{"id":28757,"date":"2024-09-19T17:27:49","date_gmt":"2024-09-19T17:27:49","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28757"},"modified":"2024-09-19T17:30:57","modified_gmt":"2024-09-19T17:30:57","slug":"brachistochrone-and-euler-lagrange-equation-with-variational-calculus","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/brachistochrone-and-euler-lagrange-equation-with-variational-calculus\/","title":{"rendered":"Brachistochrone and Euler-Lagrange Equation with Variational Calculus"},"content":{"rendered":"<h1 style=\"text-align:center;\">Variational Calculus in Classical Mechanics and the Euler-Lagrange Equation<\/h1>\n<p style=\"text-align:center;\"><em><strong>Abstract:<\/strong><\/br>In this class, we will review the derivation of the Euler-Lagrange equation of Analytical Mechanics through the use of variational calculus techniques, and from this, we will show in detail its application in solving the Brachistochrone problem.<\/em><\/p>\n<hr>\n<p style=\"text-align:center;\"><strong>Learning Objectives:<\/strong><br \/>\nBy the end of this class, the student will be able to:<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Understand<\/strong> Hamilton&#8217;s principle of least action<\/li>\n<li><strong>Demonstrate<\/strong> the Euler-Lagrange equation<\/li>\n<li><strong>Solve<\/strong> the Brachistochrone problem using the Euler-Lagrange equation.<\/li>\n<\/ol>\n<hr>\n<p style=\"text-align:center;\">\n<strong><u>TABLE OF CONTENTS<\/u>:<\/strong><br \/>\n<a href=\"#1\">WHY VARIATIONAL CALCULUS IN CLASSICAL MECHANICS<\/a><br \/>\n<a href=\"#2\">FORMULATION OF THE VARIATIONAL PROBLEM<\/a><br \/>\n<a href=\"#3\">THE EULER-LAGRANGE EQUATION<\/a><br \/>\n<a href=\"#4\">THE BRACHISTOCHRONE PROBLEM<\/a><br \/>\n<a href=\"#5\">GITHUB REPOSITORY WITH WOLFRAM ALGORITHM<\/a>\n<\/p>\n<hr>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/JyumifjGzM0\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" allowfullscreen><\/iframe><\/center><\/p>\n<hr>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Why Variational Calculus in Classical Mechanics<\/h2>\n<p style=\"text-align:justify;\">Newtonian physics presents numerous problems that can be more effectively addressed using variational calculus. This approach is fundamental in the Lagrange equations and Hamilton&#8217;s principle of least action. Essentially, this method involves finding trajectories that maximize or minimize a certain quantity. For example, one can seek the trajectory between two points that minimizes the distance traveled or the travel time. An example of this approach is Fermat&#8217;s principle, which states that light always follows the path that minimizes the travel time, which in turn leads to <a href=\"http:\/\/toposuranos.com\/material\/en\/the-refraction-of-light-and-snells-law\/\" target=\"_blank\" rel=\"noopener\">Snell&#8217;s law of light refraction<\/a>.<\/p>\n<p style=\"text-align:justify;\">Variational calculus has multiple advantages in classical mechanics. For instance, it allows obtaining exact analytical solutions for systems with symmetry, and approximate solutions through variational perturbation theory for more complex systems. Moreover, in situations where it is difficult to express the forces in terms of differential equations, the principle of least action provides a more efficient method for solving problems in classical mechanics. In summary, variational calculus is a fundamental tool that offers an alternative formulation of Newton&#8217;s laws, a unification of the laws of physics, greater efficiency in problem-solving, and higher precision in predicting experimental results.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Formulation of the Variational Problem<\/h2>\n<p style=\"text-align:justify;\">Variational calculus focuses on finding the function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span> that extremizes the value of the functional:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">J(x,y(x))=\\displaystyle \\int_{x_1}^{x_2} f\\left(x,y(x),\\frac{dy(x)}{dx}\\right)dx,<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">with the aim of finding its maximum or minimum value. In this equation, the functional <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span> depends on the function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span> and its derivative <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">dy(x)\/dx,<\/span><\/span> while the limits of integration remain fixed. To extremize the integral, variations are applied to the function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span>, seeking to obtain the function that makes the value of the functional an extremum. For example, if the integral reaches its minimum value, any function <em>within its neighborhood<\/em>, no matter how close it is to <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span>, will increase the value of the functional.