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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 $c_1,c_2,c_3$

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

with

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

and

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

Also

\begin{equation} \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}) \end{equation}

Which is a coupled ODE

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2 Answers 2

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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 $x$.

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$^1$ 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).

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  • $\begingroup$ 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 $\dot{x}$? $\endgroup$ Feb 4, 2021 at 10:52
  • $\begingroup$ The latter: No $\dot{x}$. $\endgroup$
    – Qmechanic
    Feb 4, 2021 at 11:46
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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.

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  • $\begingroup$ Im confused, what is "not possible" are you saying there is no such minimising curve (x(t),y(t))? $\endgroup$ Feb 3, 2021 at 19:39
  • $\begingroup$ another question : don't we get more from Euler Lagrange? by this I mean we get two ODE from it. $\endgroup$ Feb 3, 2021 at 19:40
  • $\begingroup$ see my edit :) good point about degeneracy of kinetic energy of $L$. $\endgroup$ Feb 3, 2021 at 19:54
  • $\begingroup$ 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 $y,\dot y$ $\endgroup$
    – Cesareo
    Feb 3, 2021 at 19:55
  • $\begingroup$ you will obtain 2 ODE from E.L? $\endgroup$ Feb 3, 2021 at 19:57

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