Solving a Coupled Second-Order Differential Equation

I'm trying to solve the real-valued differential equation

$$xy\ddot{y}+2y\dot{x}\dot{y}+\alpha\ddot{x}+x\dot{y}^2=0$$

where $\alpha\in\mathbb{R}$ and $\alpha>0$ for both $x(t)$ and $y(t)$. How would I even approach this problem?

EDIT: Sorry, I forgot a conservation law:

$$\alpha^2\dot{x}^2+x^4\dot{y}^2 = P^2$$

where $P$ is a conserved quantity.

EDIT 2: The original Lagrangian this comes from is

$$\mathcal{L}=\frac{1}{2}\left(-\alpha\dot{x}^2+x^2\dot{y}^2+x^2y^2\dot{\theta}^2+x^2y^2\sin^2\theta\dot{\phi}^2\right)$$

• Unless you can find a conservation law, go for numerics or asymptotics.
– Ian
Sep 20, 2016 at 13:09
• Isn't there another equation? Where did you find this one in the first place? What do you need it for? Did you have something like $$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 1$$ Sep 20, 2016 at 15:54
• See my edit (thanks!) Sep 20, 2016 at 19:03

As I said before, knowing more detail of the problem may lead to its solution. So you are studying the motion of a free point-mass in a mildly curved space-time with Lorenzian metric $$d\sigma^2 = - \alpha^2 \, dx^2 + x^2\Big(dy^2 + y^2 \big(d\theta^2 + \sin(\theta) \, d\phi^2\big)\Big)$$ Let me rename the variables $x=t$ and $y = r$. Then $$d\sigma^2 = - \alpha^2 \, dt^2 + t^2\Big(dr^2 + r^2 \big(d\theta^2 + \sin(\theta) \, d\phi^2\big)\Big)$$ If you consider $(t,r,\theta, \phi)$ as spherical coordinates plus time, the Cartesian coordinates are \begin{align} x &= r \, \cos(\phi) \, \sin(\theta)\\ y &= r \, \sin(\phi) \, \sin(\theta)\\ z &= r \, \cos(\theta)\\ \end{align} so the space part of the metric becomes $$g_E^2 = dr^2 + r^2 \big(d\theta^2 + \sin(\theta) \, d\phi^2\big) = dx^2 + dy^2 + dz^2$$ so your space-time metric is basically $$d\sigma^2 = -\alpha^2 \, dt^2 + t^2 \, g^2_E = -\alpha^2 \, dt^2 + t^2 \, \big( dx^2 + dy^2 + dz^2\big)$$ So your Lagrangian is technically the Lorenzian archlength $$\mathcal{L} = \sqrt{ -\alpha^2 \, \dot{t}^2 + t^2 \, \big( \dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}$$ i.e. $$\mathcal{L}^2 = { -\alpha^2 \, \dot{t}^2 + t^2 \, \big( \dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}$$ where $\dot{t} = \frac{dt}{ds}, \,\, \dot{x} = \frac{dx}{ds}, \, \,\dot{y} = \frac{dy}{ds}, \,\, \dot{x} = \frac{dz}{ds}$ with respect to some parameter $s$ which will be determined later. The Euler-Lagrange equations look like this \begin{align} \frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{t}}\right) & = \frac{\partial \mathcal{L}}{\partial t}\\ \frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) & = \frac{\partial \mathcal{L}}{\partial x}\\ \frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{y}}\right) & = \frac{\partial \mathcal{L}}{\partial y}\\ \frac{d}{ds} \left(\frac{\partial \mathcal{L}}{\partial \dot{z}}\right) & = \frac{\partial \mathcal{L}}{\partial z} \end{align} Now \begin{align} \frac{1}{2} \, \frac{\partial}{\partial \dot{t}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial \dot{t}} = - \alpha^2 \, \dot{t} \\ \frac{1}{2} \,\frac{\partial}{\partial \dot{x}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial \dot{x}} = t^2 \, \dot{x}\\ \frac{1}{2} \,\frac{\partial}{\partial \dot{y}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial \dot{y}} = t^2 \, \dot{y}\\ \frac{1}{2} \, \frac{\partial}{\partial \dot{z}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial \dot{z}}= t^2 \, \dot{z} \end{align} leading to \begin{align} \frac{\partial \mathcal{L}}{\partial \dot{t}} &= - \frac{\alpha^2 \, \dot{t}}{ \mathcal{L} } \\ \frac{\partial \mathcal{L}}{\partial \dot{x}} &= \frac{t^2 \, \dot{x}}{ \mathcal{L} }\\ \frac{\partial \mathcal{L}}{\partial \dot{x}} &= \frac{t^2 \, \dot{y}}{ \mathcal{L} }\\ \frac{\partial \mathcal{L}}{\partial \dot{x}} &= \frac{t^2 \, \dot{z}}{ \mathcal{L} }\\ \end{align} Similarly \begin{align} \frac{1}{2} \, \frac{\partial}{\partial {t}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial {t}} = t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)\\ \frac{1}{2} \,\frac{\partial}{\partial {x}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial {x}} = 0\\ \frac{1}{2} \,\frac{\partial}{\partial {y}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial {y}} =0\\ \frac{1}{2} \, \frac{\partial}{\partial {z}} \big( \mathcal{L}^2\big) & = \mathcal{L} \, \frac{\partial \mathcal{L}}{\partial {z}}= 0 \end{align} Finally, the raw Euler-Lagrange equations look as follows

