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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 \(\mathscr{L}y=\lambda y\)

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

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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>

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>

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
---|---|---|---|---|---|---|---|

repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

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