Theorem: Sturm-Liouville Problem

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:
:$\ds \int_0^\pi \sin k x \sin l x \rd x = 0$
:$\ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x = \dfrac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^2 = 1$
:$\ds \map {J_n} {\alpha_1, \ldots, \alpha_n} = \int_o^\pi \paren {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin k x}'^2 + \map Q {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2} \rd x$
{{explain|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}}
:$\ds \map {y_n^{\paren 1} } x = \sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin k x$
:$\set {\lambda_n^{\paren 1} }$
:$\set {y_n^{\paren 1} }$
:$\map {J_n} {\alpha_1, \ldots, \alpha_n} = \map {J_{n + 1} } {\alpha_1, \ldots, \alpha_n, 0}$
:$\lambda_{n + 1}^{\paren 1} \le \lambda_n^{\paren 1}$
:$\lambda^{\paren 1} = \lim_{n \mathop \to \infty} \lambda_n^{\paren 1}$
:$\ds \lambda_n^{\paren 1} = \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x$
:$\ds \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x \le M$
:$\ds \int_0^\pi Py_n'^2 \rd x \le M + \size {\int_0^\pi Q y_n^2 \rd x} \le M + \max \limits_{a \mathop \le x \mathop \le b } \size {\map Q x} = M_1$
:$\ds \int_0^\pi \map {y_n'^2} x \rd x \le \dfrac {M_1} {\min \limits_{a \mathop \le x \mathop \le b} \map P x} = M_2$
:$\map {y_n} 0 = 0$
:$\ds \size {\map {y_n} x}^2 = \size {\int_0^x \map {y_n'} \zeta \rd \zeta}^2 \le \int_0^x \map {y_n'^2} \zeta \rd \zeta \int_0^x \rd \zeta \le M_2 \pi$
:$\ds \size {\map {y_n} {x_2} - \map {y_n} {x_1} }^2 = \size {\int_{x_1}^{x_2} \map {y_n'} x \rd x}^2 \le \int_{x_1}^{x_2} y_n'^2 \rd x \size {\int_{x_1}^{x_2} \rd x}^2 \le M_2 \size {x_2 - x_1}$
:$\map {y^{\paren 1} } x = \lim_{m  \mathop \to \infty} \map {y_{n_m} } x$
:$\int_0^\pi \paren {-\paren {P h'}' + Q_1 h} y \rd x = 0$
:$\map h x \in \map {\DD_2} {0, \pi}$
:$\map h 0 = \map h \pi = \map {h'} 0 = \map {h'} \pi = 0$
:$\map y x \in \map {\DD_2} {0, \pi}$
:$-\paren {P y'}' + Q_1 y = 0$
:$\ds \int_0^\pi \paren {-\paren {P y'}' + Q_1 y} y \rd x = -\int_0^\pi P h' ' y \rd x - \int_0^\pi P' h' y \rd x + \int_0^\pi Q_1 h y \rd x = - \int_0^\pi \paren {- P y + \int_0^x P' y \rd \zeta + \int_0^x \paren {\int_0^\zeta Q_1 y \rd t} \rd \zeta} \rd x = 0$
:$-\paren {P y}' + P' y + \ds \int_0^x Q_1 y \rd \zeta = c_1$
:$-P y' + \ds \int_0^x Q_1 y \rd \zeta = c_1$
:$-\paren {P y'}' + Q_1 y = 0$
:$-\paren {P {y^{\paren 1} }'}' + Q y^{\paren 1} = \lambda^{\paren 1} y^{\paren 1}$
:$\dfrac \partial {\partial \alpha_r} \paren {\map {J_n} {\alpha_1, \ldots, \alpha_n} - \lambda_n^{\paren 1} \ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x} = 0$
{{ProofWanted}}
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