This question is strictly related to this one. Let us consider the differential system with constant coefficients

$$\left(\begin{array}{ccc} B_{11} & B_{12} & 0\\ -B_{11} & 0 & 0\\ 0 & -B_{12} & 0 \end{array} \right)\left(\begin{array}{c} \partial_{z_1}a_1\\ \partial_{z_1}a_2\\ \partial_{z_1}a_3\\ \end{array}\right)+ \left(\begin{array}{ccc} 0 & 0 & -B_{23}\\ 0 & B_{22} & B_{23}\\ 0 & -B_{22} & 0 \end{array} \right)\left(\begin{array}{c} \partial_{z_2}a_1\\ \partial_{z_2}a_2\\ \partial_{z_2}a_3\\ \end{array}\right)+ \left(\begin{array}{ccc} 0 & 0 & -B_{33}\\ -B_{31} & 0 & 0\\ B_{31} & 0 & B_{33} \end{array} \right)\left(\begin{array}{c} \partial_{z_3}a_1\\ \partial_{z_3}a_2\\ \partial_{z_3}a_3\\ \end{array}\right)= \left(\begin{array}{c} b_1\\ b_2\\ -b_1-b_2 \end{array}\right) $$

and add this time the boundary conditions $a_i=b_i$, where we mean $a_3=-b_1-b_2$, on the hypersurface $z_1+z_2+z_3=0$. I know that the differential system has a solution (e.g. Hormander an introduction to Complex Analysis in several complex variables, ch. 7), however the additional boundary data modifies things a lot.

Is the Cauchy problem still well posed? If not, What conditions are to be added?

Is it still true that if the $b_i$ are smooth around a point $x$, then we can find $a_1,a_2,a_3$ smooth in a possibly smaller neighorhood of $x$?

What is the regulrity we can expect and what is the dependence on the initial data?

The difficulty I'm encoutering is because I am not able to let my situation fall in a classical one (i.e. either elliptic or hyperbolic) so I would like to know standard references to treat such overdetermined Cauchy Problems.

Many thanks.


  • $\begingroup$ If you are willing to work in the real analytic category, you could apply the Cartan-Kaehler theorem, if the initial data surface is not characteristic. $\endgroup$
    – Ben McKay
    Jan 7, 2015 at 14:25
  • $\begingroup$ @BenMcKay What about the smooth category? $\endgroup$ Jan 7, 2015 at 14:30
  • $\begingroup$ You might take a look at Deane Yang's book Involutive Hyperbolic Differential Systems, which proves a version of the Cartan-Kaehler theorem in the smooth category. (I haven't yet even looked at your pdes, so I don't know if this would apply.) $\endgroup$
    – Ben McKay
    Jan 7, 2015 at 14:51
  • $\begingroup$ Your problem is very concrete. You can get much more out of writing down explicit formulas first and proving theorems about them later, than the other way around. In particular, you can take the Fourier transform in the directions parallel to the initial value surface and solve the problem as an (overdetermined) ODE, where most of the conceptual difficulties are absent. $\endgroup$ Jan 7, 2015 at 15:14
  • $\begingroup$ I can post an answer assuming that $z_1, z_2, z_3$ are real. But are you assuming that? $\endgroup$
    – Deane Yang
    Jan 7, 2015 at 18:42

1 Answer 1


Here is a second attempt to answer the question:

There are three equations for three unknown functions. As Igor points out, if you add the three equations together, you get $0 = 0$. Therefore, the third equation is a consequence of the first two. It therefore suffices to solve the first two equations only. This is now an underdetermined system, so there is a lot of freedom in what to do. Basically, you can set one function to anything you want and solve for the other two.

For certain values of the coefficients, the system could still be degenerate. If so, then there is another compatibility condition that must be satisfied by $b_1$ and $b_2$ and, if this condition is assumed to hold, the system reduces in a simiilar fashion to a single first order equation, which is essentially an ODE and therefore has a well-posed Cauchy problem.

If the system is nondegenerate, then you look for a constant linear transformation of the unknown functions $a_1, a_2, a_3$ into new unknown functions $u_1, u_2, u_3$ such that if you fix the function $u_3$ (i.e., set it to anything you want), the resulting $2$-by-$2$ system is hyperbolic. Recall that a $2$-by-$2$ first order differential operator in $3$ variables $$ L_1\partial_1 + L_2\partial_2 + L_3\partial_3 $$ is hyperbolic if and only if the quadratic form $Q(x,y,z) = \det (xL_1 + yL_2 + zL_3)$ is nondegenerate and has indefinite signature, say $(-,+,+)$. Then the Cauchy problem is well-posed for an initial hypersurface if and only if $Q(v_1, v_2, v_3) > 0$ for any nonzero vector $(v_1,v_2,v_3)$ tangent to the hypersurface.

