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Consider the following ODEs: $\phi^2=\phi''\sqrt{1-\phi'^2}$, or $\phi^2=-\phi''\sqrt{1-\phi'^2}$.

Is there any theory (e.g. comparison theorems) which analyzes solutions of the above ODEs? I am only interested in nonnegative solutions, i.e. $\phi\geq 0$. Actually one can write down a solution $\phi(t)=\cos t,\ -\frac{\pi}{2}\leq t\leq \frac{\pi}{2}$ to the second equation. I want to know whether one can find explicit solutions to the first equations, as well as other solutions to the second one (of course excluding those coming from time translations of $\cos t$). If one can not find explicit formula for the solution, I am also interested in asymptotic properties/existence time of the solution. In particular, I want to know if there exist solutions defined on the entire real line.

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    $\begingroup$ Strictly speaking, the equation(s) you are analyzing seems more an algebraic differential equation (ADE) than an ODE: $$ \phi^4=\phi^{\prime\prime 2}(1-\phi^{\prime2}). $$ The properties of the solutions of this kind of equations depend heavily on the properties of the algebraic function involved. You'll probably find more useful information by looking on google for ADEs instead of ODEs $\endgroup$ – Daniele Tampieri Apr 28 at 6:57
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    $\begingroup$ One thing that's not clear from your question: are you considering each equation individually? Or are you asking, like @DanieleTampieri suggested, for a $\phi$ that satisfies at least one of the two equations? If $\phi$ is allowed to follow exactly one (but not the other) equation, then by convexity there cannot be any global in time solutions. $\endgroup$ – Willie Wong Apr 28 at 12:01
  • $\begingroup$ @WillieWong Yes, I am considering each equation individually. $\endgroup$ – Yuhang Liu Apr 29 at 1:22
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Edited on May 2, 2020: The OP pointed out that I had not addressed a special case (namely $C=1$ below), so I am amending my answer to address this and reorganizing so that the $C=1$ case gets addressed naturally when it comes up. — RLB

We can assume that $\phi$ is not constant, since the only constant solution is $\phi\equiv0$.

Thus, assume that the solution has $\phi'\not=0$ on some interval $I$. Multiply the equation $\phi^2 = \pm\phi''\sqrt{1-\phi'^2}$ by $\phi'$, then integrate both sides to get that there is a constant $C$ such that $$ \phi^3 = C^3 \mp (1-\phi'^2)^{3/2}. $$ The case $C=0$ corresponds to $\phi^2 = 1-\phi'^2$, which, since $\phi'$ is assumed nonzero on $I$ implies that $\phi = \cos(t-t_0)$ for some constant $t_0$. We can thus set $C=0$ aside and assume that $C\not=0$.

We have $C^3{-}1\le \phi^3\le C^3{+}1$, so $C\ge -1$, otherwise there cannot be any nonnegative solutions. If $C = -1$, then $\phi\le0$ and, since we are only interested in non-negative solutions, the only solution in this case is $\phi\equiv0$, so we can assume henceforth that $C>-1$.

Both equations (with either sign) can be studied as special cases of the polynomial differential equation $$ (\phi^3-C^3)^2 - (1 - \phi'^2)^3 = 0, $$ so that is what we will do. Set $\phi^3-C^3 = u^3$ where $|u|\le 1$ and note that this implies $(\phi')^2 = 1-u^2$. We then have $$ \pm 1 = \frac{u^2u'}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}, $$ so $$ t_1-t_0 = \pm\int_{u(t_0)}^{u(t_1)} \frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}. $$ For $C\not=0,1$ (remember that $C>-1$), the integral $$ \int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}} = \int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C+u)^{2/3}(C^2-Cu+u^2)^{2/3}} $$ converges. However, when $C=1$, the denominator contains a factor of $(1+u)^{1/2+2/3} = (1+u)^{7/6}$, so the integral diverges at $u=-1$.

In the case $C=1$, we can write down a parametrization of the graph $\bigl(t,\phi(t)\bigr)$ in the form $\bigl(t(v),\phi(t(v))\bigr)$ where $|v|<\sqrt2$ with $$ \phi(t(v)) = \bigl(1+(1-v^2)^3\bigr)^{1/3} \quad\text{and}\quad t(v) = \int_0^v \frac{2(1-s^2)^2\,ds}{(2-s^2)^{7/6}(1-s^2+s^4)^{2/3}}. $$ Note that, while $t$ is a strictly increasing function of $v$ and $|t|\to\infty$ as $|v|\to\sqrt2$, $t'(\pm1) = 0$, and $\phi$ is not a smooth function of $t$ where $v = \pm 1$. (It is, however, continuously once differentiable there, see below.)

Meanwhile, when $C\not=1$, a solution $\phi$ exists for all time and is periodic, as $\phi$ oscillates between $(C^3{-}1)^{1/3}$ and $(C^3{+}1)^{1/3}$. The period of $\phi$ is $$ p(C) = 2\int_{(C^3{-}1)^{1/3}}^{(C^3{+}1)^{1/3}}\frac{d\xi}{\sqrt{1-(\xi^3-C^3)^{2/3}}} = \int_{-1}^1\frac{2u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}. $$ Of course, $p(0) = 2\pi$ and $p(C)$ has a series expansion $2\pi\,C^{-2}+\tfrac{175}{576}\pi\,C^{-8}+\cdots$ when $C>1$.

Near a minimum at $t=t_0$, the solution has a series expansion of the form $$ \phi(t) = (C^3{-}1)^{1/3}\left(1+\tfrac12(C^3{-}1)^{1/3}(t-t_0)^2+\tfrac1{24}(C^3+1)(C^3{-}1)^{2/3}(t-t_0)^4 + \cdots\right) $$ Near a maximum at $t=t_1$ the solution has a series expansion of the form $$ \phi(t) = (C^3{+}1)^{1/3}\left(1-\tfrac12(C^3{+}1)^{1/3}(t-t_1)^2-\tfrac1{24}(C^3-1)(C^3{+}1)^{2/3}(t-t_1)^4 + \cdots\right) $$

Note that, when $C\not=0$, $\phi$ is not $C^2$ when it attains the intermediate value $C$. In fact, if $\phi(t_2) = C$ and $\phi'(t_2) = 1$, then $\phi$ has a series expansion in powers of $(t{-}t_2)^{1/3}$: $$ \phi(t) = C +(t{-}t_2) - \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3} - \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots. $$ Meanwhile, if $\phi(t_3) = C$ and $\phi'(t_3) = -1$, then $\phi$ has a series expansion in powers of $(t{-}t_3)^{1/3}$: $$ \phi(t) = C -(t{-}t_2) + \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3} + \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots. $$ Finally, note that $\phi''<0$ when $C<\phi<(C^3+1)^{1/3}$ while $\phi''>0$ when $(C^3-1)^{1/3}<\phi<C$.

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  • $\begingroup$ Hi, Robert: Thanks for your answer! However I'd like to add something: it seems when C=1, the minimum $\phi(t_0)=0$, and the integral expressing $t-t_0$ diverges (the order of integrand is $\xi^{-3/2}$). Does that mean that $t_0=-\infty$ when $C=1$? When $C\neq 1$, the integral converges as the main term becomes $(\xi-(C^3-1)^{1/3})^{-1/2}$. $\endgroup$ – Yuhang Liu May 2 at 10:14
  • $\begingroup$ @YuhangLiu: You are right, of course. I didn't look at the $C=1$ case, and I should have. I'll fix that. $\endgroup$ – Robert Bryant May 2 at 13:10

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