Understanding the shape of $\phi''(x)=F(\phi(x))$

Question

Hello I've got a question and no idea to get a solution. Maybe someone can give me an advice.

The following problem is given:

There is given a function $\phi \in C^2([a,b])$. Furthermore there is a $C^2$-function F (sufficiently well-defined). And there is the following relation: $\phi''(x)=F(\phi(x))$ (1)

The statement is: By differential equation (1) it is clear that every function $\phi$ that satisfies these requirements will have the following described shape:

$\phi(x)$ is even with respect to any local maximum or minimum, so that its graph is symmetric with respect to vertical lines through such maxima and minima. (image)

So, I've got no idea how to prove that statement. I would be grateful for every hint.

Answer 1

This is the equation of motion of a physical system. You multiply with the derivative and integrate (this only works for space dimension one) to obtain $$ \tfrac12ϕ'(x)^2+V(ϕ(x))=E=const. $$ where $V$ is an anti-derivative of $-F$, or $F(y)=-\nabla V(y)$.

Now the curves $$p=\pm\sqrt{2E-2V(y)}$$ parametrize the level sets of constant energy $E$ in the $(y,p)$ phase space, and any trajectory will be bound to stay on one energy level, which is fixed by the initial conditions.

Answer 2

We are given the second order ODE $$y''(x)=F\bigl(y(x)\bigr)\ ,\tag{1}$$ where $F$ is continuous on a suitable $y$-interval. Assume that $x\mapsto\phi(x)$ is a solution of $(1)$ in some open $x$-interval, and that $\phi$ has a local extremum at $x_0$. Then by the general existence and uniqueness theorem for ODEs this $\phi$ is the solution of the initial problem $(1)\wedge(2)$, where the initial conditions are given by $$y(x_0)=\phi(x_0),\quad y'(x_0)=0\ .\tag{2}$$ Consider now the function $\psi$ obtained by reflecting the graph of $\phi$ at the vertical $x=x_0$: $$\psi(x):=\phi(2x_0 -x)\qquad\bigl(|x-x_0|<h\bigr)\ .$$ One computes $\psi(x_0)=\phi(x_0)$, then $\psi'(x)=-\phi'(2x_0-x)$, and in particular $\psi'(x_0)=0$. Finally one has $$\psi''(x)=\phi''(2x_0-x)=F\bigl(\phi(2x_0-x)\bigr)=F\bigl(\psi(x)\bigr)\ ,$$ which establishes $\psi$ as "another" solution of $(1)\wedge(2)$. It follows that in fact $\psi(x)\equiv\phi(x)$, in other words, that the graph of $\phi$ is symmetric with respect to the vertical $x=x_0$.