Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to understand the discussion around equation (2.1) of the paper It says that the linear PDE $M(\partial_x,\partial_y)q=0$ with constant coefficients has the Lax pair $\mu_x+ik\mu=q$ and $M(\partial_x,\partial_y)\mu=0$, where $k$ is any complex number and $\mu$ is a function.

The way I'm used to thinking of Lax pairs is as operators $L$ and $B$ such that $\dot{L}+[L,B]=0$ is equivalent to the original PDE. This is equivalent to requiring that the eigenvalue equations $L\phi=\lambda \phi$ and $\dot{\phi}=B\phi$ are compatible, where $\lambda$ is a fixed parameter and $\phi$ is any function. Can anyone explain how this connects with the discussion in the paper? What are $L$ and $B$ in the above case?


share|cite|improve this question

1 Answer 1

up vote 2 down vote accepted

This answer may be a bit late, but it looks like the author is using "Lax pair" to mean a pair of equations in $\mu$, one of which is independent of $t$ and one of which is not. Here, "independent of $t$" means the same as it does in the traditional Lax pair approach - $L\phi=\lambda\phi$ where $\lambda_t=0$, so $L$ doesn't change with $t$. $B$, on the other hand, is inherently dependent on $t$, since $\phi_t=B\phi$.

Later in the paper (e.g., equation 6.4), the author returns to the traditional concept of a Lax pair (albeit in matrix form: a pair of matrices $X$ and $T$ such that $\phi_x=X\phi$, $\phi_t=T\phi$, which gives the compatibility condition $X_t-T_x+[X,T]=0$ for the usual commutator $[X,T]=XT-TX$).

If that doesn't seem to help, the group which innovated the Unified Transform Method keeps a very nice collection of results and papers related to the technique here:

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.