
greens functions
i am asked to find the greens function for
U' +kU=f(x)
with u(0)=0 and x>=0
i found the homogenous solution and used the 4 conditions...
my homogen soln was u=constant * exp(kU)....
is it possible that my G(x,s)= 0 for x<s and X>s?
also how do i show that u"= f(x) is not self adjoint with bc
u(0)=0 and u'(0) +u(1)=0
i have the greens function but the bc's dont hold for it?
is that right?

My understanding is if your equation is
$\displaystyle Lu(x) = f(x),$
then the Green's function is the solution to
$\displaystyle Lg(x;s) = \delta(xs)$.
So when you solve this equation you have in the first case
$\displaystyle g'(x;s) +kg(x;s) = \delta(xs)$
multiply by the integrating factor $\displaystyle \exp(kx)$ you have
$\displaystyle \frac{d}{dx}\left[g(x;s)\exp(kx)\right] = \delta(xs)\exp(kx)$,
and integrating
$\displaystyle g(x;s)\exp(kx) = H(xs)\exp(ks) + C$
Enforcing the $\displaystyle g(0;s)=0$ boundary condition and assuming $\displaystyle x,s > 0$ we can deduce $\displaystyle C=0$ and therefore
$\displaystyle g(x;s) = H(xs)\exp[k(xs)].$
Hope this helps. You might need to give more complete details to answer the second part.