Solving a Coupled Second-Order Differential Equation

Question

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

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

Answer 1

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}

Answer 2

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 $$\alpha^2 \, \dot{x}^2 + x^4 y^2 \dot{y}^2 = P^2 \, ? :)$$