Stability from Phase Lines - Differential Equations

The phase line for an autonomous equation $y' = f(y)$ shows that $f(y) \ge C > 0$ for some constant $C$ and for all real $y$ . Which of the following conclusions about any non-constant solution $y(t)$ is necessarily false?
Answer: The graph of the solution $y(t)$ has a horizontal asymptote. The condition $f(y) > 0$ for all $y$ means that $y'(t)$ is always positive. This implies that any solution $y(t)$ must be strictly increasing (making A and D true). Since the equation is autonomous and has no equilibrium points (since $f(y)$ is never zero), a solution $y(t)$ cannot approach a finite limit $L$ as $t \to \infty$ . If it did, $L$ would have to be an equilibrium point. A horizontal asymptote corresponds to a finite limit as $t \to \infty$ . Therefore, the graph of $y(t)$ cannot have a horizontal asymptote, making statement C necessarily false. Statement B can be true; for example, the solution to $y' = 1+y^2$ is $y(t) = \tan(t+c)$ , which becomes unbounded in finite time. This equation satisfies $f(y) \ge 1 > 0$ .
The phase line for $y' = f(y)$ has a semi-stable equilibrium at $y=2$ (arrows point towards 2 from above, away from 2 from below) and an unstable equilibrium at $y=0$ . Consider the perturbed equation $y' = f(y) + \epsilon$ for a small constant $\epsilon > 0$ . How many equilibrium points does the perturbed equation have?
Answer: One The equation $y' = f(y) + \epsilon$ corresponds to a small upward vertical shift of the graph of $f(y)$ by $\epsilon$ . We analyze what happens to the roots of $f(y)$ , which are the equilibrium points. 1. At the unstable equilibrium $y=0$ : For $f(y)$ to have an unstable equilibrium, the graph of $f(y)$ must cross the y-axis at $y=0$ with a positive slope. Shifting this graph up by a small $\epsilon$ will move the root slightly to the left, but a root will still exist near $y=0$ . 2. At the semi-stable equilibrium $y=2$ : The arrows point towards 2 from above (so $f(y)<0$ for $y>2$ ) and away from 2 from below (so $f(y)>0$ for $y<2$ ). This means the graph of $f(y)$ must touch the y-axis at $y=2$ from above and have a local maximum at $y=2$ . For example, $f(y)$ could look like $-y(y-2)^2$ . When we shift this graph up by $\epsilon$ , the local maximum at $(2,0)$ becomes $(2, \epsilon)$ . The function is now positive around $y=2$ , so the root at $y=2$ vanishes. Therefore, the original two equilibrium points are reduced to one equilibrium point after the perturbation.
The phase line for $y' = f(y)$ has a stable equilibrium at $y=2$ and a semi-stable equilibrium at $y=0$ . What is the stability of the equilibrium points for the new equation $y' = -f(y)$ ?
Answer: $y=2$ becomes unstable and $y=0$ remains semi-stable. The transformation from $y' = f(y)$ to $y' = -f(y)$ multiplies the rate of change by -1, which reverses the direction of all arrows on the phase line. 1. An equilibrium at $y=2$ is stable for $f(y)$ . This means for $f(y)$ , the arrow above $y=2$ points down, and the arrow below $y=2$ points up. For $-f(y)$ , these arrows are reversed: the arrow above $y=2$ will point up, and the arrow below $y=2$ will point down. This configuration corresponds to an unstable equilibrium. 2. An equilibrium at $y=0$ is semi-stable for $f(y)$ . This means arrows on one side point toward $y=0$ and on the other side point away. For example, the arrow above points away (up) and the arrow below points away (up). When reversed for $-f(y)$ , the arrow above will point toward $y=0$ (down) and the arrow below will point toward $y=0$ (down). This is still a semi-stable configuration. The same logic applies if the arrows were originally both pointing down. Therefore, the semi-stable equilibrium remains semi-stable.
