{"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":"THE HEAT EQUATION - SOUL OF MATHEMATICS","type":"rich","width":600,"height":338,"html":"<blockquote class=\"wp-embedded-content\" data-secret=\"jIkreMHq34\"><a href=\"https:\/\/soulofmathematics.com\/index.php\/the-heat-equation\/\">THE HEAT EQUATION<\/a><\/blockquote><iframe sandbox=\"allow-scripts\" security=\"restricted\" src=\"https:\/\/soulofmathematics.com\/index.php\/the-heat-equation\/embed\/#?secret=jIkreMHq34\" width=\"600\" height=\"338\" title=\"&#8220;THE HEAT EQUATION&#8221; &#8212; SOUL OF MATHEMATICS\" data-secret=\"jIkreMHq34\" 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\/2020\/11\/LiquidSociableLadybug-small.gif?fit=273%2C205&ssl=1","thumbnail_width":273,"thumbnail_height":205,"description":"The heat equation is among the most widely studied topics in&nbsp;pure mathematics, and its analysis is regarded as fundamental to the broader field of&nbsp;partial differential equations. Solutions of the heat equation are sometimes known as&nbsp;caloric functions. The theory of the heat equation was first developed by&nbsp;Joseph Fourier&nbsp;in 1822 for the purpose of modeling how a quantity such as&nbsp;heat&nbsp;diffuses through a given region. The heat equation, along with variants thereof, is also important in many fields of science and&nbsp;applied mathematics. In&nbsp;probability theory, the heat equation is connected with the study of&nbsp;random walks&nbsp;and&nbsp;Brownian motion&nbsp;via the&nbsp;Fokker\u2013Planck equation. The infamous&nbsp;Black\u2013Scholes equation&nbsp;of&nbsp;financieal mathematics&nbsp;is a small variant of the heat equation, and the&nbsp;Schr\u00f6dinger equation&nbsp;of&nbsp;quantum mechanics&nbsp;can be regarded as a heat equation in&nbsp;imaginary time. For the better mathematical and physical understanding of the Heat Equation we should start by studying diffusion of heat. The Heat Equation For A Finite Rod We will consider the heat equation Ut = k Uxx where k is the heat constant. Now we take the initial conditions and boundary conditions for this situation. Initial Conditions Boundary Conditions U(x,0) = f(x) U(0,t) = 0 = U(l,t) U is a function of time and distance. U = X(x).T(t) Ut = XT&#8217; Uxx = X&#8221;T XT&#8217; = kX&#8221;T = \u03bb X&#8221; &#8211; \u03bbX = 0 and T&#8217; &#8211; k\u03bbT = 0 There shall be three cases &#8211; 1. \u03bb = 0 2. \u03bb = \u03b12 3. \u03bb = &#8211;\u03b12 Now we shall delve into a detailed discussion of each case. Starting with the boundary conditions. U(0,t) = 0, from this we can imply, X(0).T(t) = 0 X(0) = 0 U(l,t) = 0, from this we can imply, X(l).T(t) = 0 X(l) = 0 CASE 1- \u03bb = 0 X&#8221; &#8211; \u03bbX = 0 and T&#8217; &#8211; k\u03bbT = 0 So we can imply, X&#8221; = 0 and T&#8217; = 0 X = C1x + C2 X(0) = 0 , X(l) = 0 C1 = C2 = 0 , X = 0 , U = 0 is trivial solution. CASE 2- \u03bb = \u03b12 X&#8221; &#8211; \u03b12X = 0 , T&#8217; &#8211; k\u03b12T = 0 X = C1e\u03b1x + C2e&#8211;\u03b1x X(0) = 0 , X(l) = 0 C1 = C2 = 0 , U = 0 is trivial solution. CASE 3- \u03bb = -\u03b12 X&#8221; + \u03b12X = 0 , T&#8217; + k\u03b12T = 0 X = A cos (\u03b1x) + B sin (\u03b1x) When, X(0) = 0 , A = 0 When, X(l) = 0 , B sin (\u03b1l) = 0 , B is not equal to zero. sin (\u03b1l) = 0, \u03b1l = n\u03c0, \u03b1 = n\u03c0\/l STAY TUNED&#8230;"}