Calculus of variations changing variables

In summary: Memorizing the conditions for a minimum is a good idea, but omitting negative multiples changes the direction of optimization.Memorizing the conditions for a minimum is a good idea, but omitting negative multiples changes the direction of optimization.
  • #1
bobred
173
0

Homework Statement


Hi
I am given the functional

x?S[y]=%5Cint_%7Ba%7D%5E%7Bb%7D%5Cfrac%7Bx%5E%7B3%7Dy%5E%7B%5Cprime2%7D%7D%7By%5E%7B4%7D%7D%20dx.png


I am asked to show that if
png.png
and with an appropriate value for
png.png
that

png.png


Homework Equations


[/B]

The Attempt at a Solution


So I get

du%7D=%5Cfrac%7By%5E%7B%5Cprime%7D%28u%29%7D%7B%5Cbeta%20u%5E%7B%5Cbeta-1%7D%7D.png


png.png

png.png


If I set
2.png
then I get

png.png

I think that it is correct but what about the factor of 2?
 
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  • #2
Your work looks correct. Unless you do another variable transformation, you will not get rid of the 2.
 
  • #3
I imagine that since you are trying to find stationary values of this integral, they are not affected by a constant multiple, so you can drop the factor. I think it should be -2, but that's a minor point.
 
  • #4
PeroK said:
I think it should be -2, but that's a minor point.

This would depend on the order of the integration limits (note that ##a## was the lower integration limit in ##x## but ##A## is the upper integration limit in ##u## - of course I am just making the arbitrary inference that ##a## corresponds to ##A## here ...).
 
  • #5
Orodruin said:
This would depend on the order of the integration limits (note that ##a## was the lower integration limit in ##x## but ##A## is the upper integration limit in ##u## - of course I am just making the arbitrary inference that ##a## corresponds to ##A## here ...).

Yes, didn't notice that.
 
  • #6
Thanks guys.
I may have missed this in my notes PeroK but if we are trying to find stationary points of a functional constant multiples can be ignored?
James
 
  • #7
bobred said:
but if we are trying to find stationary points of a functional constant multiples can be ignored?

Yes, but there is nothing strange about this. It works this way for functions as well, if ##f(x)## has the stationary point ##x=2##, then so does ##2f(x)##.
 
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  • #8
Yes, thanks for the clarification.
James
 
  • #9
bobred said:
Thanks guys.
I may have missed this in my notes PeroK but if we are trying to find stationary points of a functional constant multiples can be ignored?
James

Positive multiples can be ignored, but omitting negative multiples changes the direction of optimization.
 
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  • #10
Ray Vickson said:
Positive multiples can be ignored, but omitting negative multiples changes the direction of optimization.

But it does not change the fact that the point is stationary. Just exchanges minima for maxima.
 
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  • #11
Orodruin said:
But it does not change the fact that the point is stationary. Just exchanges minima for maxima.

Of course, but that is another issue. It certainly changes the second-order tests, so that instead of looking to see if the Hessian is positive-definite, we look instead to see if it is negative-definite. However, when I taught this stuff I recommended that students just "memorize" the conditions for a minimum, then switch the sign of the objective if the problem was a maximization; that eliminates the need for a whole raft of special cases, etc.
 

1. What is calculus of variations?

Calculus of variations is a branch of mathematics that deals with finding the optimal solution to a functional, which is a mathematical expression involving a function. It involves finding the function that minimizes or maximizes the value of the functional, often subject to certain constraints.

2. What does it mean to "change variables" in calculus of variations?

In calculus of variations, changing variables refers to the process of transforming the original functional into an equivalent one using a different set of independent variables. This is often done to simplify the problem or to make it easier to solve.

3. How is changing variables beneficial in calculus of variations?

Changing variables can make the problem easier to solve by reducing the complexity of the functional or by making it more amenable to existing methods and techniques. It can also reveal hidden relationships between different functions and provide a deeper understanding of the problem.

4. What are some commonly used techniques for changing variables in calculus of variations?

Some commonly used techniques for changing variables in calculus of variations include the substitution method, the integration by parts method, and the Legendre transformation method. These methods can help transform the functional into a simpler form or reveal new relationships between the variables.

5. Are there any limitations to changing variables in calculus of variations?

While changing variables can be a powerful tool in solving problems in calculus of variations, it does have its limitations. In some cases, the transformation may not result in a simpler functional, or it may introduce new constraints that make the problem more difficult to solve. Additionally, the choice of variables can greatly affect the outcome of the problem, so it is important to carefully consider which variables to use for the transformation.

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