\begin{align} - \alpha^2 \, \frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &= \frac{t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}{\mathcal{L}}\\ \frac{d}{ds}\left( \frac{t^2 \, \dot{x}}{ \mathcal{L}} \right) &= 0 \\ \frac{d}{ds}\left( \frac{t^2 \, \dot{y}}{ \mathcal{L}} \right) &= 0 \\ \frac{d}{ds}\left( \frac{t^2 \, \dot{z}}{ \mathcal{L}} \right) &= 0 \end{align} The last three equations can be immediately integrated once \begin{align} \frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &= - \frac{t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}{\alpha^2 \, \mathcal{L}}\\ \frac{t^2 \, \dot{x}}{ \mathcal{L}} &= u_0 \\ \frac{t^2 \, \dot{y}}{ \mathcal{L}} &= v_0 \\ \frac{t^2 \, \dot{z}}{ \mathcal{L}} &= w_0 \end{align} leading to \begin{align} \frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &= - \frac{t \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big)}{\alpha^2 \, \mathcal{L}}\\ \dot{x} &= \frac{u_0}{t^2} \,\mathcal{L}\\ \dot{y} &= \frac{v_0}{t^2} \,\mathcal{L} \\ \dot{z} &= \frac{w_0}{t^2} \,\mathcal{L} \end{align} Square the last three equations and plug them in the first one \begin{align} \frac{d}{ds}\left(\frac{\dot{t}}{ \mathcal{L} } \right) &= - \frac{t}{\alpha^2 \, \mathcal{L}} \, \big(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\big) = - \frac{t}{\alpha^2 \, \mathcal{L}} \, \frac{\big({u_0}^2 + {v_0}^2 + {w_0}^2\big)}{t^4} \, \mathcal{L}^2\\ &= - \, \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2 \, t^3} \, \mathcal{L} \end{align} or rewrite it as $$\frac{1}{\mathcal{L}} \, \frac{d}{ds} \left(\frac{1}{ \mathcal{L} } \, \frac{dt}{ds}\right) = - \, \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2 \, t^3}$$ The proper time is $d\tau = \mathcal{L}ds$ so the equations turn into $$\frac{d}{d\tau} \left( \frac{dt}{d\tau }\right) = - \, \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2 \, t^3}$$ $$\frac{d^2 t}{d\tau^2} = - \, \frac{k_0}{t^3}$$ where $k_0 = \frac{{u_0}^2 + {v_0}^2 + {w_0}^2}{\alpha^2}$ is a constant. Finally, in terms of proper time, the system of Euler-Lagrange equations describing the geodesics become \begin{align} \frac{d^2 t}{d\tau^2} &= - \, \frac{k_0}{t^3}\\ \frac{dx}{d\tau} &= \frac{u_0}{t^2}\\ \frac{dy}{d\tau} &= \frac{v_0}{t^2} \\ \frac{dz}{d\tau}&= \frac{w_0}{t^2} \end{align} So you solve the first equation, which is completely decoupled from the rest, and obtain the function $t = t(\tau)$. Then \begin{align} x(\tau) &= x_0 + \int \frac{u_0}{t(\tau)^2} \, d\tau\\ y(\tau) &= y_0 + \int \frac{v_0}{t(\tau)^2} \, d\tau \\ z(\tau)&= z_0 + \int \frac{w_0}{t(\tau)^2} \, d\tau \end{align}

