Step-wise finite element formulations: can this be done?

Question

Given the functional: $$ F[\mathbf{x}]=\frac{1}{2}[\mathbf{x}^{\text{T}} * D(\mathbf{x})]-\frac{1}{2}[\mathbf{x}^{\text{T}} * \mathbf{Ax}]-\frac{1}{2}\mathbf{x}^{\text{T}}(0)\mathbf{x}(t) $$

Where $\mathbf{A}$ is symmetric, $\mathbf{x}(0)$ being the initial condition, $\mathbf{x}(t)$ is continuous, and: $$ [\mathbf{f}^{\text{T}} * \mathbf{g}]=\int^{t}_0 \mathbf{f}^{\text{T}}(t-\tau)\mathbf{g}(\tau)\,\text{d}\tau $$ and: $$ D\left(\cdot\right)=\frac{\text{d}}{\text{d}t}\left(\cdot\right) $$

We take the variation as: $$ \delta F\left[\mathbf{x}\right]=\left.\frac{\partial}{\partial \varepsilon} F\left[\mathbf{x}+\varepsilon\delta\mathbf{x}\right]\right|_{\varepsilon=0} $$ Functional at the varied function is then: $$ F\left[\mathbf{x}+\varepsilon\delta\mathbf{x}\right]=\frac{1}{2}[\mathbf{x}^{\text{T}}+\varepsilon\delta\mathbf{x}^{\text{T}} * D(\mathbf{x}+\varepsilon\delta\mathbf{x})]-\frac{1}{2}[\mathbf{x}^{\text{T}}+\varepsilon\delta\mathbf{x}^{\text{T}} * \mathbf{A}\left(\mathbf{x}+\varepsilon\delta\mathbf{x}\right)]-\frac{1}{2}\left(\mathbf{x}^\text{T}(0)+\varepsilon\delta\mathbf{x}^\text{T}(0)\right)\left(\mathbf{x}(t)+\varepsilon\delta\mathbf{x}(t)\right) $$ The individual terms being: $$ \frac{1}{2}[\mathbf{x}^{\text{T}}+\varepsilon\delta\mathbf{x}^{\text{T}} * \mathbf{A}\left(\mathbf{x}+\varepsilon\delta\mathbf{x}\right)]=\frac{1}{2}[\mathbf{x}^{\text{T}}* D(\mathbf{x})]+\frac{1}{2}\varepsilon[\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})]+\frac{1}{2}\varepsilon[\delta\mathbf{x}^{\text{T}}* D(\mathbf{x})]+\frac{1}{2}\varepsilon^2[\delta\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})] $$ $$ \frac{1}{2}[\mathbf{x}^{\text{T}}+\varepsilon\delta\mathbf{x}^{\text{T}} * \mathbf{A}\left(\mathbf{x}+\varepsilon\delta\mathbf{x}\right)]=\frac{1}{2}[\mathbf{x}^{\text{T}} * \mathbf{A}\mathbf{x}]+\frac{1}{2}\varepsilon[\mathbf{x}^{\text{T}}* \mathbf{A}\delta\mathbf{x}]+\frac{1}{2}\varepsilon[\delta\mathbf{x}^{\text{T}} * \mathbf{A}\mathbf{x}]+\frac{1}{2}\varepsilon^2[\delta\mathbf{x}^{\text{T}} * \mathbf{A}\delta\mathbf{x}] $$ $$ -\frac{1}{2}\left(\mathbf{x}^\text{T}(0)+\varepsilon\delta\mathbf{x}^\text{T}(0)\right)\left(\mathbf{x}(t)+\varepsilon\delta\mathbf{x}(t))\right)=-\frac{1}{2}\mathbf{x}^\text{T}(0)\mathbf{x}(t)-\frac{1}{2}\varepsilon\mathbf{x}^\text{T}(0)\delta\mathbf{x}(t)-\frac{1}{2}\varepsilon\delta\mathbf{x}^\text{T}(0)\mathbf{x}(t)-\frac{1}{2}\varepsilon^2\delta\mathbf{x}^\text{T}(0)\delta\mathbf{x}(t) $$ We can differentiate these with respect to $\varepsilon$ and then set it to zero, adding them up leads to: $$ \delta