MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for an analytical solution for the linear PDE

$(1)\qquad\qquad \qquad f_t+ A f_x + B f = 0, $

Where $A$ and $B$ are constant matrices and $f=f(x,t)$ is a vector.

Clearly each one of $f_t + A f_x = 0,$ and $f_t+ B f = 0,$ has an analytical solution involving eigenvectors and eigenvalues of $A$ and $B$ separately, but I'm not sure about equation (1).

My question is:

What are the minimal conditions on $A$ and $B$ such that there is an analytical solution to the problem in (1), (clearly if they share eigenvectors, I can write a solution...) and what is the solution in that case?

share|cite|improve this question
@jc: 1st order scalar PDEs admit solutions via the method of characteristics. The same is not true for first order systems. (Any higher order PDE can be converted to a system of first order equations. I challenge you to solve $\triangle u = 0$ using the method of characteristics.) – Willie Wong Mar 15 '11 at 12:25
Thanks Willie Wong, I withdraw my earlier comment. – j.c. Mar 16 '11 at 20:34
up vote 7 down vote accepted

By analytical, I presume that you mean explicit, or in close form. The known case so far is when $A$ and $B$ can be diagonalized in the same basis. Notice that the case where some eigenvalues come as complex conjugate pairs is subtle.

To see the importance of the condition $[A,B]=0_n$, let us make the Fourier transform of your equation: $$\partial_t\hat f=(B-i\xi A)\hat f(t,\xi).$$ Even if you can compute the exponential of $t(B-i\xi A)$ (unlikely), it will be a complicated function of $\xi$, whose backward Fourier transform is not explicit at all. When $A$ and $B$ commutte, we have $$\exp t(B-i\xi A)=e^{tB}e^{it\xi A},$$ and this explains that everything works well in this case. A less obvious good situation is when the commutator $[A,B]$ is a linear combination of $A$ and $B$. This argument was used (in an infinite dimensional setting) to compute explicitly the solution of the Cauchy problem for the harmonic oscillator: $$\partial_tu-\Delta u+|x|^2u=0.$$

If you accept a less explicit formula, and if you assume that $A$ is diagonalisable with real eigenvalues (we say that the operator $\partial_t+A\partial_x$ is hyperbolic), then the Cauchy problem is well posed. You can construct the solution by using the Laplace transorm in time and Fourier transform in space. You obtain an integral formula involving the inverses $(B-i\xi A)^{-1}$.

share|cite|improve this answer
Thanks for this answer. I would like to understand in more details why having a pair of degenerate characteristics is a problem for such a system. I've asked something similar there : , though a bit more complicated equation. If you would be kind enough to have a look, I'll be glad to have good references about this topic. Thanks in advance. – FraSchelle Jun 17 '14 at 12:37

Use the ansatz $f(x,t) = exp(a x) exp(c t) v$ where v is a constant vector. You have a solution whenever $(c + a A + B) v = 0$, i.e. a nontrivial solution if v is an eigenvector of $ a A + B $ for eigenvalue -c.

share|cite|improve this answer

We have been working on a special case of your system (1) recently in an economic problem. Special case means that we have a two-dimensional system and A has a special structure. We obtain examples for closed form solutions whose structure resembles gamma-distributions, i.e. the structure is $f(x,t) = (c_0 + c_1x+c_2t)exp(c_3x+c_4t)$. For more details, see our pdf-file, also available at

We would expect that there are many generalizations of this solution. We also believe that it would work for more general structures of A than we used so far. (In fact, any hint to the mathematical literature on generalizations of this solution would be very welcome.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.