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 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.
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 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.
Because the integrand is independent of y, hence, when I use Euler equation, it gives me,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.
Well . . . can't say the Calculus of Variation is my strong suit but when I solve the Euler equation, I get: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]
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?
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?Hurkyl said:That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?
How can the derivative be continuous when the derivative takes on discrete values of either -1 or -2 ?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...)
What's wrong with: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?
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?"
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.Hurkyl said:That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?