Problem 13 - Calculus for Biology and Medicine

Question

For which value of \(a\) has $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{2}\left(x_{1}+a\right) \\ \frac{d x_{2}}{d t}=x_{2}^{2}+x_{2}-x_{1} \end{array} $$ a unique equilibrium? Characterize its stability.

Step-by-Step Solution

Answer

The value of \(a\) that yields a unique equilibrium is \(\frac{1}{4}\); this equilibrium is locally stable.

Step 1 - Identify Equilibrium Points

To find the equilibrium points, we need to set the system of differential equations to zero:\[ \frac{dx_1}{dt} = x_2(x_1 + a) = 0 \] \[ \frac{dx_2}{dt} = x_2^2 + x_2 - x_1 = 0 \]The first equation implies either \(x_2 = 0\) or \(x_1 + a = 0\).

Step 2 - Analyze First Equation Solutions

For the first case, if \(x_2 = 0\), substitute into the second equation:\[ 0^2 + 0 - x_1 = 0 \Rightarrow x_1 = 0 \]In this scenario, the equilibrium point is \((x_1, x_2) = (0, 0)\).

Step 3 - Solve for Second Case

For the second case, \(x_1 + a = 0\), or \(x_1 = -a\), substitute into the second equation:\[ x_2^2 + x_2 + a = 0 \]To ensure a unique equilibrium, this quadratic equation for \(x_2\) should have a single solution.

Step 4 - Determine Conditions for Unique Solution

A quadratic equation \(ax^2 + bx + c = 0\) has a unique solution if its discriminant is zero:\[ b^2 - 4ac = 0 \]For our quadratic \(x_2^2 + x_2 + a = 0\), \(b = 1, a = 1, c = a\), resulting in the discriminant:\[ 1^2 - 4 \cdot 1 \cdot a = 0 \]\[ 1 - 4a = 0 \Rightarrow a = \frac{1}{4} \]

Step 5 - Check Equilibrium for a = 1/4

Substitute \(a = \frac{1}{4}\) back into the equation:The quadratic becomes \(x_2^2 + x_2 + \frac{1}{4} = 0\). The discriminant is 0, indicating a unique solution \(-\frac{1}{2}\).The equilibrium is at \((x_1, x_2) = (-\frac{1}{4}, -\frac{1}{2})\).

Step 6 - Characterize Stability at Equilibrium

To analyze the stability, evaluate the Jacobian matrix at the equilibrium:\[ J = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix} \]\(f_1 = x_2(x_1 + \frac{1}{4})\) and \(f_2 = x_2^2 + x_2 - x_1\).Evaluate at \((-\frac{1}{4}, -\frac{1}{2})\) to compute eigenvalues and determine stability.

Step 7 - Compute Jacobian Matrix

The elements of the Jacobian matrix at the equilibrium are: - \( \frac{\partial f_1}{\partial x_1} = x_2 \), - \( \frac{\partial f_1}{\partial x_2} = x_1 + \frac{1}{4} \), - \( \frac{\partial f_2}{\partial x_1} = -1 \), - \( \frac{\partial f_2}{\partial x_2} = 2x_2 + 1 \).At \((-\frac{1}{4}, -\frac{1}{2})\), the Jacobian is:\[ J = \begin{bmatrix} -\frac{1}{2} & 0 \ -1 & 0 \end{bmatrix} \]

Step 8 - Analyze Stability Using Eigenvalues

The eigenvalues of the Jacobian \(J\) are solutions of the equation:\[ \text{det}(J - \lambda I) = 0 \]\[ \begin{vmatrix} -\frac{1}{2} - \lambda & 0 \ -1 & -\lambda \end{vmatrix} = 0 \]\[ (-\frac{1}{2} - \lambda)(-\lambda) - 0 = 0 \]Solving the characteristic equation, the real part of both eigenvalues is negative, indicating local stability of the equilibrium point.

Key Concepts

Equilibrium Points

In the context of differential equations, equilibrium points, also known as steady states, are critical values

• These points are where the system does not change over time.
• This means that the derivatives (or rates of change) in the system become zero.

The search for equilibrium points involves setting each differential equation to zero and solving for the variables. This gives us specific values where the system can remain in balance.
In the exercise, equilibrium points are found by solving the two given equations:\[ \frac{dx_1}{dt} = x_2(x_1 + a) = 0 \]\[ \frac{dx_2}{dt} = x_2^2 + x_2 - x_1 = 0 \]The solutions to these equations tell us when the system does not change, hence it identifies where equilibrium occurs. Identifying equilibrium is crucial as it lays the foundation for understanding stability in complex systems.

Jacobian Matrix

The Jacobian matrix plays a vital role in determining the stability of equilibrium points in systems of differential equations. It is essentially a matrix of first-order partial derivatives, providing a local linear approximation of the system near the equilibrium points.
For our system, the Jacobian matrix is constructed by calculating the partial derivatives of the functions \( f_1 \) and \( f_2 \):

• \( \frac{\partial f_1}{\partial x_1} = x_2 \)
• \( \frac{\partial f_1}{\partial x_2} = x_1 + a \)
• \( \frac{\partial f_2}{\partial x_1} = -1 \)
• \( \frac{\partial f_2}{\partial x_2} = 2x_2 + 1 \)

At the equilibrium point \((-\frac{1}{4}, -\frac{1}{2})\), these derivatives fill the Jacobian matrix as follows:\[ J = \begin{bmatrix} -\frac{1}{2} & 0 \ -1 & 0 \end{bmatrix} \]This matrix is fundamental in analyzing how small perturbations around the equilibrium point will evolve over time.

Eigenvalues

Eigenvalues are crucial in determining the behavior near equilibrium points and are derived from the Jacobian matrix. The eigenvalues provide insights into the local stability of a system.

• If all the eigenvalues have negative real parts, the equilibrium is locally stable.
• If any eigenvalue has a positive real part, the equilibrium is unstable.

For our Jacobian matrix:\[ J = \begin{bmatrix} -\frac{1}{2} & 0 \ -1 & 0 \end{bmatrix} \]the eigenvalues are calculated by solving the characteristic equation:\[ \text{det}(J - \lambda I) = 0 \]This involves finding the determinant of \( J - \lambda I \), where \( I \) is the identity matrix. Solving:\[ (-\frac{1}{2} - \lambda)(-\lambda) - 0 = 0 \]will provide the solutions for \( \lambda \), the eigenvalues. In our exercise, the real parts of both eigenvalues are negative, indicating the local stability of the equilibrium point. Understanding eigenvalues allows us to predict whether small disturbances will die out or grow, thus determining the long-term behavior of the system.