Problem 47 - 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. $$ \frac{d x}{d t}=x\left(x^{2}-1\right)-h $$

Step-by-Step Solution

Answer

Equilibria exist when \(-\frac{2}{3\sqrt{3}} \leq h \leq \frac{2}{3\sqrt{3}}\). Outer equilibria are stable, center one is unstable.

Step 1: Define Equilibrium

An equilibrium occurs when the derivative of x with respect to time is zero. So, to find equilibrium points, set \( \frac{dx}{dt} = 0 \). Therefore, we have the equation \( x(x^2 - 1) - h = 0 \).

Step 2: Rearrange for X Values

Rearrange the equation to isolate terms involving x: \( x(x^2 - 1) = h \). This can be rewritten as \( x^3 - x = h \) to solve for x given h.

Step 3: Analyze as a Cubic Equation

\( x^3 - x = h \) is a cubic equation in x. The number of equilibria is determined by the intersection points of the curve \( x^3 - x \) with the horizontal line at height h. Since \( x^3 - x \) is a cubic function, it can have up to three real roots.

Step 4: Determine Values of h for Equilibria

The maximum and minimum values of \( x^3 - x \) are points where its derivative is zero. Calculate its derivative, \( d(x^3-x)/dx = 3x^2 - 1 \), and set it to zero to find critical points: \( 3x^2 - 1 = 0 \), or \( x^2 = \frac{1}{3} \), giving \( x = \pm\frac{1}{\sqrt{3}} \).

Step 5: Calculate Corresponding Values of h

Evaluate \( x^3 - x \) at \( x = \pm\frac{1}{\sqrt{3}} \) to find respective h-values. For \( x = \frac{1}{\sqrt{3}} \), \( \left(\frac{1}{\sqrt{3}}\right)^3 - \frac{1}{\sqrt{3}} = -\frac{2}{3\sqrt{3}} \). Similarly, for \( x = -\frac{1}{\sqrt{3}} \), it also evaluates to \( \frac{2}{3\sqrt{3}} \). This leads to two critical h-values, approximately \( \pm\frac{2}{3\sqrt{3}} \).

Step 6: Consider Graphical Argument for Stability

Plots of \( x^3 - x \) show that when \( h \) is between these critical values, it has three intersection points. The stability depends on the derivative \( 3x^2 - 1 \). The equilibria at the outer roots where the slope of \( x^3 - x \) crosses from positive to negative are stable, while the center one, where it crosses from negative to positive, is unstable.

Key Concepts

Cubic Function

Understanding cubic functions is vital as they frequently appear in differential equations and real-world problems. A cubic function is any function that can be expressed as a polynomial of degree three, like \(f(x) = ax^3 + bx^2 + cx + d\). In the provided exercise, the equation \(x^3 - x = h\) is a key part of the analysis, representing a typical cubic equation.

• Characteristics: Cubic functions can have up to three real roots. They exhibit an S-shaped curve, which means the graph can change direction twice, allowing for multiple crossing points with any horizontal line.
• Roots and Intersections: To find where the function intersects ==i.e.*, where equilibria exist, it is essential to solve for roots in terms of the constant \(h\). These root-intersect points are critical for understanding stability in systems described by cubic functions.

Grasping how the cubic function behaves is an essential step before moving to analyze equilibria and their stabilities in a more advanced context.

Stability Analysis

Stability analysis is a crucial procedure used to determine whether the system returns to equilibrium after a small disturbance. In the context of the exercise, the equilibrium points of \(x^3 - x = h\) must be analyzed to verify which are stable or unstable.

• Graphical Approach: The graphical method is a valuable tool for stability analysis. By plotting \(x^3 - x\), you can compare it against the horizontal line representing \(h\). Where these curves intersect signifies an equilibrium point.
• Slope and Stability: At equilibrium points, the function's slope (or derivative) helps determine stability. △When slope is negative (descending left to right), the equilibrium is stable (returns to equilibrium if perturbed). Conversely, a positive slope results in instability.

Through this graphical and derivative-based method, it becomes clear which equilibria are stable (outside ones) and unstable (middle intersection). This delineation is key for comprehending real-world stability in applied differential equations.

Critical Points

Critical points play a significant role when analyzing cubic functions. They occur where the first derivative of the function equals zero. In other words, the slope of the tangent to the curve is horizontal at these points.In the exercise, the critical points of the cubic function \(x^3-x\) are essential for determining the equilibrium points with respect to \(h\).

• Deriving Critical Points: By finding the derivative, \(3x^2 - 1\), and setting it to zero, critical points are revealed: \(x = \pm\frac{1}{\sqrt{3}}\).
• Evaluating Function Values: At these critical values, evaluating if \(x^3 - x\) leads to the critical \(h\) values: approximately \(\pm\frac{2}{3\sqrt{3}}\). This information verifies multiple equilibria occur only when \(h\) is within these limits.

Knowing the critical points helps in understanding both where these turning points lie and how critical values of \(h\) guide the behavior and stability of equilibria in the differential equation.