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If you form the Euler-Lagrange equation for some calculus of variations problem in x, f(x) and f'(x), and the resulting expression is always non-negative over the domain of x (because the expression is independent of f(x) and f'(x)), does that mean any perturbation that increases f(x) increases the functional? This seems natural given it should be some generalisation of what one gets when the first derivative of a function is always non-negative...

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1 Answer 1

Given a functional $L(x,f(x),{f(x)}')$, the Euler-Lagrange equation provides the solutions which make $L$ stationary (not extremum). It is given by:

$\frac{\partial L(x,f(x),{f(x)}')}{\partial f(x)}-\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial L(x,f(x),{f(x)}')}{\partial {f(x)}'}=0$

The resulting expression is thus always zero. If $L$ is independent of $f(x), {f(x)}'$, then the derivatives in the equation vanish and you have the identity $0=0$. Maybe you should clarify a bit more your question.

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