For the S-L system \(\frac{\mathrm{d}}{\mathrm{d} x}\left[p(x) \frac{\mathrm{d} y}{\mathrm{~d} x}\right]+q(x) y=-\lambda w(x) y\), we can define a symmetric/self-adjoint operator of S-L system: [...]
Answer
\(\mathscr{L}=\frac{-1}{r}\left[\frac{\mathrm{d}}{\mathrm{d} x}\left(p(x) \frac{\mathrm{d}}{\mathrm{~d} x}\right)+q(x)\right]\) and transform the original equation to \(\mathscr{L}y=\lambda y\)
Question
For the S-L system \(\frac{\mathrm{d}}{\mathrm{d} x}\left[p(x) \frac{\mathrm{d} y}{\mathrm{~d} x}\right]+q(x) y=-\lambda w(x) y\), we can define a symmetric/self-adjoint operator of S-L system: [...]
Answer
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Question
For the S-L system \(\frac{\mathrm{d}}{\mathrm{d} x}\left[p(x) \frac{\mathrm{d} y}{\mathrm{~d} x}\right]+q(x) y=-\lambda w(x) y\), we can define a symmetric/self-adjoint operator of S-L system: [...]
Answer
\(\mathscr{L}=\frac{-1}{r}\left[\frac{\mathrm{d}}{\mathrm{d} x}\left(p(x) \frac{\mathrm{d}}{\mathrm{~d} x}\right)+q(x)\right]\) and transform the original equation to \(\mathscr{L}y=\lambda y\)
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Defintion of the symmetric/self-adjoint operator of S-L system For the S-L system \(\frac{\mathrm{d}}{\mathrm{d} x}\left[p(x) \frac{\mathrm{d} y}{\mathrm{~d} x}\right]+q(x) y=-\lambda w(x) y\), we can define a symmetric/self-adjoint operator of S-L system: <span>\(\mathscr{L}=\frac{-1}{r}\left[\frac{\mathrm{d}}{\mathrm{d} x}\left(p(x) \frac{\mathrm{d}}{\mathrm{~d} x}\right)+q(x)\right]\) and transform the original equation to \(\mathscr{L}y=\lambda y\) <span>
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