This is a generic example of a Poisson Equation with homogeneous BCs which I am reading the solution to in Haberman's Applied Partial Differential Equations with Fourier Series and Boundary Value Problems. We started with:
$$\nabla^2 u(x,y) = Q(x,y)$$
And we want a Green's function in the form: $$G(x,y,x',y') = \sum_{n=1}^\infty a_n(y)\phi_n(x)$$
Plugging in $G(x,y,x',y')$ we get $$\nabla^2 G = \delta(x-x')\delta(y-y')$$
We use an eigenfunction in 1-direction: $\phi(x)$ with $\lambda = (\frac{n\pi}{L})^2$ and we plug this into the PDE and using orthogonality of sines to set up an ODE:
$$a_n(y)'' - \lambda_na_n(y) = \frac{2}{L}\delta(y-y')\phi_n(x')$$
I followed along perfectly fine up to this point. It's the next following parts that I don't understand and no workings or steps are explicitly shown which lead up to what I saw next and that's what I need help with figuring out.
According to the text I'm reading from which is going over this, we're supposed to utilize the fact that "the differential equation is homogeneous if $y \ne y'$. In addition if we utilize the boundary conditions we obtain: $$a_n(y) = c_n(x)\cases{\sinh(\sqrt{\lambda}y)\sinh(\sqrt{\lambda}(y'-H)) & y<y' \\ \sinh(\sqrt{\lambda}(y-H))\sinh(\sqrt{\lambda}(y')) & y>y'}$$ ". And the jump condition evaluates $c_n$:
$$c_n(x) = \frac{d a_n}{dy}|_{y'-}^{y'+} = \frac{2}{L}\sin(\sqrt{\lambda} x')$$
(which is odd because I thought the jump condition was a difference of derivatives the evaluation of a definite integral). It's then asserted that Green's function is represented as a sum in the form: $\boxed{G(x,y,x',y') = \sum_{n=1}^\infty \phi_n(x) c_n(x) a_n(y)}$.
I am honestly completely lost here. I have no idea where $c_n(x)$ is coming from or how $a_n(x)$ was formulated as I assumed it was simply the solution to the ODE for $a_n$ but when I solve it, I have no functions of x nor do I see where or why we insert $x'$ here.
My work starts below so you can see what I tried:
To summarize, I thought I could use Green's function solution to the ODE for $a_n(y)$ and plug this solution into the eigenfunction expansion solution to $u(x,y)$. I thought maybe that's how Green's function was arrived to but, I don't think that was the case.
I followed the 1-D methodology from my work here except instead of integrating the forcing term, I used the dirac-delta function which is obtained from plugging into the ODE $\nabla^2 G(\mathbf{x},\mathbf{x'}) = \delta(\mathbf{x}-\mathbf{x'})$.
I used these 3 properties of Green's functions to set up the boundary conditions
i. $G(a,y') = G(b,y')$
ii. $G(y'+,y') = G(y'-,y')$
iii. jump condition: $G_{y}(y'+,y')-G_{y}(y'-,y') = 1$
Using (i), I get:
$$0 = \cases{A + B = 0 & y<y' \\ Ce^{\sqrt{\lambda}y} + De^{-\sqrt{\lambda}y} & y>y'}$$
Which gives us:
$$G = \cases{A\sinh(\sqrt{\lambda}y) & y<y' \\ -2C e^{\sqrt{\lambda}L}\sinh(\sqrt{\lambda} (L-y)) & y>y'} $$
Using (ii) I have $$A\sinh(\sqrt{\lambda}y') = -2C e^{\sqrt{\lambda}L}\sinh(\sqrt{\lambda}(L-y')) \\ \rightarrow C = -A\frac{\sinh(\sqrt{\lambda}y')}{2e^{\sqrt{\lambda}L}\sinh(\sqrt{\lambda}(L-y'))}$$
And lastly, using (iii) (instead of the integral variant I mentioned earlier) I get: $$+2\sqrt{\lambda}Ce^{\sqrt{\lambda}L}\cosh(\sqrt{\lambda}(L-y))-\sqrt{\lambda}A\cosh(\sqrt{\lambda}y) = 1 \\ \text{plug in for C: } -\sqrt{\lambda}\Big(\frac{\cosh(\sqrt{\lambda}(L-y))\sinh(\sqrt{\lambda}y')}{\sinh(\sqrt{\lambda}(L-y'))}+\cosh(\sqrt{\lambda}y)\Big) = \frac{1}{A} \\ \rightarrow A = -\frac{1}{\sqrt{\lambda}\Big(\frac{\cosh(\sqrt{\lambda}(L-y'))\sinh(\sqrt{\lambda}y')}{\sinh(\sqrt{\lambda}(L-y'))}+\cosh(\sqrt{\lambda}y')\Big)}$$
$$\rightarrow C = \frac{\sinh(\sqrt{\lambda}y')}{2e^{\sqrt{\lambda}L}\sqrt{\lambda}\Big(\frac{\cosh(\sqrt{\lambda}(L-y'))\sinh(\sqrt{\lambda}y')}{\sinh(\sqrt{\lambda}(L-y'))}+\cosh(\sqrt{\lambda}y')\Big)\sinh(\sqrt{\lambda}(L-y'))} = \frac{\sinh(\sqrt{\lambda}y')}{2e^{\sqrt{\lambda}L}\sqrt{\lambda}\Big(\cosh(\sqrt{\lambda}(L-y'))\sinh(\sqrt{\lambda}y')+\cosh(\sqrt{\lambda}y')\sinh(\sqrt{\lambda}(L-y'))\Big)}$$
So I get for Green's function, $$G = \cases{-\frac{\sinh(\sqrt{\lambda}y')}{\sqrt{\lambda}\Big(\frac{\cosh(\sqrt{\lambda}(L-y'))\sinh(\sqrt{\lambda}y')}{\sinh(\sqrt{\lambda}(L-y'))}+\cosh(\sqrt{\lambda}y')\Big)} & y<y' \\ - \frac{\sinh(\sqrt{\lambda} (L-y))\sinh(\sqrt{\lambda}y')}{\sqrt{\lambda}\Big(\cosh(\sqrt{\lambda}(L-y'))\sinh(\sqrt{\lambda}y')+\cosh(\sqrt{\lambda}y')\sinh(\sqrt{\lambda}(L-y')\Big)}& y>y'} \\ = \sum_{n=1}^\infty -\phi_n(x)\frac{\sinh(\sqrt{\lambda}y')}{\sqrt{\lambda}}\cases{\frac{1}{\Big(\frac{\cosh(\sqrt{\lambda}(L-y'))\sinh(\sqrt{\lambda}y')}{\sinh(\sqrt{\lambda}(L-y'))}+\cosh(\sqrt{\lambda}y')\Big)} & y<y' \\ \frac{\sinh(\sqrt{\lambda} (L-y))}{\cosh(\sqrt{\lambda}(L-y'))\sinh(\sqrt{\lambda}y')+\cosh(\sqrt{\lambda}y')\sinh(\sqrt{\lambda}(L-y')}& y>y'}$$
So in regards to the summary, I can immediately notice in my wording that it doesn't allow for y to fit in, we're only using a Green's function solution to the ODE for $a_n(y)$. I would like to see how this is done as it is not shown in the textbook I am reading.
