# Confusion over the definition of a Green's function

1. May 7, 2014

### TheFerruccio

This is how I learned about Green's functions:

For the 1-D problem with the linear operator L and the inner product,
$(\cdot,\cdot)$,
$Lu(x) = f(x) \rightarrow u=(f(x),G(\xi,x))$

if the Green's function G is defined such that

$L^*G(\xi,x) = \delta(\xi-x)$

I understand how to arrive at this algebraically. However, most articles I read define the Green's function backwards (?) like this:

$LG(x,\xi)=\delta(x-\xi)$

How do I arrive at this definition? As in, how do I work through the algebra to show that the green's function can be defined like this? I am assuming that the swapping of the variables indicates an equivalence between the two definitions, but I do not immediately see it, and it has been confusing me for quite a bit. Does it have something to do with whether we're on the interval [a,b] in $\xi$ vs $x$? Could someone walk me through the steps? No, this is not a homework or coursework question. I'm just confused with the definition.

2. May 7, 2014

### strangerep

Part of the difficulty might be that you haven't indicated explicitly which variable $L$ acts on in each case.

So re-write your earlier equation as
$$u(\xi) ~=~ \Big(f(x),G(\xi,x)\Big)$$
and then apply your operator $L$ to it. But take care: the "x" is a dummy integration variable in the inner product, so your $L$ would need to act on the $\xi$ variable.

You'll still have to work through it to get understanding...

3. May 7, 2014

### TheFerruccio

How do you specify what variable an operator acts on? I thought an operator was implicit, and how it behaves is defined based on the function it's operating on, for example, if

$L=\frac{\partial}{\partial x}$

Then wouldn't the result of that operation be apparent depending on what u is? If u is independent of x, then the Lu would just be 0. That's why I didn't think that I would need to define what variables L is operating on. Also, in the definitions and examples I've worked through, I've never had to be explicit about what variables L was acting on. I had to be explicit about the functions L was acting on, but not the variables, since that depends on a further restriction of L. Wouldn't I lose generality if I explicitly define L like that?

Also, my first equation should be $$u(x) ~=~ \Big(f(x),G(\xi,x)\Big)$$ with $x$ instead of $\xi$ (Line 3 in my first post). So, it's all in functions of x, with the dummy variable of integration in the inner product being $\xi$

Last edited: May 8, 2014
4. May 9, 2014

### strangerep

It depends on the details on the operator. See below.

That depends on the details of the operator. Integral operators are a bit trickier...
Suppose L is just the operator of differentiation. Then you can of course write $f' = L f$ in abstract notation. You could also use concrete notation, e.g., $$f'(x) ~=~ \frac{d}{dx} \; f(x)$$ which contains essentially the same information as $$f'(z) ~=~ \frac{d}{dz} \; f(z) ~,$$ (functionally speaking).

That wouldn't make sense, since in that case you could pull $f(x)$ outside the integral, i.e.,
$$\int d\xi \, f(x) G(\xi,x) ~=~ f(x) \int d\xi\, G(\xi,x) ~.$$But what's actually needed is to "contract" $G$ with $f$.

To explain what I mean by "contract", here's another way to think about Green's functions is as a continuous-index generalization of ordinary matrices. Consider a column vector with components $a_i$, and a 2x2 matrix with components $M_{ij}$. The action of M on $a$ produces a new vector $b$ as follows:$$b_i~=~ \sum_j M_{ij} a_j ~.$$Now think of a function $f$ as a column vector with a continuous index $x$. So I'll write $f_x := f(x)$, etc. The action of the Green's function is $$u_x \equiv u(x) ~=~ \int d\xi \, G_{x\xi} f_\xi ~\equiv~ \int d\xi \, G(x,\xi) \, f(\xi) ~,$$where I hope you can see that the Greens function is analogous to the earlier matrix M, but here we're doing a "contraction" over the "index" $\xi$, implemented as integration instead of discrete summation. A purist might even think of G as an integral operator (instead of just the kernel function of an integral operator as above), and write $u = Gf$ and then the notation exactly parallels the matrix/vector case -- provided one can keep track of the types of all the symbols.

5. May 9, 2014

### TheFerruccio

Oh, that's just a brilliant explanation. Yes, I've dealt with tensors in indicial notation a ton when talking about solid mechanics. What you just explained to me thoroughly unified so many concepts in my head. Thanks! I see where I was erring in my thought process now. I need to make sure to define my contracted variable when specifying the inner product. Thanks!

6. May 9, 2014

### strangerep

I felt the same way when I first saw this. It means that we can think of
$$L u = f$$ as being solved by $$u = L^{-1} f ~,$$and $G$ is just a concrete implementation of $L^{-1}$.

Happy gardening.