Will my Green's function approximation be any good?

Question

I would like to use the Greens Function $G(t,t')$, satisfying \begin{equation} \int G^{-1}(t,s)G(s,t') ds = \left( \begin{array}{cc} \delta(t-t') & 0 \\ 0 & \delta(t-t') \end{array} \right) \end{equation}

where $$ G^{-1}(t,t') = \begin{pmatrix} -(3+\frac{d}{dt})\delta(t-t') & - 2 \pi t \delta(t-t') \\ 2 \pi t \delta(t-t') & -(\frac{1}{2}+\frac{d}{dt}) \delta(t-t') \\ \end{pmatrix}. $$

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(detail added in response to a comment)

I can't find a closed form expression for $G$, but if I approximate $G^{-1} \approx H$ (changing only the upper left component) there is a closed form approximate expression for $G$, as follows:

Choosing $$ H(t,t') = \begin{pmatrix} -(\frac{1}{2}+\frac{d}{dt})\delta(t-t') & - 2 \pi t \delta(t-t')\\ 2 \pi t \delta(t-t') & -(\frac{1}{2}+\frac{d}{dt}) \delta(t-t')\\ \end{pmatrix} $$ We then require \begin{equation} \int H(t,s)G(s,t') ds = \left( \begin{array}{cc} \delta(t-t') & 0 \\ 0 & \delta(t-t') \end{array} \right) \end{equation} Columns of $G = \left(\begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \right) = \Theta \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ (where $\Theta$ is a step function) may be solved for separately.

$A$ and $C$ must jointly satisfy (the 0,0) identity, integrating out the $s$ variable leaves: \begin{equation} -\Big( \frac{1}{2} + \frac{d}{dt}\Big) A - 2 \pi t C = \delta \end{equation} $$\implies$$ \begin{equation} \left(-\frac{1}{2} a - a' - 2\pi t c\right)\Theta = (1+a)\delta\end{equation} and the (0,1) identity which amounts to: \begin{equation} \left(-\frac{1}{2} c - c' - 2\pi t a \right)\Theta = c\delta\end{equation}

I find $A$ and $B$ to be \begin{equation} A = \Theta (t-s) \left(-e^{\frac{s-t}{2 }}\right) \cos (\pi (s-t) (s+t)) \end{equation} \begin{equation} C = \Theta (t-s) e^{\frac{s-t}{2}} \sin (\pi (s-t) (s+t)) \end{equation}

Plugging these in, and noticing that some of the terms proportional to $\delta$ will be identically zero, shows they should work as solutions.

I then use the same argument for $B$ and $D$.

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Assuming the above solution for $G$ works, is there any way to know how good an approximate $G$, derived from $H$, would be? And if this approximation idea is no good, how can I deal with finding a useable $G$?

Answer 1

You've made it hard to see by gratuitously complicating and misframing the problem. Your first duty is to integrate out the useless dummy variable, given your effective Ansatz for $G(s-t') = \left(\begin{smallmatrix} A(t') & B(t') \\ C((t') & D(t') \end{smallmatrix} \right) \delta(s-t')$, $$\int\!\!ds~~ H(t,s)G(s,t') \\ = \int\!\!ds~~\begin{pmatrix} -(\frac{1}{2}+\frac{d}{dt}) & - 2 \pi t \\ 2 \pi t & -(\frac{1}{2}+\frac{d}{dt}) \end{pmatrix}\delta (t-s) \begin{pmatrix} A(t') & B(t')\\ C(t') & D(t') \end{pmatrix}\delta(s-t') \\ =\begin{pmatrix} -( 1/2+\tfrac{d}{dt}) & - 2 \pi t \\ 2 \pi t & -( 1/2+\tfrac{d}{dt}) \end{pmatrix} \begin{pmatrix} A(t) & B(t)\\ C(t) & D(t) \end{pmatrix}\delta(t-t'). $$

You now need to solve the four operator equations in Heaviside operator calculus (e.g. $\partial C =[\partial,C]+C\partial$) for the four operators, A,B,C,D, $$ -( 1/2+\tfrac{d}{dt})A - 2 \pi t C=1,\\ -( 1/2+\tfrac{d}{dt})C + 2 \pi t A=0,\\ -( 1/2+\tfrac{d}{dt})B - 2 \pi t D=0,\\ -( 1/2+\tfrac{d}{dt})D + 2 \pi t B=1 . $$ Using the two middle ones you collapse to two equations for two variables, B,C, whose symmetry you should note, so one equation.