<\/p>\n<p style=\"text-align:justify;\">To establish the concept of a <em>neighboring function<\/em>, we can assign a parametric representation <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(\\alpha,x)<\/span><\/span> to all possible functions <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span>, so that if <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha=0<\/span><\/span>, then <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(0,x)=y(x)<\/span><\/span> is the function that extremizes <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span>. This can be expressed as follows:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(\\alpha, x) = y(x) + \\alpha \\eta(x),<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x)<\/span><\/span> is some function of class <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathcal{C}^1<\/span><\/span> that vanishes at <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2<\/span><\/span>, so that the function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(\\alpha,x)<\/span><\/span> that includes this variation is identical to <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span> at the initial and final points of the integration path.<\/p>\n<p style=\"text-align:justify;\">By substituting the function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(\\alpha,x)<\/span><\/span> that includes the variation <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x)<\/span><\/span> instead of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x)<\/span><\/span> in the integral that defines the functional <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span>, we obtain a new functional that depends on the parameter <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span>:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">J(x,y(\\alpha, x)) = \\displaystyle \\int_{x_1}^{x_2} f\\left(x,y(\\alpha,x), \\dfrac{d}{dx}y(\\alpha,x)\\right)dx<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">For local extrema to exist, it is necessary that the condition holds:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left.\\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha}\\right|_{\\alpha=0} = 0<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">for any function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x).<\/span><\/span><\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>The Euler-Lagrange Equation<\/h2>\n<p style=\"text-align:justify;\">By analyzing the derivative <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\partial J(x,y(\\alpha,x))\/\\partial \\alpha<\/span><\/span>, we obtain:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n{}\\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha}\n\n&amp;=&amp;\\dfrac{\\partial}{\\partial \\alpha} \\displaystyle \\int_{x_1}^{x_2} f\\left(x,y(\\alpha,x),\\dfrac{dy(\\alpha, x)}{dx}\\right)dx \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\int_{x_1}^{x_2} \\left(\\dfrac{\\partial f}{\\partial x}\\dfrac{\\partial x}{\\partial  \\alpha} + \\dfrac{\\partial f}{\\partial y(\\alpha, x)}\\dfrac{\\partial y(\\alpha, x)}{\\partial  \\alpha}  + \\dfrac{\\partial f }{ \\partial \\frac{dy(\\alpha,x)}{dx}} \\dfrac{\\partial \\frac{dy(\\alpha,x)}{dx}}{\\partial \\alpha}\\right)dx \\\\\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">From this point, it&#8217;s important to note that:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n\\dfrac{\\partial x}{\\partial \\alpha} &amp;=&amp; 0 \\\\ \\\\\n\n\\dfrac{\\partial y(\\alpha,x)}{\\partial \\alpha} &amp;=&amp; \\dfrac{\\partial}{\\partial \\alpha} \\left(y(x) + \\alpha \\eta(x) \\right) = \\eta(x) \\\\ \\\\\n\n\\dfrac{\\partial}{\\partial \\alpha}\\left( \\dfrac{dy(\\alpha, x)}{dx} \\right)&amp;=&amp; \\dfrac{\\partial}{\\partial \\alpha} \\left(\\dfrac{dy(x)}{dx} + \\alpha\\dfrac{d\\eta(x)}{dx} \\right) = \\dfrac{d\\eta}{dx}\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Therefore, the expression reduces as shown below:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}\n\n{} \\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha} &amp;=&amp; \\displaystyle \\int_{x_1}^{x_2} \\left(\\dfrac{\\partial f}{\\partial y(\\alpha,x)}\\eta(x) + \\dfrac{\\partial