Can a Scalar Equation Be Transformed into Lagrangian Form?

In summary: It is nice to see how this problem can be approached in different ways and how one can bring in different concepts and theorems to solve it. It certainly shows the depth and richness of classical mechanics. In summary, the conversation discussed a problem from a Russian textbook in classical mechanics, where a scalar equation was given and it was required to show that it can be multiplied by a function to obtain the Lagrangian form. The Cauchy-Kowalewski theorem was mentioned as a possible approach, but the conversation also delved into the concept of Helmholtz conditions and the technique of Jacobi's Last Multiplier as alternative methods to solve the problem.
  • #1
wrobel
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There is a problem from a Russian textbook in classical mechanics.

Consider a scalar equation $$\ddot x=F(t,x,\dot x),\quad x\in\mathbb{R}.$$ Show that this equation can be multiplied by a function ##\mu(t,x,\dot x)\ne 0## such that the resulting equation
$$\mu\ddot x=\mu F(t,x,\dot x)$$ has the Lagrangian form
$$\frac{d}{dt}\frac{\partial L}{\partial \dot x}-\frac{\partial L}{\partial x}=0.$$
I can only say that by the Cauchy-Kowalewski theorem it can be done locally provided ##F## is an analytic function.

But some relatively elementary way is supposed. Sure it is a local assertion.

What do you think?
 
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  • #2
I've only seen the Lagrangian written like that when trying to maximize an integral. How is ##L## defined here? My brief googling didn't find anything useful.
 
  • #3
Office_Shredder said:
How is L defined here?
from the Cauchy-Kowalewski theorem. I do not have other idea.
 
  • #4
Sorry, I think I misunderstood the question. Is it supposed to be show there exists ##\mu## such that ##\mu\ddot x - \mu F## can be written as ##\frac{d}{dt} \frac{\partial L}{\partial \dot x} -\frac{\partial L}{\partial x}## for some ##L## which is presumably formed by ##F## and ##\mu##?
 
  • #5
Office_Shredder said:
Sorry, I think I misunderstood the question. Is it supposed to be show there exists μ such that μx¨−μF can be written as ddt∂L∂x˙−∂L∂x for some L which is presumably formed by F and μ?
yes and ##\mu(t,x,\dot x)\ne 0##
 
  • #6
The obvious course is to expand the total time derivative and substitute [itex]\ddot x = F[/itex], which yields [tex]
F\frac{\partial^2L}{\partial \dot x^2} + \dot x \frac{\partial^2L}{\partial x \,\partial \dot x} + \frac{\partial^2L}{\partial t\,\partial \dot x} - \frac{\partial L}{\partial x} = 0[/tex] as an equation for [itex]L[/itex]; [itex]\mu[/itex] is then equal to [itex]\dfrac{\partial^2L}{\partial \dot x^2}.[/itex] I haven't attempted to solve this.

Another approach is to write [itex]\mu \ddot x = \frac{d}{dt}(\mu \dot x) - \frac{d\mu}{dt}\dot x[/itex] so that [tex]
\frac{d}{dt}(\mu\dot x) - \frac{d\mu}{dt}\dot x - \mu F = 0[/tex] and try to solve [tex]
\begin{split}
\frac{\partial L}{\partial \dot x} &= \mu\dot x \\
\frac{\partial L}{\partial x} &= \frac{d\mu}{dt}\dot x + \mu F\end{split}[/tex]
 
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  • #7
pasmith said:
The obvious course is to expand the total time derivative and substitute x¨=F, which yields F∂2L∂x˙2+x˙∂2L∂x∂x˙+∂2L∂t∂x˙−∂L∂x=0 as an equation for L;
yes and that is why I referred to Cauchy-Kowalewski
 
  • #8
wrobel said:
There is a problem from a Russian textbook in classical mechanics.
Does the textbook mention the "Helmholtz conditions" in The Inverse Problem in Lagrangian Mechanics, (or similar)?

In brief, the Helmholtz conditions on a given equation of motion (EoM) are necessary and sufficient for a corresponding Lagrangian to exist.

For 1D problems, 2 of the 3 Helmholtz conditions are trivially satisfied.

Write the EoM as ##\,G(x,\dot x, \ddot x, t) = 0\,##. Then the 3rd Helmholtz condition boils down to$$\frac{\partial G}{\partial \dot x} ~=~ \frac{d}{dt}\, \frac{\partial G}{\partial \ddot x}~.$$If this is not immediately satisfied, one can use the technique of "Jacobi's Last Multiplier", and write ##H := \mu(\dot x, x,t)\, G## which obviously satisfies the EoM. The Helmholtz condition on ##H## says $$\frac{\partial \mu}{\partial \dot x} \, G ~+~ \mu \, \frac{\partial G}{\partial \dot x} ~=~ \frac{d\mu}{dt}~,$$ (since we're assuming ##G## is 1st order in ##\ddot x##). Then the idea is to choose ##\mu## to be independent of ##\dot x## and solve what's left to get something like $$\mu ~=~ \exp\left(\int\! \frac{\partial G}{\partial \dot x} \, dt\right) ~.$$
wrobel said:
[...] But some relatively elementary way is supposed.
Elementary, yes -- but relying on a very nontrivial theorem about the Helmholtz conditions. :oldwink:

References:

1) Anton Almen's paper: https://jfuchs.hotell.kau.se/kurs/amek/prst/19_heco.pdf
which contains the nontrivial proof, but also works through the 1D harmonic oscillator as an example. Partially, a shorter version of the next reference:

2) Nigam & Banerjee, "A Brief Review of Helmholtz Conditions",
Available as: arxiv:1602.01563

Google gives many references about Helmholtz conditions and the technique of "Jacobi's Last Multiplier".

I hope that helps. :oldsmile:
 
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  • #9
Thanks a lot. That is interesting indeed. I have never heard about the Helmholtz conditions
 
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Related to Can a Scalar Equation Be Transformed into Lagrangian Form?

1) What is the Lagrangian form of an equation?

The Lagrangian form of an equation is a mathematical formalism used to describe the dynamics of a physical system. It is based on the principle of least action, which states that the actual path taken by a system is the one that minimizes the action, a quantity that represents the total energy of the system.

2) How is the Lagrangian form of an equation different from other forms of equations?

The Lagrangian form of an equation is different from other forms of equations, such as Newton's laws or Hamilton's equations, because it takes into account the entire history of a system's motion rather than just its current state. This allows for a more comprehensive understanding of the dynamics of a system.

3) What are the advantages of using the Lagrangian form of an equation?

There are several advantages to using the Lagrangian form of an equation. It allows for a more elegant and concise representation of the dynamics of a system, it is invariant under coordinate transformations, and it can be used to derive the equations of motion for complex systems with multiple degrees of freedom.

4) Are there any limitations to the Lagrangian form of an equation?

While the Lagrangian form of an equation is a powerful tool for describing the dynamics of a system, it does have some limitations. It is not always easy to determine the Lagrangian for a given system, and it may not be the most efficient method for solving certain types of problems.

5) In what fields of science is the Lagrangian form of an equation commonly used?

The Lagrangian form of an equation is commonly used in fields such as physics, engineering, and mathematics. It is particularly useful for studying classical mechanics, electromagnetism, and quantum mechanics, among others.

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