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The Gateaux derivative

The derivative for functionals: how much does the energy $\Pi$ change when we nudge the whole displacement function $u$ a small step $\theta$ in the direction $v$ ? Expand $\Pi(u + \theta v)$ and take the term proportional to $\theta$ – that is $\delta\Pi(u, v)$ .

\delta\Pi(u, v) = \lim_{\theta\to0}\frac{\Pi(u + \theta v) - \Pi(u)}{\theta}

definition

\delta\Pi(u, v) = \int_0^L EA\,u'v'\,dx - \int_0^L F_x\,v\,dx - P\,v(L)

for the bar

Symbols

$u$	the point (displacement) we differentiate at
$v$	the direction (variation, test function), $v(0) = 0$
$\theta$	small number

Example

$\Pi(u) = \int_0^1(\tfrac12(u')^2 - u)\,dx$ .

The term with $\theta$ in $\Pi(u + \theta v)$ is $\theta\int_0^1(u'v' - v)\,dx$ .

So $\delta\Pi(u, v) = \int_0^1(u'v' - v)\,dx$ .

Shortcut:

\delta\Pi(u, v) = \frac{d}{d\theta}\Pi(u + \theta v)

\theta = 0

. Differentiate under the integral sign.

Practise weak form and shape functions for free →

← Boundary conditions and Ku = f · The weak form for the bar →

Part of Finite Element Method: Weak form and shape functions.