f}{\\partial  \\frac{dy(\\alpha,x)}{dx}} \\dfrac{d\\eta(x)}{dx} \\right)dx \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\int_{x_1}^{x_2} \\dfrac{\\partial f}{\\partial y(\\alpha,x)}\\eta(x) dx +  \\int_{x_1}^{x_2} \\dfrac{\\partial f}{\\partial  \\frac{dy(\\alpha,x)}{dx}} \\dfrac{d\\eta(x)}{dx} dx\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">Then, if we observe the second integral, we will see that it can be simplified using integration by parts:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n\\displaystyle \\int_{x_1}^{x_2} \\dfrac{\\partial f}{\\partial  \\frac{dy(\\alpha,x)}{dx}} \\dfrac{d\\eta}{dx} dx\n\n&amp;=&amp; \\left. \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha,x)}{dx}} \\eta(x)\\right|_{x_1}^{x_2} - \\displaystyle \\int_{x_1}^{x_2}\\eta(x) \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha, x)}{dx}} \\right) dx\\\\ \\\\\n\n&amp;=&amp; - \\displaystyle \\int_{x_1}^{x_2}\\eta(x) \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha, x)}{dx}} \\right)dx\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">And therefore<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n{} \\dfrac{\\partial J(x,y(\\alpha,x))}{\\partial \\alpha}\n\n&amp;=&amp; \\displaystyle \\int_{x_1}^{x_2} \\left[ \\eta(x) \\dfrac{\\partial f}{\\partial y(\\alpha, x)}  - \\eta(x) \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha,x)}{dx}} \\right) \\right]dx \\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\int_{x_1}^{x_2} \\left[ \\dfrac{\\partial f}{\\partial y(\\alpha, x)}  -  \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(\\alpha,x)}{dx}} \\right) \\right] \\eta(x) dx\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">So, by the condition that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left.\\dfrac{\\partial J (x,y(\\alpha, x))}{\\partial \\alpha}\\right|_{\\alpha=0} = 0,<\/span><\/span> and since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\eta(x)<\/span><\/span> is any function subject only to the condition of vanishing at <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2,<\/span><\/span> we have:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\dfrac{\\partial f}{\\partial y(0, x)}  -  \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(0,x)}{dx}}\\right) = \\dfrac{\\partial f}{\\partial y(x)}  -  \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial \\frac{dy(x)}{dx}}\\right) = 0.\n\n<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">Finally, \u00abunloading the notation\u00bb in this last expression, we arrive at what is known as the Euler-Lagrange Equation:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\dfrac{\\partial f}{\\partial y}= \\dfrac{d}{dx}\\left( \\dfrac{\\partial f}{\\partial y^\\prime} \\right)},<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">and this represents in a much simpler way the necessary condition for the functional <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span> to reach an extreme value.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>The Brachistochrone Problem<\/h2>\n<h3>Problem Formulation<\/h3>\n<p style=\"text-align:justify;\">The brachistochrone problem is a classic in mechanical physics that is solved using variational calculus. The situation presented is as follows: Suppose we have a material object moving under the effect of a constant force field, traveling from an initial point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_1,y_1)<\/span><\/span> to a final point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span><\/span>, where the initial point is at a higher elevation than the final point. The question posed is: What is the trajectory the particle should follow to reach the final point in the shortest possible time?<\/p>\n<h3>Solution Formulation<\/h3>\n<p style=\"text-align:justify;\">To solve the brachistochrone problem, it&#8217;s useful to consider the situation in a simple form. Therefore, we can set the starting point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_1, y_1)<\/span><\/span> at the origin of coordinates, while the endpoint <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span><\/span> is to the right of the origin and below the <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}<\/span><\/span> axis.