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I am trying to understand the mathod of characteristics for solving partial differential equations. However, all the examples I found over the internet are for first order PDEs or for second order PDEs in two variables.

I could not find a single example where the solutions of a second order PDE in three variables are found via the method of characteristics. Could anyone give me a reference where I could find such an example?

For instance, how can I find all the solutions $u = u(x_1,x_2,x_3)$ of this PDE

$$ \frac{\partial^2 u}{\partial x_1^2}-\frac{\partial^2 u}{\partial x_2\partial x_3} = 0 $$ via the method of characteristics?

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    $\begingroup$ To OP: when you say more than two variables to you mean having a domain with more than two dimensions, or having more than 2 unknowns? It is a bit of folklore that method of characteristics is impossible for vector-valued equations in higher dimensions, and similarly for higher order scalar equations in higher dimensions; I'll try to see if I can find a reference or a heuristic argument for you. $\endgroup$ Commented Oct 30, 2023 at 22:59
  • $\begingroup$ I added an example to my question to make it more clear. $\endgroup$
    – Puzzled
    Commented Oct 30, 2023 at 23:10
  • $\begingroup$ @ThomasKojar. In the reference you gave there are just examples with first order PDEs. $\endgroup$
    – Puzzled
    Commented Oct 30, 2023 at 23:13
  • $\begingroup$ In general, there is no way to solve a 2nd order PDE in 3 variables using the method of characteristics. If there were, you would have learned about it, and we would not be struggling so much with PDE theory. The question is whether there are examples of 2nd order PDEs in 3 variables that can be solved by such a method, and how does one detect such a PDE. $\endgroup$
    – Deane Yang
    Commented Oct 31, 2023 at 0:55
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    $\begingroup$ If you set $y_1 = x_1$, $y_2 = x_2 + x_3$ and $y_3 = x_2 - x_3$, the specific example you asked about can be seen to just be the wave equation (with $y_2$ being the time direction). The fundamental solution of this is well-known, and you can see based on it that a method of characteristics solution is not really possible. $\endgroup$ Commented Oct 31, 2023 at 3:16

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I hope to use this answer to convince you that in general the method of characteristics cannot work for higher order PDEs in more than 2 variables. Nevertheless, there are some ideas in PDEs that are, in some sense, the next best thing, and maybe that will satisfy you.

First though: we will in this answer focus only on linear (systems) of differential equations, the philosophy being that if a tool works in general on nonlinear problems, then it should also work on linear problems.

Secondly, we can perform a reduction of order to focus only on first order systems of differential equations. A precise description of what you need to do is a bit annoying to write down, so I should just give an example instead: say you want to study the linear wave equation on $\mathbb{R} \times\mathbb{R}^2$, which we can write $$ \partial^2_{tt} u - \partial^2_{xx} u - \partial^2_{yy} u = 0 $$ Then consider the vector-valued function $$ U = (\partial_t u, \partial_x u, \partial_y u, u) $$ we see that it must solve a first order system $$ \partial_t U = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix} \partial_x U + \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix} \partial_y U + \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\end{pmatrix} U $$

So, now, let us think in general about linear, constant coefficient partial differential systems of first order. Our domain is $\mathbb{R}^n$ (or a subset thereof). We shall assume $n \geq 2$ since otherwise we are already dealing with a system of ODEs. Our unknown is $U$ which takes values in $\mathbb{R}^k$. Since we have $k$ unknowns, we expect to have (for a well-determined problem) $k$ equations. Our system can be written in the following form: $$ 0 = \sum_{i = 1}^n \sum_{\beta = 1}^k A_{\alpha\beta}^i \partial_i U_\beta + \sum_{\beta = 1}^k B_{\alpha\beta} U_\beta $$

The method of characteristics, if it were to succeed, would reduce this system to a set of $k$ ordinary differential equations for components of $U$. In other words, there would be a basis $v_1, \ldots, v_k$ of $\mathbb{R}^k$, such that, if we were to write $U = \sum u_\alpha v_\alpha$ in the basis, the coefficients would satisfy ODEs of the form $$ \sum_{i = 1}^n \xi_i^{(\alpha)} \partial_i u_\alpha = \sum_{\beta = 1}^k \tilde{B}_{\alpha\beta} u_\beta $$ where for each $\alpha$, the object $\sum_{i} \xi_i^{(\alpha)} \partial_i$ is regarded as a vector field.

