<p>In this paper, we consider the following nonlinear Schrödinger system with three wave interaction: <Equation ID="Equ38"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_Equ38.gif" Format="GIF" Height="65" Rendition="HTML" Resolution="72" Type="Linedraw" Width="386" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} {\left\{ \begin{array}{ll} - \varepsilon ^2 \Delta u_1 + V_1(x) u_1 = \mu _1 |u_1|^{p-1} u_1 + \alpha u_2 u_3,\quad \text {in}\ \mathbb {R}^N,\\ - \varepsilon ^2 \Delta u_2 + V_2(x) u_2 = \mu _2 |u_2|^{p-1} u_2 + \alpha u_1 u_3,\quad \text {in}\ \mathbb {R}^N,\\ - \varepsilon ^2 \Delta u_3 + V_3(x) u_3 = \mu _3 |u_3|^{p-1} u_3 + \alpha u_1 u_2,\quad \text {in}\ \mathbb {R}^N, \end{array}\right. } \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mo>-</mo> <msup> <mi>ε</mi> <mn>2</mn> </msup> <mi mathvariant="normal">Δ</mi> <msub> <mi>u</mi> <mn>1</mn> </msub> <mo>+</mo> <msub> <mi>V</mi> <mn>1</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>u</mi> <mn>1</mn> </msub> <mo>=</mo> <msub> <mi>μ</mi> <mn>1</mn> </msub> <msup> <mrow> <mo stretchy="false">|</mo> <msub> <mi>u</mi> <mn>1</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msub> <mi>u</mi> <mn>1</mn> </msub> <mo>+</mo> <mi>α</mi> <msub> <mi>u</mi> <mn>2</mn> </msub> <msub> <mi>u</mi> <mn>3</mn> </msub> <mo>,</mo> <mspace width="1em" /> <mi mathvariant="normal">in</mi> <mspace width="4pt" /> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mi mathvariant="normal">N</mi> </msup> <mo>,</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mo>-</mo> <msup> <mi>ε</mi> <mn>2</mn> </msup> <mi mathvariant="normal">Δ</mi> <msub> <mi>u</mi> <mn>2</mn> </msub> <mo>+</mo> <msub> <mi>V</mi> <mn>2</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>u</mi> <mn>2</mn> </msub> <mo>=</mo> <msub> <mi>μ</mi> <mn>2</mn> </msub> <msup> <mrow> <mo stretchy="false">|</mo> <msub> <mi>u</mi> <mn>2</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msub> <mi>u</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>α</mi> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>u</mi> <mn>3</mn> </msub> <mo>,</mo> <mspace width="1em" /> <mi mathvariant="normal">in</mi> <mspace width="4pt" /> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mi mathvariant="normal">N</mi> </msup> <mo>,</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mo>-</mo> <msup> <mi>ε</mi> <mn>2</mn> </msup> <mi mathvariant="normal">Δ</mi> <msub> <mi>u</mi> <mn>3</mn> </msub> <mo>+</mo> <msub> <mi>V</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <msub> <mi>u</mi> <mn>3</mn> </msub> <mo>=</mo> <msub> <mi>μ</mi> <mn>3</mn> </msub> <msup> <mrow> <mo stretchy="false">|</mo> <msub> <mi>u</mi> <mn>3</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msub> <mi>u</mi> <mn>3</mn> </msub> <mo>+</mo> <mi>α</mi> <msub> <mi>u</mi> <mn>1</mn> </msub> <msub> <mi>u</mi> <mn>2</mn> </msub> <mo>,</mo> <mspace width="1em" /> <mi mathvariant="normal">in</mi> <mspace width="4pt" /> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mi mathvariant="normal">N</mi> </msup> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>where <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="48" /> </InlineMediaObject> <EquationSource Format="TEX">\(N \le 5\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>N</mi> <mo>≤</mo> <mn>5</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq2.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="107" /> </InlineMediaObject> <EquationSource Format="TEX">\(1&lt; p &lt; 2^* - 1\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>1</mn> <mo>&lt;</mo> <mi>p</mi> <mo>&lt;</mo> <msup> <mn>2</mn> <mo>∗</mo> </msup> <mo>-</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq3.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="119" /> </InlineMediaObject> <EquationSource Format="TEX">\(2^* = \infty \ (N \le 2)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mn>2</mn> <mo>∗</mo> </msup> <mo>=</mo> <mi>∞</mi> <mspace width="4pt" /> <mrow> <mo stretchy="false">(</mo> <mi>N</mi> <mo>≤</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq4.