I'm a physicist and during my research work, I found a system of linear pde with non constant coefficients that I have to study, since I have totally no experience about systems of pde and I have even problems to find some references. I would be really great to have some advice about how to face the problem. What I would like to do is of course find a solution of this system but even proving the existence of the solution can be a first great step. So thank you in advance for any possible help and some good reference. Here in the link you can find the system: System of pde

$$\begin{pmatrix} 0 & f_1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & f_1 \end{pmatrix}\partial_x \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} + \begin{pmatrix} f_2 & 0 & 0 \\ 0 & 0 & f_2 \\ 0 & 0 & 0 \end{pmatrix}\partial_y \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 \\ 0 & f_3 & 0 \\ f_3 & 0 & 0 \end{pmatrix}\partial_z \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix}= \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3\end{pmatrix}$$

Where we have three independent variables $(x,y,z)$ and three dependent variables $(N_1,N_2,N_3)$. The functions $\omega_i:R^3\rightarrow R$ $f_i: R^3\rightarrow R $, $N_i:R^3\rightarrow R $ are all smooth functions. Given that the matrix of coefficients are all singular I don't know how to proceed even about the classification of this system. Like I said I would be interested in a local solution or at least the proof of the existence of the solution. Any reference or help about this system will be really appreciated.

*Edit, sorry but for some reason the editor was not allowing me to write the matrices that I need in a good way, so I put a link to the image of the system that I create with latex.

  • $\begingroup$ The way you posted this originally, the equation with matrices was not rendered. Please, check whether this is what you were trying to get. $\endgroup$ – Martin Sleziak Nov 28 '19 at 13:41
  • $\begingroup$ Maybe I am missing something, but I do not actually see what is the difference between the formula in the linked picture and the formula which you have removed from the post (and I have now included it back). $\endgroup$ – Martin Sleziak Nov 28 '19 at 14:23
  • $\begingroup$ Yes the problem is that with the formula that I previous wrote for some reason I was getting matrices with just 1 row and 0 columns, but now everything is perfect, so thank you very much for your help, and sorry about my incompetence but is the first time I'm using this forum. Thank you again for the support! $\endgroup$ – reduced team Nov 28 '19 at 14:26
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    $\begingroup$ Consider first the constant coefficient case, use Fourier and convert the problem into an algebraic one. Once the algebra is clear it should be possible to understand what's going on $\endgroup$ – Piero D'Ancona Nov 28 '19 at 15:43
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    $\begingroup$ Being able to solve a special case is always a good first step. Just do it. See whether you notice any patterns that might be useful also in the more general case. There likely are - your problem clearly has some symmetries. $\endgroup$ – Michael Engelhardt Nov 28 '19 at 16:26

Recall that a first order system $A_{ij}^\mu \partial_\mu \phi^j + B_{ij} \phi^j = 0$ (the right hand-side could also be inhomogeneous) is symmetric hyperbolic when there exists at least one covector $p_\mu$ such that $A_{ij}^\mu p_\mu > 0$, the contraction is a positive definite symmetric matrix. If the coefficients $A_{ij}^\mu = A_{ij}^\mu(x)$ are $x$-dependent, then the positivity condition should be satisfied for all $x$. There are no special conditions imposed on $B_{ij}$, which could also be $x$-dependent. This is all standard. And the initial value problem for such systems is well posed, at least on those domains where the causal structure of formed by the cones of $p_\mu$'s satisfying the above positivity conditions is globally hyperbolic (the domain admits a foliation by Cauchy surfaces). See for instance the excellent account in Chapter 7 of [1].

Let me now introduce some non-standard terminology. I call a linear system $$A_{ij}^{\mu_1\cdots \mu_k} \partial_{\mu_1} \cdots \partial_{\mu_k} \phi^j + B(\phi,\partial\phi, \cdots \partial^{k-1}\phi) = 0 \tag{*}$$ higher order symmetric hyperbolic when it can be put in symmetric hyperbolic form by reduction to first order. Next, I call a linear system (using the notation $A_{ij}(\partial) = A_{ij}^{\mu_1\cdots \mu_k} \partial_{\mu_1} \cdots \partial_{\mu_k}$ and writing l.o.t for what I wrote as $B_{ij}(\cdots)$ above) $$A_{ij}(\partial) \phi^j + l.o.t = 0$$ generalized symmetric hyperbolic when there exists a complementary operator $C_k^k(\partial) = (C_k^i)^{\mu_1\cdots \mu_l} \partial_{\mu_1} \cdots \partial_{\mu_l}$ such that applying it to (*) gives a system $$ C_k^i(\partial) A_{ij}(\partial) \phi^j + l.o.t = 0$$ that is higher order symmetric hyperbolic.

The point is that a generalized symmetric hyperbolic system has an equally well-posed initial value problem, with or without a source term, as a symmetric hyperbolic one. Unfortunately, I don't have a reference for this, except for the parallel discussion that I gave for generalized normally hyperbolic systems in the recent paper [2] (just replace "normally hyperbolic" by "symmetric hyperbolic" everywhere).

Claim: Your PDE system is generalized symmetric hyperbolic.

Let me first illustrate by a relevant example, what it means for a system to be higher order symmetric hyperbolic. Take the equation $$\delta_{ij} \partial_x \partial_y \partial_z N^j + l.o.t = 0 , \tag{**}$$ where $\delta_{ij}$ is just the Kronecker delta. By introducing the auxiliary variables $N_z^j = \partial_z N^j$ and $N_{yz}^j = \partial_y N_z^j$, it can be reduced to the first order system $$\begin{pmatrix} \delta_{ij}\partial_z & 0 & 0 \\ 0 & \delta_{ij}\partial_y & 0 \\ 0 & 0 & \delta_{ij}\partial_x \end{pmatrix} \begin{pmatrix} N^j \\ N_z^j \\ N_{yz}^j \end{pmatrix} + l.o.t = 0 ,$$ where the desired positivity is satisfied by any covector from the first octant, $p_x,p_y,p_z>0$. This shows that (**) is indeed higher order symmetric hyperbolic.

