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What I am trying to find is also referred to as $\tilde{G}$ and I was trying to verify what @Gonçalo had found for it in their answer to MY QUESTION. I'm using the exact same inhomogeneous differential equation with inhomogeneous boundary conditions (BCs) from the question, but I'm focusing on a single part of the problem which is, again, obtaining $\tilde{G}$ as it is defined in the answer. I can't figure out exactly where I'm screwing up and I'm beginning to suspect that my error is systematic because I've gone over the work several times now and I don't see any silly mistakes.

Here is the definition of $\tilde{G}$ and its conditions as was provided by @Gonçalo:

$\tilde{G}_{xx} = \delta(x-x')$ is satisfied by

$\tilde{G} = \cases{\frac{1}{2}(x+1)(x'-1) & x<x' \\ \frac{1}{2}(x-1)(x'+1) & x>x'}$ which is verified using the following conditions: $\text{(i)} \tilde{G}(±1,x′)=0, \text{(ii)} \tilde{G}(x′+0,x′)=\tilde{G}(x′−0,x′), \text{and (iii)} \tilde{G}_x(x′+0,x′)−\tilde{G}_x(x′−0,x′)=1$

My work starts here:

$$\text{starting with homogeneous eqn for }\tilde{G}:\\ \tilde{G}_{xx} = 0 \\ \text{solve: }\int\int\tilde{G}_{xx} = 0 \rightarrow \tilde{G}_{xx} = Ax+B \\ \text{from (i), the solution satisfying BC1: } A_1(-1) + B_1 = 0 \rightarrow B_1 = A_1 \\ \text{from (i), the solution satisfying BC2: } A_2(1) + B_2 = 0 \rightarrow B_2 = -A_2 \\ \text{gives 2 solutions: }\tilde{G} = \cases{A_1x+A_1 = A_1(x+1) \\ A_2x-A_2 = A_2(x-1)} \\ \text{apply (ii): } A_1(x'+1) = A_2(x'-1) \\ A_1 = A_2 \frac{(x'-1)}{(x'+1)} \\ \text{apply (iii): } A_1 - A_2 = 1 \rightarrow A_1 = 1 + A_2 \\ \text{plug into (ii) and solve for constants: } 1+A_2 = A_2\frac{(x'-1)}{(x'+1)} \\ \frac{1+A_2}{A_2} = \frac{(x'-1)}{(x'+1)} \rightarrow \frac{1}{A_2} = \frac{(x'-1) - (x'+1)}{(x'+1)} \rightarrow \frac{1}{A_2} = -\frac{2}{(x'+1)} \\ A_2 = -\frac{1}{2}(x'+1) \\ A_1 = 1+A_2 = \frac{2}{2} - \frac{1}{2}(x'+1) \rightarrow \frac{1}{2}(-x'+1) \\ A_1 = -\frac{1}{2}(x'-1)$$

This gives: $$\tilde{G} = \cases{-\frac{1}{2}(x+1)(x'-1) & x<x'\\ -\frac{1}{2}(x-1)(x'+1) & x>x'}$$

I've gone over the work several times already, and I can't figure out how to get rid of this pesky negative sign, so I I'm starting to think the error I'm making is systematic and I don't know what step it is that I'm messing up or glossing over. I'm thinking my inequality might have been set up wrong such that it would instead be $$\tilde{G} = \cases{-\frac{1}{2}(x+1)(x'-1) & x>x'\\ -\frac{1}{2}(x-1)(x'+1) & x<x'} \\ \text{negative sign flips the inequality: } \tilde{G} = \cases{\frac{1}{2}(x+1)(x'-1) & x<x'\\ \frac{1}{2}(x-1)(x'+1) & x>x'}$$. I don't actually understand how the inequality is supposed to be set up to begin with, so I'm thinking that could be it. Any help/suggestions would be appreciated!

Researcher R
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    You are using the indices $1$ and $2$ for the regions $x<x'$ and $x>x'$, respectively. Therefore, the equation after "apply (iii)" should be $A_2-A_1=1$. – Gonçalo Nov 04 '23 at 03:42
  • @Gonçalo Oh that's right! You just helped me realize the physical meaning of this form of what you had written: $\big(\lim_{x \rightarrow x_0^+} \frac{\partial^{n-1}G}{\partial x^{n-1}}\big) - \big(\lim_{x \rightarrow x_0^-} \frac{\partial^{n-1}G}{\partial x^{n-1}}\big) = 1$. The first term takes the derivative as $x$ goes to $x_0^+$ meaning that (if we're on a number line) we're approaching $x_0$ from the right, and from the piecewise definition of $\tilde G$, we would take the derivative of the part defined on $x>x_0$. Conversely we would do the same for the part defined on $x<x_0$. – Researcher R Nov 04 '23 at 22:16
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    @Gonçalo When I see some of this notation come back up from Calculus I, I often feel intimidated and lose my head dealing with something that would otherwise be trivial. I ended up psyching myself out on this one. Thanks again! – Researcher R Nov 04 '23 at 22:17

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