This question was previously posted on MSE at Change of variable for differential equations.

Given the following differential equation \begin{equation} -y(\zeta) \left(\frac{d^2 y(\zeta)^{-1}}{d \zeta^2}+2 j c \frac{d y(\zeta)^{-1}}{d \zeta}\right)+c^2 (1+y(\zeta)^{2})=\\ \frac{y''(\zeta)}{y(\zeta)}-\frac{2 y'(\zeta)^2}{y(\zeta)^2}+\frac{2 c j y'(\zeta)}{y(\zeta)}+c^2 y(\zeta)^2+c^2=\\ \frac{d^{2}}{d \zeta^{2}} \log{y(\zeta)}-\left(\frac{d}{d \zeta} \log{y(\zeta)}\right)^{2}+2jc \frac{d}{d \zeta} \log{y(\zeta)}+c^2 y(\zeta)^2+c^2 =f(\zeta), \end{equation} where $c$ is a constant, while $j=\sqrt{-1}$, is there a change of variables that allows it to be transformed into the following form? \begin{equation} -\frac{1}{2 a(\zeta)}\left(\frac{d^2 a(\zeta)}{d \zeta^2}-\frac{1}{2 a(\zeta)}\left(\frac{d a(\zeta)}{d \zeta}\right)^2 \right)+\frac{c}{a(\zeta)^2}=\\ -\frac{d^{2}}{d \zeta^{2}} \log{\sqrt{a(\zeta)}}-\left(\frac{d}{d \zeta} \log{\sqrt{a(\zeta)}}\right)^{2}+\frac{c_{1}}{a(\zeta )^2}=\\ \frac{d^{2}}{d \zeta^{2}} \log{\frac{1}{\sqrt{a(\zeta)}}}-\left(\frac{d}{d \zeta} \log{\frac{1}{\sqrt{a(\zeta)}}}\right)^{2}+\frac{c_{1}}{a(\zeta)^2}= f(\zeta). \end{equation} If this change of variables exists, how is it possible to find it?

A first attempt is to use a generic change of variables to identify the function $F$ such that $a(\zeta)=F(y(\zeta))$. Proceeding in this way it is possible to transform the second equation in the following way \begin{equation} -y''(\zeta)\frac{F'(y(\zeta))}{2 F(y(\zeta))}+y'(\zeta)^2 \left(\frac{F'(y(\zeta))^2}{4 F(y(\zeta))^2}-\frac{F''(y(\zeta))}{2 F(y(\zeta))}\right)+\frac{c}{F(y(\zeta))^2}=f(\zeta). \end{equation} The initial problem then reduces to identifying the function $F$ such that the first and the third equations are equal. However, as of this point, I have no idea how to proceed. Also, I am not sure that changing variables $a(\zeta)=F(y(\zeta))$ is enough to solve the problem. The math.stackexchange user Sal pointed out that the equation involving $F$ has no $y'$ term, while the first equation does. Since $F$ is presumed to be only a function of $y$ they can't be equal. For instance we could have $F=F(y,y')$, but also $F=F(\zeta,y)$, $F=F(\zeta,y,y')$. For instance for $F=F(\zeta,y)$, I have \begin{equation} -\frac{y''(\zeta) F^{(0,1)}(\zeta,y(\zeta))}{2 F(\zeta,y(\zeta))}+y'(\zeta)^2 \left(\frac{F^{(0,1)}(\zeta,y(\zeta))^2}{4 F(\zeta,y(\zeta))^2}-\frac{F^{(0,2)}(\zeta,y(\zeta))}{2 F(\zeta,y(\zeta))}\right)+y'(\zeta) \left(\frac{F^{(0,1)}(\zeta,y(\zeta)) F^{(1,0)}(\zeta,y(\zeta))}{2 F(\zeta,y(\zeta))^2}-\frac{F^{(1,1)}(\zeta,y(\zeta))}{F(\zeta,y(\zeta))}\right)+\frac{F^{(1,0)}(\zeta,y(\zeta))^2}{4 F(\zeta,y(\zeta))^2}-\frac{F^{(2,0)}(\zeta ,y(\zeta))}{2 F(\zeta,y(\zeta))}+\frac{c}{F(\zeta,y(\zeta))^2}=f(\zeta). \end{equation} Assuming that $F=F(\zeta,y)$ is the right change of variable, what is the next step?

