Sign of the Green's Function? I have a fairly simply question, but my lecture notes don't give me a specific yes or no answer and my intuition isn't really helping me at all.
$$y''-8y'+16y=0,\ \ \ \ \ \ \ \ \ \ \ \ \ \  0<x<1$$
$$y(0)=y(1)=0$$
I have been given the above homogeneous differential equation and am trying to find the Green's Function. The question gives the form of the Green's Function as:
$$G_1(x,t)=(t-1)e^{-4t}xe^{4x},\ \ \ \ \ \ \ \ \  x<t$$
$$G_2(x,t)=te^{-4t}(x-1)e^{4x}, \ \ \ \ \ \ \ \ \ x>t $$
However my answer is exactly $-1$ times the solution in both regions; my parentheses are just the other way round, so $(1-t)$ and $(1-x)$ rather than $(t-1)$ and $(x-1)$.
I think it's because when I was calculating the GF I did the following:
$$G_1(x,t)=e^{4x}(\alpha(t)+x\beta(t))$$
$$G_2(x,t)=e^{4x}(\gamma(t)+x\delta(t))$$
By matching BC's and the continuity/discontinuity conditions I found:
$$\gamma(t)=-\delta(t)$$
I proceeded to eliminate $\delta$ from my equations. If, on the other hand, I had chosen to eliminate $\gamma$ (I think) I would get the same answer as the question provides.
Do I need to even worry about this detail? Is it the case that, since both parts of the GF are multiplied by a factor of $-1$ it doesn't actually matter?
 A: It does make a difference: one is the response of the system to a forcing of $\delta(x)$, the other to $-\delta(x)$.
The only place in a homogeneous system that the coefficients are determined to be different from zero is in the discontinuity, or "jump" condition, obtained by integrating the equation over the discontinuity of the delta function:
$$ \left[ G_x(x,t) \right]_{t^{-}}^{t^+} = 1. $$
(This is intuitively obvious: if the delta function weren't there, the solution would have to be identically zero to match both sets of boundary conditions.)
In practical terms, for your example this means that
$$ \lim_{x \to t} \partial_x G_2(x,t)-\partial_x G_1(x,t) = 1 $$
(one should be careful in general to make the limits one-sided in the region of definition, but in this case $G_1$ and $G_2$ are regular enough to get away with being less precise, and this makes the notation easier).
The boundary condition at $x=0$ gives $\alpha=0$, that at $x=1$ gives $\gamma+\delta=0$, and we are left with
$$ G_1(x,t) = xe^{4x} \beta(t), \quad G_2(x,t) = (1-x)e^{4x}\gamma(t). $$
Continuity at $x=t$ gives
$$ te^{4t}\beta(t)=(1-t)e^{4t}\gamma(t), $$
and the jump condition is
$$ (3-4t)e^{4t}\gamma(t)-(1+4t)e^{4t}\beta(t)=1. $$
These are two linear equations for $\beta,\gamma$, and the Wronskian doesn't vanish for the two solutions we started with, so the corresponding matrix is invertible and the solution is unique: substituting in $\beta(t)=(t-1)e^{-4t}$ and $\gamma(t)=-te^{-4t}$ shows that the question's expression is this solution.
