Roots of linearly independent solutions of homogeneous ODE 2

Question

If $y_1$ and $y_2$ are linearly independent solutions (set of fundamental solutions) of homogeneous ODE 2 $$ y'' + p(t) y' + q(t) y = 0 $$ prove that between 2 consecutive roots of $y_1$ there exists one and only one root of $y_2$

I believe Abels theorem is required in addition to the Wronskian formula, and the fact that since both solutions are LI, the Wronskian is non zero

I don´t have a clue how to approach this problem, any advice as well as a solution is welcome

Thanks in advance

Answer 1

I don't know what the precise minimal conditions are for this proof but it will suffice to assume that $y_1$, $y_2$, $p$ and $q$ are smooth for all $t$ of interest.

The Wronskian $W(t)$ for these solutions is given by $$ W(t) = y_1(t) y_2'(t) - y_2(t) y_1'(t)$$ and is smooth (from our assumptions).

If $y_1$ and $y_2$ are linearly dependent then $W(t) \neq 0$ for all $t$ (this also depends on the smoothness of $p$ and $q$ I think). Since $W(t)$ is smooth this implies that $W(t)$ is positive for all $t$ or negative for all $t$. We will assume w.l.o.g. that $$W(t) > 0, \quad \forall t$$

Now suppose that $t_1$ and $t_2$ are two consecutive roots of $y_1$ with $t_1 < t_2$. We have then that $y_1(t_1) = y_1(t_2) = 0$. But this means that $$W(t_1) = y_1(t_1) y_2'(t_1) - y_2(t_1) y_1'(t_1) = - y_2(t_1) y_1'(t_1) > 0.$$

We can easily see here that if $t_1$ is a root of $y_1$ it cannot also be a root of $y_2$. Also we see that any root of $y_1$ must have non-zero slope and must therefore result in a sign-change for $y_1$. Likewise for $t_2$ we must have $$ - y_2(t_2) y_1'(t_2) > 0. $$ so this is also not a root of $y_2$ and represents a sign-change for $y_1$.

Now if $t_1$ and $t_2$ are two consecutive sign-changing roots of $y_1$ is follows that the slopes $y_1'(t_1)$ and $y_1'(t_2)$ must have opposite sign. But from our last two inequalities that implies that $y_2(t_1)$ and $y_2(t_2)$ must also have opposite sign. The intermediate value theorem gives that there is at least one root $t_3$ of $y_2$ with $t_1 < t_3 < t_2$.

We've proved that there is at least one root for $y_2$ between any consecutive roots of $y_1$. To see that there is exactly one, reverse the argument above: between any two consecutive roots of $y_2$ there must be at least one root of $y_1$.

Answer 2

Let $a$ and $b$ denote the two consecutive roots of $y_2$. We will show the existence and uniqueness of zero of $y_1$ between $a$ and $b$.

For existence, assume for the sake of contradiction that $y_1$ does not vanish on $(a,b)$. Then $z:=y_2/y_1$ is differentiable there, and your calculation shows $z'=W/y_1^2$. By your observation that neither $a$ nor $b$ are zeros of $y_1$, we have $z(a)=z(b)=0$. By Rolle's theorem, there is some place in $(a,b)$ where $z'$, and thus $W$, vanishes. This contradicts the fact that $W$ never vanishes.

For uniqueness, assume $y_1$ has two roots on $(a,b)$, say $c$ and $d$. Apply the same argument above to the reciprocal $z^{-1}=y_1/y_2$ to conclude that $y_2$ must vanish somewhere in $(c,d)\subset(a,b)$, contradicting the fact that $a$ and $b$ are consecutive roots of $y_2$.

Source: https://math.stackexchange.com/a/2689725/581242 Source: