# greens functions

• Jan 25th 2009, 08:17 PM
pepe
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?
• Jan 26th 2009, 04:50 AM
Rincewind
My understanding is if your equation is

$Lu(x) = f(x),$

then the Green's function is the solution to

$Lg(x;s) = \delta(x-s)$.

So when you solve this equation you have in the first case

$g'(x;s) +kg(x;s) = \delta(x-s)$

multiply by the integrating factor $\exp(kx)$ you have

$\frac{d}{dx}\left[g(x;s)\exp(kx)\right] = \delta(x-s)\exp(kx)$,

and integrating

$g(x;s)\exp(kx) = H(x-s)\exp(ks) + C$

Enforcing the $g(0;s)=0$ boundary condition and assuming $x,s > 0$ we can deduce $C=0$ and therefore

$g(x;s) = H(x-s)\exp[-k(x-s)].$

Hope this helps. You might need to give more complete details to answer the second part.