What is the Extremum of the Integral for Calculus of Variations?

In summary: I think there is something wrong if you factorise the equation and let y' be equal to a constant. If y' does equal a constant, then, y is a straight line having a slope equals to that constant.But in the original question, there are boundary conditions, ie, P=(0,0) and Q=(1,2)
  • #1
touqra
287
0
I am facing a difficult integral here for calculus of variations. The question reads:
Find the extremum to the integral:
[tex]
I[y(x)] = \int_{Q}^{P} (dy/dx)^2(1+dy/dx)^2 dx [/tex]
[tex] where [/tex][tex] P = (0,0) [/tex] [tex] and [/tex] [tex]Q = (1,2)[/tex]
 
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  • #2
How about using Euler's equation:

[tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0[/tex]

where:

[tex]F(x,y,y^{'})=\left(y^{'}\right)^2(1+y^{'})^2[/tex]

However, that's usually on an interval and not from point to point as you have it defined above.

This of course results in a highly non-linear ODE. Push come to shove, I'd solve it numerically.
 
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  • #3
saltydog said:
How about using Euler's equation:
[tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0[/tex]
where:
[tex]F(x,y,y^{'})=\left(y^{'}\right)^2(1+y^{'})^2[/tex]
However, that's usually on an interval and not from point to point as you have it defined above.
This of course results in a highly non-linear ODE. Push come to shove, I'd solve it numerically.

Since this is about extremum, we clearly have to start with the Euler-Lagrange equation. But, you are right. We will obtain a non-linear ODE. It is this ODE I am having headache with.
 
  • #4
touqra said:
Since this is about extremum, we clearly have to start with the Euler-Lagrange equation. But, you are right. We will obtain a non-linear ODE. It is this ODE I am having headache with.

Can you post the resulting ODE? I'd like to check it against mine which is really not too bad: second-degree, first order ODE.
 
  • #5
saltydog said:
Can you post the resulting ODE? I'd like to check it against mine which is really not too bad: second-degree, first order ODE.

Because the integrand is independent of y, hence, when I use Euler equation, it gives me,

[tex] 2y'^3 + 3y'^2 + y' = constant [/tex]
 
  • #6
That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?
 
  • #7
touqra said:
Because the integrand is independent of y, hence, when I use Euler equation, it gives me,
[tex] 2y'^3 + 3y'^2 + y' = constant [/tex]

Well . . . can't say the Calculus of Variation is my strong suit but when I solve the Euler equation, I get:

[tex]6(y^{'})^2+6y^{'}+1=0[/tex]

Can someone rectify this discrepancy please?
 
  • #8
saltydog said:
Well . . . can't say the Calculus of Variation is my strong suit but when I solve the Euler equation, I get:
[tex]6(y^{'})^2+6y^{'}+1=0[/tex]
Can someone rectify this discrepancy please?

Hurkyl said:
That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?

Errr... I guess there is a way to solve this non-linear differential equation which I am not aware of its existence. What should you do after you factorise a differential equation?
Let say for example,

[tex] (y'+1)(y'+2) = 0 [/tex]

Then, [tex] y'+1 = 0 [/tex] or [tex] y'+2 = 0 [/tex]

That means to say that y is a function whereby its first derivative can take only the value -1 or -2 ?

What's the next step?
 
  • #9
Well, there aren't many functions whose derivative is always either -1 or -2, are there? (Even fewer if you assume the derivative is continuous...)
 
  • #10
Hurkyl said:
Well, there aren't many functions whose derivative is always either -1 or -2, are there? (Even fewer if you assume the derivative is continuous...)
How can the derivative be continuous when the derivative takes on discrete values of either -1 or -2 ?
IMO, the only function whose derivative is either -1 or -2, are somewhat like stepwise functions with straight lines. I suppose you are saying that this applies to the original question. Are there any other functions?
 
  • #11
touqra said:
How can the derivative be continuous when the derivative takes on discrete values of either -1 or -2 ?
IMO, the only function whose derivative is either -1 or -2, are somewhat like stepwise functions with straight lines. I suppose you are saying that this applies to the original question. Are there any other functions?

