A mimetic discretization of the macroscopic Maxwell equations in Hamiltonian form

Let $(\Omega,g)$ be a Riemannian manifold of dimension $n$. Throughout this paper, we let our manifold be the $n$-torus, $\Omega=\mathbb{T}^{n}$, as all considerations of boundaries are neglected. Let $\{(V^{k},\mathsf{d}_{k})\}_{k=0}^{n}$ be the vector spaces of differential forms on $\Omega$. That is, we define

$$V^{k}=H^{1}\Lambda^{k}(\Omega)=\left\{\eta^{k}\in L^{2}\Lambda^{k}(\Omega):\mathsf{d}_{k}\eta^{k}\in L^{2}\Lambda^{k+1}(\Omega)\right\}.$$

(1)

This sequence of vector spaces forms a Hilbert complex [2]. For a discussion of the properties of differential forms and definitions of the standard operations on differential forms, e.g. the wedge product and the Hodge star operator, see [22] [2]. We may construct a second complex, $\{(\tilde{V}^{k},\tilde{\mathsf{d}}_{k})\}_{k=0}^{n}$, called the complex of twisted forms. This complex is identical to the first in manner of definition, but differs in that twisted forms change sign under orientation changing transformations. These two complexes are related to each other through the Hodge star operator. Diagrammatically, this is given by

(2)

This structure is called the double de Rham complex. In most finite element treatments of exterior calculus, explicit reference to the complex of twisted forms is avoided. Rather, the codifferential operator, the formal $L^{2}$ adjoint of the exterior derivative, is introduced and those equations which are naturally expressed on the twisted complex are formulated weakly. However, in this work, we elect to explicitly consider both the twisted and straight forms in tandem.

The Hodge star is typically constructed in a local manner, however for the purposes of this paper it is more convenient to construct the Hodge star as a secondary structure that arises from two different notions of duality on the double de Rham complex. First, we define the $L^{2}$ inner product of $k$-forms $(\cdot,\cdot):V^{k}\times V^{k}\to\mathbb{R}$ as follows: if $g_{x}$ is the pointwise inner product on $k$-forms induced by the Riemannian metric and $\bm{\mathsf{{vol}}}$ is the volume form, then

$$(\omega^{k},\eta^{k})=\int_{\Omega}g_{x}(\omega^{k},\eta^{k})\bm{\mathsf{{vol}}}.$$

(3)

The second notion of duality is Poincaré duality $\langle\cdot,\cdot\rangle:V^{k}\times\tilde{V}^{n-k}\to\mathbb{R}$ which is defined

$$\left\langle\omega^{k},\tilde{\eta}^{n-k}\right\rangle=\int_{\Omega}\omega^{k}\wedge\tilde{\eta}^{n-k}.$$

(4)

The $L^{2}$ inner product is metric dependent while Poincaré duality is metric independent. This feature will be conserved at the discrete level. So, all quantities in a physical theory which are naturally metric free should be expressed in a manifestly metric free manner with a Poincaré duality structure and those which are metric dependent with $L^{2}$ duality. The Hodge star is defined to be the operator $\star:V^{k}\to\tilde{V}^{n-k}$ such that

$$\left(\omega^{k},\eta^{k}\right)=\left\langle\omega^{k},\star\eta^{k}\right\rangle.$$

(5)

The Hodge star is not a single operator, but rather a family of operators. A more precise notation might be $\star_{n-k,k}:V^{k}\to\tilde{V}^{n-k}$, however we opt for the more concise notation where confusion is unlikely.

As the construction of the finite element spaces on each complex is taken from [22], we shall be brief in our exposition including only enough detail to establish notation. Suppose that $\mathcal{T}_{h}$ is a complex of cells defining a discrete representation of the reference element, $[-1,1]^{n}$. It is defined by dividing $[-1,1]$ into subintervals defined by the nodes $-1=\xi_{0}<\xi_{2}<\cdots<\xi_{N}=1$, and taking tensor products of this $1$-dimensional grid. We let $\mathcal{C}_{0}$ represent vertices, $\mathcal{C}_{1}$ edges, $\mathcal{C}_{2}$ faces, and so on within this complex. We let $|\mathcal{C}_{k}|=N_{k}$. We interpret $\bm{\mathsf{c}}=(\mathsf{c}_{i})_{1\leq i\leq N_{k}}\in\mathcal{C}_{k}$ as a vector of numbers which associates numerical coefficients to the geometric entities in the complex. Let $\{V^{k}_{h}\}_{k=0}^{n}$ be the finite element spaces of differential forms as described in [22], such that $V^{k}_{h}\subset V^{k}$ and $\text{dim}(V^{k}_{h})=N_{k}$.

