Confusion about Green's functions for inhomogeneous boundary conditions but no forcing. I have recently seen Green's functions for the first time but for some reason can't seem to get my head around them. I have been told we can use them to solve differential equations up to an integral, however I am having difficulties seeing why they should work (probably from my inexperience).
For example if we are asked to solve $\mathcal{L}(u(x)) = f(x)$ on say $[0,1]$ with $u(0) = u(1) = 1$ where $\mathcal{L}$ is any second order Sturm-Liouville operator (say take $\mathcal{L} = \frac{d^2}{dx^2}$) then with Green's functions we would have $G(x,\xi)$ satisfies $\mathcal{L}(G) = \delta(x-\xi)$ and our solution for $u$ is $u(x) = \int\limits_0^1 G(x,\xi) f(\xi)\ d\xi$. 
Now consider the case where $f = 0$ everywhere. This means that in our integral we have $f(\xi) = 0$ everywhere so we get $u(x) = 0$ everywhere as $G$ has to be bounded (at least that is what I think it has to based on all the examples I have seen so far) which is wrong.
I will be glad if someone could point out to me where I am making my mistake. Thank you in advance.
 A: The formula $u(x) = \int\limits_0^1 G(x,\xi) f(\xi)\ d\xi$ is only valid for homogeneous boundary conditions (the same kind of boundary conditions that $G$ itself satisfies).  To use Green's function for inhomogeneous boundary conditions you have two options: 


*

*Pick a function $u_0$ that satisfies the boundary conditions, and write $u = u_0 +w$. Now $w$ satisfies $\mathcal L(w) = \tilde f$ where $\tilde f = f - \mathcal L(u_0)$ so it can be found as $w(x) = \int\limits_0^1 G(x,\xi) \tilde f(\xi)\ d\xi$, and then you get $u$.

*Modify the integral formula so it works for inhomogeneous boundary conditions, by performing integration by parts in 
$$
u(x) = \int_0^1 u(\xi)\delta_{x=\xi}\,d\xi =  \int_0^1 u(\xi)\mathcal L_\xi(G(x, \xi))\,d\xi = \cdots
$$
For example, if $\mathcal L = d^2/dx^2$, then we get 
$$
\int_0^1 u(\xi) \frac{d^2}{d\xi^2} G(x, \xi)\,d\xi 
= u(\xi)\frac{d}{d\xi} G(x, \xi)\bigg|_{\xi = 0}^{\xi=1} - \int_0^1 u'(\xi) \frac{d}{d\xi} G(x, \xi)\,d\xi  \\ 
 = u(\xi)\frac{d}{d\xi} G(x, \xi)\bigg|_{\xi = 0}^{\xi=1} +  \int_0^1 f(\xi) G(x, \xi)\,d\xi 
$$
If $f\equiv0$, the last integral vanishes. The term $u(1)\frac{d}{d\xi} G(1, \xi) -  u(0)\frac{d}{d\xi} G(0, \xi) $ reflects the contribution of boundary conditions. Green's function is still involved, but now in the form of its derivative on the boundary. 
This works in higher dimensions too, where we get an integral over the boundary, involving the normal derivative of $G$.
