I am trying to understand why substitution destroys the information in original problem. I have the following second order nonlinear differential equation,

\begin{equation} \frac{d^2 Y}{d x^2} + Y\sqrt{1 - Y^2} =0 \label{eq:xdef} \end{equation}

for Y(0) = C and Y(0)' = 0

When I plot the above equation using numerical solver, I get a cosine function. I am trying to get rid of squareroot function in nonlinear equation. I am trying to avoid using trigonometric functions or exponential function in substitution. Therefore my attempt in substitution is to use:

\begin{equation} z=\sqrt{1 - Y^2} \label{} \end{equation}

then the derivatives of the above substitution is, \begin{equation} \frac{dY}{dx}=-z\frac{dz}{dx}\frac{1}{Y} \label{} \end{equation}

\begin{equation} \frac{d^2 Y}{d x^2}=(-(\frac{dY}{dx})^2-(\frac{dz}{dx})^2-z\frac{d^2 z}{d x^2})\frac{1}{Y} \label{} \end{equation} then the nonlinear ODE simplifies to

\begin{equation} \frac{d^2 z}{d x^2}+(\frac{dz}{dx})^2\frac{1}{z(1-z^2)}-(1-z^2)=0 \label{} \end{equation}

\begin{equation} z(0)=\sqrt{1 - C^2} \label{} \end{equation} and z(0)' = 0

I think that the above results are correct. However when I try to plot it using numerical solver, instead a cosine curve shifted upward by constant 1 I get either error message (singularity) or some other curve (nonoscillating). Does that means that my substitution is invalid because information is lost? Or is my approach to plot it is incorrect? Is there a better substitute that will eliminate squareroot without destroying the information?

  • 2
    $\begingroup$ Multiplying through your original ODE by $y'$, integrating and manipulating gives us $$y' = \pm \sqrt{\frac{2}{3}} (1-y^{2})^{3/4}$$ which is a separable ODE, but the solution is in terms of the hypergeometric function, so I'm not sure how you are getting $A \cos(\cdot)$ is a solution? $\endgroup$ Commented Aug 15, 2016 at 15:27
  • $\begingroup$ What value did you use for $C$ ? If too close to 1 you are likely to get a singularity problem for $z$... $\endgroup$
    – H. H. Rugh
    Commented Aug 15, 2016 at 15:28
  • $\begingroup$ I think btw that in your expression for $z''$, in the second term the factor $z'$ should be squared? $\endgroup$
    – H. H. Rugh
    Commented Aug 15, 2016 at 15:45
  • $\begingroup$ I am sorry folks. I made a typo error. In my notebook its correct, but when typed on PC I've got ODE equation (after substitution) incorrect. Therefore I corrected it again. $\endgroup$
    – Aschoolar
    Commented Aug 16, 2016 at 1:01

1 Answer 1


You seem to have an error in your substitution formulas. Your first and second derivative formulas are correct. But the final equation in variable $z$ is not right. The right equation after substitution is

$$\frac{d^2z}{dx^2} + \frac{1}{z(1-z^2)}\left(\frac{dz}{dx}\right)^2 + z^2-1 = 0.$$

Is this substitution important or you need to find information on how to solve the equation? Because the original equation is a Hamiltonian system with one degree of freedom and as such is integrable, i.e. it is (more or less) explicitly solvable. One can write the system as a planar system (a system of two equations but involving only first derivatives)

$$\begin{align} \frac{dY}{dx} &= V\\ \frac{dV}{dx} &= -Y\sqrt{1-Y^2} \end{align}$$

Now, one way to go about this is to write the system as $$\frac{dY}{V} = dx = - \frac{dV}{Y\sqrt{1-Y^2}}$$ $$\frac{dY}{V} = - \frac{dV}{Y\sqrt{1-Y^2}}$$ which after cross-multiplying leads to $$Y\sqrt{1-Y^2}dY = -VdV$$ $$VdV + Y\sqrt{1-Y^2}dY = 0.$$ Integrating the last identity on both sides leads to $$H(Y,V) = \frac{V^2}{2} + \int Y\sqrt{1-Y^2} dY = \frac{V^2}{2} - \frac{1}{3}(1-Y^2)^{\frac{3}{2}}=E_0$$ where $E_0$ is a constant (energy level). This means that a solution $(Y(x),V(x))$ to the system leaves the function $H$ invariant, i.e. $H(Y(x),V(x)) = E_0$ for all $x$. $H$ is the total energy of the system, where $\frac{V^2}{2}$ is the kinetic energy, while the therm $U(Y) = -\frac{1}{3}(1-Y^2)^{3/2}$ is the potential energy. To understand the trajectories of the system in the $(Y,V)$ plane, one can draw a graph of the potential $U(Y)$ and based on it one can reconstruct the dynamics of the system. On the other hand, since $V = \frac{dY}{dx}$, then the Hamiltonian gives us the equation $$\left(\frac{dY}{dx}\right)^2 = 2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}$$ which turns into $$\frac{dY}{dx} = \pm \sqrt{2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}}$$ and so you can write it as

