# On the solutions of $y''(t)=f(y(t))$.

This topic has been inspired by some time spending on trying to refining my knowledge about PDE's in general. Then everybody knows how to solve $$y''(t)=\pm y(t).$$ Then I tried to slightly modify the question and I focused on $$(\therefore)\;y''(t)=f(y(t)),\; f\in C^1(\mathbb R,\mathbb R).$$ What I was trying to do was to derive some general properties about the solutions to this equation. In particular I ask to you, since I was not able to answer myself:

Must a solution of $(\therefore)$, not identically zero, have necessarily a finite number of zeroes on $[0,1]$? My idea was to derive an estimate like

$$|y(\eta)-y(\xi)|\leq C|\eta-\xi|^p,\; p>1;$$ Moreover, if a solution $y$ were to have an infinite number of zeroes in $[0,1]$, the the set of zeroes should have an accumulation point, and in this point all the derivatives should be equal to $0$ by continuity, then maybe the function should remain to much squeezed to be different from zero.

Hope you can help me because this interests me a lot.

(I tried to post this on mathlinks as well but nobody answered me yet)

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Try an analysis in the phase plane: the equation $$\frac{1}{2} \frac{d}{dx}p^2 = \frac{d}{dx} F(y(x))$$ where $p=y'$ and $F'=f$. –  Siminore Aug 18 '12 at 15:26
Sorry, the word "is" was in my brain but not in my fingers! The equation IS... –  Siminore Aug 18 '12 at 15:32
What do you mean by a finite number of zeros? Do you mean that the set $\{ t | y(t) = 0 \}$ is finite? The equation $y'' = -y$, $y(0) = 0, y'(0)=1$ has a solution $y(t) = \sin t$ which has an infinite number of zeros. –  copper.hat Aug 18 '12 at 16:41
Yes.. in $[0, 1]$.. not on the whole real line.. –  uforoboa Aug 18 '12 at 16:43
Sorry, I missed that in the question. –  copper.hat Aug 18 '12 at 16:46

This is a familiar type of ODE. In the following I shall deal with the "purely formal" aspect of it. Questions of sign under the square root or behavior near special points have to be treated "at runtime".

Let $F$ be a primitive of the given function $y\mapsto f(y)$. Multiplying the given differential equation by $y'(t)$ we get $$y''(t)y'(t)=f\bigl(y(t)\bigr)\,y'(t)$$ or $${d\over dt}\left({1\over2} y'^2(t)-F\bigl(y(t)\bigr)\right)=0\ .$$ Therefore there is a constant $C$ such that $$y'(t)=\sqrt{2F\bigl(y(t)\bigr)+C}\ .$$ Now the variables can be separated: $${dy\over\sqrt{2F(y)+C}}= dt\ .$$ This shows that ODEs of the considered type can be solved by quadratures. (The trick of multiplying by $y'$ is absolutely standard in mechanics, where it leads to the principle of "conservation of energy".)

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But how to conclude on the finiteness of the zeroes? –  uforoboa Aug 18 '12 at 17:00
An analysis of the curves $p^2 = 2F(y)+C$ in the $(y,p)$-space might help. What happens when $y=0$? Is $p \neq 0$? And if $p=0$, does it change sign nearby? –  Siminore Aug 18 '12 at 17:18
How does this answer address the zeros? –  copper.hat Aug 18 '12 at 18:46

Another way to do what Christian did.

The equation does not explicitly mention $t$ (only $y(t)$), so we can reduce to a first-order equation. Say $u = y'$ and get an equation for $u$ as a function of $y$ like this: \begin{align} \frac{du}{dt} &= \frac{du}{dy}\cdot\frac{dy}{dt} = \frac{du}{dy}\cdot u \\ \text{so}\qquad y'' &= f(y)\qquad\text{becomes} \\ u \frac{du}{dy} &= f(y) \end{align} a separable first order linear equation. Solve it. Once you get $u$ as a function of $y$, you have an integration problem to get (implicitly) $y$ as a function of $t$.

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Still i cannot see how this implies the finiteness of the number of zeroes.. –  uforoboa Aug 18 '12 at 16:34

As you said, if the set of zeroes has an accumulation point $x$, then the solution and its derivatives should vanish there. In particular, this implies that $f(0)=f(y(x))=y''(x)=0$. Then by ODE uniqueness theory the solution must vanish everywhere, because $y\equiv0$ is a solution.

A detailed treatment of such equations can be found in Cazenave's notes An introduction to semilinear elliptic equations. Google can find it easily for you.

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I don't see how uniqueness follows.. I mean I cannot apply the standard theorems.. can you detail your argument? Thanks.. –  uforoboa Aug 18 '12 at 16:42
Why must the derivative vanish at an accumulation point? –  copper.hat Aug 18 '12 at 16:48
@uforoboa: You can replace the equation by an equivalent first order system, then you can apply standard theory. –  timur Aug 18 '12 at 16:50
@copper.hat: It is not? I am not sure. Let us think. –  timur Aug 18 '12 at 16:51
Can you show me how? If you're right I will accept your answer because is what I am looking for.. @copper.. continuity and mean value theorems do the job.. –  uforoboa Aug 18 '12 at 16:53