The decrees of freedom are denoted $\bm{\sigma}^{k}=(\sigma^{k}_{i})_{1\leq i\leq N_{k}}:V^{k}\to\mathcal{C}_{k}$. The degrees of freedom is a linear operator which associates a continuous differential $k$-form with coefficients on the chain complex. For $0$-forms, we accomplish this by evaluating pointwise at the vertices. For $1$-forms, we integrate over edges. For $2$-forms, we integrate over faces, and so on. These operators are sometimes called de Rham operators. Moreover, these may be thought of as finite element degrees of freedom. Interpolation is defined such that $\mathcal{I}^{k}=\left(\left.\bm{\sigma}^{k}\right|_{V_{h}^{k}}\right)^{-1}:\mathcal{C}_{k}\to V^{k}_{h}$. We denote the basis functions $\{\Lambda_{i}^{k}\}_{i=1}^{N_{k}}$ so that $\forall\bm{\mathsf{c}}=(\mathsf{c}_{i})_{1\leq i\leq N_{k}}\in\mathcal{C}_{k}$,

$$\mathcal{I}^{k}\bm{\mathsf{c}}=\sum_{i=1}^{N_{k}}\mathsf{c}_{i}\Lambda^{k}_{i}.$$

(6)

Finally, we define the projection operator $\Pi^{k}=\mathcal{I}^{k}\bm{\sigma}^{k}:V^{k}\to V^{k}_{h}$. Hence,

$$\Pi^{k}\phi=\sum_{i=1}^{N_{k}}\sigma^{k}_{i}(\phi)\Lambda^{k}_{i}.$$

(7)

We require $\sigma^{k}_{i}(\Pi^{k}\phi)=\sigma^{k}_{i}(\phi)$ so that the operator is indeed a projection. We assume that

$$\|\phi-\Pi^{k}\phi\|=O(h^{p+1})$$

(8)

where $p\in\mathbb{N}$ is the degree of polynomials used for interpolation. The essential features of the finite element de Rham complex may be summarized in the following commuting diagram:

(9)

That the diagram commutes means

$$\mathbbm{d}_{k}\bm{\sigma}^{k}=\bm{\sigma}^{k+1}\mathsf{d}_{k},\quad\mathsf{d}_{k}\mathcal{I}^{k}=\mathcal{I}^{k+1}\mathbbm{d}_{k},\quad\mathsf{d}_{k}\Pi^{k}=\Pi^{k+1}\mathsf{d}_{k}.$$

(10)

Note, the final expression follows from the first two:

$$\mathsf{d}_{k}\Pi^{k}=\mathsf{d}_{k}\mathcal{I}^{k}\bm{\sigma}^{k}=\mathcal{I}^{k+1}\mathbbm{d}_{k}\bm{\sigma}^{k}=\mathcal{I}^{k+1}\bm{\sigma}^{k+1}\mathsf{d}_{k}=\Pi^{k+1}\mathsf{d}_{k}.$$

(11)

We proceed in the same manner beginning from a dual grid, $\tilde{\mathcal{T}}_{n}$, defined over nodes, $\{\tilde{\xi}_{i}\}\subset[-1,1]$, which have been staggered with respect to those of the complex of straight forms. The dual complex is written

(12)

with all objects defined in a like manner to their analogs in the straight complex.

