<p>For all <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13661_2025_2137_Article_IEq1.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="47" /> </InlineMediaObject> <EquationSource Format="MATHML"><math> <mi>k</mi> <mo>∈</mo> <mi mathvariant="bold">N</mi> </math></EquationSource> <EquationSource Format="TEX">$k\in {\mathbf{N}}$</EquationSource> </InlineEquation>, we consider the following two-point boundary value problems: <Equation ID="Equa"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="13661_2025_2137_Article_Equa.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="496" /> </MediaObject> <EquationSource Format="MATHML"><math> <msup> <mi>u</mi> <mo>″</mo> </msup> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mo>+</mo> <msup> <mi>k</mi> <mn>2</mn> </msup> <mi>u</mi> <mo stretchy="false">(</mo> <mi>π</mi> <mo>−</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mo>+</mo> <mi>g</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo>,</mo> <mi>u</mi> <mo stretchy="false">(</mo> <mi>π</mi> <mo>−</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>=</mo> <mi>h</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mtext> in </mtext> <mo stretchy="false">(</mo> <mn>0</mn> <mo>,</mo> <mi>π</mi> <mo stretchy="false">)</mo> <mo>,</mo> <mi>u</mi> <mo stretchy="false">(</mo> <mn>0</mn> <mo stretchy="false">)</mo> <mo>=</mo> <mn>0</mn> <mo>=</mo> <mi>u</mi> <mo stretchy="false">(</mo> <mi>π</mi> <mo stretchy="false">)</mo> </math></EquationSource> <EquationSource Format="TEX">$ u''(x) +k^{2}u(\pi -x)+g(x,u(\pi -x))=h(x) \text{ in } (0,\pi ), u(0)=0 =u(\pi ) $</EquationSource> </Equation> and <Equation ID="Equb"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="13661_2025_2137_Article_Equb.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="484" /> </MediaObject> <EquationSource Format="MATHML"><math> <msup> <mi>u</mi> <mo>″</mo> </msup> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mo>+</mo> <msup> <mi>k</mi> <mn>2</mn> </msup> <mi>u</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mo>−</mo> <mi>g</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo>,</mo> <mi>u</mi> <mo stretchy="false">(</mo> <mi>π</mi> <mo>−</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>=</mo> <mo>−</mo> <mi>h</mi> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> <mtext> in </mtext> <mo stretchy="false">(</mo> <mn>0</mn> <mo>,</mo> <mi>π</mi> <mo stretchy="false">)</mo> <mo>,</mo> <mi>u</mi> <mo stretchy="false">(</mo> <mn>0</mn> <mo stretchy="false">)</mo> <mo>=</mo> <mn>0</mn> <mo>=</mo> <mi>u</mi> <mo stretchy="false">(</mo> <mi>π</mi> <mo stretchy="false">)</mo> <mo>,</mo> </math></EquationSource> <EquationSource Format="TEX">$ u''(x) +k^{2}u(x)-g(x,u(\pi -x))=-h(x) \text{ in } (0,\pi ), u(0)=0 =u( \pi ), $</EquationSource> </Equation> where <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13661_2025_2137_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="140" /> </InlineMediaObject> <EquationSource Format="MATHML"><math> <mi>g</mi> <mo>:</mo> <mo stretchy="false">(</mo> <mn>0</mn> <mo>,</mo> <mi>π</mi> <mo stretchy="false">)</mo> <mo>×</mo> <mi mathvariant="bold">R</mi> <mo stretchy="false">→</mo> <mi mathvariant="bold">R</mi> </math></EquationSource> <EquationSource Format="TEX">$g:(0,\pi )\times {\mathbf{R}}\to {\mathbf{R}}$</EquationSource> </InlineEquation> is a Carathéodory function that grows linearly in <i>u</i> as <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13661_2025_2137_Article_IEq3.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="68" /> </InlineMediaObject> <EquationSource Format="MATHML"><math> <mo stretchy="false">|</mo> <mi>u</mi> <mo stretchy="false">|</mo> <mo stretchy="false">→</mo> <mi mathvariant="normal">∞</mi> </math></EquationSource> <EquationSource Format="TEX">$|u|\to \infty $</EquationSource> </InlineEquation> and <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="13661_2025_2137_Article_IEq4.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="88" /> </InlineMediaObject> <EquationSource Format="MATHML"><math> <mi>h</mi> <mo>∈</mo> <msup> <mi>L</mi> <mn>1</mn> </msup> <mo stretchy="false">(</mo> <mn>0</mn> <mo>,</mo> <mi>π</mi> <mo stretchy="false">)</mo> </math></EquationSource> <EquationSource Format="TEX">$h\in L^{1}(0,\pi )$</EquationSource> </InlineEquation> satisfies a generalized Landesman–Lazer condition. In particular, the Leray–Schauder continuation method is used to prove the existence of solutions to these problems.</p>

Further study on solvability of two-point boundary value problems with reflection of the argument

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2025年9月29日

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Nai-Sher Yeh

摘要

For all $k \in N$ $k\in {\mathbf{N}}$ , we consider the following two-point boundary value problems: $u^{″} (x) + k^{2} u (π - x) + g (x, u (π - x)) = h (x) in (0, π), u (0) = 0 = u (π)$ $ u''(x) +k^{2}u(\pi -x)+g(x,u(\pi -x))=h(x) \text{ in } (0,\pi ), u(0)=0 =u(\pi ) $ and $u^{″} (x) + k^{2} u (x) - g (x, u (π - x)) = - h (x) in (0, π), u (0) = 0 = u (π),$ $ u''(x) +k^{2}u(x)-g(x,u(\pi -x))=-h(x) \text{ in } (0,\pi ), u(0)=0 =u( \pi ), $ where $g : (0, π) \times R \to R$ $g:(0,\pi )\times {\mathbf{R}}\to {\mathbf{R}}$ is a Carathéodory function that grows linearly in u as $| u | \to \infty$ $|u|\to \infty $ and $h \in L^{1} (0, π)$ $h\in L^{1}(0,\pi )$ satisfies a generalized Landesman–Lazer condition. In particular, the Leray–Schauder continuation method is used to prove the existence of solutions to these problems.