Is this a homework problem or is it some sort of practice question? If not, any possible symmetries coming from the formulation of the problem? If not, I don't know how far one can go, but let's see. It may help if you tell us how you derived the equation, weather it is an Euler-Lagrange equation of some sort. Maybe it's coming from a problem in differential geometry related to curves on surfaces?

Invert the function $x=x(t)$ and write it as a function $t=t(x)$. Then $$\frac{d}{dt} = \frac{dx}{dt} \, \frac{d}{dx}$$ Form the metric $$\alpha^2 \left(\frac{dx}{dt}\right)^2 + x^4 \, \left(\frac{dy}{dt}\right)^2 = P^2$$ one gets $$P^2 = \alpha^2 \left(\frac{dx}{dt}\right)^2 + x^4 \, \left(\frac{dx}{dt}\right)^2\left(\frac{dy}{dx}\right)^2 = \left(\frac{dx}{dt}\right)^2 \, \left(\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2\right)$$ and so $$\left(\frac{dx}{dt}\right)^2 = \frac{P^2}{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}$$ $$\frac{dx}{dt}= \frac{P}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}$$ Thus $$\frac{d}{dt} = \left(\frac{P}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} \right)\,\, \frac{d}{dx}$$ Now the equation $$x y \, \frac{d^2 y}{dt^2} + 2 \, y \, \frac{dx}{dt}\frac{dy}{dt} + \alpha \frac{d^2x}{dt^2} + x \left(\frac{dy}{dt}\right)^2 = 0$$ can be written as

$$\frac{d}{dt}\left(x y \, \frac{dy}{dt} + \alpha \frac{dx}{dt}\right) + y \, \frac{dx}{dt}\frac{dy}{dt} = 0$$ Writing the latter equation in terms of $x$ as an independent variable $$\left(\frac{P^2}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} \right)\,\, \frac{d}{dx} \left(\frac{x y \frac{dy}{dx} + \alpha}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}\right) + \frac{P^2 \, y \, \frac{dy}{dx}}{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2} = 0$$ $$\frac{d}{dx} \left(\frac{x y \frac{dy}{dx} + \alpha}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}}\right) + \frac{y \, \frac{dy}{dx}}{\sqrt{\alpha^2 + x^4 \, \left(\frac{dy}{dx}\right)^2}} = 0$$ In addition to that, one can also write $z(x) = \frac{1}{2} y(x)^2$ and so the equation becomes $$\frac{d}{dx} \left(\frac{x \frac{dz}{dx} + \alpha}{\sqrt{\alpha^2 + \frac{x^4}{2 \, z} \, \left(\frac{dz}{dx}\right)^2}}\right) + \frac{ \frac{dz}{dx}}{\sqrt{\alpha^2 + \frac{x^4}{2 \, z} \, \left(\frac{dz}{dx}\right)^2}} = 0$$ Alternatively, you can multiply the initial equation by $x$ and get $$x^2 y \ddot{y} + 2xy \, \dot{x} \dot{y} + x^2 \dot{y}^2 + \alpha \, x \ddot{x} = 0$$ which becomes $$0 = \frac{d}{dt} \big(x^2 y \dot{y}\big) + \alpha \, x \ddot{x} = \frac{d}{dt} \big(x^2 y \dot{y}\big) + \alpha \,\big( x \ddot{x} + \dot{x}^2\big) - \alpha \, \dot{x}^2 = \frac{d}{dt} \big(x^2 y \dot{y} + \alpha \, x \dot{x} \big) - \alpha \, \dot{x}^2$$ Now had your second equation been $$\alpha^2 \, \dot{x}^2 + x^4 y^2 \dot{y}^2 = P^2$$ then a miracle could have happened and since $x^2y\dot{y} = \sqrt{P^2 - \alpha^2 \, \dot{x}^2}$ the equation would have become $$\frac{d}{dt} \Big(\sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\Big) - \alpha \, \dot{x}^2 = 0$$ which would have been a whole new story. This equation would have had much more potential.

