Stability of Equilibrium Points in Differential Equations with Methods and Examples
In mathematics, a stability solution of equations refers to a situation in which a minor disturbance in a system does not have a significant impact on the system. A function f(x) is said to be stable in terms of the solution of a differential equation if every other solution starts fairly close to it. When x = 0, it stays similar to it for subsequent x values. As x increases, the difference between the solutions reaches zero, and the solution is said to be asymptotically stable. A solution is said to be unstable if it lacks any of these properties.
In this article, we will discuss the stability of equilibrium solutions and conditions for asymptotically stable differential equations.
Stability of Equilibrium Solutions
The behaviour of integral curves near an equilibrium solution is used to classify its stability – they represent the graphs of specific solutions satisfying initial conditions whose initial values,y0, vary only slightly from the equilibrium value.
The equilibrium solution is said to be stable, or asymptotically stable if the nearby integral curves all converge towards an equilibrium solution as t increases. The long-term behaviour of such a solution is unaffected by minor (or often large) changes in its initial state.
When the integral curves near an equilibrium solution all diverge away from it as t increases, the equilibrium solution is said to be unstable. Such a solution is highly sensitive to even minor changes in its initial state, as shown by the previous example, where the smallest change in initial value results in completely different behaviours (in both long- and short-term)
As a result, the solution y = 0 is an unstable equilibrium solution in the logistic equation example, whereas y = K is an (asymptotically) stable equilibrium solution.
Plotting y'=f(y)versus y is an alternative graphical form. This graph is easier to draw, but it contains the same amount of information as the direction field. It's close to calculus's First Derivative Test for local extrema. Any interval where f(y) > 0 will rise, as indicated by a rightward arrow (they are separated by equilibrium solutions / critical points, which are the graph's horizontal intercepts). (As we know y is the horizontal axis in this map, and its coordinates increase from left to right, from (- to +). Similarly, on any interval where f(y) < 0, y is decreasing.
This fact will be indicated by a leftward arrow. To summarise: f(y)>0, y increases, resulting in a rightward arrow; f(y)0, y decreases, resulting in a leftward arrow. The following is how the outcome can be interpreted: If y=c is an equilibrium solution with f(y)=0
The equilibrium solution y=c is unstable if f(y)0 is to the left of c and f(y)>0 is to the right of c. (The arrows on both sides are visually moving away from c.)
The equilibrium solution y=c is asymptotically stable if f(y)>0 on the left of c and f(y)0 on the right of c. (The arrows on both sides are visually heading towards c.)
A leftward arrow indicates that y is decreasing as t increases. It refers to the direction field's downward-sloping arrows. A rightward arrow indicates that y grows as t grows. On the direction field, it corresponds to upward-sloping arrows.
Asymptotically Stable Differential Equations
Stable solution of differential equation, for example, the solution \[y = ce^{-x}\] of the equation \[y = -y\]is asymptotically stable, because the difference of any two solutions \[c_{1}e^{-x}\] and \[c_{2}e^{-x}\] is \[(c_{1} - c_{2})e^{-x}\]which always approaches zero as x increases.
Stable Semistable Unstable Differential Equations
If there are other solutions on both sides of an equilibrium equation, it is said to be asymptotically stable. If there are other solutions that approach this equilibrium solution on one side of the equilibrium solution, and other solutions diverge from this equilibrium solution on the other side of the equilibrium solution, the equilibrium solution is said to be Semi-Stable. If other solutions diverge from this equilibrium solution on both sides of the equilibrium solution, it is said to be unstable.
If \[\frac{dy}{dx} = f(t,y)\]is a differential equation, then the equilibrium solutions can be obtained by setting \[\frac{dy}{dx} = 0\]. For example, if \[\frac{dy}{dx} = y(y+2)\] then the equilibrium solutions can be obtained by solving y(y+2)=0 for y. We hence see that y=0 and y=-2 are the equilibrium solution.
