Laplace’s Equation<\/h2>\n\n\n\n
Suppose we have a thin, flat (two-dimensional) plate of homogeneous material and we measure the temperature at the border. What is the temperature inside the material? The solution to this problem is described by Laplace’s equation<\/a>, one of the most ubiquitous partial differential equations in physics. Let (1) <\/span> <\/span> Laplace’s equation (1) and the specification of the temperature on the boundary form a well-posed<\/a> mathematical problem in the sense that the temperature is uniquely determined at each point <\/span> <\/span> are known as Dirichlet boundary conditions<\/a>.<\/p>\n\n\n\n Another area of physics where the Laplace equation (1) appears is the study of electrostatics. In this case, The electrostatic application motivates a different way of thinking about the Laplace equation. Consider the following question:<\/p>\n\n\n\n How would I place electric charges on the boundary to produce the electric potential This is a deliciously clever question. If I were able to find an arrangement of charges answering the question, then I could calculate the potential We shall actually use a slight variant of this charge distribution idea which differs in two ways:<\/p>\n\n\n\n With these modifications, our new question is as follows:<\/p>\n\n\n\n How would I place a density of “charge” dipoles We call this function (2) <\/span> <\/span> where the integral is taken over the surface There’s one last ingredient: How do we compute the electric potential (3) <\/span> <\/span> The integral equation (2) is certainly nothing to sneeze at. Rather than trying to comprehend it in its full glory, we shall focus on a special case for the rest of our discussion. Suppose the region <\/span> <\/span> Substituting in (2) then gives<\/p>\n\n\n (4) <\/span> <\/span> We are interested in solving the integral equation (4) to obtain an expression for the double-layer potential Suppose we want to solve the system of linear equations<\/p>\n\n\n (5) <\/span> <\/span> where Standing back, observe that we now have a system of <\/span> <\/span> which combined with the second equation<\/p>\n\n\n <\/span> <\/span> gives the (6) <\/span> <\/span> The original equation for <\/span> <\/span> Solving this, we deduce that<\/p>\n\n\n <\/span> <\/span> From the equation <\/span> <\/span> Since this formula holds for every vector <\/span> <\/span> This example shows how it can be conceptually useful to lift a linear system of equations by adding additional variables and equations and then “do Gaussian elimination in a different order”. The same insight shall be useful in solving integral equations like (4).<\/p>\n\n\n\n Let’s try repeating the playbook we executed for the rank-one-updated linear system (5) and apply it to the integral equation (4). We are ultimately interested in computing <\/span> <\/span> Substituting this into (4) gives the system of equations<\/p>\n\n\n (7) <\/span> <\/span> (8) <\/span> <\/span> In order to obtain (4) from (7) and (8), we add <\/span> <\/span> Adding <\/span> <\/span> Thus plugging this expression for <\/span> <\/span> We’ve solved for our double layer potential!<\/p>\n\n\n\n As promised, the double layer potential can be used to give a solution formula<\/em> (known as the Poisson integral formula<\/a>) for the Laplace Dirichet problem. The details are a mechanical, but also somewhat technical, exercise in vector calculus identities. We plug through the details in the following extra section.<\/p>\n\n\n (9) <\/span> <\/span> <\/p>\n\n\n<\/p>\n We now need to do a quick calculation which is somewhat technical and not particularly enlightening. We evaluate <\/span> <\/span> We denote <\/span> <\/span> \n\n\n <\/p>\n Computing the boundary derivative for the spherical region centered at the origin with radius <\/span> <\/span> We’ve succeeded at deriving a solution formula for Having achieved our main goal of deriving a solution formula for the 2D Laplace Dirichlet problem for a circular domain, I want to take a step back to present the approach from two sections ago in more generality. Consider a more general integral equation of the form<\/p>\n\n\n (10) <\/span> <\/span> where <\/span> <\/span> With the circular domain, the Laplace Dirichlet integral equation (2) is separable with rank (11) <\/span> <\/span> Let’s try and write this equation in a way that’s more similar to the linear system of equation (5). To do this, we make use of linear operators<\/a> defined on functions:<\/p>\n\n\n\n With these notations, equation (11) can be written as<\/p>\n\n\n <\/span> <\/span> Using the same derivation which led to the Sherman\u2013Morrison formula for linear systems of equations, we can apply the Sherman\u2013Morrison formula to this integral equation in operator form, yielding<\/p>\n\n\n <\/span> <\/span> Therefore, the solution to the integral equation (11) is<\/p>\n\n\n <\/span> <\/span> This can be interpreted as a kind of Sherman\u2013Morrison formula for the integral equation (11).