The phase line for an autonomous equation $y' = f(y)$ has an unstable equilibrium at $y=3$ and a stable equilibrium at $y=-1$ . What is the stability of the equilibrium points for the new equation $y' = [f(y)]^2$ ?
Answer: $y=3$ becomes semi-stable and $y=-1$ becomes semi-stable. The equilibrium points for $y' = [f(y)]^2$ are the same as for $y' = f(y)$ , since $f(y)=0$ if and only if $[f(y)]^2=0$ . However, the dynamics between the equilibrium points change. Since $[f(y)]^2 \ge 0$ for all $y$ , the derivative $y'$ is always non-negative. This means that on the new phase line, all arrows must point upward, except at the equilibrium points where the rate of change is zero. For the equilibrium at $y=3$ : The arrow just below it will point up, and the arrow just above it will also point up. This is the definition of a semi-stable equilibrium. For the equilibrium at $y=-1$ : The arrow just below it will point up, and the arrow just above it will also point up. This is also a semi-stable equilibrium. Therefore, both the originally unstable and stable equilibria become semi-stable under this transformation.
Consider the differential equation $\frac{dy}{dt} = y(y-a)(y-b)$ where $0 < a < b$ . If the phase line analysis reveals that exactly two of the three equilibrium points are stable, which of the following must be true about the parameters?
Answer: This scenario is impossible since alternating stability requires exactly one stable point among three equilibria For $f(y) = y(y-a)(y-b)$ with $0 < a < b$ , the equilibria are at $y = 0, a, b$ . Analyzing signs: $f(y) < 0$ for $y \in (-\infty, 0) \cup (a, b)$ and $f(y) > 0$ for $y \in (0, a) \cup (b, \infty)$ . This gives: $y = 0$ unstable (arrows away), $y = a$ stable (arrows toward), $y = b$ unstable (arrows away). The alternating pattern of stability for consecutive equilibria is a fundamental property of autonomous equations with simple roots - stability must alternate. Having two stable points among three consecutive equilibria violates this principle.
The phase line for a differential equation $\frac{dy}{dt} = f(y)$ indicates that $y = 1$ is a semistable equilibrium point. If $f(y) = (y-1)^n g(y)$ where $g(1) \neq 0$ and $g(y)$ does not change sign near $y = 1$ , what constraints exist on the integer $n$ ?
Answer: $n$ must be even and $n \geq 2$ to prevent sign changes across the equilibrium For semistable behavior at $y = 1$ , solutions must approach from one side and move away from the other side, which requires $f(y)$ to maintain the same sign on both sides of $y = 1$ (no sign change). Since $f(y) = (y-1)^n g(y)$ and $g(y)$ doesn't change sign near $y = 1$ , the factor $(y-1)^n$ must not change sign either. This occurs when $n$ is even. If $n$ were odd, $(y-1)^n$ would change sign across $y = 1$ , making it either stable or unstable, not semistable. We need $n \geq 2$ for an equilibrium to exist.
The phase line for the differential equation $\frac{dy}{dt} = f(y)$ shows arrows pointing toward $y = 2$ from both directions, arrows pointing away from $y = -1$ in both directions, and arrows pointing toward $y = 4$ only from the left while pointing away from $y = 4$ on the right. If $f(y)$ has exactly these three zeros, what can be concluded about the behavior of solutions?
Answer: Solutions with initial conditions $-1 < y_0 < 2$ will approach $y = 2$ as $t \to \infty$ From the phase line description: $y = -1$ is unstable (arrows away), $y = 2$ is stable (arrows toward from both sides), and $y = 4$ is semistable (arrows toward from left, away from right). For solutions in $(-1, 2)$ , arrows point toward $y = 2$ , so these solutions approach 2. Choice A is wrong because solutions with $y_0 > 4$ move away from 4. Choice B is wrong because arrows in $(2, 4)$ point left toward 2, not toward 4. Choice C is wrong because we don't know the behavior as $y \to \infty$ without more information about $f(y)$ for large $y$ .
