[proofplan] The proof uses the principal Jacobi solution to factor the second-variation quadratic form. Absence of conjugate points makes this solution non-vanishing on $(a,b]$, so the logarithmic derivative $u'/u$ is available away from the left endpoint. A Riccati identity then turns the quadratic form into a square plus a boundary term, and the boundary term vanishes by the endpoint conditions and the normalization of $u$ at $a$. Coercivity follows in the regular Sturm-Liouville setting from the positive first Dirichlet eigenvalue, and the strict weak local minimum follows by comparing the uniform Taylor remainder with the coercive second variation. [/proofplan] [step:Rewrite the second variation in Jacobi normal form] Let $R:[a,b]\to \mathbb{R}$ and $S:[a,b]\to \mathbb{R}$ be the coefficient functions \begin{align*} R(x):=L_{sp}(x,y(x),y'(x)) \end{align*} and \begin{align*} S(x):=L_{ss}(x,y(x),y'(x)). \end{align*} Since $L\in C^3$ and $y\in C^2([a,b];\mathbb{R})$, the functions $P,R:[a,b]\to\mathbb{R}$ are continuously differentiable and $S:[a,b]\to\mathbb{R}$ is continuous. Hence $Q=S-R'$ is continuous on $[a,b]$. For $h\in C^1([a,b];\mathbb{R})$ with $h(a)=h(b)=0$, the classical second variation is \begin{align*} \int_a^b \left(P(x)(h'(x))^2+2R(x)h(x)h'(x)+S(x)h(x)^2\right)\,d\mathcal{L}^1(x). \end{align*} Because $(h^2)'=2hh'$ and $h(a)=h(b)=0$, [integration by parts](/theorems/210) with respect to $\mathcal{L}^1$ gives \begin{align*} \int_a^b 2R(x)h(x)h'(x)\,d\mathcal{L}^1(x)=-\int_a^b R'(x)h(x)^2\,d\mathcal{L}^1(x). \end{align*} Thus \begin{align*} \delta^2J[y;h,h]=\int_a^b \left(P(x)(h'(x))^2+Q(x)h(x)^2\right)\,d\mathcal{L}^1(x), \end{align*} where $Q=S-R'$. [/step] [step:Use absence of conjugate points to define the Riccati coefficient] Let \begin{align*} u:[a,b]\to \mathbb{R} \end{align*} be the principal Jacobi solution \begin{align*} -(P(x)u'(x))'+Q(x)u(x)=0,\qquad u(a)=0,\qquad u'(a)=1. \end{align*} By the no-conjugate-point hypothesis, $u(x)\neq 0$ for every $x\in (a,b]$. Since $u'(a)=1$, there exists $\eta>0$ such that $u(x)>0$ for every $x\in (a,a+\eta)$. The non-vanishing on $(a,b]$ then implies, after fixing this sign, that $u(x)>0$ for every $x\in (a,b]$. Define \begin{align*} w:(a,b]\to \mathbb{R},\qquad w(x):=\frac{P(x)u'(x)}{u(x)}. \end{align*} For $x\in (a,b]$, differentiating $w$ and using $(P u')'=Q u$ gives \begin{align*} w'(x)=Q(x)-\frac{w(x)^2}{P(x)}. \end{align*} This is the Riccati equation associated with the Jacobi equation. [/step] [step:Complete the square and remove the singular endpoint by a limit] Fix $h\in C^1([a,b];\mathbb{R})$ with $h(a)=h(b)=0$. For each $\varepsilon\in (a,b)$, the functions appearing below are continuous on $[\varepsilon,b]$, and the product rule gives \begin{align*} \left(w(x)h(x)^2\right)'=w'(x)h(x)^2+2w(x)h(x)h'(x). \end{align*} Using $w'=Q-w^2/P$ and $w=P u'/u$, we obtain on $[\varepsilon,b]$ \begin{align*} P(x)(h'(x))^2+Q(x)h(x)^2=P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2+\left(w(x)h(x)^2\right)'. \end{align*} Integrating over $[\varepsilon,b]$ with respect to $\mathcal{L}^1$ yields \begin{align*} \int_\varepsilon^b \left(P(x)(h'(x))^2+Q(x)h(x)^2\right)\,d\mathcal{L}^1(x)=\int_\varepsilon^b