F\left[\mathbf{x}\right]=\frac{1}{2}[\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})]+\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* D(\mathbf{x})]-\frac{1}{2}[\mathbf{x}^{\text{T}}* \mathbf{A}\delta\mathbf{x}]-\frac{1}{2}[\delta\mathbf{x}^{\text{T}} * \mathbf{A}\mathbf{x}]-\frac{1}{2}\mathbf{x}^\text{T}(0)\delta\mathbf{x}(t)-\frac{1}{2}\delta\mathbf{x}^\text{T}(0)\mathbf{x}(t) $$ We also can say: $$ \frac{1}{2}[\mathbf{x}^{\text{T}}* \mathbf{A}\delta\mathbf{x}]=\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* \mathbf{A}^{\text{T}}\mathbf{x}] $$ Since $\mathbf{A}$ is symmetric, this just means that: $$ \frac{1}{2}[\mathbf{x}^{\text{T}}* \mathbf{A}\delta\mathbf{x}]=\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* \mathbf{A}\mathbf{x}] $$ Substituting this back into the varied functional yields: $$ \delta F\left[\mathbf{x}\right]=\frac{1}{2}[\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})]+\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* D(\mathbf{x})]-\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* \mathbf{A}\mathbf{x}]-\frac{1}{2}[\delta\mathbf{x}^{\text{T}} * \mathbf{A}\mathbf{x}]-\frac{1}{2}\mathbf{x}^\text{T}(0)\delta\mathbf{x}(t)-\frac{1}{2}\delta\mathbf{x}^\text{T}(0)\mathbf{x}(t) $$ Notice, now we have two similar terms which can be grouped, doing so yields: $$ \delta F\left[\mathbf{x}\right]=\frac{1}{2}[\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})]+\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* D(\mathbf{x})]-[\delta\mathbf{x}^{\text{T}}* \mathbf{A}\mathbf{x}]-\frac{1}{2}\mathbf{x}^\text{T}(0)\delta\mathbf{x}(t)-\frac{1}{2}\delta\mathbf{x}^\text{T}(0)\mathbf{x}(t) $$ We can lump this into the other term which a variation: $$ \delta F\left[\mathbf{x}\right]=\frac{1}{2}[\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})]+\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* \left(D(\mathbf{x})-2\mathbf{A}\mathbf{x}\right)]-\frac{1}{2}\mathbf{x}^\text{T}(0)\delta\mathbf{x}(t)-\frac{1}{2}\delta\mathbf{x}^\text{T}(0)\mathbf{x}(t) $$ Now, if we say that $\delta\mathbf{x}(0)=0$, we get: $$ \delta F\left[\mathbf{x}\right]=\frac{1}{2}[\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})]+\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* \left(D(\mathbf{x})-2\mathbf{A}\mathbf{x}\right)]-\frac{1}{2}\mathbf{x}^\text{T}(0)\delta\mathbf{x}(t) $$ This is the form I would use for discretizing, but we can show which system it is stationary with respect to. Taking integration by parts on the first term yields: $$ \frac{1}{2}[\mathbf{x}^{\text{T}}* D(\delta\mathbf{x})]=\frac{1}{2}\mathbf{x}^{\text{T}}(0)\delta\mathbf{x}(t)-\frac{1}{2}\mathbf{x}^{\text{T}}(t)\delta\mathbf{x}(0)+\frac{1}{2}[\delta\mathbf{x}^{\text{T}}* D(\mathbf{x})] $$ Substituting this back in yields (with $\delta\mathbf{x}(0)=0$): $$ \delta F\left[\mathbf{x}\right]=[\delta\mathbf{x}^{\text{T}}* \left(D(\mathbf{x})-\mathbf{A}\mathbf{x}\right)] $$ Which implies: $$ D(\mathbf{x})-\mathbf{A}\mathbf{x}=0 $$ Or: $$ \frac{d\mathbf{x}(t)}{dt}= \mathbf{Ax}(t) $$

