In my research (in differential geometry) I recently came across the following nonlinear second order ode: $$\frac{f''(x)}{f'(x)}-\frac{2}{x}+\frac{f'(x)+1}{2f(x)-x-1}+\frac{f'(x)-1}{2f(x)+x}=0$$ It actually arose from the symmetry reduction of some pde. I know from an analysis of the equation that there exists a 1-parameter family of solutions. Moreover I also know two explicit solutions; $$f(x)=x+\frac{1}{2}$$ $$f(x)=\frac{1}{4}+\frac{1}{4}(1+3x)\sqrt{(1+2x)}$$ The existence of these 2 solutions, expressible in terms of elementary functions, makes me wonder if one can in fact find more (if not all) explicit solutions to this ode. Note that both these solutions are well defined at $x=0$, although the ode itself is singular at that point! It is not too hard to show that any solution well-defined at $x=0$ requires $f(0)=\frac{1}{2}$ and $f'(0)=1$.

As far as I am aware there are no standard tricks for these type of fully nonlinear odes. I have been trying to simplify the ode by various substitutions but without any success.

I was hoping that someone might be able to spot a clever transformation, or even argue that it is impossible to find any other explicit solutions. I would also be interested to know of any references where such a class of ode that might have been studied.

This also leads me to ask if there is any general theory known about when can a solution to an ode (say of second order) be expressed in terms of elementary functions, or is it just a case-by-case study? Thanks!

  • 1
    $\begingroup$ It appears that $f(x) = -x$ is another solution. Note that this solution is also "globally regular" but does not satisfy your condition that $f(0) = 1/2$ and $f'(0) = 1$. $\endgroup$ Apr 28 '20 at 13:38
  • 2
    $\begingroup$ Another solution is $$ f(x) = \frac14 - \frac14(1+3x) \sqrt{1+2x} $$ Note: my two solutions are obtained from yours by the symmetry transformation $$ f(x) \mapsto 1/2 - f$$ $\endgroup$ Apr 28 '20 at 13:54
  • $\begingroup$ @WillieWong Thanks for pointing out that $\mathbb{Z}_2$ symmetry $\endgroup$
    – u184
    Apr 29 '20 at 8:29
  • 1
    $\begingroup$ If you use the ansatz $f(x)=a_0+a_1x+a_2 x^2 \dots$, put $a_0=1/2$, $a_1=1$ and multiply everything out you get that $a_2$ can be chosen arbitrary whereas then $a_3=-2a_2/5$, $a_4=a_3a_2+3/5$. You can define a formal power series by iteratively solving for the next coefficient. However I am not sure if that power series converges for every $a_2$. The solution with the square root seems to be the case $a_2=5/8$. I am not sure if the other power series can be simplified for other $a_2$'s. $\endgroup$
    – user35593
    May 7 '20 at 14:50
  • $\begingroup$ it should be $a_4=a_3a_2+3/2a_3$ above. $\endgroup$
    – user35593
    May 7 '20 at 15:14

This ODE has some very interesting properties. If one clears fractions and writes it out as $$ x(x+2y)(x-2y+1)\,y'' = (4x^2-8y^2+3x+4y)\,y' + x(4y-1)\,(y')^2, \tag1 $$ one recognizes this as the equation for geodesics of a projective connection in the complement $D$ (which has $7$ components) of the three lines $x = 0$, $x+2y=0$, and $x-2y+1=0$ in the $xy$-plane. Moreover, because the right hand side of (1) has no terms of degree $0$ or $3$ in $y'$, it follows that the lines $x=x_0$ and $y=y_0$ are geodesics of this projective connection in $D$ and should be regarded as 'solutions' of the equation. This is probably most evident if one writes the equation in parametric form for a curve $\bigl(x(t),y(t)\bigr)$, in which case, the equation becomes $$ x(x{+}2y)(x{-}2y{+}1)\,\bigl(\dot y\,\ddot x-\dot x\,\ddot y\bigr) + x(4y{-}1)\,\dot x\,\dot y^2 + (4x^2{-}8y^2{+}3x{+}4y)\,\dot x^2\,\dot y =0 \tag2 $$

Contrary to what the OP claims, there is a $2$-parameter family of solutions that are regular at $x=0$. If one looks for a formal power series solution in the form $$ y(x) = a_0 + a_1\,x + a_2\,x^2 + a_3\, x^3 + \cdots,\tag3 $$
then one finds, by examining the three lowest terms in the equation, that one must have either $$ (i)\ \ a_1 = a_2 = 0,\qquad (ii)\ \ a_0 = 0,\ a_1 = -1, \quad\text{or}\quad (iii)\ \ a_0=\tfrac12,\ a_1 = 1. $$ Since $y(x)$ is a solution if and only if $\tfrac12 - y(x)$ is a solution, the second and third cases are essentially the same, so I will treat only the first two cases henceforth.

In the first case, one finds that there is a formal power series solution of the form $$ y(x) = \tfrac14(1{+}a) + \frac{(a^2{-}1)b}{12} x^3 -\frac{b}{4}\,x^4 - \frac{b}{5}\,x^5 - \frac{a(a^2{-}1)b^2}{72}\,x^6 + \cdots + p_k(a,b)\,x^k + \cdots,\tag4 $$ where $p_k(a,b)$ is a (unique) polynomial in constants $a$ and $b$. Moreover, this series has a positive radius of convergence for each $(a,b)$. [Proofs that these and similar series listed below have positive radii of convergence can be based on techniques in the book Singular Nonlinear Partial Differential Equations by R. Gérard and H. Tahara.] Note that the symmetry $y(x)\mapsto \tfrac12 - y(x)$ corresponds to the symmetry $(a,b)\mapsto (-a,-b)$.

In the second case (and, similarly, via the symmetry $y\mapsto \tfrac12 - y$, the third case), one finds that there is a formal power series solution $$ y(x) = - x + \frac{b}{2}\,x^2 - \frac{b}{5}\,x^3 + \frac{b(b{+}3)}{10}\,x^4 - \frac{4b(13b{+}25)}{175}\,x^5 + \cdots + q_k(b)\,x^k + \cdots, \tag5 $$ where $q_k(b)$ is a (unique) polynomial in $b$ of degree at most $\tfrac12 k$. This series has a positive radius of convergence for every $b$. The value $b=0$ gives the solution $y(x) = -x$ and the value $b=-5/4$ gives the solution $y(x) = \tfrac14 - \tfrac14(1+3x)(1+2x)^{1/2}$. Note that (4) with $a=-1$ and (5) give two distinct $1$-parameter families of solutions passing through the point $(x,y)=(0,0)$, where the two singular lines $x=0$ and $x+2y=0$ meet.

As for analytic solutions meeting the singular line $x+2y=0$, there exist two distinct $2$-parameter families of series solutions: The first is given in parametric form by a formal power series $$ \begin{aligned} x(t) &= a + a(2a{+}1)\,t\,,\\ y(t) &= -\frac{a}{2} + a(2a{+}1)b\,t^2\left(1 + \frac{2(5a{-}4b{+}2)}{3}\,t +\cdots + p_k(a,b)\,t^k + \cdots \right), \end{aligned} \tag6 $$ where $p_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. The second $2$-parameter family can be written in the form $$ \begin{aligned} x(t) &= a + a(2a{+}1)\,t^3\,,\\ y(t) &= -\frac{a}{2} + (2a{+}1)\,t^2\left(b + a\,t + \frac{2b^2}{5}\,t^2 +\cdots + q_k(a,b)\,t^k + \cdots \right), \end{aligned} \tag7 $$ where $q_k(a,b)$ is a (unique) polynomial in $a$ and $b$ and where the $y$-series in $t$ has a positive radius of convergence for every $(a,b)$. Interestingly, these solutions with $b\not=0$ have a cusp singularity at $t=0$, while, when $b=0$, only the terms involving $t^{3k}$ remain, so that $x(t)$ and $y(t)$ are analytic functions of $t^3$.

Note, however, that these two series solutions degenerate at the special values $a=0$ and $a=-\tfrac12$. The value $a = 0$ corresponds to the point $(x,y)=(0,0)$, where the singular lines $x=0$ and $x+2y=0$ cross, while the value $a=-\tfrac12$ corresponds to the point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross.

Finally, through the singular point $(x,y) = (-\tfrac12,\tfrac14)$, where the singular lines $x+2y=0$ and $x-2y+1=0$ cross, there are two convergent series solutions with one parameter: The first is $$ \begin{aligned} x(t) &= -\frac{1}{2} + t\,,\\ y(t) &= +\frac{1}{4} + b\,t^3 -3b\,t^4 + \cdots + f_k(b)\,t^k + \cdots , \end{aligned} \tag8 $$ where $f_k(b) = -f_k(-b)$ is a polynomial in $b$. The second series is $$ \begin{aligned} x(t) &= -\frac{1}{2} + b\,t^2 + \frac{b^2(5b{+}32)}{16}\,t^4 + \cdots + g_k(b)\,t^{2k} + \cdots\,,\\ y(t) &= +\frac{1}{4} + t\, , \end{aligned} \tag9 $$ where $g_k(b)$ is a polynomial in $b$. Note that the value $b=2$ in this latter series corresponds to the known solution(s) represented by solving $(y-\tfrac14)^2-(1+3x)^2(1+2x) = 0$ for $y$ as a function of $x$.

It is very interesting that through every point of the $xy$-plane, there passes at least one $1$-parameter family of solution curves, and, along two of the singular lines, there can be two distinct $1$-parameter families of solution curves.

One more interesting feature should be noted: Because the equation defines a projective structure on $D$, each solution curve in $D$ comes equipped with a canonical projective structure, i.e., a 'developing map' to $\mathbb{RP}^1$ that is unique up to linear fractional transformation and provides a local parametrization of the curve. This developing map extends analytically across the points where such a curve crosses one of the three singular lines, but the developing map is no longer a local diffeomorphism at such places; its differential vanishes to second or third order at such places.


Not a solution just the octave code related to the comment above

function nonlinear_ode

%number of new coefficients (starting at a_3)

%initial guess

%values of a_2 that are considered

for i=1:length(as)
  x(i,:)=fsolve(@(x) fun(x,as(i)),x0);

  for i=1:n


function y = fun(x,a)

f=[1/2 1 a x];


d{2}=[0 1];
d{3}(1:2)=d{3}(1:2)-[1 1];

for k=1:4

for j=1:4
  if j~=k

if k==1


%restrict to first terms



ans =

-1.1592e-18 -1.0962e-17 6.4646e-17 -4.0000e-01 -3.0447e-16

ans =

6.2764e-17 2.5432e-17 -4.0000e-01 6.0000e-01 3.3331e-15

ans =

-1.2474e-09 1.8926e-09 1.1886e+00 -1.1429e+00 -4.3242e-08

ans =

-2.8979e-09 5.0286e-01 -3.3600e+00 2.4286e+00 -1.2213e-07


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