Proving a certain nonlinear ODE has a solution on $t \geq 0$

Question

I'm trying to prove that the following ODE admits a solution on the interval $[0, \infty )$:

$\begin{align*} \begin{cases} x_1^\prime &= x_2 - x_1^3 \\ x_2^\prime &= \frac{1}{2}x_1 - x_2 + d \sin t \end{cases} \end{align*}$

With initial condition $x(0) = x_0$.

The usual approaches seem to fail: the derivative isn't bounded on $\mathbb{R}^2$ and the function does not seem Lipschitz on all of that domain either. Using energy functions doesn't seem to help either.

Any hints would be highly appreciated!

Answer 1

We use the comment by Hans Engler to prove your result.

Let $w(t) = x_1^2(t) + x_2^2(t) = || x ||_2^2$. Then, $$\begin{align*} w^\prime (t) &= 3x_1 x_2 - 2x_1^4 - 2x_2^2 + d x_2 \sin t \\ &\leq 3x_1 x_2 - 2x_2^2 + d |x_2| \end{align*}$$

Using the elementary inequality $ab \leq \frac{1}{2} (a^2 + b^2)$, we get that: \begin{align*} w^\prime (t) &\leq \frac{3}{2} ( x_1^2 + x_2^2 ) - 2x_2^2 + \frac{1}{2} (x_2^2 + d^2) \\ &\leq \frac{3}{2} \left( w(t) + \frac{1}{3}d^2 \right) \end{align*}

It follows by Grönwall's that $w(t)$ cannot blow up in finite time, which means that the norm of $x$, and hence $x$, cannot blow up in finite time either.

Answer 2

Hint.

We have

$$ x_2'=x_1'' + 3x_1^2x_1' $$

substituting into the second DE we have

$$ x_1''+\left(3x_1^2+1\right)x_1'+\left(x_1^2-\frac 12\right)x_1=d\sin t $$

In this equation, $|x_1|$ can't grow without limit, being stable for $x_1^2>\frac 12 $ and consequently $x_2$ is also stable.