=== Outline === Firstly, an equivalence between the Sturm-Liouville equation and minimisation of functional $\displaystyle J \sqbrk y = \int_a^b \paren {P y'^2 + Q y^2} \rd x$ problems is established. Then, the lower bound of the functional $J$ is found, thus allowing $J$ to have a finite minimisation.  Afterwards, a trial minimizing sequence is chosen, and $J$ becomes a function of expansion coefficients. The more coefficients, (possibly) the lower value. Sequences of trial mappings $\sequence {y_n^{\paren 1} }$ and values $\sequence {\lambda_n^{\paren 1} }$ of $J$ are introduced.  Convergence of $\sequence {\lambda_n^{\paren 1} }$ is shown at once. As for $\sequence {y_n^{\paren 1} }$, convergence to $y^{\paren 1}$ is proved for its subsequence. The rest of arguments rest upon this weaker result. Furthermore, $\lambda^{\paren 1}$ and $y^{\paren 1}$ are observed to satisfy Sturm-Liouville equation. Convergence of the original sequence $\sequence {y_n^{\paren 1} }$ is secured. Finally, construction of the rest of eigenfunctions and eigenvalues is described.  {{finish|Maybe add a diagrammatical presentation of the whole proof with Lemmas and somehow enumerated intermediate passages}} === Lemma 1 === The given Sturm-Liouville equation is an Euler equation of the following functional: :$\ds J \sqbrk y = \int_a^b \paren {P y'^2 + Qy^2} \rd x$ constrained by a subsidiary condition: :$\ds \int_a^b y^2 \rd x = 1$ === Proof === According to Simplest Variational Problem with Subsidiary Conditions, the following equation must hold: :$F_y - \dfrac \rd {\rd x} F_{y'} + \lambda \paren {G_y - \dfrac \rd {\rd x} G_{y'} } = 0$ where: :$F = P y'^2 + Q y^2$ :$G = y^2$ Then the Euler equation reads: :$2 Q y - 2 \paren {P y'}' + 2 \lambda y = 0$ Division by $2$ and rearrangement of terms yields the desired result. {{qed|lemma}} By Necessary Condition for Integral Functional to have Extremum for given function, if $y$ is an extremum of $J$, it is also a solution of the Sturm-Liouville equation. === Lemma 2 === $J$ is bounded from below. === Proof === Since $Q$ is continuous on an interval, it is bounded. Since $P > 0$, it holds that: :$\displaystyle \int_a^b \paren {P y'^2 + Q y^2} \rd x > \int_a^b Q y^2 \rd x \ge M \int_a^b y^2 \rd x = M$ where: :$\displaystyle M = \min_{a \mathop \le x \mathop \le b} \map Q x$ Therefore, $J$ is bounded from below. {{qed|lemma}} Introduce a new variable: :$t := \pi \dfrac {x - a} {b - a}$ Then the interval of consideration $\closedint a b$ is mapped onto $\closedint 0 \pi$. Choose Ritz sequence $\sequence {\map {\phi_n} t} = \sequence {\sin n t}$, where $n \in \N$. === Lemma 3 === The elements of the sequence $\sequence {\sin n t}$ are orthogonal on the interval $\closedint 0 \pi$: :$\displaystyle \int_0^\pi \map \sin {k t} \map \sin {l t} \rd x = \frac \pi 2 \delta_{k l}$ === Proof === The product involves two elements of the sequence $\sequence {\map \sin {n t} }$. Their indices either match each other or not. Suppose $k = l$. Then: {{begin-eqn}} {{eqn | l = \int_0^\pi \map {\sin^2} {k t} \rd t       | r = \int_0^\pi \frac {1 - \map \cos {2 k t} } 2 \rd t       | c = Square of Sine }} {{eqn | r = \intlimits {\frac \pi 2 - \frac {\map \sin {2 k t} } {4 k} } 0 \pi }} {{eqn | r = \frac \pi 2       | c = Sine of Multiple of Pi }} {{end-eqn}} Suppose $k \ne l$. Then: {{begin-eqn}} {{eqn | l = \int_0^\pi \map \sin {l t} \map \sin {k t} \rd t       | r = \int_0^\pi \frac {\map \cos {\paren {l - k} t} - \map \cos {\paren {l + k} t} } 2 \rd t       | c = Simpson's Formula for Sine by Sine }} {{eqn | r = \intlimits {\frac 1 2 \paren {\frac {\map \sin {\paren {l - k} t} } {l - k} - \frac {\map \sin {\paren {l + k} t} } {l + k} } } 0 \pi }} {{eqn | r = 0       | c = Sine of Multiple of Pi }} {{end-eqn}} By Proof by Cases, the statement is proved. {{qed|lemma}} Let the trial solution be of the following form: :$\displaystyle \map y x = \sum_{k \mathop = 1}^n \alpha_k \map \sin {k \map t x}$ Trial solution has to satisfy boundary and subsidiary conditions. Boundary conditions are satisfied without further requirements. Subsidiary condition results into an additional constraint on coefficients $\alpha_k$: {{begin-eqn}} {{eqn | l = \int_a^b \paren {\sum_{k \mathop = 1}^n \alpha_k \map \sin {k \map t x} }^2 \rd x       | r = \frac \pi {b - a} \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }^2 \rd t }} {{eqn | r = \frac \pi {b - a} \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} } \paren {\sum_{l \mathop = 1}^n \alpha_l \sin l x} \rd t       | c = Second summation label $k$ relabelled as $l$ }} {{eqn | r = \frac \pi {b - a} \sum_{k, l \mathop = 1}^n \alpha_k \alpha_l \int_0^\pi \map \sin {k t} \map \sin {l t} \rd t }} {{eqn | r = \frac \pi {b - a} \sum_{k, l \mathop = 1}^n \alpha_k \alpha_l \frac \pi 2 \delta_{k l}       | c = as $\displaystyle \int_0^\pi \map \sin {k t} \map \sin {l t} \rd x = \frac \pi 2 \delta_{k l}$ }} {{eqn | r = \frac \pi {b - a} \frac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^2 }} {{eqn | r = 1 }} {{end-eqn}}  All the points $\boldsymbol \alpha$ constitute a set $\sigma_n$ which is a surface of an $n$-dimensional sphere, defined by the subsidiary condition. For the assumed trial mapping the functional $\map {J_n} {\boldsymbol \alpha}$ reads as: :$\ds \map {J_n} {\boldsymbol \alpha} = \frac \pi {b - a} \int_0^\pi \sqbrk {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }'^2 + \map Q {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }^2} \rd t$ The integrand is a second order polynomial {{WRT}} the components of $\boldsymbol \alpha$. Hence, $J$ is continuous {{WRT}} the components of $\boldsymbol \alpha$. The components of $\boldsymbol \alpha$ constitute a closed and bounded set. By definition, $\sigma_n$ is a compact set. Thus, $\map {J_n} {\boldsymbol \alpha}$ is continuous on $\sigma_n$. By Continuous Function on Compact Subspace of Euclidean Space is Bounded, $\map {J_n} {\boldsymbol \alpha}$ has a minimum on $\sigma_n$.  