I am trying to write a computer program which computes the action of the exponential of a differential operator on a function, for any given differential operator.


$\exp(\varepsilon \partial_x) f(x) = f(x + \varepsilon)$

$\exp(\varepsilon x \partial_x) f(x) = f(x \exp(\varepsilon) )$

$\exp(\varepsilon x^2 \partial_x) f(x) = f\left( \frac{x}{1-\varepsilon x}\right)$

$\exp\Big(\varepsilon (x \partial_y + y \partial_x) \Big) f(x, y) = f\Big(\cosh(\varepsilon) x + \sinh(\varepsilon) y, \sinh(\varepsilon) x + \cosh(\varepsilon) y\Big) $

All these equalities are varifiable by Taylor expanding the variable $\varepsilon$ around zero.

My ideas:

  • Taylor expand the exponential of differential and try to guess the function yielding the same expansion (think this requires going through differential equations, which my program cannot do).

  • Recognize Lie algebras equivalent to the differential form, contruct a Lie algebra matrix and exponentiate.

I'm stuck because I do not know precisely how to proceed, except for particular cases in which the Lie algebra identification is obvious.

I also tried to look for papers both on Google and Google Scholar to get something out of it, but I didn't manage to find an explanation for such an algorithm.

  1. Do you know such an algorithm to find the action of such an exponential map on a function?
  2. Do you have any useful references for this?
  3. Is the Lie-algebra/Lie-group approach correct and valid for all types of differential operators?


I identified the case of linear operators, such as $ \exp\Big( \epsilon ( \sum_{i,j} a_{ij} x_i \partial_{x_j} ) \Big ) f(x_1, \ldots, x_N) $ which would be easily solved by constructing a matrix $M_{ij} = a_{ij}$ and then exponentiating it, using well known algorithms for matrix exponentiation, then make the matrix act on the function parameters $x_1, \ldots, x_N$ and substitute the resulting vector as new parameters.

e.g.: the hyperbolic rotation mentioned earlier: $\exp\Big(\varepsilon (x \partial_y + y \partial_x) \Big) f(x, y) = f\Big(\cosh(\varepsilon) x + \sinh(\varepsilon) y, \sinh(\varepsilon) x + \cosh(\varepsilon) y\Big) $

has matrix:

$ \begin{pmatrix} 0 & \varepsilon \end{pmatrix}$

$\begin{pmatrix} \varepsilon & 0 \end{pmatrix} $

which results by exponentiation in:

$ \begin{pmatrix} \cosh(\varepsilon) & \sinh(\varepsilon) \end{pmatrix} $

$ \begin{pmatrix} \sinh(\varepsilon) & \cosh(\varepsilon) \end{pmatrix} $

The problem concerns general differential operators, such as

$\exp\Big(\varepsilon x^n \partial_{x}\Big) f(x) $

or maybe even multivariable non-homogeneous differential operators, such as: $\exp\Big(\varepsilon ( x^2 y^3 \partial_x + x y^5 \partial_y ) \Big) f(x, y) $

How do I find a non-infinite formula for the action on $f(x, y)$?

4) Is the matrix algebra approach a good way?

5) Is an analysis of the structure of the Taylor expansion a good way?


Given a differential operator $D$, the exponential action $\exp(t \, D) f(x_1,\ldots)$ is given by the partial differential equation:

$\partial_t g(t, x_1, \ldots, x_n) = D [g(t, x_1, \ldots, x_n)] $

$ g(0, x_1,\ldots,x_n) = f(x_1, \ldots, x_n)$

Then $g(1, x_1, \ldots, x_n)$ is the result of the exponential action.

  • $\begingroup$ These are just formal solutions of first order partial differential equations... $\endgroup$
    – H. Arponen
    May 18, 2013 at 18:40
  • $\begingroup$ The formula for the action on $f(x,y)$ is given in detail in the Engel-Nagel reference below. Does it help? $\endgroup$ May 19, 2013 at 17:49
  • $\begingroup$ I had a look at Engel-Nagel, it works! It's just a first order differential equation to solve to find the flow of the vector field. Thanks! $\endgroup$
    – user34129
    May 19, 2013 at 23:21

1 Answer 1


A partial answer: What you call the "exponential function" is the so-called flow semigroup, see Engel-Nagel, Section II.3.28.

Another reference on the Lie derivative is the monograph by Chicone and Swanson.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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