<\/p>\n\n\n\n One can also generalize to provide a solution formula for the second-kind integral equation (10) for a separable kernel with rank My goal in writing this post was to discuss two topics which are both near and dear to my heart, integral equations and the Sherman\u2013Morrison formula. I find the interplay of these two ideas to be highly suggestive. It illustrates the power of the analogy between infinite-dimensional linear equations, like differential and integral equations, and finite-dimensional ones, which are described by matrices. Infinite dimensions certainly do have their peculiarities and technical challenges, but it can be very useful to first pretend infinite-dimensional linear operators (like integral operators) are matrices, do calculations to derive some result, and then justify these computations rigorously post hoc<\/em>.7<\/a><\/sup>The utility of this technique is somewhat of an open secret among some subset of mathematicians and scientists, but such heuristics are usually not communicated to students explicitly, at least in rigorous mathematics classes.<\/span>\n","protected":false},"excerpt":{"rendered":" In this post, I want to discuss two of my favorite topics in applied mathematics: the Sherman\u2013Morrison formula and integral equations. As a bridge between these ideas, we’ll use an integral equation analog of the Sherman\u2013Morrison formula to derive the solution for the Laplace equation with Dirichlet boundary conditions in the 2D disk. Laplace’s EquationRead more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-935","post","type-post","status-publish","format-standard","hentry","category-expository"],"_links":{"self":[{"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/posts\/935","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/comments?post=935"}],"version-history":[{"count":61,"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/posts\/935\/revisions"}],"predecessor-version":[{"id":1184,"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/posts\/935\/revisions\/1184"}],"wp:attachment":[{"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/media?parent=935"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/categories?post=935"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.ethanepperly.com\/index.php\/wp-json\/wp\/v2\/tags?post=935"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"u(x,y)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-6c72e52ae99bf12dadb1364d9ef3f929_l3.png\" "\\"Rendered")
denote the temperature of the material at point ![\"(x,y)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-bd3d5e5a67f33b4e889ddca1e20db1b7_l3.png\" "\\"Rendered")
. Laplace’s equation states that, at any point on the interior of the material,<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-9d9ef40d57b2a93aca1e8a169ba2124c_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n.1<\/a><\/sup>A well-posed problem is also required to depend continuously on the input data which, in this case, are the boundary temperatures. Indeed, the Laplace problem with boundary data is well-posed in this sense.<\/span> We call this problem the Laplace Dirichlet problem<\/em> since the boundary conditions<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-ff144f6e2539ff8015dd6fe52897a1b2_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\nThe Double Layer Potential<\/h2>\n\n\n\n
represents the electric potential<\/a> at the point . The Laplace Dirichlet problem is to find the electric potential in the interior of the region with knowledge of the potential on the boundary.<\/p>\n\n\n\n at each point on the boundary?<\/em><\/p><\/blockquote>\n\n\n\n at each point in the interior by adding up the contribution to the electric potential of each element of charge on the boundary. Thus, I can reduce the problem of finding the electric potential at each point in the 2D region to finding a charge distribution<\/em> on the 1D boundary<\/em> to that region.<\/p>\n\n\n\n induced by a unit “charge” at the origin is given by ![\"1/2\pi \cdot \ln \sqrt{x^2+y^2}\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-d4f27bc7528db2c1e29fde3e861933c1_l3.png\" "\\"Rendered")