P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2\,d\mathcal{L}^1(x)-w(\varepsilon)h(\varepsilon)^2, \end{align*} because $h(b)=0$. It remains to show that the endpoint term tends to $0$ as $\varepsilon\downarrow a$. Since $u(a)=0$ and $u'(a)=1$, Taylor expansion at $a$ gives \begin{align*} u(\varepsilon)=(\varepsilon-a)+o(\varepsilon-a) \end{align*} and \begin{align*} u'(\varepsilon)=1+o(1). \end{align*} Since $h(a)=0$ and $h\in C^1([a,b];\mathbb{R})$, \begin{align*} h(\varepsilon)=h'(a)(\varepsilon-a)+o(\varepsilon-a). \end{align*} Therefore \begin{align*} w(\varepsilon)h(\varepsilon)^2=\frac{P(\varepsilon)u'(\varepsilon)}{u(\varepsilon)}h(\varepsilon)^2\to 0. \end{align*} The function $x\mapsto P(x)(h'(x))^2+Q(x)h(x)^2$ is continuous on $[a,b]$, hence integrable with respect to $\mathcal{L}^1$. By the absolute continuity of the [Lebesgue integral](/page/Lebesgue%20Integral), \begin{align*} \int_a^\varepsilon \left(P(x)(h'(x))^2+Q(x)h(x)^2\right)\,d\mathcal{L}^1(x)\to 0 \end{align*} as $\varepsilon\downarrow a$. Combining this with the identity on $[\varepsilon,b]$ and the limit $w(\varepsilon)h(\varepsilon)^2\to 0$, then letting $\varepsilon\downarrow a$, gives the Weierstrass-Jacobi identity \begin{align*} \delta^2J[y;h,h]=\int_a^b P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2\,d\mathcal{L}^1(x). \end{align*} [guided] The obstacle is that the natural logarithmic derivative $u'/u$ is singular at $a$, because the principal Jacobi solution satisfies $u(a)=0$. We therefore first work on a shortened interval $[\varepsilon,b]$, where $\varepsilon>a$. On this interval $u$ has no zeros, so the function \begin{align*} w:(\varepsilon,b]\to \mathbb{R},\qquad w(x):=\frac{P(x)u'(x)}{u(x)} \end{align*} is well-defined and continuously differentiable. The Jacobi equation is \begin{align*} -(P(x)u'(x))'+Q(x)u(x)=0. \end{align*} Equivalently, $(Pu')'=Qu$. Differentiating $w=Pu'/u$ gives \begin{align*} w'(x)=\frac{(P(x)u'(x))'}{u(x)}-\frac{P(x)(u'(x))^2}{u(x)^2}. \end{align*} Substituting $(Pu')'=Qu$ into the first term gives \begin{align*} w'(x)=Q(x)-\frac{P(x)(u'(x))^2}{u(x)^2}. \end{align*} Since $w^2/P=P(u')^2/u^2$, this is \begin{align*} w'(x)=Q(x)-\frac{w(x)^2}{P(x)}. \end{align*} Now we complete the square. For $x\in [\varepsilon,b]$, \begin{align*} P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2=P(x)(h'(x))^2-2w(x)h(x)h'(x)+\frac{w(x)^2}{P(x)}h(x)^2. \end{align*} Also, \begin{align*} \left(w(x)h(x)^2\right)'=w'(x)h(x)^2+2w(x)h(x)h'(x). \end{align*} Adding these two identities cancels the mixed terms. Using $w'=Q-w^2/P$, the remaining terms are exactly \begin{align*} P(x)(h'(x))^2+Q(x)h(x)^2. \end{align*} Thus \begin{align*} P(x)(h'(x))^2+Q(x)h(x)^2=P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2+\left(w(x)h(x)^2\right)'. \end{align*} Integrating from $\varepsilon$ to $b$ with respect to $\mathcal{L}^1$ gives \begin{align*} \int_\varepsilon^b \left(P(x)(h'(x))^2+Q(x)h(x)^2\right)\,d\mathcal{L}^1(x)=\int_\varepsilon^b P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2\,d\mathcal{L}^1(x)+w(b)h(b)^2-w(\varepsilon)h(\varepsilon)^2. \end{align*} The endpoint condition $h(b)=0$ removes the term at $b$. It remains to justify that the term at $\varepsilon$ vanishes as $\varepsilon\downarrow