MediaWiki API result

{
    "batchcomplete": "",
    "continue": {
        "gapcontinue": "Requests",
        "continue": "gapcontinue||"
    },
    "warnings": {
        "main": {
            "*": "Subscribe to the mediawiki-api-announce mailing list at <https://lists.wikimedia.org/mailman/listinfo/mediawiki-api-announce> for notice of API deprecations and breaking changes."
        },
        "revisions": {
            "*": "Because \"rvslots\" was not specified, a legacy format has been used for the output. This format is deprecated, and in the future the new format will always be used."
        }
    },
    "query": {
        "pages": {
            "2259": {
                "pageid": 2259,
                "ns": 0,
                "title": "Reaction-Diffusion Systems",
                "revisions": [
                    {
                        "contentformat": "text/x-wiki",
                        "contentmodel": "wikitext",
                        "*": "{{nonlinear waves course\n | chapter title = Reaction-Diffusion Systems\n | next chapter = [[Burgers Equation]]\n | previous chapter = [[Example Calculations for the KdV and IST]]\n}}\n\n{{complete pages}}\n\nWe present here a brief theory of reaction diffusion waves.\n\n== Law of Mass Action ==\n\nThe law of mass action states that equation rates are proportional to the concentration\nof reacting species and the ratio in which they combined. It is discussed in detail in \n[[Billingham and King 2000]]. We will present here a few simple examples.\n\n=== Example 1: Simple Decay ===\n\nSuppose we have of chemical <math>P</math> which decays to <math>A</math>, i.e.\n<center>\n<math>P \\to A</math>\n</center>\nwith rate <math>k[P]</math> where <math>[P]</math> denotes concentration.  Then if we\nset <math>p=[P]</math> and <math>a = [A] </math> we obtain the equations\n<center>\n<math> \\frac{\\mathrm{d}p}{\\mathrm{d}t} = -kp\\,\\,\\,\\textrm{and}\\,\\,\\,  \\frac{\\mathrm{d}a}{\\mathrm{d}t} = kp</math>\n</center>\nwhich has solution\n<center>\n<math> \np = p_0 e^{-kt}\\,\\,\\,\\textrm{and}\\,\\,\\, a = a_0 + p_0(1-e^{-kt})\n</math>\n</center>\nwhere <math>a_0</math> and <math>p_0</math> are the values of <math>a</math>\n<math>p</math> repectively at <math>t=0</math>.\n\n=== Example 2: Quadratic Autocatalysis ===\n\nThis example will be important when we consider reaction diffusion problems.\nWe consider the reaction \n<center>\n<math>A + B \\to 2B</math>\n</center>\nwith rate proportional to <math>k[A][B]</math>.  If we define <math>a = [A]</math>\nand <math>b = [B]</math> we obtain the following equations\n<center>\n<math> \\frac{\\mathrm{d}a}{\\mathrm{d}t} = -kab\\,\\,\\,\\textrm{and}\\,\\,\\,  \\frac{\\mathrm{d}b}{\\mathrm{d}t} = kab</math>\n</center>\nWe can solve these equations by observing that\n<center>\n<math> \n\\frac{\\mathrm{d}(a+b)}{\\mathrm{d}t} = 0\n</math>\n</center>\nso that <math>a + b = a_0 + b_0</math>. We can then eliminate <math>a</math> to obtain\n<center>\n<math> \n\\frac{\\mathrm{d}b}{\\mathrm{d}t} = k(a_0 + b_0 - b)b\n</math>\n</center>\nwhich is separable with solution \n<center>\n<math> \nb = \\frac{b_0(a_0 + b_0)e^{k(a_0 + b_0)t}}{a_0 + b_0e^{k(a_0 + b_0)t}}\n</math>\n</center>\nand\n<center>\n<math> \na = \\frac{a_0(a_0 + b_0)}{a_0 + b_0e^{k(a_0 + b_0)t}}\n</math>\n</center>\nNote that <math>a\\to 0</math> and <math>b\\to a_0 + b_0</math>\nas <math>t\\to \\infty.