In a try to solve a physical problem, I've faced a system of first-order partial differential equations of the form


with $\left(x,y,t\right)$-independent matrix $\mathbf{A}$ and $t$-dependent / $\left(x,y\right)$-independent $\mathbf{B}$. Also, $t\in\left[0,2\pi\right]$ is not really a time, but an angle, and $\mathbf{B}\left(t+2\pi\right)=\mathbf{B}\left(t\right)$ is periodic, so I guess the solutions will be periodic as well. Actually $\mathbf{A}$ is real and diagonal, so my system of equations seems to be in a canonical form from the beginning. Note nevertheless that two eigenvalues of $$\mathbf{A}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)$$ are degenerate, so I wonder if the system is still hyperbolic (I know it is no more strictly hyperbolic according to Courant (see references below), but Wikipedia defines it as hyperbolic / I've been unable to find the Wikipedia definition in any textbook I've put an hand on). I would like (if possible) to obtain analytic (in a close form or as a series expansion) solutions of this problem, or understand a bit better which perturbation scheme I could use. It seems to me this system is extraordinary simple, and at the same time I'm paralysed by my ignorance in powerful enough methods.

So my first and general question would be : which literature would you recommend to me ? about this problem.

Giving more details: for the moment I've found

R. Courant and P. Lax On nonlinear partial differential equations with two independent variables Commun. Pure Appl. Math. 2, 255 (1949). (beyond a paywall)

where section 3 seems of interest for me, it shortly discusses an old method by

O. Perron Über Existenz und Nichtexistenz von Integralen partieller Differentialgleichungssysteme im reellen Gebiet Math. Zeitschrift 27, 549 (1928). (also beyond a paywall)

which seems to answer my problem, giving analytical solution of system of semi-linear partial differential equation. The book by

R. Courant and D. Hilbert, Methods of Mathematical Physics, Volume II: Partial Differential Equation (John Wiley and Sons, 1962).

seems too much related to non-linear systems, and I've been unable to figure out what to do with my problem. Finally a short lecture note by

E. Kersalé, Analytic Solutions of Partial Differential Equations http://www1.maths.leeds.ac.uk/~kersale/teaching.html (2003)

ends up with a short discussion of my problem, but I still do not figure out what to do with the system of ordinary differential equations once fond the characteristics lines / surfaces... (actually, two characteristics are circles, and two straight lines are degenerate). So are there other (perhaps better or more specific for physicists) textbook/notes you would recommend ?

Thanks in advance for any remark aiming at improving this question.

  • 2
    $\begingroup$ If you can find a time independent similarity transformation that simultaneously diagonalizes both $\mathbf{A}$ and $\mathbf{B}(t)$, then your system decouples into four first order, scalar, linear equations. Each of those can be solved by the method of characteristics. Perhaps that is enough. If not, one possibility is that $B(t)$ could be split into two parts, one part would allow for simultaneous diagnalization, the other might be treated as a perturbation. If the equations do decouple, perhaps that will give you an idea of what kind of boundary/initial conditions would be allowed. $\endgroup$ – Igor Khavkine Jun 16 '14 at 19:26
  • $\begingroup$ @IgorKhavkine Thanks for your comment. Actually, it is impossible to find a $t$-independent diagonalisation of $B$ which will keep $A$ diagonal. It seems to me that Courant and Lax propose a solution for this problem in section 3 of the cited paper, in a form which seems to me a kind of Dyson series. But I have difficulties to see if I'm correct about that. Also, in my problem the boundary conditions are known, they are some initial boundary conditions. $\endgroup$ – FraSchelle Jun 17 '14 at 12:51
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    $\begingroup$ Without looking at the Courant-Lax article, I would guess that the Dyson series arises by treating $\mathbf{B}(t)$ as a perturbation. If that term is absent, then the system decouples and can be solved by the methods I described above. I don't know if you'd be happy with a series solution, but unless $\mathbf{B}(t)$ has some very special algebraic structure, this might be the best you can do. $\endgroup$ – Igor Khavkine Jun 17 '14 at 14:53

For simplicity, consider solutions where $ u$ does not depend on $x, y$: $A u_t + B(t) u = 0$. If $y^T A = 0$, that says $y^T B(t) u = 0$, so $u$ is restricted to belong to a certain (possibly $t$-dependent) subspace.
Thus for your example $$A = \pmatrix{1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & -1\cr}$$ $(B(t) u)_2 = (B(t) u)_3 = 0$. If the appropriate $2 \times 2$ submatrix of $B(t)$ is invertible, this lets you express $u_2$ and $u_3$ in terms of $u_1$ and $u_4$, and you get a periodic linear system for $u_1$ and $u_4$.
The solutions are usually not periodic in $t$. Rather, the linear operator $u(0) \to u(2\pi)$ will have eigenvalues $\lambda$ corresponding to solutions where $u(2\pi) = \lambda u(0)$ (see Floquet theory).

