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Consider the following equation where $p(u)$ is a probability distribution and where $g$ is the unknown : \begin{equation} g = \int_{-\infty}^\infty p(u)f(u,g) \mathrm{d} u \end{equation}

Using the definition of a functional derivative : $$ \frac{\delta F[f(x)]}{\delta f(y)}=\lim _{\epsilon \rightarrow 0} \frac{F[f(x)+\epsilon \delta(x-y)]-F[f(x)]}{\epsilon}$$

Am I allowed to say that the first equation is equivalent to :

$$ \frac{\delta g[p(u)]}{\delta p(u_0)} = f(u_0,g)$$

If not, where is my mistake ?

I am trying to understand how to invert definite integral equations. Any reference or advice is always welcome.

Thank you.

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  • $\begingroup$ Is $g$ a real number and $f$ a mapping from $\mathbb{R}^2$ to $\mathbb{R}$? Do you know the probability distribution $p(u)$? And finally, is your ultimate goal solving the integral equation numerically for $g$? $\endgroup$
    – J. Heller
    Jun 24, 2020 at 20:22

1 Answer 1

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If you define the functional $F$ of the function $p$ as $$ F[p] = \int_{-\infty}^{\infty} L(u,p(u)) \, du = \int_{-\infty}^{\infty} p(u) f(u,g) \,du $$ then the functional derivative $\delta F / \delta p$ is $f(u,g)$. It seems that you are viewing your integral equation as a recursively defined functional with $g=F$. This seems to me to be over-complicating things. It is simpler to just solve the integral equation for the unknown real number $g$ using numerical integration and Newton's method. Define $$ I(g) = \int_{-\infty}^{\infty} p(u) f(u,g) \,du. $$ Note that $I(g)$ is simply a function from $\mathbb{R}$ to $\mathbb{R}$. Then your integral equation is $$ g - I(g) = 0. $$ The Newton's method iteration for this equation is $$ g_{n+1} = g_n - \frac{g_n - I(g_n)}{1 - I'(g_n)} $$ where $$ I'(g_n) = \int_{-\infty}^{\infty} p(u) f_y(u, g_n) \,du. $$ If there is no closed form for $I(g)$ or $I'(g)$, you need to use some form of numerical integration (Gaussian quadrature would probably be the most accurate and efficient) to evaluate $I(g_n)$ or $I'(g_n)$.

Consider one of the simplest probability distributions, $p(u) = \delta(u - u_0)$, where $\delta$ is the Dirac delta function. In this case, $I(g)$ has the closed form $f(u_0,g)$. The equation to solve is then $g - f(u_0,g) = 0$. In general, you still need to use Newton's method to solve this. The obvious exception is for $f$ such that there is some closed form solution.

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