Consider the system \\[ \\dot{x}=\\lambda_1 x\\\\ \\dot{y}=\\lambda_2y ,\\] where $\\lambda_1 < 0 < \\lambda_2$ and $D:=\\lambda_1+\\lambda_2 < 0$.
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", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "information", "useCustomName": true, "customName": "questions", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [{"label": "Hint to 1.", "rawLabel": "", "otherPart": 1, "variableReplacements": [], "availabilityCondition": "", "penalty": "", "penaltyAmount": 0, "showPenaltyHint": true, "lockAfterLeaving": false}, {"label": "Hint to 2 and 3", "rawLabel": "", "otherPart": 2, "variableReplacements": [], "availabilityCondition": "", "penalty": "", "penaltyAmount": 0, "showPenaltyHint": true, "lockAfterLeaving": false}, {"label": "Solution", "rawLabel": "", "otherPart": 3, "variableReplacements": [], "availabilityCondition": "", "penalty": "", "penaltyAmount": 0, "showPenaltyHint": true, "lockAfterLeaving": false}], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "The flow naturally defines a mapping from the half line $L_1 = \\{(1,y):y>0\\}$, to the half line $L_2=\\{(x,1):x>0\\}$. Indeed given a point $(0,\\eta) \\in L_1$ the orbit through $(1,\\eta)$ intersects $L_2$ in one and only one point $(\\xi(\\eta), 1) \\in L_2$. In this way define a mapping $\\xi : \\mathbb{R_+}\\to \\mathbb{R_+}$.
\nThere are several ways to proceed, here's one.
\nYou can assume that for a solution in the first quadrant you can write $y$ as a function of $x$ (this follows since in this case $x(t)$ is injective). By the chain rule
\\[y'(x)=\\frac{\\dot{y}}{\\dot{x}}.\\]
This should give you solutions depending on an arbitrary constant we could call $c$.
The solution we are interested in, is the one where $y=\\eta>0$ when $x=1$. Use this to determine $c$.
\nKnowing $c$, you can find $\\xi=\\xi(\\eta)$ such that $y(\\xi)=1$.
"}, {"type": "information", "useCustomName": true, "customName": "Hint to 2 and 3", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [{"label": "questions", "rawLabel": "", "otherPart": 0, "variableReplacements": [], "availabilityCondition": "", "penalty": "", "penaltyAmount": 0, "showPenaltyHint": true, "lockAfterLeaving": false}], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "In the first part you should've shown that $\\xi(\\eta)=\\eta^a$, with $a = -\\frac{\\lambda_1}{\\lambda_2}$.
\nShow that $a > 0$, this should give you 2.
\nShow that $a > 1$ (since $D = \\lambda_1 + \\lambda_2 <0$), this should give you 3.
"}, {"type": "information", "useCustomName": true, "customName": "Solution", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Since for a solution in the first quadrant $\\dot{x}<0$, $x(t)$ is an injective mapping, so a solution curve can be written $(x,y(x))$.
\nUsing the chain rule
\\[ y'(x)=\\frac{\\dot{y}}{\\dot{x}} = \\frac{\\lambda_2}{\\lambda_1} \\frac{y}{x}=-\\frac{1}{a}\\frac{y}{x} ,\\]
where we have set $a = -\\lambda_1 / \\lambda_2$.
Since $\\lambda_1 < 0 < \\lambda_2$, we have $a > 0$. Since $\\lambda+\\lambda_2 < 0$ we have $-\\lambda_1 > \\lambda_2$, so $a > 1$.
\nSeparating the variables, we find that $y(x)= c x^{-1/a}$, where $c$ is a constant.
\nWe look at the particular solution that satisfies $y(1)=\\eta$. This corresponds to $c=\\eta$, so the particular solution is $y(x)=\\eta x^{-1/a}$.
\nWhen does the solution curve cross $L_2$? It does so when $(y(x), x) = (1, \\xi)$ for some $\\xi>0$. So we look at the equation $y(x) = 1$: \\[\\eta x^{-1/a} = 1 \\Rightarrow x=\\eta^a .\\]
\nThis means that the function we are looking for is $\\xi(\\eta)=\\eta^a$.
\nSince $a > 0$, we have $\\xi(\\eta) \\to 0$ when $\\eta \\to 0$. This shows 2.
\nWe also have $\\xi'(\\eta) = a \\eta^{a-1} $, and since $a>1$ this is equal to zero when $\\eta = 0$. This shows 3.
\nEnd of exercise.
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