$$\frac{dY}{ \sqrt{2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}}} = \pm dx$$ and by integrating both sides, one gets $$\int_{Y_0}^{Y(x)} \frac{d\tilde{Y}}{ \sqrt{2E_0 + \frac{2}{3}(1-\tilde{Y}^2)^{\frac{3}{2}}}} = \pm(x-x_0),$$

so basically you can try to solve the integral $$\int \frac{d{Y}}{ \sqrt{2E_0 + \frac{2}{3}(1-{Y}^2)^{\frac{3}{2}}}},$$ although the solutions are probably restricted to the real line holomorphic functions that uniformize an algebraic curve of genus 2 (a hyper-elliptic Riemann surface of genus 2), which is a bit tough.
Alternativly, you can try to run some simulations of the equation $$\frac{dY}{dx} = \pm \sqrt{2E_0 + \frac{2}{3}(1-Y^2)^{\frac{3}{2}}},$$ having in mind that whenever the solution's derivative $Y'(x)$ becomes $0$, the sign of the equation switches (the sign $\pm$ in front of the square root). The behavior, for certain energy levels is oscillatory, kind of like sine and cosine, but definitely not exactly like them. And do not forget $-1 \leq Y \leq 1$.

Now if one performs the change $z=\sqrt{1-Y^2}$, then this is a type of canonical transformation of the Hamilton system, so the Hamiltonian of the old system, in terms of $(Y,V)$ turns into the Hamiltonian of the new system in terms of $(z,w)$ (which I wrote first), where the 2D transformation is \begin{align} z &= \sqrt{1-Y^2}\\ w &= - \frac{Y}{\sqrt{1-Y^2}}V \left(= \frac{dz}{dx}\right) \end{align}
and the inverse is \begin{align} Y &= \sqrt{1-z^2}\\ V &= - \frac{z}{\sqrt{1-z^2}}w \left(= - \frac{z}{\sqrt{1-z^2}} \frac{dz}{dx}\right) \end{align} Consequently, $$\frac{dY}{dx} = - \frac{z}{\sqrt{1-z^2}} \frac{dz}{dx}$$ and when we substitute it in the first-order equation (which would be equivalent to differentiating the substitution one more time, substituting it in the original equation and getting the equation I wrote first): $$-\frac{z}{\sqrt{1-z^2}}\frac{dz}{dx} = \pm \sqrt{2E_0 + \frac{2}{3} z^3}.$$ If we square on both sides we get $$\left(z\frac{dz}{dx}\right)^2 = \frac{2}{3}(1-z)(1+z)(3E_0 + z^3)$$ where $z=z(x), w = \frac{dz}{dx}(x)$ is a parametrization (after allowing $x \in \mathbb{C}$) of the hyper-elliptic Riemann surface $$z^2 w^2 = \frac{2}{3}(1-z)(1+z)(3E_0 + z^3).$$ The general solution to the equation in terms of $z$ is then

$$ \int_{z_0}^{z(x)} \frac{\tilde{z} d\tilde{z}}{\sqrt{(1-\tilde{z})(1+\tilde{z})(3E_0 + \tilde{z}^3)}} = \pm \sqrt{\frac{2}{3}}(x-x_0).$$

  • $\begingroup$ I am sorry.. I made a typo error. In my notebook its correct, but when typed on PC I've got ODE equation (after substitution) incorrect. I corrected it. I believe z should be a numerator (where derivative of z squared is) $\endgroup$
    – Aschoolar
    Commented Aug 16, 2016 at 1:02
  • $\begingroup$ The reason I want to get rid of squareroot is because I want to solve the ODE using series approximation method( let Y = sum of An times x^n). Since I have squareroot there, expressing it in terms of series will lead to infinite number of terms (a1,a2...etc) which will make series approximation method useless. Also z(0) = C and C is less than 1 and greater than zero. $\endgroup$
    – Aschoolar
    Commented Aug 16, 2016 at 1:07
  • $\begingroup$ I don't think the $z$ is in the numerator. It should be in the denominator the way I wrote it. $\endgroup$ Commented Aug 16, 2016 at 3:07
  • $\begingroup$ To Futurologist, you are correct about z in the denominator. I am horrified at my sloppiness. Just horrified. I understand your last five equation. I am still not clear about equation where you let Y and V equal to. I particularly do not get V=-z/sqrt(1-z^2)w. Where it comes from? Should it not be V=1/(z*sqrt(1-z^2)w). I will review my equation and see if I can plot my equation and yours numerically. Note that I am not attempting to solve ODE. I just want to get rid of squareroot in original ode so that I can employ series expansion method. Thanks Futurologist. $\endgroup$
    – Aschoolar
    Commented Aug 17, 2016 at 18:24
  • $\begingroup$ Isn't the series expansion an attempt to solve the ODE? Anyway, I introduced an extra variable $V=V(x)$ defined as $V = \frac{dY}{dx}$ in order to write down your original second order nonlinear differential equation as a system of two nonlinear differential equations of two functions $(Y(x), V(x))$ of first order. This is very standard, turns the equation into a Hamiltonian form and gives you a way too construct a conserved quantity -- the total energy, aka the Hamiltonian function. This allows you to integrate the equation and bring it into a functional equation of integral form. $\endgroup$ Commented Aug 17, 2016 at 19:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .