| Summary: | In this article, we consider the following fractional <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo>(</mo><mi>p</mi><mo>,</mo><mi>q</mi><mo>)</mo></mrow></semantics></math></inline-formula>-Laplacian problem <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msubsup><mrow><mo>(</mo><mo>−</mo><mo>Δ</mo><mo>)</mo></mrow><mrow><mi>p</mi></mrow><msub><mi>s</mi><mn>1</mn></msub></msubsup><mi>u</mi><mo>+</mo><msubsup><mrow><mo>(</mo><mo>−</mo><mo>Δ</mo><mo>)</mo></mrow><mrow><mi>q</mi></mrow><msub><mi>s</mi><mn>2</mn></msub></msubsup><mi>u</mi><mo>+</mo><msup><mrow><mi>V</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>(</mo><mo>|</mo><mi>u</mi><mo>|</mo></mrow><mrow><mi>p</mi><mo>−</mo><mn>2</mn></mrow></msup><mi>u</mi><mo>+</mo><msup><mrow><mo>|</mo><mi>u</mi><mo>|</mo></mrow><mrow><mi>q</mi><mo>−</mo><mn>2</mn></mrow></msup><msup><mrow><mi>u</mi><mo>)</mo><mo>=</mo><mi>f</mi><mrow><mo>(</mo><mi>u</mi><mo>)</mo></mrow><mo>+</mo><mi>λ</mi><mo>|</mo><mi>u</mi><mo>|</mo></mrow><mrow><mi>r</mi><mo>−</mo><mn>2</mn></mrow></msup><mi>u</mi><mo>,</mo></mrow></semantics></math></inline-formula> where <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>x</mi><mo>∈</mo><msup><mi mathvariant="double-struck">R</mi><mi>N</mi></msup></mrow></semantics></math></inline-formula>, <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><msubsup><mrow><mo>(</mo><mo>−</mo><mo>Δ</mo><mo>)</mo></mrow><mrow><mi>p</mi></mrow><msub><mi>s</mi><mn>1</mn></msub></msubsup></semantics></math></inline-formula> is the fractional <i>p</i>-Laplacian operator (<inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><msubsup><mrow><mo>(</mo><mo>−</mo><mo>Δ</mo><mo>)</mo></mrow><mrow><mi>q</mi></mrow><msub><mi>s</mi><mn>2</mn></msub></msubsup></semantics></math></inline-formula> is similar), <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mn>0</mn><mo><</mo><msub><mi>s</mi><mn>1</mn></msub><mo><</mo><msub><mi>s</mi><mn>2</mn></msub><mo><</mo><mn>1</mn><mo><</mo><mi>p</mi><mo><</mo><mi>q</mi><mo><</mo><mfrac><mi>N</mi><msub><mi>s</mi><mn>2</mn></msub></mfrac></mrow></semantics></math></inline-formula>, <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msubsup><mi>q</mi><mrow><msub><mi>s</mi><mn>2</mn></msub></mrow><mo>*</mo></msubsup><mo>=</mo><mfrac><mrow><mi>N</mi><mi>q</mi></mrow><mrow><mi>N</mi><mo>−</mo><msub><mi>s</mi><mn>2</mn></msub><mi>q</mi></mrow></mfrac></mrow></semantics></math></inline-formula>, <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>r</mi><mo>≥</mo><msubsup><mi>q</mi><mrow><msub><mi>s</mi><mn>2</mn></msub></mrow><mo>*</mo></msubsup></mrow></semantics></math></inline-formula>, <i>f</i> is a <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><msup><mi>C</mi><mn>1</mn></msup></semantics></math></inline-formula> real function and <i>V</i> is a coercive function. By using variational methods, we prove that the above problem admits a sign-changing solution if <inline-formula><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>λ</mi><mo>></mo><mn>0</mn></mrow></semantics></math></inline-formula> is small.
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