Your PDE system has the special form $$\begin{pmatrix} f_2 \partial_y & f_1 \partial_x & 0 \\ 0 & f_3 \partial_z & f_2 \partial_y \\ f_3 \partial_z & 0 & f_1 \partial_x \end{pmatrix} \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} = \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{pmatrix} .$$ As mentioned in Deane Yang's answer, this equation cannot be put into symmetric hyperbolic form. That is, there exists no matrix $C_k^i$ such that the principal symbol $C_k^i A_{ij}^\mu p_\mu$ satisfies both the symmetry and positivity conditions. You can find $C_k^i$ such that the symbol becomes symmetric, but not positive. At least one eigenvalue of the resulting $p$-dependent matrix will always remain negative.

However, assuming that $f_1,f_2,f_3 \ne 0$ are nowhere vanishing, one can multiply this system by a second order matrix differential operator (which you can read off from the formula below) such that the system becomes $$\partial_x \partial_y \partial_z \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} + l.o.t = \frac{1}{2 f_1 f_2 f_3} \begin{pmatrix} f_1 f_3 \partial_x \partial_z & -f_1^2 \partial_x^2 & f_1 f_2 \partial_x \partial_y \\ f_2 f_3 \partial_y \partial_z & f_1 f_2 \partial_x \partial_y & -f_2^2 \partial_y^2 \\ -f_3^2 \partial_z^2 & f_1 f_3 \partial_x \partial_z & f_2 f_3 \partial_y \partial_z \end{pmatrix} \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{pmatrix} .$$ From the example (**) it should now be obvious that this system is higher order symmetric hyperbolic, verifying the claim that your system is generalized symmetric hyperbolic. Therefore, it has a well-posed initial-value problem on any hyperplane whose conormal $p_\mu$ lies in the first octant.

[1] Ringström, Hans, The Cauchy problem in general relativity., ESI Lectures in Mathematics and Physics. Zürich: European Mathematical Society (EMS) (ISBN 978-3-03719-053-1/pbk). xiii, 294 p. (2009). ZBL1169.83003.

[2] See the discussion around Lemma 3 in García-Parrado, Alfonso and Khavkine, Igor, Conformal Killing Initial Data, Journal of Mathematical Physics 60 122502 (2019). [arXiv:1905.01231]

  • $\begingroup$ First of all thank you very much for such a precise answer. I'm really grateful for your help and time. I will read carefully the refeferences that you suggest. $\endgroup$ – reduced team Nov 29 '19 at 16:45
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    $\begingroup$ Many thanks for posting this. I had never thought this through carefully enough. Once we had figured out the invariance of the system under changes of individual coordinates $x = a(x'), y = b(y'), z = c(z')$, I had always assumed that this implied that the initial value problem had to be ill-posed. But your answer shows that I'm wrong. I also confirmed that my original reasoning was faulty. $\endgroup$ – Deane Yang Nov 29 '19 at 17:20
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    $\begingroup$ @DeaneYang Thanks! And sorry, I didn't mean to rain on anyone's parade. :-) BTW, do you happen to recognize this trick of composing an equation with a differential operator to get another equation with ostensibly better analytical properties? The only systematic application of it that I've seen is for the Dirac equation (the Dirac operator is composed with itself to get the Laplacian or the d'Alembertian, depending on the signature). But it seems such a natural thing to do that it has probably found applications in less obvious places. $\endgroup$ – Igor Khavkine Nov 29 '19 at 20:07
  • $\begingroup$ @IgorKhavkine, you certainly shouldn $\endgroup$ – Deane Yang Nov 30 '19 at 6:32

If you switch the second and third rows of your system, the differential operator is the same as the linearization of equation (4.3) in Existence of elastic deformations with prescribed principal strains and triply orthogonal systems at a point where $$ g_{\alpha\beta} = \delta_{\alpha\beta} \text{ and }\frac{\partial x^\alpha}{\partial y^i} = \delta_{\alpha i}. $$ As mentioned immediately after the equation, this system is not (symmetric) hyperbolic. It also can never be elliptic. It follows that this system is difficult to solve. I don't know of any work in this direction.

In the paper cited, we observe that the obstruction is due to a type of gauge invariance and reformulate the problem in a way that eliminates this issue, resulting in a symmetric hyperbolic system that can be solved.

Perhaps your problem can be treated similarly?

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    $\begingroup$ First of all thank you for such an usefull answer, to understand if we can reformulate the problem as in the paper that you cited I think that I have to read the paper, because that system has a precise physical meaning so I have to understand what does it mean reformulate it into another system, which, if I understood correctly has different mathematical properties at least. I will read the paper and I will let you know as soon as possible. Thank you very much for your time and help! $\endgroup$ – reduced team Nov 28 '19 at 18:52
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    $\begingroup$ I just give a fast read to the paper and I understand why the system that we have is essentially the same. Basically we obtained our system using the hamiltonian formulation of General relativity, because we would like to fix a coordinates system in which the metric is diagonal using three diffeomorphisms, whose generator in the Hamiltonian formulation are the functions $N_i$ and the coefficients that I call $f_i$ are exactly the diagonal metric components. Your reference was like a gift, I searched a lot about this problem but I found basically nothing. Thank you again $\endgroup$ – reduced team Nov 28 '19 at 20:07

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