{"id":1085,"date":"2014-10-13T12:00:20","date_gmt":"2014-10-13T17:00:20","guid":{"rendered":"https:\/\/blogs.mathworks.com\/cleve\/?p=1085"},"modified":"2016-04-11T08:47:39","modified_gmt":"2016-04-11T13:47:39","slug":"mathworks-logo-part-one-why-is-it-l-shaped","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/cleve\/2014\/10\/13\/mathworks-logo-part-one-why-is-it-l-shaped\/","title":{"rendered":"MathWorks Logo, Part One. Why Is It L Shaped?"},"content":{"rendered":"<div class=\"content\">\r\n\r\n<!--introduction-->MathWorks is the only company in the world whose logo satisfies a partial differential equation. Why is the region for this equation shaped like a capital letter L?\r\n\r\n<!--\/introduction-->\r\n<h3>Contents<\/h3>\r\n<div>\r\n<ul>\r\n\t<li><a href=\"#86af6264-5609-4231-8deb-9fa44c28eda2\">Wave Equation<\/a><\/li>\r\n\t<li><a href=\"#29f24b28-fb78-482b-acf4-3c81790ec68b\">Initial and Boundary Values<\/a><\/li>\r\n\t<li><a href=\"#23df62e3-d0b6-4702-a0cb-751e00fbc568\">Eigenvalues and Eigenfunctions<\/a><\/li>\r\n\t<li><a href=\"#8c2c226f-ea57-4098-a54b-bdeb4b10b69d\">One Dimension<\/a><\/li>\r\n\t<li><a href=\"#d728ea3f-b4d2-4e4d-a8ee-2df3efd32dc7\">A Square<\/a><\/li>\r\n\t<li><a href=\"#16f4f3c9-0c01-473e-96a4-70431ddfb896\">A Circular Disc<\/a><\/li>\r\n\t<li><a href=\"#b2dabb03-14ef-4bae-abc3-cc314f39a4c0\">A Circular Sector<\/a><\/li>\r\n\t<li><a href=\"#6e7d580a-2950-4ced-a202-e34679984bfd\">The L-shaped Membrane<\/a><\/li>\r\n\t<li><a href=\"#e32cad28-f489-4f3d-85b4-34951b1b6288\">Microwave Waveguide<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h4>Wave Equation<a name=\"86af6264-5609-4231-8deb-9fa44c28eda2\"><\/a><\/h4>\r\nThe wave equation describes how a disturbance travels through matter. If the units are chosen so that the wave propagation speed is equal to one, the amplitude of a wave satisfies\r\n\r\n$$ {{\\partial^2 u} \\over {\\partial t^2}} = \\triangle u $$\r\n\r\nThe $\\triangle$ denotes Laplace's operator\r\n\r\n$$ \\triangle = {\\partial^2 \\over {\\partial x^2}}\r\n+ {\\partial^2 \\over {\\partial y^2}} $$\r\n<h4>Initial and Boundary Values<a name=\"29f24b28-fb78-482b-acf4-3c81790ec68b\"><\/a><\/h4>\r\nGeometry plays a crucial role here. Initial values of the amplitude and velocity of the wave are prescribed on a certain region. Values of the amplitude or its normal derivative are also prescribed on the boundary of the region. If the wave vanishes outside the region, these boundary values are zero.\r\n<h4>Eigenvalues and Eigenfunctions<a name=\"23df62e3-d0b6-4702-a0cb-751e00fbc568\"><\/a><\/h4>\r\nSeparating out periodic time behavior leads to solutions of the form\r\n\r\n$$ u(x,y,t) = \\cos{(\\sqrt{\\lambda}\\,t)} v(x,y) $$\r\n\r\nThe functions $v(x,y)$ also depend upon $\\lambda$ and the region. They satisfy the differential equation\r\n\r\n$$ \\triangle v + \\lambda v = 0 $$\r\n\r\nand are zero on the boundary of the region. The quantities $\\lambda$ that lead to nonzero solutions are the <i>eigenvalues<\/i>, and the corresponding functions $v(x,y)$ are the <i>eigenfunctions<\/i> or <i>modes<\/i>. They are determined by the physical properties of the medium and the geometry of the region. The square roots of the eigenvalues are resonant frequencies. A periodic external driving force at one of these frequencies generates an unboundedly strong response in the medium.\r\n\r\nAny solution of the wave equation can be expressed as a linear combination of these eigenfunctions. The coefficients in the linear combination are obtained from the initial conditions.