已知 $E(r)=\int_{0}^{2\pi}(\alpha\sqrt{r^{2}(\theta)+r'^{2}(\theta)}+\beta r(\theta))\,d\theta+\eta S^{-\nu}(r),$其中
\[
S(r)=\frac{1}{2}\int_{0}^{2\pi}r^{2}(\theta)\,d\theta,
\]
且 $\alpha,\beta,\eta,\nu>0,$ $r\in C^{m+2}(2\pi),$ $r>0.$ 求 $E$
关于 $r$ 的 Frechet 导数$.$

记
\begin{eqnarray*}
E(r) & = & E_{1}(r)+E_{2}(r)+E_{3}(r)\\
& := & \alpha\int_{0}^{2\pi}\sqrt{r^{2}(\theta)+r'^{2}(\theta)}\,d\theta+\beta\int_{0}^{2\pi}r(\theta)\,d\theta+\eta S^{-\nu}(r),
\end{eqnarray*}
那么对于 $h,r\in C^{m+2}(2\pi),$ 有
\begin{eqnarray*}
E_{1}'(r)h & = & \lim_{\varepsilon\rightarrow0}\frac{E_{1}(r+\varepsilon h)-E_{1}(r)}{\varepsilon}\\
& = & \alpha\lim_{\varepsilon\rightarrow0}\frac{\int_{0}^{2\pi}\sqrt{(r+\varepsilon h)^{2}+(r'+\varepsilon h'){}^{2}}\,d\theta-\int_{0}^{2\pi}\sqrt{r^{2}+r'^{2}}\,d\theta}{\varepsilon}\\
& = & \alpha\int_{0}^{2\pi}\lim_{\varepsilon\rightarrow0}\frac{\sqrt{(r^{2}+r'^{2})+2\varepsilon(rh+r'h')+\varepsilon^{2}(h^{2}+h'^{2})}-\sqrt{r^{2}+r'^{2}}}{\varepsilon}\,d\theta\\
& = & \alpha\int_{0}^{2\pi}\frac{rh+r'h'}{\sqrt{r^{2}+r'^{2}}}\,d\theta\\
& = & \alpha\int_{0}^{2\pi}\frac{rh}{\sqrt{r^{2}+r'^{2}}}\,d\theta+\alpha\int_{0}^{2\pi}\frac{r'h'}{\sqrt{r^{2}+r'^{2}}}\,d\theta\\
& = & \alpha\int_{0}^{2\pi}\frac{rh}{\sqrt{r^{2}+r'^{2}}}\,d\theta+\left[\alpha\frac{r'h}{\sqrt{r^{2}+r'^{2}}}\right]_{\theta=0}^{\theta=2\pi}-\alpha\int_{0}^{2\pi}\frac{r''h}{\sqrt{r^{2}+r'^{2}}}\,d\theta\\
& = & \alpha\int_{0}^{2\pi}\frac{r-r''}{\sqrt{r^{2}+r'^{2}}}h\,d\theta\\
& = & \alpha\int_{0}^{2\pi}\frac{(r-r'')(r^{2}+r'^{2})}{(r^{2}+r'^{2})^{3/2}}h\,d\theta\\
& = & \alpha\int_{0}^{2\pi}\frac{r^{3}+rr'^{2}-r^{2}r''-r''r'^{2}}{(r^{2}+r'^{2})^{3/2}}h\,d\theta\\
& = & \alpha\int_{0}^{2\pi}\frac{r^{3}+2rr'^{2}-r^{2}r''}{(r^{2}+r'^{2})^{3/2}}h\,d\theta,
\end{eqnarray*}
其中最后一步用到了奇函数的特性$,$ 即
\[
\int_{0}^{2\pi}\frac{rr'^{2}+r''r'^{2}}{(r^{2}+r'^{2})^{3/2}}h\,d\theta=0,
\]
而 $E_{2}(r)$ 关于 $r$ 是线性的$,$ 所以立即有
\[
E_{2}'(r)h=E_{2}(h)=\beta\int_{0}^{2\pi}h(\theta)\,d\theta,
\]
最后$,$
\begin{eqnarray*}
E_{3}'(r)h & = & \lim_{\varepsilon\rightarrow0}\frac{E_{3}(r+\varepsilon h)-E_{3}(r)}{\varepsilon}\\
& = & \eta\lim_{\varepsilon\rightarrow0}\frac{\left(\frac{1}{2}\int_{0}^{2\pi}(r+\varepsilon h)^{2}\,d\theta\right)^{-\nu}-\left(\frac{1}{2}\int_{0}^{2\pi}r^{2}\,d\theta\right)^{-\nu}}{\varepsilon}\\
& = & -\nu\eta\lim_{\varepsilon\rightarrow0}\left(\frac{1}{2}\int_{0}^{2\pi}(r+\varepsilon h)^{2}\,d\theta\right)^{-\nu-1}\frac{d}{d\varepsilon}\left[\frac{1}{2}\int_{0}^{2\pi}(r+\varepsilon h)^{2}\,d\theta\right]\\
& = & -\nu\eta\lim_{\varepsilon\rightarrow0}\left(\frac{1}{2}\int_{0}^{2\pi}(r+\varepsilon h)^{2}\,d\theta\right)^{-\nu-1}\int_{0}^{2\pi}(r+\varepsilon h)h\,d\theta\\
& = & -\nu\eta\left(\frac{1}{2}\int_{0}^{2\pi}r^{2}\,d\theta\right)^{-\nu-1}\int_{0}^{2\pi}rh\,d\theta\\
& = & -\frac{\nu\eta}{S^{\nu+1}(r)}\int_{0}^{2\pi}rh\,d\theta.
\end{eqnarray*}
综上$,$ 由 $E'(r)h=E_{1}'(r)h+E_{2}'(r)h+E_{3}'(r)h$ 得所需结果$.$