How to calculate this Frechet derivative?

Question

$F:C^2[0,1] \rightarrow R $ be defined by $F(u)=\int_{0}^{1}\sqrt{1+(u')^2}dx $. Find the Frechet derivative of F.

I have no idea how to solve this.

Answer 1

Given that $F$ has Fréchet derivative, you can compute directional derivatives (Gâteaux derivatives) instead:

$$ (DF)_u (\eta) = \lim_{\epsilon \to 0} \frac{F(u+\epsilon\eta) - F(u)}{\epsilon}. $$

The result is that

\begin{align*} \lim_{\epsilon \to 0} \frac{F(u+\epsilon\eta) - F(u)}{\epsilon} &= \left. \frac{\partial}{\partial\epsilon}\right|_{\epsilon=0} \int_{0}^{1} \sqrt{1+(u' + \epsilon\eta')^2} \, dx \\ &= \int_{0}^{1} \left. \frac{\partial}{\partial\epsilon}\right|_{\epsilon=0} \sqrt{1+(u' + \epsilon\eta')^2} \, dx \\ &= \int_{0}^{1} \frac{u'\eta'}{\sqrt{1+u'^2}} \, dx \end{align*}

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If it is required to establish Fréchet-differentiability, notice that for $\|\eta\| = \|\eta\|_{C^2([0,1])} \ll 1$, Taylor's theorem gives

\begin{align*} F(u+\eta) &= \int_{0}^{1} \sqrt{1+(u'+\eta')^2} \, dx \\ &= \int_{0}^{1} \sqrt{1+(u')^2}\left( 1 + \frac{2u'\eta' + (\eta')^2}{1+(u')^2} \right)^{1/2} \, dx \\ &= \int_{0}^{1} \sqrt{1+(u')^2}\left( 1 + \frac{u'\eta'}{1+(u')^2} + \mathcal{O}\left(\|\eta\|^2\right) \right) \, dx \\ &= F(u) + \int_{0}^{1} \frac{u'\eta'}{\sqrt{1+(u')^2}} \, dx + \mathcal{O}\left(\|\eta\|^2\right) \end{align*}

with implicit bounds depending only on $\|u\|$. This shows that $(DF)_u$ exists and is given by a linear functional

$$(DF)_u \ : \ \eta \mapsto \int_{0}^{1} \frac{u'\eta'}{\sqrt{1+(u')^2}} \, dx. $$