\r\n<h4>One Dimension<a name=\"8c2c226f-ea57-4098-a54b-bdeb4b10b69d\"><\/a><\/h4>\r\nIn one dimension, the eigenvalues and eigenfunctions are easily determined. The simplest example is a violin string, held fixed at the ends of an interval. The eigenfunctions are trig functions.\r\n\r\n$$ v_k(x) = \\sin{(k x)} $$\r\n\r\nIf the length of the interval is $\\pi$, the eigenvalues are determined by the boundary condition, $v_k(k \\pi) = 0$. Hence, $k$ must be an integer and\r\n\r\n$$ \\lambda_k = k^2 $$\r\n\r\nIf the initial condition is expanded in a Fourier sine series,\r\n\r\n$$ u(x,0) = \\sum_k a_k \\sin{(k x)} $$\r\n\r\n(And the initial velocity is zero), then the solution to the wave equation is\r\n\r\n$$ u(x,t) = \\sum_k a_k \\cos{(\\sqrt{\\lambda_k}\\,t)} v_k(x) $$\r\n\r\nHere are graphs of the first nine eigenfunctions in one dimension. The corresponding eigenvalues are the squares of integers.\r\n<pre class=\"codeinput\">   eigenvals = (1:9).^2\r\n   plot_modes(<span class=\"string\">'1d'<\/span>)\r\n<\/pre>\r\n<pre class=\"codeoutput\">eigenvals =\r\n     1     4     9    16    25    36    49    64    81\r\n<\/pre>\r\n<img decoding=\"async\" src=\"https:\/\/blogs.mathworks.com\/images\/cleve\/logo_pde_01.png\" alt=\"\" hspace=\"5\" vspace=\"5\" \/>\r\n<h4>A Square<a name=\"d728ea3f-b4d2-4e4d-a8ee-2df3efd32dc7\"><\/a><\/h4>\r\nThe simplest region in two dimensions is a square. The eigenfunctions are again trig functions.\r\n\r\n$$ v_{k,j}(x,y) = \\sin{(k x)}\\,\\sin{(j y)} $$\r\n\r\nIf the sides have length $\\pi$, the boundary conditions imply that $k$ and $j$ must be integers. Here are the first nine eigenvalues and eigenfunctions.\r\n<pre class=\"codeinput\">   [k,j] = meshgrid(1:3); e = k.^2+j.^2;  eigenvals = e(:)'\r\n   plot_modes(<span class=\"string\">'square'<\/span>)\r\n<\/pre>\r\n<pre class=\"codeoutput\">eigenvals =\r\n     2     5    10     5     8    13    10    13    18\r\n<\/pre>\r\n<img decoding=\"async\" src=\"https:\/\/blogs.mathworks.com\/images\/cleve\/logo_pde_02.png\" alt=\"\" hspace=\"5\" vspace=\"5\" \/>\r\n<h4>A Circular Disc<a name=\"16f4f3c9-0c01-473e-96a4-70431ddfb896\"><\/a><\/h4>\r\nIf the region is a circular disc, we switch to polar coordinates, $r$ and $\\theta$. Trig functions are replaced by Bessel functions. The eigenfunctions become\r\n\r\n$$ v_{k,j}(r,\\theta) = B_j(\\mu_k r) \\,\\sin{(j \\theta)} $$\r\n\r\nwhere $B_j$ is the $j$ -th order Bessel function and $\\mu_k = \\sqrt{\\lambda_k}$. To find the eigenvalues we need to have the eigenfunctions vanish on the boundary of the disc. If the radius is one, we require\r\n\r\n$$ B_j(\\mu_k) = 0 $$\r\n\r\nIn other words, we need to compute zeros of Bessel functions. Here are the first nine eigenvalues and eigenfunctions of the circular disc. The violin string has become a tambourine.\r\n<pre class=\"codeinput\">   eigenvals = [bjzeros(0,3) bjzeros(1,3) bjzeros(2,3)].^2\r\n   plot_modes(<span class=\"string\">'disc'<\/span>)\r\n<\/pre>\r\n<pre class=\"codeoutput\">eigenvals =\r\n  Columns 1 through 7\r\n    5.7832   30.4713   74.8870   14.6820   49.2185  103.4995   26.3746\r\n  Columns 8 through 9\r\n   70.8500  135.0207\r\n<\/pre>\r\n<img decoding=\"async\" src=\"https:\/\/blogs.mathworks.com\/images\/cleve\/logo_pde_03.png\" alt=\"\" hspace=\"5\" vspace=\"5\" \/>\r\n<h4>A Circular Sector<a name=\"b2dabb03-14ef-4bae-abc3-cc314f39a4c0\"><\/a><\/h4>\r\nReplace the full circular disc by a three-quarter circular sector. The angle at the origin is $270^\\circ$ or $\\frac{3}{2}\\pi$ radians. We can make our eigenfunctions adapt to this angle. Take\r\n\r\n$$ v_{k,j}(r,\\theta) = B_{\\alpha_j}(\\mu_k r) \\,\\sin{(\\alpha_j \\theta)} $$\r\n\r\nwith\r\n\r\n$$ \\alpha_j = \\frac{2}{3} j $$\r\n\r\nand fractional order Bessel functions. The eigenfunctions satisfy their differential equation and also satisfy the boundary conditions on both sides of the angle.