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Show that any nontrivial solution to $-u'' + q(x)u = 0$ has finitely many zeros

1

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I am looking through an example and need some help to understand it.

For the problem $-u'' + q(x)u = 0$, where $0<x<1$, $q in mathcalC(0,1]$ and $x^2q(x) rightarrow q_0$ as $x rightarrow 0^+$, the goal is to show that any non-trivial solution only has finitely many zeros if $q_0 > -1/4$.

If $q_0 > -1/4$, then we can use Sturm's comparison theorem, and compare our equation with $-u'' - frac14x^2u = 0$, and the solution $sqrtx$. We see that $sqrtx$ has no zero in $(0,1]$, and since $q(x) > -1/4$ for sufficiently small $x$, we have that any non-trivial solution $u$ to $-u'' + q(x)u = 0$ can't have an accumulation of zeros at $x=0$. This part I understand.

Next, it says that it can also only have finitely many zeros in $(0,1]$, because otherwise the zeros would accumulate at some point $x_0$, and this would imply that $u(x_0) = u'(x_0)$ (which in turn implies that $u(x) equiv 0$).

The part that I don't quite understand is why an accumulation of zeros at $x_0$ implies that $u(x_0) = u'(x_0)$.

I had a thought that if we didn't have $u(x) equiv 0$, we would have a solution that kept oscillating for all future times around $x_0$ (if there were infinitely many zeros there) but I can't quite piece it together. Can such a solution not exist? Or does it have to do with something else entirely?

ordinary-differential-equations sturm-liouville

share|cite|improve this question

asked Mar 19 at 22:17

RickSRickS

778

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you missed part of that conclusion, it is that $u(x_0)=0$ as accumulation point and $u'(x_0)=0$ by the mean value theorem. As both are zero, one may shortly write $u(x_0)=u'(x_0)=0$.
  $endgroup$
  – LutzL
  Mar 20 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is clear to me now. Thank you. :)
  $endgroup$
  – RickS
  Mar 21 at 10:22
  

  
  
  
  

add a comment | 

1

$begingroup$

I am looking through an example and need some help to understand it.

For the problem $-u'' + q(x)u = 0$, where $0<x<1$, $q in mathcalC(0,1]$ and $x^2q(x) rightarrow q_0$ as $x rightarrow 0^+$, the goal is to show that any non-trivial solution only has finitely many zeros if $q_0 > -1/4$.

If $q_0 > -1/4$, then we can use Sturm's comparison theorem, and compare our equation with $-u'' - frac14x^2u = 0$, and the solution $sqrtx$. We see that $sqrtx$ has no zero in $(0,1]$, and since $q(x) > -1/4$ for sufficiently small $x$, we have that any non-trivial solution $u$ to $-u'' + q(x)u = 0$ can't have an accumulation of zeros at $x=0$. This part I understand.

Next, it says that it can also only have finitely many zeros in $(0,1]$, because otherwise the zeros would accumulate at some point $x_0$, and this would imply that $u(x_0) = u'(x_0)$ (which in turn implies that $u(x) equiv 0$).

The part that I don't quite understand is why an accumulation of zeros at $x_0$ implies that $u(x_0) = u'(x_0)$.

I had a thought that if we didn't have $u(x) equiv 0$, we would have a solution that kept oscillating for all future times around $x_0$ (if there were infinitely many zeros there) but I can't quite piece it together. Can such a solution not exist? Or does it have to do with something else entirely?

ordinary-differential-equations sturm-liouville

share|cite|improve this question

asked Mar 19 at 22:17

RickSRickS

778

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you missed part of that conclusion, it is that $u(x_0)=0$ as accumulation point and $u'(x_0)=0$ by the mean value theorem. As both are zero, one may shortly write $u(x_0)=u'(x_0)=0$.
  $endgroup$
  – LutzL
  Mar 20 at 6:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it is clear to me now. Thank you. :)
  $endgroup$
  – RickS
  Mar 21 at 10:22
  

  
  
  
  

add a comment | 

$begingroup$

I am looking through an example and need some help to understand it.

For the problem $-u'' + q(x)u = 0$, where $0<x<1$, $q in mathcalC(0,1]$ and $x^2q(x) rightarrow q_0$ as $x rightarrow 0^+$, the goal is to show that any non-trivial solution only has finitely many zeros if $q_0 > -1/4$.

If $q_0 > -1/4$, then we can use Sturm's comparison theorem, and compare our equation with $-u'' - frac14x^2u = 0$, and the solution $sqrtx$. We see that $sqrtx$ has no zero in $(0,1]$, and since $q(x) > -1/4$ for sufficiently small $x$, we have that any non-trivial solution $u$ to $-u'' + q(x)u = 0$ can't have an accumulation of zeros at $x=0$. This part I understand.

Next, it says that it can also only have finitely many zeros in $(0,1]$, because otherwise the zeros would accumulate at some point $x_0$, and this would imply that $u(x_0) = u'(x_0)$ (which in turn implies that $u(x) equiv 0$).

The part that I don't quite understand is why an accumulation of zeros at $x_0$ implies that $u(x_0) = u'(x_0)$.

I had a thought that if we didn't have $u(x) equiv 0$, we would have a solution that kept oscillating for all future times around $x_0$ (if there were infinitely many zeros there) but I can't quite piece it together. Can such a solution not exist? Or does it have to do with something else entirely?

ordinary-differential-equations sturm-liouville

share|cite|improve this question

asked Mar 19 at 22:17

RickSRickS

778

$endgroup$

share|cite|improve this question

asked Mar 19 at 22:17

RickSRickS

778

share|cite|improve this question

asked Mar 19 at 22:17

RickSRickS

778

asked Mar 19 at 22:17

RickSRickS

778

asked Mar 19 at 22:17

RickSRickS

778