1
$\begingroup$

If the functions $y_1$ and $y_2$ are linearly independent solutions of $$y'' + p(t)y' + q(t)y = 0$$, show that between consecutive zeros of $y_1$ there is one and only one zero of $y_2$. Note that this

result is illustrated by the solutions $y_1(t) = cos t$ and $y_2(t) = sin t$ of the equation $y'' + y = 0$.

What i tried

I dont quite get what the question means when it says between consecutive zeros of $y_1$

there is one and only one zero of $y_2$. What i do know is that since $y_1$ and $y_2$ are

linearly independent, then the linear combination of $y_1$ and $y_2$ can be written as

$$c_1y_1+c_2y_2=0$$

Where $c_1=c_2=0$

Could anyone explain. Thanks

$\endgroup$
  • $\begingroup$ Consecutive zeros mean between the points of $\cos t =0$ namely $$x= \left(n-\frac{1}{2}\right)\pi$$ plus what is the shift between $\cos t$ and $\sin t$? $\endgroup$ – Chinny84 Feb 20 '15 at 16:42
  • $\begingroup$ cost=sin(t+$\pi/2$) $\endgroup$ – ys wong Feb 20 '15 at 16:53
  • $\begingroup$ i still dont get it could anyone explain. thanks $\endgroup$ – ys wong Feb 20 '15 at 17:25
  • $\begingroup$ i posted an explanation. check it out. $\endgroup$ – abel Feb 20 '15 at 20:04
3
$\begingroup$

i think you can show this interlacing property by considering the wronskian $w.$

suppose we have two linearly independent solutions $y_1$ and $y_2.$ suppose too, that there are two points $a < b$ such that $y_1(a) = 0, y_1(b) = 0, y_1(x) > 0 \text{ for } a < x < b \text{ and wlog } y_2(a) > 0.$

let the wronskian $w$ be defined by $$w(x) = y_1'(x)\,y_2(x) - y_2'(x)y_1(x).$$ we will show that $y_2$ has a zero in $(a, b)$ by showing that $y_2(a)y_2(b) < 0.$ note that by uniqueness theorem $y'_1(a) \neq 0$ and by the positivity of $y_1,$ we have in fact $y'_1(a) > 0$ and in the same way $y'_1(b) < 0.$ that is $$y'_1(a) > 0 \text{ and } y'_1(b) < 0. \tag 1 $$

by abel's theorem, $$w = w(a)e^{-\int_a^x p(t) \, dt}.$$ that is $w(x)$ has the same sign as $w(a)$ for $a < x < b.$ now $$w(a) = y'_1(a)y_2(a)-y'_2(a)y_1(a)=y'_1(a)y_2(a) > 0. \tag2$$

and $$w(b) = y'_1(b)y_2(b)-y'_2(b)y_1(b)=y'_1(b)y_2(b) > 0.\tag3$$

$(1), (2)$ and $(3)$ proves the claim.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.