7
$\begingroup$

I tried asking a similar question in SE.Physics, and I got some information regarding the abstract side of this, but I figured I should post here to get more complete information about the numerical benefits of variational formalisms.

Assuming I am able derive a functional representation for any dynamical system (dissipative, nonlinear, fractional, PDE, ODE, discontinuous, etc), why would such a result or capability be useful? What are some practical consequences?

In essence, what are functionals/variational formalisms used for in practice?

Keep in mind, the functionals I'm referring to might be of non-standard form (i.e, non-conservative systems), for example:

Any system with that is a potential or conservative will have a functional representation as: $$ F[\mathbf{x}]=\int^{t}_0\left(\frac{1}{2}m\dot{\mathbf{x}}(\tau)^2-V(\mathbf{x}(\tau))\right)\,\text{d}\tau $$

Where 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).

But, you can construct other functionals for non-conservative systems, such as this convolutional 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: $$ \frac{d\mathbf{x}(t)}{dt}= \mathbf{Ax}(t) $$

Point being that one should also consider the implications of functionals which are not inner product based.

I know that these functionals are useful in the context of FEM (but how?), but I'm also interested in whether or not they may be useful in other contexts (parameter estimation, simulation, data assimilation, etc).

$\endgroup$
  • $\begingroup$ I'm not sure of understanding your question. Do you want some applications of these formulations, so you understand their importance? $\endgroup$ – nicoguaro Mar 4 '15 at 15:16
  • $\begingroup$ @nicoguaro: that's exactly what I'm asking, yes. Also, why can these functional relationships be used liked this? $\endgroup$ – Ron Mar 4 '15 at 15:25
  • $\begingroup$ Since I don't understand your question completely, let me know if the answer is what you wanted. $\endgroup$ – nicoguaro Mar 5 '15 at 14:32
  • $\begingroup$ @nicoguaro: Your answer is well thought out and very good! I actually just wanted some time to read it and properly digest it first before accepting it, hope that's OK! You cited some sources and made some points I wanted to carefully understand. One of the other things I was wondering about is the relationship between data or parameters with respect to these functionals, but I think I might ask a different question about those specifically. $\endgroup$ – Ron Mar 5 '15 at 15:04
  • $\begingroup$ It is correct. I'm not pushing you to accept my answer, I'm just asking if it was clear. I tried to be clear... But was not as easy as I expected in the first place. So, if you want me to elaborate in some specific point, just comment it. $\endgroup$ – nicoguaro Mar 5 '15 at 15:56
7
$\begingroup$

Let us start with FEM. We have a differential equation of the form

$$\mathcal{A}u=f\quad \forall x \text{ in } \Omega \enspace ,$$

where $\mathcal{A}$ is a linear operator, and this problem is subject to some boundary conditions

$$\mathcal{H}u=g(x)\quad \forall x \text{ in } \partial\Omega \enspace .$$

We are interested in finding an approximate solution to the PDE, $\hat{u}$ (let's assume that this approximate solution satisfy BCs), if we plug it in the first equation we do not get an equality. The difference

$$R = \mathcal{A}\hat{u} - f$$

is the residual, and it is the zero function for the solution. We would like to make this $R$ as close to zero as possible, for this we can try to minimize the maximum norm, or the two norm of $R$ (the latter leads to Least Squares Finite Elements).

One way to go is to compute an average over the domain, i.e., compute the weighted residual

$$\int\limits_{\Omega} wR\, d\Omega = \int\limits_{\Omega} w\mathcal{A}\hat{u}\, d\Omega - \int\limits_{\Omega} wf\, d\Omega \enspace .$$

For positive-definite operators we can write $\mathcal{A} = \mathcal{T}^*\mathcal{T}$ ($\mathcal{T}^*$ is the adjoint), thus

$$\int\limits_{\Omega} wR\, d\Omega = \int\limits_{\Omega} \mathcal{T}^*w\mathcal{T}\hat{u}\, d\Omega - \int\limits_{\Omega} wf\, d\Omega \enspace ,$$

this last step is the Green's theorem. Depending on the constraints that you impose on your functions $w$ and $u$ you can have different version of the FEM: Galerkin, Petrov-Galerkin, Bubnov-Galerkin. The most common case is the Galerkin FEM, when the two functions are the same and you end up with a variational formulation of your differential equation.

In the case of a (static) membrane, we can compute the Lagrangian for the system as $$\Pi[u] = \int\limits_\Omega (\nabla u)^2\, d\Omega - \int\limits_\Omega uf\, d\Omega$$ the solution is then given for the $u$ that makes $$\delta \Pi[u] = 0$$ and for a given set of approximate solutions $\hat{u}$ (a base) we can find the best one in the same way.

Other methods where we can find these functionals are Hartree-Fock Methods and Density functional theory (DFT). In these methods one changes the Schrödinger equation for a variational formulation where the electron-electron interaction is replaced for the interaction of the electrons with a mean field. Something similar is done here, you propose a set of functions (a base) and then find the coefficients for the combinations of your functions.

In the case of DFT, the functional looks like $$E[n] = T[n] + U[n] + \int\limits_{\mathbb{R}^3} V(\mathbf{r})n(\mathbf{r})\, d^3\mathbf{r}$$ $n$ is the density function (unknown), $T$ is the kinetic energy functional, $U$ is the potential energy functional and the last term is the one for the interaction with the mean field. In this case you should solve in a self-consistent way because the function $n$ is not known before hand (what makes the problem non-linear).

$\endgroup$
  • $\begingroup$ This is a great explanation of FEM! I don't have much experience with the QM aspect of your answer, but it is interesting to know as well. Thank's for the great answer! $\endgroup$ – Ron Mar 6 '15 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.