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

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?

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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.

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Do we call those equations real principal type which do not fall into the standard classification (elliptic/hyperbolic/parabolic) or are they of some specific form that makes the qualitative analysis easier? I just have gone through the definition of a real principal type operator in the books you have mentioned and I would not say that I have understood it fully. Could you kindly explain me what a real principal type operator is and how did you conclude that the system of our concern is of that type? –  Ayan Aug 4 '12 at 19:05
Respected Sir, I have been trying to obtain a solution of my first order linear system. I was wondering whether there is any standard method to obtain the fundamental solution of any arbitrary first order liner system. I have been looking into standard PDO books but I could nowhere find a 'method' to obtain the same. Everyone discusses only the analysis of the PDO. thanks, Ayan. –  Ayan Aug 16 '12 at 5:19
Alas, no. A linear first order system might not have any solutions. And even if it does, there is usually no way to write down a fundamental solution explicitly. –  Deane Yang Aug 16 '12 at 13:37
I would suggest that you start with the special cases described by Robert, where the system uncouples and can be solved. You should identify what those cases correspond to in your mechanical system. If you successfully analyze those cases, then you might be able to study systems "close" to those using perturbation methods. –  Deane Yang Aug 16 '12 at 14:05

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.

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It appears to me that you have made use of characteristics to perform the change of variables. Have you followed any general method? –  Ayan Aug 6 '12 at 15:05
@AyanVivek: Yes, of course, the point is that, when the characteristics are distinct, one can diagonalize the leading order term (in the linear, first order case in $2$ independent variables). When you do this, then the controlling terms are the 'off-diagonal' terms in the 0th-order terms. Unless your $\phi$, $f$, $g$, and $h$ are special, though, you probably won't have either $B=0$ or $c=0$ in your case, although these would the cases in which the behavior of the solutions is reasonably simple and straightforward. –  Robert Bryant Aug 6 '12 at 17:46
Robert, thanks for providing an honestly helpful analysis of the PDE! –  Deane Yang Aug 6 '12 at 20:03
In this case, having complex roots is a good thing in the sense that it means that in those directions the PDE is effectively elliptic. The only singularities possible are in the direction of the eigenvector corresponding to the real root, and they propagate along the integral curves of the corresponding Hamiltonian vector field in the cotangent bundle. If you're still interested in this kind of analysis, I recommend this paper: Dencker, Nils, On the propagation of polarization sets for systems of real principal type. J. Funct. Anal. 46 (1982), no. 3, 351–372. –  Deane Yang Aug 16 '12 at 17:42
Deane's comments are good ones. The only thing I can add is that, sometimes, you can get better control. For example, in the above normal form that I gave, if $B$=0, then the $w$-component of the solution will be as regular as the coefficients $A$ and $D$, so that all of the 'singularities' will be in the $v$-component (and there can be some because it's just a linear first order PDE for $v$ once $w$ is chosen. In the case, $c=0$, though, you can see that the 'singularities' in $v$ will show up in some form in the $w$-component because $v$ is part of the inhomogeneous part of that equation. –  Robert Bryant Aug 16 '12 at 18:17

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.

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My equations (domain is $\mathbb{R}^2$) have variable coefficients (the roots of the characteristic equation are $i$, $-i$ and $c\,\cot f$, where $f$ is a real valued function in $\mathbb{R}^2$) and I am not looking for an explicit solution right away. I want to know the qualitative behaviour of the solutions e.g. the discontinuities of the derivatives of the dependent variable along the characteristic, the domain of dependence etc. –  Ayan Aug 4 '12 at 8:22
Could you just write your equation explicitly. What you wrote above doesn't give me much information, except that since you have variable coefficients (which was not stated in your question originally) the Fourier transform may not be of much use. –  Igor Khavkine Aug 4 '12 at 10:40
The system is:\\ $3\,u_{1,1}+u_{3,1}+2\,u_{2,2}=0$\\ $u_{1,2}+2\,u_{2,1}+3\,u_{3,2}=0$\\ $\cos\phi\,(u_{1,2}-u_{2,1})-\sin\phi\,(u_{1,1}+u_{2,2})-f\,u_2+g\,u_3+h=0$,\\ where $\phi$, $f$, $g$ and $h$ 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. $()_{,1}$ and $()_{,2}$ denote partial derivatives with respect to $x_1$ and $x_2$ respectively.\\ Roots of the characteristic polynomial of this system can be found out to be $i$, $-i$ and $3\,\cot\phi$. –  Ayan Aug 4 '12 at 15:02