[proofplan] We use the Sturm-Liouville Wronskian adapted to the coefficient $P$. Define \begin{align*} W(x)=P(x)\bigl(u'(x)v(x)-u(x)v'(x)\bigr) \end{align*} and prove that it is constant. First we show that this constant cannot be zero because a zero Wronskian would force linear dependence by uniqueness for the associated first-order system. Between two consecutive zeros of $u$, the function $v$ cannot vanish at the endpoints; if it had no zero inside, the quotient $u/v$ would have derivative of one fixed nonzero sign but equal endpoint values. The same quotient argument, now applied to $v/u$, rules out two distinct zeros of $v$ inside the interval. [/proofplan] [step:Differentiate the adapted Wronskian and show it is constant] Define the adapted Wronskian $W:[a,b]\to\mathbb{R}$ by \begin{align*} W(x)=P(x)\bigl(u'(x)v(x)-u(x)v'(x)\bigr) \end{align*} for $x \in [a,b]$. Equivalently, \begin{align*} W(x)=P(x)u'(x)v(x)-P(x)u(x)v'(x). \end{align*} So $W \in C^1([a,b];\mathbb{R})$, because $u,v \in C^1([a,b];\mathbb{R})$ and $Pu',Pv' \in C^1([a,b];\mathbb{R})$. For each $x \in [a,b]$, the differential equation gives $(Pu')'(x)=Q(x)u(x)$ and $(Pv')'(x)=Q(x)v(x)$. Differentiating the displayed formula for $W$ and substituting these identities gives \begin{align*} W'(x)=(Pu')'(x)v(x)+P(x)u'(x)v'(x)-u'(x)P(x)v'(x)-u(x)(Pv')'(x). \end{align*} The middle two terms cancel, so \begin{align*} W'(x)=Q(x)u(x)v(x)-u(x)Q(x)v(x)=0. \end{align*} for every $x \in [a,b]$. Hence $W$ is constant on $[a,b]$. [/step] [step:Use linear independence to prove the Wronskian constant is nonzero] We prove that the constant value of $W$ is not zero. Suppose, to the contrary, that $W(c)=0$ for some $c \in [a,b]$. Define the phase vector of a solution $w$ by $Y_w:[a,b]\to\mathbb{R}^2$, \begin{align*} Y_w(x)=\bigl(w(x),(Pw')(x)\bigr) \end{align*} The condition $W(c)=0$ says that the two vectors $Y_u(c)$ and $Y_v(c)$ are linearly dependent, since their coordinate determinant satisfies \begin{align*} u(c)(Pv')(c)-v(c)(Pu')(c)=-W(c)=0 \end{align*} Thus there exist $\lambda,\mu \in \mathbb{R}$, not both zero, such that the function $z:[a,b]\to\mathbb{R}$ defined by $z=\lambda u+\mu v$ satisfies $z(c)=0$ and $(Pz')(c)=0$. The function $Z:[a,b]\to\mathbb{R}^2$ defined by $Z(x)=(z(x),(Pz')(x))$ satisfies the first-order system \begin{align*} Z'(x)=A(x)Z(x) \end{align*} where $A:[a,b]\to\mathbb{R}^{2\times 2}$ is the continuous matrix-valued map with entries $A_{11}(x)=0$, $A_{12}(x)=1/P(x)$, $A_{21}(x)=Q(x)$, and $A_{22}(x)=0$. Indeed, the first component equation is $z'(x)=(Pz')(x)/P(x)$, and the second component equation is $(Pz')'(x)=Q(x)z(x)$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let $|\cdot|$ denote the Euclidean norm on $\mathbb{R}^2$, and let $\|\cdot\|_{\mathrm{op}}$ denote the corresponding operator norm on $\mathbb{R}^{2\times 2}$. Since $A$ is continuous on the compact interval $[a,b]$, define \begin{align*} M=\sup_{x\in[a,b]}\|A(x)\|_{\mathrm{op}}<\infty \end{align*} For $x \ge c$, the [fundamental theorem of calculus](/theorems/632) applied componentwise gives \begin{align*} Z(x)=\int_c^x A(t)Z(t)\,d\mathcal{L}^1(t) \end{align*} because $Z(c)=0$. Hence \begin{align*} |Z(x)|\le \int_c^x M|Z(t)|\,d\mathcal{L}^1(t) \end{align*} Choose $\delta>0$ with $M\delta<1$ if $M>0$, and choose arbitrary $\delta>0$ if $M=0$. On any interval $I=[r,s]\subset [c,b]$ of length at most $\delta$, if $Z(r)=0$, then \begin{align*} \sup_{x\in I}|Z(x)|\le M(s-r)\sup_{x\in I}|Z(x)| \end{align*} so $\sup_{x\in I}|Z(x)|=0$. Covering $[c,b]$ by finitely many such intervals shows $Z=0$ on $[c,b]$. For the left side, if $x\le c$, then the fundamental theorem of calculus and $Z(c)=0$ give \begin{align*} Z(x)=-\int_x^c A(t)Z(t)\,d\mathcal{L}^1(t) \end{align*} and hence \begin{align*} |Z(x)|\le \int_x^c M|Z(t)|\,d\mathcal{L}^1(t) \end{align*} The same interval estimate, now on intervals $[r,s]\subset[a,c]$ with $Z(s)=0$ and $s-r\le\delta$, gives $Z=0$ on $[a,c]$. Thus $z=0$ on $[a,b]$, so $\lambda u+\mu v=0$ with $(\lambda,\mu)\neq(0,0)$, contradicting the [linear independence](/page/Linear%20Independence) of $u$ and $v$. Therefore $W$ is a nonzero constant on $[a,b]$. [guided] The important point is that a zero Wronskian at one point gives equal initial data for a nontrivial linear combination. We make this precise. For any solution $w$, define its phase vector $Y_w:[a,b]\to\mathbb{R}^2$ by \begin{align*} Y_w(x)=\bigl(w(x),(Pw')(x)\bigr) \end{align*} This is the correct phase vector because the Sturm-Liouville equation controls $(Pw')'$, not necessarily $w''$. If $W(c)=0$ at some point $c \in [a,b]$, then the coordinate determinant of $Y_u(c)$ and $Y_v(c)$ satisfies \begin{align*} u(c)(Pv')(c)-v(c)(Pu')(c)=-W(c)=0 \end{align*} so the two vectors $Y_u(c)$ and $Y_v(c)$ are linearly dependent. Hence there are $\lambda,\mu \in \mathbb{R}$, not both zero, such that $z=\lambda u+\mu v$ satisfies \begin{align*} z(c)=0 \end{align*} and \begin{align*} (Pz')(c)=0 \end{align*} Now define $Z:[a,b]\to\mathbb{R}^2$ by $Z(x)=(z(x),(Pz')(x))$. Since $z$ also solves the same linear equation, $Z$ solves \begin{align*} Z'(x)=A(x)Z(x) \end{align*} with \begin{align*} A_{11}(x)=0,\quad A_{12}(x)=1/P(x),\quad A_{21}(x)=Q(x),\quad A_{22}(x)=0. \end{align*} The entries of $A$ are continuous because $P\in C^1([a,b];\mathbb{R})$, $P>0$ on $[a,b]$, and $Q\in C([a,b];\mathbb{R})$. Thus $A$ is bounded. Let \begin{align*} M=\sup_{x\in[a,b]}\|A(x)\|_{\mathrm{op}} \end{align*} where $\|\cdot\|_{\mathrm{op}}$ is the operator norm induced by the Euclidean norm on $\mathbb{R}^2$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Since $Z(c)=0$, the fundamental theorem of calculus gives, for $x\ge c$, \begin{align*} Z(x)=\int_c^x A(t)Z(t)\,d\mathcal{L}^1(t) \end{align*} Taking Euclidean norms and using the operator norm estimate $|A(t)Z(t)|\le \|A(t)\|_{\mathrm{op}}|Z(t)|$ gives \begin{align*} |Z(x)|\le \int_c^x M|Z(t)|\,d\mathcal{L}^1(t) \end{align*} Choose $\delta>0$ so that $M\delta<1$ when $M>0$; if $M=0$, any positive $\delta$ works. Suppose $Z(r)=0$ on the left endpoint of an interval $[r,s]\subset[c,b]$ with $s-r\le\delta$. Then for every $x\in[r,s]$, \begin{align*} |Z(x)|\le M(s-r)\sup_{y\in[r,s]}|Z(y)| \end{align*} Taking the supremum over $x\in[r,s]$ gives \begin{align*} \sup_{x\in[r,s]}|Z(x)|\le M(s-r)\sup_{x\in[r,s]}|Z(x)| \end{align*} Because $M(s-r)<1$, this forces $\sup_{x\in[r,s]}|Z(x)|=0$. Thus $Z$ vanishes on that interval. Repeating over finitely many intervals covers $[c,b]$. To move left from $c$, take $x\le c$. Since $Z(c)=0$, the fundamental theorem of calculus gives \begin{align*} Z(x)=-\int_x^c A(t)Z(t)\,d\mathcal{L}^1(t) \end{align*} Taking norms gives \begin{align*} |Z(x)|\le \int_x^c M|Z(t)|\,d\mathcal{L}^1(t) \end{align*} If $[r,s]\subset[a,c]$, $Z(s)=0$, and $s-r\le\delta$, the same supremum estimate gives $Z=0$ on $[r,s]$. A finite chain of such intervals from $c$ to $a$ gives $Z=0$ on $[a,c]$. Therefore $Z=0$ on $[a,b]$, so $z=0$ on $[a,b]$. But $z=\lambda u+\mu v$ and $(\lambda,\mu)\neq(0,0)$, which says exactly that $u$ and $v$ are linearly dependent. This contradicts the hypothesis. Therefore $W$ cannot vanish at any point, and since it is constant, its constant value is nonzero. [/guided] [/step] [step:Show the second solution cannot vanish at the endpoint zeros of the first] Let $\alpha,\beta \in [a,b]$ be consecutive zeros of $u$. We claim that $v(\alpha)\neq 0$ and $v(\beta)\neq 0$. If $v(\alpha)=0$, then using $u(\alpha)=0$ gives \begin{align*} W(\alpha)=P(\alpha)\bigl(u'(\alpha)v(\alpha)-u(\alpha)v'(\alpha)\bigr)=0 \end{align*} which contradicts that $W$ is a nonzero constant. Hence $v(\alpha)\neq 0$. The same computation at $\beta$, using $u(\beta)=0$, gives $v(\beta)\neq 0$. [/step] [step:Prove existence of a zero by applying the quotient monotonicity argument] Assume, for contradiction, that $v(x)\neq 0$ for every $x\in(\alpha,\beta)$. By the previous step, $v(\alpha)\neq 0$ and $v(\beta)\neq 0$, so $v$ is nonzero on all of $[\alpha,\beta]$. Define the quotient map $r:[\alpha,\beta]\to\mathbb{R}$ by \begin{align*} r(x)=\frac{u(x)}{v(x)} \end{align*} Since $u,v\in C^1([a,b];\mathbb{R})$ and $v$ is nonzero on $[\alpha,\beta]$, the map $r$ belongs to $C^1([\alpha,\beta];\mathbb{R})$. For each $x\in[\alpha,\beta]$, \begin{align*} r'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}=\frac{W}{P(x)v(x)^2} \end{align*} The denominator $P(x)v(x)^2$ is strictly positive on $[\alpha,\beta]$, while $W$ is a fixed nonzero real number. Therefore $r'$ has one fixed nonzero sign on $[\alpha,\beta]$. Since $u(\alpha)=u(\beta)=0$ and $v(\alpha),v(\beta)\neq 0$, we have $r(\alpha)=r(\beta)=0$. The fundamental theorem of calculus gives \begin{align*} 0=r(\beta)-r(\alpha)=\int_\alpha^\beta r'(x)\,d\mathcal{L}^1(x) \end{align*} But the [continuous function](/page/Continuous%20Function) $r'$ has one fixed nonzero sign on $[\alpha,\beta]$, so the integral cannot be zero. This contradiction proves that there exists at least one $\gamma\in(\alpha,\beta)$ such that $v(\gamma)=0$. [/step] [step:Prove uniqueness by applying the quotient argument with the roles reversed] Suppose, for contradiction, that $v$ has two distinct zeros $\gamma_1,\gamma_2\in(\alpha,\beta)$ with $\gamma_1<\gamma_2$. Since $\alpha$ and $\beta$ are consecutive zeros of $u$, we have $u(x)\neq 0$ for every $x\in(\alpha,\beta)$, and therefore $u$ is nonzero on $[\gamma_1,\gamma_2]$. Define $s:[\gamma_1,\gamma_2]\to\mathbb{R}$ by \begin{align*} s(x)=\frac{v(x)}{u(x)} \end{align*} Then $s\in C^1([\gamma_1,\gamma_2];\mathbb{R})$, and for each $x\in[\gamma_1,\gamma_2]$, \begin{align*} s'(x)=\frac{v'(x)u(x)-v(x)u'(x)}{u(x)^2}=-\frac{W}{P(x)u(x)^2} \end{align*} Again $P(x)u(x)^2>0$ on $[\gamma_1,\gamma_2]$ and $W\neq 0$, so $s'$ has one fixed nonzero sign. Since $v(\gamma_1)=v(\gamma_2)=0$ and $u$ is nonzero on $[\gamma_1,\gamma_2]$, we have $s(\gamma_1)=s(\gamma_2)=0$. Thus \begin{align*} 0=s(\gamma_2)-s(\gamma_1)=\int_{\gamma_1}^{\gamma_2} s'(x)\,d\mathcal{L}^1(x) \end{align*} which is impossible because $s'$ is continuous and has one fixed nonzero sign. Hence $v$ has at most one zero in $(\alpha,\beta)$. Combining this with existence gives exactly one zero of $v$ between the consecutive zeros $\alpha$ and $\beta$ of $u$. [/step]