Just as the Hodge star couples together the twisted and straight complexes in the continuous setting, so too does a suitably defined discrete Hodge star operator in the discrete context. The construction of this discrete Hodge star operator proceeds analogously to that of the continuous Hodge star operator. The mass matrices associated with the $L^{2}$-pairing are:

$$\left(\mathbb{M}_{k}\right)_{ij}=\left(\Lambda^{k}_{i},\Lambda^{k}_{j}\right)\quad\text{and}\quad\left(\tilde{\mathbb{M}}_{k}\right)_{ij}=\left(\tilde{\Lambda}^{k}_{i},\tilde{\Lambda}^{k}_{j}\right).$$

(13)

The Poincaré duality structure is built from the wedge product: $\wedge:\Lambda^{k}\times\Lambda^{l}\to\Lambda^{k+l}$ (or $\wedge:\tilde{\Lambda}^{k}\times\tilde{\Lambda}^{l}\to\tilde{\Lambda}^{k+l}$). In the discrete setting the wedge product is often associated with the cup product from algebraic topology [30]. In general, one should exclusively form the wedge product of forms with like orientation: twisted with twisted and straight with straight. This is because, at the discrete level, twisted forms and straight forms are associated with distinct cell complexes. Hence, mixing these objects haphazardly is ill-advised. However, by considering a Galerkin representation of our discrete twisted and straight forms, we circumvent the usual difficulty coupling the two cell complexes to form a product of twisted and straight forms. That is, we define a family of mass matrices which discretely express Poincaré duality:

$$\left(\tilde{\mathbb{M}}_{n-k,k}\right)_{ij}=\left\langle\tilde{\Lambda}^{n-k}_{i},\Lambda^{k}_{j}\right\rangle\quad\text{and}\quad\left(\mathbb{M}_{n-k,k}\right)_{ij}=\left\langle\Lambda^{n-k}_{i},\tilde{\Lambda}^{k}_{j}\right\rangle$$

(14)

where $\{\Lambda^{k}_{i}\}_{i=1}^{N_{k}}$ are the basis functions for $V^{k}_{h}$ and $\{\tilde{\Lambda}^{n-k}_{i}\}_{i=1}^{\tilde{N}_{n-k}}$ are the basis functions for $\tilde{V}^{n-k}_{h}$. We shall frequently call this the Poincaré mass matrix.

We may interpret the $L^{2}$ mass matrix as arising from a dual degrees of freedom operator [19]. In duality to the primal bases $\{\Lambda^{k}_{i}\}_{i=1}^{N_{k}}\subset V^{k}$ and $\{\tilde{\Lambda}^{k}_{i}\}_{i=1}^{\tilde{N}_{k}}\subset\tilde{V}^{k}$, we define dual bases of $k$-forms $\{\Sigma_{i}^{k}\}_{i=1}^{N_{k}}\subset V^{k}$ and $\{\tilde{\Sigma}_{i}^{k}\}_{i=1}^{\tilde{N}_{k}}\subset\tilde{V}^{k}$ such that

$$\left(\Lambda_{i}^{k},\Sigma_{j}^{k}\right)=\delta_{ij}\quad\text{and}\quad\left(\tilde{\Lambda}_{i}^{k},\tilde{\Sigma}_{j}^{k}\right)=\delta_{ij}.$$

(15)

The corresponding dual degrees of freedom, $\bm{\mu}_{i}^{k}:V^{k}\to\mathcal{C}_{k}^{*}$ and $\tilde{\bm{\mu}}_{i}^{k}:\tilde{V}^{k}\to\tilde{\mathcal{C}}_{k}^{*}$, are defined such that

$$\bm{\mu}_{i}^{k}\left(\Sigma_{j}^{k}\right)=\delta_{ij}\quad\text{and}\quad\tilde{\bm{\mu}}_{i}^{k}\left(\tilde{\Sigma}_{j}^{k}\right)=\delta_{ij}.$$

(16)

Hence, it follows that

$$\bm{\mu}_{i}^{k}\left(v^{k}\right)=\left(\Lambda^{k}_{i},v^{k}\right)\quad\text{and}\quad\tilde{\bm{\mu}}_{i}^{k}\left(\tilde{u}^{k}\right)=\left(\tilde{\Lambda}^{k}_{i},\tilde{u}^{k}\right).$$

(17)

