Intuitive meaning of variational extremum

69 Views Asked by At

Suppose we have a functional $F$ for which the first variation vanishes for some function $f(\vec{x})$. Suppose further that the quadratic form (matrix) associated to the second variation, evaluated at $f$ has only, say, positive eigenvalues.

We can then assume, if I understand correctly, that $f$ is a (weak) local minimum of $F$ in function space.

I'm a little confused conceptually about how parameters of $f$ can affect this conclusion (or how strong it is). For instance, is it still possible $f$ has some fixed parameter (i.e. one that was not optimized over or considered in taking the variational derivatives) that could (say) make $F$ arbitrarily large at $f$? In other words what do these conclusions mean for classes of parameterized functions, intuitively, i.e. are they somehow constrained by being extrema?