</math>\n\n== Diffusion ==\n\nThe equation for spatially homogeneous diffusion of a chemical with concentration\n<math>c</math> is\n<center>\n<math> \n\\partial_t c = D\\nabla^2 c\n</math>\n</center>\nwhich is the [http://en.wikipedia.org/wiki/Heat_equation heat equation]. We will consider this in\nonly one spatial dimension. Consider it on the boundary  <math>-\\infty < x < \\infty</math>. In this case\nwe can solve by the [http://en.wikipedia.org/wiki/Fourier_transform Fourier transform] and obtain \n<center>\n<math> \n\\partial_t \\hat{c} = -D k^2 \\hat{c}\n</math>\n</center>\nwhere <math>\\hat{c}</math> is the Fourier transform of <math>c</math>. This has solution\n<center>\n<math> \n\\hat{c} = \\hat{c}_0 e^{-D k^2 t}\n</math>\n</center>\nWe can find the inverse transform using convolution and obtain\n<center>\n<math> \nc(x,t) = \\frac{1}{\\sqrt{4\\pi D t}} \\int_{-\\infty}^{\\infty} c_0(x) e^{(x-s)^2/4Dt}\\mathrm{d}s\n</math>\n</center>\n\n=== Solution of the dispersion equation using FFT ===\n\nWe can solve the dispersion equation using the discrete Fourier transform and\nits closely related numerical implementation the FFT (Fast Fourier Transform). \nWe have already met the FFT [[Numerical Solution of the KdV]] but we consider it here in more detail. \nWe consider the  concentration \non the finite domain <math>-L \\leq x \\leq  L</math> and use a \n[http://en.wikipedia.org/wiki/Fourier_series Fourier series] expansion \n<center>\n<math> \nc(x,t) = \\sum_{n=-\\infty}^{\\infty} \\hat{c}_n(t) e^{\\mathrm{i} k_n x}\n</math>\n</center>\nwhere <math>k_n = \\pi n /L </math>. If we substitute this into the diffusion equation we obtain\n<center>\n<math> \nc(x,t) = \\sum_{n=-\\infty}^{\\infty} \\hat{c}_n(0)e^{-k_n^2 D t} e^{\\mathrm{i} k_n x}\n</math>\n</center>\nNote that this is not the same solution as we obtained on the infinite domain because\nof the boundary conditions on the finite domain. The coefficients <math>\\hat{c}_n(0)</math>\nare found using the initial conditions so that \n<center>\n<math> \n\\hat{c}_n(0) = \\frac{1}{2L} \\int_{-L}^{L} e^{-\\mathrm{i} k_n x} c_0(x) \\mathrm{d}x \n</math>\n</center>\n\nThe key to the numerical solution of this equation is the use of the FFT. We begin by discretising the\ndomain into a series of <math>N</math> points <math>x_m = -L + 2Lm/N </math>. We then use this to\napproximate the integral above and obtain\n<center>\n<math> \n\\hat{c}_n(0) = \\frac{1}{N} \\sum_{m=0}^{N-1} e^{-\\mathrm{i} k_n x_m} c_0(x_m)\n</math>\n</center>\n<center>\n<math> \n= \\frac{1}{N} \\sum_{m=0}^{N-1} e^{-2\\mathrm{i} \\pi nm/N} e^{\\mathrm{i} \\pi n} c_0(x_m)\n</math>\n</center>\n\nWe also get\n<center>\n<math> \nc(x_m,t) = \\sum_{n=-\\infty}^{\\infty} \\hat{c}_n(0) e^{-k_n^2 D t} e^{\\mathrm{i} k_n x_m}\n</math>\n</center>\n<center>\n<math> \n = \\sum_{n=-\\infty}^{\\infty} \\hat{c}_n(0) e^{-k_n^2 D t} e^{2\\mathrm{i} \\pi nm/N} e^{-\\mathrm{i} \\pi n} \n</math>\n</center>\nbut we know that \n<center>\n<math> \n\\hat{c}_n(0) e^{-\\mathrm{i} \\pi n}  = \\frac{1}{N} \\sum_{m=0}^{N-1} e^{-2\\mathrm{i} \\pi nm/N} c_0(x_m)\n</math>\n</center>\n\n=== The discrete Fourier transform ===\nThe\n[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]\nof a sequence of ''2N'' complex numbers ''c''<sub>0</sub>, ..., ''c''<sub>''2N''\u22121</sub> is transformed into the  