<\/p>\n<div style=\"text-align:center;\"><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona.png\" alt=\"variational calculus - brachistochrone problem\" width=\"711\" height=\"505\" class=\"aligncenter size-full wp-image-28729 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona.png\" alt=\"variational calculus - brachistochrone problem\" width=\"711\" height=\"505\" class=\"aligncenter size-full wp-image-28729 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona.png 711w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/probbraquistocrona-300x213.png 300w\" sizes=\"(max-width: 711px) 100vw, 711px\" \/><\/noscript><\/div>\n<p style=\"text-align:justify;\">In this situation, we can consider a force field acting downward (in the <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-\\hat{y}<\/span><\/span> direction) generated by gravity, and assume the motion occurs without friction. In this context, the particle is constrained to follow different trajectories connecting the departure and arrival points to find which minimizes the travel time.<\/p>\n<h3>Examining the Energy<\/h3>\n<p style=\"text-align:justify;\">To solve this problem, we can utilize the conservation of energy in the gravitational system. The total energy of the system will remain constant, considering both the kinetic energy <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">E_{kin}=\\frac{1}{2}mv^2<\/span><\/span> and the gravitational potential energy <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">E_{pot,g}<\/span><\/span>, where <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> is the particle&#8217;s mass and <span class=\"katex-eq\" data-katex-display=\"false\">v<\/span> is its velocity. For the potential energy, the origin is taken as the reference, so <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">E_{pot,g}(y=0)=0<\/span><\/span>, while at any other height <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span>, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">E_{pot,g}(y)=mgy.<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">Since the particle starts from rest at the origin, its total energy equals zero. Therefore, we have:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">E_{kin} + E_{pot,g}=0<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">As the particle falls below the reference point, its potential energy will be negative, and its kinetic energy will be positive. Thus, we can solve for the velocity <span class=\"katex-eq\" data-katex-display=\"false\">v<\/span> using the energy conservation equation to obtain:<\/p>\n<p style=\"text-align:center\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} &amp;\\dfrac{1}{2}mv^2 + (-mgy) = 0 \\\\ \\\\\n\n\\vdash &amp;\\dfrac{1}{2}mv^2 = mgy \\\\ \\\\\n\n\\vdash &amp;v^2 = 2gy \\\\ \\\\\n\n\\vdash &amp;v = \\sqrt{2gy}\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">In this way, we can calculate the particle&#8217;s velocity at any point along its trajectory as a function of the height <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> at that point.<\/p>\n<h3>Examining the Travel Time<\/h2>\n<p style=\"text-align:justify;\">Once we have obtained the speed of movement, we can construct the travel time element using the displacement element <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">ds=\\sqrt{dx^2 + dy^2}<\/span><\/span> as follows:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} dt &amp;= \\dfrac{ds}{v} = \\dfrac{\\sqrt{dx^2 + dy^2}}{\\sqrt{2gy}}\\\\ \\\\\n\n&amp;= \\sqrt{\\dfrac{dx^2 + dy^2}{2gy} }\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">Therefore, the travel time between points <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_1,y_1)<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span><\/span> can be obtained by integrating<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} t &amp;= \\displaystyle \\int_{(x_1,y_1)}^{(x_2,y_2)} dt \\\\ \\\\\n\n&amp;= \\displaystyle \\int_{(x_1,y_1)}^{(x_2,y_2)} \\sqrt{\\dfrac{dx^2 + dy^2}{2gy}} \\\\ \\\\\n\n&amp;= \\displaystyle \\dfrac{1}{\\sqrt{2g}}\\int_{y_1}^{y_2} \\sqrt{\\dfrac{1+ \\left(\\dfrac{dx}{dy}\\right)^2 }{y}}dy \\\\ \\\\\n\n\\end{array}<\/span><\/span><\/p>\n<h3>Variational Problem Formulation<\/h3>\n<p style=\"text-align:justify;\">With this last expression, we have managed to express the time as a functional of the form<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n{}t = J(y,x(y)) = \\displaystyle \\int_{y_1}^{y_2} f\\left(y,x(y),\\dfrac{dx(y)}{dy} \\right) dy\n\n<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">where<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f\\left(y,x(y), \\dfrac{dx(y)}{dx}\\right) = \\sqrt{\\dfrac{1+ \\left(\\dfrac{dx(y)}{dy} \\right)^2}{y}} <\/span><\/span><\/p>\n<p style=\"text-align:justify;\">At this point, we can overlook the factor <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{2g},<\/span><\/span> because optimizing <span class=\"katex-eq\" data-katex-display=\"false\">J<\/span> is the same as optimizing <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{2g}J<\/span><\/span>.