\r\n\r\n$$ v_{k,j}(r,\\theta) = 0 $$\r\n\r\nat $\\theta = 0$ and at $\\theta = \\frac{3}{2}\\pi$.\r\n\r\nBy finding the zeros of the Bessel functions we can also have the eigenfunctions satisfy the boundary conditions on the outer circular portion of the boundary. Here are the first nine eigenvalues and eigenfunctions of the three-quarter circular sector.\r\n<pre class=\"codeinput\">   eigenvals = [bjzeros(2\/3,3) bjzeros(4\/3,3) bjzeros(6\/3,3)].^2\r\n   plot_modes(<span class=\"string\">'sector'<\/span>)\r\n<\/pre>\r\n<pre class=\"codeoutput\">eigenvals =\r\n  Columns 1 through 7\r\n   11.3947   42.6442   93.6362   18.2785   56.1131  113.6860   26.3746\r\n  Columns 8 through 9\r\n   70.8500  135.0207\r\n<\/pre>\r\n<img decoding=\"async\" src=\"https:\/\/blogs.mathworks.com\/images\/cleve\/logo_pde_04.png\" alt=\"\" hspace=\"5\" vspace=\"5\" \/>\r\n\r\nThese eigenfunctions have another important property. Most of them are singular; the derivatives of the fractional order Bessel functions are unbounded at the origin. You can see the black concentration of grid lines in the plots. If you tried to make a tambourine with this sector shape, it would rip at the sharp corner. This singular behavior is needed to model the solution to the wave equation on this region.\r\n<h4>The L-shaped Membrane<a name=\"6e7d580a-2950-4ced-a202-e34679984bfd\"><\/a><\/h4>\r\nFor all the regions we have discussed so far it is possible to express the eigenvalues as zeros of analytic functions. For the interval and the square, the eigenvalues are integers, which are the zeros of $\\sin{\\pi x}$. For the circular disc and sector, the eigenvalues are zeros of Bessel functions. Once we have the eigenvalues, it is easy to compute the eigenfunctions using sines and Bessel functions.\r\n\r\nSo, an L-shaped region formed from three unit squares is interesting for at least two reasons. It is the simplest geometry for which solutions to the wave equation cannot be expressed analytically; numerical computation is necessary. Furthermore, the 270 degree nonconvex corner causes a singularity in the solution. Mathematically, the gradient of the first eigenfunction is unbounded near the corner. Physically, a membrane stretched over such a region would rip at the corner. This singularity limits the accuracy of finite difference methods with uniform grids.\r\n\r\nI used the L-shaped region as the primary example in my doctoral thesis fifty years ago. MathWorks has adopted a modified surface plot of the first eigenfunction as the company logo. I am going to devote a series of blog posts to the L.\r\n\r\nHere are the first nine eigenvalues and eigenfunctions, computed by a function from <a href=\"https:\/\/www.mathworks.com\/matlabcentral\/fileexchange\/37976-numerical-computing-with-matlab\">Numerical Computing with MATLAB<\/a>, which I will discuss in a future posting. Compare these eigenfunctions with the ones from the circular sector, which has the same reentrant corner and resulting singularity.