I'm really interested in solving this problem, so if anything is unclear, please don't hesitate to let me know so that I can improve the post. Any suggestion is welcome.

If the general solution of this problem is difficult to determine, if it exists at all, it is probably possible to determine a particular solution. For example if $y=\frac{1}{\sqrt{a(\zeta)}}$ the first and second equations are satisfied if \begin{equation} 2jc \frac{d}{d \zeta} \log{y(\zeta)}+c^2 y(\zeta)^2+c^2 =c_{1} y(\zeta)^4 \end{equation} Is there a way to find an analytical solution of this equation?

  • 3
    $\begingroup$ What led you to believe such a change of variable exists? Is this claimed in a paper? $\endgroup$ Mar 10, 2022 at 15:42
  • $\begingroup$ I doubt that a solution exists. I have rearranged the equations so that they can be compared. $\endgroup$
    – Upax
    Mar 13, 2022 at 14:56

1 Answer 1


The equivalence of (in general) non-linear 2nd order ordinary differential equations $y'' = Q(x,y,y')$ under various types of transformations (including the ones you are considering) is a classic problem. Whether two equations can be transformed into each other can have different answers depending on the allowed transformations. The most general local transformations are contact transformations $x=X(x,y,y')$, $y=Y(x,y,y')$, $y'=P(x,y,y')$, with some conditions on the functions $X$, $Y$, $P$ to make the transformation make sense. There are also point transformations $x=X(x,y)$, $y=Y(x,y)$, and fiber preserving transformations, $x=X(x)$, $y=Y(x,y)$. This book is a classic reference on the so called Cartan approach to equivalence problems, where the case of 2nd order ODEs is treated as an example:

Olver, Peter J., Equivalence, invariants, and symmetry, Cambridge: Cambridge University Press (ISBN 978-0-521-10104-2/pbk). xvi, 525 p. (2009). ZBL1156.58002.

The equivalence problem is set up within the framework of Cartan's equivalence method in [Olver, Ex.9.3,9.6]. The basic classic result is that every 2nd order ODE of the form $y'' = Q(x,y,y')$ is equivalent by a contact transformation to $y''=0$ [Olver, Thm.11.11]. Unfortunately, the transformation that would do the job is highly implicit in the proof and you shouldn't expect to be able to easily find its explicit form. For point and fiber preserving transformations, there are distinct equivalence classes, which can be distinguished by differential invariants of the function $Q(x,y,y')$. Details can be found in the last section of [Olver, Ch.12]. If you're looking for very explicit formulas for the invariants that can help you distinguish your two equations, then you might want to follow some of the references that Olver gives in that section.

  • $\begingroup$ The question seems to be more about solving the equation, and positing a possible equivalence that might allow solution, rather than about testing whether the possible equivalence is realised. (Nonetheless, a reference to Olver is always welcome.) $\endgroup$
    – LSpice
    Apr 16, 2022 at 0:57
  • $\begingroup$ @LSpice Sure, but I doubt very much that there is any explicit solution (that goes for the solutions of the original equation, as well as for a trivializing contact transformation). Moreover, if the point transformation invariants of the OP's two equation can be shown to be different, it will also give a reason to stop looking for a point equivalence between them. $\endgroup$ Apr 16, 2022 at 10:29
  • $\begingroup$ @Igor. As pointed out by LSpice the question is about solving the equation. However, I really appreciated your suggestion to have a look to the Olver's book. Any help is welcome. $\endgroup$
    – Upax
    Apr 17, 2022 at 9:24

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