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Duhamel's formula and iterates.

  1. Apr 20, 2007 #1

    I am having trouble understanding how to use Duhamel's formula as a contraction to give existence uniqueness theorems for certain semi linear PDE.
    To be more precise, have a look at the PDE and corresponding Duhamel formula in the wikipedia link given below:


    How is the exp(tL) even rigorously, defined? I can't find this anywhere.
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Apr 20, 2007 #2


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    L here is a differential operator. Exponentials of operators, like exponentials of matrices, are defined through the Taylor's series for ex:
    [tex]e^L= I+ L+ \frac{1}{2}L^2+ \cdot\cdot\cdot + \frac{1}{n!}L^n+ \cdot\cdot\cdot[/tex]
    where the powers mean repeated application of the operator and "I" is the identity operator. Of course, that should be applied to some function:
    [tex]e^L(f)= f+ L(f)+ \frac{1}{2}L^2(f)+ \cdot\cdot\cdot + \frac{1}{n!}L^n(f)+ \cdot\cdot\cdot[/tex]
    That is typically very difficult (or impossible) to evaluate for all but self-adjoint operators.
  4. Apr 20, 2007 #3
    I see. So for example, that thing would make sense if the operator L was a Laplacian?

    Moreover, if that's the case, could you possibly give me an example of using that Duhamel formula as a contraction to solve a PDE problem? I've seen this sort of thing before for PDEs like the one above with u being a function of two variables and L the corresponding (spatial) laplacian, but I didn't understand how to show that the duhamel formula has a fixed point iff the PDE problem had a solution.
  5. Apr 29, 2007 #4
    Last edited by a moderator: Apr 22, 2017
  6. Apr 29, 2007 #5
    HallsofIvy, that is not difficult like you see! by semigroupe theory the expression [tex]e^{tL}f [/tex] is the solution of the differential equation on Banach space : [tex]u'+Lu=0, u(0)=f[/tex], so for example for the laplacian,
    [tex]e^{t\Delta}f [/tex] is the solution of the Heat equation (for Dirichlet or Neumann conditions) wich is the gaussian kernel:wink:
  7. Mar 19, 2008 #6
    Thank you, I think I got it.
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