Let $\map {y_n^{\paren 1} } x$ be defined as: :$\displaystyle \map {y_n^{\paren 1} } x = \sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin {k \map t x}$ for which $\map {J_n} {\boldsymbol \alpha}$ achieves the minimum $\lambda_n^{\paren 1}$, unrelated to $\lambda$. Then the $n$-th element of the sequence $\sequence {y_n^{\paren 1} }$ corresponds to the $n$-th element of the sequence of minima $\sequence {\lambda_n^{\paren 1} }$ of $\map {J_n} {\boldsymbol \alpha}$. Since $\sigma_n \subset \sigma_{n + 1}$, where $\sigma_n$ has $\alpha_{n + 1} = 0$, it holds that: :$\map {J_n} {\alpha_1, \ldots, \alpha_n} = \map {J_{n + 1} } {\alpha_1, \ldots, \alpha_n, 0}$ By Ritz Method implies Not Worse Approximation with Increased Number of Functions: :$\lambda_{n + 1}^{\paren 1} \le \lambda_n^{\paren 1}$ Therefore, by increasing the domain of definition of $y_n^{\paren 1}$ through additional summands, the minima of $\map {J_n} {\boldsymbol \alpha}$ cannot increase. {{explain|Define "domain of definition"}} From the last inequality and $J$ being bounded from below it follows, that the following limit exists: :$\displaystyle \lambda^{\paren 1} = \lim_{n \mathop \to \infty} \lambda_n^{\paren 1}$ === Lemma 4 === The sequence $\sequence {y_n^{\paren 1} }$ contains a uniformly convergent subsequence. === Proof === The sequence  :$\displaystyle \lambda_n^{\paren 1} = \frac \pi {b - a} \int_0^\pi \paren {P {y_n^{\paren 1} }'^2 + Q {y_n^{\paren 1} }^2} \rd t$ is convergent with its limit being $\lambda^{\paren 1}$. Hence, it is bounded: :$\displaystyle \frac \pi {b - a} \int_0^\pi \paren {P {y_n^{\paren 1} }'^2 + Q {y_n^{\paren 1} }^2} \rd t \le M $ Furthermore: {{begin-eqn}} {{eqn | l = \frac \pi {b - a} \int_0^\pi P {y_n^{\paren 1} }'^2 \rd t       | o = \le       | r = M - \frac \pi {b - a} \int_0^\pi Q {y_n^{\paren 1} }^2 \rd t }} {{eqn | o = \le       | r = M + \frac \pi {b - a} \size {\int_0^\pi Q {y_n^{\paren 1} }^2 \rd t} }} {{eqn | o = \le       | r = M + \max_{a \mathop \le x \mathop \le b} \size {\map Q x} \size {\int_a^b {y_n^{\paren 1} }^2 \rd x} }} {{eqn | r = M + \max_{a \mathop \le x \mathop \le b} \size {\map Q x}       | c = as $\displaystyle \int_a^b y^2 \rd x = 1$, where $y = y _n^{\paren 1}$ minimizes $\map {J_n} {\boldsymbol \alpha}$ }} {{eqn | r = M_1 }} {{end-eqn}} Consequently: :$\displaystyle \frac \pi {b - a} \min_{a \mathop \le x \mathop \le b} \map P x \int_0^\pi \map { {y_n^{\paren 1} }'^2}  t \rd t \le \frac \pi {b - a} \int_0^\pi P \map { {y_n^{\paren 1} }'^2} t \rd t \le M_1$ For positive $P$, division by $P$ does not affect the direction of inequality. It follows that: :$\displaystyle \int_0^\pi \map { {y_n^{\paren 1} }'^2} t \rd t \le \frac {b - a} \pi \frac {M_1} {\min_{a \mathop \le x \mathop \le b} } \map P x = M_2$ Consider squared absolute value of $y_n^{\paren 1}$. Then, for $0 \le t \le \pi$: {{begin-eqn}} {{eqn | l = \size {\map {y_n^{\paren 1} } t}^2       | r = \size {\map {y_n^{\paren 1} } t - 0}^2       | c = as $\map {y_n^{\paren 1} } {t = 0} = 0$ }} {{eqn | r = \size {\map {y_n^{\paren 1} } t - \map {y_n^{\paren 1} } 0}^2 }} {{eqn | r = \size {\int_0^t \map { {y_n^{\paren 1} }'} \zeta \rd \zeta}^2 }} {{eqn | o = \le       | r = \int_0^t \map { {y_n^{\paren 1} }'^2} \zeta \rd \zeta \int_0^t \rd \zeta       | c = Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals: }} {{eqn | o = \le       | r = \int_0^\pi \map { {y_n^{\paren 1} }'^2} \zeta \rd \zeta \int_0^\pi \rd \zeta       | c = as all integrands are non-negative }} {{eqn | o = \le       | r = M_2 \pi }} {{end-eqn}} In other words: :$\forall t \in \closedint 0 \pi, n \in \N : \size {\map {y_n^{\paren 1} } t - \map {y_n^{\paren 1} } 0} \le \sqrt {M_2 \pi}$ Thus $\sequence {y_n^{\paren 1} }$ is uniformly bounded. In addition to this, for $0 \le t_1, t_2 \le \pi$: {{begin-eqn}} {{eqn | l = \size {\map {y_n^{\paren 1} } {t_2} - \map {y_n^{\paren 1} } {t_1} }^2       | r = \size {\int_{t_1}^{t_2} \map { {y_n^{\paren 1} }'} t \rd t}^2       | c = Fundamental Theorem of Calculus }} {{eqn | o = \le       | r = \int_{t_1}^{t_2} {y_n^{\paren 1} }'^2 \rd x \size {\int_{t_1}^{t_2} \rd t}       | c = Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals }} {{eqn | o = \le       | r = \int_0^\pi {y_n^{\paren 1} }'^2 \rd x \size {\int_{t_1}^{t_2} \rd t} }} {{eqn | o = \le       | r = M_2 \size {t_2 - t_1} }} {{end-eqn}} Let $\epsilon$ be any strictly positive real number such that $\epsilon = \sqrt {M_2 \delta}$, where $\delta$ is a strictly positive real number. Suppose $\delta$ is such that $\size {t_2 - t_1} < \delta$. Then: {{begin-eqn}} {{eqn | l = \size {\map {y_n^{\paren 1} } {t_2} - \map {y_n^{\paren 1} } {t_1} }        | o = \le       | r = \sqrt {M_2 \size {t_2 - t_1} }  }} {{eqn | o = <        | r = \sqrt {M_2 \delta}  }} {{eqn | r = \epsilon }} {{end-eqn}} In other words: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall n \in \N: \forall t_1, t_2 \in \closedint 0 \pi: \size {t_2 - t_1} < \delta \implies \size {\map {y_n^{\paren 1} } {t_2} - \map {y_n^{\paren 1} } {t_1} } < \epsilon$ where metric is induced by norm. Thus $\sequence {y_n^{\paren 1} }$ is uniformly equicontinuous. By Arzela's Theorem, there exists a uniformly convergent subsequence $\sequence {y_{n_m}^{\paren 1} }$ from $\sequence {y_n^{\paren 1} }$. {{qed|lemma}} Denote: :$\displaystyle \map {y^{\paren 1} } x = \lim_{m \mathop \to \infty} \map {y_{n_m}^{\paren 1} } x$ Now a proposition is established, needed for the upcoming Lemma 6. === Lemma 5 === Let $\map y t$ be continuous in $\closedint 0 \pi$. Suppose: :$\displaystyle \forall h \in C^2 \openint 0 \pi : \map h 0 = \map h \pi = \map {h'} 0 = \map {h'} \pi = 0 : \int_0^\pi \sqbrk {-\paren {P h'}' + Q_1 h} y \rd t = 0$ Then $\map y t \in C^2 \openint 0 \pi$ and: :$-\paren {P y'}' + Q_1 y = 0$ === Proof === By Integration by parts, Product Rule for Derivatives, boundary conditions for $h$, and noticing that: {{begin-eqn}} {{eqn | l = \map \d {\int_0^t \map {f'} \xi \rd \xi}       | r = \map \d {\map f t - \map f 0}       | c = Fundamental Theorem of Calculus }} {{eqn | r = \map {f'} t \rd t }} {{end-eqn}}   {{begin-eqn}} {{eqn | l = \map \d {\int_0^t \paren {\int_0^\xi \map g \zeta \rd \zeta} \rd \xi}        | r = \paren {\int_0^t \map g \zeta \rd \zeta} \rd t }} {{eqn | r = \paren {\int_0^t \map g \xi \rd \xi} \rd t }} {{end-eqn}} the previous integral can be rewritten as: {{begin-eqn}} {{eqn | l = \int_0^\pi \paren {-\paren {P h'}' + Q_1 h} y \rd t       | r = -\int_0^\pi P h'' y \rd t - \int_0^\pi P' h' y \rd t + \int_0^\pi Q_1 h y \rd t }} {{eqn | r = -\int_0^\pi P h'' y \rd t - \int_0^\pi h' \rd \paren {\int_0^t P' y \rd \xi} + \int_0^\pi h \rd \paren {\int_0^t Q_1 y \rd \xi} }} {{eqn | r = -\int_0^\pi P h'' y \rd t - \paren {\intlimits {h' \int_0^t P' y \rd \xi} {t = 0} {t = \pi} - \int_0^\pi h'' \paren {\int_0^t P' y \rd \xi} \rd t} + \paren {\intlimits {h \int_0^t Q_1 y \rd \xi} {t \mathop = 0} {t \mathop = \pi} - \int_0^\pi h' \paren {\int_0^t Q_1 y \rd \xi} \rd t} }} {{eqn | r = -\int_0^\pi P h'' y \rd t + \int_0^\pi h'' \paren {\int_0^t P' y \rd \xi} \rd t - \int_0^\pi h' \paren {\int_0^t Q_1 y \rd \xi } \rd t }} {{eqn | r = -\int_0^\pi P h'' y \rd t + \int_0^\pi h'' \paren {\int_0^t P' y \rd \xi} \rd t - \paren {\int_0^\pi h' \rd \paren {\int_0^t \paren {\int_0^\xi Q_1 y \rd \zeta} \rd \xi} } }} {{eqn | r = -\int_0^\pi P h'' y \rd t + \int_0^\pi h'' \int_0^t P' y \rd \xi - \paren {\intlimits {h' \int_0^t \paren {\int_0^\xi Q_1 y \rd \zeta} \rd \xi} {t \mathop = 0} {t \mathop = \pi} - \int_0^\pi h'' \paren {\int_0^t \paren {\int_0^\xi Q_1 y \rd \zeta} \rd \xi} \rd t} }} {{eqn | r = -\int_0^\pi P h'' y \rd t + \int_0^\pi h'' \int_0^t P' y \rd \xi + \int_0^\pi h'' \paren {\int_0^t \paren {\int_0^\xi Q_1 y \rd \zeta} \rd \xi} \rd t  }} {{eqn | r = \int_0^\pi \paren {- P y + \int_0^t P' y \rd \xi + \int_0^t \paren {\int_0^\xi Q_1 y \rd \zeta} \rd \xi} h'' \rd t }} {{eqn | r = 0 }} {{end-eqn}} From the lemma: :$\displaystyle - P y + \int_0^t P' y \rd \zeta + \int_0^x \paren {\int_0^t Q_1 y \rd t} \rd \zeta = c_0 + c_1 t$ The {{RHS}} as well as the second and third terms on the {{LHS}} are differentiable {{WRT}} $t$. Thus $\paren {P y}'$ exists. Differentiation {{WRT|Differentiation}} $t$ leads to: :$\displaystyle -\paren {P y}' + P' y + \int_0^t Q_1 y \rd \zeta = c_1$ or: :$\displaystyle - P y' + \int_0^t Q_1 y \rd \zeta = c_1$ The {{RHS}}, the second term on the {{LHS}} and $P$ are continuous and differentiable {{WRT|Differentiation}} $t$, while $P$ is also positive. Therefore, $y'$ exists and is continuous. Hence, $\paren {P y'}'$ exists and: :$-\paren {P y'}' + Q_1 y = 0$ Furthermore, $P$ is continuous and differentiable, while $Q_1$ is continuous. Then $y''$ exists and is continuous. {{qed|lemma}} === Lemma 6 === $y^{\paren 1}$ together with $\lambda^{\paren 1}$ satisfy the Sturm-Liouville equation, where $\map w x = 1$: :$-\paren {P {y^{\paren 1} }' }' + Q y^{\paren 1} = \lambda^{\paren 1} y^{\paren 1}$ === Proof === Let $J_n$, together with its subsidiary condition $\displaystyle \int_a^b y^2 \rd x = 1$, achieve a minimum for $\boldsymbol \alpha = \boldsymbol \alpha^{\paren 1}$. Then the necessary condition for its minimum is: {{begin-eqn}} {{eqn | l = \map {\dfrac \partial {\partial \alpha_r} } {\map {J_n} {\boldsymbol \alpha} - \lambda_n^{\paren 1} \int_a^b y^2 \rd x}       | r = \map {\dfrac \partial {\partial \alpha_r} } {\map {J_n} {\boldsymbol \alpha} - \lambda_n^{\paren 1} \frac \pi {b - a} \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin {k x} }^2 \rd x} }} {{eqn | r = 0 }} {{end-eqn}} Notice, that: {{begin-eqn}} {{eqn | l = \dfrac \partial {\partial \alpha_r} \paren {\sum_{k \mathop = 1}^n \alpha_k \sin {k x} }^2       | r = \dfrac {\partial \paren {\sum_{k \mathop = 1}^n \alpha_k \sin {k x} }^2} {\partial \sum_{k = 1}^n \alpha_k \sin {k x} } \dfrac {\partial \sum_{j \mathop = 1}^n \alpha_j \sin {j x} } {\partial \alpha_r}       | c = Derivative of Composite Function }} {{eqn | r = 2 \sqbrk {\sum_{k \mathop = 1}^n \alpha_k \sin {k x} } \sum_{j = 1}^n \delta_j^r \sin {j x}       | c = as $\dfrac {\partial \alpha_j} {\partial \alpha_r} = \delta_j^r$ }} {{eqn | r = 2 \sqbrk {\sum_{k \mathop = 1}^n \alpha_k \sin {k x} } \sin {r x} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \dfrac {\partial J_n} {\partial \alpha_r}       | r = \frac {\partial} {\partial \alpha_r} \frac \pi {b - a} \int_0^\pi \sqbrk {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }'^2 + Q \paren {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }^2} \rd t }} {{eqn | r = \frac \pi {b - a} \int_0^\pi \sqbrk {P \dfrac {\partial \paren {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }'^2} {\partial \paren {\sum_{j \mathop = 1}^n \alpha_j \sin {j t} }'} \dfrac {\partial \paren {\sum_{j \mathop = 1}^n \alpha_j \sin {j t} }'} {\partial \alpha_r} + 2 Q \sqbrk {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} } \sin {r t} } \rd t       | c = Derivative of Composite Function }} {{eqn | r = \frac {2 \pi} {b - a} \int_0^\pi \sqbrk {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }' \dfrac {\partial \paren {\sum_{j \mathop = 1}^n \alpha_j j \cos {j t} } }{\partial \alpha_r} + Q \sqbrk {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} } \sin {r t} } \rd t }} {{eqn | r = \frac {2 \pi} {b - a} \int_0^\pi \sqbrk {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }' \paren {\sum_{j \mathop = 1}^n \delta_j^r j \cos {j t} } + Q \sqbrk {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} } \sin {r t} } \rd t }} {{eqn | r = \frac {2 \pi} {b - a} \int_0^\pi \sqbrk {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }' \paren {r \cos {r t} } + Q \sqbrk {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} } \sin {r t} }\rd t }} {{eqn | r = \frac {2 \pi} {b - a} \int_0^\pi \sqbrk {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} }' \paren {\sin {r t} }' + Q \sqbrk {\sum_{k \mathop = 1}^n \alpha_k \sin {k t} } \sin {r t} } \rd t }} {{end-eqn}} This leads to a system of equations: :$\displaystyle \int_0^\pi \paren {\map P t \sqbrk {\sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \paren {\sin {k t} }'} \paren {\sin {r x} }' + \sqbrk {Q - \lambda_n^{\paren 1} } \sqbrk {\sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin {k t} } \sin {r t} } \rd t = 0$ Multiplying each equation by an arbitrary constant $C_r^{\paren n}$ and summing over $r$ results in: :$\displaystyle \int_0^\pi \sqbrk {P y_n' h_n' + \paren {Q - \lambda_n^{\paren 1} y_n h_n} } \rd t = 0$ where: :$\displaystyle \map {h_n} t = \sum_{r \mathop = 1}^n C_r^{\paren n} \sin {r t}$ :$\displaystyle y_n = \sum_{k \mathop = 1}^n \alpha_k \sin {k t}$ By Integration by parts: {{begin-eqn}} {{eqn | l = \int_0^\pi \sqbrk {P y_n' h_n' + \paren {Q - \lambda_n^{\paren 1} y_n h_n} } \rd t       | r = \int_0^\pi P h_n' \map {\rd y_n} t + \int_0^\pi \sqbrk {\paren {Q - \lambda_n^{\paren 1} y_n h_n} } \rd t }} {{eqn | r = P h_n' y_n \big \vert_0^\pi - \int_0^\pi y_n \rd P \map {h_n'} t + \int_0^\pi \sqbrk {\paren {Q - \lambda_n^{\paren 1} y_n h_n} } \rd t }} {{eqn | r = \int_0^\pi \sqbrk { - \paren {P h_n'}' + \paren {Q - \lambda_n^{\paren 1} y_n h_n} } y_n \rd t }} {{end-eqn}} Consider all real mappings $h$ such that: :$\map h x \in C^2 \openint 0 \pi$ and satisfying the boundary conditions. Then $C_r^{\paren n}$ can be chosen, such that: :$\displaystyle \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n} x - \map h x}^2 \rd x = 0$ :$\displaystyle \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n'} x - \map {h'} x}^2 \rd x = 0$ :$\displaystyle \lim_{n \mathop \to \infty} \int_0^\pi \size {\map {h_n''} x - \map {h''} x}^2 \rd x = 0$ {{finish|proof of the $C_r^{\paren n}$ segment}} Due to the existence of uniformly convergent subsequence, $y_n^{\paren 1}$ converges to $y^{\paren 1}$ uniformly on $\closedint 0 \pi$: :$\displaystyle \lim_{m \mathop \to \infty} \int_0^\pi \paren {-\paren {P h_{n_m}'}' + \paren {Q - \lambda_{n_m}^{\paren 1} } h_{n_m} } y_{n_m}^{\paren 1} \rd x = \int_0^\pi \paren { -\paren {P h'}' + \paren {Q - \lambda^{\paren 1} } h} y^{\paren 1} \rd x = 0$ {{finish|prove}} By Lemma 5, where $Q_1 = Q - \lambda^{\paren 1}$, $y^{\paren 1} \in C^2 \closedint 0 \pi$ and satisfies Sturm-Liouville equation with $w = 1$. {{qed|lemma}} === Lemma 7 === $\sequence {\map {y_n^{\paren 1} } x}$ pointwise converges to $\map {y^{\paren 1} } x$. === Proof === By Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions, where $\map R x = 0$, the Sturm-Liouville equation :$-\paren {P y'}' + Q y = \lambda y$ satisfying the boundary conditions: :$\map y 0 = \map y \pi = 0$ and the subsidiary condition: :$\displaystyle \int_0^\pi \map {y^2} t = 1$ is unique up to the sign of $y$.  Let $\map {y^{\paren 1} } t$ be a solution corresponding to $\lambda = \lambda^{\paren 1}$ Due to the subsidiary condition, the condition $\map {y^{\paren 1} } t = 0$ cannot hold in the entire closed interval $\closedint 0 \pi$. Hence, the set of roots to this condition is countable. Then: :$\exists t_0 \in \closedint 0 \pi: \map {y^{\paren 1} } {t_0} \ne 0$ Choose the sign so that $\map {y^{\paren 1} } {t_0} > 0$ Similarly, let $\map {y_n^{\paren 1} } t$ be a solution corresponding to $\lambda = \lambda_n^{\paren 1}$ Choose the signs so that: :$\forall n \in \N : \map {y_n^{\paren 1} } {t_0} \ge 0$ Suppose $\map {y_n^{\paren 1} } t$ does not pointwise converge to $\map {y^{\paren 1} } t$. By Arzela's Theorem there exists another subsequence from $\sequence {\map {y_n^{\paren 1} } t}$, converging to another solution $\overline y^{\paren 1}$, where $\lambda = \lambda^{\paren 1}$. Because of the uniqueness of solutions, except for the sign, both solutions may differ only in their signs: :$\map {\overline y^{\paren 1} } x = - \map {y^{\paren 1} } t$ Therefore: :$\map {\overline y^{\paren 1} } {t_0} < 0$ This is impossible, since: :$\forall n \in N : \map {y_n^{\paren 1} } {t_0} \ge 0$  Therefore $\map {y_n^{\paren 1} } t$ pointwise converges to $\map {y^{\paren 1} } t$, provided $\map {y_n^{\paren 1} } t$ is chosen with the correct sign. {{finish|In fact, convergence is uniform; establish this result by a proof}} {{qed|lemma}} {{finish|Check how properly insert Limit of Subsequence equals Limit of Sequence as an alternative}} === Lemma 8 === Sequences $\sequence {y^{\paren n} }$ and $\sequence {\lambda^{\paren n} }$ are infinite. === Proof === Suppose, $y^{\paren r}$ and $\lambda^{\paren r}$ are known. The next eigenfunction $y^{\paren {r + 1} }$ and the corresponding eigenvalue $\lambda^{\paren {r + 1} }$ can be found by minimising  :$\displaystyle J \sqbrk y = \int_0^\pi \paren {P y'^2 + Q y^2} \rd x $ where boundary and subsidiary conditions are supplied with orthogonality conditions: :$\forall m \in \N : {1 \le m \le r} : \displaystyle \int_0^\pi \map {y^{\paren m} } t \map {y^{\paren {r + 1} } } t \rd t = 0$ The new solution of the form: :$\displaystyle \map {y_n^{\paren {r + 1} } } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$ is now also orthogonal to mappings: :$\displaystyle \map {y_n^{\paren m} } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren m} \sin {k t}$ This results into: :$\displaystyle \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \int_0^\pi \sin {k t} \paren {\sum_{l \mathop = 1}^n \alpha_l^{\paren m} \sin {l t}  } \rd t = \frac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \alpha_k^{\paren m} = 0$ These equations describe $r$ distinct $\paren {n - 1}$-dimensional hyperplanes, passing through the origin of coordinates in $n$ dimensions. These hyperplanes intersect the sphere $\sigma_n$, resulting in an $\paren {n - r}$-dimensional sphere $\hat \sigma_{n - r}$. By definition, it is a compact set. By Continuous Function on Compact Subspace of Euclidean Space is Bounded, $\map {J_n} {\boldsymbol \alpha}$ has a minimum on $\hat {\sigma}_{n - r}$.  Denote it as $\lambda_n^{\paren {r + 1} }$. By Ritz Method implies Not Worse Approximation with Increased Number of Functions: :$\displaystyle \lambda_{n + 1}^{\paren {r + 1} } \le \lambda_n^{\paren {r + 1} }$ This, together with $J$ being bounded from below, implies: :$\displaystyle \lambda^{\paren {r + 1} } = \lim_{n \mathop \to \infty} \lambda_n^{\paren {r + 1} }$ Additional constraints may or may not affect the new minimum: :$\lambda^{\paren r} \le \lambda^{\paren {r + 1} }$ Let: :$\displaystyle \map {y_n^{\paren {r + 1} } } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$ $y^{\paren {r + 1} }$ satisfies Sturm-Liouville equation together with boundary, subsidiary and orthogonality conditions. By Lemma 7, which is not affected by additional constraints, $\sequence {y_n^{\paren {r + 1} } }$ uniformly converges to $y^{\paren {r + 1} }$. Thus, $y^{\paren {r + 1} }$ is an eigenfunction of Sturm-Liouville equation with an eigenvalue $\lambda^{\paren {r + 1} }$. Orthogonal mappings are linearly independent. Each eigenvalue corresponds only to one eigenfunction, unique up to a constant factor. Thus: :$\lambda^{\paren r} < \lambda^{\paren {r + 1} }$ {{qed}}