This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.

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## 1 The dual equation with inflow

In the general cell complex case the material is still contained in $n$-dimensional cells (we assume that they form an $n$-dimensional combinatorial manifold) and each cell exchanges the material with its neighbors through its boundary that consists of its $(n-1)$-dimensional faces. The amount of material in a cell $U$ is a $(n,0)$-form, an $n$-form with respect to location (in the $n$-dimensional space) and a $0$-form with respect to time.

An alternative (dual) approach is to think of the nodes, i.e., $0$-cells, as the containers instead of the $n$-cells. The fluid is then exchanges via the pipes, i.e., edges or $1$-cells, that connect the nodes.

Let's derive the equation in this *dual setting*.

The *amount of material* $M$ is a dual $(1,0)$-form, a $1$-form with respect to location (on the grid) and a $0$-form with respect to time.

The *outflow* $F$ is a dual $(1,1)$-form: the amount of flow across an $(n-1)$-face (from the $n$-cell to its neighbor) per unit of time.

We will also include this time the *inflow* $S$ of the material to a given cell from the outside of the system which is given by a dual $(1,1)$-form. It can be written simply as $S=Pdt$, where $P$ is an $1$-form with respect to space.

Then the preservation of the material in cell $\sigma=A^\star$, where $A$ is the dual vertex, is given by $$d_t M(A,t) = −\int_{\partial A^\star} \star F(·,t) + S(A,t).$$

By the Stokes Theorem, we rewrite: $$d_t M(A,t) = −(d_x \star F)(A^\star,t) + S(A,t).$$ And, based on $\phi (\sigma ^\star)=\phi ^\star(\sigma)$, we re-write: $$d_t M = − \star d_x \star F + S.$$

Now, the flow $F$ through pipe $b$ is proportional to the difference of amounts of material at the end-points of $b$. So,
$$F(b,t) = - k(b,t)d_x M(b,t).$$
Here $k$ is a $1$-form that may represent the *thickness of the pipe*. Substitute:
$$d_t M= \star d_x \star kd_x M + S.$$

**Exercise.** Derive the primary equation with inflow.

## 2 Diffusion

It is insufficient to have the two metric tensors, for a $K$ and $K^\star$: $$\begin{array}{ccc} \langle\cdot,\cdot\rangle(\cdot): T_2(K) \to R,\\ \langle\cdot,\cdot\rangle(\cdot): T_2(K^\star) \to R, \end{array}$$ as they only give the angles between cells of the same complex. Instead of dealing with the angle itself, we assume that the inner product of any $(n-1)$-cell and its dual is given to us: $$\langle a^\star, a \rangle .$$

It is also insufficient to have the Hodge star operators, as defined previously, provided for the primary and dual complexes: $$\begin{array}{ccc} \star^k : C^k(K)\to C^{n-k}(K^\star),\\ \star^k : C^k(K^\star)\to C^{n-k}(K), \end{array}$$ as they only keep record of the volumes of the cells.

We realize now that the formula for the Hodge star operator (the ratio of the dual and primal volume) that we have used is only applicable to rectangular grids. The star operator is still just a diagonal matrix; however, it will have a new interpretation.

The necessary generalization is below.

**Definition.** The *Hodge star operator* is a linear operator
$$\star^k : C^k(K)\to C^{n-k}(K^\star),\ k=0,1,...,n,$$
the matrix of which is diagonal whose $i$th entry is the ratio of the dual volume and the projection of the primal volume of the $i$th cell. In other words, it is
$$\star ^k_{ii}:=\frac{\langle a_i^\star, a_i \rangle }{|a_i|^2},$$
where $\dim a_i =k$.

**Exercise.** Find the Hodge star operator for the regular triangular grid.

**Exercise.** What is the geometry of a complex the Hodge star of which is the identity?