If we have a second order quasilinear PDE of the form

$A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+ lower\,\, order \,\, terms=0$

where $A,B,C$ are functions of $x,y,u$,

then the equation is called elliptic if $det=\begin{vmatrix}A &C \\C & B\end{vmatrix}>0$, parabolic if $det=0$ and hyperbolic if $det<0$.

Now what happens if we have a system of two coupled PDEs of the form

$A\frac{\partial^2 u}{\partial x^2}+B\frac{\partial^2 u}{\partial y^2}+2C\frac{\partial^2 u}{\partial x\partial y}+D\frac{\partial^2 v}{\partial x^2}+E\frac{\partial^2 v}{\partial y^2}+2F\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0 \\ G\frac{\partial^2 u}{\partial x^2}+H\frac{\partial^2 u}{\partial y^2}+2K\frac{\partial^2 u}{\partial x\partial y}+L\frac{\partial^2 v}{\partial x^2}+M\frac{\partial^2 v}{\partial y^2}+2N\frac{\partial^2 v}{\partial x\partial y}+ lower\,\, order \,\, terms=0$

with $A,B,C,...$ being functions of $x,y,u,v$.

Does it make sense to construct the determinant

$det=\begin{vmatrix}A &C & D&F\\C & B&F&E\\G &K & L&N\\K &H & N&M\end{vmatrix}$

and investigate its sign, or something like that?

--------- Update ----------

In the chapter 10 of the book [Seiler, Werner M., Involution. The formal theory of differential equations and its applications in computer algebra] the author states that since ellipticity is a property defined at points and depends only on the principal symbol it then suffices to study it for linear equations. I understand that the same procedure holds for quasilinear ones, the only difference is that the coefficients are allowed to be functions of the dependent variables. Then regarding the specific example above if one implement the theory for linear equations the principal symbol should be

$\tau[\xi,\eta]=\begin{bmatrix}B \eta^2 + 2 C \eta\xi + A \xi^2& E\eta^2 + 2 F \eta\xi + D \xi^2\\ H\eta^2 + 2 K \eta\xi + G \xi^2 & M\eta^2 + 2 N \eta\xi + L \xi^2\end{bmatrix}$

Ellipticity requires $det(\tau)\neq 0$ that is

$(B M-E H) \eta^4 + 2 (C M + B N-F H - E K ) \eta^3 \xi + (B L + A M + 4 C N-E G - D H - 4 F K ) \eta^2 \xi^2 + 2 ( C L + A N-F G - D K ) \eta \xi^3 + (A L-D G ) \xi^4\neq 0$

well, when this condition is to be satisfied? Is it true that it is hyperbolic in direction $x$ or $y$ direction, if

$(B M-E H) \eta^4 + 2 (C M + B N-F H - E K ) \eta^3 \xi + (B L + A M + 4 C N-E G - D H - 4 F K ) \eta^2 \xi^2 + 2 ( C L + A N-F G - D K ) \eta \xi^3 + (A L-D G ) \xi^4 = 0$

possesses only simple real roots?

  • You are looking for the theory of the symbol of a system of partial differential equations. It is more complicated than what you have written. – Ben McKay Jul 7 at 9:21
  • @BenMcKay Can you please provide more details? – DK13 Jul 7 at 9:31
  • @BenMcKay: It seems to me that the OP's differential operator (of order $2$) acts on sections in a vector bundle of rank $2$, in which case ellipticity has a well-known definition. I do not know about hyperbolicity and parabolicity, but there is a nice answer on MSE. – Alex M. Jul 7 at 14:34
  • @AlexM.The OP did not ask about ellipticity specifically, but about the analogue determinants that one might compute. – Ben McKay Jul 7 at 14:52
  • For the hyperbolicity condition, your definition is basically the second order version of "strictly hyperbolic". For the ellipticity condition, see my updated answer. – Igor Khavkine Jul 9 at 8:57
up vote 3 down vote accepted

Consider a determined linear system of differential order $k$ of the form $\sigma^{i_1\cdots i_k}_{ab}(x) \partial_{i_1} \cdots \partial_{i_k} u^b(x) + l.o.t = 0$. The coefficients $\sigma^{i_1\cdots i_k}_{ab}$, a square matrix in the $ab$ indices, constitute its principal symbol. By replacing $\partial_i$ by an indeterminate $p_i$ (like in the Fourier transform), the symbol defines a square matrix valued function on the cotangent space at $x$, $\sigma_{ab}(x,p) = \sigma^{i_1\cdots i_k}_{ab}(x) p_{i_1} \cdots p_{i_k}$, also referred to as the symbol. The locus $\mathcal{C}(x)$ of all $p$ for which the symbol fails to be an invertible matrix is the characteristic variety of the PDE at $x$.

Further classification is based on the geometry of $\mathcal{C}(x)$, which is usually assumed to behave somewhat uniformly with respect to $x$. If $\mathcal{C}(x) = \{0\}$ then the equation is called elliptic. If $\mathcal{C}(x)$ looks like a cone which has compact intersection with some affine codimension-1 hyperplane (a "spacelike" hyperplane) then the equation is considered in a rather weak sense to be hyperbolic (the geometric optics approximation could be used to construct waves traveling at finite speed). The simplest situation is when this intersection is smooth and is the boundary of some bounded convex set (e.g., a sphere). In general, the intersection can be multi-sheeted, non-convex, self-intersecting, etc. There is a gradation of notions of hyperbolicity corresponding to these variations.

