# Relationship between integral equations and partial differential equations

In my functional analysis class, we did a lot of problems involving integral equations such as proving existence & uniqueness using spectral theory and the Banach fixed point theorem.

I've never seen integral equations outside of functional analysis, but apparently they are useful for ordinary/partial differential equations. If someone familiar with integral equation methods could give some motivation, I would really appreciate it.

Also, are there any good textbooks that discuss integral equation methods for PDE's?

• Already integral equations are used in the proof of the existence and uniqueness theorem for ODE. For a book, DiBenedetto's 'Partial Differential Equations' has a discussion of integral equations (he treats somewhat explicitly the double layer potential method for the Laplacian). Dec 8, 2012 at 20:53
• Thanks for the recommendation! I also found a nice chapter on integral equations in Partial Differential Equations by David L. Colton. Dec 8, 2012 at 22:34
• There is also a chapter on this in Handbook of Integral Equations Second Edition Andrei D . Polyanin and Alexander V . Manzhirov Chapman and Hall/CRC 2008
– kelu
Mar 24, 2014 at 6:56
• An excellent place to start is Courant-Hilbert, Volume II. This work is written as the level of Advanced Calculus. The chapter on Potential Theory is particularly well-done and deals with integral equations and methods of solution based on operator expansions. This material is presented in an elegant and classical way. It should be very accessible to someone with your background, and should prove useful in understanding classical applications. Jan 17, 2016 at 17:06

Here's a classic motivating example. Let $H$ be the Heaviside unit step function $$H(x) = \begin{cases} 1 & x > 0 \\ 0 & x \leq 0 \end{cases},$$ and suppose we want to solve the differential equation $$u' = H.$$ We attack each part of $H$ separately: \begin{align*} x > 0 &\implies u(x) = x + c_0 &\mbox{ for some } c_0 \in \mathbb R, \\ x \leq 0 &\implies u(x) = c_1 &\mbox{ for some } c_1 \in \mathbb R. \end{align*} Now our differential equation has $u'$ in it; this means that $u$ is differentiable, and hence $u$ is continuous. Continuity of $u$ at $0$ enforces the constraint $c_0 = c_1$. Hence, letting $c := c_0 = c_1$, we get $$u(x) = \begin{cases} x + c & x > 0 \\ c & x \leq 0 \end{cases}.$$
We've run into the problem: we cannot differentiate $u$ at $0$, as we get $1$ from above and $0$ from below. Hence there is no differentiable function $u$ satisfying $u' = H$. Uh oh.
Let's see the magic that happens if we turn our differential equation into an integral equation by first multiplying by a "nice" test function $\varphi$ and integrating. By nice, we require that we can take derivatives of $\varphi$, and that $\varphi$ vanishes at infinity. Our integral equation is $$\int_{\mathbb R} \varphi u' = \int_{\mathbb R} \varphi H.$$ Let's see if $u$ works in this equation. Starting with the LHS, and integrating by parts \begin{align*} \varphi u \big|_{-\infty}^\infty - \int_{\mathbb R} u \varphi' &= - \int_{\mathbb R} u \varphi ' \\ &= - \int_{-\infty}^0 c \varphi' - \int_0^\infty (x + c) \varphi' \\ &=- \int_0^\infty x \varphi' \\ &=- x \varphi \big|_{0}^\infty + \int_0^\infty \varphi \\ &= \int_0^\infty \varphi \\ &= \int_{\mathbb R} \varphi H, \end{align*} where we've integrated by parts again and used our "nice" properties of $\varphi$.
The upshot: $u$ did not work for the differential equation, but it did work for the integral equation. This is the point of using integral equations instead: they allow for greater possibility of the existence of solutions because solutions aren't required to have such stringent regularity conditions (like differentiability and continuity). We "shift" those requirements onto the "nice test function" $\varphi$.