.<\/li><\/ul>\n\n\n\n![\"\phi(x,y)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-ff756488d09cd0d8b1c254f54725e7af_l3.png\" "\\"Rendered")
on the boundary to produce the electric potential at each point on the boundary?<\/em><\/p><\/blockquote>\n\n\n\n the double layer potential <\/a>for the Laplace Dirichlet problem. One can show the double layer potential satisfies a certain integral equation<\/em>. To write down this integral equation, let’s introduce some more notation. Let ![\"R\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-d08fac616919760e7538df715d3ca0e6_l3.png\" "\\"Rendered")
be the region of interest and ![\"\partial R\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-2840f0a72b93fd12148dbe7c34ee87ec_l3.png\" "\\"Rendered")
its boundary. Denote points concisely as vectors ![\"\mathbf{r} = (x,y)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-fb3b0816211183fdd83f1f46cc4e6ded_l3.png\" "\\"Rendered")
, with the length of ![\"\mathbf{r}\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-7066244bea5735ebe6af03951fc70df7_l3.png\" "\\"Rendered")
denoted ![\"\|\mathbf{r}\| = \sqrt{x^2+y^2}\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-6a1730be18dd1095f9f02b07ded9752d_l3.png\" "\\"Rendered")
. The double layer potential satisfies<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-47b6efc035892fdeb0bd423da090be92_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n of the region ; ![\"\partial / \partial \nu_{\mathbf{r}'}\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-8a91e34fbe3676b5d0bcde98f2031952_l3.png\" "\\"Rendered")
denotes the directional derivative taken in the direction normal (perpendicular) to the surface at the point ![\"\mathbf{r}'\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-7e0d6351fe454ceb8ef68c2fa904c05e_l3.png\" "\\"Rendered")
. Note we choose a unit system for ![\"\phi\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-b889ecc5781bc4fa48f70ba23ef2cfb8_l3.png\" "\\"Rendered")
which hides physical constants particular to the electrostatic context, since we are interested in applying this methodology to the Laplace Dirichlet problem in general (possibly non-electrostatic) applications.<\/p>\n\n\n\n![\"u(\mathbf{r})\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-266315bcf38f8a18c2bbd029297ee18b_l3.png\" "\\"Rendered")
at points in the interior of the region? This is answered by the following formula:<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-0db4e9892036ad2265c1ebb5282133a6_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n is a circular disk with radius ![\"r\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-0cfb382a93bc3f68981565d49d1aeb9e_l3.png\" "\\"Rendered")
centered at ![\"0\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-54589d9b5610bf48dcf5a1b1f24a67b6_l3.png\" "\\"Rendered")
. The the partial derivative in the integrand in (2) then is readily computed for points and both on the boundary of the circle:<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-b6c2537513499f21b0f6c111116de46e_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-91ffeb95a25fca541e146f9c23044af9_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\nThe Sherman\u2013Morrison Formula<\/h2>\n\n\n\n
, as this will give us a solution formula for the Laplace Dirichlet problem. Ultimately, we accomplish this by using a clever trick. In an effort to make this trick seem more self-evident and less of a “rabbit out of a hat”, I want to draw an analogy to a seemingly unrelated problem: rank-one updates to linear systems of equations and the Sherman\u2013Morrison formula.3<\/a><\/sup>In accordance with Stigler’s law of eponymy<\/a>, the Sherman\u2013Morrison formula was actually discovered by William J. Duncan<\/a> five years before Sherman, Morrison, and Woodbury.<\/span> For a more general perspective on the Sherman\u2013Morrison formula and its generalization to the Sherman\u2013Morrison\u2013Woodbury formula<\/a>, you may be interested in the following post of mine on Schur complements and block Gaussian elimination<\/a>.<\/p>\n\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-aa137fb04cb5916b7a2cb3eeafd3117b_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"A\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-11f4f587954b361e7d78940f65b8d70d_l3.png\" "\\"Rendered")
