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Consider the second-order linear differential equation $u'' + p(x)u' + q(x)u = 0$ where $p$ and $q$ are continuous on the entire $\mathbb{R}$. Suppose that $q(x) < 0 $ everywhere. Show that if $u$ is not identically $0$ then $u$ can have at most one zero on $\mathbb{R}$.

Any suggestions on how to go about this? Anything will be deeply appreciated!

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There seems to be an error in your question: Take $p(x) = 0$, $q(x) = -1$, then a solution of your ODE is given by $u(x) = x^2-1$, which has two zeros at $\{-1,1\}$. Btw: your ODE is not "homogeneous". –  gerw Jan 21 '13 at 7:38
    
I think you want to say $u'' + p(x)u' + q(x) u= 0$. –  Paul Jan 21 '13 at 7:40
    
Yes! I will amend it! –  user44069 Jan 21 '13 at 7:45

1 Answer 1

up vote 3 down vote accepted

My suggestion for a proof by contradiction:

  • Take $x_0,x_1$ such that $u(x_0) = u(x_1) = 0$ and $u|_{(x_0,x_1)} \ne 0$. Assume wlog. that there is $x \in (x_0,x_1)$ with $u(x) > 0$.
  • Take $x \in \mathrm{argmax}_{[x_0,x_1]} u$. By optimality, we have $u'(x) = 0$ and $u''(x) \le 0$. This contradicts the ODE.
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How do you take a interval $(x_0,x_1)$ with $u\ne 0?$ –  user52188 Jan 21 '13 at 11:15
    
One is supposing, for the purposes of contradiction, that $u$ has two roots. Since this is the case, the interval between the roots satisfies that $u(x) \neq 0$, since we are assuming $u$ has only two roots. –  user44069 Jan 21 '13 at 17:28
    
what is $argmax_[x_{0},x_{1}]u$ means? –  Real Hilbert Jan 22 '13 at 15:55

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