How is exp(tL) rigorously defined in Duhamel's formula?

[Yugo]

Hello,

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:

http://tosio.math.toronto.edu/wiki/index.php/Duhamel's_formula

How is the exp(tL) even rigorously, defined? I can't find this anywhere.

 

[HallsofIvy]

L here is a differential operator. Exponentials of operators, like exponentials of matrices, are defined through the Taylor's series for e^x:
e^L= I+ L+ \frac{1}{2}L^2+ \cdot\cdot\cdot + \frac{1}{n!}L^n+ \cdot\cdot\cdot
where the powers mean repeated application of the operator and "I" is the identity operator. Of course, that should be applied to some function:
e^L(f)= f+ L(f)+ \frac{1}{2}L^2(f)+ \cdot\cdot\cdot + \frac{1}{n!}L^n(f)+ \cdot\cdot\cdot
That is typically very difficult (or impossible) to evaluate for all but self-adjoint operators.

 

[Yugo]

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.

 

[Feynman]

e^{tL} is the semi groupe of $L$, you can see this theory on http://cantor.mathematik.uni-ulm.de/m5/index.php?file=arendt/index.htmlt

 

[Feynman]

L here is a differential operator. Exponentials of operators, like exponentials of matrices, are defined through the Taylor's series for e^x:
e^L= I+ L+ \frac{1}{2}L^2+ \cdot\cdot\cdot + \frac{1}{n!}L^n+ \cdot\cdot\cdot
where the powers mean repeated application of the operator and "I" is the identity operator. Of course, that should be applied to some function:
e^L(f)= f+ L(f)+ \frac{1}{2}L^2(f)+ \cdot\cdot\cdot + \frac{1}{n!}L^n(f)+ \cdot\cdot\cdot
That is typically very difficult (or impossible) to evaluate for all but self-adjoint operators.

HallsofIvy, that is not difficult like you see! by semigroupe theory the expression e^{tL}f is the solution of the differential equation on Banach space : u'+Lu=0, u(0)=f, so for example for the laplacian,
e^{t\Delta}f is the solution of the Heat equation (for Dirichlet or Neumann conditions) which is the gaussian kernel

 

[Yugo]

Thank you, I think I got it.