Consider the differential equation $y' = y^3 - \alpha y$ , which depends on the parameter $\alpha$ . How does the number of asymptotically stable equilibrium points change as $\alpha$ increases from negative to positive values?
Answer: The number of stable equilibria changes from zero to one. We analyze the equilibrium points and their stability based on $\alpha$ . Let $f(y, \alpha) = y^3 - \alpha y = y(y^2 - \alpha)$ . The equilibria are solutions to $f(y, \alpha)=0$ . We use the linearization theorem, checking the sign of $f_y = \frac{\partial f}{\partial y} = 3y^2 - \alpha$ . Case 1: $\alpha < 0$ . The only real solution to $y(y^2 - \alpha) = 0$ is $y=0$ . At this point, $f_y(0, \alpha) = -\alpha > 0$ . Since the derivative is positive, $y=0$ is an unstable equilibrium. There are zero stable equilibria. Case 2: $\alpha = 0$ . The equation is $y'=y^3$ . The only equilibrium is $y=0$ . Since $f_y(0,0)=0$ , the test is inconclusive. However, the phase line for $y'=y^3$ has arrows pointing up for $y>0$ and down for $y<0$ , so $y=0$ is unstable. There are zero stable equilibria. Case 3: $\alpha > 0$ . There are three equilibria: $y=0$ and $y=\pm\sqrt{\alpha}$ . - At $y=0$ : $f_y(0, \alpha) = -\alpha < 0$ . Since the derivative is negative, $y=0$ is a stable equilibrium. - At $y=\pm\sqrt{\alpha}$ : $f_y(\pm\sqrt{\alpha}, \alpha) = 3(\sqrt{\alpha})^2 - \alpha = 3\alpha - \alpha = 2\alpha > 0$ . Since the derivative is positive, both $y=\pm\sqrt{\alpha}$ are unstable equilibria. Thus, for $\alpha > 0$ , there is exactly one stable equilibrium. As $\alpha$ increases from negative to positive, the number of stable equilibria changes from 0 to 1.
A differential equation $\frac{dy}{dt} = f(y)$ has equilibrium points at $y = -2, 0, 1, 3$ . Phase line analysis shows that solutions starting near $y = 0$ move toward $y = 1$ , while solutions starting near $y = 1$ also move toward $y = 1$ . Additionally, solutions near $y = 3$ move away from $y = 3$ . What can be concluded about the stability of $y = -2$ ?
Answer: $y = -2$ must be stable since the alternating pattern requires it to attract solutions When analyzing autonomous differential equations like $\frac{dy}{dt} = f(y)$ , understanding stability patterns of equilibrium points is crucial. The key insight is that for typical autonomous equations, stability must alternate between consecutive equilibrium points. Given the information, you know that solutions near $y = 0$ move toward $y = 1$ , meaning $y = 0$ is unstable and $y = 1$ is stable. You also know that solutions near $y = 1$ move toward $y = 1$ , confirming $y = 1$ is stable. Additionally, solutions near $y = 3$ move away from $y = 3$ , so $y = 3$ is unstable. This gives you the pattern: $y = 1$ (stable) and $y = 3$ (unstable). Since $y = 0$ is between $y = -2$ and $y = 1$ , and $y = 0$ is unstable (solutions move toward $y = 1$ ), the alternating stability pattern requires $y = -2$ to be stable. Looking at the wrong answers: A) incorrectly suggests that multiplicities determine stability when we already have enough information from the phase line behavior. B) makes a false claim about consecutive stable points - the alternating pattern actually supports having $y = -2$ stable followed by $y = 0$ unstable. D) incorrectly identifies $y = -2$ as semistable and mischaracterizes the overall pattern. The correct answer is C: $y = -2$ must be stable because the alternating stability pattern in autonomous equations requires it. Study tip: For phase line problems, always map out the stability pattern systematically - stability typically alternates between consecutive equilibrium points in simple autonomous equations.
