# Thread: Darn those Green's Functions

1. ## Darn those Green's Functions

Consider the differential operator $\displaystyle Lu = u_{xx} + u_x$ on $\displaystyle (-1,0)$ with boundary conditions $\displaystyle u'(-1)-3u(-1)=u'(0)=0$

Find the Green's function for $\displaystyle L$ with the boundary conditions

My attempt:
$\displaystyle G_{xx}(x,\xi) + G_x(x,\xi) = \delta(x - \xi)$
$\displaystyle G_x(-1,\xi)-3G(-1,\xi)=G(0,\xi) = 0$

For $\displaystyle x\ne\xi$
$\displaystyle G_{xx}(x,\xi) + G_x(x,\xi) = \0$
$\displaystyle G(x,xi) = A(\xi) e^{-x} + B(\xi)$

For $\displaystyle x < \xi$
$\displaystyle G_x(-1,\xi) - 3G(-1,\xi) = -A(\xi)e - 3A(\xi)e - 3B(\xi) = 0$
$\displaystyle A(\xi) = -\frac{3}{4e} B(\xi)$
$\displaystyle \Rightarrow G(x,\xi) = \left(-\frac{3}{4e} e^{-x} + 1\right) B(\xi)\; \forall x < \xi$

For $\displaystyle x > \xi$
$\displaystyle G_x(0,\xi) = -A(\xi) = 0$
$\displaystyle \Rightarrow G(x,\xi) = B(\xi)\; \forall x>\xi$

Green's functions are continuous and so for $\displaystyle x = \xi$
$\displaystyle \left(-\frac{3}{4e} e^{-\xi} + 1\right) B(\xi) - B(\xi) = 0$
$\displaystyle B(\xi) = 0$

This can't be right.

2. My attempt #2:
$\displaystyle G_{xx}(x,\xi) + G_x(x,\xi) = \delta(x - \xi)$
$\displaystyle G_x(-1,\xi)-3G(-1,\xi)=G(0,\xi) = 0$

For $\displaystyle x\ne\xi$
$\displaystyle G_{xx}(x,\xi) + G_x(x,\xi) = 0$
$\displaystyle G(x,xi) = A(\xi) e^{-x} + B(\xi)$

For $\displaystyle x < \xi$
$\displaystyle G_x(-1,\xi) - 3G(-1,\xi) = -A_1(\xi)e - 3A_1(\xi)e - 3B_(\xi) = 0$
$\displaystyle A_1(\xi) = -\frac{3}{4e} B_1(\xi)$
$\displaystyle \Rightarrow G(x,\xi) = \left(-\frac{3}{4e} e^{-x} + 1\right) B_1(\xi)\; \forall x < \xi$

For $\displaystyle x > \xi$
$\displaystyle G_x(0,\xi) = -A_2(\xi) = 0$
$\displaystyle \Rightarrow G(x,\xi) = B_2(\xi)\; \forall x>\xi$

Green's functions are continuous and so for $\displaystyle x = \xi$
$\displaystyle \left(-\frac{3}{4e} e^{-\xi} + 1\right) B_1(\xi) = B_2(\xi)$

So now I also want to ensure that the first derivative has the proper jump, i.e.
$\displaystyle \lim_{\epsilon \to 0} \int_{\xi - \epsilon}^{\xi + \epsilon}\! G_{xx}(x,\xi)\, dx = 1$
$\displaystyle 0 - B_1(\xi) \frac{3}{4} e^{-\xi - 1} = 1 \Rightarrow B_1(\xi) = -\frac{4}{3} e^{\xi + 1}$

$\displaystyle \Rightarrow 1 -\frac{4}{3} e^{\xi + 1} = B_2(\xi)$

Therefore,
$\displaystyle G(x,\xi) = \begin{cases} e^{\xi - x} - \frac{4}{3}e^{\xi + 1} & x<\xi \\ 1 -\frac{4}{3} e^{\xi + 1} & x > \xi \end{cases}$

Forgot that I was dealing with different constants.