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: [...]
$$\mathscr{L}=\frac{-1}{r}\left[\frac{\mathrm{d}}{\mathrm{d} x}\left(p(x) \frac{\mathrm{d}}{\mathrm{~d} x}\right)+q(x)\right]$$ 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: [...]
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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: [...]
$$\mathscr{L}=\frac{-1}{r}\left[\frac{\mathrm{d}}{\mathrm{d} x}\left(p(x) \frac{\mathrm{d}}{\mathrm{~d} x}\right)+q(x)\right]$$ 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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