A property of solution of ODE y''+p(x)y=0

[ipaper]

Let $f$ be a solution of the following equation $y''+p(x)y=0$, $p$ is continuous on $\mathbb{R}$ such that $p(x)\leq 0$ for all $x\in\mathbb{R}$. Suppose that $f$ is defined on $[a,+\infty)$, $f(a)>0$, $f'(a)>0$, $a\in\mathbb{R}$ .

Prove $f(x)>0$ for all $x\in[a,\infty)$.

Any help would be appreciated.

 

[chisigma]

Let $f$ be a solution of the following equation $y''+p(x)y=0$, $p$ is continuous on $\mathbb{R}$ such that $p(x)\leq 0$ for all $x\in\mathbb{R}$. Suppose that $f$ is defined on $[a,+\infty)$, $f(a)>0$, $f'(a)>0$, $a\in\mathbb{R}$ .

Prove $f(x)>0$ for all $x\in[a,\infty)$.

Any help would be appreciated.

Welcome on MHB ipaper!...

... writing the ODE as...

$\displaystyle y^{\ ''} = - p(x)\ y\ (1)$

... if $p(x) \le 0, \forall x \in \mathbb{R}$, $y(a)>0$ and $y^{\ '}(a)>0$, then y(x) must be a non decreasing function in $[a, + \infty)$ and therefore it will be $y(x)> 0\ \forall x \in [a, + \infty)$ ...

Kind regards

$\chi$ $\sigma$

 

[ipaper]

Would you please tell me more specifically? I haven't got the picture.

 

[chisigma]

Would you please tell me more specifically? I haven't got the picture.

All right!... writing again the ODE...

$\displaystyle y^{\ ''} = - p(x)\ y\ (1)$

... let's suppose $p(x) \le 0, y(a)>0, y^{\ '}(a)>0$... that means that for x=a y, y' are positive and y'' non negative... that means that for $x= a + \varepsilon$ it will be $y(a + \varepsilon) > y(a)$ and that is true for all $\varepsilon$, so that y will be a non decreasing function and that means that it will be $y(x) > 0$ for all x in $[a, + \infty)$...

Kind regards

$\chi$ $\sigma$