Extremal of a functional $I=\int\limits_0^{x_1} y^2(y')^2dx$

Question

The extremal of the function $$I=\int\limits_0^{x_1} y^2(y')^2dx$$ that passes through $(0,0)$ and $(x_1,y_1)$ is

1. a constant function

2. a linear function of x

3. part of a parabola

4. part of an ellipse.

Attempt:

Using Euler-Lagrange equation, $$\frac{\partial f(x)}{\partial y}-\dfrac{d }{dx}\left(\frac{\partial f(x)}{\partial y'}\right)=0$$ I get $$y^2y''+y(y')^2=0$$ Solving letting $v=y'$, $$\frac{y^2}2=cx+d,c=\frac{2x_1}{y_1^2}, d=0$$

Hence extremal is parabolic (right open).

Am I correct?

(CSIR June 2015, in case someone comes searching for it!)

Answer 1

Put $$\frac{\partial (y^2(y')^2)}{\partial y}-\frac{d }{dx}\left(\frac{\partial (y^2(y')^2)}{\partial y'}\right) = 0$$ $$2 y (y')^2 -\frac{d }{dx}\left(2 y^2 y'\right) = 0$$ Cancel the 2s then derivate: $$y (y')^2 -2y(y')^2-y^2y'' = 0$$ $$y \left( (y')^2 + y y'' \right) = 0$$ Assuming that the function is not everywhere zero you get: $$(y')^2 + y y'' = 0$$ Rearranging you get $$\frac{d }{dx}(y y') = 0$$ Which we can take one step further to: $$\frac{d^2}{dx^2}(y^2) = 0$$ Which indeed means that if $y_1$ is zero then $y = y_0 = y_1$ a constant (everywhere zero afterall!). Or if not then, $y = \pm\sqrt{A x + B}$ as you say.

Since $y(0) = 0$, we have $B = 0$. Then $$y(x_1) = y_1 = \pm\sqrt{A x_1}$$ So $A = \frac{y_1^2}{x_1}$ Then both cases are covered by $y = \pm y_1 \sqrt{x/x_1}$