The following image is the slope field of the differential equation \[\frac{dy}{dx} = (y-1)^{2} (y-2) (y-3)\] which has three equilibrium solutions y=1,y=2 and y=3.
(Image will be uploaded soon)
The equilibrium solution y=1 is green and is a semi-stable equilibrium solution.
The equilibrium solution y=2 is yellow and is an asymptotically stable equilibrium solution.
The equilibrium solution y=3 is a red colour, indicating that it is an unstable equilibrium solution.
FAQs on Stability Solutions of Differential Equations Explained Clearly
1. What is stability in differential equations?
In differential equations, stability refers to whether solutions remain close to an equilibrium point when slightly disturbed. For an equilibrium solution x = x₀:
- If nearby solutions stay close as t → ∞, it is stable.
- If nearby solutions move away, it is unstable.
- If solutions approach x₀ as t → ∞, it is asymptotically stable.
2. What is an equilibrium point in a differential equation?
An equilibrium point is a constant solution where the derivative equals zero. For a first-order equation dx/dt = f(x), equilibrium points satisfy f(x) = 0. For example, if dx/dt = x(1 − x), then solving x(1 − x) = 0 gives equilibrium points x = 0 and x = 1. Stability analysis is performed around these points.
3. How do you determine the stability of an equilibrium point?
You determine stability by analyzing the sign of the derivative or eigenvalues near the equilibrium point. For dx/dt = f(x):
- Compute f′(x₀).
- If f′(x₀) < 0, the equilibrium is asymptotically stable.
- If f′(x₀) > 0, it is unstable.
4. What is asymptotic stability in differential equations?
An equilibrium is asymptotically stable if solutions not only remain close but also approach the equilibrium as t → ∞. This means:
- The solution is stable (small disturbances stay small).
- limₜ→∞ x(t) = x₀.
5. What is Lyapunov stability?
A solution is Lyapunov stable if small initial disturbances produce only small deviations for all future time. Formally, for every ε > 0, there exists δ > 0 such that if |x(0) − x₀| < δ, then |x(t) − x₀| < ε for all t ≥ 0. Lyapunov’s direct method uses a Lyapunov function V(x) to test stability without solving the differential equation.
6. How do eigenvalues determine stability of linear systems?
For a linear system dx/dt = Ax, stability depends on the eigenvalues of matrix A.
- If all eigenvalues have negative real parts, the system is asymptotically stable.
- If any eigenvalue has a positive real part, the system is unstable.
- If eigenvalues have zero real parts, further analysis is needed.
7. What is the difference between stable and asymptotically stable?
The difference is that stable means solutions stay close, while asymptotically stable means they also converge to the equilibrium. In summary:
- Stable: Solutions remain near x₀.
- Asymptotically stable: Solutions approach x₀ as t → ∞.
8. Can you give an example of stability analysis for a first-order differential equation?
Yes, consider dx/dt = x − x² to analyze stability solutions of differential equations.
- Find equilibrium points: x − x² = 0 ⇒ x(1 − x) = 0.
- So equilibria are x = 0 and x = 1.
- Compute f′(x) = 1 − 2x.
- f′(0) = 1 > 0 ⇒ x = 0 is unstable.
- f′(1) = −1 < 0 ⇒ x = 1 is asymptotically stable.
9. What is a stable node and an unstable node?
A stable node occurs when all eigenvalues are real and negative, while an unstable node occurs when all eigenvalues are real and positive. For a 2×2 linear system:
- Both eigenvalues < 0 ⇒ trajectories move toward equilibrium (stable node).
- Both eigenvalues > 0 ⇒ trajectories move away (unstable node).
10. Why is stability important in differential equations?
Stability is important because it determines whether solutions behave predictably under small disturbances. In applications:
- In control systems, stability ensures safe operation.
- In population models, it predicts long-term behavior.
- In engineering systems, it prevents oscillations or failure.
