# Calculus of Variations

by touqra
Tags: calculus, variations
 P: 282 I am facing a difficult integral here for calculus of variations. The question reads: Find the extremum to the integral: $$I[y(x)] = \int_{Q}^{P} (dy/dx)^2(1+dy/dx)^2 dx$$ $$where$$$$P = (0,0)$$ $$and$$ $$Q = (1,2)$$
 Sci Advisor HW Helper P: 1,593 How about using Euler's equation: $$\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0$$ where: $$F(x,y,y^{'})=\left(y^{'}\right)^2(1+y^{'})^2$$ 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.
P: 282
 Quote by saltydog How about using Euler's equation: $$\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0$$ where: $$F(x,y,y^{'})=\left(y^{'}\right)^2(1+y^{'})^2$$ 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.

HW Helper
P: 1,593
Calculus of Variations

 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.
P: 282
 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,

$$2y'^3 + 3y'^2 + y' = constant$$
 Emeritus Sci Advisor PF Gold P: 16,100 That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?
HW Helper
P: 1,593
 Quote by touqra Because the integrand is independent of y, hence, when I use Euler equation, it gives me, $$2y'^3 + 3y'^2 + y' = constant$$
Well . . . can't say the Calculus of Variation is my strong suit but when I solve the Euler equation, I get:

$$6(y^{'})^2+6y^{'}+1=0$$

Can someone rectify this discrepancy please?
P: 282
 Quote by saltydog Well . . . can't say the Calculus of Variation is my strong suit but when I solve the Euler equation, I get: $$6(y^{'})^2+6y^{'}+1=0$$ Can someone rectify this discrepancy please?
 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,

$$(y'+1)(y'+2) = 0$$

Then, $$y'+1 = 0$$ or $$y'+2 = 0$$

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?
 Emeritus Sci Advisor PF Gold P: 16,100 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...)
P: 282
 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?
HW Helper
P: 1,593
 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:

$$y(x)=-x+b$$

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?"
P: 282
 Quote by saltydog 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?"
 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.
 Sci Advisor HW Helper P: 1,593 For the record I wish to correct a conclusion I made above: Solving the Euler equation, I obtained: $$2(1+6y^{'}+6(y^{'})^2)y^{''}=0$$ Now, either the quantity in paranthesis is zero or the second derivative. In the former case: $$1+6y^{'}+6(y^{'})^2=0$$ I obtain: $$y^{'}=\frac{-1\pm\sqrt{1-2/3}}{2}$$ Note that the slope is negative in both instances so this solution cannot meet the boundary conditions above. In the second case: $$y^{''}=0$$ 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?