<\/p>\n<p style=\"text-align:justify;\">With the above, we can now construct the Euler-Lagrange equation following the same procedure used earlier, finally arriving at:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{\\partial f}{\\partial x} = \\dfrac{d}{dy} \\dfrac{\\partial f}{\\partial x^\\prime}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">However, here we can see that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{\\partial f}{\\partial x} = 0,<\/span><\/span> so it will be<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{d}{dy}\\dfrac{\\partial f}{\\partial x^\\prime} = 0,<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">or in other words<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\dfrac{\\partial f}{\\partial x^\\prime} = \\dfrac{1}{\\sqrt{2a}},<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">where <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> is an arbitrary constant written that way because it is \u00abconvenient\u00bb for later developments.<\/p>\n<h3>Resolution of the Variational Problem<\/h3>\n<p style=\"text-align:justify;\">Substituting the function <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> into this last expression, we have:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} &amp;\\dfrac{\\partial }{\\partial x^\\prime} \\sqrt{\\dfrac{1+ x^{\\prime 2}}{y}}  = \\dfrac{1}{\\sqrt{2a}} \\\\ \\\\\n\n\\vdash &amp; \\dfrac{1}{2}\\left( \\dfrac{1 + x^{\\prime 2} }{y} \\right)^{-1\/2} \\left(\\dfrac{2x^\\prime}{y} \\right) = \\dfrac{1}{\\sqrt{2a}} \\\\ \\\\\n\n\\vdash &amp; \\dfrac{1}{2}\\sqrt{\\dfrac{y}{1 + x^{\\prime 2}}} \\left(\\dfrac{2x^\\prime}{y} \\right) = \\dfrac{1}{\\sqrt{2a}} \\\\ \\\\\n\n\\vdash &amp; \\sqrt{\\dfrac{4x^{\\prime 2} y}{4y^2 (1 + x^{\\prime 2})} }  = \\sqrt{\\dfrac{1}{2a}} \\\\ \\\\\n\n\\vdash &amp;  \\dfrac{y x^{\\prime 2} }{y^2 (1 + x^{\\prime 2})}   = \\dfrac{1}{ 2a} \\\\ \\\\\n\n\\vdash &amp; 2ayx^{\\prime 2} = y^2 + y^2 x^{\\prime 2} \\\\ \\\\\n\n\\vdash &amp;  x^{\\prime 2} (2ay - y^2) = y^2 \\\\ \\\\\n\n\\vdash &amp; \\left(\\dfrac{dx}{dy}\\right)^2 = \\dfrac{y^2}{2ay - y^2} \\\\ \\\\\n\n\\vdash &amp; \\dfrac{dx}{dy} = \\pm \\sqrt{\\dfrac{y^2}{2ay - y^2}} \\\\ \\\\\n\n\\vdash &amp; dx = \\pm \\dfrac{ydy}{\\sqrt{2ay - y^2}} \\\\ \\\\\n\n\\vdash &amp; x = \\displaystyle  \\pm \\int \\dfrac{y}{\\sqrt{2ay - y^2}}dy\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">To solve this integral, one option to consider is performing the following substitution<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}\n\n{} y &amp;=&amp; a[1-\\cos(\\theta)] \\\\\n\n dy &amp;=&amp; a\\sin(\\theta) d\\theta\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">With this we have:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n{} x= &amp; \\pm \\displaystyle \\int \\dfrac{y}{\\sqrt{2ay - y^2}}dy = \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]a\\sin(\\theta)}{\\sqrt{2a^2[1-\\cos(\\theta)] - a^2[1-\\cos(\\theta)]^2 }}d\\theta \\\\ \\\\\n\n&amp; {} = \\pm \\displaystyle \\int \\dfrac{a^2[1-\\cos(\\theta)]\\sin(\\theta)}{\\sqrt{a^2[1-\\cos(\\theta)]\\left\\{ 2 - [1-\\cos(\\theta)] \\right\\} }}d\\theta \\\\ \\\\\n\n&amp; {} = \\pm \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]\\sin(\\theta)}{\\sqrt{[1-\\cos(\\theta)]  [1 + \\cos(\\theta)]  }}d\\theta \\\\ \\\\\n\n&amp; {} = \\pm \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]\\sin(\\theta)}{\\sqrt{ 1-\\cos^2(\\theta)}}d\\theta \\\\ \\\\\n\n&amp; {} = \\pm \\displaystyle \\int \\dfrac{a[1-\\cos(\\theta)]\\sin(\\theta)}{\\sin(\\theta)}d\\theta \\\\ \\\\\n\n&amp; {} = \\pm \\displaystyle \\int a[1-\\cos(\\theta)] d\\theta \\\\ \\\\\n\n&amp; {} = \\pm a(\\theta - \\sin(\\theta)) + C\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">We can see that the brachistochrone curve can be expressed as a parametric curve in polar coordinates, which coincides with a cycloid that starts at the origin.