A question on integral equations

  • Thread starter eljose79
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Main Question or Discussion Point

Let be the integral equation:

f(x)-g(x)=Kf where K is the integral operator having the kernel

my question : is there an operator R satisfying:

g(x)-f(x)=Rg where f satisfy the original integral equation so it can be solved by mean of R operator.


Another question is can any integral equation be transformed into a differential equation?..if so how it is made?..thanks.
 

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Tom Mattson
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Originally posted by eljose79
Let be the integral equation:

f(x)-g(x)=Kf where K is the integral operator having the kernel
OK

my question : is there an operator R satisfying:

g(x)-f(x)=Rg where f satisfy the original integral equation so it can be solved by mean of R operator.
Just using algebra, I get:

1. f(x)-g(x)=Kf(x)
2. f(x)-Kf(x)=g(x)
3. (1-K)f(x)=g(x)
4. f(x)=(1-K)-1g(x)
5. g(x)-f(x)=[1-(1-K)-1]g(x)

Defining R as [1-(1-K)-1] seems to fit the bill. The only thing I am not sure of is whether (1-K) is invertible. I assumed that it was invertible in Step 4.

edit: fixed bracket
 
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