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Consider a functional $$J = \int_{a}^{b} F(x, y, y^{'}),$$ where $F(x, y, y^{'}) = \frac{1 + y^{2}}{(y^{'})^2}$ for admissible function $y(x).$ Which of the following are extremals for $J$?

  1. $y(x) = A \sin x $
    1. $y(x) = A\sinh(x) + B\cosh(x) $
    2. $y(x) = A\sinh(Ax + B) $
    3. $y(x) = A\sin(x) + B\cos(x) $

What I have done is to consider Euler Poisson condition.

$$F_y-\frac{d}{dx}F_{y '}=0$$

Here $$F_y=\frac{2y}{(y')^2}$$

$$F_{y'}=-\frac{2(1+y^2)}{(y')^3}$$

$$\frac{d}{dx}F_{y'}=-\frac{d}{dx}\large(\frac{2(1+y^2)}{(y')^3})=-\frac{2y(y')^3+3(1+y^2)y''}{(y')^5}$$

So, $$F_y-\frac{d}{dx}F_{y '}=0$$ implies

$$ \frac{2y}{(y')^2}-\frac{2y(y')^3+3(1+y^2)y''}{(y')^5}=0$$

i.e., $$2y(y')^3-2y(y')^3-3(1+y^2)y''=0$$

i.e., $$(1+y^2)y''=0$$

i.e., $$y''=0$$

This tells that no option is correct..

I am confused...

Could any one kindly take some pain in clearing this issue.

Thank You.

Martin Sleziak
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1 Answers1

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I don't follow your computation of $\frac{d}{dx}F_{\dot y}$. At the very least you lost a factor of $2$ there, but there seem to be other mistakes too. I get
$$ \frac{d}{dx} (-2 \dot y^{-3}(1+y^2)) = 6 \dot y^{-4} \ddot y (1+y^2) - 4 \dot y^{-2} y $$ (I prefer dots to primes in such calculations, because they don't get in the way of exponents). Hence, the equation is $$ 2\dot y^{-2}y- 6 \dot y^{-4} \ddot y (1+y^2) + 4 \dot y^{-2} y =0 $$ which simplifies to $$ \dot y^{2}y- \ddot y (1+y^2) =0 $$

Post No Bulls
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  • yes i got that i made some mistakes.... I do not know how do solve such equations $\dot y^{2}y- \ddot y (1+y^2) =0$ I know how to solve if coefficients of derivatives are constants... could you please help me –  Dec 25 '13 at 06:05
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    @PraphullaKoushik But do you have to solve it? You are given four choices A,B,C,D: you can simply check which one satisfies the equation by plugging them in, one at a time. – Post No Bulls Dec 25 '13 at 06:07
  • A very valid point i would say :D Thank you so much –  Dec 25 '13 at 06:12