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Calculus of Variations

by touqra
Tags: calculus, variations
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touqra
#1
Oct14-05, 12:56 AM
P: 282
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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saltydog
#2
Oct14-05, 04:34 AM
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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.
touqra
#3
Oct14-05, 04:43 AM
P: 282
Quote Quote by saltydog
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.

saltydog
#4
Oct14-05, 05:16 AM
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Calculus of Variations

Quote Quote by touqra
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.
touqra
#5
Oct14-05, 10:57 AM
P: 282
Quote Quote by saltydog
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]
Hurkyl
#6
Oct14-05, 12:56 PM
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That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?
saltydog
#7
Oct14-05, 05:50 PM
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Quote Quote by touqra
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?
touqra
#8
Oct14-05, 07:35 PM
P: 282
Quote Quote by saltydog
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?
Quote Quote by Hurkyl
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?
Hurkyl
#9
Oct14-05, 08:05 PM
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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...)
touqra
#10
Oct14-05, 09:34 PM
P: 282
Quote Quote by Hurkyl
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?
saltydog
#11
Oct15-05, 04:52 AM
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Quote Quote by touqra
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?"
touqra
#12
Oct15-05, 07:02 AM
P: 282
Quote Quote by saltydog
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?"
Quote Quote by Hurkyl
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?
Hurkyl
#13
Oct15-05, 07:34 AM
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Straight lines, even splits, et cetera are often solutions to extremal problems, so I'm not surprised.

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.
saltydog
#14
Oct15-05, 08:18 PM
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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?
touqra
#15
Oct16-05, 06:53 AM
P: 282
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.


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