A little bit of linear algebra tells you then what this would require, is that the $n\times k \times k$ tensor $A^i_{\alpha\beta}$ can be simultaneously diagonalized. More precisely, we need that the $n$ matrices $$ \mathbf{A}^{(i)}: (\mathbf{A}^{(i)})_{\alpha\beta} = A^i_{\alpha\beta} $$ can be simultaneously diagonalized, or that there exists a fixed set of left- and right-eigenvectors $r_1, \ldots, r_k$ and $\ell_1, \ldots, \ell_k$ such that each $$ \mathbf{A}^{(i)} = \sum_{\beta = 1}^k \lambda_\beta^{(i)} \ell_\beta \otimes r_\beta. $$ Furthermore, for this to be representing the method of characteristics, the eigenvalues $\lambda_\beta^{(i)}$ must all be real, as they map to the coefficients $\xi_i^{(\alpha)}$ which needs to be interpreted as the characteristic vector fields on $\mathbb{R}^n$.

Here's the punch line For generic systems of differential equations, the matrices $\mathbf{A}^{(i)}$ are a set of $n$ generic matrices, and there is no hope that one can simultaneously diagonalize them.

However, in some special cases this diagonalization is possible.

  • When the equation is scalar: in this case $k = 1$ and $1$-dimensional matrices are trivially diagonalized always. The very well developed method of characteristics for nonlinear first order scalar PDEs is what happens when you extend this as far as it takes you.
  • When the the total number of dimensions is $n = 2$. In this case we just need to simultaneously diagonalize two matrices, which we get nice sufficient conditions from spectral theorems. For example, assuming that $\mathbf{A}^{(1)}$ is invertible, then a sufficient condition is that the matrix $(\mathbf{A}^{(1)})^{-1} \mathbf{A}^{(2)}$ is symmetric. This leads to the well-developed theory for "symmetric hyperbolic systems" in "1 space and 1 time dimension".

Note that in the second case, unlike the first, diagonalizability is not guaranteed. So for example, you cannot use the method of characteristics to study fundamentally elliptic problems. An example being the Cauchy-Riemann equations $$ \partial_x U_1 = \partial_y U_2, \qquad \partial_x U_2 = - \partial_y U_1. $$ Trying to diagonalize this will lead to imaginary eigenvalues, which do not represent physical vector fields.


But all hope is not lost!

The fundamental idea of the method of characteristics is that it is easier to solve ODEs than PDEs. Given a PDE (especially a constant coefficient linear one), we can look for those solutions that are "essentially ODEs". In other words, we can pose the ansatz $$ U(x) = \phi(\xi\cdot x) $$ where $\xi$ is some unit vector in $\mathbb{R}^n$ and $\phi:\mathbb{R}\to\mathbb{R}^k$ a function. These types of solutions will satisfy $$ \sum_i \sum_{\beta} \xi_i A^i_{\alpha\beta} \phi'_\beta + \sum_\beta B_{\alpha\beta} \phi_\beta = 0 $$ which is a system of ODEs for $\phi$. Solutions of these types are called "plane waves".

A natural question then is: can we reconstitute a general solution $U$ as "linear combinations" of such plane wave solutions. And this question is essentially what Fourier theory is all about.

From this point of view, the study of partial differential equations using harmonic analysis techniques is an extension of the method of characteristics. For linear but variable coefficients, this develops into the approach using "microlocal analysis" (and if you read the literature, a lot of the guiding principles of microlocal analysis is based on wave packets traveling along characteristic curves, and is very similar to what one would expect from the method of characteristics). If you are interested in these types of developments, the standard reference is Hormander's four volume "Analysis of Linear Partial Differential Operators".