In the usual case of a single scalar equation in two variables, the sign of the $A$-$B$-$C$-determinant can be used to determine the geometry of the characteristic variety. In principle, one could try to do something similar in higher dimensions, based on some more sophisticated invariants of the principal symbol coefficients $\sigma^{i_1\cdots i_k}_{ab}(x)$. But this will obviously become increasingly difficult in higher dimensions of dependent and independent variables.

Not much else can be said at this level of generality. But if you're interested in this level of generality, you'll find more information for instance in

Seiler, Werner M., Involution. The formal theory of differential equations and its applications in computer algebra, Algorithms and Computation in Mathematics 24. Berlin: Springer (ISBN 978-3-642-01286-0/hbk; 978-3-642-01287-7/ebook). xxii, 650 p. (2010). ZBL1205.35003.


Update: The following answers the updated question and the idea is to show how cumbersome it is to check the ellipticity condition for generic coefficients.

To check the ellipticity condition, you basically want to know whether the determinant $\tau(\xi,\eta)$ has the same sign everywhere outside $(\xi,\eta)\ne(0,0)$, lets say positive. You are hugely in luck! Because, according to a theorem of Hilbert (cf Hilbert's 17th problem and positive semidefinite polynomials), the condition $\tau(\xi,\eta)\ge 0$ for a homogeneous polynomial in two variables is is one of the few cases when it is equivalent to the existence of a sum of squares representation $\tau(\xi,\eta) = \sum_i q_i(\xi,\eta)^2$. If a sum of squares representation of $\tau(\xi,\eta) = V\xi^4 + 2W\xi^3\eta + X\xi^2\eta^2 + 2Y\xi\eta^3 + Z\eta^4$ exists, then it must come from the diagonalization of a quadratic form.

Namely, there is a 1-parameter family of representations of $\tau(\xi,\eta)$ as a quadratic form $$ \tau(\xi,\eta) = \begin{pmatrix} \xi^2 & \xi\eta & \eta^2 \end{pmatrix} \begin{pmatrix} V & W & -\lambda \\ W & X+2\lambda & Y \\ -\lambda & Y & Z \end{pmatrix} \begin{pmatrix} \xi^2 \\ \xi\eta \\ \eta^2 \end{pmatrix} , $$ with $\lambda$ the only redundancy in this representation. So $\tau(\xi,\eta)$ is a sum of squares if and only if the above form is positive semidefinite for some value of $\lambda$. By Sylvester's criterion, the necessary and sufficient condition is the non-negativity of all of its principal minors: $$\begin{aligned} 0 &\le V , \\ 0 &\le X+2\lambda , \\ 0 &\le Z , \\ 0 &\le VX-W^2+2V\lambda , \\ 0 &\le VZ - \lambda^2 , \\ 0 &\le XZ-Y^2 +2Z\lambda , \\ 0 &\le V X Z-W^2 Z - V Y^2 + 2 (V Z - 2 W Y) \lambda -X \lambda^2 -2 \lambda^3 . \end{aligned}$$

If one wants, each of the above polynomial inequalities on $\lambda$ may be converted into an interval bound on $\lambda$, with the end points of the intervals explicitly expressed in terms of the coefficients of $\tau$, but the presence of quadratic and cubic polynomials will make these formulas rather cumbersome, involving several special cases depending on the discriminants of these polynomials.

Of course, once you succeed in writing $\tau(\xi,\eta) = \sum_i q_i(\xi,\eta)^2$, you also have to check that the $q_i(\xi,\eta)$ have no common roots except for $(\xi,\eta)=(0,0)$.

  • Thank you @IgorKhavkine, but what about Douglis-Nirenberg ellipticity? see onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160080406 – DK13 Jul 7 at 14:33
  • I'm afraid that the Douglis-Nirenberg approach works only for linear PDEs, not for quasilinear – DK13 Jul 7 at 14:58
  • @Dimitri, Douglis-Nirenberg slightly generalize what one means by the principal symbol, by allowing the "highest differential order" to vary depending on which component of the PDE system you are considering. Otherwise, the classification is the same. I'm sorry, I didn't notice that you referred to quasi-linear, not just linear, equations in your question. But not much changes in the definition $\mathcal{C}(x,u)$, except that it is allowed to depend on $u$ (possibly up to $\partial^{k-1} u$). But even if the same definitions apply, there might not be nice theorems to go along with them. – Igor Khavkine Jul 7 at 15:25
  • 1
    I have no doubt that quasi linear Douglis Nirenberg systems can be studied using the same tools as standard quasi linear elliptic systems. It’s just that someone needs to work through the details. Perhaps someone already has, but I don’t know where. – Deane Yang Jul 7 at 16:11
  • 1
    @DK13, the classification of quasi linear systems is the same. The only difference is that, since the coefficients depend on the unknown functions, the type might, too. – Deane Yang Jul 7 at 18:06

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