How is green's function a right inverse to the operator L?

[ENgez]

the definition of a green's function is:
$LG(x,s)=δ(x-s)$

the definition of a right inverse of a function f is:
$h(y)=x,f(x)=y→f°h=y$

how does it add up?

 

[homeomorphic]

You are confusing two different (but related) meanings of the word inverse.

You want to say that the delta function is the identity with respect to the convolution.

In the discrete version of this, L would be a difference operator, which could be represented by a matrix. The delta function would be replace by Kronecker delta, which has the same components as the identity matrix.

 

[ENgez]

you mean ∫G(x,s)δ(x-s)ds=1, with integration over all domain of s?

 

[homeomorphic]

you mean ∫G(x,s)δ(x-s)ds=1, with integration over all domain of s?

No. What I mean is that L * G (* means convolution) is equal to the identity with respect to convolution, which is the delta function. The delta function is the identity because convolving it with a function just gives you the function.

 

[Charles49]

Let's say you want to find a function y so that Ly=f. Then first solve Ly=δ and call that solution G. Since LG=δ, L(G*f)=f because L(G)=δ and δ*f=f. So the solution would be y=G*f.

Example: Solve y"+y=cos(x).

Solution: The operator here is L=D^2+1, where D=d/dx. The wikipedia entry on greens function tells you that the Green's function of D^2+1 is sin(x). So y=sin(x)*cos(x) would solve the above example.

 

[Charles49]

Here the convolution is the one used in Laplace transforms.