Classification of a certain System of Linear First Order PDEs whose characteristic polynomial has one real and two complex conjugate zeros

Question

In my research work, I recently have come across the following system of three linear first order pde's whose characteristic polynomial has one real and two complex conjugate zeros.

$3u_{1,1}+u_{3,1}+2u_{2,2}=0,$

$u_{1,2}+2u_{2,1}+3u_{3,2}=0,$

$\cos \phi\ (u_{1,2}-u_{2,1})-\sin \phi\ (u_{1,1}+u_{2,2})-fu_2+gu_3+h=0$

Here, $f$, $g$, $h$, $\phi$ are real valued functions of the independent variables $x_1$ and $x_2$. $(u_1\ u_2\ u_3 )$ is the vector containing the three dependent variables. $(\cdot)_{,i}$ denotes partial derivative with respect to $x_i$. Roots of the characteristic polynomial of this system can be found out to be $i$, $−i$ and $3\ \cot \phi$. Therefore it does not fall into any of the classifications viz. elliptic, parabolic, hyperbolic. Could someone kindly direct me to any research that has been done on the analytic solution of such a pde?

Answer 1

Your system of PDE's appears to be of real principal type, as defined by Hormander. Hormander studied the singularities of distributional solutions to such a PDE and how they propagate. This in turn leads to an a regularity theorem for a compactly supported distributional solution on a bounded open domain. Since the adjoint of a PDO of real principal type is also of real principal type, the regularity theorem in turn implies a local existence theorem. If you google "hormander propagation of singularities" you should be able to find references. Books that discuss this include ones written in the 80's or earlier by Hormander, Treves, Chazarain-Piriou, and Michael Taylor.

Beyond that, as far as I know, very little is known about such PDE's. You might be able to find a way to use ad hoc techniques adapted to the specific PDE to do better than these rather general results.

ADDED: The determinant of the first order term of the PDO defines a function on the cotangent bundle of $R^2$, which is a symplectic manifold. The zero set of this function is called the characteristic variety. The function also defines a Hamiltonian vector field $H$ on the cotangent bundle. Any integral curve of the Hamiltonian vector field that lies in the characteristic variety is called a null bicharacterstic. Hormander defines an operator to be of real principal type relative to a domain $\Omega$, if there are no null bicharacteristics trapped over the domain.

It is easy to see that given a point where the function defined above has nonvanishing gradient at any point in the characteristic variety over that point is of real principal type with respect to any sufficiently small open neighborhood of that point. Your PDO appears to me to satisfy this.

Hormander's propagation of singularities theorem says that the wavefront set (which lies in the cotangent bundle and is a refinement of the singular set of a distribution) of a solution lies in the characteristic variety and is invariant under the flow of the Hamiltonian vector field.

So this indicates what you could try to do: Find the Hamiltonian vector field associated with your system of PDE's and study its flow. This would act as a guide to figuring out what the "right" domain for solving your PDE is. You would then try to prove an a priori regularity theorem, maybe through energy integral estimates, directly from your PDE, rather than using the full machinery of microlocal analysis and Fourier integral operators. Actually, you should probably also study the full matrix symbol of the PDE. I'm less familiar with the details of what to do here. You might want to consult early work of Nils Dencker on systems of real principal type.

Answer 2

The following device might help in applying Deane's suggestions. At least the calculations on the leading order terms will be simpler: Consider the change of variables $$ \begin{align} v &= u^1 + u^3\ ( = \bar v), \\\ w &= \bigl(9\cos\phi - i\ \sin\phi\bigr)\ u^1 +2i\bigl(3\cos\phi + i\ \sin\phi\bigr)\ u^2 + 3\bigl(\cos\phi - i\ \sin\phi\bigr)\ u^3 ,\\\ \overline{w} &= \bigl(9\cos\phi + i\ \sin\phi\bigr)\ u^1 -2i\bigl(3\cos\phi - i\ \sin\phi\bigr)\ u^2 + 3\bigl(\cos\phi + i\ \sin\phi\bigr)\ u^3.\\\ \end{align} $$ then the system takes the following form, (where the first equation is real and the second is complex) $$ \begin{align} \sin\phi\frac{\partial v}{\partial x^1} + 3\cos\phi\frac{\partial v}{\partial x^2} &= a + b\ v + c\ w + \bar c\ \overline{w}\\\ & \\\ \frac{\partial w}{\partial x^1} + i\frac{\partial w}{\partial x^2} &= A + B\ v + C\ w + D\ \overline{w} \end{align} $$ and where the functions $a = \bar a$, $b = \bar b$, $c$, $A$, $B$, $C$, and $D$ can be explicitly computed in terms of the given coefficient functions $\phi$, $f$, $g$, and $h$.

One can simplify this further as follows: Let $\lambda$ and $\mu$ be solutions to the equations $$ \sin\phi\frac{\partial \lambda}{\partial x^1} + 3\cos\phi\frac{\partial \lambda}{\partial x^2} = b \qquad\text{and}\qquad \frac{\partial\mu}{\partial x^1} + i\frac{\partial\mu}{\partial x^2} = C $$ (which are uncoupled linear equations that are easily solved, in theory), then replacing $v$ by $e^{-\lambda}v$ and $w$ by $e^{-\mu}w$ reduces to the case $b=C=0$, so that the equations take the simpler form $$ \begin{align} \sin\phi\frac{\partial v}{\partial x^1} + 3\cos\phi\frac{\partial v}{\partial x^2} &= a + c\ w + \bar c\ \overline{w}\\\ & \\\ \frac{\partial w}{\partial x^1} + i\frac{\partial w}{\partial x^2} &= A + B\ v + D\ \overline{w} \end{align} $$

For geometric reasons, this is as far 'uncoupled' as you can get, in general. From this, you can see, for example, that if $c=0$ or $B=0$, then one or the other of the two equations uses only $v$ or $w$ and so is solvable by standard techniques, and then the other can be solved by the standard technique for it. The interesting case is when $Bc\not=0$, and I suspect that the analysis will need to proceed differently in this case.

Answer 3

The answer depends strongly on what you mean by "analytic solution". If your equation has constant coefficients, then the general solution can be obtained via the Fourier transform, independent of what class it falls into. This basic method explained in essentially all PDE books. Your ability to explicitly evaluate the resulting Fourier integrals will control the amount of explicit formulas you'll obtain. If you are worried about issues of mathematical analysis (i.e., various inequalities and estimates) you may want to look into the Malgrange-Ehrenpreis theorem.

If your equation does not have constant coefficients (though, then it would be confusing what you mean by "characteristic polynomial"), then obtaining a few explicit solutions of high symmetry may be the best you can hope for.