<p>In this paper, we are concerned with the following mixed order conformally invariant system with exponential Hartree nonlinearity and cubic nonlinearity: <Equation ID="Equ71"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_Equ71.gif" Format="GIF" Height="55" Rendition="HTML" Resolution="72" Type="Linedraw" Width="337" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{\frac{1}{2}} u(x)=\left( \frac{1}{|x|^2}*e^{2pv(x)}\right) e^{pv(x)},~~~&amp; x\in \mathbb {R}^3,\\ ~~~~~~~~~~(-\Delta )^{\frac{3}{2}} v(x)= u^{3}(x),~~~&amp; x\in \mathbb {R}^3, \end{array}\right. } \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <msup> <mrow> <mo stretchy="false">(</mo> <mo>-</mo> <mi mathvariant="normal">Δ</mi> <mo stretchy="false">)</mo> </mrow> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> </msup> <mi>u</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mfenced close=")" open="("> <mfrac> <mn>1</mn> <msup> <mrow> <mo stretchy="false">|</mo> <mi>x</mi> <mo stretchy="false">|</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mrow /> <mo>∗</mo> <msup> <mi>e</mi> <mrow> <mn>2</mn> <mi>p</mi> <mi>v</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> </msup> </mfenced> <msup> <mi>e</mi> <mrow> <mi>p</mi> <mi>v</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> </msup> <mo>,</mo> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mi>x</mi> <mo>∈</mo> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mn>3</mn> </msup> <mo>,</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <msup> <mrow> <mo stretchy="false">(</mo> <mo>-</mo> <mi mathvariant="normal">Δ</mi> <mo stretchy="false">)</mo> </mrow> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> </msup> <mi>v</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <msup> <mi>u</mi> <mn>3</mn> </msup> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> <mspace width="3.33333pt" /> </mrow> </mtd> <mtd columnalign="left"> <mrow> <mi>x</mi> <mo>∈</mo> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mn>3</mn> </msup> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>where <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\(p&gt;0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>p</mi> <mo>&gt;</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_IEq5.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="42" /> </InlineMediaObject> <EquationSource Format="TEX">\(u\ge 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>u</mi> <mo>≥</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, <i>v</i> may change sign and <i>u</i> satisfies the finite total curvature condition <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_IEq6.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="137" /> </InlineMediaObject> <EquationSource Format="TEX">\(\int _{\mathbb {R}^3} u^3(x)\textrm{d}x&lt;+\infty \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mo>∫</mo> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mn>3</mn> </msup> </msub> <msup> <mi>u</mi> <mn>3</mn> </msup> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mtext>d</mtext> <mi>x</mi> <mo>&lt;</mo> <mo>+</mo> <mi>∞</mi> </mrow> </math></EquationSource> </InlineEquation>. Under extremely mild assumptions, we prove that, the classical solution (<i>u</i>,&#xa0;<i>v</i>) must take the unique form: <Equation ID="Equ72"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_Equ72.gif" Format="GIF" Height="78" Rendition="HTML" Resolution="72" Type="Linedraw" Width="416" /> </MediaObject> <EquationSource Format="TEX">\( u(x)=\frac{2^{\frac{4}{3}}p^{-\frac{1}{3}}\mu }{1+\mu ^2|x-x_0|^2},\qquad v(x)=\frac{1}{p}\ln \Bigg [\frac{\left( \frac{2^{\frac{7}{3}}p^{-\frac{1}{3}}\pi ^2}{I^2(1)}\right) ^{\frac{1}{3}}{\mu }}{1+\mu ^2|x-x_0|^2}\Bigg ] \)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mi>u</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msup> <mn>2</mn> <mfrac> <mn>4</mn> <mn>3</mn> </mfrac> </msup> <msup> <mi>p</mi> <mrow> <mo>-</mo> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> </mrow> </msup> <mi>μ</mi> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msup> <mi>μ</mi> <mn>2</mn> </msup> <msup> <mrow> <mo stretchy="false">|</mo> <mi>x</mi> <mo>-</mo> <msub> <mi>x</mi> <mn>0</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>,</mo> <mspace width="2em" /> <mi>v</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <mi>p</mi> </mfrac> <mo>ln</mo> <mrow> <mo maxsize="2.470em" minsize="2.470em" stretchy="true">[</mo> </mrow> <mfrac> <mrow> <msup> <mfenced close=")" open="("> <mfrac> <mrow> <msup> <mn>2</mn> <mfrac> <mn>7</mn> <mn>3</mn> </mfrac> </msup> <msup> <mi>p</mi> <mrow> <mo>-</mo> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> </mrow> </msup> <msup> <mi>π</mi> <mn>2</mn> </msup> </mrow> <mrow> <msup> <mi>I</mi> <mn>2</mn> </msup> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </mfrac> </mfenced> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> </msup> <mi>μ</mi> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msup> <mi>μ</mi> <mn>2</mn> </msup> <msup> <mrow> <mo stretchy="false">|</mo> <mi>x</mi> <mo>-</mo> <msub> <mi>x</mi> <mn>0</mn> </msub> <mo stretchy="false">|</mo> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mrow> <mo maxsize="2.470em" minsize="2.470em" stretchy="true">]</mo> </mrow> </mrow> </math></EquationSource> </Equation>for some <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_IEq7.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="43" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mu &gt;0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>μ</mi> <mo>&gt;</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_IEq8.gif" Format="GIF" Height="18" Rendition="HTML" Resolution="72" Type="Linedraw" Width="55" /> </InlineMediaObject> <EquationSource Format="TEX">\({{x}}_0\in \mathbb {R}^{3}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>x</mi> <mn>0</mn> </msub> <mo>∈</mo> <msup> <mrow> <mi mathvariant="double-struck">R</mi> </mrow> <mn>3</mn> </msup> </mrow> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="30_2025_1089_Article_IEq9.gif" Format="GIF" Height="36" Rendition="HTML" Resolution="72" Type="Linedraw" Width="97" /> </InlineMediaObject> <EquationSource Format="TEX">\(I(1):=\frac{\pi ^{\frac{3}{2}}\Gamma (\frac{1}{2})}{\Gamma (2)}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>I</mi> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> <mo>:</mo> <mo>=</mo> <mfrac> <mrow> <msup> <mi>π</mi> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> </msup> <mi mathvariant="normal">Γ</mi> <mrow> <mo stretchy="false">(</mo> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mo stretchy="false">)</mo> </mrow> </mrow> <mrow> <mi mathvariant="normal">Γ</mi> <mo stretchy="false">(</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> </mfrac> </mrow> </math></EquationSource> </InlineEquation>.</p>

