What can we say about the zero's of any non-trivial solution of the linear differential equation$$y^{''}+q(x)y=0$$ (where $q(x)$ is positive monotonically increasing continuous function of $x$). Can we say that it must have infinitely many zeros in $\mathbb{R}?$ For example $y^{''}+y=0$ has $sin(x)$ as a non trivial solution, which has infinitely many zero's. Please help me to prove the general result about infinitely many zero's. Thanks in advance.
[Math] Zero’s of non trivial solution of an second order ordinary differential equation
ordinary differential equations
Related Solutions
In fact it belongs to an Emden-Fowler equation.
First, according to http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf or http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=6 , all Emden-Fowler equations can be transformed into Abel equation of the second kind.
Let $\begin{cases}u=\dfrac{x^3}{y_n^\frac{3}{2}}\\v=\dfrac{x}{y_n}\dfrac{dy_n}{dx}\end{cases}$ ,
Then $\dfrac{dv}{du}=\dfrac{\dfrac{dv}{dx}}{\dfrac{du}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{1}{y_n}\dfrac{dy_n}{dx}-\dfrac{x}{y_n^2}\left(\dfrac{dy_n}{dx}\right)^2}{\dfrac{3x^2}{y_n^\frac{3}{2}}-\dfrac{3x^3}{2y_n^\frac{5}{2}}\dfrac{dy_n}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{v}{x}-\dfrac{v^2}{x}}{\dfrac{3u}{x}-\dfrac{3uv}{2x}}=\dfrac{\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2}{3u\left(1-\dfrac{v}{2}\right)}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}=\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2$
$\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}=3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v$
$\dfrac{d^2y_n}{dx^2}=\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)$
$\therefore\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)-\dfrac{nx}{\sqrt{y_n}}=0$
$\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)=\dfrac{nx}{\sqrt{y_n}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=\dfrac{nx^3}{y_n^\frac{3}{2}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=nu$
$3u\left(\dfrac{v}{2}-1\right)\dfrac{dv}{du}=v^2-v-nu$
Let $w=\dfrac{v}{2}-1$ ,
Then $v=2w+2$
$\dfrac{dv}{du}=2\dfrac{dw}{du}$
$\therefore6uw\dfrac{dw}{du}=(2w+2)^2-(2w+2)-nu$
$6uw\dfrac{dw}{du}=4w^2+6w+2-nu$
$w\dfrac{dw}{du}=\dfrac{2w^2}{3u}+\dfrac{w}{u}+\dfrac{2-nu}{6u}$
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{2}{3uz^2}+\dfrac{1}{uz}+\dfrac{2-nu}{6u}$
$\dfrac{dz}{du}=\dfrac{(nu-2)z^3}{6u}-\dfrac{z^2}{u}-\dfrac{2z}{3u}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2.
We can rewrite the ODE as $y''=q(x)y$. When $x=0$, both $y$ and $y'$ are positive. The ODE tells us that $y''$ is also positive. Therefore, the function is increasing, and its derivative is also increasing. As the function gets bigger, $q(x)$ also gets bigger. This leads to $y''$ also getting bigger, which means that $y'$ gets bigger too! There is nothing to stop the growth process; the solution just grows without bound forever. This should give you enough information to determine which of those options are correct.
Best Answer
Let $0<c\le\sqrt{q(x)}$. The infinitesimal angle increment of the complexified phase vector $y'+icy$ is the imaginary part of $$ \frac{d}{dt}Ln(y'+icy)=\frac{y''+icy'}{y'+icy}=\frac{(-qy+icy')(y'-icy)}{y'^2+c^2y^2} \\ =\frac{-(q-c^2)yy'+ic(y'^2+qy^2)}{y'^2+c^2y^2} $$ Thus the angular velocity around the origin is $$ c+\frac{q-c^2}{y'^2+c^2y^2}cy^2\ge c, $$ so there are infinitely many revolutions around the origin, each with time smaller $\frac{2\pi}c$, and each revolution gives two zeros of $y$.
Or, to state compatibility with the Sturm–Picone comparison theorem (where the second equation is $y''+c^2y=0$), each interval of length $\frac{\pi}c$ contains at least a half-revolution and thus at least one zero of $y$.