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Treatment of Integral as an Operator?

  1. Nov 26, 2013 #1


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    My tutor showed me something today, and I still can't completely wrap my head around why it makes sense. Consider the following integral equation:

    ##\int f(x) = f(x) - 1##


    ##\int (f(x)) - f(x) = -1 \implies f(x) \left( \int (\text{Id}) - 1\right) = -1##

    so we get the geometric series

    ##f(x) = \frac{1}{1 - \int(Id)} = 1 + \int + \int\int + \int\int\int + ... = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + ... = e^x##

    which satisfies our original equation if we choose the right constant of integration. Similarly, take the differential equation (using D as the differential operator):

    ##Df = f - 1 \implies f\left(D(\text{Id}) - 1\right) = -1##

    ##f = 1 + D + (D \circ D) + (D \circ D \circ D) + ... = 1##

    which also satisfies our original equation.

    Is this mathematically valid, or was he pulling my leg? And if so, could someone help me make sense of why this is valid?

    Trying this on some other differential equations seems to show that this doesn't always work -- I was wondering if anyone could point me to an article that says more about the idea behind this.
    Last edited: Nov 26, 2013
  2. jcsd
  3. Nov 26, 2013 #2


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    I would have kept the ##f(x)## on the RHS so that the integral operator acts on it, but there is a way to make the formalism mathematically sound. The right terminology to use when searching on this is the resolvent method for (Fredholm) integral equations. In particular, this seems to be an example of a Liouville-Neumann series solution to the particular equation of interest.

    I don't have any great recommendations for sources to learn more about this, but I have probably encountered integral equations years ago in Whittaker and Watson's Course of Modern Analysis. There are various lecture notes available via google that might be worthwhile as well.
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