Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Symbiotic relationships are common and persist for long periods. It is curious that there are no known or very few symbiotic relationships between mammals. There are, however, many symbiotic relationships between mammals and other organisms, Escherichia coli, for example. Analysis of the equations for symbiosis, Equations 18.63: $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ a. Show that the equilibrium point without zeros is $$ x_{1}=\frac{b+d}{a d-b c}, \quad y_{1}=\frac{a+c}{a d-b c} $$ $$ \text { if } a d-b c \neq 0 $$ b. \(x_{1}\) and \(y_{1}\) are positive only if \(a d-b c>0 .\) This is a surprise to us. Set \(b=c=d=1\) and examine the equilibrium point for \(a>1\) and \(a<1\). c. Assume \(a d-b c>0\) so that \(x_{1}\) and \(y_{1}\) are positive. Either work it out (no!) or accept our analysis that the Jacobian at \(\left(x_{1}, y_{1}\right)\) is $$ J=\left[\begin{array}{cc} -a \frac{b+d}{a d-b c} & b \frac{b+d}{a d-b c} \\ c \frac{a+c}{a d-b c} & -d \frac{a+c}{a d-b c} \end{array}\right]=\frac{1}{a d-b c}\left[\begin{array}{cc} -a(b+d) & b(b+d) \\ c(a+c) & -d(a+c) \end{array}\right] $$ Argue that if \(a d-b c>0,\left(x_{1}, y_{1}\right)\) is an asymptotically stable equilibrium. d. With persistence you might show that the characteristic roots are not complex. It requires showing that the discrimiant $$ \begin{array}{l} (a(b+d)+d(a+c))^{2}-4(a(b+d) d(a+c)-c(a+c) d(b+d))= \\ (a(b+d)-d(a+c))^{2}+4 c(a+c) d(b+d) \geq 0 \end{array} $$

Step-by-Step Solution

Answer

Equilibrium \((x_1, y_1)\) is positive and stable if \(ad-bc>0\). It is real and non-complex if the discriminant is non-negative.

Step 1: Set Up Equilibrium Conditions

To find the equilibrium points \((x_1, y_1)\) of the given equations, set \(x'(t) = 0\) and \(y'(t) = 0\). So we solve:\[ r_x \times x(t) \times (1 - ax(t) + by(t)) = 0 \] and \[ r_y \times y(t) \times (1 + cx(t) - dy(t)) = 0 \].

Step 2: Solve for Non-zero Equilibrium

Assuming \(x(t) eq 0\) and \(y(t) eq 0\), simplify the equilibrium conditions to: \[ 1 - ax + by = 0 \] and \[ 1 + cx - dy = 0 \]. Solve these linear equations to obtain \(x_1 = \frac{b+d}{ad-bc}\) and \(y_1 = \frac{a+c}{ad-bc}\).

Step 3: Examine Positivity Conditions

Given that \(x_1\) and \(y_1\) are positive only if \(ad-bc > 0\), setting \(b = c = d = 1\), explore for \(a > 1\) and \(a < 1\). For \(a > 1\), \(ad-bc = a - 1 > 0\), making \(x_1\) and \(y_1\) positive. For \(a < 1\), \(x_1\) and \(y_1\) are not positive.

Step 4: Analyze Stability Using Jacobian

The Jacobian matrix \(J\) is given, and the condition \(ad-bc > 0\) implies \((x_1, y_1)\) is asymptotically stable. Since the determinant of \(J\) is positive and trace is negative, it indicates stability.

Step 5: Confirm Non-complex Roots

Compute the discriminant \((a(b+d)-d(a+c))^2 + 4c(a+c)d(b+d)\), which is derived as non-negative \(\geq 0\), ensuring that the characteristic roots are not complex.

Key Concepts

Symbiotic Relationships

Symbiotic relationships describe interactions between two different biological organisms that are often mutually beneficial. The given exercise explores a mathematical model that represents such interdependencies between species. Using differential equations, it reflects the growth rates of two interacting organisms, expressed in terms of their respective populations over time. The presence of terms like \(ax(t)\) and \(by(t)\) in the equations suggests interactions that either inhibit or promote growth. These equations model the complex dynamics inherent in symbiotic relationships, such as those observed between mammals and bacteria like Escherichia coli.
Understanding these interactions in mathematical terms helps us predict the conditions under which one organism might benefit another and vice versa. Such equations are simplified reflections of real-world scenarios, highlighting the importance of positive interactions (competition, facilitation) or negative effects (inhibition) in these pairings.
When analyzing symbiotic relationships mathematically, it is crucial to identify equilibrium points where neither species’ population changes, meaning any immediate pressures are balanced. For biologists and ecologists, these insights are pivotal in understanding ecosystem stability and species survival.

Jacobian Matrix

The Jacobian matrix represents the cornerstone of determining the nature of equilibrium in dynamic systems modeled by differential equations. In the given exercise, the Jacobian matrix \( J \) is used to assess the behavior of small disturbances around equilibrium points for the given symbiotic model.

• The matrix \( J \) contains partial derivatives that describe how each population influences and is influenced by the other.
• This matrix helps predict whether a small change will grow, shrink, or oscillate over time.

For the symbiotic equations, the Jacobian's entries involve parameters like \( a, b, c, \) and \( d \), which are indicative of interaction strengths. When analyzing at equilibrium, these parameters define the stability conditions needed to maintain the balance between species.
A Jacobian matrix with positive determinant and negative trace in its simplified form typically signals asymptotic stability, meaning the system will return to equilibrium after minor deviations. This assessment is essential before predicting long-term outcomes for the population dynamics modeled by the differential equations.

Stability Analysis

Stability analysis is central for understanding and predicting the behavior of systems at or near equilibrium points. Using the Jacobian matrix, one can determine stability by examining the eigenvalues or characteristic roots.

• If the equilibrium is stable, slight deviations from this point will cause the system to naturally return to equilibrium over time.
• An unstable equilibrium means that small changes will cause the system to deviate further, potentially leading to wildly different outcomes.

In the context of the exercise, stability is checked by ensuring that the determinant of the Jacobian is positive and the trace is negative. This guarantees that small disturbances fade away rather than amplify.
Such stability guarantees are important in maintaining predictable population sizes, crucial for ecosystems where symbiotic relationships are fundamental. By understanding these stability conditions, biologists can make important predictions about how environmental changes might affect symbiotic pairs, ensuring better conservation and management strategies for fragile ecosystems.

Differential Equations

Differential equations are mathematical expressions that describe the rate of change for a given function. In this exercise, they represent the growth dynamics of two interacting populations over time.
The exercise uses differential equations to define how each species’ population size evolves due to internal growth rates and interactions with the other species. The terms in these equations include growth rates \(r_x\) and \(r_y\), and interaction coefficients \(a, b, c,\) and \(d\), which represent the influence of one species on the other's growth rate.

• A key idea in using these equations is to reach an equilibrium, where the dynamic interactions equilibrate so that the population sizes remain constant over time.
• The condition \(x'(t) = 0\) and \(y'(t) = 0\) at equilibrium assures that the populations do not change, revealing the long-term sustainability of the symbiotic pair.

Differential equations like these form the backbone for studies involving ecological interactions and growth models, offering insight into how populations respond to various influences in their environment. A strong understanding of these systems aids ecologists in making predictions about future trends and potential threats to symbiotic relationships.