// Numbas version: exam_results_page_options {"name": "Dynamical system 9: Saddle.", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "parts": [{"stepsPenalty": 0, "showCorrectAnswer": true, "variableReplacements": [], "showFeedbackIcon": true, "prompt": "

The system can be written in the form $\\dot{\\boldsymbol{x}}=\\mathsf{A}\\boldsymbol{x}$, where $\\boldsymbol{x}=\\pmatrix{x,y}^\\mathsf{T}$.

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Input the components of the matrix $\\mathsf{A}$ in order to obtain a saddle.

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In order to achieve this you have to supply the diagonal elements of the matrix, the entries for $b,\\;c$ are given. In this case the eigenvalues of $A$ are real and of opposite signs. 

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You are given that $b=\\var{f},\\;c=\\var{g}$.

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$\\mathsf{A}=\\Bigg($[[0]]$\\var{f}$$\\Bigg)$
$\\var{g}$[[1]]
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Once you have input appropriate values into the matrix, the diagram below shows the plot of $(x(t),y(t))$.

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At $t=0$ we have initally $x=-5,\\;\\;y=5$. Moving the point gives phase diagrams for the following initial values at $t=0$:

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$x=\\;$       $y=\\;$

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You can click on Steps to see the solutions for $x(t),\\;y(t)$ after you have input values into the matrix.

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Graph of $x(t),\\;y(t)$

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$x(t)$ is in black, $y(t)$ in blue.

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You can use the navigation bar to zoom in and out of the graph.

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Asking users to input coefficients of a system of diff equations so that the phase space is a saddle. All systems input by the user are graphed together with immediate feedback. Also included in the Steps are the graphs of the solutions for $x(t),\\; y(t);\\; x(0)=-5,\\;y(0)=5.$

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Consider the following two-dimensional dynamical system . You have to find values for $a,\\;b\\;c,\\;d$ such that the system's phase space is a saddle.

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\\[\\begin{align}\\dot{x}&=\\simplify[std]{a*x+b*y},\\\\\\dot{y}&=\\simplify[std]{c*x+d*y}.\\end{align}\\]

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Note that this question is purely formative and for experimenting with. No marks are given.

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Given the system of differential equations:

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\\[\\begin{align}\\dot{x}&=\\simplify[std]{a*x+b*y},\\\\\\dot{y}&=\\simplify[std]{c*x+d*y}.\\end{align}\\]

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It can be written in the form $\\dot{\\boldsymbol{x}}=\\mathsf{A}\\boldsymbol{x}$, where $\\boldsymbol{x}=\\pmatrix{x,y}^\\mathsf{T}$ and 

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\\[\\mathsf{A}=\\pmatrix{a& b\\\\c & d}\\]

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In this case, with $b=\\var{f},\\;c=\\var{g}$  we want to enter values for $a,\\;d$ such that the system gives a saddle for its phase space.

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For this to happen we need the eigenvalues of $A$ to be real with one positive and one negative.

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The characteristic polynomial for $A$ is given by 

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\\[\\det\\left(\\mathsf{A}-\\lambda\\mathsf{I}\\right)=0,\\]

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i.e.  $(a-\\lambda)(d-\\lambda)-bc=0$.  This leads to $\\lambda^2-(a+d)\\lambda+(ad-bc)=0$.

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So in order to get real eigenvalues with opposite signs we need :

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1. Real roots. $(a+d)^2 \\gt 4(ad-bc) \\Rightarrow (a-d)^2 \\gt -4bc$. 

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Hence we have one condition is that $(a-d)^2 \\gt \\simplify[all,!collectNumbers]{-4*{f}*{g}={-4*f*g}}$

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2. Opposite signs. The roots are given by $\\displaystyle \\frac{(a+d) \\pm \\sqrt{(a+d)^2-4(ad-bc)}}{2}$.

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Hence if $\\operatorname{det}(A)=ad-bc \\lt 0$ we see that the roots are opposite in sign.

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So this is the other condition: $ad-bc \\lt 0 \\Rightarrow ad \\lt \\var{f}\\times \\var{g}=\\var{f*g}$

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