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Find all continuous functions satisfying a given integral equation

by RoRi

Find all functions which are continuous on the positive real axis and satisfy the equation

    \[ f(x) = 1 - x \int_1^x f(t) \, dt \qquad \text{for all } x > 0. \]

One such function is

    \[ f(x) = xe^{\frac{1-x^2}{2}} - xe^{-\frac{x^2}{2}} \int_1^x t^{-2} e^{\frac{t^2}{2}} \, dt. \]

We differentiate both sides of the equation the functions we are looking for must satisfy,

    \[ f(x) = 1-x \int_1^x f(t) \, dt \quad \implies \quad f'(x) = -xf(x) - \int_1^x f(t) \, dt. \]

Since x> 0 we know we can solve the original equation for \int_1^x f(t) \, dt,

    \[ f(x) = 1 - x \int_1^x f(t) \, dt \quad \implies \quad \int_1^x f(t) \, dt = \frac{1-f(x)}{x}. \]

Substituting this into the first equation we then have

    \[ f'(x) = -xf(x) - \frac{1-f(x)}{x} \quad \implies \quad f'(x) + \left( x - \frac{1}{x} \right)f(x) = -\frac{1}{x}. \]

Now this is a differential equation of the form y' + P(x) y = Q(x) so by Theorem 8.3 we know the solutions for any given initial condition f(a) = b exist and are unique. Taking a = b = 1 we have that the given function f(x) is the only one satisfying both conditions of continuity and the given equation. \qquad \blacksquare

2 comments

  1. Anonymous says:

    Why is a=b=1 neccessary for the solution to be continuous? Every a > 0 can be used to express the solution in terms of arbitrary b.

    • Mohammad Azad says:

      Continuity allowed us to differentiate the integral at the start and the given equation gave us the condition f(1)=1 which guarantees the uniqueness of the function

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