I am asked whether any solution for $y''+y'+y+y^3 =0$ exists for all $t \in \mathbb{R}$.
Here is what I am thinking:
First, introduce a change of variables $z_1 = y$, $z_2 = y'$ so that the equation given becomes the 1st order system:
\begin{equation} \pmatrix{z'_1\\z_2'}=\pmatrix{0&1\\-1&-1}\pmatrix{z_1\\ z_2} +\pmatrix{0\\-z_1^3} \end{equation}
After computing the Jacobian, we get that:
$\frac{\partial F}{\partial y} = \pmatrix{0&1\\-1-3z_1^2&-1}$
Now, we need to check whether $F(z_1,z_2)=\pmatrix{0&1\\-1&-1}\pmatrix{z_1\\ z_2} +\pmatrix{0\\-z_1^3}$ is Lipschitz.
Clearly, $F$ is continuous (component wise). Note then if we compute the operator norm of the Jacobian, we get (using the sup norm):
$\left\lVert\frac{\partial F}{\partial y}\right\rVert_\infty = \max${$|1|$, $|1+3z_1^2|+|1|$}
Since $3x^2 +1 \neq 0$ for $x \in \mathbb{R}$, then the above expression is just \begin{equation} 1 +|3z_1^2 +1| \leq 2 +3|z_1|^2 \end{equation}
Because of the $z_1$ term above, then the Jacobian is not bounded. Thus, $F$ cannot be (globally) Lipschitz.
If we did assume that $z_1 = y$ was bounded beforehand, then it might work. However, we're not given any further assumptions.
I am aware that there are alternate methods to solving this problem, maybe finding a potential function $V(y)$ and checking if $V'(y) \leq 0$. However, I don't want to go that route, as we can only use material we've covered so far. We just got to linear systems last lecture.
Does this work? I feel there's something I am overlooking.
Multiply the equation by $y'$ and integrate between $0$ and $x$ to get $$ \frac12\,(y'(x))^2+\int_0^x(y'(t))^2\,dt+\frac12\,y(x)^2+\frac14\,y(x)^4=C. $$ It follows easily that $y$ and $y'$ are bounded on any interval $[0,a]$ with $a>0$, so that the solution is defined on $[0,\infty)$. If $x<0$ we have $$ \frac12\,(y'(x))^2+\frac12\,y(x)^2+\frac14\,y(x)^4=C+\int_x^0(y'(t))^2\,dt, $$ and things are not clear. However, we have $$ \frac12\,(y'(x))^2\le C+\int_x^0(y'(t))^2\,dt, $$ and we apply Gronwall's lemma to $(y'(x))^2$ and show that it is bounded on bounded intervals $[-a,0]$ with $a>0$. The same is true then of $y$, showing that the solution is defined on $(-\infty,0)$. Observe that in general the solution will be unbounded.