I encountered the following ODE in order to find a solution for Einstein equation $$y''+f(x)y=g(x)y^3.$$ It seems to me that it is not among the solvable nonlinear second order differential equations. Is there any known method to solve that? Any help will be appreciated.
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$\begingroup$ It is difficult to say something if you do not say anything on the structure of coefficients $f(x)$ and $g(x)$: could you say more on them? $\endgroup$– Daniele TampieriCommented Jul 24, 2019 at 6:03
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5$\begingroup$ With $f$ and $g$ constant, this equation has exact solutions in terms of Jacobi elliptic functions. $\endgroup$– JonCommented Jul 24, 2019 at 6:14
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$\begingroup$ The functions $f$ and $g$ are sufficiently smooth and not constant. $\endgroup$– MasoudCommented Jul 24, 2019 at 6:31
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4$\begingroup$ You can take $f,g\in\mathbb{R}$, excluding the trivial case $f=g=0$, and your solution, when $g\ne 0$ is expressed using the Jacobi function ${\rm sn}(x,k)$. Indeed, this equation is a definition of that function by itself. When $f$ and $g$ are not constant, you are in trouble to get a closed form solution unless for special cases or numerically. Already the linear case with $g(x)=0$ can be generally hard. Finally, a perturbative WKB approach could be helpful. $\endgroup$– JonCommented Jul 24, 2019 at 8:04
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4 Answers
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When $g(x) = 2$ (which can be achieved by a transformation in $x$) and $f(x)$ is linear in $x$, then this equation is a special case of the second Painlevé equation, which also appears in other areas of mathematical physics. It is known to be integrable by the isomonodromy method via its Lax representation and it has many other interesting properties.
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The paper by Norbert Euler (1997) ``Transformation Properties of $\ddot{x}+f_1(t)\dot{x} + f_2(t)x + f_3(t) x^n = 0$" Journal of Nonlinear Mathematical Physics, 4:3-4, 310-337
describes in detail the "the most general conditions on the functions $f_1$, $f_2$ and $f_3$, by which the equation may be integrable, as well as conditions for the existence of Lie point symmetries."
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While closed-form solutions will be very rare, you can get series solutions. For example, the solution with $y(0)=0,\; y'(0)=v$ has Maclaurin series
$$ y = v x - \frac{f(0)}{6} v x^3 - \frac{f'(0)}{12} v x^4 + \left(\frac{f(0)^2 - 3 f''(0)}{120} v + \frac{g(0)}{20} v^3 \right) x^5 + \ldots $$
answered Jul 24, 2019 at 12:27
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1$\begingroup$ This is produced by the command of Maple dsolve({diff(y(x), x, x) + f(x)*y(x) = g(x)*y(x)^3, y(0) = 0, D(y)(0) = v}, y(x), series). $\endgroup$ Commented Jul 24, 2019 at 13:11
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Consider the ODE \begin{align} y'+\frac{A'(x)}{3A(x)}y&=A(x)y^2, \end{align} with derivative \begin{align} y''+\left[\left(\frac{A'}{3A}\right)'-\left(\frac{A'}{3A}\right)^2\right]y=2A^2y^3. \end{align} This is equivalent to your equation if \begin{align} \left(\frac{A'}{3A}\right)'-\left(\frac{A'}{3A}\right)^2=f,\quad\text{and}\quad2A^2=g. \end{align} In other words, if $f(x)$ and $g(x)$ satisfy the equation \begin{align} \left(\frac{g'}{6g}\right)'-\left(\frac{g'}{6g}\right)^2=f, \end{align} your problem is reduced to an easily integrable Bernoulli equation.
EDIT
Taking $y=u+\mu(x)$ we have that \begin{align} u''=gu^3+3g\mu u^2+(3g\mu^2-f)u+g\mu^3-f\mu-\mu''. \end{align} Now consider the Riccati equation \begin{align} u'=r_0(x)+r_1(x)u+r_2(x)u^2, \end{align} with derivative \begin{align} u''=2r_2^2u^3+(3r_1r_2+r_2')u^2+(2r_0r_2+r_1^2+r_1')u+r_0r_1+r_0'. \end{align} Our transformed equation is equivalent to the derivative of the Riccati equation if the four following equations hold \begin{align} 2r_2^2=g,&\qquad 3r_1r_2+r_2'=3g\mu,\\ 2r_0r_2+r_1^2+r_1'=3g\mu^2-f,&\qquad r_0r_1+r_0'=g\mu^3-f\mu-\mu''. \end{align} These give that \begin{align} r_2&=\sqrt{\frac{g}{2}},\qquad r_1=\sqrt{2g}\mu-\frac{g'}{6g},\\ r_0&=\sqrt{2g}\mu^2-\frac{\sqrt2}{3\sqrt g}\left(f+\left(\sqrt{2g}\mu-\frac{g'}{6g}\right)'+\left(\sqrt{2g}\mu-\frac{g'}{6g}\right)^2\right), \end{align} and finally \begin{align}\tag{*} \mu''-g\mu^3+f\mu=r_0(g,g',\mu,\mu')'+\left(\sqrt{2g}\mu-\frac{g'}{6g}\right)r_0(g,g',\mu,\mu'). \end{align} So given a particular solution $\mu$ to equation (*), your equation $y''+fy=gy^3$ is reducible to a Riccati equation, (or a linear second order equation if you'd like).
I have not tried to write equation (*) out to see if any nice simplifications occur. Though this result is within expectations, as a particular case of $f$ and $g$ result in a Bernoulli equation, a special case of the Riccati equation.
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$\begingroup$ I take it your situation requires $y\in\mathbb R$? Is it possible for $g(x)<0$ in your situation? $\endgroup$ Commented Feb 7, 2023 at 8:03
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$\begingroup$ In my problem $g<0$, but there is no restriction on $y$. $\endgroup$– MasoudCommented Feb 7, 2023 at 22:17
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$\begingroup$ For $f$ and $g$ satisfying this relation with $g<0$ I find $y$ must be complex. $\endgroup$ Commented Feb 9, 2023 at 9:25