Darn those Green's Functions

**lvleph** · August 3rd 2010, 10:03 AM

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

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

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

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

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

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

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

This can't be right.

**lvleph** · August 3rd 2010, 04:19 PM

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

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

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

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

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

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

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

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

Forgot that I was dealing with different constants.

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