Again, the first variation of this functional (with no IBP) is:

$$ \delta F[\mathbf{x}]=\frac{1}{2}[\mathbf{x}^{\text{T}} * D(\delta\mathbf{x})]+\frac{1}{2}[\delta\mathbf{x}^{\text{T}} * \left(D(\mathbf{x})-2\mathbf{Ax}\right)]-\frac{1}{2}\mathbf{x}^{\text{T}}(0)\delta\mathbf{x}(t) $$

I now discretize this over the interval $\left[0,\Delta\right]$, using the approximation: $$ \mathbf{x}(t)=\mathbf{x}_0 N_0(t)+\mathbf{x}_1 N_1(t) $$ $$ \delta\mathbf{x}(t)=\delta\mathbf{x}_0 N_0(t)+\delta\mathbf{x}_1 N_1(t) $$ Where: $$ N_0(t)=1-\frac{t}{\Delta} $$ And $$ N_1(t)=\frac{t}{\Delta} $$

The result is: $$ \delta F[\mathbf{x}] \approx \delta\mathbf{x}_1^{\text{T}}\left(\left(\frac{1}{2}\mathbf{I}-\frac{\Delta}{6}\mathbf{A}\right)\mathbf{x}_1-\frac{\Delta}{3}\mathbf{A}\mathbf{x}_0-\frac{1}{2}\mathbf{x}(0)\right) $$

Given that $\delta\mathbf{x}_1$ is some arbitrary discrete variation, we can say that it's coefficient must be zero: $$ \left(\frac{1}{2}\mathbf{I}-\frac{\Delta}{6}\mathbf{A}\right)\mathbf{x}_1-\frac{\Delta}{3}\mathbf{A}\mathbf{x}_0-\frac{1}{2}\mathbf{x}(0)=0 $$

My question is, is the following also true?:

$$ \left(\frac{1}{2}\mathbf{I}-\frac{\Delta}{6}\mathbf{A}\right)\mathbf{x}_{i+1}-\frac{\Delta}{3}\mathbf{A}\mathbf{x}_i-\frac{1}{2}\mathbf{x}(0)=0 $$

Answer 1

(too many comments, so I'll continue here)

I assume you're referring to equation (4.40) in http://www.dic.units.it/perspage/tonti/DEPOSITO/Initial.pdf. In this case the answer is yes, it is possible: Due to the specific integral formulation based on convolution, the first variation of (4.40) leads to the initial value problem $$ x'(t) = Ax(t)$$ with $x(0)$ given. You can then discretize this equation with a standard Galerkin method; the one you are trying to get is equivalent to a Crank-Nicolson time stepping method.

In principle, you can also go the other way around as you are proposing: Discretize the functional, and then take discrete variations. Depending on the discretisation, you end up with different time stepping methods, but often there's one choice that leads to a standard time stepping method such as Crank-Nicolson. Whether that is possible in this case, I'm not sure -- the Crank-Nicolson method is equivalent to a Petrov-Galerkin method (i.e., trial and test functions come from different spaces; continuous piecewise linear and discontinuous piecewise constant, in this case), which means there can be no variational formulation (since the test functions come from variations of the trial functions, which means they have to live in the same space).

As regards your specific question: No, this is unlikely to be true, since the initial condition shouldn't appear there. To get the next time step, you need to explicitly compute variations with respect to $\delta x_2$ (and so forth).

(Note that in the definition of your functional, $t$ should really be $T$ (the end time), otherwise the functional makes no sense. (This means that $f^T*g$ is not really the convolution of $f$ and $g$ (which would be a function), but a related bilinear form (and hence a real number).) That means you need to be careful when taking variations not to mix up $t$ and $T$. Also, the $x(0)$ in your formulation is a fixed initial condition, better denoted by $x_0$. Since $x(0)=x_0$ is fixed, there can be no variation at $t=0$, i.e., $\delta x(0) = 0$.)