Find the Euler- Lagrange equation for $L = (v-yu)^2+u^2$ where $v=\frac{dy}{dt}$ and $u=\frac{dx}{dt}$

Question

Find the Euler- Lagrange equation for $$L = \left(\frac{dy}{dt}-y\frac{dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2.$$

I tried following manner.

$$\frac{d}{dt}\begin{bmatrix}\frac{dL}{dy/dt} & \frac{dL}{dx/dt}\end{bmatrix} =\begin{bmatrix} \frac{dL}{dx} & \frac{dL}{dy}\end{bmatrix}.$$

Is this the way to solve this?

Answer 1

Supposing an unactuated system, we have

$$ L(x(t),y(t),x'(t),y'(t),t) = \left(y'(t)-y(t) x'(t)\right)^2-x'(t)^2 $$

then

$$ L_X -\frac{d}{dt}L_{X'} = 0 = \cases{-\frac{d}{dt}L_{x'}\\ L_{y}-\frac{d}{dt}L_{y'}} \Rightarrow \left\{ \begin{array}{l} y(t) \left(y'(t)-y(t) x'(t)\right)+x'(t)= C_0 \\ 2 y(t)\left(x'(t)^2+x''(t)\right)-2 y''(t)=0 \end{array} \right. $$

NOTE

To solve the DEs, deriving the first, we can form

$$ \cases{ x''(t)= 2 y(t) x'(t) y'(t)-y(t)^2 x'(t)^2-y'(t)^2\\ y''(t)= 2 y(t)^2 x'(t) y'(t)-y(t)^3 x'(t)^2+y(t) x'(t)^2-y(t) y'(t)^2 } $$

Answer 2

1. Let's try to solve OP's 2 coupled 2nd-order ODEs. Let's for starters multiply OP's Lagrangian with an overall conventional factor $1/2$. This normalization does of course not affect the EL equations. OP's Lagrangian then reads $$ L(x,y,\dot{x},\dot{y})~=~\frac{1}{2}\left(\dot{x}^2+ (\dot{y}-y\dot{x})^2\right) ~=~\frac{1}{2}\sum_{i,j=1}^2g_{ij}\dot{x}^i\dot{x}^j,\tag{1}$$ where $$ g_{ij}~=~\begin{pmatrix} 1+y^2 & -y \cr -y & 1 \end{pmatrix}. \tag{2}$$ The inverse matrix is $$ (g^{-1})^{ij}~=~\begin{pmatrix} 1 & y \cr y & 1+y^2 \end{pmatrix}. \tag{3}$$ The momenta are $$ p_i~=~\frac{\partial L}{\partial \dot{x}^i}~=~\sum_{j=1}^2g_{ij}\dot{x}^j, $$ $$p_x~=~(1+y^2)\dot{x}-y\dot{y}, \qquad p_y~=~\dot{y}-y\dot{x}. \tag{4}$$ The Hamiltonian becomes $$ H(x,y,p_x,p_y)~=~\sum_{i=1}^2p_i\dot{x}^i-L ~=~\frac{1}{2}\sum_{i,j=1}^2(g^{-1})^{ij}p_ip_j~=~\frac{1}{2}\left((p_x+yp_y)^2+ p_y^2 \right) .\tag{5}$$

2. The coordinate $x$ is a cyclic coordinate and hence the corresponding momentum $$ p_x\text{ is a constant of motion}, \tag{6}$$ cf. a comment by user Semiclassical. Hence we should instead consider the Routhian $$ -R(y,p_x,\dot{y})~=~L-p_x\dot{x}~=~\frac{\dot{y}^2-p_x^2-2p_xy\dot{y}}{2(1+y^2)} .\tag{7}$$ Now there is effectively only 1 DOF. The EL equation becomes a single 2nd-order ODE $$\frac{d}{dt} \frac{\partial R}{\partial \dot{y}}~=~\frac{\partial R}{\partial y}.\tag{8}$$ See also this related Phys.SE post.

3. However we can do even better. Since there is no explicit time dependence, the corresponding energy$^1$ $$ E~=~h(y,p_x,\dot{y})~=~H\left(x,y,p_x,p_y=\frac{\dot{y}-yp_x}{1+y^2}\right)~=~\frac{\dot{y}^2+p_x^2}{2(1+y^2)}\text{ is a first integral}.\tag{9}$$ The full solution is $$ y(t)~=~A\sinh(\sqrt{2E}\Delta t), \qquad A~:=~\pm \sqrt{\frac{2E-p_x^2}{2E}}, \qquad \Delta t~:=~t-t_0.\tag{10} $$

4. By integrating $$ \dot{x}~\stackrel{(4)}{=}~\frac{p_x+y\dot{y}}{1+y^2} ~\stackrel{(10)}{=}~\frac{p_x+A^2\sqrt{2E} \sinh(\sqrt{2E}\Delta t)\cosh(\sqrt{2E}\Delta t)}{1+A^2\sinh^2(\sqrt{2E}\Delta t)}, \tag{11}$$ we find the full solution $$ x(t)~=~\sqrt{2E} \ln\left(\cosh(\sqrt{2E}\Delta t)+\frac{p_x}{\sqrt{2E}}\sinh(\sqrt{2E}\Delta t)\right) +x_0. \tag{12}$$

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$^1$ From Hamilton's equations the $y$-velocity is $$ \dot{y}~=~\frac{\partial H}{\partial p_y}~\stackrel{(5)}{=}~yp_x+(1+y^2)p_y\qquad\Leftrightarrow\qquad p_y~=~\frac{\dot{y}-yp_x}{1+y^2}.\tag{13} $$