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- Homework Statement
- Let ## a<b, A ## and ## B ## be constants.

Find the Gateaux differential of the functional ## S[y]=\int_{a}^{b}(y'^{2}+\omega^{2}y^{2}+2yx^{4})dx, y(a)=A, y(b)=B ##, where ## \omega ## is a positive constant.

- Relevant Equations
- Suppose ## X ## and ## Y ## are locally convex topological vector spaces, ## U\subseteq X ## is open, and ## F: X\rightarrow Y ##. The Gateaux differential ## dF(u; \psi) ## of ## F ## at ## u\in U ## in the direction ## \psi\in X ## is defined as ## dF(u; \psi)=\lim_{\tau\rightarrow 0}\frac{F(u+\tau\psi)-F(u)}{\tau}=\frac{d}{d\tau}F(u+\tau\psi)\biggr\rvert_{\tau=0} ##. If the limit exists for all ## \psi\in X ##, then one says that ## F ## is Gateaux differentiable at ## u ##.

I am not sure if this is correct, but here is my work by using the definition of the Gateaux differential:

\begin{align*}

&dS(y; \psi)=\lim_{\tau\rightarrow 0}\frac{S(y+\tau\psi)-S(y)}{\tau}=\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}\\

&S(y+\tau\psi)=\int_{a}^{b}[(y+\tau\psi)'^{2}+\omega^{2}(y+\tau\psi)^{2}+2(y+\tau\psi)x^{4}]dx\\

&=[(y+\tau\psi)'^{2}x+\omega^{2}(y+\tau\psi)^{2}x+\frac{2}{5}(y+\tau\psi)x^{5}]_{a}^{b}\\

&=(y+\tau\psi)'^{2}(b-a)+\omega^{2}(y+\tau\psi)^{2}(b-a)+[\frac{2}{5}(y+\tau\psi)](b^{5}-a^{5})\\

\end{align*}

Thus ## S(y+\tau\psi)\biggr\rvert_{\tau=0}=y'^{2}(b-a)+\omega^{2}y^{2}(b-a)+\frac{2}{5}y(b^{5}-a^{5}) ##.

So ## \frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}=0 ##.

Can anyone please check/verify/confirm this?

\begin{align*}

&dS(y; \psi)=\lim_{\tau\rightarrow 0}\frac{S(y+\tau\psi)-S(y)}{\tau}=\frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}\\

&S(y+\tau\psi)=\int_{a}^{b}[(y+\tau\psi)'^{2}+\omega^{2}(y+\tau\psi)^{2}+2(y+\tau\psi)x^{4}]dx\\

&=[(y+\tau\psi)'^{2}x+\omega^{2}(y+\tau\psi)^{2}x+\frac{2}{5}(y+\tau\psi)x^{5}]_{a}^{b}\\

&=(y+\tau\psi)'^{2}(b-a)+\omega^{2}(y+\tau\psi)^{2}(b-a)+[\frac{2}{5}(y+\tau\psi)](b^{5}-a^{5})\\

\end{align*}

Thus ## S(y+\tau\psi)\biggr\rvert_{\tau=0}=y'^{2}(b-a)+\omega^{2}y^{2}(b-a)+\frac{2}{5}y(b^{5}-a^{5}) ##.

So ## \frac{d}{d\tau}S(y+\tau\psi)\biggr\rvert_{\tau=0}=0 ##.

Can anyone please check/verify/confirm this?