mriteyl

2022-10-02

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

Abigayle Lynn

Beginner2022-10-03Added 12 answers

The functional is

$J[f]={\int}_{0}^{1}L(x,f(x),{f}^{\prime}(x))\phantom{\rule{thinmathspace}{0ex}}dx={\int}_{0}^{1}xf(x){f}^{\prime}(x)\phantom{\rule{thinmathspace}{0ex}}dx$

The Euler-Lagrange equation for this functional, with $L=xf{f}^{\prime}$ is given by

$\begin{array}{rl}\frac{\mathrm{\partial}}{\mathrm{\partial}f}L(x,f,{f}^{\prime})-\frac{d}{dx}\frac{\mathrm{\partial}}{\mathrm{\partial}{f}^{\prime}}L(x,f,{f}^{\prime})& =\frac{\mathrm{\partial}}{\mathrm{\partial}f}(xf{f}^{\prime})-\frac{d}{dx}\frac{\mathrm{\partial}}{\mathrm{\partial}{f}^{\prime}}(xf{f}^{\prime})\\ \\ & =x{f}^{\prime}-\frac{d}{dx}(xf)\\ \\ & =-f\end{array}$

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 ${f}_{2}(x)=1-x$).

$J[f]={\int}_{0}^{1}L(x,f(x),{f}^{\prime}(x))\phantom{\rule{thinmathspace}{0ex}}dx={\int}_{0}^{1}xf(x){f}^{\prime}(x)\phantom{\rule{thinmathspace}{0ex}}dx$

The Euler-Lagrange equation for this functional, with $L=xf{f}^{\prime}$ is given by

$\begin{array}{rl}\frac{\mathrm{\partial}}{\mathrm{\partial}f}L(x,f,{f}^{\prime})-\frac{d}{dx}\frac{\mathrm{\partial}}{\mathrm{\partial}{f}^{\prime}}L(x,f,{f}^{\prime})& =\frac{\mathrm{\partial}}{\mathrm{\partial}f}(xf{f}^{\prime})-\frac{d}{dx}\frac{\mathrm{\partial}}{\mathrm{\partial}{f}^{\prime}}(xf{f}^{\prime})\\ \\ & =x{f}^{\prime}-\frac{d}{dx}(xf)\\ \\ & =-f\end{array}$

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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