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Classification of solutions to mixed order system with exponential Hartree nonlinearity and cubic nonlinearity in \(\mathbb {R}^{3}\)

  • Wei Dai,
  • Zhenping Feng

摘要

In this paper, we are concerned with the following mixed order conformally invariant system with exponential Hartree nonlinearity and cubic nonlinearity: \(\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^{\frac{1}{2}} u(x)=\left( \frac{1}{|x|^2}*e^{2pv(x)}\right) e^{pv(x)},~~~& x\in \mathbb {R}^3,\\ ~~~~~~~~~~(-\Delta )^{\frac{3}{2}} v(x)= u^{3}(x),~~~& x\in \mathbb {R}^3, \end{array}\right. } \end{aligned}\) ( - Δ ) 1 2 u ( x ) = 1 | x | 2 e 2 p v ( x ) e p v ( x ) , x R 3 , ( - Δ ) 3 2 v ( x ) = u 3 ( x ) , x R 3 , where \(p>0\) p > 0 , \(u\ge 0\) u 0 , v may change sign and u satisfies the finite total curvature condition \(\int _{\mathbb {R}^3} u^3(x)\textrm{d}x<+\infty \) R 3 u 3 ( x ) d x < + . Under extremely mild assumptions, we prove that, the classical solution (uv) must take the unique form: \( u(x)=\frac{2^{\frac{4}{3}}p^{-\frac{1}{3}}\mu }{1+\mu ^2|x-x_0|^2},\qquad v(x)=\frac{1}{p}\ln \Bigg [\frac{\left( \frac{2^{\frac{7}{3}}p^{-\frac{1}{3}}\pi ^2}{I^2(1)}\right) ^{\frac{1}{3}}{\mu }}{1+\mu ^2|x-x_0|^2}\Bigg ] \) u ( x ) = 2 4 3 p - 1 3 μ 1 + μ 2 | x - x 0 | 2 , v ( x ) = 1 p ln [ 2 7 3 p - 1 3 π 2 I 2 ( 1 ) 1 3 μ 1 + μ 2 | x - x 0 | 2 ] for some \(\mu >0\) μ > 0 and \({{x}}_0\in \mathbb {R}^{3}\) x 0 R 3 , where \(I(1):=\frac{\pi ^{\frac{3}{2}}\Gamma (\frac{1}{2})}{\Gamma (2)}\) I ( 1 ) : = π 3 2 Γ ( 1 2 ) Γ ( 2 ) .