Let $y=c$ be an equilibrium point for the autonomous differential equation $y' = f(y)$ , where $f$ is a smooth function. If $f(c)=0$ , $f'(c)=0$ , and $f''(c) > 0$ , what is the stability of the equilibrium point $y=c$ ?
Answer: Semi-stable Since $f'(c)=0$ , the standard first-derivative test for stability is inconclusive. We must examine the behavior of $f(y)$ near $y=c$ using higher-order derivatives. The Taylor expansion of $f(y)$ around $y=c$ is $f(y) = f(c) + f'(c)(y-c) + \frac{f''(c)}{2!}(y-c)^2 + O((y-c)^3)$ . Given the conditions, this simplifies to $f(y) \approx \frac{f''(c)}{2}(y-c)^2$ for $y$ near $c$ . Since we are given that $f''(c) > 0$ , the term $(y-c)^2$ ensures that $f(y)$ is positive for $y$ on both sides of $c$ (for $y \neq c$ ). This means that $y' = f(y)$ is positive for $y$ just above $c$ and also for $y$ just below $c$ . On a phase line, the arrows on both sides of the equilibrium point $c$ would point upward (away from $c$ ). This is the definition of a semi-stable equilibrium. (If $f''(c)<0$ , it would also be semi-stable, with both arrows pointing down.)
A population $P(t)$ is modeled by the logistic equation with constant harvesting, $P' = P(5-P) - 4$ . Based on the phase line for this model, what is the long-term behavior of the population if the initial population is $P(0)=3$ ?
Answer: The population thrives and approaches a carrying capacity of $4$ . When you encounter a logistic equation with harvesting, you need to analyze the equilibrium points and their stability using a phase line. This tells you how populations behave over time from different starting points. First, find the equilibrium points by setting $P' = 0$ : $P(5-P) - 4 = 0$ , which gives $5P - P^2 - 4 = 0$ , or $P^2 - 5P + 4 = 0$ . Factoring yields $(P-1)(P-4) = 0$ , so the equilibrium points are $P = 1$ and $P = 4$ . To determine stability, check the sign of $P'$ in each interval. For $P < 1$ : choose $P = 0$ , then $P' = 0(5) - 4 = -4 < 0$ . For $1 < P < 4$ : choose $P = 2$ , then $P' = 2(3) - 4 = 2 > 0$ . For $P > 4$ : choose $P = 5$ , then $P' = 5(0) - 4 = -4 < 0$ . This means $P = 1$ is unstable (populations move away from it) and $P = 4$ is stable (populations approach it). Since $P(0) = 3$ lies between these equilibrium points where $P' > 0$ , the population increases toward the stable equilibrium at $P = 4$ . Answer D is correct because the population thrives (increases) and approaches carrying capacity 4. Answer A is wrong because the population doesn't decrease first from $P = 3$ . Answer B is incorrect because the population doesn't approach 0 from this initial condition. Answer C is wrong because $P = 3$ isn't an equilibrium point. Remember: always find equilibrium points first, then test intervals to build your phase line and predict long-term behavior.
Consider two differential equations: (I) $\frac{dy}{dt} = y(1-y)(2-y)$ and (II) $\frac{dy}{dt} = -y(1-y)(2-y)$ . How do their stability patterns compare at the shared equilibrium points?
Answer: The stability of each equilibrium point is exactly reversed between the two equations For equation (I), $f_1(y) = y(1-y)(2-y)$ : $y = 0$ unstable, $y = 1$ stable, $y = 2$ unstable. For equation (II), $f_2(y) = -y(1-y)(2-y) = -f_1(y)$ : since the right-hand side is negated, the direction of all arrows on the phase line reverses. This means $y = 0$ becomes stable, $y = 1$ becomes unstable, and $y = 2$ becomes stable. The stability of each equilibrium point is exactly opposite between the two equations. Choice A is wrong because equilibrium locations don't determine stability patterns. Choice C is wrong because neither equation has uniform stability.

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