\r\n<pre class=\"codeinput\">   <span class=\"keyword\">for<\/span> k = 1:9\r\n      [~,eigenvals(k)] = membranetx(k);\r\n   <span class=\"keyword\">end<\/span>\r\n   eigenvals\r\n   plot_modes(<span class=\"string\">'L'<\/span>)\r\n<\/pre>\r\n<pre class=\"codeoutput\">eigenvals =\r\n  Columns 1 through 7\r\n    9.6397   15.1973   19.7392   29.5215   31.9126   41.4745   44.9485\r\n  Columns 8 through 9\r\n   49.3480   49.3480\r\n<\/pre>\r\n<img decoding=\"async\" src=\"https:\/\/blogs.mathworks.com\/images\/cleve\/logo_pde_05.png\" alt=\"\" hspace=\"5\" vspace=\"5\" \/>\r\n<h4>Microwave Waveguide<a name=\"e32cad28-f489-4f3d-85b4-34951b1b6288\"><\/a><\/h4>\r\nSimple model problems involving waves on an L-shaped region include an L-shaped membrane, or L-shaped tambourine, and a beach towel blowing in the wind, constrained by a picnic basket on one fourth of the towel.\r\n\r\nA more practical example involves ridged microwave waveguides. One such device is a waveguide-to-coax adapter. The active region is the channel with the H-shaped cross section visible at the end of the adapter. The ridges increase the bandwidth of the guide at the expense of higher attenuation and lower power-handling capability. Symmetry of the H about the dotted lines shown in the contour plot of the electric field implies that only one quarter of the domain needs to be considered and that the resulting geometry is our L-shaped region. The boundary conditions are different than our membrane problem, but the differential equation and the solution techniques are the same.\r\n\r\nThe photo is courtesy of Advanced Technical Materials, Inc. See their website, <a href=\"http:\/\/atmmicrowave.com\/\">&lt;http:\/\/atmmicrowave.com\/<\/a>&gt;, for lots of devices like this.\r\n<pre class=\"codeinput\">   waveguide\r\n<\/pre>\r\n<img decoding=\"async\" src=\"https:\/\/blogs.mathworks.com\/images\/cleve\/logo_pde_06.png\" alt=\"\" hspace=\"5\" vspace=\"5\" \/><script>\/\/ <![CDATA[\r\nfunction grabCode_75d5fc0f321244f8987800b367f35da4() {\r\n        \/\/ Remember the title so we can use it in the new page\r\n        title = document.title;\r\n\r\n        \/\/ Break up these strings so that their presence\r\n        \/\/ in the Javascript doesn't mess up the search for\r\n        \/\/ the MATLAB code.\r\n        t1='75d5fc0f321244f8987800b367f35da4 ' + '##### ' + 'SOURCE BEGIN' + ' #####';\r\n        t2='##### ' + 'SOURCE END' + ' #####' + ' 75d5fc0f321244f8987800b367f35da4';\r\n    \r\n        b=document.getElementsByTagName('body')[0];\r\n        i1=b.innerHTML.indexOf(t1)+t1.length;\r\n        i2=b.innerHTML.indexOf(t2);\r\n \r\n        code_string = b.innerHTML.substring(i1, i2);\r\n        code_string = code_string.replace(\/REPLACE_WITH_DASH_DASH\/g,'--');\r\n\r\n        \/\/ Use \/x3C\/g instead of the less-than character to avoid errors \r\n        \/\/ in the XML parser.\r\n        \/\/ Use '\\x26#60;' instead of '<' so that the XML parser\r\n        \/\/ doesn't go ahead and substitute the less-than character. \r\n        code_string = code_string.replace(\/\\x3C\/g, '\\x26#60;');\r\n\r\n        copyright = 'Copyright 2014 The MathWorks, Inc.';\r\n\r\n        w = window.open();\r\n        d = w.document;\r\n        d.write('\r\n\r\n\r\n\r\n<pre>\\n');\r\n        d.write(code_string);\r\n\r\n        \/\/ Add copyright line at the bottom if specified.\r\n        if (copyright.length > 0) {\r\n            d.writeln('');\r\n            d.writeln('%%');\r\n            if (copyright.length > 0) {\r\n                d.writeln('% _' + copyright + '_');\r\n            }\r\n        }\r\n\r\n        d.write('<\/pre>\r\n\r\n\r\n\r\n\r\n\\n');\r\n\r\n        d.title = title + ' (MATLAB code)';\r\n        d.close();\r\n    }\r\n\/\/ ]]><\/script>\r\n<p style=\"text-align: right; font-size: xx-small; font-weight: lighter; font-style: italic; color: gray;\"><a><span style=\"font-size: x-small; font-style: italic;\">Get\r\nthe MATLAB code<noscript>(requires JavaScript)<\/noscript><\/span><\/a><\/p>\r\nPublished with MATLAB\u00ae R2014b\r\n\r\n<\/div>\r\n<!