a$. The normalization $u(a)=0$ and $u'(a)=1$ implies \begin{align*} u(\varepsilon)=(\varepsilon-a)+o(\varepsilon-a). \end{align*} Similarly, since $h(a)=0$ and $h$ is $C^1$, \begin{align*} h(\varepsilon)=h'(a)(\varepsilon-a)+o(\varepsilon-a). \end{align*} Hence $w(\varepsilon)$ grows like $P(a)/(\varepsilon-a)$, while $h(\varepsilon)^2$ is of order $(\varepsilon-a)^2$. Their product is of order $\varepsilon-a$, and therefore \begin{align*} w(\varepsilon)h(\varepsilon)^2\to 0. \end{align*} The remaining passage to the full interval is justified by absolute continuity of the Lebesgue integral. Indeed, $x\mapsto P(x)(h'(x))^2+Q(x)h(x)^2$ is continuous on the compact interval $[a,b]$, hence belongs to $L^1([a,b])$, and therefore its integral over $[a,\varepsilon]$ tends to $0$ as $\varepsilon\downarrow a$. Combining this with the identity on $[\varepsilon,b]$ and the endpoint limit proves \begin{align*} \delta^2J[y;h,h]=\int_a^b P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2\,d\mathcal{L}^1(x). \end{align*} [/guided] [/step] [step:Conclude positive definiteness from the square identity] Since $P(x)>0$ on $[a,b]$, the Weierstrass-Jacobi identity implies \begin{align*} \delta^2J[y;h,h]\geq 0. \end{align*} If $\delta^2J[y;h,h]=0$, then for every $\eta\in (0,b-a)$ the non-negative integrand \begin{align*} x\mapsto P(x)\left(h'(x)-\frac{u'(x)}{u(x)}h(x)\right)^2 \end{align*} is continuous on $[a+\eta,b]$ and has integral $0$ there. Hence \begin{align*} h'(x)-\frac{u'(x)}{u(x)}h(x)=0 \end{align*} for every $x\in (a,b]$. Thus the function \begin{align*} \varphi:(a,b]\to \mathbb{R},\qquad \varphi(x):=\frac{h(x)}{u(x)} \end{align*} satisfies $\varphi'(x)=0$ on $(a,b]$. Hence $\varphi$ is constant on $(a,b]$, so there exists $C\in \mathbb{R}$ such that $h(x)=C u(x)$ for every $x\in (a,b]$. Evaluating at $b$ gives \begin{align*} 0=h(b)=C u(b). \end{align*} The no-conjugate-point hypothesis gives $u(b)\neq 0$, hence $C=0$. Therefore $h=0$ on $(a,b]$, and by continuity $h(a)=0$ as well. This proves positive definiteness on non-zero fixed-endpoint variations. [/step] [step:Upgrade positivity to coercivity in the regular Sturm-Liouville setting] Define the quadratic form \begin{align*} q:H^1_0(a,b)\to \mathbb{R}, \qquad q[h]:=\int_a^b \left(P(x)(h'(x))^2+Q(x)h(x)^2\right)\,d\mathcal{L}^1(x). \end{align*} Since $P,Q\in C([a,b])$, the form is continuous on $H^1_0(a,b)$. Let $H^{-1}(a,b):=(H^1_0(a,b))^*$ denote the dual [Sobolev space](/page/Sobolev%20Space). Define the weak Sturm-Liouville operator associated to $q$ by \begin{align*} A:H^1_0(a,b)\to H^{-1}(a,b) \end{align*} and \begin{align*} A h(\phi):=\int_a^b \left(P(x)h'(x)\phi'(x)+Q(x)h(x)\phi(x)\right)\,d\mathcal{L}^1(x) \end{align*} for $h,\phi\in H^1_0(a,b)$. The coefficients satisfy the regular Dirichlet Sturm-Liouville hypotheses: the interval $[a,b]$ is compact, $P\in C^1([a,b])$ is strictly positive, $Q\in C([a,b])$ is real-valued, and the domain is the Dirichlet Sobolev space $H^1_0(a,b)$. The previous step proves $q[h]\geq 0$ for every $h\in C^1([a,b];\mathbb{R})$ with $h(a)=h(b)=0$. Since $C_c^\infty(a,b)$ is dense in $H^1_0(a,b)$ and $q:H^1_0(a,b)\to\mathbb{R}$ is continuous, this non-negativity extends to every $h\in