The dual degrees of freedom, when acting on $v^{k}_{h}\in V^{k}_{h}\subset V^{k}$ and $\tilde{u}^{k}_{h}\in\tilde{V}^{k}_{h}\subset\tilde{V}^{k}$, take the form

$$\bm{\mu}^{k}\left(v^{k}_{h}\right)=\left(\bm{\Lambda}^{k},v^{k}_{h}\right)=\mathbb{M}_{k}\bm{\mathsf{v}}^{k}\quad\text{and}\quad\tilde{\bm{\mu}}^{k}\left(\tilde{u}^{k}_{h}\right)=\left(\tilde{\bm{\Lambda}}^{k},\tilde{u}^{k}_{h}\right)=\tilde{\mathbb{M}}_{k}\tilde{\bm{\mathsf{u}}}^{k}.$$

(18)

Hence, we may interpret $\mathbb{M}_{k}:\mathcal{C}_{k}\to\mathcal{C}_{k}^{*}$ and $\tilde{\mathbb{M}}_{k}:\tilde{\mathcal{C}}_{k}\to\tilde{\mathcal{C}}_{k}^{*}$.

Likewise, we may interpret the Poincaré mass matrix as arising from a dual degrees of freedom operator. We define dual bases of $(n-k)$-forms $\{\tilde{\Sigma}_{i}^{n-k,k}\}_{i=1}^{N_{n-k}}\subset\tilde{V}^{k}$ and $\{\Sigma_{i}^{n-k,k}\}_{i=1}^{\tilde{N}_{n-k}}\subset V^{k}$ such that

$$\left\langle\Lambda_{i}^{n-k},\tilde{\Sigma}_{j}^{n-k,k}\right\rangle=\delta_{ij}\quad\text{and}\quad\left\langle\tilde{\Lambda}_{i}^{n-k},\Sigma_{j}^{n-k,k}\right\rangle=\delta_{ij}.$$

(19)

The corresponding dual degrees of freedom, $\tilde{\bm{\mu}}^{n-k,k}_{i}:\tilde{V}^{k}\to\mathcal{C}_{n-k}^{*}$ and $\bm{\mu}^{n-k,k}_{i}:V^{k}\to\tilde{\mathcal{C}}_{n-k}^{*}$, are defined such that

$$\tilde{\bm{\mu}}^{n-k,k}_{i}\left(\tilde{\Sigma}_{j}^{n-k,k}\right)=\delta_{ij}\quad\text{and}\quad\bm{\mu}^{n-k,k}_{i}\left(\Sigma_{j}^{n-k,k}\right)=\delta_{ij}.$$

(20)

Hence, it follows that

$$\tilde{\bm{\mu}}^{n-k,k}_{i}\left(\tilde{u}^{k}\right)=\left\langle\Lambda^{n-k}_{i},\tilde{u}^{k}\right\rangle\quad\text{and}\quad\bm{\mu}^{n-k,k}_{i}\left(v^{k}\right)=\left\langle\tilde{\Lambda}^{n-k}_{i},v^{k}\right\rangle.$$

(21)

The dual degrees of freedom, when acting on $\tilde{u}^{k}_{h}\in\tilde{V}^{k}_{h}\subset\tilde{V}^{k}$ and $v^{k}_{h}\in V^{k}_{h}\subset V^{k}$, take the form

	$\displaystyle\tilde{\bm{\mu}}^{n-k,k}\left(\tilde{u}^{k}_{h}\right)$	$\displaystyle=\left\langle\bm{\Lambda}^{n-k},\tilde{u}^{k}_{h}\right\rangle=\mathbb{M}_{n-k,k}\tilde{\bm{\mathsf{u}}}^{k}$		(22)
	$\displaystyle\bm{\mu}^{n-k,k}\left(v^{k}_{h}\right)$	$\displaystyle=\left\langle\tilde{\bm{\Lambda}}^{n-k},v^{k}_{h}\right\rangle=\tilde{\mathbb{M}}_{n-k,k}\bm{\mathsf{v}}^{k}.$

Hence, these operators act as a projection from the space of $k$-forms onto the degrees of freedom of the twisted $(n-k)$-forms (and vice versa). Hence, $\mathbb{M}_{n-k,k}:\tilde{\mathcal{C}}_{k}\to\mathcal{C}_{n-k}^{*}$ and $\tilde{\mathbb{M}}_{n-k,k}:\mathcal{C}_{k}\to\tilde{\mathcal{C}}_{n-k}^{*}$. Whereas the $L^{2}$ mass matrices are symmetric positive definite, and hence bijections, the Poincaré mass matrices are not square since in general $N_{k}\neq\tilde{N}_{n-k}$.

