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I need to find the cases where the solution to the unidimensional ODE: $$\frac{dx}{dt}=x-x^3+1$$ is an increasing function.

My attempt was solving it like a separable ODE and then looking at how it behaves. But I couldn't do the integral, so I looked for it and it gives weird functions. Then I tried to see it like a Bernoulli equation (ignoring the $+1$) which gave me a cleaner result and an idea of the behavior of the solution but I don't know if my approach is correct. Could you please tell me if my approach is good or wether I can do something else? Thanks for the help.

Jennifer Avila
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    Note that if $x$ is unbounded, then $\dot{x}$ will eventually be negative. I suspect the DE has been chosen to be hard to solve, and that the problem will not require solving it. – preferred_anon Oct 19 '22 at 20:27
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    Assuming $x$ is defined on all of $\mathbb{R}$, if it is bounded and increasing then $\dot{x} \to 0$ as $t \to \pm \infty$, so the limiting values of $x$ must be roots of $x-x^3+1=0$. – preferred_anon Oct 19 '22 at 20:32
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    Note that $x-x^3+1 = 0$ only has on real solution at $x = x_c \approx 1.3247$. When $x < x_c$, $dx/dt > 0$ and when $x > x_c$, $dx/dt < 0$. Consequently, whatever initial condition you choose for $x$, it will converge to $x = x_c$. – Philip Winchester Oct 19 '22 at 20:32
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    Descarte's principle shows that there are one positive solution to $-x^3+x+1=0$ and two negative solutions to $-x^3+x+1=0$ (replacing $x$ by $-x$ yields $x^3-x+1$ ) depending on which subinterval your initial condition is you may determine the sign of the derivative. – Mittens Oct 19 '22 at 20:39
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    @OliverDíaz Two or zero negative solutions. In this case, it is the latter. – eyeballfrog Oct 19 '22 at 20:47
  • @eyeballfrog: seems that I should go back to high school...:) – Mittens Oct 19 '22 at 20:49
  • Thanks a lot for the comments, they helped me a lot. It never came to my mind analyzing the polynomial. It's something I had never seen before and I find it an amazing technique to approach ODEs. – Jennifer Avila Oct 19 '22 at 20:52

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Let $p(x)=x-x^3+1$. As mentioned in the comments on the question, this real polynomial has a single real root that we'll call $x_c$. $p$ is negative for $x \lt x_c$ and positive for $x \gt x_x$.

Also notice that $p$ being a polynomial is locally Lipschitz for any $x \in \mathbb R$. According to Picard-Lindelöf theorem, the Initial Value Problem - IVP $x^\prime(t) = p(x(t))$, $x(t_0)=x_0$ has a unique maximal solution.

Now coming back to the question, which is essentially: for which $x_0$ is the maximal solution of the IVP increasing, notice that we must have $x_0 \lt x_c$ as otherwise $x^\prime(t_o) \lt 0$.

We have however to answer the question: is $x_0 \lt x_c$ a sufficient condition? The answer is positive. Let's see why. Suppose that the solution of the IVP with the initial condition $x(t_0) \lt x_c$ reaches the value $x_c$ in a finite time $t_c$. We would have at $t_c$, $x^\prime(t_c) = 0$ and the constant map equal to $x_c$ would be a solution. A contradiction with the uniqueness statement of Picard-Lindelöf theorem.

Conclusion: $x_0 \lt x_c$ is a necessary and sufficient condition for the solution to be increasing. In that case, it would be possible to prove that the solution is defined for $t \gt t_0$ as a solution that is only defined for finite times, blows up to infinite.

mathcounterexamples.net
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