The Strengthened Jacobi Condition and Conjugate Points

Derives Lemma A40.4. The system Equation (A40.3) is linear with a continuous coefficient matrix on the compact interval \([a,b]\), so Corollary 13.9 applies verbatim with \(\vect{g}=\vect{0}\): there is exactly one solution on all of \([a,b]\) for each choice of \(s\) and \(\vect{Y}_{0}\).

(1) If \(u(x_{0})=0\) and \(\left(P\,\dd u/\dd x\right)(x_{0})=0\) then \(\vect{Y}(x_{0})=\vect{0}\); but \(\vect{Y}\equiv\vect{0}\) is a solution with those data, so uniqueness forces \(\vect{Y}\equiv\vect{0}\) and hence \(u\equiv0\), which was excluded.

(2) Let \(x_{0}\) be a zero. By (1) and \(P>0\) we have \(\left(\dd u/\dd x\right)(x_{0})\neq0\), so by the definition of the derivative there is a punctured neighbourhood of \(x_{0}\) in which \(u(x)/(x-x_{0})\) keeps the sign of \(\left(\dd u/\dd x\right)(x_{0})\) and in particular \(u(x)\neq0\): the zero is isolated. If there were infinitely many zeros in \([a,b]\), Bolzano–Weierstrass (Theorem 11.7) would give a sequence of distinct zeros converging to some \(x_{\ast}\in[a,b]\); continuity of \(u\) makes \(x_{\ast}\) a zero as well, and it is not isolated — a contradiction.

Derives Lemma A40.5. Take first \(x\ge x_{0}\) and set \(\Phi(x)=C+L\int_{x_{0}}^{x}\varphi(t)\,\dd t\), which is of class \(C^{1}\) with \(\Phi'=L\varphi\) by the fundamental theorem of calculus (Theorem 11.42). Hypothesis Equation (A40.5) reads \(\varphi\le\Phi\) on \([x_{0},b]\), so \(\Phi'=L\varphi\le L\Phi\) there, and therefore

By the mean value theorem (Theorem 11.35), a function whose derivative is nowhere positive on an interval satisfies \(g(x)-g\left(x_{0}\right)=g'(\xi)\left(x-x_{0}\right)\le0\) for some interior \(\xi\) whenever \(x>x_{0}\); applied to \(g(x)=\Phi(x)\,\ee^{-L\left(x-x_{0}\right)}\) this gives \(\Phi(x)\,\ee^{-L\left(x-x_{0}\right)}\le\Phi(x_{0})=C\), that is \(\varphi(x)\le\Phi(x)\le C\,\ee^{L\left(x-x_{0}\right)}\).

For \(x\le x_{0}\) apply what has just been proved to \(\tilde\varphi(t)=\varphi\left(x_{0}-t\right)\) on \(\left[0,x_{0}-a\right]\), which satisfies the same hypothesis with the same constants because the substitution \(t\longmapsto x_{0}-t\) turns \(\left|\int_{x_{0}}^{x}\varphi\right|\) into \(\left|\int_{0}^{x_{0}-x}\tilde\varphi\right|\).

Derives Theorem A40.6. Write \(\vect{Y}=\vect{Y}\left(\cdot\,;s,\vect{Y}_{0}\right)\) and \(\vect{Z}=\vect{Y}\left(\cdot\,;\tilde s,\tilde{\vect{Y}}_{0}\right)\). Both are continuous and, by the equivalence of the initial value problem with its integral form Equation (13.7) — which is the first step of the proof of Theorem 13.8 and uses only the fundamental theorem of calculus —

A bound for \(\vect{Z}\) alone. Taking norms in the second identity of Equation (A40.7),

so Lemma A40.5 with \(\varphi=\norm{\vect{Z}}\), \(C=\norm{\tilde{\vect{Y}}_{0}}\) and \(x_{0}=\tilde s\) gives

The difference. Subtracting the two identities of Equation (A40.7) and splitting the second integral at \(s\),

The last term is bounded in norm by \(L\,\abs{s-\tilde s}\,\max_{[a,b]}\norm{\vect{Z}}\), which Equation (A40.8) bounds by \(L\,\ee^{L\left(b-a\right)}\norm{\tilde{\vect{Y}}_{0}}\, \abs{s-\tilde s}\). Hence, writing \(\varphi=\norm{\vect{Y}-\vect{Z}}\) and \(C=\norm{\vect{Y}_{0}-\tilde{\vect{Y}}_{0}} +L\,\ee^{L\left(b-a\right)}\norm{\tilde{\vect{Y}}_{0}}\, \abs{s-\tilde s}\),

and Lemma A40.5 gives Equation (A40.6).