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  • $\begingroup$ Thank you very much for your answer. At this stage I guess the quetion is the following: If in general we can not find the solutions of a PDE via characteristics is there anything we can say about the PDE by knowing the characteristics of the associated differential operator? $\endgroup$
    – Puzzled
    Commented Oct 31, 2023 at 8:02
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    $\begingroup$ Yes! There's a lot you can say about the characteristics of the associated differential operator. Some key ideas: (a) in microlocal analysis there is the notion of "propagation of singularities", which states roughly that singularities (including finite but not infinite differentiability) propagate along characteristics for solutions to a PDE. A key related concept is the "wave front set". This underlies a lot of regularity results. (b) Understanding the characteristics structure is key to understanding local solvability of a nonlinear PDE. Key results including the Cauchy-Kovalevskaya Theorem $\endgroup$ Commented Oct 31, 2023 at 11:49
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    $\begingroup$ ... and a theorem (which you can find in Hormander) equating hyperbolicity of a linear PDO with solvability of its corresponding inhomogeneous problem. The keyword here is "hyperbolicity of a differential operator". $\endgroup$ Commented Oct 31, 2023 at 11:52
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    $\begingroup$ @Puzzled The (rare) exceptional cases where characteristics exist for second order PDEs in three independent variables are described in D. H. Parsons, The Extension of Darboux's Method, Mémorial des Sciences Mathématiques, fasc 142 (1960) (numdam.org/issue/MSM_1960__142__1_0.pdf). $\endgroup$ Commented Mar 22 at 5:30
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Check Harry Bateman’s book, Partial Differential Equations, section 2.24, pg. 133. The book is on the internet archive web site. He gives an equation for the characteristics for a second order PDE in any number of variables. The formula for the characteristics follows his equation II for the PDE. I won’t give his derivation here because there are much clearer derivations in Edouard Goursat's, Course in Mathematical Analysis, Vol. 3, section 25, translated by Howard Bergmann, 1964, and also Jacques Hadamard, Lecons sur la Propagation des Ondes et les Equations de l'Hydrodynamique, chapter 7, both books also available on the internet archive web site.

However, his formula works for the simple case in two variables for the PDE
$$ U_{xx} - U_{yy} = 0, $$ where $U$ is the unknown function and $U_{xx}$, $U_{yy}$ refer to the partial derivatives with respect to the variables $x$ and $y$. The characteristic surface in that case is the lines $$ x + y = constant $$ $$ x - y = constant $$

The basic idea is that if the function $U$ is calculated on the characteristic surface, only the function $U$ and its first derivatives $U_x$ and $U_y$ are needed to find the function $U$ at other points. This is seen in his equation I for the PDE for $U$ because the term in equation I for the second derivatives of $U$ is zero on the characteristic surface. E. H. Twizell’s book, Computational Methods for Partial Differential Equations, section 4.1, makes this idea clear. Also see Bateman's book, Differential Equations, also on the internet archive, pg. 171, the equation following equation 9. When the differentials $dp$ and $dq$ in the equation following equation 9 are replaced by finite differences in the first derivatives $p=U_x$ and $q=U_y$, then the solution $U$ at any point $(x, y)$ can be found from the boundary values for $U_x$ and $U_y$ by working along the characteristic lines to get to the point $(x, y)$. The second and higher order differences are not needed.

Explicit examples for three dimensions are given in a British Government report by C. K. Thornhill, The Numerical Method of Characteristics for Hyperbolic Problems in Three Independent Variables, on the Cranfield University website. Equations 1.12 and 1.121 are solved along the (bi)characteristics using only first order differences.

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  • $\begingroup$ Welcome to mathOverflow. The answers are processed using MarkDown formatting, a feature of which is that paragraph/lines that are indented by four spaces are treated as "computer code". I've removed the paragraph indentation you included in your original posting to improve readability. Additionally, this forum supports latex-style mathematics through MathJax. Just put latex-style formulae such as U_{xx} between dollars signs $..$ to get $U_{xx}$. This may make it easier to read your answer. $\endgroup$ Commented Mar 22 at 4:30
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    $\begingroup$ On the mathematics: while finding the (bi)characteristics surface is certainly related to the method of characteristics, they are not quite the same thing. $\endgroup$ Commented Mar 22 at 4:32

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