sequence of ''N'' complex numbers <math>\\hat{c}</math><sub>0</sub>, ..., <math>\\hat{c}</math><sub>''N''\u22121</sub> by the DFT according to the formula:\n<center>\n<math>\\hat{c}_m = \\sum_{n=0}^{N-1} c_n e^{-2\\pi \\mathrm{i}mn/N} \\quad \\quad m = 0, \\dots, N-1</math>   \n</center>           \n\nWe denote the transform by the symbol <math>\\mathcal{F}</math>, as in <math>\\mathbf{X} = \\mathcal{F} \\left \\{ \\mathbf{x} \\right \\} </math> or <math>\\mathcal{F} \\left ( \\mathbf{x} \\right )</math> or <math>\\mathcal{F} \\mathbf{x}</math>.  \n\nThe '''inverse discrete Fourier transform (IDFT)''' is given by \n<center>\n<math>c_n = \\frac{1}{2N} \\sum_{m=0}^{N-1} \\hat{c}_m e^{2\\pi \\mathrm{i}mn/N} \\quad \\quad n = 0,\\dots,N-1.</math>\n</center>\n\nTherefore we can write\n<center>\n<math> \nc(x_m,t) =  \\mathcal{F} \\left\\{ e^{-k_n^2 D t} \\mathcal{F}^{-1} \\left\\{ c_0(x_m) \\right\\}\n\\right\\} \n</math>\n</center>\n\nThe only difficulty is that we need to define carefully the values of \n<math>k_n</math>\n\nThe real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]\nor '''FFT''' algorithm. A naive implementation of the discrete Fourier transform above (or its inverse) \nwill involve order <math>N^2</math> operations. Using FFT algorithms, this can be reduced to  order <math>N  \\log(N)</math>. This is an incredible speed up, for example\nif N = 1024, FFT algorithms are more efficient by a factor of 147. This is the reason FFT\nalgorithms are used so extensively.\n\n== Reaction Diffusion Equations ==\n\nWe consider an auto catalytic reaction where the chemical species also diffuse. In this \ncase the equations are\n<center>\n<math>\\partial_t a = D\\partial_x^2 a - kab</math>\n</center>\n<center>\n<math>\\partial_t b = D\\partial_x^2 b + kab</math>\n</center>\nWe can non-dimensionalise these equations scaling the variables as\n<center>\n<math>\nz = x/x^*\\,\\,\\,\\tau = t/t^*\\,\\,\\,\\alpha = a/a_0\\,\\,\\,\\beta = b/a_0\n</math>\n</center>\nSo that the equations become\n<center>\n<math>\n\\frac{1}{t^*}\\partial_\\tau \\alpha = \\frac{a_0}{(x^*)^2}D\\partial_z^2 \\alpha \n- k a_0^2 \\alpha\\beta\n</math>\n</center>\n<center>\n<math>\n\\frac{1}{t^*}\\partial_\\tau \\beta = \\frac{a_0}{(x^*)^2}D\\partial_z^2 \\beta\n+ k a_0^2 \\alpha\\beta\n</math>\n</center>\nIf we choose \n<center>\n<math>\nx^* = \\sqrt{\\frac{D}{ka_0}}\\,\\,\\,t^*=\\frac{1}{ka_0}\n</math>\n</center>\nthen we obtain the system \n<center>\n<math>\n\\partial_\\tau \\alpha = \\partial_z^2 \\alpha \n-\\alpha\\beta\n</math>\n</center>\n<center>\n<math>\n\\partial_\\tau \\beta =\\partial_z^2 \\beta\n+ \\alpha\\beta\n</math>\n</center>\n\n=== Solution via split step method ===\n\n[[Image:Reaction diffusion.gif|thumb|right|500px|Solutions for <math>\\alpha(z,0) =1\n\\, \\beta(z,0) = \\exp(-10z^2)</math>]]\n\nWe can solve this equations numerically using a \n[http://en.wikipedia.org/wiki/Split-step_method split step method]. We assume\nthat at time <math>\\tau</math> we know <math>\\alpha(z,\\tau)</math>\nand <math>\\beta(z,\\tau)</math>. We then solve first the following equation\n<center>\n<math>\n\\partial_\\tau \\alpha = \\partial_z^2 \\alpha \n</math>\n</center>\nfrom <math>\\tau</math> to <math>\\tau + \\Delta\\tau</math>\n(which we can do exactly using the spectral methods just discussed for\nthe dispersion equation). We write