How to find the minimal function of this functional using Euler Method for extremals?

Question

The original functional was $\ J[x]= \int_{0}^{1} x \cdot f(x) \cdot f'(x) dx $ with $\ h(x) =x $. Solved for the Gateau derivative: $\ \partial J[f,h]= \int_{0}^{1} x \cdot f(x) \cdot h'(x) dx + \int_{0}^{1} x \cdot f'(x) \cdot h(x) dx$. To use the Euler method, $\ L = x \cdot f \cdot x + x \cdot f' \cdot x$ such that $\ L_{f} = x^2, L_{f'} = x^2, \frac{d}{dx}L_{f'}=2x $. However, I must have made a mistake that I can't see with that because it yields $\ x^2-2x =0 $ which doesn't tell me anything about what $\ f $ must be.

Answer 1

The functional is

$$J[f]=\int_0^1 L(x,f(x),f'(x))\,dx=\int_0^1 xf(x)f'(x)\,dx$$

The Euler-Lagrange equation for this functional, with $L=xff'$ is given by

$$\begin{align} \frac{\partial }{\partial f}L(x,f,f')-\frac{d}{dx}\frac{\partial}{\partial f'}L(x,f,f')&=\frac{\partial }{\partial f}(xff')-\frac{d}{dx}\frac{\partial}{\partial f'}(xff')\\\\ &=xf'-\frac{d}{dx}(xf)\\\\ &=-f \end{align}$$

Therefore, the functional has a stationary point for $f(x)\equiv 0$.

At $f=0$, $J[f]=0$, which is not necessarily an extreme value. That is to say, there exist $f_1(x)$ and $f_2(x)$ for which $L[f_1]>0$ and $L[f_2]<0$ (e.g., $f_1(x)=x$ and $f_2(x)=1-x$).

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