# Lagrange multipliers

1. Sep 12, 2012

### aaaa202

Normally lagrange multipliers are used in the following sense.

Suppose we are given a function f(x,y.z..,) and the constraint g(x,y,z,...,) = c
Define a lagrange function:
L = f - λ(g-c)

And find the partial derivatives with respect to all variables and λ. This gives you the extrema since for an extrema ∇f = λ∇g

However, I find that in my mechanics book this is used differently. I have attached a picture of the place where lagrange multipliers are used. I don't see how they in this case are used to maximize a quantity (which should be L) under a constraint like the above. Can anyone show me how they are and which gradients are to be parallel?

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2. Sep 12, 2012

### clamtrox

I'm not sure if I understand; how are they used differently? You have a set of constraints $f_\alpha(\mathbf{q},\dot{\mathbf{q}},t) = 0$ and then you just write $L = L + \sum_\alpha \lambda_\alpha f_\alpha$

3. Sep 12, 2012

### aaaa202

maybe I just don't see it. But my point is that you usually want to make the gradients of f and g=y parallel. How is that done in the example? There are no gradients being taken...

4. Sep 12, 2012

### clamtrox

Aah sorry, I somehow missed the whole point of your question :) They are not maximizing L. They are minimizing (or maximizing) the integral of L. This is entirely different.

So do you understand how you get from $$S = \int dt L$$ by varying the action to
$$\delta S = \int dt \left( \frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} \right) \delta q = 0 \rightarrow \frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0$$?

5. Sep 12, 2012

### aaaa202

Yes, I know that the above implies the Euler-lagrange equation.
So they want to make the lagrangian stationary. Normally if the qi's are independent you that leads to n independent lagrange equations. But now they are instead using lagrange multipliers. But how exactly are they used? In analogy to my original example, what is f and what is g and how does the above correspond to the problem of making the gradient of f and g parallel. Maybe it is completely obvious, but I can't quite see it.

6. Sep 12, 2012

### Ray Vickson

Imagine that you approximate the integral by a finite sum of a large number of terms (which we actually do in numerical work), and you replace the continuum of constraints $f(q,\dot{q},t) = 0$ by a large number of constraints $f(q_i,v_i ,t_i) = 0, \; i=1,2, \ldots, N.$ Here, the $q_i$ are you estimate of $q(t_i)$ and the $v_i$ are your estimate of $\dot{q}(t_i).$ (Again, this is often how such problems ARE dealt with using numerical methods---we replace a continuous problem by a disctete problem involving a large number of small time steps.) Now your solution to
$$\min \int_a^b L(q,\dot{q},t)\, dt, \: \text{ subject to } f(q,\dot{q},t) = 0 \text{ for all } t \in [a,b]$$ would be replaced by that of the problem
$$\min \sum_{i=1}^N \Delta t_i f(q_i,v_i,t_i),\\ \text{subject to}\\ v_i = (q_i - q_{i-1})/ \Delta_i \;(\text{ or } v_i = (q_{i+1} - q_i)/ \Delta_i), i=1, \ldots N\\ f(q_i,v_i ,t_i) = 0, \; i=1, \ldots, N.$$
Now, of course, you would have a Lagrange multiplier $\lambda_i$ for each of the f = 0 constraints, so your Lagrangian would contain
$$\sum_{i=1}^N \lambda_i f(q_i,v_i,t_i).$$
and this would go over into an integral in the limit of infinite N.

Well, that is the intuition, anyway; rigorous justification is another matter.

RGV

7. Sep 12, 2012

### aaaa202

Thanks, just the type of argument I was looking for.