# Turning a definite integral equation into a differential one.

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

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.

• 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 $g$? Jun 24, 2020 at 20:22

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.