this solution as \n<math>\\tilde{\\alpha}(z,\\tau + \\Delta\\tau)</math>\nThen we solve \n<center>\n<math>\n\\partial_\\tau \\alpha = -\\alpha\\beta\n</math>\n</center>\nby assuming that <math>\\beta</math> is constant and subject to the boundary\ncondition that <math>\\alpha(z,\\tau) = \\tilde{\\alpha}(z,\\tau + \\Delta\\tau)</math>. \nThis gives\n<center>\n<math>\n\\alpha(z,\\tau + \\Delta\\tau) = e^{-\\beta(z,\\tau) \\Delta\\tau} \\tilde{\\alpha}(z,\\tau+ \\Delta\\tau)\\,\n</math>\n</center>\nand we do likewise for the equation for <math>\\beta</math>. Note that\nwhile both steps are exact the result from the split step method is an \napproximation with error which becomes smaller as the step size becomes\nsmaller.\n\nWe can easily implement this split step method in matlab and we obtain\na pair of travelling waves.\n\n== Travelling Waves solution ==\n\nWhen we solve the equations we found the solution formed travelling waves and\nwe now consider this phenomena in detail. \n\nWe define a new coordinate <math>y = z - v\\tau</math> (so we will consider only\nwaves travelling to the right, although we could analyse waves travelling to\nthe left in a similar fashion).  We seek stationary solutions in \n<math>\\alpha(y)</math> and <math>\\beta(y)</math> which satisfy\n<center>\n<math>\n\\frac{\\mathrm{d}^2 \\alpha}{\\mathrm{d}y^2} + v \\frac{\\mathrm{d} \\alpha}{\\mathrm{d}y} = \\alpha\\beta\n</math>\n</center>\nand\n<center>\n<math>\n\\frac{\\mathrm{d}^2 \\beta}{\\mathrm{d}y^2} + v \\frac{\\mathrm{d} \\beta}{\\mathrm{d}y} = -\\alpha\\beta\n</math>\n</center>\nIf we add these equations we obtain\n<center>\n<math>\n\\frac{\\mathrm{d}^2 (\\alpha+\\beta)}{\\mathrm{d}y^2} + v \\frac{\\mathrm{d} (\\alpha+\\beta)}{\\mathrm{d}y} = 0\n</math>\n</center>\nso that <math>\\alpha + \\beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions\nare that as <math>y\\to\\infty </math> <math>\\alpha = 1</math> and \n<math>\\beta = 0</math> and if <math>y\\to-\\infty</math> then <math>\\alpha = 0</math> and \n<math>\\beta = 1</math>. Therefore <math>\\alpha + \\beta = 1</math>. \nThis means that, since <math>\\alpha \\geq 0</math>, we must have\n<math>0\\leq \\beta \\leq 1</math>.\nWe can then obtain the following equation\n<center>\n<math>\n\\frac{\\mathrm{d}^2 \\beta}{\\mathrm{d}y^2} + v \\frac{\\mathrm{d} \\beta}{\\mathrm{d}y} + \\beta(1-\\beta)= 0\n</math>\n</center>\nwhich we can write as the system of first order equations.\n\n[[Image:R_d_phase_portrait.jpg|thumb|right|500px|Phase portrait for out system showing\nthe equilibrium points and the heteroclinic connection]]\n\nWe define the variable <math>\\gamma = \\frac{\\mathrm{d}\\beta}{\\mathrm{d}y}</math>\nand we obtain \n<center><math>\n\\begin{align}\n\\frac{\\mathrm{d}\\beta}{\\mathrm{d} y} &= \\gamma&\\\\\n\\frac{\\mathrm{d}\\gamma}{\\mathrm{d} y} &= -v\\gamma + \\beta(\\beta -1)& \\\\\n\\end{align}\n</math></center>\nThis dynamical system has equilibrium points at <math>(0,0)</math>\nand <math>(1,0)</math>. We can analyse these equilibrium points by \nlinearization. The Jacobian matrix is \n<center>\n<math>\n J =\\begin{pmatrix}\n0 & 1 \\\\\n-1 + 2\\beta & -v\n\\end{pmatrix}\n</math>\n</center>\n\nWe can easily see that the Jacobian evaluated at our first equilibrium point is\n<center>\n<math>\n J_{(0,0)} =\\begin{pmatrix}\n0 & 1 \\\\\n-1 & -v\n\\end{pmatrix}\n</math>\n</center>\nwhich has eigenvalues <math>\\mu_{\\pm} = -1/2 (v \\mp \\sqrt{v^2-4})</math>.  