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="188" /> </InlineMediaObject> <EquationSource Format="TEX">\(2^* = 2N/(N-2)\ (N \ge 3)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mn>2</mn> <mo>∗</mo> </msup> <mo>=</mo> <mn>2</mn> <mi>N</mi> <mo stretchy="false">/</mo> <mrow> <mo stretchy="false">(</mo> <mi>N</mi> <mo>-</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> <mspace width="4pt" /> <mrow> <mo stretchy="false">(</mo> <mi>N</mi> <mo>≥</mo> <mn>3</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq5.gif" Format="GIF" Height="13" Rendition="HTML" Resolution="72" Type="Linedraw" Width="40" /> </InlineMediaObject> <EquationSource Format="TEX">\(\varepsilon &gt; 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>ε</mi> <mo>&gt;</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq6.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="70" /> </InlineMediaObject> <EquationSource Format="TEX">\(V_j(x)&gt;0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>V</mi> <mi>j</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>&gt;</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq7.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="136" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mu _j &gt; 0\ (j=1,2,3)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>μ</mi> <mi>j</mi> </msub> <mo>&gt;</mo> <mn>0</mn> <mspace width="4pt" /> <mrow> <mo stretchy="false">(</mo> <mi>j</mi> <mo>=</mo> <mn>1</mn> <mo>,</mo> <mn>2</mn> <mo>,</mo> <mn>3</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq8.gif" Format="GIF" Height="13" Rendition="HTML" Resolution="72" Type="Linedraw" Width="43" /> </InlineMediaObject> <EquationSource Format="TEX">\(\alpha &gt; 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>α</mi> <mo>&gt;</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>. We construct a peak solution that is concentrating at a local minimum point of a function <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq9.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="150" /> </InlineMediaObject> <EquationSource Format="TEX">\(c(V_1(x),V_2(x),V_3(x))\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo stretchy="false">(</mo> <msub> <mi>V</mi> <mn>1</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>V</mi> <mn>2</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <msub> <mi>V</mi> <mn>3</mn> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>. Here <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq10.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="85" /> </InlineMediaObject> <EquationSource Format="TEX">\(c(\lambda _1,\lambda _2,\lambda _3)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo stretchy="false">(</mo> <msub> <mi>λ</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>λ</mi> <mn>2</mn> </msub> <mo>,</mo> <msub> <mi>λ</mi> <mn>3</mn> </msub> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> is a mountain pass value of the following limit system <Equation ID="Equ39"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_Equ39.gif" Format="GIF" Height="65" Rendition="HTML" Resolution="72" Type="Linedraw" Width="334" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} {\left\{ \begin{array}{ll} - \Delta v_1 + \lambda _1 v_1 = \mu _1 |v_1|^{p-1} v_1 + \alpha v_2 v_3\quad \text {in}\ \mathbb {R}^N,\\ - \Delta v_2 + \lambda _2 v_2 = \mu _2 |v_2|^{p-1} v_2 + \alpha v_1 v_3\quad \text {in}\ \mathbb {R}^N,\\ - \Delta v_3 + \lambda _3 v_3 = \mu _3 |v_3|^{p-1} v_3 + \alpha v_1 v_2\quad \text {in}\ \mathbb {R}^N. \end{array}\right. } \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mo>-</mo> <mi mathvariant="normal">Δ</mi> <msub> <mi>v</mi> <mn>1</mn> </msub> <mo>+</mo> <msub> <mi>λ</mi> <mn>1</mn> </msub> <msub> <mi>v</mi> <mn>1</mn> </msub> <mo>=</mo> <msub> <mi>μ</mi> <mn>1</mn> </msub> <msup> <mrow> <mo stretchy="false">|</mo> <msub> <mi>v</mi> <mn>1</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msub> <mi>v</mi> <mn>1</mn> </msub> <mo>+</mo> <mi>α</mi> <msub> <mi>v</mi> <mn>2</mn> </msub> <msub> <mi>v</mi> <mn>3</mn> </msub> <mspace width="1em" /> <mi mathvariant="normal">in</mi> <mspace width="4pt" /> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mi mathvariant="normal">N</mi> </msup> <mo>,</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mo>-</mo> <mi