)
$$ \bigl\Vert \mathcal{B}_{k_{j}}u(x_{k_{j}}) \bigr\Vert \geq \varepsilon . $$
(2.6)
Set
So we assume that
holds, where \(k\in K\).
$$ K=\bigl\{ k| \bigl\Vert \mathcal{B}_{k}u(x_{k}) \bigr\Vert \geq \varepsilon \bigr\} . $$
$$ \bigl\Vert u(x_{k}) \bigr\Vert \geq \varepsilon $$
It follows from the definition of Algorithm and Lemma 2.1 that
$$ \sum_{k\in K}\bigl[\nu (x_{k})-\nu (x_{k+1})\bigr]\geq -\sum_{k\in K}\rho \mathcal{P}\tau _{k}\bigl(\tau _{k}^{p_{k}}\bigr) \geq \sum_{k\in K}\rho \frac{1}{2}\min \biggl\{ c^{p_{k}}\varepsilon , \frac{\varepsilon }{M_{l}^{2}}\biggr\} \varepsilon . $$
Lemma 2.2 gives us that the sequence \(\{\nu (x_{k})\}\) is convergent, which yields that
$$ \sum_{k\in K}\rho \frac{1}{2}\min \biggl\{ c^{p_{k}}\varepsilon ,\frac{ \varepsilon }{M_{l}^{2}}\biggr\} \varepsilon < +\infty . $$
Then \(p_{k} \rightarrow +\infty \) when \(k\rightarrow +\infty \) and \(k\in K\). It follows that
is unacceptable.
$$ \begin{aligned} \min\quad & q_{k}(\tau ) =\frac{1}{2} \biggl\Vert u(x_{k})+\mathcal{B}_{k}\tau +\frac{3}{2}\bigl(s _{k-1}^{\mathcal{T}}\tau \bigr)^{2}s_{k-1} \biggr\Vert ^{2}, \\ &\mbox{s.t.} \quad \Vert \tau \Vert \leq c^{p_{k}-1} \bigl\Vert u(x_{k}) \bigr\Vert \end{aligned} $$
(2.7)
If we put \(x_{k+1}'=x_{k}+\tau _{k}'\), then we have
$$ \frac{\nu (x_{k})-\nu (x_{k+1}')}{-\mathcal{P}\tau _{k}(\tau _{k}')}< \rho . $$
(2.8)
By applying Lemma 2.1 and the definition of \(\triangle _{k}\), we know that
$$ -\mathcal{P}\tau _{k}\bigl(\tau _{k}'\bigr) \geq \frac{1}{2}\min \biggl\{ c^{p_{k}-1} \varepsilon , \frac{\varepsilon }{M_{l}^{2}}\biggr\} \varepsilon . $$
By applying Lemma 2.2, we know that
$$ \nu \bigl(x_{k+1}'\bigr)-\nu (x_{k})- \mathcal{P}\tau _{k}\bigl(\tau _{k}'\bigr)=O \bigl( \bigl\Vert \tau _{k}' \bigr\Vert ^{2} \bigr)=O\bigl(c^{2(p_{k}-1)}\bigr). $$
So
$$ \biggl\vert \frac{\nu (x_{k+1}')-\nu (x_{k})}{\mathcal{P}\tau _{k}(\tau _{k}')}-1 \biggr\vert \leq \frac{O(c^{2(p_{k}-1)})}{0.5\min \{c^{p_{k}-1}\varepsilon ,\frac{ \varepsilon }{M_{l}^{2}}\}\varepsilon +O(c^{2(p_{k}-1)}\varepsilon ^{2})}. $$
By applying \(p_{k}\rightarrow +\infty \) as \(k\rightarrow +\infty \), we know that
which also gives a contradiction to (2.8). □
$$ \frac{\nu (x_{k})-\nu (x_{k+1}')}{-\mathcal{P}\tau _{k}(\tau _{k}')} \rightarrow 1, \quad k\in K, $$
3 Numerical results
This section reports some numerical results of Algorithm.
