Question about Lagrange multipliers

In summary, the conversation is about understanding a proof involving Lagrange multipliers. The problem is to find the stationary value of ##a_{ik}q_iq_k## under the condition ##b_{ik}q_iq_k = 1##. The author drops this condition and minimizes ##a_{ik}q_iq_k - \frac 1{\lambda} b_{ik}q_iq_k##, but the other person questions if it should be ##a_{ik}q_iq_k - \frac 1{\lambda} (b_{ik}q_iq_k - 1)## instead. The first person then explains that the constant term can't be neglected and
  • #1
dRic2
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I'm having some trouble understanding the following proof (##a_{ik}## and ##b_{ik}## are constants)

Our problem is to find the stationary value of ##a_{ik}q_iq_k## under the auxiliary condition the we move on the surface ##b_{ik}q_iq_j = 1##.

We drop the auxiliary condition and minimize ##a_{ik}q_iq_k - \frac 1 {\lambda} b_{ik}q_iq_k##.

Shouldn't it be ##a_{ik}q_iq_k - \frac 1 {\lambda} (b_{ik}q_iq_k-1)## ?

(Summation convention is used)

Thanks Ric
 
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  • #2
One thing I don't understand is Lagrange multipliers. However, one thing I do follow is that adding a constant to a Lagrangian doesn't change the equations obtained by the variation. Isn't the term you've added, ##-\frac{1}{\lambda}##, held constant in the variation?
 
  • #3
Yes I see you point and in fact I am a bit puzzled myself, but then what would be the difference between the condition ##b_{ik}q_iq_k = 1## and some other condition ##b_{ik}q_iq_k = c## or even ##b_{ik}q_iq_k = 0## since I can always neglect the constant term ?
 
  • #4
You can not neglect constant. Loosely speaking, you have a condition ##g(x)=c## and your task is to find a critical point of a function ##f=f(x)## on the manifold ##\{x\in\mathbb{R}^m\mid g(x)=c\}##. To do that you must solve the following system
$$g(x)=c,\quad \frac{\partial f}{\partial x^i}=\lambda \frac{\partial g}{\partial x^i},\quad i=1,\ldots,m$$
you have m+1 equations and m+1 unknowns: ##\lambda,x^1,\ldots,x^m##
 
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  • #5
I know, that's why I'm asking. I don't understand the argument given in that proof. I have to specify that I do not need to evaluate the minimum of that function, but the values of ##q_i## such that the function has minimum, maybe in that sense the constant can be neglected... I don't know
 
  • #6
Why do not you take a regular textbook in analysis? V. Zorich Mathematical Analysis for example
 
  • #7
Maybe I was not expressing myself very well. I was reading a book. The book is not about math nor about Lagrange multipliers, it is about classical mechanics. While reading the book I ran into the argument posted in the first post which I don't understand.

PS: maybe the confusing part is the word "proof". I didn't mean that the proof is about Lagrange multipliers. The first line should read "I'm having some trouble understanding the following statement"
 
  • #8
I also do not understand that argument and I don't care because I understand what Lagrange multipliers are and I can solve the problem independently on that book.
 
Last edited:
  • #9
wrobel said:
I can solve the problem independently on that book.
Ok, so let's try to solve it. Do you agree with me that if ##x_i = q_i##, ##f(x_i) = a_{ik}q_iq_k##, ##g(x_i) = b_{ik}q_iq_k## and ##c = 1## then the system becomes:
$$b_{ik}q_i = \lambda a_{ik} q_i$$
plus the condition ##b_{ik}q_i q_k = 1##. (Note that I used ##\frac 1 {\lambda}## as a multiplier to stick to the convention introduced by the author of the book)
 
  • #10
dRic2 said:
Note that I used 1λ\frac 1 {\lambda} as a multiplier to stick to the convention introduced by the author of the book)
it is not a matter of convention, ##\lambda## must be situated as I said
 
  • #11
You can always define a new variable ##\lambda' = \frac 1 {\lambda}## any get the same result
 
  • #12
dRic2 said:
You can always define a new variable λ′=1λ\lambda' = \frac 1 {\lambda} any get the same result
what about ##\lambda=0##? :)
 
  • #13
Ok you are right.

But in the end I solved my problem. In this particular case the formal procedure and the one used by the author are the same, he probably used this "trick" to get a faster result.
 

1. What are Lagrange multipliers and how are they used in optimization?

Lagrange multipliers are a mathematical tool used in optimization to find the maximum or minimum value of a function subject to a set of constraints. They involve creating a new function, called the Lagrangian, by adding the constraints to the original function using a set of unknown variables called multipliers. The optimal solution can then be found by taking the partial derivatives of the Lagrangian and setting them equal to zero.

2. When should Lagrange multipliers be used in optimization problems?

Lagrange multipliers should be used when a function needs to be optimized subject to a set of constraints. This can include problems in economics, engineering, and physics, among others. Lagrange multipliers are particularly useful when the constraints are nonlinear or when there are multiple constraints that cannot be easily combined into a single equation.

3. What is the relationship between Lagrange multipliers and the method of undetermined multipliers?

The method of undetermined multipliers is a specific application of Lagrange multipliers, where the constraints are equations that must be satisfied exactly. In this case, the Lagrangian can be simplified and the optimal solution can be found by solving a system of equations. The method of undetermined multipliers is often used in engineering and physics problems.

4. Are there any limitations or drawbacks to using Lagrange multipliers?

One limitation of Lagrange multipliers is that they can only be used for constrained optimization problems, where the constraints are in the form of equations. They also require the constraints to be differentiable and continuous. Additionally, Lagrange multipliers can be computationally intensive for problems with a large number of constraints or variables.

5. Can Lagrange multipliers be used for non-differentiable functions?

No, Lagrange multipliers cannot be used for non-differentiable functions, as they rely on taking the partial derivatives of the Lagrangian to find the optimal solution. If the function is not differentiable, then the partial derivatives do not exist and the method cannot be applied. In this case, alternative methods such as subgradients or convex optimization may be used.

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