mriteyl

2022-10-02

The original functional was with . Solved for the Gateau derivative: . To use the Euler method, such that . However, I must have made a mistake that I can't see with that because it yields which doesn't tell me anything about what must be.

Abigayle Lynn

The functional is
$J\left[f\right]={\int }_{0}^{1}L\left(x,f\left(x\right),{f}^{\prime }\left(x\right)\right)\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{1}xf\left(x\right){f}^{\prime }\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx$
The Euler-Lagrange equation for this functional, with $L=xf{f}^{\prime }$ is given by
$\begin{array}{rl}\frac{\mathrm{\partial }}{\mathrm{\partial }f}L\left(x,f,{f}^{\prime }\right)-\frac{d}{dx}\frac{\mathrm{\partial }}{\mathrm{\partial }{f}^{\prime }}L\left(x,f,{f}^{\prime }\right)& =\frac{\mathrm{\partial }}{\mathrm{\partial }f}\left(xf{f}^{\prime }\right)-\frac{d}{dx}\frac{\mathrm{\partial }}{\mathrm{\partial }{f}^{\prime }}\left(xf{f}^{\prime }\right)\\ \\ & =x{f}^{\prime }-\frac{d}{dx}\left(xf\right)\\ \\ & =-f\end{array}$
Therefore, the functional has a stationary point for $f\left(x\right)\equiv 0$.
At $f=0$, $J\left[f\right]=0$, which is not necessarily an extreme value. That is to say, there exist ${f}_{1}\left(x\right)$ and ${f}_{2}\left(x\right)$ for which $L\left[{f}_{1}\right]>0$ and $L\left[{f}_{2}\right]<0$ (e.g., ${f}_{1}\left(x\right)=x$ and ${f}_{2}\left(x\right)=1-x$).

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