3.1 Problems
Define
$$ u(x)=\bigl(\upsilon _{1}(x),\upsilon _{2}(x),\ldots , \upsilon _{n}(x)\bigr)^{ \mathcal{T}}. $$
Problem 1
The Schrödinger differential function (see [12])
where \(l=1,2,3,\ldots ,n\).
$$ \upsilon _{l}(x)=2\Biggl(n+l(1-\cos x_{l})-\sin x_{l}-\sum_{j=1}^{n} \cos x _{j}\Biggr) (2\sin x_{l}-\cos x_{l}), $$
Initial guess:
$$ x_{0}=\biggl(\frac{101}{100n},\frac{101}{100n},\ldots , \frac{101}{100n}\biggr)^{ \mathcal{T}}. $$
Problem 2
Logarithmic function
where \(l=1,2,3,\ldots ,n\).
$$ \upsilon _{l}(x)=\ln (x_{l}+1)-\frac{x_{l}}{n}, $$
Initial points:
$$ x_{0}=(1,1,\ldots ,1)^{\mathcal{T}}. $$
Problem 3
Schrödinger differential function (see [5, pp. 471–472])
where \(l=1,2,3,\ldots ,n\).
$$\begin{aligned}& \upsilon _{1}(x) = (2-0.2x_{1})x_{1}-x_{2}+1, \\& \upsilon _{l}(x) = (2-0.2x_{l})x_{l}-x_{i-1}+x_{i+1}+1, \\& \upsilon _{n}(x) = (2-0.2x_{n})x_{n}-x_{n-1}+1, \end{aligned}$$
Initial points:
$$ x_{0}=(-1,-1,\ldots ,-1)^{\mathcal{T}}. $$
Problem 4
Trigexp function (see [5, p. 473])
where \(l=1,2,3,\ldots ,n\).
$$\begin{aligned}& \upsilon _{1}(x) = 3x_{1}^{3}+x_{2}-4+2 \sin (x_{1}-x_{2})\sin (x_{1}+x _{2}), \\& \upsilon _{l}(x) = -2x_{i-1}e^{x_{i-1}-x_{l}}+3x_{l} \bigl(4+3x_{l}^{2}\bigr)+2x _{i+1} \\& -\sin (x_{l}-x_{i+1})\sin (x_{l}+x_{i+1})-2, \\& \upsilon _{n}(x) = -2x_{n-1}e^{x_{n-1}-x_{n}}+3x_{n}-2, \end{aligned}$$
Initial guess:
$$ x_{0}=(0,0,\ldots ,0)^{\mathcal{T}}. $$
Problem 5
Let \(u(x)\) be the gradient of
Then
where \(l=1,2,3,\ldots ,n\).
$$ h(x)=\sum_{l=1}^{n}\bigl(e^{x_{l}}-x_{l} \bigr). $$
$$ \upsilon _{l}(x)=e^{x_{l}}-1, $$
Initial points:
$$ x_{0}=\biggl(\frac{1}{n},\frac{2}{n},\ldots ,1 \biggr)^{\mathcal{T}}. $$
Parameters: \(c=0.2\), \(\epsilon =10^{-2}\), \(\rho =0.03\), \(p=3\), \(m=6\)
\(\mathcal{H}_{0}\) is the unit matrix.
Code experiments: run on a PC with Intel Pentium(R) Xeon(R) E5507 CPU 2.27 GHz, 6.00 GB of RAM, and Windows 7 operating system.
Code software: MATLAB r2017a.
Stop rules: the program stops if \(\|u(x)\|\leq 1e-4\) holds.
Other cases: we will stop the program if the iteration number is larger than ten hundred.
3.2 Results and discussion
The column meaning in the following tables:
Dim: the dimension. NI: the number of iterations.
NG: the norm function number. Time: the CPU-time in seconds.