H^1_0(a,b)$. By the regular Dirichlet Sturm-Liouville spectral theorem, the first Dirichlet eigenvalue $\lambda_1$ of $A$ is characterized by the Rayleigh quotient \begin{align*} \lambda_1=\inf_{0\neq h\in H^1_0(a,b)}\frac{q[h]}{\int_a^b h(x)^2\,d\mathcal{L}^1(x)}. \end{align*} The extended non-negativity of $q$ gives $\lambda_1\geq 0$. The same spectral theorem also gives that if this infimum were $0$, then $0$ would be a Dirichlet eigenvalue, so there would exist a non-zero $h\in H^1_0(a,b)$ satisfying $Ah=0$ with zero Dirichlet trace. Because $P\in C^1([a,b])$ is strictly positive and $Q\in C([a,b])$, one-dimensional regularity for the equation $(P h')'=Qh$ upgrades such an eigenfunction to a non-zero $C^2$ solution of the Jacobi equation with $h(a)=h(b)=0$. This solution cannot have $h'(a)=0$, because uniqueness for the initial value problem with data $h(a)=0$ and $h'(a)=0$ would force $h\equiv 0$. Hence $h'(a)\neq 0$, and the rescaled function $\tilde h:=h/h'(a)$ satisfies the same initial conditions as the principal Jacobi solution $u$. By uniqueness, $\tilde h=u$, so $u(b)=\tilde h(b)=0$, making $b$ conjugate to $a$. The no-conjugate-point hypothesis excludes this. Therefore $\lambda_1>0$. Let \begin{align*} p_0:=\min_{x\in [a,b]}P(x)>0 \end{align*} and \begin{align*} M:=\max_{x\in [a,b]}|Q(x)|. \end{align*} For every $h\in H^1_0(a,b)$, \begin{align*} q[h]\geq p_0\int_a^b (h'(x))^2\,d\mathcal{L}^1(x)-M\int_a^b h(x)^2\,d\mathcal{L}^1(x). \end{align*} Also, by the eigenvalue lower bound, \begin{align*} q[h]\geq \lambda_1\int_a^b h(x)^2\,d\mathcal{L}^1(x). \end{align*} Choose $\theta\in (0,1)$ so close to $1$ that \begin{align*} \theta\lambda_1-(1-\theta)M>0. \end{align*} Taking the convex combination of the two lower bounds gives \begin{align*} q[h]\geq (1-\theta)p_0\int_a^b (h'(x))^2\,d\mathcal{L}^1(x)+\left(\theta\lambda_1-(1-\theta)M\right)\int_a^b h(x)^2\,d\mathcal{L}^1(x). \end{align*} Therefore, with \begin{align*} \gamma:=\min\{(1-\theta)p_0,\theta\lambda_1-(1-\theta)M\}>0, \end{align*} we obtain \begin{align*} q[h]\geq \gamma\|h\|_{H^1_0(a,b)}^2 \end{align*} for every $h\in H^1_0(a,b)$, where \begin{align*} \|h\|_{H^1_0(a,b)}^2:=\int_a^b h(x)^2\,d\mathcal{L}^1(x)+\int_a^b (h'(x))^2\,d\mathcal{L}^1(x). \end{align*} [/step] [step:Use the uniform Taylor expansion to obtain a strict weak local minimum] Let $\gamma>0$ be the coercivity constant from the previous step. By the assumed uniform Taylor expansion, choose $\rho_0\in (0,\rho]$ such that \begin{align*} \varepsilon(r)\leq \frac{\gamma}{4} \end{align*} for every $r\in (0,\rho_0]$. If $h\in C^1([a,b];\mathbb{R})$ satisfies $h(a)=h(b)=0$, $0<\|h\|_{C^1([a,b])}\leq \rho_0$, and $y+h\in \mathcal{A}_{\alpha,\beta}$, then \begin{align*} J[y+h]-J[y]\geq \frac{1}{2}\delta^2J[y;h,h]-\frac{\gamma}{4}\|h\|_{H^1_0(a,b)}^2. \end{align*} Using coercivity, \begin{align*} J[y+h]-J[y]\geq \frac{\gamma}{2}\|h\|_{H^1_0(a,b)}^2-\frac{\gamma}{4}\|h\|_{H^1_0(a,b)}^2. \end{align*} Hence \begin{align*} J[y+h]-J[y]\geq \frac{\gamma}{4}\|h\|_{H^1_0(a,b)}^2>0. \end{align*} Thus every sufficiently small non-zero fixed-endpoint $C^1$ variation increases $J$. This is exactly the assertion that $y$ is a strict weak local minimum in the $C^1$ topology. [/step]