--\r\n75d5fc0f321244f8987800b367f35da4 ##### SOURCE BEGIN #####\r\n%% MathWorks Logo, Part One.  Why Is It L Shaped?\r\n% MathWorks is the only company in the world with logo that satisfies a\r\n% partial differential equation.\r\n% Why is the region for this equation shaped like a capital letter L?\r\n\r\n%% Wave Equation\r\n% The wave equation describes how a disturbance travels through matter.\r\n% If the units are chosen so that the wave\r\n% propagation speed is equal to one, the amplitude of a wave satisfies\r\n%\r\n% $$ {{\\partial^2 u} \\over {\\partial t^2}} = \\triangle u $$\r\n%\r\n% The $\\triangle$ denotes Laplace's operator\r\n%\r\n% $$ \\triangle = {\\partial^2 \\over {\\partial x^2}}\r\n%           + {\\partial^2 \\over {\\partial y^2}} $$\r\n%\r\n\r\n%% Initial and Boundary Values\r\n% Geometry plays a crucial role here.\r\n% Initial values of the amplitude and velocity of the wave are prescribed\r\n% on a certain region.  Values of the amplitude or its normal derivative\r\n% are also prescribed on the boundary of the region.  If the wave vanishes\r\n% outside the region, these boundary values are zero.\r\n\r\n%% Eigenvalues and Eigenfunctions\r\n% Separating out periodic time behavior leads to solutions of the form\r\n%\r\n% $$ u(x,y,t) = \\cos{(\\sqrt{\\lambda}\\,t)} v(x,y) $$\r\n%\r\n% The functions $v(x,y)$ also depend upon $\\lambda$ and the region.\r\n% They satisfy the differential equation\r\n%\r\n% $$ \\triangle v + \\lambda v = 0 $$\r\n%\r\n% and are zero on the boundary of the region.  The quantities $\\lambda$\r\n% that lead to nonzero solutions are the _eigenvalues_, and\r\n% the corresponding functions $v(x,y)$ are the _eigenfunctions_\r\n% or _modes_.  They are determined by the physical\r\n% properties of the medium and the geometry of the region.\r\n% The square roots of the eigenvalues are resonant frequencies.\r\n% A periodic external driving force at one of these frequencies\r\n% generates an unboundedly strong response in the medium.\r\n\r\n%%\r\n% Any solution of the wave equation can be expressed as a linear\r\n% combination of these eigenfunctions.  The coefficients in the linear\r\n% combination are obtained from the initial conditions.\r\n\r\n%% One Dimension\r\n% In one dimension, the eigenvalues and eigenfunctions are\r\n% easily determined.  The simplest example is a violin string, held fixed\r\n% at the ends of an interval. The eigenfunctions are trig functions.\r\n%\r\n% $$ v_k(x) = \\sin{(k x)} $$\r\n%\r\n% If the length of the interval is $\\pi$, the eigenvalues are determined\r\n% by the boundary condition, $v_k(k \\pi) = 0$.  Hence, $k$ must be an integer\r\n%  and\r\n%\r\n% $$ \\lambda_k = k^2 $$\r\n\r\n%%\r\n% If the initial condition is expanded in a Fourier sine series,\r\n%\r\n% $$ u(x,0) = \\sum_k a_k \\sin{(k x)} $$\r\n%\r\n% (And the initial velocity is zero),\r\n% then the solution to the wave equation is\r\n%\r\n% $$ u(x,t) = \\sum_k a_k \\cos{(\\sqrt{\\lambda_k}\\,t)} v_k(x) $$\r\n\r\n%%\r\n% Here are graphs of the first nine eigenfunctions in one dimension.\r\n% The corresponding eigenvalues are the squares of integers.\r\n\r\neigenvals = (1:9).^2\r\nplot_modes('1d')\r\n\r\n%% A Square\r\n% The simplest region in two dimensions is a square.\r\n% The eigenfunctions are again trig functions.\r\n%\r\n% $$ v_{k,j}(x,y) = \\sin{(k x)}\\,\\sin{(j y)} $$\r\n%\r\n% If the sides have length $\\pi$, the boundary conditions imply that\r\n% $k$ and $j$ must be integers.  Here are the first nine eigenvalues\r\n% and eigenfunctions.