For the final assertion, fix \(\left(x_{1},s_{1},\vect{Y}_{1}\right)\). The estimate Equation (A40.6) shows that \(\vect{Y}\left(\cdot\,;s,\vect{Y}_{0}\right)\) converges uniformly in \(x\) to \(\vect{Y}\left(\cdot\,;s_{1},\vect{Y}_{1}\right)\) as \(\left(s,\vect{Y}_{0}\right)\longrightarrow \left(s_{1},\vect{Y}_{1}\right)\), and each solution is continuous in \(x\); a uniform limit of continuous functions is continuous, and the triangle inequality

makes both contributions small.

Derives Lemma A40.7. Equation (A40.1) is the Sturm–Liouville equation \(L[u]=0\) of Lemma 13.56 with the data \(p=P\) and \(q=-Q\). The Lagrange identity Equation (13.56) therefore gives

so \(W\) is constant. In the notation of Definition A40.3, \(W\) is the determinant of the matrix with columns \(\vect{Y}_{1}=\left(u_{1},v_{1}\right)\) and \(\vect{Y}_{2}=\left(u_{2},v_{2}\right)\), since \(u_{1}v_{2}-u_{2}v_{1}=P\left(u_{1}u_{2}'-u_{2}u_{1}'\right)\). If \(W=0\), the two vectors are parallel at every point; picking a point \(x_{0}\) at which \(\vect{Y}_{1}\left(x_{0}\right)\neq\vect{0}\) — one exists unless \(u_{1}\equiv0\), in which case the dependence is trivial — gives \(\vect{Y}_{2}\left(x_{0}\right) =\kappa\,\vect{Y}_{1}\left(x_{0}\right)\) for a scalar \(\kappa\), and uniqueness in Lemma A40.4 then forces \(\vect{Y}_{2}=\kappa\vect{Y}_{1}\) identically, hence \(u_{2}=\kappa u_{1}\). Conversely \(u_{2}=\kappa u_{1}\) makes Equation (A40.9) vanish outright.

Derives Theorem A40.8. By Lemma A40.7 the constant \(W=P\left(u_{1}u_{2}'-u_{2}u_{1}'\right)\) is nonzero. Neither solution vanishes identically, so Lemma A40.4 applies to both.

No common zero. If \(u_{1}\left(x_{0}\right) =u_{2}\left(x_{0}\right)=0\) then both terms of Equation (A40.9) vanish at \(x_{0}\) and \(W=0\), a contradiction.

At least one zero in between. Let \(x_{1}<x_{2}\) be consecutive zeros of \(u_{1}\) in \([a,b]\), so that \(u_{1}\neq0\) on \(\left(x_{1},x_{2}\right)\). Evaluating Equation (A40.9) at the two ends, where \(u_{1}\) vanishes,

Both derivatives are nonzero by Lemma A40.4(1), and they have opposite signs: \(u_{1}\) keeps one sign, say positive, throughout \(\left(x_{1},x_{2}\right)\), so \(\left(\dd u_{1}/\dd x\right)\left(x_{1}\right) =\lim_{x\to x_{1}^{+}}u_{1}(x)/\left(x-x_{1}\right)\ge0\) and hence is \(>0\), while \(\left(\dd u_{1}/\dd x\right)\left(x_{2}\right) =\lim_{x\to x_{2}^{-}}u_{1}(x)/\left(x-x_{2}\right)\le0\) and hence is \(<0\); if \(u_{1}<0\) in between, both signs reverse. Since \(W\neq0\) is the same number at both ends of Equation (A40.10) and \(P>0\), the values \(u_{2}\left(x_{1}\right)\) and \(u_{2}\left(x_{2}\right)\) are both nonzero and of opposite sign. The intermediate value theorem (Theorem 11.23) gives a zero of \(u_{2}\) in \(\left(x_{1},x_{2}\right)\).

At most one. Suppose \(u_{2}\) had two zeros \(y_{1}<y_{2}\) in \(\left(x_{1},x_{2}\right)\); choosing them consecutive among the zeros of \(u_{2}\) — possible because by Lemma A40.4(2) there are finitely many — and applying the previous paragraph with the roles of \(u_{1}\) and \(u_{2}\) exchanged produces a zero of \(u_{1}\) in \(\left(y_{1},y_{2}\right)\subset\left(x_{1},x_{2}\right)\), contradicting the assumption that \(x_{1}\) and \(x_{2}\) are consecutive zeros of \(u_{1}\).

Proof of Theorem A40.1. Derives Theorem A40.1. The three implications are proved in the cycle \((1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(1)\).