<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}\n\n{} &amp; x(\\theta) &amp;=&amp; \\pm a(\\theta - \\sin(\\theta)) \\\\\n\n   &amp; y(\\theta) &amp;=&amp; a(1-\\cos(\\theta))\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">The integration constant <span class=\"katex-eq\" data-katex-display=\"false\">C<\/span> has been nullified to satisfy the initial condition that the trajectory starts at the origin. Moreover, we can observe that there are a couple of equations that provide possible solutions to the problem, where the constant <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> can be adjusted so that the curve passes through the point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)<\/span><\/span> at the end of the path. These equations are:<\/p>\n<p style=\"text-align:center;\"><strong>Option 1:<\/strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\begin{array}\n\n{} &amp; x(\\theta) &amp;=&amp; a(\\theta - \\sin(\\theta)) \\\\\n\n   &amp; y(\\theta) &amp;=&amp; a(1-\\cos(\\theta))\n\n\\end{array}}<\/span><\/span><\/p>\n<p style=\"text-align:center;\"><strong>Option 2:<\/strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\begin{array}\n\n{} &amp; x(\\theta) &amp;=&amp; - a(\\theta - \\sin(\\theta)) \\\\\n\n   &amp; y(\\theta) &amp;=&amp; a(1-\\cos(\\theta))\n\n\\end{array}}<\/span> <\/span><\/p>\n<p style=\"text-align:justify;\">The viable solution for this problem is given by the second option, and by adjusting the constant <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> as a negative value, we obtain a curve that meets the necessary conditions to be the solution.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis.png\" alt=\"Possible solution example, a cycloid arc\" width=\"497\" height=\"329\" class=\"aligncenter size-full wp-image-28731 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis.png\" alt=\"Possible solution example, a cycloid arc\" width=\"497\" height=\"329\" class=\"aligncenter size-full wp-image-28731 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis.png 497w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/solbraquis-300x199.png 300w\" sizes=\"(max-width: 497px) 100vw, 497px\" \/><\/noscript><\/center><\/p>\n<h3>Final Adjustment of the Solution<\/h3>\n<p style=\"text-align:justify;\">After the latest adjustments, the brachistochrone curve has the parametric form:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}\n\n{} x(\\theta) &amp;= b(\\theta - \\sin(\\theta)) \\\\\n\n   y(\\theta) &amp;= -b(1-\\cos(\\theta))\n\n\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align:justify;\">It was substituted <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a=-b,<\/span><\/span> where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">0\\lt b.<\/span><\/span> The curve has a period <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2b\\pi<\/span><\/span> and must satisfy the condition <span class=\"katex-eq\" data-katex-display=\"false\">x_2 \\in ]0,2b\\pi[<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">y_2 \\in ]-2b,0[.<\/span> This last part is crucial, as it requires that the brachistochrone curve be represented as a single arc of a cycloid, since the solution will cease to be valid if the particle returns to rest when returning to a point of zero height.<\/p>\n<p style=\"text-align:justify;\">To adjust these equations to the problem, we need to find the values of <span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> that satisfy the system:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}\n\n{} x_2 &amp;= b(\\theta - \\sin(\\theta))\\\\\n\ny_2 &amp;= - b(1-\\cos(\\theta))\n\n\\end{array}<\/span>\n<p style=\"text-align:justify;\">This nonlinear system does not seem to have analytical solutions, so we will use numerical methods in Wolfram Mathematica. Below is a series of steps to solve the problem:<\/p>\n<p><\/br><\/p>\n<h4>Step 1: Set up the system<\/h4>\n<p style=\"text-align:justify;\">Set the equations that form the system to be solved<\/p>\n<p><span dir=\"ltr\"><code>eq1 = x2 == b*(theta - Sin[theta])<br \/>\n eq2 = y2 == -b*(1 - Cos[theta])<\/code><\/span><br \/>\n<\/br><\/p>\n<h4>Step 2: Define the arrival point<\/h4>\n<p style=\"text-align:justify;\">Set the point the particle will reach at the end of its path. In this case, we will set it at <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)=(1,-2).<\/span><\/span> You can modify these values to test other similar configurations.<\/p>\n<p><span dir=\"ltr\"><code>x2val = 1; y2val = -2;<\/code><\/span><br \/>\n<\/br><\/p>\n<h4>Step 3: Calculate the sought values numerically<\/h4>\n<p style=\"text-align:justify;\">Use the \u00abFindRoot\u00bb function to numerically compute the solution to the problem<\/p>\n<p><span dir=\"ltr\"><code>sol = FindRoot[{eq1, eq2} \/. {x2 -> x2val, y2 -> y2val}, {{b,1}, {theta, 1}}]<\/code><\/span><\/p>\n<p style=\"text-align:justify;\">Here, the values <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b=1<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta=1<\/span><\/span> have been used as a starting point for the numerical approximation of the solution. This gives us the solution <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b\\approx 2.4056<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta \\approx 1.40138<\/span><\/span><\/p>\n<p><\/br><\/p>\n<h4>Step 4: Verification of Results<\/h4>\n<p style=\"text-align:justify;\">Remember that for these answers to make physical sense, it is necessary that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2 \\in ]0,2b\\pi[<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y_2 \\in ]-2b, 0[.