is an ![\"n\times n\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-41c0efe7f3a82ce93dc2542200956ad7_l3.png\" "\\"Rendered")
square matrix and ![\"u\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-0a4093b232696c87a445d0fa1a862a1b_l3.png\" "\\"Rendered")
, ![\"v\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-1ccb53cb35fc73370b29b241e3fa4bcd_l3.png\" "\\"Rendered")
, and ![\"b\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-3ab6f6c64ce17f786a81f8cc8fdcec90_l3.png\" "\\"Rendered")
are length-![\"n\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-75a5652acadcd645b180a972b75a9d09_l3.png\" "\\"Rendered")
vectors. We are ultimately interested in finding ![\"x\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-b312d649591164b7149ed0756f694a76_l3.png\" "\\"Rendered")
from . To gain insight into this problem, it will be helpful to first carefully considered the problem in reverse: computing from . We could, of course, perform this computation by forming the matrix ![\"A + uv^\top\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-45806932eed43e976ef5fe61230996e2_l3.png\" "\\"Rendered")
in memory and multiplying it with , but there is a more economical way:<\/p>\n\n\n\n![\"\alpha = v^\top x\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-fc596b3afa5a8f780ebcc929802445f6_l3.png\" "\\"Rendered")
.<\/li>![\"b = \alpha u + Ax\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-fd8d95361a320233df949844170443c7_l3.png\" "\\"Rendered")
.<\/li><\/ol>\n\n\n\n![\"n+1\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-cd47f9141b069463374b650a53568ac2_l3.png\" "\\"Rendered")
equations for unknowns and ![\"\alpha\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-4e8e0f9c0be9d91a4d8967ea0d3374c7_l3.png\" "\\"Rendered")
. Specifically, our first equation can be rewritten as<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-f733a59480d3659ca1316f9b07728682_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-2ea47659df6cf62ff0e37750922a87de_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n by system4<\/a><\/sup>This “state space approach” of systematically writing out a matrix\u2013vector multiply algorithm and then realizing this yields a larger system of linear equations was initially taught to be by my mentor Shiv Chandrasekaran<\/a>; this approach has much more powerful uses, such as in the theory of rank-structured matrices<\/a>.<\/span>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-3c3baa6d2c48a869a2e6612ffae047bb_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n (5) can be derived from the “lifted” equation (6) by applying Gaussian elimination<\/a> and eliminating the first row of the linear system (6). But now that we have the lifted equation (6), one can naturally wonder what would happen if we instead used Gaussian elimination to eliminate the last rows of (6); this will give us an equation for which we can solved without first computing <\/em>. Doing this so-called block Gaussian elimination<\/a> yields<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-d7556530bd80a39e38b57f7e463d714c_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-a882e9c9d9587659e381f2d8226eb63b_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\alpha u + Ax = b\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-39ae046450605a89cc13d2ac01170c2d_l3.png\" "\\"Rendered")
, we have that<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-f7bf24ae3df859e420126c4ba736599b_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n, we deduce the famous Sherman\u2013Morrison formula<\/a><\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-5d476322e58cb65d31d8a9505062d5d9_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\nSolving for the Double Layer Potential<\/h2>\n\n\n\n
![\"\phi(\cdot)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-8c9fd10da8a4f65408c2da4bf7a3dca3_l3.png\" "\\"Rendered")
from ![\"u(\cdot)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-0e3bb85a18690c20afc80c60e68f9aa6_l3.png\" "\\"Rendered")
but, as we did last section, let’s first consider the reverse. To compute from , we first evaluate the integral <\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-3474aca27d3ddba430a93163bc46e2ef_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-4816dc0c9e593f889847229c31413403_l3.png\" "\\"Rendered")
<\/p><\/p>\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-e2c706009eab6d8c1ab8e1283060cfc9_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"1/4\pi r\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-34502bcd40b32bad394890e27cabc45b_l3.png\" "\\"Rendered")
