# I don't understand integral operators

1. Dec 16, 2012

### fred_91

1. The problem statement, all variables and given/known data

I am struggling to understand the meaning behind: an operator of an operator:
Let A and B be 2 volterra integral operators.
If A and B are 2 integral operators, what does the following mean:

$(\textbf{A}) (\textbf{B}^{-1}(f))(x)$

2. Relevant equations
We have the following definition of the operator:
$(\textbf{B}f)(x)=\int_0^x K(t,x)f(t)dt$ (1)

3. The attempt at a solution

I guess:
The operator A will act on the inverse of operator B.

I don't understand what the inverse operator means.
If the operator $\textbf{B}$ is given by equation (1), what is its inverse $\textbf{B}^{-1}(f)(x)$?

Thank you very much in advance.

2. Dec 16, 2012

### lurflurf

There is such a thing as an operator acting on an operator, but I think you are talking about a composition of operator.

so A B^-1 f
means do A to what you get when you undo B from f

Depending on the situation one might solve your equation (1) to compute B^-1 f
Other times one knows that
$(B^{-1}f)(x)=\int_0^x K_1 (t,x)f(t)dt$
For some K1
That can be quite difficult at time to compute from K, sometimes we need only to know that the function exists.

3. Dec 16, 2012

### fred_91

Thank you very much.

''Depending on the situation one might solve your equation (1) to compute B^-1 f''

OK, So, let's say K(t,x)=sin(t-x).
I will solve:

$\int_0^x K(t,x)f(t) dt=g(x)$ (2)

for some function g(x).

Once I have solved this, I will have a solution for f(x) in terms of the function g(x).
How will I have the inverse operator $(\textbf{B^{-1}}f)(x)$?
Will the inverse operator be the solution of the integral equation (2)?

4. Dec 16, 2012

### Kreizhn

I think that lurllurf is saying that there are times when you are fortunate enough to know that the inverse of the integral operator is also an integral operator which may be prescribed by some integrability kernel. In such a case, there are things you can do you track down the form of the kernel. However, this is certainly not always true.

I am slightly confused by your above example though. If you have the kernel corresponding to the inverse, you do not need to do any solving of one function in terms of the other: you already have a closed form of the inverse!

5. Dec 16, 2012

### fred_91

We already have a closed form of the inverse?

So, if the operator $(\textbf{B}f)(x)$
is given by
$(\textbf{B}f)(x)=\int_0^x sin(t-x)f(t)dt$,

the what would the inverse be?
$(\textbf{B^{-1}}f)(x)$

Sorry if this is a stupid question, I think I'm confused about what an inverse integral operator is.

6. Dec 16, 2012

### Ray Vickson

If $B(g) = f$, then $g = B^{-1}(f).$ Essentially, $B^{-1}(f)$ is the solution of the integral equation $B(g) = f$. Of course, there will be issues of existence and uniqueness, etc.

7. Dec 17, 2012

### lurflurf

If you have done some linear algebra it is quite analogous.
In linear algebra one might need to compute
AB-1x
the operation Ax can (given a basis) be found by a sum (analogous to the integral in integral operators)
$$Ax=\sum_{k=1}^n A_{jk}x_k$$
we might compute the inverse of B then we do A to B-1 done to x
If the inverse cannot be found accurately and inexpensively we might prefer to first solve
By=x for y
There are many methods we might use like iterative and direct methods
and then
AB-1x=Ay

So back to integral operators,
often to compute
AB-1x
we solve
Bg=f for g then
AB-1f=Ag
Using whatever method we like
On a lucky day we might know the general expression for B-1 from software, a table, a usual method or whatever.
Then
AB-1f is easily computed
find y=B-1f
then AB-1f=Ay

No examples came to me so I got this one
http://eqworld.ipmnet.ru/en/solutions/ie/ie0124.pdf
here
http://eqworld.ipmnet.ru/en/solutions/ie/ie-toc1.htm
$$By(x)=\int_a^x \cos [ \lambda (x-t) ] y(t) \mathop{\text{dt}}=f(x)$$
with f(a)=0 of course, then
$$B^{-1}f(x)=f ^\prime (x)+\lambda^2 \int_a^x f(t) \mathop{\text{dt}}=y(x)$$
a particular example
$$B^{-1}x^2=(x^2) ^\prime +3^2 \int_0^x t^2 \mathop{\text{dt}}=2x+3x^3$$
so
$$B(2x+3x^3)=\int_0^x \cos [ 3 (x-t) ] (2t+3t^3) \mathop{\text{dt}}=x^2$$

Last edited: Dec 17, 2012