ODE solution existence and uniqueness

Question

my question is: can i apply the theorem I mentioned in the title to assert that there is only a solution(given initial conditions) to a energy conservation equation?

I'm talking about an equation like this:

$ E= m \dot x^2 + kx $

My doubts come from the fact that the equation isn't linear, but I can obtain a form like this:

$ \dot x = \pm \sqrt{ f(x)} $

So we have two equations that singularly seem to respect the hypotesis of the theorem cited, but I'm not sure they do that when together.

Answer 1

No, your problem $\dot{x} = \pm\sqrt{f(x)}$ doesn't satisfy the assumpitons of the theorem, because $\pm\sqrt{f(x)}$ is not even a function, while Picard's theorem deals with problems

$\dot{x}(t) = F(x(t),t)\quad , \qquad x(t_0) = x_0$

where $F(x(t),t)$ is a function that, beyond being a function, must be continuous in $t$, and Lischitz-continuous in $x$.

Answer 2

I don't think you could apply the Picard-Lindelöf theorem directly, as your expression looks like $\dot x=\sqrt{E-kx}$ and fails to be defined for its entire domain (square root of a negative number).

What you instead would want to do in a physics class is to apply the Picard-Lindelöf theorem to the Hamilton’s equations for the energy expression, that should be more tractable.

Answer 3

As long as the initial condition is such that $E-kx_0>0$ then the Cauchy-Lipschitz theorem does appply and so you can assert that there exists a unique solution to the ODE defined for $t$ close to $t_0$, providing two (local) solutions of the energy conservation equation. The solutions should also be explicitely computable, no ?

Answer 4

The equation $$ 2E=m\dot x^2+kx^2 $$ can be locally uniquely solved for $\dot x$ if $\dot x\ne 0$. This local explicit form is then also continuously differentiable under this restriction. This gives monotonic solution segments.

It also has constant solutions at $\dot x=0$, that is, at $x=\pm\sqrt{2E/k}$. At these constant solutions there is no Lipschitz condition, the derivative is unbounded around it. More, one can easily construct piecewise solutions that alternate monotonous and constant segments.

Only by requiring that the second derivative be continuous one gets the "physical" solutions.