times equation (8) to equation (7). Following last section, we now instead eliminate ![\"\phi(\mathbf{r})\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-03d6f4968e6c852d1fbd4ae5d7cf68e7_l3.png\" "\\"Rendered")
from equation (8) using equation (7). To do this, we need to integrate equation (7) in order to cancel the integral in equation (8):<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-a5cc3ce9c557b620b127d851cada63d2_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"2\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-b5fed14f2144ce9f7a2e7c6c1a9858d9_l3.png\" "\\"Rendered")
times this integrated equation to equation (8) yields<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-e04f8efa732a6073cd2171031d47f937_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n into equation (7) yields<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-fc839728fe2a437fd7a20021e679b176_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n at a point in the interior of the region. To do this, we use equation (3):<\/p>\n![\"\begin{align*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-3af6238ad0c9e6f4c9d6058fcfab7be3_l3.png\" "\\"Rendered")
<\/p><\/p>\n![\"-\frac{1}{4\pi}\int_{\partial R} \frac{\partial}{\partial \nu_{\mathbf{r}'}} \frac{1}{|\mathbf{r}-\mathbf{r}'|} \, dS(\mathbf{r}')\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-b4c3857b1904b0b0000a468811432e64_l3.png\" "\\"Rendered")
using the divergence theorem<\/a>:<\/p>\n![\"\begin{align*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-218da09c77aa5773391a58dd21590f32_l3.png\" "\\"Rendered")
<\/p><\/p>\n![\"\nabla\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-ff101821d793c89cda5f3eed68d2c99c_l3.png\" "\\"Rendered")
for the gradient<\/a>, ![\"\nu_{\mathbf{r}'}\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-863cf7af8d4efaff0b76ba09af4f0732_l3.png\" "\\"Rendered")
for the normal vector to at the point , ![\"\cdot\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-51d8e2c739f72ff29db877230311f73f_l3.png\" "\\"Rendered")
for the dot product<\/a>, ![\"\nabla^2\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-686c1b3ef54470fa75cc9b021ffee805_l3.png\" "\\"Rendered")
for the Laplace operator<\/a>, and ![\"\delta\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-50f767701973c5f8f4cb8cc6ed459bad_l3.png\" "\\"Rendered")
for the Dirac delta “function”<\/a>. The last equality holds because the function ![\"v(\mathbf{r}') = \tfrac{1}{2\pi} \ln \| \mathbf{r} - \mathbf{r}'\|\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-377e7b81642c961c1d8893d67283bea1_l3.png\" "\\"Rendered")
is a so-called fundamental solution<\/a> for Laplace equation in the sense that ![\"\nabla_{\mathbf{r}'}^2 v(\mathbf{r}') = \delta(\mathbf{r}'-\mathbf{r})\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-18cab6a6ab0c00c620c819ead655d3d3_l3.png\" "\\"Rendered")
. Therefore, (9) simplifies to<\/p>\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-67e33517ef6cb4a83974cc4942323dc1_l3.png\" "\\"Rendered")
<\/p><\/p>\n, we obtain the formula<\/p>\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-12a62591678230bc5e3ff189a5af09df_l3.png\" "\\"Rendered")
<\/p><\/p>\n for points in the interior of the disk in terms of ![\"u(\mathbf{r}')\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-0917b427344740887e06ab36000b0c38_l3.png\" "\\"Rendered")
for points on the boundary of the disk. This is known as the Poisson integral formula<\/a> for the disk in two dimensions. This formula can be generalized to balls in higher dimensions<\/a>, though this proof technique using “Sherman\u2013Morrison” fails to work in more than two dimensions.<\/div><\/div>\n\n\nSherman\u2013Morrison for Integral Equations<\/h2>\n\n\n\n
![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-5f167f2b3959e87053cfc2691846274f_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\Omega\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-0e42a68047144196ffa9763b964056ac_l3.png\" "\\"Rendered")
is some region in space, ![\"K(\cdot,\cdot)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-f364c8993052acfc9eddb366fbbb8aef_l3.png\" "\\"Rendered")
, ![\"f(\cdot)\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-725a8141dc45270bed45aab5f9865f93_l3.png\" "\\"Rendered")
, and are functions of one or two arguments on , and ![\"a\ne 0\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-0f7d40a1d536380d1a02024d1b05a991_l3.png\" "\\"Rendered")
