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I want to take a Gateaux differential of a functional

$$I(u)=\int_{\Omega} \left[\frac{1}{2}(\frac{du}{dx})^2+\frac{1}{2}u^2\right]d{\Omega}$$

so the Gateaux differential is defined as follow: $$D_hI(u)=[\frac{d}{dw}I(u+hw)]_{w=0}$$ My attempt: \begin{align} D_hI(u)&=\frac{d}{dw}\int_{\Omega} \left[\frac{1}{2}(\frac{du+hw}{dx})^2+\frac{1}{2}(u+hw)^2\right]d{\Omega}\\ &=\frac{d}{dw}\int_{\Omega} \left[\frac{1}{2}((\frac{du}{dx})^2+2*\frac{du}{dx}(\frac{dhw}{dx})+(\frac{dwh}{dx})^2)+\frac{1}{2}(u^2+2uhw+hw^2)\right]d{\Omega}\\ &= \int_{\Omega}\frac{d}{dw} \left[\frac{1}{2}((\frac{du}{dx})^2+2*\frac{du}{dx}(w\frac{dh}{dx})+(w\frac{dh}{dx})^2)+\frac{1}{2}(u^2+2uhw+hw^2)\right]d{\Omega}\\ &=\int_{\Omega}\left[\frac{1}{2}(2*\frac{du}{dx}(\frac{dh}{dx})+2w(\frac{dh}{dx})^2)+\frac{1}{2}(2uh+2hw)\right]d{\Omega} \\ &=\int_{\Omega}\left[\frac{du}{dx}(\frac{dh}{dx})+uh\right]d{\Omega} . \end{align}

Does in the last integral $(\frac{dh}{dx}) = 0?$

Arctic Char
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John G.
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  • No, not in general. What is the domain of $I$, or, put differently, do you have any boundary conditions? – MaoWao Jun 30 '20 at 13:27

1 Answers1

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From the last integral one normally does an integration by parts: $$ \int_\Omega \left[ \frac{du}{dx}(\frac{dh}{dx}) + uh \right] d\Omega = \int_\Omega \frac{du}{dx}(\frac{dh}{dx}) \, d\Omega + \int_\Omega uh \, d\Omega \\ = \int_\Omega \frac{d}{dx}\left[\frac{du}{dx} h\right] \, d\Omega - \int_\Omega \frac{d^2u}{dx^2} h \, d\Omega + \int_\Omega uh \, d\Omega \\ = \oint_{\partial\Omega} \frac{du}{dx} h \, d(\partial\Omega) - \int_\Omega \frac{d^2u}{dx^2} h \, d\Omega + \int_\Omega uh \, d\Omega \\ = \int_\Omega \left[ u - \frac{d^2u}{dx^2} \right] h \, d\Omega $$ if $\oint_{\partial\Omega} \frac{du}{dx} h \, d(\partial\Omega) = 0$ which is the case if we for example do not vary $u$ on the boundary of $\Omega$ so that $h\equiv 0$ on $\partial\Omega.$

md2perpe
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