\r\n\r\n[k,j] = meshgrid(1:3); e = k.^2+j.^2;  eigenvals = e(:)'\r\nplot_modes('square')\r\n\r\n%% A Circular Disc\r\n% If the region is a circular disc, we switch to polar coordinates, $r$ and\r\n% $\\theta$.  Trig functions are replaced by Bessel functions.\r\n% The eigenfunctions become\r\n%\r\n% $$ v_{k,j}(r,\\theta) = B_j(\\mu_k  r) \\,\\sin{(j \\theta)} $$\r\n%\r\n% where $B_j$ is the $j$ -th order Bessel function and\r\n% $\\mu_k = \\sqrt{\\lambda_k}$.\r\n% To find the eigenvalues we need to have the eigenfunctions vanish\r\n% on the boundary of the disc.  If the radius is one, we require\r\n%\r\n% $$ B_j(\\mu_k) = 0 $$\r\n%\r\n% In other words, we need to compute zeros of Bessel functions.\r\n% Here are the first nine eigenvalues and eigenfunctions of the circular disc.\r\n% The violin string has become a tambourine.\r\n%\r\neigenvals = [bjzeros(0,3) bjzeros(1,3) bjzeros(2,3)].^2\r\nplot_modes('disc')\r\n\r\n%% A Circular Sector\r\n% Replace the full circular disc by a three-quarter circular sector.\r\n% The angle at the origin is $270^\\circ$ or $\\frac{3}{2}\\pi$ radians.\r\n% We can make our eigenfunctions adapt to this angle.\r\n% Take\r\n%\r\n% $$ v_{k,j}(r,\\theta) = B_{\\alpha_j}(\\mu_k  r) \\,\\sin{(\\alpha_j \\theta)} $$\r\n%\r\n% with\r\n%\r\n% $$ \\alpha_j = \\frac{2}{3} j $$\r\n%\r\n% and fractional order Bessel functions.\r\n% The eigenfunctions satisfy their differential equation and also\r\n% satisfy the boundary conditions on both sides of the angle.\r\n%\r\n% $$ v_{k,j}(r,\\theta) = 0 $$\r\n%\r\n% at $\\theta = 0$ and at $\\theta = \\frac{3}{2}\\pi$.\r\n%\r\n\r\n%%\r\n% By finding the zeros of the Bessel functions we can also have the\r\n% eigenfunctions satisfy the boundary conditions on the outer circular\r\n% portion of the boundary.\r\n% Here are the first nine eigenvalues and eigenfunctions of the\r\n% three-quarter circular sector.\r\n\r\neigenvals = [bjzeros(2\/3,3) bjzeros(4\/3,3) bjzeros(6\/3,3)].^2\r\nplot_modes('sector')\r\n\r\n%%\r\n% These eigenfunctions have another important property.  Most of them\r\n% are singular; the derivatives of the fractional order Bessel functions\r\n% are unbounded at the origin.  You can see the black concentration of\r\n% grid lines in the plots.  If you tried to make a tambourine with this\r\n% sector shape, it would rip at the sharp corner.  This singular behavior\r\n% is needed to model the solution to the wave equation on this region.\r\n\r\n%% The L-shaped Membrane\r\n% For all the regions we have discussed so far it is possible to express\r\n% the eigenvalues as zeros of analytic functions.  For the interval and\r\n% the square, the eigenvalues are integers, which are the zeros of\r\n% $\\sin{\\pi x}$.  For the circular disc and sector, the eigenvalues are\r\n% zeros of Bessel functions.  Once we have the eigenvalues, it is\r\n% easy to compute the eigenfunctions using sines and Bessel functions.\r\n\r\n%%\r\n% So, an L-shaped region formed from three unit squares is interesting\r\n% for at least two reasons.  It is the simplest geometry for which\r\n% solutions to the wave equation cannot be expressed analytically;\r\n% numerical computation is necessary.  Furthermore, the 270 degree\r\n% nonconvex corner causes a singularity in the solution.  Mathematically,\r\n% the gradient of the first eigenfunction is unbounded near the corner.\r\n% Physically, a membrane stretched over such a region would rip at the\r\n% corner.  This singularity limits the accuracy of finite difference\r\n% methods with uniform grids.\r\n\r\n%%\r\n% I used the L-shaped region as the primary example in my doctoral thesis\r\n% fifty years ago.  