Greens functions inspection method

1. Jun 25, 2012

center o bass

Hi! A nice method for solving differential equations that occur in physics utilize greens functions as well as the fourier transform of the delta function to solve them partially by inspection. For example in electromagnetism Poisson's equation reads
$$\nabla^2 \phi = - \rho/\varepsilon_0$$
which can be solved by using the greens function
$$\nabla^2 G(\vec r,\vec r') = \delta(\vec r-\vec r') = \frac{1}{(2\pi)^3} \int e^{\vec k \cdot(\vec r-\vec r')}d^3\vec k$$
where the right hand side is the fourier integral of the 3 dimensional delta function (integration is understood to be over all space). Now if we inspect the equation above one sees that
$$G(\vec r,\vec r') = -\frac{1}{(2\pi)^3} \int \frac{e^{\vec k \cdot(\vec r-\vec r')}}{k^2}d^3\vec k$$
satisfies it. This is the 'inspection step'.
Now by performing the integration one gets
$$G = - \frac{ 1}{4\pi |\vec r - \vec r'|}$$
and by substituting back into
$$\phi = \int G(\vec r, \vec r') (\frac{-\rho(\vec r')}{\varepsilon_0}) d^3\vec r' = \frac{1}{4\pi \varepsilon_0 }\int \frac{\rho(\vec r')}{|\vec r - \vec r'|} d^3\vec r'$$
one get the correct equation for the electrostatic potential. These kind of arguments inspired me to try the same thing out for one dimensional equations like for example

$$(\frac{d^2}{dx^2} + 1)f(x) = \frac{1}{sin(x)}$$
Using the same procedure
$$(\frac{d^2}{dx^2} + 1)G(x,x') = \delta(x-x')= \frac{1}{2\pi}\int e^{ik(x-x')}dk$$
inspection indicates that
$$G(x,x') = \frac{1}{2\pi}\int\frac{ e^{ik(x-x')}}{1-k^2}dk$$
But solving this equation by complex contour integration I get an imaginary greens function which is nonsense.

Specifically I did it like
$$G(x,x') = \frac{1}{2\pi}\int\frac{ e^{ik(x-x')}}{1-k^2}dk = \frac{1}{2\pi (x-x')} \int_{-\infty}^{\infty} \frac{e^{iy}}{1-(\frac{y}{x-x'})^2}dy = \frac{2 \pi (x-x')}2\pi i = \frac{i}{x-x'}$$
which would give a solution
$$f(x) = i \int \frac{1}{x-x'} \frac{1}{\sin x'} dx'$$

I see that we get some problems above with divergences when x=x', however I do not see that we do _not_ get these problems in the higher dimensional examples.

So the question is if it's something special with my electromagnetic example above or numerous other examples (from physics) which allows them to be solved by this method? And what is it about the one dimensional example above which does not allow it to be solved by this method?

Is there some way that allowes it to be solved?

2. Jul 7, 2012

jasonRF

Up until this point you are fine. It is in the contour integration that you made a mistake. I did it this way: define $$f(k)=\frac{ e^{ik(x-x')}}{2\pi(1-k^2)}$$ and consider
$$\int_\gamma dk \, f(k)$$
where $\gamma$ has several parts: 1) goes along the real axis from $-R$ to $-1-\epsilon_1$, 2) is a ccw semicircle of radius $\epsilon_1$ about $k=-1$, 3) along the real axis from $-1+\epsilon_1$ to $1-\epsilon_2$, 4) a ccw semicircle of radius $\epsilon_2$ about $k=1$, 5) along the real axis from $1+\epsilon_2$ to $R$, and 6) is a ccw semicircle of radius $R$. Take limit as $R\rightarrow\infty, \, \epsilon_1\rightarrow 0,\epsilon_2\rightarrow 0$ and use residue theorem. I end up with,
$$G(x,x') = \frac{\sin( x' - x)}{2}$$

I may have made a mistake, but at least this may help.

jason

3. Jul 11, 2012

center o bass

Thank you Jason. I checked it up and it turned out that the answer was sin(x-x'). I then get an answer

$$f(x) = \int_a^b \frac{\sin(x-x')}{\sin x'} dx' = \left[\sin x \ln(\sin x') - \cos x x'\right]_a^b$$

if I now chose a = x then this produces the correct answer. However I am unsure of an argument on how to choose the suitable integration limits. The method seem to produce correct special solutions, but not without somehow setting x=x'.