is a nonzero constant. Such an integral equation is said to be of the second kind<\/a>. The integral equation for the Laplace Dirichlet problem (2) is of this form with ![\"\Omega = \partial R\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-c990f4385674ef0110a6f2f2489e3b4b_l3.png\" "\\"Rendered")
, ![\"a = 1/2\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-b8d3c51336f3b0c36035451ebccc20db_l3.png\" "\\"Rendered")
, ![\"K(\mathbf{x},\mathbf{y}) = \tfrac{\partial}{\partial \nu_{\mathbf{y}}} \ln \|\mathbf{x} - \mathbf{y}\|\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-53e65663eb899d9882b7c9a4f84b82ff_l3.png\" "\\"Rendered")
, and ![\"f(\mathbf{x}) = u(\mathbf{x})\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-4932523db5a64257149e81567cbc3233_l3.png\" "\\"Rendered")
. We say the kernel is separable<\/em> with rank ![\"k\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-f65286b751f121928913d4aa91d94ee9_l3.png\" "\\"Rendered")
if can be expressed in the form<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-75a645168a961f02c36dc2aaa2514a7b_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"k = 1\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-b66c50807a61cc34e2c14a3373a1f4e0_l3.png\" "\\"Rendered")
.5<\/a><\/sup>E.g., set ![\"g_1(\mathbf{x}) = 1\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-c3b0464174e8bf4d15bbf9566cbfe37e_l3.png\" "\\"Rendered")
and ![\"h_1(\mathbf{y}) = 1/4\pi r\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-451d36605cc4ad054d9029977d99d88d_l3.png\" "\\"Rendered")
.<\/span> We shall focus on the second kind integral equation (10) assuming the kernel is separable with rank ![\"1\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-27e3598fd2d4f0491067f1afaced92e9_l3.png\" "\\"Rendered")
(for simplicity, we set ![\"a = 1\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-8075db9c3116604450b7d43da788966c_l3.png\" "\\"Rendered")
):<\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-89428ad5af359d014840e00faec9a3dc_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\operatorname{Id}\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-104233b5ab878658755f9eca8bf8a446_l3.png\" "\\"Rendered")
denote the identity operator on functions: It takes as inputs function and outputs the function unchanged.<\/li>![\"I_h\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-039841f6c3678949badc8fda5134eaf9_l3.png\" "\\"Rendered")
denote the “integration against ![\"h\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-471140fbd20a10d0f411d52aa20652e4_l3.png\" "\\"Rendered")
operator”: It takes as input a function and outputs the number ![\"\int_\Omega h(\mathbf{y})\phi(\mathbf{y}) \, d\mathbf{y}\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-2ce249b915af61cf9e52df9bb497451c_l3.png\" "\\"Rendered")
.<\/li><\/ul>\n\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-091e70628a0b9e4a7a3aeb2521ef4b8b_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-a6e41172ccf6362fa89e2d9b042c3849_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n![\"\begin{equation*}](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-9f75242a54f69e60d849ff456bbe1feb_l3.png\" "\\"Rendered")
<\/p><\/p>\n\n\n; in this case, the natural matrix analog is now the Sherman\u2013Morrison\u2013Woodbury identity<\/a> rather than the Sherman\u2013Morrison formula. Note that this solution formula requires the solution of a ![\"k\times k\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-979373ae59303770cd68d74797bc73f9_l3.png\" "\\"Rendered")
system of linear equations. One can use this as a numerical method to solve second-kind integral equations: First, we approximate<\/a> ![\"K\"](\"https:\/\/www.ethanepperly.com\/wp-content\/ql-cache\/quicklatex.com-8bc9819614418a525aa5b16965b18391_l3.png\" "\\"Rendered")
by a separable kernel of a modest rank and then compute the exact solution of the resulting integral equation with the approximate kernel.6<\/a><\/sup>A natural question is why one might want to solve an integral equation formulation of a partial differential equations like the Laplace or Helmholtz equation. An answer is that that formulations based on second-kind integral equations tend to lead to systems of linear equations which much more well-conditioned<\/a> as compared to other methods like the finite element method<\/a>. They have a number of computational difficulties as well, as the resulting linear systems of equations are dense and may require elaborate quadrature rules to accurately compute.<\/span>\n\n\n\n