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

[touqra]

I am facing a difficult integral here for calculus of variations. The question reads:
Find the extremum to the integral:
<br /> I[y(x)] = \int_{Q}^{P} (dy/dx)^2(1+dy/dx)^2 dx
whereP = (0,0) and Q = (1,2)

 

[saltydog]

How about using Euler's equation:

\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{&#039;}}\right)=0

where:

F(x,y,y^{&#039;})=\left(y^{&#039;}\right)^2(1+y^{&#039;})^2

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]

How about using Euler's equation:
\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{&#039;}}\right)=0
where:
F(x,y,y^{&#039;})=\left(y^{&#039;}\right)^2(1+y^{&#039;})^2
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]

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]

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,

2y&#039;^3 + 3y&#039;^2 + y&#039; = constant

 

[Hurkyl]

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

 

[saltydog]

Because the integrand is independent of y, hence, when I use Euler equation, it gives me,
2y&#039;^3 + 3y&#039;^2 + y&#039; = constant

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

6(y^{&#039;})^2+6y^{&#039;}+1=0

Can someone rectify this discrepancy please?

 

[touqra]

Well . . . can't say the Calculus of Variation is my strong suit but when I solve the Euler equation, I get:
6(y^{&#039;})^2+6y^{&#039;}+1=0
Can someone rectify this discrepancy please?

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,

(y&#039;+1)(y&#039;+2) = 0

Then, y&#039;+1 = 0 or y&#039;+2 = 0

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]

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]

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]

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:

y(x)=-x+b

or:

y(x)=-2x+b

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]

What's wrong with:
y(x)=-x+b
or:
y(x)=-2x+b
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?"

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]

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]

For the record I wish to correct a conclusion I made above:

Solving the Euler equation, I obtained:

2(1+6y^{&#039;}+6(y^{&#039;})^2)y^{&#039;&#039;}=0

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

1+6y^{&#039;}+6(y^{&#039;})^2=0

I obtain:

y^{&#039;}=\frac{-1\pm\sqrt{1-2/3}}{2}

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

y^{&#039;&#039;}=0

The solution is:

y(x)=ax+b

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

y(x)=2x

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]

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.