Do not enter decimals in your answer.
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", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}, {"answer": "{c1}/{d1}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "notallowed": {"message": "Do not enter decimals in your answer.
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "all", "marks": 1, "vsetrangepoints": 5}, {"answer": "{a1}/{b1}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "notallowed": {"message": "Do not enter decimals in your answer.
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "all", "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\\[\\begin{align}\\dot{x}&=x(\\simplify[std]{{a1}-{b1}*y}),\\\\\\dot{y}&=y(\\simplify[std]{{c1}-{d1}*x}).\\end{align}\\]
\nThere are two fixed points. Enter the fixed point with the smallest $x$-component in the first set of boxes. Do not enter decimals in your answers.
\nFixed point 1: $\\boldsymbol{x}_0=($[[0]]$,$[[1]]$)^\\mathsf{T}$.
\nFixed point 2: $\\boldsymbol{x}_0=($[[2]]$,$[[3]]$)^\\mathsf{T}$.
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\nOne fixed point always exists; enter it in the first set of boxes.
\nFixed point 1: $\\boldsymbol{x}_0=($[[0]]$,$[[1]]$)^\\mathsf{T}$.
\nFour additional fixed points can exist, depending on the coefficients in the system. For this particular set of coefficients, decide whether these additional fixed points exist. If they do, enter $x_0$ and $y_0$ in the boxes below. If no additional fixed points exist, enter the value $0$ for both components.
\nAdditional fixed points: $\\boldsymbol{x}_0=(\\pm$[[2]]$,\\pm$[[3]]$)^\\mathsf{T}$. (Enter your answers to 3d.p.)
", "showCorrectAnswer": true, "marks": 0}], "statement": "Find the fixed points $\\boldsymbol{x}_0=\\pmatrix{x_0,y_0}^\\mathsf{T}$ of the following two-dimensional dynamical systems $\\dot{\\boldsymbol{x}}=\\boldsymbol{f}(\\boldsymbol{x})$, where $\\boldsymbol{x}=\\pmatrix{x,y}^\\mathsf{T}$.
", "tags": ["checked2015", "MAS2106"], "rulesets": {"std": ["all", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "", "licence": "Creative Commons Attribution 4.0 International", "description": "Fixed points of 2D dynamical systems.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "The fixed points of a dynamical system
\n\\[\\begin{align}\\dot{x}&=f(x,y),\\\\\\dot{y}&=g(x,y),\\end{align}\\]
\nare given by $\\dot{x}=0$ and $\\dot{y}=0$, i.e. when $f(x,y)=0$ and $g(x,y)=0$.
\na)
\nWe have
\n\\[\\begin{align}\\dot{x}&=x(\\simplify[std]{{a1}-{b1}*y}),\\\\\\dot{y}&=y(\\simplify[std]{{c1}-{d1}*x}).\\end{align}\\]
\nFor $\\dot{x}=0$, either $x=0$ or $y=\\simplify{{a1}/{b1}}$.
\nIf $x=0$ then $\\dot{y}=0$ requires that $y=0$.
\nIf $y=\\simplify{{a1}/{b1}}$, then $\\dot{y}=0$ requires $x=\\simplify{{c1}/{d1}}$.
\nHence, there are two fixed points $\\boldsymbol{x}_0=\\pmatrix{0,0}^\\mathsf{T}$ and $\\boldsymbol{x}_0=\\pmatrix{\\simplify{{c1}/{d1}},\\simplify{{a1}/{b1}}}^\\mathsf{T}$.
\n \nb)
\nWe have
\n\\[\\begin{align}\\dot{x}&=x(\\simplify[std]{{a3}-{b3}*y^2}),\\\\\\dot{y}&=y(\\simplify[std]{{c3}-{d3}*x^2}).\\end{align}\\]
\nTo begin with, write the system in the general form
\n\\[\\begin{align}\\dot{x}&=x(a-by^2),\\\\\\dot{y}&=y(c-dx^2).\\end{align}\\]
\nIt is clear that $\\boldsymbol{x}_0=\\pmatrix{0,0}^\\mathsf{T}$ is always a fixed point.
\nFor $\\dot{x}=0$, then $x\\ne 0\\implies y=\\pm\\sqrt{\\frac{a}{b}}$, which is possible only if $\\frac{a}{b}>0$.
\nAssuming this is true, then $\\dot{y}=0$ requires $x=\\pm\\sqrt{\\frac{c}{d}}$, which is possible only if $\\frac{c}{d}>0$.
\nIf this is also true, then there are four additional fixed points $\\boldsymbol{x}_0=\\pmatrix{\\pm\\sqrt{\\frac{c}{d}},\\pm\\sqrt{\\frac{a}{b}}}^\\mathsf{T}$.
\nIn this case $\\frac{a}{b}=\\simplify{{a3}/{b3}}$ and $\\frac{c}{d}=\\simplify{{c3}/{d3}}$, {fixedpoints}.
\nThe answers in the boxes should therefore be $\\boldsymbol{x}_0=\\pmatrix{0,0}^\\mathsf{T}$ and $\\boldsymbol{x}_0=\\pmatrix{\\pm\\var{ans3a},\\pm\\var{ans3b}}^\\mathsf{T}${todp}
", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}