I know this is a dumb question, but I'm having some trouble with the notation.
In 1-D, the solution to a BVP with some linear differential operator $$Lu(x) = f(x)$$ is of the form: $$u(x)=\int_{-\infty}^{\infty} f(\zeta)G(x,\zeta)d\zeta$$ where $f(\zeta)$ is the Admissible Function and $G(x,\zeta)$ is referred to as the Test Function as well as Green's Function. I'm focusing on the Admissible function as trying to compute the integral with it has been rather troublesome.
is $f(\zeta)$ simply the RHS of the PDE with $\zeta$ plugged in? I ask because in some cases, the RHS is a function which does not allow the integral to converge and now I'm second guessing myself on whether I've understood the sources I've read on it. Is it NOT always an infinite integral perhaps?
EDIT: I tried to do this for a 1-D Diffusion Equation: $$u_t = u_{xx}; t\ge0, -\infty < x <\infty \\ u(x,0) = \delta(x)$$ According to the solution I found, we simply take the Green's Function $G(x,\zeta,t) = \frac{1}{2\sqrt{\pi Dt}}e^{-(x^2)/4Dt}$ (you can find this is the table of Green's Functions on the wiki page for Green's Function) with D = 1 for our problem, and we plug into $$u(x,t) = \int_{-\infty}^{\infty}f(\zeta)G(x,\zeta,t)d\zeta$$ the initial condition for $f(\zeta)$ and Green's Function to get: $$u(x,t) = \frac{1}{2\sqrt{\pi t}}\int_{-\infty}^{\infty} \delta(\zeta)e^{-(x^2)/4t}d\zeta = \frac{1}{2\sqrt{\pi t}}e^{-(x^2)/4t}$$. This is the problem that has me confused for what we plug in for $f(\zeta)$ because we aren't plugging in the RHS like I saw in other examples like $$u_{xx} = -2$$ where in the solution -2 was plugged in (This was worked out here on slide 22). Also in that example it was a spatially limited domain of $0\le x \le L$
EDIT 2: What I'd like to compute would be the same 1-D Heat Diffusion Equation if the Initial Condition wasn't a Dirac Delta Function and was instead $u(x,0) = x^2$? Would I have to do a special step or would it be impossible using this Green's Function.