# An Euler-Lagrange Equation

I have an action with a Lagrangian which I would like to apply the Euler-Lagrange equations to https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation but have spent hours really struggling with it. That is I define

$$L\Big(\begin{pmatrix}x \\ y\end{pmatrix},\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} \Big):= \Big\| A \Big( \begin{pmatrix}x \\ y\end{pmatrix} + \begin{pmatrix} -\dot{y} \\ \beta\dot{y}\end{pmatrix} \Big) \Big\|^2$$

where

$$A=\begin{pmatrix} c_1 & c_2 \\ c_2 & c_3 \end{pmatrix}, \text{for some }~c_i~\text{for which A is invertible}.$$

Which $$x,y:[0,T]\to \mathbb{R}$$ solve the Euler -Lagrange equation

$$\nabla_{\begin{pmatrix}x \\ y\end{pmatrix}} L-\partial_t\nabla_{\begin{pmatrix} \dot{x} \\ \dot{y}\end{pmatrix} }L=0$$ with initial and terminal conditions $$x(0)=q,x(T)=q',y(0)=p,y(T)=p'$$. ? Please help !

$$\Big($$would more information on how the constants $$c_1,c_2,c_3$$ relate to eachother be useful? because infact $$c_1=\sigma^2+\gamma^2, c_2=-\gamma(\sigma+\alpha),c_3=\alpha^2+\gamma^2$$, where these constants $$\alpha,\sigma,\gamma$$ are positive and $$\alpha\sigma>\gamma^2$$ $$\Big)$$.

EDIT : I got the E.L as with the above choices for $$c_1,c_2,c_3$$

$$$$\nabla_{(x,y)} L = (\frac{1}{\alpha\sigma-\gamma^2})^2\begin{pmatrix} 0 \\ \nabla_{y} B_1+\nabla_{y} B_2 \end{pmatrix}$$$$

with

$$$$\nabla_{y} B_1=2(\sigma+\gamma \beta) y + \frac{1}{\sigma+\gamma\beta}(\gamma \dot{y}-\sigma \dot{x})$$$$

and

$$$$\nabla_{y} B_2=2(\gamma + \alpha \beta) y + \frac{1}{\gamma+\alpha \beta}(\alpha \dot{x}-\gamma \dot{y}).$$$$

Also

$$$$\partial_t \nabla_{(\dot{x},\dot{y})} L=2 (A^{-1})^2(\begin{pmatrix}\ddot{x}\\ \ddot{y} \end{pmatrix}-\begin{pmatrix} \dot{y} \\ -\beta \dot{y} \end{pmatrix})$$$$

Which is a coupled ODE

Here is a sketched derivation.

1. First of all, notice that the Lagrangian $$L$$ does not depend $$\dot{x}$$. Then we don't need the 2 Dirichlet boundary conditions (BCs) for $$x$$ to deduce the EL equation for $$x$$. The EL equation for $$x$$ reduces to $$0~=~\chi(x,y,\dot{y})~:=~\frac{1}{2}\frac{\partial L(x,y,\dot{y})}{\partial x}~=~c_1(x-\dot{y}) +c_2(y+\beta\dot{y}) \tag{1}$$ from which we can determine $$x$$.

2. If we eliminate$$^1$$ $$x$$ in the Lagrangian $$L$$, the Lagrangian becomes (up to an overall non-zero multiplicative normalization) $$L_0~=~\frac{1}{2}(y+\beta\dot{y})^2. \tag{2}$$ The momentum is $$p ~=~ \frac{\partial L_0}{\partial \dot{y}} ~=~\beta(y+\beta\dot{y}).\tag{3}$$ The energy function $$E~=~p\dot{y}-L_0~=~\frac{1}{2}(\beta\dot{y}-y)(\beta\dot{y}+y) ~=~\frac{1}{2}(\beta^2\dot{y}^2-y^2) \tag{4}$$ is a constant, since there is no explicit $$t$$-dependence.

3. In other words, we get a 1st-order ODE $$\beta^2\dot{y}^2~=~2E +y^2 \tag{5}$$ Eq. (5) can easily be solved via separation of variables. This produces 1 integration constant. Together with $$E$$ we then have 2 integration constants, which can be matched with the 2 Dirichlet BCs for $$y$$.

4. Generically, it is impossible to match 4 boundary conditions (BCc), as already noted in Cesareo's answer. It would be natural to discard the 2 Dirichlet BCs for $$x$$.

--

$$^1$$ Normally, one is not allowed to use a EL equation (1) inside the Lagrangian, but one may show that it is okay for a quadratic $$x$$-dependence. We can complete the square $$L~=~L_0+L_2 \tag{6}$$ where $$L_n ~\propto~ \chi^n .\tag{7}$$ It is not difficult to see that $$L$$ and $$L_0$$ lead to the same EL equation for $$y$$ modulo the constraint (1).

• Hi can you help me understand what it is about this specific Lagrangian which means I cannot match 4 boundary conditions? Is is that theres no dependence on $t$ or $\dot{x}$? Commented Feb 4, 2021 at 10:52
• The latter: No $\dot{x}$. Commented Feb 4, 2021 at 11:46

The Lagrangian doesn't depends on $$t$$ so it obeys Betrami's identity. With $$p=(x,y)$$

$$L - \dot p\nabla_{\dot p}L = c_0$$

or

$$c_2^2 \left(x^2+y^2-\left(\beta ^2+1\right) \dot y^2\right)+c_3^2 \left(y^2-\beta ^2 \dot y^2\right)+2 c_2 c_1 \left(\beta \dot y^2+x y\right)+2 c_2 c_3 \left(\beta \dot y^2+x y\right)+c_1^2 \left(x^2-\dot y^2\right)=c_0$$

From Euler-Lagrange's equations we obtain

$$x = \frac{\left(c_2 \left(c_2-\beta c_3\right)+c_1^2-\beta c_2 c_1\right) \dot y-c_2 \left(c_1+c_3\right) y}{c_1^2+c_2^2}$$

now substituting into the former ODE we obtain a new ODE now depending only on $$y,\dot y$$. Concluding, we have two independent constants to fix: $$c_0$$ an one additional boundary from the last obtained ODE, and four independent boundary conditions. This is not feasible.

NOTE

The lagrangian is kind of degenerate concerning the kinetic energy because

$$\frac 12\nabla_{\dot p}\left(\nabla_{\dot p}L\right) = \left( \begin{array}{cc} 0 & 0 \\ 0 & \left(c_1-\beta c_2\right){}^2+\left(c_2-\beta c_3\right){}^2 \\ \end{array} \right)$$

which is not positive definite.

EDIT

Corrected some equations.

• Im confused, what is "not possible" are you saying there is no such minimising curve (x(t),y(t))? Commented Feb 3, 2021 at 19:39
• another question : don't we get more from Euler Lagrange? by this I mean we get two ODE from it. Commented Feb 3, 2021 at 19:40
• see my edit :) good point about degeneracy of kinetic energy of $L$. Commented Feb 3, 2021 at 19:54
• There is not such a minimizing curve which should attend to four independent boundary conditions. $x(0), x(T), y(0), y(T)$. From Euler-Lagrange we obtain one ODE and a algebraic relationship between $x$ and $y,\dot y$ Commented Feb 3, 2021 at 19:55
• you will obtain 2 ODE from E.L? Commented Feb 3, 2021 at 19:57