Consider the two differential equations \begin{align*} \ddot{x}_{A} - \gamma(x_{A} + x_{B}) &= 0, \\ \ddot{x}_{B} + \gamma(x_{A} + x_{B}) &= 0 \end{align*} where $\gamma$ is a constant. I am looking either for a Lagrangian $L = L(x, \dot{x}, t)$ that can reproduce these equations or a proof that shows this is impossible.

Unfortunately, my initial question was framed too narrowly, so I will leave two questions (they're not exactly the same as you will see in the answers). The first one was my original one. The second one is my new one.

Question 1. Is there a function $L = L(x, \dot{x}, t)$ such that \begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{A}} - \frac{\partial L}{\partial x_{A}} = \ddot{x}_{A} - \gamma(x_{A} + x_{B}), \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{B}} - \frac{\partial L}{\partial x_{B}} = \ddot{x}_{B} + \gamma(x_{A} + x_{B})? \end{align*}

Question 2. Is there a function $L = L(x, \dot{x}, t)$ such that when setting the Euler-Lagrange expressions to zero we obtain differential equations that yield the same solutions as my original differential equations?

After looking around, I found the inverse problem for Lagrangians giving the Helmholtz conditions. Let us define $f^{A}(x^{A}, x^{B}) = \gamma\cdot (x^{A} + x^{B})$ and $f^{B}(x^{A}, x^{B}) = -\gamma\cdot (x^{A} + x^{B})$ as in the wiki page, and let us define $$\Phi_{j}^{i} = \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial f^{i}}{\partial \dot{x}^{j}} - \frac{\partial f^{i}}{\partial x^{j}} - \frac{1}{4} \frac{\partial f^{i}}{\partial \dot{x}^{k}} \frac{\partial f^{k}}{\partial \dot{x}^{j}}.$$ Here we easily obtain $$ (\Phi_{j}^{i}) = \begin{pmatrix} -\gamma & -\gamma \\ \gamma & \gamma \end{pmatrix}. $$ Now the wiki page says that we need to find a non-singular symmetric matrix $g = (g_{ij})$ such that conditions $(\text{H}1)$-$(\text{H}3)$ in the wiki page are satisfied (see image at the bottom for conditions). I found that $$ g = \begin{pmatrix} 1 & \frac{3}{2} \\ \frac{3}{2} & 2 \end{pmatrix} $$ works, so I concluded that there should be a Lagrangian for my differential equations. However, I have not been able to find such any desired function that works.


  • Was my above reasoning in determining that a Lagrangian exists correct? Did I make a correct application of Douglas's theorem?
  • Can anyone help me find a Lagrangian? I tried various functions, yet they all failed to produce the differential equations.
  • If no one can give me an explicit answer, then can anyone tell me what steps I should take to construct a Lagrangian?

Some references:

  • The wikipedia page is here.
  • This is the link to Douglas's paper on the inverse problem for Lagrangians.
  • This post asks about the inverse problem, and the answers provide some references.
  • The thesis referenced in the MSE post provides an interesting so-called explicit condition on page 67, but it is only stated for the 1D one particle case. NOTE: It seems their condition only works for problems similar to Question 1. In Example IV.1 they conclude that a damped oscillator doesn't have a Lagrangian for the general case. This is true in the context of Question 1, but false in the context of Question 2 as this post demonstrates. If you multiply the differential equation by an exponential, then a Lagrangian can reproduce the differential equation.
  • The MSE post referenced Olver's Applications of Lie groups to differential equations.
  • Relevant wikipedia image as of 12/6/2022 in case it gets edited: enter image description here
  • $\begingroup$ Irrelevant now, but a good approach would have been the ansatz $L(x, \dot{x}) = \sum_{j = A, B} \sum_{k = A, B} M_{j k} \dot{x}_j \dot{x}_k + U_{jk} x_j x_k$. Differentiate to determine $M$ and $U$ by equating like terms. $\endgroup$ Commented Dec 21, 2021 at 22:12
  • $\begingroup$ Please refrain from addressing more than one question, and asking three subsequent questions. $\endgroup$
    – amWhy
    Commented Jan 3, 2022 at 17:39

2 Answers 2


To answer Question 2: The role of $g$ in Douglas's theorem appears to be to "symmetrize" the potential terms. (I gleaned this from the condition (H1) in the Wikipedia article.) In particular, the fact that $g \Phi = (g \Phi)^T$ implies that multiplying a column vector containing the given equations by $g$ would be fruitful. Doing this yields \begin{align*} \ddot{x}_{A} + \frac{3}{2} \ddot{x}_B + \frac{1}{2}\gamma(x_{A} + x_{B}) &= 0 \\ \frac{3}{2} \ddot{x}_A + 2 \ddot{x}_{B} + \frac{1}{2} \gamma(x_{A} + x_{B}) &= 0 \end{align*} and so it seems that the desired Lagrangian is $$ L = \frac{1}{2} \dot{x}_A^2 + \frac{3}{2} \dot{x}_A \dot{x}_B + \dot{x}_B^2 - \frac{1}{4} \gamma (x_A + x_B)^2. $$

As an aside, note that the above "kinetic term" is not positive definite. This may or may not be relevant for your purposes.