Currently, you can end up with $$\frac{d}{dt} \Big( y \sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\Big) - \alpha \, \dot{x}^2 = 0$$ Maybe now one can write $y=y(t)$ as a function $t=t(y)$ and $x=x(y)$. Then $$\frac{d}{dt} = \dot{y} \frac{d}{dy}$$ so $$\dot{y} \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, \dot{x}^2} + \alpha \, x \dot{x}\right) - \alpha \, \dot{x}^2 = 0$$ $$\dot{y} \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, \dot{y}^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x \dot{y} \, \frac{dx}{dy}\right) - \alpha \, \dot{y}^2 \left(\frac{dx}{dy}\right)^2 = 0$$ Express $$\dot{y} = \frac{P}{\sqrt{\alpha^2\left(\frac{dx}{dy}\right)^2 + x^4}} = f\left(x,\frac{dx}{dy}\right)$$ so we get $$f \frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x f \, \frac{dx}{dy}\right) - \alpha \, f^2 \left(\frac{dx}{dy}\right)^2 = 0$$ $$\frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x f \, \frac{dx}{dy}\right) - \alpha \, f \, \left(\frac{dx}{dy}\right)^2 = 0$$

$$\frac{d}{dy}\left( y \sqrt{P^2 - \alpha^2 \, f\left(x,\frac{dx}{dy}\right)^2 \left(\frac{dx}{dy}\right)^2} + \alpha \, x f\left(x,\frac{dx}{dy}\right) \, \frac{dx}{dy}\right) - \alpha \, f\left(x,\frac{dx}{dy}\right) \, \left(\frac{dx}{dy}\right)^2 = 0$$ This is again a $y$ inhomogeneous second order ODE of type $F\big(y, x, x'(y), x''(y)\big) = 0$.

Are you sure that the restriction is not $$\alpha^2 \, \dot{x}^2 + x^4 y^2 \dot{y}^2 = P^2 \, ? :)$$

• You are correct this is part of a (much larger) Lagrangian problem. Here's a question which might simplify things... am I free to parametrize the Lagrangian by something other than time? Could I, for instance, use a radial or azimuthal coordinate as my independent variable? Sep 21, 2016 at 13:08
• Well, you should write the Lagrangian then. Knowing the Lagrangian can allow us to switch to Hamiltonian formulation, where you have double the equations but only first derivatives. Moreover, you may spot other conservation laws. The Hamiltonian itself is a conserved quantity. Knowing the Lagrangian formulation can allow you to design good numerical methods for solution. Sep 21, 2016 at 13:23
• In general you can use as an independent variable whatever is most convenient and whatever does the job. You just have to keep track of the changes. The question is whether this equation is integrable or solvable at all. If you do not have enough symmetries/conservation laws finding closed form solution is hopeless. Sep 21, 2016 at 13:28
• Which brings me back to my original question: is this a practice/homework problem that is guaranteed to have a closed form solution or is this a something you have derived while trying to model something? Sep 21, 2016 at 13:31
• My intuition tells me, by looking at the equations, that this is a geometric problem. It is about a curve on a two-manifold (surface, two dimensional space), which is parametrized by constant arch-length with respect to a Riemannian metric given by the second equation. Sep 21, 2016 at 13:37