Can variational formulations be solved using series solutions?

What I specifically mean is, given some functional $F\left[\mathbf{x}\right]$ which is stationary with respect to $\dot{\mathbf{x}}=f(\mathbf{x})$ and some boundary or initial conditions, can one choose: $$\mathbf{x}(t)=\mathbf{x}_0+\mathbf{x}_1t+\mathbf{x}_2t^2+\dots$$ And substitute this into the functional (or it's first variation) and some how solve for the coefficients?

How would the variations be expressed in this case, would they also be: $$\delta\mathbf{x}(t)=\delta\mathbf{x}_0+\delta\mathbf{x}_1t+\delta\mathbf{x}_2t^2+\dots$$ And then I would collect with respect to them if I was substituting into the first variation?

• Sounds like a global Galerkin method, to me, assuming the same spaces are used. – Bill Barth Mar 12 '15 at 21:19
• @BillBarth: I looked that up and it does seem to fit. This would just be a kind of version of global Galerkin for powers of $t$. Do you know of any good sources on this? I saw something about element-free Galerkin as well, so that might be related. – Ron Mar 12 '15 at 21:28
• There are a whole host of methods that try to not have elements but they all have different features. Mesh-free methods tend to use overlapping functions with compact support at arbitrary locations. Element-free appears to use a moving LS interpolant. Global (polynomial) methods use fully-global functions and lead to dense systems and tend to have bad condition numbers. Global spectral methods often use trig functions which give better conditioning but are still dense unless you find a globally orthogonal basis for your operator (which you usually won't be able to do). – Bill Barth Mar 12 '15 at 21:39
• @BillBarth: Yes, seems like using trig functions would be better, as polynomials have some undesirable characteristics inherently. I was just curious about whether or not the coefficients would could get closer and closer to the Taylor series coefficients. – Ron Mar 12 '15 at 21:48