The following commuting diagram summarizes the projection from the twisted forms into the straight forms:

(23)

That this diagram commutes is an expression of the discrete integration by parts formula:

$$\mathbb{M}_{k,n-k}\tilde{\mathbbm{d}}_{n-(k+1)}=(-1)^{n-k}\mathbbm{d}_{k}^{T}\tilde{\mathbbm{M}}_{k+1,n-(k+1)}$$

(24)

which in turn follows from the integration by parts formula:

$$\left\langle\tilde{\mathsf{d}}_{n-(k+1)}\tilde{\phi}^{n-(k+1)},\psi^{k}\right\rangle=(-1)^{n-k}\left\langle\tilde{\phi}^{n-(k+1)},\mathsf{d}_{k}\psi^{k}\right\rangle.$$

(25)

A similar diagram describes projection from the straight forms into the twisted forms.

This motivates our definition of the discrete Hodge star operator. At the coefficient level, we define $\text{\char 73}_{k,n-k}:\tilde{\mathcal{C}}_{n-k}\to\mathcal{C}_{k}$ and $\tilde{\text{\char 73}}_{k,n-k}:\mathcal{C}_{n-k}\to\tilde{\mathcal{C}}_{k}$ as follows:

	$\displaystyle\bm{\mathsf{u}}^{k}=\text{\char 73}_{k,n-k}\tilde{\bm{\mathsf{u}}}^{n-k}=\mathbb{M}_{k}^{-1}\mathbb{M}_{k,n-k}\tilde{\bm{\mathsf{u}}}^{n-k}$		(26)
	$\displaystyle\tilde{\bm{\mathsf{v}}}^{k}=\tilde{\text{\char 73}}_{k,n-k}\bm{\mathsf{v}}^{n-k}=\tilde{\mathbb{M}}_{k}^{-1}\tilde{\mathbb{M}}_{k,n-k}\bm{\mathsf{v}}^{n-k}.$

At the level of the finite element spaces, we see that the discrete Hodge star operator is identical to that of the continuous Hodge star operator with the trial and test functions restricted to the appropriate finite element spaces:

	$\displaystyle u^{k}_{h}=\star_{h}\tilde{u}^{n-k}_{h}\iff\left\langle\eta^{k},\tilde{u}^{n-k}_{h}\right\rangle$	$\displaystyle=\left(\eta^{k},u^{k}_{h}\right)\quad\forall\eta^{k}\in V^{k}_{h}$		(27)
	$\displaystyle\tilde{v}^{k}_{h}=\tilde{\star}_{h}v^{n-k}_{h}\iff\left\langle\tilde{\chi}^{k},v^{n-k}_{h}\right\rangle$	$\displaystyle=\left(\tilde{\chi}^{k},\tilde{v}^{k}_{h}\right)\quad\forall\tilde{\chi}^{k}\in\tilde{V}^{k}_{h}$

where the notation $\star_{h}$ refers to one of many different discrete Hodge star operators depending on the context. Hence, we see that the discrete Hodge star operator is simply a Galerkin projection.

In the following, we shall show that the reduction and interpolation operators are approximately related through the adjoint operation.

Now, let $\psi\in V^{k}$ be arbitrary. Then the above tells us

$$\tilde{\bm{\mathsf{u}}}^{T}\tilde{\mathbb{M}}_{n-k,k}\bm{\sigma}_{k}(\psi)=\left\langle\tilde{\mathcal{I}}_{n-k}\tilde{\bm{\mathsf{u}}},\Pi_{k}\psi\right\rangle.$$