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    $\begingroup$ Wow, truly amazing! $\endgroup$ Commented Dec 21, 2021 at 22:57
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    $\begingroup$ @MaximalIdeal: You did the hard work by finding $g$; all I did was figure out how to use it. $\endgroup$ Commented Dec 22, 2021 at 1:27

For future reference I recite your differential equations \begin{align} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{A}} - \frac{\partial L}{\partial x_{A}} = \ddot{x}_{A} - \gamma(x_{A} + x_{B}) \tag{EL1}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{B}} - \frac{\partial L}{\partial x_{B}} = \ddot{x}_{B} + \gamma(x_{A} + x_{B})\tag{EL2} \end{align}

Unfortunately I am not familiar with Douglas theorem. However I was able to find the following proof which gives an answer to Question 1:

If we assume our Lagrangian to be of the most general form $L=L(\dot {x}_A,\dot{x}_B,x_A,x_B,t)$ and we plug it in the first equation $(EL1)$ collecting the $\ddot{x}_A$ terms yields $$ \frac{\partial^2 L}{\partial \dot{x}_{A}^2}=1 $$ Thus, $L$ has to be a polynomial of second order in $\dot{x}_{A}$, or more specifically $$ L(\dot {x}_A,\dot{x}_B,x_A,x_B,t)=\frac{\dot{x}_A^2}{2}+\dot{x}_A\cdot L_1(\dot{x}_B,x_A,x_B,t)+L_0(\dot{x}_B,x_A,x_B,t) $$ Now we plug this into the second equation $(EL2)$ and collect the $\ddot{x}_A$ and $\ddot{x}_B$ terms and see that $$ \ddot{x}_A:\qquad\frac{\partial L_1}{\partial \dot{x}_{B}}=0\\ \ddot{x}_B\dot{x}_A: \qquad\frac{\partial^2 L_1}{\partial \dot{x}_{B}^2}=0\\ \ddot{x}_B:\qquad\frac{\partial^2 L_0}{\partial \dot{x}_{B}^2}=1 $$ which yields more refined representations of the functions $L0$ and $L1$, namely $$ L(\dot {x}_A,\dot{x}_B,x_A,x_B,t)=\frac{\dot{x}_A^2+\dot{x}_B^2}{2}+\dot{x}_A\cdot \alpha(x_A,x_B,t)+\dot{x}_B\cdot \beta(x_A,x_B,t)+\lambda(x_A,x_B,t) $$ Once again, plugging in the adjusted form in the two equations $(EL1)$ and $(EL2)$ we end up with three equations for the unknown functions $\alpha,\beta$ and $\lambda$. \begin{align} \frac{\partial \alpha}{\partial x_B}&=\frac{\partial \beta}{\partial x_A}\tag{1}\\ \frac{\partial \alpha}{\partial t}-\frac{\partial \lambda}{\partial x_A}&=-\gamma(x_A+x_B)\tag{2}\\ \frac{\partial \beta}{\partial t}-\frac{\partial \lambda}{\partial x_B}&=\gamma(x_A+x_B)\tag{3} \end{align} These equations look promising at first, however if we consider $$ \frac{\partial}{\partial x_B}\big(2)\big)-\frac{\partial}{\partial x_A}\big((3)\big)\\ \Leftrightarrow \frac{\partial^2\alpha}{\partial t\partial x_B}-\frac{\partial^2\beta}{\partial t\partial x_A}=-2\gamma\\ \stackrel{(1)}{\Rightarrow}0=-2\gamma $$ we see that a solution can only exist for the trivial case of $\gamma=0$.

edit: for the last step I considered $\alpha,\beta$ and $\gamma$ to be differentiable twice. If we relax this condition we might find a solution to the system $(1)-(3)$.

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    $\begingroup$ I think your proof is correct here that the equations as given can't come from a Lagrangian; but that leaves open the possibility that some non-singular transformation of the equations can come from a Lagrangian. In other words, the fact that there is no $\ddot{x}_B$ term in (EL2) does not imply that $\partial L_1/\partial\dot{x}_B = 0$. See my answer below. $\endgroup$ Commented Dec 21, 2021 at 22:04
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    $\begingroup$ @MichaelSeifert Yeah, this is my fault more than his, because I realized just now that I formulated my problem too narrowly. I'll try to edit my question to accommodate this. $\endgroup$ Commented Dec 21, 2021 at 22:05
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    $\begingroup$ Nonetheless, this post provides valuable information, so please don't delete it. $\endgroup$ Commented Dec 21, 2021 at 22:05
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    $\begingroup$ Thanks for the comments, it seems I was not versed enough in the topic to see the true question behind the problem! :-) However, thanks for editing the problem in a way that leaves my answer valid. I made an edit to clarify it is only an answer to Question 1. $\endgroup$
    – maxmilgram
    Commented Dec 21, 2021 at 22:23

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