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?


1 Answer 1


The classification of linear PDEs with constant coefficients is based on the principle that any linear transformation of coordinates should preserve the classification. That is, we define $$ \xi = ax+by+c, \quad \eta = dx+ey+f. $$ where $ae \neq bd$ so that lines of constant $\xi$ and $\eta$ are not parallel. Substituting in this change of coordinates, you will find a new system of PDEs in $\xi, \eta$ which have a new set of coefficients. A classification of the system should be the same whether the coefficients of the $x, y$ or $\xi, \eta$ system are used. However, the classes of your system will be more numerous than those of the classical second order PDE.

With the general approach out of the way, I suspect that your system will be classified by the sign of the following 4 determinants: $$ \left|\begin{array} {ccc} A & C \\ C & B \end{array} \right| $$ $$ \left|\begin{array} {ccc} D & F \\ F & E \end{array} \right| $$ $$ \left|\begin{array} {ccc} G & K \\ K & H \end{array} \right| $$ $$ \left|\begin{array} {ccc} L & N \\ N & M \end{array} \right| $$ and should be classified by four words, based on the sign of each. You should verify that the signs of these determinants are invariant under the change of coordinates.

Edit: For the case where the coefficients are not constant, we use exactly the same classification system pointwise. Defining the change of coordinates $$ \xi = \xi(x, y), \quad \eta = \eta(x, y) $$ which has a nonzero and finite Jacobian at all points, then local to any point $(x_0,y_0)$ we have that $$ \xi(x, y) \sim \xi(x_0, y_0) + (x-x_0) \cdot \frac{\partial \xi} {\partial x}(x_0, y_0) + (y-y_0) \cdot \frac{\partial \xi} {\partial y}(x_0, y_0) $$ $$ \eta(x, y) \sim \eta(x_0, y_0) + (x-x_0) \cdot \frac{\partial \eta} {\partial x}(x_0, y_0) + (y-y_0) \cdot \frac{\partial \eta} {\partial y}(x_0, y_0) $$ which gives us the local linear transform around each point. The coefficients at each point are locally given by there value at each point e.g. $$ A(x, y, u,v) \sim A(x_0, y_0, u(x_0, y_0),v(x_0, y_0)). $$ Thus, from an understand of the linear constant coefficient case, the quasi-linear variable coefficient case can be investigated in a pointwise manner.

  • $\begingroup$ the coefficients A,B,C... are not constants but functions of the indepependent and dependent variables $\endgroup$
    – SpaceChild
    Jul 6, 2018 at 12:36
  • $\begingroup$ Edited for quasi linear case $\endgroup$
    – Eddy
    Jul 6, 2018 at 12:51
  • $\begingroup$ and how sure can we be that the classification depends on the sign of the four determinants you mention above? $\endgroup$
    – SpaceChild
    Jul 6, 2018 at 13:02
  • $\begingroup$ Like I said, investigate the linear system yourself and check that those determinants have the same sign irrespective of the linear transform used, but I'm almost certain that will be the case. There may be other expressions that are invariant under the transformation, and independent of those given, this I'm less certain of. $\endgroup$
    – Eddy
    Jul 6, 2018 at 13:05
  • $\begingroup$ Specifically, linear combinations of your two equations produce new equations of the same form. You may want to investigate the simultaneous transformation of coordinates and linear combination of the two equations, and what functions of the coefficients are invariant under this combined transformation. $\endgroup$
    – Eddy
    Jul 6, 2018 at 13:09

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