{"version":"1.0","provider_name":"SOUL OF MATHEMATICS","provider_url":"https:\/\/soulofmathematics.com","author_name":"Rajarshi Dey","author_url":"https:\/\/soulofmathematics.com\/index.php\/author\/rajarshidey1729gmail-com\/","title":"GREEN'S FUNCTION - SOUL OF MATHEMATICS","type":"rich","width":600,"height":338,"html":"<blockquote class=\"wp-embedded-content\" data-secret=\"v4jjFZOrSy\"><a href=\"https:\/\/soulofmathematics.com\/index.php\/greens-function\/\">GREEN&#8217;S FUNCTION<\/a><\/blockquote><iframe sandbox=\"allow-scripts\" security=\"restricted\" src=\"https:\/\/soulofmathematics.com\/index.php\/greens-function\/embed\/#?secret=v4jjFZOrSy\" width=\"600\" height=\"338\" title=\"&#8220;GREEN&#8217;S FUNCTION&#8221; &#8212; SOUL OF MATHEMATICS\" data-secret=\"v4jjFZOrSy\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" class=\"wp-embedded-content\"><\/iframe><script type=\"text\/javascript\">\n\/* <![CDATA[ *\/\n\/*! This file is auto-generated *\/\n!function(d,l){\"use strict\";l.querySelector&&d.addEventListener&&\"undefined\"!=typeof URL&&(d.wp=d.wp||{},d.wp.receiveEmbedMessage||(d.wp.receiveEmbedMessage=function(e){var t=e.data;if((t||t.secret||t.message||t.value)&&!\/[^a-zA-Z0-9]\/.test(t.secret)){for(var s,r,n,a=l.querySelectorAll('iframe[data-secret=\"'+t.secret+'\"]'),o=l.querySelectorAll('blockquote[data-secret=\"'+t.secret+'\"]'),c=new RegExp(\"^https?:$\",\"i\"),i=0;i<o.length;i++)o[i].style.display=\"none\";for(i=0;i<a.length;i++)s=a[i],e.source===s.contentWindow&&(s.removeAttribute(\"style\"),\"height\"===t.message?(1e3<(r=parseInt(t.value,10))?r=1e3:~~r<200&&(r=200),s.height=r):\"link\"===t.message&&(r=new URL(s.getAttribute(\"src\")),n=new URL(t.value),c.test(n.protocol))&&n.host===r.host&&l.activeElement===s&&(d.top.location.href=t.value))}},d.addEventListener(\"message\",d.wp.receiveEmbedMessage,!1),l.addEventListener(\"DOMContentLoaded\",function(){for(var e,t,s=l.querySelectorAll(\"iframe.wp-embedded-content\"),r=0;r<s.length;r++)(t=(e=s[r]).getAttribute(\"data-secret\"))||(t=Math.random().toString(36).substring(2,12),e.src+=\"#?secret=\"+t,e.setAttribute(\"data-secret\",t)),e.contentWindow.postMessage({message:\"ready\",secret:t},\"*\")},!1)))}(window,document);\n\/* ]]> *\/\n<\/script>\n","thumbnail_url":"https:\/\/i0.wp.com\/soulofmathematics.com\/wp-content\/uploads\/2021\/02\/gapEta.gif?fit=800%2C504&ssl=1","thumbnail_width":800,"thumbnail_height":504,"description":"Green&#8217;s functions are named after the British&nbsp;mathematician George Green, who first developed the concept in the 1820s. In the modern study of linear&nbsp;partial differential equations, Green&#8217;s functions are studied largely from the point of view of&nbsp;fundamental solutions&nbsp;instead. The term is also used in&nbsp;physics, specifically in&nbsp;quantum field theory,&nbsp;aerodynamics,&nbsp;aeracoustics,&nbsp;electrodynamics, seismology&nbsp;and&nbsp;statistical field theory, to refer to various types of&nbsp;correlation functions, even those that do not fit the mathematical definition. In quantum field theory, Green&#8217;s functions take the roles of&nbsp;propagators. Preliminary ideas and motivation The delta function Definition- The \u03b4-function is defined by the following three properties, where f is continuous at x = a. The last is called the shifting property of the \u03b4-function. To make proofs with the \u03b4-function more