Somewhat more generally, solutions of the form $u(x,y,t) = \exp(\alpha x + \beta y) v(t)$ lead to the same type of system (with $\alpha \cos(t) + \beta \sin(t)$ added to $B$).

EDIT: If the submatrix of $B(t)$ is not invertible for some $t$, you may find that some or all of the nontrivial solutions have singularities at those $t$.

  • $\begingroup$ @RobertIsreal Thanks so much. I'll try your ansatz and the associated method asap, and tell you if it works (I guess it is). Still, is there references about that, or is it mathematical skills and intuitions acquired along years? Thanks again. $\endgroup$ – FraSchelle Jun 18 '14 at 10:30
  • $\begingroup$ Thanks again. I verified and it works pretty well. So in fact a separation of variable is sufficient to resolve this problem. Is there a way to prove that we exhaust the solutions by this Ansatz you proposed ? Thanks again for your clear and insightful answer. $\endgroup$ – FraSchelle Jun 18 '14 at 14:41
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    $\begingroup$ If, for example, your initial data at $t=0$ are in $L^2$, you can do a Fourier transform, and then you're looking at a "generalized linear combination" of solutions of the form $u(x,y,t) = \exp(\alpha x + \beta y) v(t)$ for imaginary $\alpha$ and $\beta$. $\endgroup$ – Robert Israel Jun 18 '14 at 14:59

Well, assuming that the matrix $\mathbf{B}$ is a real-analytic function of $t$, the local real-analytic theory gives you this result, which may or may not be useful to you:

Start with a real-analytic, $2$-dimensional surface $S\subset\mathbb{R}^3$ (coordinates $x,y,t$) that is nowhere tangent to any one of the three vector fields $$ E_c = \cos(t)\ \frac{\partial\ }{\partial x}+\sin(t)\ \frac{\partial\ }{\partial x} + c\ \frac{\partial\ }{\partial t} $$ where $c = 0,1,-1$ and choose a real-analytic function $\phi:S\to\mathbb{R}^4$. Then there will exist an open neighborhood $U\subset\mathbb{R}^3$ of $S$ on which there will exist a unique real-analytic function $\mathbf{u}:U\to\mathbb{R}^4$ that satisfies your equation and has the property that $\mathbf{u}(p) = \phi(p)$ for all $p\in S$.

The point is that the data $(S,\phi)$ are noncharacteristic initial data for the Cauchy problem, and, because of real-analyticity, the Cauchy-Kowalewski theorem applies.

  • $\begingroup$ Thanks a lot for your answer. I believed my problem is indeed Cauchy's, since the initial data are non-characteristic. But I didn't care so much about all the details. Actually, B is Hermitian (still continuous, differentiable in t, ... all the good physical properties of a Hamiltonian somehow), does your theorem still apply ? I guess so, since the eigenvalues are then real (and analytic, ...) or is there a trick I don't see ? Thanks again. $\endgroup$ – FraSchelle Jun 18 '14 at 10:32
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    $\begingroup$ The Cauchy-Kowalewski Theorem will still apply if you allow $\mathbf{u}$ and $\mathbf{B}$ to be complex valued, as long as they are real-analytic functions of their arguments. If you only have smoothness (or weaker regularity) of the coefficients, then the C-K Theorem may not apply. By the way, I now realize that you are using 'analytic' in a different sense from what I mean when I use 'analytic' (which is really 'real-analytic'), you probably mean 'analytic' to be some concept that is in opposition to 'numerical' and somehow related to 'explicit'. Is that right? $\endgroup$ – Robert Bryant Jun 18 '14 at 11:59
  • $\begingroup$ Well analytic in my sense means continuous, differentiable, having series expansion... In short, the $t$-dependency of the B-matrix is $B \propto e^{it}$. Is that analytical enough ;-) ? $\endgroup$ – FraSchelle Jun 18 '14 at 12:17
  • $\begingroup$ It's hard to tell until you specify what you mean by 'series expansion'. For example, since $\mathbf{B}$ is $2\pi$-periodic, you might mean Fourier series expansion, which every continuous, differentiable, $2\pi$-periodic function has. What I mean by 'analytic' is that the function equals its Taylor series expansion (which is assumed to have a positive radius of convergence) at every point. Is that what you mean? $\endgroup$ – Robert Bryant Jun 18 '14 at 12:52
  • $\begingroup$ Yes :-), a Taylor expansion. Sorry to have missed the point when I said series. It is always good to be precise indeed. $\endgroup$ – FraSchelle Jun 18 '14 at 12:56

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