What's wrong with:

[tex]y(x)=-x+b[/tex]

or:

[tex]y(x)=-2x+b[/tex]

Thus, being a variational problem, I'd substitute them both into the integral, evaluate it, and see which one yields the extremum I seek.
Or should I say, "is that what I SHOULD do?"
 
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  • #12
saltydog said:
What's wrong with:
[tex]y(x)=-x+b[/tex]
or:
[tex]y(x)=-2x+b[/tex]
Thus, being a variational problem, I'd substitute them both into the integral, evaluate it, and see which one yields the extremum I seek.
Or should I say, "is that what I SHOULD do?"

Hurkyl said:
That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?

I think there is something wrong if you factorise the equation and let y' be equal to a constant. If y' does equal a constant, then, y is a straight line having a slope equals to that constant.
But in the original question, there are boundary conditions, ie,
P=(0,0) and Q=(1,2)

If it is a straight line, obviously we can get the function y straight away from just the end points P and Q.
What is your opinion?
 
  • #13
Straight lines, even splits, et cetera are often solutions to extremal problems, so I'm not surprised. :smile:

I'm pretty sure my approach is valid -- if you've proven that the derivative must satisfy that cubic, then at all points it must be equal to one of its roots, and if the derivative is continuous, it must be a constant.

I don't know if you should assume the derivative is continuous or not.

If it's not, then the simpler solutions for y are piecewise-linear... I'm not sure if there are pathological continuous solutions.
 
  • #14
For the record I wish to correct a conclusion I made above:

Solving the Euler equation, I obtained:

[tex]2(1+6y^{'}+6(y^{'})^2)y^{''}=0[/tex]

Now, either the quantity in paranthesis is zero or the second derivative. In the former case:

[tex]1+6y^{'}+6(y^{'})^2=0[/tex]

I obtain:

[tex]y^{'}=\frac{-1\pm\sqrt{1-2/3}}{2}[/tex]

Note that the slope is negative in both instances so this solution cannot meet the boundary conditions above. In the second case:

[tex]y^{''}=0[/tex]

The solution is:

[tex]y(x)=ax+b[/tex]

Using the boundary conditions: y(0)=0 and y(1)=2, I obtain:

[tex]y(x)=2x[/tex]

Toqura, I assume you're finished with this problem by now right?

Edit: Oh yea, can someone tell me how to prove y(x) is a solution to this functional equation?
 
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  • #15
Thanks.

I think I will write assumptions whether the first derivative is continuous or not, first case and second case.
And then, go on to solve it by factorising like what saltydog did.
 

1. What is the purpose of the Calculus of Variations?

The Calculus of Variations is a mathematical tool used to find the optimal path or function that minimizes or maximizes a certain quantity, such as the area or length of a curve. It is commonly used in physics, engineering, and economics to solve optimization problems.

2. What are the key concepts in the Calculus of Variations?

The key concepts in the Calculus of Variations include functionals, variations, and Euler-Lagrange equations. A functional is a function that takes in other functions as its input and produces a real number as its output. Variations are small changes made to a function, and the Euler-Lagrange equations are used to find the critical points of a functional, which correspond to the optimal solution.

3. How is the Calculus of Variations applied in real-world problems?

The Calculus of Variations is applied in a wide range of real-world problems, including finding the shortest distance between two points, the shape of a soap film between two wire frames, and the path of a rocket that minimizes fuel consumption. It is also used in variational methods for solving differential equations and in the calculus of variations in economics.

4. What are the limitations of the Calculus of Variations?

One of the main limitations of the Calculus of Variations is that it can only find local minima and maxima, rather than global ones. This means that there is no guarantee that the solution found is the absolute best solution. Additionally, it can be challenging to solve complex problems using the Calculus of Variations, and it may require advanced mathematical techniques.

5. How does the Calculus of Variations relate to other mathematical concepts?

The Calculus of Variations is closely related to other mathematical concepts, such as optimization, functional analysis, and differential equations. It also has applications in physics, engineering, and economics, making it a versatile tool in many areas of mathematics and science.

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