Numerical results of Table 1 show the performance of these two algorithms about NI, NG, and Time. It is not difficult to see that both of these algorithm can successfully solve all these ten nonlinear problems.
Table 1
Experiment results
Nr | Dim | Algorithm | Algorithm YL | Nr | Dim | Algorithm | Algorithm YL | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Ni | NG | Time | NI | NG | Time | NI | NG | Time | NI | NG | Time | ||||
1 | 300 | 9 | 18 | 10.93567 | 11 | 22 | 1.778411 | 6 | 300 | 3 | 6 | 1.279208 | 5 | 6 | 0.7176046 |
800 | 9 | 18 | 52.46314 | 11 | 22 | 7.176046 | 800 | 3 | 6 | 5.397635 | 5 | 16 | 2.88601 | ||
1600 | 8 | 14 | 215.453 | 11 | 22 | 42.57267 | 1600 | 3 | 6 | 29.88979 | 5 | 16 | 16.39571 | ||
2 | 300 | 4 | 10 | 11.27887 | 6 | 7 | 1.185608 | 7 | 300 | 5 | 14 | 3.790824 | 12 | 49 | 1.435209 |
800 | 4 | 10 | 45.94229 | 6 | 7 | 4.071626 | 800 | 5 | 14 | 22.52654 | 12 | 49 | 4.69563 | ||
1600 | 4 | 10 | 251.38 | 6 | 7 | 22.58894 | 1600 | 5 | 14 | 102.0403 | 17 | 83 | 19.23492 | ||
3 | 300 | 4 | 10 | 2.808018 | 64 | 125 | 8.642455 | 8 | 300 | 1 | 2 | 1.294808 | 3 | 6 | 0.2808018 |
800 | 4 | 10 | 10.74847 | 78 | 129 | 52.26034 | 800 | 1 | 2 | 5.694037 | 3 | 6 | 0.8580055 | ||
1600 | 4 | 10 | 70.80885 | 68 | 99 | 262.5653 | 1600 | 1 | 2 | 31.091 | 3 | 6 | 3.775224 | ||
4 | 300 | 2 | 2 | 0.9112052 | 6 | 17 | 1.092007 | 9 | 300 | 13 | 19 | 11.01367 | 12 | 15 | 1.60681 |
800 | 2 | 2 | 2.839218 | 6 | 22 | 3.08882 | 800 | 9 | 15 | 40.95026 | 11 | 17 | 7.191646 | ||
1600 | 2 | 2 | 14.08689 | 6 | 22 | 13.27569 | 1600 | 10 | 19 | 299.3191 | 10 | 16 | 38.07984 | ||
5 | 300 | 3 | 6 | 1.731611 | 6 | 7 | 0.936006 | 10 | 300 | 3 | 9 | 4.558416 | 40 | 50 | 12.44888 |
800 | 3 | 6 | 5.616036 | 6 | 7 | 3.650423 | 800 | 3 | 9 | 11.62207 | 40 | 50 | 49.43672 | ||
1600 | 3 | 6 | 30.32659 | 6 | 7 | 22.44854 | 1600 | 3 | 9 | 73.07087 | 41 | 53 | 365.7911 | ||
It is easy to see that the NI and the NG of Algorithm have won since their performance profile plot is on top right. And the Time of Algorithm YL has superiority over Algorithm. Both of these two algorithms have good robustness.
4 Conclusions
In this paper, we considered the numerical method for solving the Schrödinger equations via Phragmén–Lindelöf inequalities under the order induced by a symmetric cone with the function involved being monotone. Based on the Phragmén–Lindelöf inequalities, the underlying system of inequalities was reformulated as a system of smooth equations, and a Schrödinger-type method was proposed to solve it iteratively so that a solution of the system of the Schrödinger equations was found. By means of the Schrödinger type inequalities, the algorithm was proved to be well defined and to be globally convergent under weak assumptions and locally quadratically convergent under suitable assumptions. Preliminary numerical results indicate that the algorithm was effective.
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