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Igor Khavkine
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The idea of the weights used in the ADN definition of the principal symbol is quite natural in the context of graded vector spaces or more generally graded modules.

In the equation $u = Dv$, $u=(u_1,\ldots,u_n)$ and $v=(v_1,\ldots,v_n)$ are functions valued in vector spaces, say $U$ and $V$ respectively. A decomposition of $U = \bigoplus_a U_a$ and $V = \bigoplus_b V_b$ where $a$ and $b$ range over the integers $\mathbb{Z}$ endows these spaces with an integer grading. For simplicity, let us suppose that the components $(u_i)$ and $(v_j)$ are with respect to a basis that is compatible with the grading, which is the same as saying that each of the vector components has a well-defined integer degree. Since our vectors have only finitely many components, only finitely many of the $U_a$ or $V_b$ vector spaces will be non-trivial (have positive dimension). The punchline will be that the degree $a_i$ of the component $u_i$ could be identified with the ADN weight $s_i = a_i$, while the degree $b_j$ of the component $v_j$ could be identified with the ADN weight $-t_j = b_j$.

There is a notion of a homogeneous map between graded vector spaces. Namely $T \colon V \to U$ is homogeneous of degree $c$ if $T(V_b) \subseteq U_{a=b+c}$. Written as a matrix $T = (T_{ij})$, only those components $T_{ij}$ can be non-zero for which $a_i - b_j = c$. Of course, any linear map can be written as a sum of homogeneous pieces, $T = \sum_c T_{(c)}$. When the components of $T$ are not just numbers but polynomials in $\xi_l$, conventionally the variables $\xi_l$ are all given weight 1 and the notion of homogeneity changes slightly. Namely, $T$ is (polynomially) homogeneous of degree $c$ if its components $T_{ij}$ are homogeneous polynomials, say of degrees $d_{ij}$, and are non-vanishing only when $a_i - b_j - d_{ij} = c$. An arbitrary linear operator with polynomial entries with can again be expanded in homogeneous pieces, $T = \sum_{c} T_{[c]}(\xi)$, where each polynomially homogeneous pieces can again be expanded as $T_{[c]}(\xi) = \sum_{|\alpha|=c} T_{(c-|\alpha|)} \xi^\alpha$ with respect to the first notion of homogeneity. What I ad-hoc called polynomial homogeneity coincides with the notion of homogeneity when we consider $U$ and $V$ are graded modules over the polynomial ring $\mathbb{R}[\xi]$, where the $\xi_l$ variables all have weight 1.

Now, suppose that some integer grading is fixed on $U$ and $V$ and you have a linear operator between them with polynomial coefficients that has a homogeneous expansion $T = \sum_{c\le d} T_{[c]}(\xi)$, then $T_{[d]}(\xi)$ is its leading homogeneous term with respect to the grading. If you are free to adjust the grading then without loss of generality we can set $d=0$ by the shift $a_i \mapsto a_i + d$ or $b_i \mapsto b_i - d$. Now, getting back to a linear differential equation $u = Dv$, the components of the operator $D$ are polynomials in the variables $\xi_l = \partial/\partial x^l$. It will have a leading $\xi$-homogeneous term $D_{[d]}(\xi)$ of some degree $d$.

The conclusion is that the ADN weighted principal symbol $\sigma_{\text{pr},w}(\xi) = D_{[d]}$ is just the leading homogeneous term with respect to a grading given by the weights $a_i = s_i$, $b_j = -t_j$ and $d=0$. But as we have seen earlier, the $d=0$ condition is easily relaxed just by shifting the weights. What is interesting about the definition of the weighted principal symbol as the leading homogeneous term is that, for a fixed choice of weights, it is easy to see that it is well-defined and transforms covariantly in $\xi$ under changes of the $(x^l)$ coordinates (subleading terms in the transformation of highest order derivatives in each component of $D_{ij}$ only contribute to subleading terms $D_{[c]}$, while the highest derivatives transform covariantly).

The usual definition of the principal symbol can be recovered by setting all the weights to zero, $a_i = b_j = 0$. However, the simple example $$ \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} \Delta^3 & 0 & 0 \\ 0 & \Delta^2 & 0 \\ 0 & 0 & \Delta \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} $$ shows that such a default choice picks up only the highest order $\Delta^3$ term, making the principal symbol degenerate and hence, strictly speaking, non-elliptic. However, we can plainly see that the system is just a combination of three independent elliptic equations of different orders. Selecting the weights/gradings to be $b_j = 0$ and $a_1 = 6$, $a_2 = 4$ and $a_3 = 2$ allows the weighted principal symbol to pick up all the components (in this example there are no subleading terms) and become non-degenerate. The ADN definition of ellipticity simply generalizes this idea. The operator $D$ is ADN elliptic if there exists a choice of weights/gradings such that its weighted principal symbol (leading homogeneous term) is invertible for $\xi\ne 0$.

Finally, to the question whether there exists some automatic way of choosing the weights, in general I don't know. Certainly, it's possible that an operator is not elliptic with respect to any weights. Perhaps there is a way to decide when an operator can be made elliptic with respect to some choice of weights, but I don't know it of the top of my head. Possibly this question can be answered within commutative algebra and the theory of graded modules.

Igor Khavkine
  • 21.5k
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  • 60
  • 113