mathvariant="normal">Δ</mi> <msub> <mi>v</mi> <mn>2</mn> </msub> <mo>+</mo> <msub> <mi>λ</mi> <mn>2</mn> </msub> <msub> <mi>v</mi> <mn>2</mn> </msub> <mo>=</mo> <msub> <mi>μ</mi> <mn>2</mn> </msub> <msup> <mrow> <mo stretchy="false">|</mo> <msub> <mi>v</mi> <mn>2</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msub> <mi>v</mi> <mn>2</mn> </msub> <mo>+</mo> <mi>α</mi> <msub> <mi>v</mi> <mn>1</mn> </msub> <msub> <mi>v</mi> <mn>3</mn> </msub> <mspace width="1em" /> <mi mathvariant="normal">in</mi> <mspace width="4pt" /> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mi mathvariant="normal">N</mi> </msup> <mo>,</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mo>-</mo> <mi mathvariant="normal">Δ</mi> <msub> <mi>v</mi> <mn>3</mn> </msub> <mo>+</mo> <msub> <mi>λ</mi> <mn>3</mn> </msub> <msub> <mi>v</mi> <mn>3</mn> </msub> <mo>=</mo> <msub> <mi>μ</mi> <mn>3</mn> </msub> <msup> <mrow> <mo stretchy="false">|</mo> <msub> <mi>v</mi> <mn>3</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mrow> <mi>p</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msub> <mi>v</mi> <mn>3</mn> </msub> <mo>+</mo> <mi>α</mi> <msub> <mi>v</mi> <mn>1</mn> </msub> <msub> <mi>v</mi> <mn>2</mn> </msub> <mspace width="1em" /> <mi mathvariant="normal">in</mi> <mspace width="4pt" /> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mi mathvariant="normal">N</mi> </msup> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>When <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="42985_2025_312_Article_IEq11.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="68" /> </InlineMediaObject> <EquationSource Format="TEX">\(p\in (1,2)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>∈</mo> <mo stretchy="false">(</mo> <mn>1</mn> <mo>,</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>, this limit system does not necessarily have a ground state. Hence a key of the construction is to use a local mountain pass approach.</p>

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A construction of peak solutions by a local mountain pass approach for a nonlinear Schrödinger system with three wave interaction

  • Yuki Osada,
  • Yohei Sato

摘要

In this paper, we consider the following nonlinear Schrödinger system with three wave interaction: \(\begin{aligned} {\left\{ \begin{array}{ll} - \varepsilon ^2 \Delta u_1 + V_1(x) u_1 = \mu _1 |u_1|^{p-1} u_1 + \alpha u_2 u_3,\quad \text {in}\ \mathbb {R}^N,\\ - \varepsilon ^2 \Delta u_2 + V_2(x) u_2 = \mu _2 |u_2|^{p-1} u_2 + \alpha u_1 u_3,\quad \text {in}\ \mathbb {R}^N,\\ - \varepsilon ^2 \Delta u_3 + V_3(x) u_3 = \mu _3 |u_3|^{p-1} u_3 + \alpha u_1 u_2,\quad \text {in}\ \mathbb {R}^N, \end{array}\right. } \end{aligned}\) - ε 2 Δ u 1 + V 1 ( x ) u 1 = μ 1 | u 1 | p - 1 u 1 + α u 2 u 3 , in R N , - ε 2 Δ u 2 + V 2 ( x ) u 2 = μ 2 | u 2 | p - 1 u 2 + α u 1 u 3 , in R N , - ε 2 Δ u 3 + V 3 ( x ) u 3 = μ 3 | u 3 | p - 1 u 3 + α u 1 u 2 , in R N , where \(N \le 5\) N 5 , \(1< p < 2^* - 1\) 1 < p < 2 - 1 , \(2^* = \infty \ (N \le 2)\) 2 = ( N 2 ) , \(2^* = 2N/(N-2)\ (N \ge 3)\) 2 = 2 N / ( N - 2 ) ( N 3 ) , \(\varepsilon > 0\) ε > 0 , \(V_j(x)>0\) V j ( x ) > 0 , \(\mu _j > 0\ (j=1,2,3)\) μ j > 0 ( j = 1 , 2 , 3 ) and \(\alpha > 0\) α > 0 . We construct a peak solution that is concentrating at a local minimum point of a function \(c(V_1(x),V_2(x),V_3(x))\) c ( V 1 ( x ) , V 2 ( x ) , V 3 ( x ) ) . Here \(c(\lambda _1,\lambda _2,\lambda _3)\) c ( λ 1 , λ 2 , λ 3 ) is a mountain pass value of the following limit system \(\begin{aligned} {\left\{ \begin{array}{ll} - \Delta v_1 + \lambda _1 v_1 = \mu _1 |v_1|^{p-1} v_1 + \alpha v_2 v_3\quad \text {in}\ \mathbb {R}^N,\\ - \Delta v_2 + \lambda _2 v_2 = \mu _2 |v_2|^{p-1} v_2 + \alpha v_1 v_3\quad \text {in}\ \mathbb {R}^N,\\ - \Delta v_3 + \lambda _3 v_3 = \mu _3 |v_3|^{p-1} v_3 + \alpha v_1 v_2\quad \text {in}\ \mathbb {R}^N. \end{array}\right. } \end{aligned}\) - Δ v 1 + λ 1 v 1 = μ 1 | v 1 | p - 1 v 1 + α v 2 v 3 in R N , - Δ v 2 + λ 2 v 2 = μ 2 | v 2 | p - 1 v 2 + α v 1 v 3 in R N , - Δ v 3 + λ 3 v 3 = μ 3 | v 3 | p - 1 v 3 + α v 1 v 2 in R N . When \(p\in (1,2)\) p ( 1 , 2 ) , this limit system does not necessarily have a ground state. Hence a key of the construction is to use a local mountain pass approach.