Problem 46 - Calculus for Biology and Medicine

Question

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d x}{d t}=x^{3}-h x $$

Step-by-Step Solution

Answer

The equilibria exist if \(h > 0\) with \(x=0\) as stable and \(x=\pm\sqrt{h}\) as unstable.

Step 1: Setting Up Equilibrium Condition

For the system to be at equilibrium, the rate of change of \(x\) must be zero. Therefore, we set the equation \(\frac{d x}{d t} = x^3 - hx\) to zero. This gives the equilibrium condition:\[ x^3 - h x = 0 \]

Step 2: Factoring the Equation

To solve \(x^3 - h x = 0\), factor out an \(x\):\[ x(x^2 - h) = 0 \]The solutions to this equation, which are the equilibria, are when each factor equals zero.

Step 3: Solving for Equilibria

From the factored equation \(x(x^2 - h) = 0\), solve for \(x\):- \(x = 0\) is one equilibrium solution.- For \(x^2 - h = 0\), solving gives \(x^2 = h\), thus \(x = \pm\sqrt{h}\).

Step 4: Analyzing the Equilibrium Points

The equilibrium points are \(x = 0\), \(x = \sqrt{h}\), and \(x = -\sqrt{h}\). Next, determine the stability:- Consider the derivative \(f'(x) = \frac{d}{dx}(x^3 - hx) = 3x^2 - h\).- For stability, \(f'(x) < 0\) indicates stability.

Step 5: Stability Analysis of Each Equilibrium

Analyze stability using the derivative test:- At \(x = 0\), \(f'(0) = -h\). Stable only if \(h > 0\).- At \(x = \sqrt{h}\), \(f'(\sqrt{h}) = 2h\). Unstable for any \(h > 0\).- At \(x = -\sqrt{h}\), \(f'(-\sqrt{h}) = 2h\). Also unstable for any \(h > 0\).

Step 6: Graphical Interpretation

Graphically, the function \(f(x) = x^3 - hx\) has a shape that changes with \(h\). For \(h > 0\), \(x = 0\) is a local stable point (a minimum of the curve), while \(x = \pm\sqrt{h}\) are unstable (local maxima). As \(h\) changes, so do the transformations of the curve, highlighting the stability via the shape of these critical points.

Key Concepts

Stability Analysis

Stability analysis is a crucial part of understanding differential equations like our problem here. It tells us whether small disturbances or changes around equilibrium points will grow or die out over time. For this to be possible, we need to look at the derivative of the function.First, we found our equilibrium points by setting the differential equation to zero, giving us solutions at\[ x = 0, \ x = \sqrt{h}, \ x = -\sqrt{h} \]The stability of these points depends on the sign of the derivative at each point.- At \( x = 0 \), the derivative is negative when \( h > 0 \), meaning small disturbances will diminish, making it a stable point.- At \( x = \sqrt{h} \) and \( x = -\sqrt{h} \), the derivative is positive for any \( h > 0 \), so any disturbance grows, making these points unstable.Understanding which points are stable helps predict how a system behaves when not at equilibrium.

Graphical Analysis

Graphical analysis offers a visual representation to complement our mathematical work. By plotting the function \( f(x) = x^3 - h x \), we can see the behavior of the equilibrium points we found.For different values of \( h \), the shape of the graph changes:- When \( h > 0 \), the graph has a distinct shape where \( x = 0 \) is a local minimum, which correlates to stability.- The points \( x = \pm\sqrt{h} \) are local maxima, illustrating their unstableness as any small deviations will roll the system away from these points.By analyzing the curve, you can see how the stability and instability of equilibrium points visually manifest as valleys or peaks, making the graphical approach easily relatable.

Differential Equations

Differential equations are powerful tools used to describe how things change over time. They often model real-world dynamic systems.The equation \( \frac{d x}{d t} = x^3 - h x \) describes such a system. Here, \( \frac{d x}{d t} \) indicates how fast \( x \) changes with time. The equilibrium points where this rate is zero indicate where the system is balanced.Understanding differential equations involves not just finding these points but also determining their stability. It requires:- Setting the derivative to zero to find equilibrium points- Analyzing the sign of the derivative around these points to infer stabilityThese techniques allow us to predict whether a system will stay in a stable state, or if external factors can easily disturb it. Mastery of differential equations paves the way to understanding the underlying dynamics of various scientific and engineering problems.