Although this questions is very much math related, I posted it in Physics since it is related to variational (Lagrangian/Hamiltonian) principles for dynamical systems. If I should migrate this elsewhere, please tell me.
Often times, in graduate and undergraduate courses, we are told that we can only formulate the Lagrangian (and Hamiltonian) for "potential" systems, where in the dynamics satisfy the condition that: $$ m\ddot{\mathbf{x}}=-\nabla V $$ If this is true, we can formulate a functional which is stationary with respect to the system as: $$ F[\mathbf{x}]=\int^{t}_0\left(\frac{1}{2}m\dot{\mathbf{x}}(\tau)^2-V(\mathbf{x}(\tau))\right)\,\text{d}\tau $$
Taking the first variation of this functional yields the dynamics of the system, along with a condition that effectively states that the initial configuration should be similar to the final configuration (variation at the boundaries is zero).
Now, given the functional: $$ F[\mathbf{x}]=\frac{1}{2}[\mathbf{x}^{\text{T}} * D(\mathbf{x})]+\frac{1}{2}[\mathbf{x}^{\text{T}} * \mathbf{Ax}]-\frac{1}{2}\mathbf{x}'(0)\mathbf{x}(t) $$ With $\mathbf{A}$ symmetric and $\mathbf{x}(0)$ being the initial condition, and: $$ [\mathbf{f}^{\text{T}} * \mathbf{g}]=\int^{t}_0 \mathbf{f}^{\text{T}}(t-\tau)\mathbf{g}(\tau)\,\text{d}\tau $$
If we take the first variation and assume only that the initial variation is zero, the functional is stationary with respect to: $$ \frac{d\mathbf{x}(t)}{dt}= \mathbf{Ax}(t) $$
This is a functional derived by Tonti and Gurtin, it represents a variational principle for linear initial value problems with symmetric state matrices and shows, as a proof of concept, that functionals can be derived for non-potential systems, initial value or dissipative systems.
My question is, is it possible to derive these functionals for arbitrary nonlinear systems which do not have similar initial and final configurations (and cannot have similar initial and final configurations due to dissipation)?
What sorts of conditions would exists on the dynamics of these systems?
In this example, $\mathbf{A}$ must be symmetric which already implies all of it's eigenvalues are real and thus it is a non-potential system, but there is still a functional which can be derived for it.
Any related sources, information, or answers regarding specific cases would be appreciated. If anyone needs clarification, or a proof of any result I presented here, let me know.
Edit: Also, a related question anyone seeing this: I'm currently just interested in the abstract aspect of the problem (solving/investigating it for the sake of it), but why are functional representations such as these useful? I know there are some numerical application, but if I have a functional which attains a minimum for a certain system, what can I do with it?
Answer
Comments to the question (v3):
I) The Gurtin-Tonti bi-local method [which OP mentions in an example; see also Section II below] of pairing opposite times $t\leftrightarrow (t_f-t_i)-t$ (hidden inside a convolution) is an artificial trick from a fundamental physics point of view, unless further justified. Why would such correlations into the past/future take place?
In fact, it may have non-local quantum mechanical consequences if such non-local action is supposed to be used in a path integral formalism.
Also the Gurtin-Tonti convolution method does not work for a non-compact time interval $[t_i,t_f]$, i.e. if $t_i=-\infty$ or $t_f=\infty$.
Most fundamental physics models typically obey locality, but there are various non-local proposals on the market.
The question of whether a certain set of equations of motions $E_i(t)$ has an action principle (or not!) can be very difficult to answer, and is often an active research area, cf. e.g. this Phys.SE post.
Also what constitutes an acceptable action principle? E.g. can we just introduce some Lagrange multipliers $\lambda^{i}(t)$ and an action $S=\int\! dt ~\lambda^i(t) E_i(t)$ so that $\delta S/ \delta\lambda^i(t) = E_i(t)$, and call it a day? Or are we not allowed to introduce auxiliary variables or non-locality? Should it satisfy a minimum principle rather than a stationary principle? And so forth.
II) Example. Let us for simplicity consider the unit time interval $[t_i,t_f]=[0,1]$. A symmetrized version of the Gurtin-Tonti model is the following bi-local action
$$S[q]~:=~ \frac{1}{4}\iint_{[0,1]^2} \!dt~du~\left\{ q^i(t) \left(\frac{dq^i(u)}{du}- A_{ij}(t,u) q^j(u)\right)+(t\leftrightarrow u) \right\}\delta(t+u-1) $$ $$~=~\frac{1}{2}\int_{[0,1]} \!dt~\left\{\frac{1}{2} q^i(1\!-\!t) \frac{dq^i(t)}{dt}-\frac{1}{2}q^i(t) \frac{dq^i(1\!-\!t)}{dt}- q^i(1\!-\!t)A_{ij}(1-t,t) q^j(t) \right\} $$ $$~=~\frac{1}{2}\int_{[0,1]} \!dt~\left\{ q^i(1\!-\!t) \frac{dq^i(t)}{dt}- q^i(1\!-\!t)A_{ij}(1-t,t) q^j(t) \right\} \tag{1}$$
with symmetric matrix
$$\tag{2} A_{ij}(t,u) ~=~A_{ji}(u,t) .$$
Interestingly, the boundary contributions in the variation $\delta S$ cancel without imposing any boundary conditions (BC). In other words, as far as the finding stationary solutions, we may assume that the variables $q^i$ are free at both end points. (However, there might be other reasons to impose BCs.)
The functional derivative
$$\tag{3} \frac{\delta S[q]}{\delta q^i(t)}~=~\left.\left\{\frac{dq^i(u)}{du}- A_{ij}(t,u) q^j(u)\right\}\right|_{u=1-t}. $$
Hence the equations of motion become
$$\tag{4} \frac{dq^i(t)}{dt}~\approx~A_{ij}(1\!-\!t,t) q^j(t). $$
References:
- V. Berdichevsky, Variational Principles of Continuum Mechanics: I. Fundamentals, 2009; Appendix B.
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