Calculating Dyadic Green's Function Expression

[Karthiksrao]

Dear all,
Need your suggestions as to how I can arrive at the expression for the Dyadic Green's function.

The scalar case is simple:
Consider the standard equation of motion in Fourier space: $\omega^2 \hat{x}(\omega) = \omega_0^2 \hat{x}(\omega) - i \delta \omega \hat{x}(\omega)+ F(\omega)$, where $F$ is the forcing function, and $\delta$ the damping parameter. Defining the scalar Green's function from the equation: $\hat{x}(\omega) = G(\omega) F(\omega)$, and substituting this in the above equation of motion, we get the expression for the scalar green's function as:
$$ G(\omega) = \frac{1}{\omega^2 - \omega_0^2 + i \delta \omega };$$

Now consider going to the vector form for the displacement, and consequently the dyadic form for the Green's function - as seen for coupled harmonic oscillator problems. The equations of motion can be reduced to the form:
$$ \hat{\bf{x}}(\omega) = \bar{\bar{\bf{A}}} \hat{\bf{x}}(\omega) + \bf{F}(\omega) $$
where $\bar{\bar{\bf{A}}}$ is a matrix.

Now defining the Green's function as before: $\hat{\bf{x}}(\omega) = \bar{\bar{\bf{G}}}(\omega) \bf{F}(\omega)$, is it possible to arrive at an expression for the Green's function? The problem is complicated (as compared to the scalar problem) due to the inherent summations that exist in the matrix multiplication.

Can anybody suggest how I can proceed ?

Thanks!

 

[Orodruin]

You really should include $Ax$ from the RHS in the LHS instead. You would obtain
$$
(1-A)x = F.
$$
The Green's function would be $(1-A)^{-1}$, i.e., just the inverse of $1-A$.

 

[Karthiksrao]

Thanks for getting back. A follow up query: if I know the eigenvalues and the eigenvectors of the matrix A, would I be in a position to construct the Green's function from just this information ?

I have attached a document where this has been done, and I haven't been able to follow the procedure. Any help would be appreciated.

Thanks!