Sturm-Liouville Problem
Jump to navigation
Jump to search
![]() | This article needs to be tidied. In particular: under way Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code. |
![]() | This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
![]() | It has been suggested that this page be renamed. In particular: here conditions differ from the main article; once the full proof is done, choose appropriate title To discuss this page in more detail, feel free to use the talk page. |
![]() | Work In Progress In particular: long proof using Ritz method You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{WIP}} from the code. |
Theorem
Let:
- $P \in C^\infty \map P x > 0$
- $Q \in C^0$
- $-\paren {P y'}' + Q y = \lambda y$
- $\map y a = \map y b = 0$
Then the Sturm-Liouville problem has an infinite sequence of eigenvalues $\set {\lambda^{\paren n} }$, and to each $\lambda^{\paren n}$ corresponds an eigenfunction $y^{\paren n}$, unique up to a constant factor.
Proof
- $J \sqbrk y = \ds \int_a^b \paren {P y'^2 + Q y^2} \rd x$
- $\ds \int_a^b y^2 \rd x = 1$
- $\ds \int_a^b \paren {P y'^2 + Q y^2} \rd x > \int_a^b Q y^2 \rd x \ge M \int_a^b y^2 \rd x = M$
- $M = \min \limits_{a \mathop \le x \mathop \le b} \map Q x$
Assume $a = 0$, $b = \pi$.
Choose $\set {\map {\phi_n} x} = \set {\sin n x}$
For $k \ne l$ we have:
- $\ds \int_0^\pi \sin k x \sin l x \rd x = 0$
- $\ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x = \dfrac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^2 = 1$
- $\ds \map {J_n} {\alpha_1, \ldots, \alpha_n} = \int_o^\pi \paren {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin k x}'^2 + \map Q {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2} \rd x$
![]() | This article, or a section of it, needs explaining. In particular: The scope of the $'$ and ${}^2$ in $\ds \map P {\sum_{k \mathop = 1}^n \alpha_k \sin k x}'^2$ needs to be clarified in the above You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
- $\ds \map {y_n^{\paren 1} } x = \sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin k x$
- $\set {\lambda_n^{\paren 1} }$
- $\set {y_n^{\paren 1} }$
- $\map {J_n} {\alpha_1, \ldots, \alpha_n} = \map {J_{n + 1} } {\alpha_1, \ldots, \alpha_n, 0}$
- $\lambda_{n + 1}^{\paren 1} \le \lambda_n^{\paren 1}$
- $\lambda^{\paren 1} = \lim_{n \mathop \to \infty} \lambda_n^{\paren 1}$
- $\ds \lambda_n^{\paren 1} = \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x$
- $\ds \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x \le M$
- $\ds \int_0^\pi Py_n'^2 \rd x \le M + \size {\int_0^\pi Q y_n^2 \rd x} \le M + \max \limits_{a \mathop \le x \mathop \le b } \size {\map Q x} = M_1$
- $\ds \int_0^\pi \map {y_n'^2} x \rd x \le \dfrac {M_1} {\min \limits_{a \mathop \le x \mathop \le b} \map P x} = M_2$
- $\map {y_n} 0 = 0$
- $\ds \size {\map {y_n} x}^2 = \size {\int_0^x \map {y_n'} \zeta \rd \zeta}^2 \le \int_0^x \map {y_n'^2} \zeta \rd \zeta \int_0^x \rd \zeta \le M_2 \pi$
- $\ds \size {\map {y_n} {x_2} - \map {y_n} {x_1} }^2 = \size {\int_{x_1}^{x_2} \map {y_n'} x \rd x}^2 \le \int_{x_1}^{x_2} y_n'^2 \rd x \size {\int_{x_1}^{x_2} \rd x}^2 \le M_2 \size {x_2 - x_1}$
- $\map {y^{\paren 1} } x = \lim_{m \mathop \to \infty} \map {y_{n_m} } x$
- $\int_0^\pi \paren {-\paren {P h'}' + Q_1 h} y \rd x = 0$
- $\map h x \in \map {\DD_2} {0, \pi}$
- $\map h 0 = \map h \pi = \map {h'} 0 = \map {h'} \pi = 0$
- $\map y x \in \map {\DD_2} {0, \pi}$
- $-\paren {P y'}' + Q_1 y = 0$
- $\ds \int_0^\pi \paren {-\paren {P y'}' + Q_1 y} y \rd x = -\int_0^\pi P h' ' y \rd x - \int_0^\pi P' h' y \rd x + \int_0^\pi Q_1 h y \rd x = - \int_0^\pi \paren {- P y + \int_0^x P' y \rd \zeta + \int_0^x \paren {\int_0^\zeta Q_1 y \rd t} \rd \zeta} \rd x = 0$
- $-\paren {P y}' + P' y + \ds \int_0^x Q_1 y \rd \zeta = c_1$
- $-P y' + \ds \int_0^x Q_1 y \rd \zeta = c_1$
- $-\paren {P y'}' + Q_1 y = 0$
- $-\paren {P {y^{\paren 1} }'}' + Q y^{\paren 1} = \lambda^{\paren 1} y^{\paren 1}$
- $\dfrac \partial {\partial \alpha_r} \paren {\map {J_n} {\alpha_1, \ldots, \alpha_n} - \lambda_n^{\paren 1} \ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x} = 0$
![]() | This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |