By the continuity of function $\map P x$ and the solution of Jacobi's equation: :$\exists \epsilon > 0: \paren {\forall x \in \closedint a {b + \epsilon}:\map P x > 0} \land \paren {\tilde a \notin \closedint a {b + \epsilon} }$ Consider the quadratic functional: :$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x - \alpha^2 \int_a^b h'^2 \rd x$ together with Euler's equation: :$-\dfrac \rd {\rd x} \paren{\paren {P - \alpha^2} h'} + Q h = 0$ The Euler's equation is continuous {{WRT}} $\alpha$. Thus the solution of the Euler's equation is continuous {{WRT}} $\alpha $. {{ProofWanted|solution to continuous differential equation is continuous}} Since: :$\forall x \in \closedint a {b + \epsilon}: \map P x > 0$  $\map P x$ has a positive lower bound in $\closedint a {b + \epsilon}$. Consider the solution with $\map h a = 0$, $\map {h'} 0 = 1$. Then :$\exists \alpha \in \R: \forall x \in \closedint a b: \map P x - \alpha^2 > 0$ Also: :$\forall x \in \hointl a b: \map h x \ne 0$ {{Stub|seems to be Jacobi's condition where $P \to P - \alpha^2$}} By Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite: :$\ds \int_a^b \paren {\paren {P - \alpha^2} h'^2 + Q h^2} \rd x > 0$ In other words, if $c = \alpha^2$, then: :$(1): \quad \exists c > 0: \displaystyle \int_a^b \paren {P h'^2 + Q h^2} \rd x > c \int_a^b h'^2 \rd x$  Let $y = \map y x$ be an extremal. Let $y = \map y x + \map h x$ be a curve, sufficiently close to $y = \map y x$. By expansion of $\Delta J \sqbrk {y; h}$ from lemma $1$ of Legendre's Condition: :$\ds J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$ where: :$\ds \forall x \in \closedint a b: \lim_{\size h_1 \mathop \to 0} \set {\xi,\eta} = \set {0, 0}$ {{explain|$\size h_1$}} and the limit is uniform. {{Stub|why?}} By Schwarz inequality: {{begin-eqn}} {{eqn | l = \map {h^2} x       | r = \paren {\int_a^x \map {h'} x \rd x}^2 }} {{eqn | r = \int_a^x 1^2 d y \int_a^x \map {h'^2} x \rd x }} {{eqn | o = \le       | r = \paren {x - a} \int_a^x \map {h'^2} x \rd x }} {{eqn | o = \le       | r = \paren {x - a} \int_a^b \map {h'^2} x \rd x       | c = $h'^2 \ge 0$ }} {{end-eqn}} Notice that the integral on the right does not depend on $x$. Integrate the inequality {{WRT|Integration}} $x$: {{begin-eqn}} {{eqn | l = \int_a^b \map{h^2} x \rd x       | o = \le       | r = \int_a^b \paren {x - a} \rd x \int_a^b \map {h'^2} x \rd x }} {{eqn | r = \frac {\paren {b - a}^2} 2 \int_a^b \map {h'^2} x \rd x }} {{end-eqn}} Let $\epsilon \in \R_{>0}$ be a constant such that: :$\size \xi \le \epsilon$, $\size \eta \le \epsilon$ Then: {{begin-eqn}} {{eqn | l = \size {\int_a^b \paren {\xi h^2 + \eta h'^2} \rd x}       | o = \le       | r = \int_a^b \size \xi h^2 \rd x + \int_a^b \size \eta h'^2 \rd x       | c = Absolute Value of Definite Integral, Absolute Value Function is Completely Multiplicative }} {{eqn | o = \le       | r = \epsilon \int_a^b h'^2 \rd x + \epsilon \frac {\paren {a - b}^2} 2 \int_a^b h'^2 \rd x }} {{eqn | n = 2       | r = \epsilon \paren {1 + \frac {\paren {b - a}^2} 2} \int_a^b h'^2 \rd x }} {{end-eqn}} Thus, by $(1)$: :$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x > 0$  while by $(2)$: :$\ds \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$  can be made arbitrarily small. Thus, for all sufficiently small $\size h_1$, which implies sufficiently small $\size \xi$ and $\size \eta$, and, consequently, sufficiently small $\epsilon$: :$J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x > 0$ Therefore, in some small neighbourhood $y = \map y x$ there exists a weak minimum of the functional. {{qed}}