rigorous, we consider a \u03b4-sequence, that is, a sequence of functions that converge to the \u03b4-function, at least in a pointwise sense. Consider the sequence Note that, The 2D \u03b4-function is defined by the following three properties, Finding the Green\u2019s function To find the Green\u2019s function for a 2D domain D, we first find the simplest function that satisfies \u22072v = \u03b4(r). Suppose that v (x, y) is axis-symmetric, that is, v = v (r). Then For r &gt; 0, Integrating gives For simplicity, we set B = 0. To find A, we integrate over a disc of radius \u03b5 centered at(x, y),D\u03b5, From the Divergence Theorem, we have where C\u03b5 is the boundary of D\u03b5, i.e. a circle of circumference 2\u03c0\u03b5. Combining the previous two equations gives Hence This is called the fundamental solution for the Green\u2019s function of the Laplacian on 2D domains. For 3D domains, the fundamental solution for the Green\u2019s function of the Laplacian is \u22121\/(4\u03c0r), where r = (x \u2212 \u03be)2 + (y \u2212 \u03b7)2 + (z \u2212 \u03b6)2. The Green\u2019s function for the Laplacian on 2D domains is defined in terms of the corresponding fundamental solution, The term \u201cregular\u201d means that h is twice continuously differentiable in(\u03be, \u03b7)on D. Finding the Green\u2019s function G is reduced to finding a C2 function h on D that satisfies The definition of G in terms of h gives the BVP for G. Thus, for 2D regions D, finding the Green\u2019s function for the Laplacian reduces to finding h. Examples Plot of the Green\u2019s function G(x,y;\u03be,\u03b7)for the Laplacian operator in the upper half plane, for(x, y)=(\u221a2,\u221a2). (i) Full plane D = R2. There are no boundaries so h =0 will do, and (ii)Half plane D = {(x, y): y&gt; 0}. We find G by introducing what is called an \u201cimage point\u201d (x, \u2212y) corresponding to(x, y). Let r be the distance from (\u03be,\u03b7) to (x, y)and r \u2032 the distance from(\u03be,\u03b7) to the image point(x,\u2212y), Conformal mapping and the Green\u2019s function Conformal mapping allows us to extend the number of 2D regions for which Green\u2019s functions of the Laplacian \u22072u can be found. We use complex notation, and let \u03b1 = x + iy be a fixed point in D and let z = \u03be + i\u03b7 be a variable point in D (what we\u2019re integrating over). If D is simply connected (a definition from complex analysis), then by the Riemann Mapping Theorem, there is a conformal map w (z)(analytic and one-to-one) from D into the unit disk, which maps \u03b1 to the origin, w (\u03b1) =0 and the boundary of D to the unit circle, |w (z)|=1for z \u2208\u2202D and0 \u2264|w (z)|&lt; 1 for z \u2208D\/\u2202D. The Greens function G is then given by To see this, we need a few results from complex analysis. First, note that for z \u2208\u2202D, w (z)=0 so that G = 0. Also, since w (z)is1-1, w (z)&gt; 0for z = \u03b1. Thus, wecan write w (z) =(z \u2212\u03b1)n H(z) where H(z) is analytic and nonzero in D. Since w (z)is1-1, w \u2032 (z)&gt; 0 on D. Thus n =1. Hence w (z)=(z \u2212\u03b1)H(z) and where, Since H(z) is analytic and nonzero in D, then (1\/2\u03c0) lnH(z) is analytic in D and hence its real part is harmonic, i.e. h = \u211c ((1\/2\u03c0) lnH(z)) satisfies \u22072h =0in D. Thus by our definition above, G is the Green\u2019s function of the Laplacian on D. NO COPYRIGHT INFRINGEMENT INTENDED"}