Hence,

$$\left\langle\bm{\sigma}_{k}^{*}(\tilde{\bm{\mathsf{u}}}),\psi\right\rangle-\left\langle\tilde{\mathcal{I}}_{n-k}\tilde{\bm{\mathsf{u}}},\psi\right\rangle=-\left\langle\tilde{\mathcal{I}}_{n-k}\tilde{\bm{\mathsf{u}}},(I-\Pi_{k})\psi\right\rangle$$

which implies

	$\displaystyle\left|\left\langle\bm{\sigma}_{k}^{*}(\tilde{\bm{\mathsf{u}}}),\psi\right\rangle-\left\langle\tilde{\mathcal{I}}_{n-k}\tilde{\bm{\mathsf{u}}},\psi\right\rangle\right|$	$\displaystyle=\left|\left\langle\tilde{\mathcal{I}}_{n-k}\tilde{\bm{\mathsf{u}}},(I-\Pi_{k})\psi\right\rangle\right|$
		$\displaystyle\leq\|I-\Pi_{k}\|\|\tilde{\mathcal{I}}_{n-k}\tilde{\bm{\mathsf{u}}}\|\|\psi\|$
		$\displaystyle\leq\|I-\Pi_{k}\|\|\tilde{\mathcal{I}}_{n-k}\|\|\tilde{\bm{\mathsf{u}}}\|\|\psi\|.$

Because $\|I-\Pi_{k}\|=O(h^{p+1})$ and $\tilde{\mathcal{I}}_{n-k}$, being an operator between finite dimensional spaces, is bounded, (29) follows. ∎

A similar result holds for the adjoint with respect to the $L^{2}$ inner product.

As we defined $\mathcal{I}^{k}=\left(\left.\bm{\sigma}^{k}\right|_{V_{h}^{k}}\right)^{-1}$, this implies that the degrees of freedom operator is approximately unitary.

The natural Hodge star operator is often used in the mimetic discretization literature [5]:

$$\tilde{\mathbb{H}}_{k}=\tilde{\bm{\sigma}}_{n-k}\star\mathcal{I}_{k}\quad\text{and}\quad\mathbb{H}_{k}=\bm{\sigma}_{n-k}\star\tilde{\mathcal{I}}_{k}.$$

(32)

Consider a functional $K:\Lambda^{k}\to\mathbb{R}$. We define two kinds of functional derivatives:

$$DK[\phi]\delta\phi=\left(\frac{\delta K}{\delta\phi},\delta\phi\right)=\left\langle\frac{\tilde{\delta}K}{\delta\phi},\delta\phi\right\rangle$$

(36)

where $DK[\phi]$ is the standard Fréchet derivative. Hence, $\delta K/\delta\phi\in\Lambda^{k}$ whereas $\tilde{\delta}K/\delta\phi\in\tilde{\Lambda}^{n-k}$. We call this the latter the twisted functional derivative. See [3] and [13] for more details regarding functional derivatives with respect to this duality pairing.

Suppose that $\bm{\mathsf{u}}$ only appears in $\mathsf{K}$ as $\tilde{\mathbb{M}}_{n-k,k}\bm{\mathsf{u}}$. Then

$$\frac{\partial\mathsf{K}}{\partial\bm{\mathsf{u}}}=\tilde{\mathbb{M}}_{n-k,k}^{T}\frac{\partial\mathsf{K}}{\partial\tilde{\mathbb{M}}_{n-k,k}\bm{\mathsf{u}}}.$$

(40)

Hence, it follows that in this case,

$$\frac{\tilde{\delta}\mathsf{K}}{\delta\bm{\mathsf{u}}}=\frac{\partial\mathsf{K}}{\partial\tilde{\mathbb{M}}_{n-k,k}\bm{\mathsf{u}}}=:\frac{\partial\mathsf{K}}{\partial\bm{\mathsf{u}}_{*}}$$

(41)

where $\bm{\mathsf{u}}_{*}=\tilde{\mathbb{M}}_{n-k,k}\bm{\mathsf{u}}\in\tilde{\mathcal{C}}_{n-k}^{*}$. Because $\bm{\mathsf{u}}_{*}\in\tilde{\mathcal{C}}_{n-k}^{*}$, it follows that $\partial\mathsf{K}/\partial\bm{\mathsf{u}}_{*}\in\tilde{\mathcal{C}}_{n-k}$. Note, we could instead alternatively shown that the discrete functional derivative can be obtained as follows:

$$\frac{\tilde{\delta}(K\circ\tilde{\mathcal{I}}_{n-k}\circ\tilde{\bm{\mu}}^{n-k,k})}{\delta u}=\tilde{\mathcal{I}}_{n-k}\frac{\partial\mathsf{K}}{\partial\bm{\mathsf{u}}_{*}}+O(h^{p+1})$$