Therefore\nthis point is a nodal sink (possibly a spiral)\n\n\n[[Image:R_d_wave.gif|right|Travelling wave solution for v=2 and the position on\nthe heteroclinic connection]]\n\n\nAdditionally,\n<center>\n<math>\n J_{(1,0)} =\\begin{pmatrix}\n0 & 1 \\\\\n1 & -v\n\\end{pmatrix}\n</math>\n</center>\nwhich has eigenvalues  <math>\\lambda_{\\pm} = -1/2 (v \\mp \\sqrt{v^2+4})</math>. \nThis is a a saddle point. The unstable and stable\nseparatrices leave the equilibrium point at <math>(1,0)</math> in the directions  \n<math> \\begin{pmatrix}\\lambda_{\\pm} \\\\ 1\\end{pmatrix}</math>.  The only path on which <math>\\beta</math> is bounded\nas <math>y\\to-\\infty</math> are the unstable separatrices. Also, only the \nunstable separatrix which enters the region <math>\\beta<1</math> is physically meaningful. \n \nTo find a travelling wave we need to find a heteroclinic connection \nbetween the two equilibrium points which also has to satisfy the conditions \nthat <math>0\\leq \\beta \\leq 1</math>.\n\nWe need to show that the heteroclinic connection does not cross the <math>\\beta</math> axis.\nConsider the region \n<center>\n<math>\nR = \\left\\{(\\beta,\\gamma)\\,|\\, \\beta<1,\\,-k\\beta<\\gamma<0\\right\\}\n</math>\n</center>\nOn the line <math>\\beta = 1,d\\beta/dy<0</math> and hence all flow it into <math>R</math>.\nOn the line <math>\\gamma = 0, d\\gamma/dy < 0</math> for <math>0<\\beta<1</math>.  On the line\n<math>\\gamma = -k \\beta</math> we know that <math>d\\beta/dy < 0</math> so that integral paths\nenter the region if and only if <math>d\\gamma/d\\beta < \\gamma/\\beta</math>. We know that\n<center>\n<math>\nd\\gamma/d\\beta - \\gamma/\\beta = -v - \\frac{\\beta(1-\\beta)}{\\gamma} -\\frac{\\gamma}{\\beta}\n= \\frac{1}{k} (k^2 - vk +1 -\\beta)\n</math>\n</center>\nwhen <math>\\gamma = -k\\beta</math>.  Therefore we need to find a value of <math>k</math>\nso that <math>k^2 - vk +1 < 0</math>, which is possible provided <math>v\\geq 2</math>, for example\n<math>k = \\dfrac{v}{2}</math>.\n\n== Lecture Videos == \n\n=== Part 1 ===\n\n{{#ev:youtube|rF4X42jP0v8}}\n\n=== Part 2 ===\n\n{{#ev:youtube|D0NwYlM-uOg}}\n\n=== Part 3 ===\n\n{{#ev:youtube|5IEZJtJaDHk}}\n\n=== Part 4 ===\n\n{{#ev:youtube|t_OjTSwVgdo}}\n\n=== Part 5 ===\n\n{{#ev:youtube|kTHVZaYezLk}}\n\n=== Part 6 ===\n\n{{#ev:youtube|MVuSg5_sfYI}}\n\n=== Part 7 ===\n\n{{#ev:youtube|hxKMOHyy6Bw}}\n\n=== Part 8 ===\n\n{{#ev:youtube|yQ-O2KIqu44}}\n\n\n\n[[Category:Simple Nonlinear Waves]]"
                    }
                ]
            },
            "1322": {
                "pageid": 1322,
                "ns": 0,
                "title": "Removing the Depth Dependence",
                "revisions": [
                    {
                        "contentformat": "text/x-wiki",
                        "contentmodel": "wikitext",
                        "*": "{{complete pages}}\n\nWe are considering the [[Frequency Domain Problem]] for linear wave waves. \n\n{{standard linear problem notation}}\n\n{{boundary value problem for a fixed body}}\n\n{{incident plane wave}}\n\n{{sommerfeld radiation condition two dimensions}}\n\n{{removing the depth dependence}}\n\n[[Category:Linear Water-Wave Theory]]"
                    }
                ]
            }
        }
    }
}