\((1)\Rightarrow(2)\). Let \(w\) be a solution with no zero on \([a,b]\) and suppose, for contradiction, that some \(c\in\left(a,b\right]\) is conjugate to \(a\), i.e.\ \(u_{a}(c)=0\). The solution \(u_{a}\) does not vanish identically, since \(\left(P\,\dd u_{a}/\dd x\right)(a)=1\neq0\). Moreover \(u_{a}\) and \(w\) are linearly independent: a multiple of \(w\) cannot vanish at \(a\) as \(u_{a}\) does, unless the multiple is zero. By Lemma A40.4(2) the zeros of \(u_{a}\) in \([a,b]\) are finite in number, so among those lying in \(\left(a,c\right]\) there is a smallest one, \(c_{1}\); then \(a\) and \(c_{1}\) are consecutive zeros of \(u_{a}\). The Sturm separation theorem Theorem A40.8 places a zero of \(w\) in \(\left(a,c_{1}\right)\subset[a,b]\), contradicting the hypothesis on \(w\).

\((2)\Rightarrow(3)\). Since \(\left(P\,\dd u_{a}/\dd x\right)(a)=1\) and \(P(a)>0\), we have \(\left(\dd u_{a}/\dd x\right)(a)=1/P(a)>0\), and with \(u_{a}(a)=0\) this makes \(u_{a}(x)>0\) for \(x>a\) close enough to \(a\). By (2), \(u_{a}\) has no zero in \(\left(a,b\right]\); were \(u_{a}\) negative at some point of \(\left(a,b\right]\), the intermediate value theorem (Theorem 11.23) would produce a zero between that point and one where \(u_{a}>0\). Hence \(u_{a}>0\) throughout \(\left(a,b\right]\).

\((3)\Rightarrow(1)\). Let \(z\) be the solution of Equation (A40.1) with

which exists by Lemma A40.4, and consider the one-parameter family of solutions

each of which solves Equation (A40.1) by linearity. We show that \(w_{\delta}>0\) on \([a,b]\) for all small \(\delta>0\).

Write \(v_{a}=P\,\dd u_{a}/\dd x\), a continuous function with \(v_{a}(a)=1\), and recall \(z(a)=1\) with \(z\) continuous. Choose \(\eta>0\) so small that

On that subinterval \(\dd u_{a}/\dd x=v_{a}/P>0\), so \(u_{a}(x)=\int_{a}^{x}\left(v_{a}/P\right)\dd t\ge0\) by Theorem 11.43, and therefore

If \(a+\eta\ge b\) this already proves the claim for every \(\delta>0\). Otherwise, on the compact interval \(\left[a+\eta,b\right]\) the continuous function \(u_{a}\) is strictly positive by (3), so by the extreme value theorem (Theorem 11.24) it has a minimum \(m>0\) there; let \(M=\max_{[a,b]}\abs{z}\), also finite by Theorem 11.24. Choosing any \(\delta\) with \(0<\delta<m/\left(M+1\right)\) gives, for \(x\in\left[a+\eta,b\right]\),

So \(w=w_{\delta}\) has no zero on \([a,b]\), which is (1).

Derives Corollary A40.10. Sufficiency is immediate: \(u_{a-\varepsilon}\) restricted to \([a,b]\) is a solution of Equation (A40.1) without zeros there, which is statement (1) of Theorem A40.1.

Necessity. Assume the strengthened Jacobi condition, hence statement (3): \(u_{a}>0\) on \(\left(a,b\right]\). Work on the extended interval \(I=\left[a-\varepsilon_{0},b\right]\), on which Theorem A40.6 holds with the constant \(L\) of Equation (A40.4) computed over \(I\). Write \(v_{s}=P\,\dd u_{s}/\dd x\), so that the initial vector is \(\vect{Y}_{0}=\left(0,1\right)\) for every \(s\) and only the initial point varies; Equation (A40.6) then reads

where the norm on \(\R^{2}\) has been taken to be the sum of the moduli of the components and \(\norm{\vect{Y}_{0}}=1\).

Since \(v_{a}\) is continuous on \(I\) with \(v_{a}(a)=1\), fix \(\eta>0\) with \(\eta<\varepsilon_{0}\) and

If \(a+\eta<b\), the function \(u_{a}\) is continuous and strictly positive on the compact interval \(\left[a+\eta,b\right]\) and so has a minimum \(m>0\) there (Theorem 11.24); if \(a+\eta\ge b\) put \(m=+\infty\) and ignore the second condition below. Now choose \(\varepsilon>0\) subject to

By Equation (A40.13) the solution \(u=u_{a-\varepsilon}\) and its momentum \(v=v_{a-\varepsilon}\) then satisfy

On \(\left[a-\varepsilon,a+\eta\right]\) the first inequality gives \(\dd u/\dd x=v/P>0\), so \(u\) is strictly increasing there and, since \(u\left(a-\varepsilon\right)=0\), it is strictly positive on \(\left(a-\varepsilon,a+\eta\right]\) — in particular on \(\left[a,a+\eta\right]\), because \(a>a-\varepsilon\). Together with the second inequality, \(u>0\) on all of \([a,b]\), so \(u_{a-\varepsilon}\) has no zero there. The argument used only the three smallness conditions on \(\varepsilon\), each of which persists when \(\varepsilon\) is decreased, which proves the last clause.