<\/span><\/span> We can quickly check if this is the case using the following procedure<\/p>\n<p style=\"text-align:justify;\">First, extract the values of <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span> obtained as the solution<\/p>\n<p><code>bval = sol[[1, 2]]; thetaval = sol[[2, 2]];<\/code><\/p>\n<p style=\"text-align:justify;\">Then, command the confirmation<\/p>\n<p><code>If[0 < x2val < 2*Pi*bval &#038;&#038; -2*bval < y2val < 0 \"Valid Values\", \"Invalid Values\"]\n<\/code><\/p>\n<p style=\"text-align:justify;\">If everything went well, we should get \"Valid Values\" in the output. This piece of code will help you check if the physical situation is correctly modeled.<\/p>\n<p style=\"text-align:justify;\">With these procedures, we finally have our completely adjusted solution curve that connects the points <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_1,y_1)=(0,0)<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_2,y_2)=(1,-2).<\/span><\/span> The resulting curve is:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}\n\n{} x(\\theta) &amp;\\approx 2.4056(\\theta - \\sin(\\theta)) \\\\\n\ny(\\theta) &amp;\\approx -2.4056(1-\\cos(\\theta))\n\n\\end{array}\\;\\;;\\theta\\in [0, 1.40138]<\/span>\n<p style=\"text-align:justify;\">Which graphically looks like this:<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis.png\" alt=\"\" width=\"296\" height=\"345\" class=\"aligncenter size-full wp-image-28733 lazyload\" \/><noscript><img decoding=\"async\" src=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis.png\" alt=\"\" width=\"296\" height=\"345\" class=\"aligncenter size-full wp-image-28733 lazyload\" srcset=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis.png 296w, http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/sol2braquis-257x300.png 257w\" sizes=\"(max-width: 296px) 100vw, 296px\" \/><\/noscript><\/center><\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h2>GitHub Repository with Wolfram Algorithm<\/h2>\n<p style=\"text-align:justify;\">The complete code for the solution to the brachistochrone problem, including the algorithm developed in Wolfram Mathematica, is available for download and reference in my GitHub repository. This repository includes a `.nb` file with the code in an interactive notebook format, as well as a plain text version `.m` for those who prefer to view the code directly.<\/p>\n<p style=\"text-align:justify;\"><strong>You can download the repository from GitHub <a href=\"https:\/\/github.com\/girebz\/Braquist-crona\" target=\"_blank\" rel=\"noopener\">here<\/a>.<\/strong><\/p>\n<p style=\"text-align:justify;\">In addition to the code, the repository contains a \"README\" file with detailed instructions on how to use and understand the algorithm, as well as a step-by-step explanation of the solution to the brachistochrone problem. I hope you find it useful!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Variational Calculus in Classical Mechanics and the Euler-Lagrange Equation Abstract:In this class, we will review the derivation of the Euler-Lagrange equation of Analytical Mechanics through the use of variational calculus techniques, and from this, we will show in detail its application in solving the Brachistochrone problem. Learning Objectives: By the end of this class, the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28745,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":40,"footnotes":""},"categories":[936,635],"tags":[],"class_list":["post-28757","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-classical-mechanics","category-physics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Brachistochrone and Euler-Lagrange Equation with Variational Calculus - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Learn to derive the Euler-Lagrange equation using variational calculus and the principle of least action to solve the brachistochrone problem.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, 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