MathWorks has adopted a modified surface plot of the\r\n% first eigenfunction as the company logo.  I am going to devote a series\r\n% of blog posts to the L.\r\n\r\n%%\r\n% Here are the first nine eigenvalues and eigenfunctions, computed by a\r\n% function from <https:\/\/www.mathworks.com\/moler.htmlncmfilelist.html % Numerical Computing with MATLAB>, which I will discuss in a future posting.\r\n% Compare these eigenfunctions with the ones from the circular sector,\r\n% which has the same reentrant corner and resulting singularity.\r\n\r\nfor k = 1:9\r\n[~,eigenvals(k)] = membranetx(k);\r\nend\r\neigenvals\r\nplot_modes('L')\r\n\r\n%% Microwave Waveguide\r\n% Simple model problems involving waves on an L-shaped region include\r\n% an L-shaped membrane, or L-shaped tambourine, and a beach towel blowing\r\n% in the wind, constrained by a picnic basket on one fourth of the towel.\r\n%\r\n% A more practical example involves ridged microwave waveguides.  One such\r\n% device is a waveguide-to-coax adapter.  The active region is the channel\r\n% with the H-shaped cross section visible at the end of the adapter.\r\n% The ridges increase the bandwidth of the guide at the expense of higher\r\n% attenuation and lower power-handling capability.  Symmetry of the H about\r\n% the dotted lines shown in the contour plot of the electric field implies\r\n% that only one quarter of the domain needs to be considered and that the\r\n% resulting geometry is our L-shaped region.  The boundary conditions are\r\n% different than our membrane problem, but the differential equation and\r\n% the solution techniques are the same.\r\n%\r\n% The photo is courtesy of Advanced Technical Materials, Inc.\r\n% See their website, <http:\/\/atmmicrowave.com\/ http:\/\/atmmicrowave.com\/>,\r\n% for lots of devices like this.\r\n\r\nwaveguide\r\n\r\n##### SOURCE END ##### 75d5fc0f321244f8987800b367f35da4\r\n-->","protected":false},"excerpt":{"rendered":"<div class=\"overview-image\"><img src=\"https:\/\/blogs.mathworks.com\/cleve\/files\/feature_image\/logo_pde_05.png\" class=\"img-responsive attachment-post-thumbnail size-post-thumbnail wp-post-image\" alt=\"\" decoding=\"async\" loading=\"lazy\" \/><\/div><!--introduction-->MathWorks is the only company in the world whose logo satisfies a partial differential equation. Why is the region for this equation shaped like a capital letter L?\r\n\r\n<!--\/introduction-->... <a class=\"read-more\" href=\"https:\/\/blogs.mathworks.com\/cleve\/2014\/10\/13\/mathworks-logo-part-one-why-is-it-l-shaped\/\">read more >><\/a><\/p>","protected":false},"author":78,"featured_media":1116,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[24,13,4,25],"tags":[],"_links":{"self":[{"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/posts\/1085"}],"collection":[{"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/users\/78"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/comments?post=1085"}],"version-history":[{"count":11,"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/posts\/1085\/revisions"}],"predecessor-version":[{"id":1409,"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/posts\/1085\/revisions\/1409"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/media\/1116"}],"wp:attachment":[{"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/media?parent=1085"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/categories?post=1085"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.mathworks.com\/cleve\/wp-json\/wp\/v2\/tags?post=1085"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}