There are easier invariant ways to detect whether the system is hyperbolic or not, but I'll let someone else explain how to do that.

ADDED: Here is another way to describe what I said above.

1) If the left side of the three equations are scalar multiples of each other, then there is obviously only one PDE to solve and $b_1$, $b_2$ have to satisfy obvious consistency requirements.

2) Otherwise, two of the equations are not scalar multiples of each other but the third is a linear combination the other two. Let $E_1$ and $E_2$ be the two linearly independent equations and $E_3$ the third. To solve the whole system (i.e., all 3 equations), it suffices to solve $E_1$ and $E_2$ only. More generally, any two constants $p$ and $q$, it suffices to solve the equations $E_1 + pE_3$ and $E_2 + qE_3$. So the question reduces to whether you can find values of these constants so that this system is hyperbolic.

CORRECTION: If $p + q = 1$, then equations $E_1 + pE_3$ and $E_2 + qE_3$ are linearly dependent and therefore a solution to them is not necessarily a solution to the original system. Therefore, you must restrict to $p + q \ne 1$.

Below is my original incorrect answer:

The system is hyperbolic and has a well-posed Cauchy problem, if you can do the following: Find a linear change of co-ordinates $(z_1, z_2, z_3) = M(t, x, y)$, a linear change of basis for the unknown functions $(a_1, a_2, a_3) = L(u_1, u_2, u_3)$, and a linear change of basis for the eqautions $(b_1, b_2, b_3) = K(c_1, c_2, c_3)$ such that the system can be written in the form $$ \partial_t u + A_1\partial_x u + A_2\partial_y u = c, $$ such that one of the following holds (If so, then the Cauchy problem with initial data specified on the hypersurface $t = 0$ is well-posed):

  1. (strict hyperbolicity) There exists a matrix $Q(\xi,\eta)$ depending smoothly on $(\xi,\eta) \in \mathbb{R}^2$ such that the matrix $Q(\xi,\eta)(\xi A_1 + \eta A_2)Q^{-1}(\xi,\eta)$ is diagonal for all $(\xi,\eta)$.
  2. (symmetric hyperbolicity) The matrices $A_1$ and $A_2$ are symmetric.

A sufficient condition for strict hyperbolicity is the following: The matrix $\xi A_1 + \eta A_2$ has real, distinct eigenvalues for all $(\xi,\eta) \ne (0,0)$.

  • $\begingroup$ In case you hadn't noticed, the system is indeed overdetermined. The coefficient matrices share a common cokernel. Namely, the sum of each column is zero. So your approach won't work directly. The kernel/cokernel of the principle symbol is more obvious from the formulas I gave in my answer to the OP's previous question. $\endgroup$ Jan 7, 2015 at 21:36
  • $\begingroup$ Ach. Thanks for pointing that out. This is what happens when I try to do math in my head. $\endgroup$
    – Deane Yang
    Jan 7, 2015 at 22:28
  • 1
    $\begingroup$ I may have misspoken. I'm not sure an invariant way has been worked out. If it has, it would be in the reference cited by Ben McKay above (yes, I wrote it but I don't have a copy handy and I don't remember whether I dealt with this case). I'll look at it when I get a chance. $\endgroup$
    – Deane Yang
    Jan 13, 2015 at 3:19
  • 1
    $\begingroup$ Yes, this case is handled in section 1.7 of "Involutive Hyperbolic Differential Systems" (Memoirs of the AMS, #370). In the generic case, the so-called reduced Cartan characters are, I believe, $3, 2, 2$. Theorem (1.26) gives an invariant way to identify whether the procedure yields a hyperbolic system of PDE's or not. Alas, as proud as I am of this paper, it's rather difficult to read. My advice is to play around with the calculations described above. I've appended some additional information. $\endgroup$
    – Deane Yang
    Jan 13, 2015 at 15:46
  • 1
    $\begingroup$ Sorry but if $E_1$ and $E_2$ are linearly independent, you need to solve both equations. In particular, if $p + q = 1$, then $E_1 + pE_3$ and $E_1 + qE_3$ are linearly dependent and therefore do not imply solutions to $E_1$ and $E_2$. So you have to assume that $p + q \ne 1$ in order to get two linearly independent equations. $\endgroup$
    – Deane Yang
    Jan 14, 2015 at 13:00

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