(42)

where $\mathsf{K}=K\circ\tilde{\mathcal{I}}_{n-k}$ and we interpret $\tilde{\mathcal{I}}_{n-k}\circ\tilde{\bm{\mu}}^{n-k,k}$ as a kind of dual projection. This illuminates the need to define our discrete functional derivatives with respect to the dual variable. Whenever we write derivatives with respect to the variable $\bm{\mathsf{u}}_{*}$, we assume that the functional $\mathsf{K}$ is a functional of $\bm{\mathsf{u}}_{*}$ only.

Similarly, the discrete functional derivative with respect to the $L^{2}$ duality pairing is given by

$$\frac{\delta K\circ\Pi_{k}}{\delta u}=\mathcal{I}_{k}\frac{\partial\mathsf{K}}{\partial\mathbb{M}_{k,k}\bm{\mathsf{u}}}+O(h^{p+1})=\mathcal{I}_{k}\mathbb{M}_{k,k}^{-1}\frac{\partial\mathsf{K}}{\partial\bm{\mathsf{u}}}+O(h^{p+1}).$$

(43)

This is similar to the discrete twisted functional derivative, however we may write this discretized functional derivative with greater flexibility due to the invertibility of the $L^{2}$ mass matrix.

To fully discretize the functional derivative, we simply apply the degrees of freedom operator:

$$\tilde{\bm{\sigma}}_{n-k}\left(\frac{\tilde{\delta}K\circ\Pi_{k}}{\delta u}\right)=\frac{\partial\mathsf{K}}{\partial\bm{\mathsf{u}}_{*}}+O(h^{p+1})\quad\text{and}\quad\bm{\sigma}_{k}\left(\frac{\delta K\circ\Pi_{k}}{\delta u}\right)=\mathbb{M}_{k}^{-1}\frac{\partial\mathsf{K}}{\partial\bm{\mathsf{u}}}+O(h^{p+1})$$

(44)

since interpolation is the inverse of the degrees of freedom operator.

For the purposes of studying Casimir invariants of the discretized dynamics later, it is helpful to consider variational derivatives of functionals of the form $K[\phi]=\hat{K}[\mathsf{d}\phi]$. At the continuous level, we have

$$\left\langle\frac{\tilde{\delta}K}{\delta u},\delta u\right\rangle=\left\langle\frac{\tilde{\delta}\hat{K}}{\delta(\mathsf{d}u)},\mathsf{d}\delta u\right\rangle=(-1)^{n-k}\left\langle\mathsf{d}\frac{\tilde{\delta}\hat{K}}{\delta(\mathsf{d}u)},\delta u\right\rangle+\int_{\partial M}\left(\frac{\tilde{\delta}\hat{K}}{\delta(\mathsf{d}u)}\wedge\delta u\right).$$

(45)

Hence, under homogeneous boundary conditions or on a manifold without boundary,

$$\frac{\tilde{\delta}K}{\delta u}=(-1)^{n-k}\mathsf{d}\frac{\tilde{\delta}\hat{K}}{\delta(\mathsf{d}u)}.$$

(46)

Applying the previous results on discretizing functional derivatives, we find Let $\mathsf{K}=K\circ\mathcal{I}_{k}$ and $\hat{\mathsf{K}}=\hat{K}\circ\mathcal{I}_{k+1}$. Then, recalling that $\bm{\mathsf{u}}_{*}=\tilde{\mathbb{M}}_{n-k,k}\bm{\mathsf{u}}$ and $(\mathbbm{d}_{k}\bm{\mathsf{u}})_{*}=\tilde{\mathbb{M}}_{n-(k+1),k+1}\mathbbm{d}_{k}\bm{\mathsf{u}}$, we find

$$\frac{\partial\mathsf{K}}{\partial\bm{\mathsf{u}}_{*}}=(-1)^{n-k}\tilde{\mathbbm{d}}_{n-(k+1)}\frac{\partial\hat{\mathsf{K}}}{\partial(\mathbbm{d}_{k}\bm{\mathsf{u}})_{*}}.$$

(47)