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Chapter 5: Problem 23

The following equations were given by J. S. Griffith \((1971, \mathrm{pp} .118-122)\), as a model for the interactions of messenger RNA \(M\) and protein \(E\) : $$ \dot{M}=\frac{a K E^{m}}{1+K E^{m}}-b M, \quad \dot{E}=c M-d E . $$ (See problem 25 in Chapter 7 for an interpretation.) (a) Show that by changing units one can rewrite these in terms of dimensionless variables, as follows $$ \dot{M}=\frac{E^{m}}{1+E^{m}}-\alpha M, \quad \dot{E}=M-\beta E . $$ Find \(\alpha\) and \(\beta\) in terms of the original parameters. (b) Show that one steady state is \(E=M=0\) and that others satisfy \(E^{m-1}=\alpha \beta\left(1+E^{m}\right)\). For \(m=1\) show that this steady state exists only if \(\alpha \beta \leq 1\). (c) Case 1. Show that for \(m=1\) and \(\alpha \beta>1\), the only steady state \(E=M=0\) is stable. Draw a phase-plane diagram of the system. (d) Case 2. Show that for \(m=2\), at steady state $$ E=\frac{1 \pm\left(1+4 \alpha^{2} \beta^{2}\right)^{1 / 2}}{2 \alpha \beta} $$ Conclude that there are two solutions if \(2 \alpha \beta<1\), one if \(2 \alpha \beta=1\), and none if \(2 \alpha \beta>1\). (e) Case 2 continued. For \(m=2\) and \(2 \alpha \beta<1\), show that there are two stable steady states (one of which is at \(E=M=0\) ) and one saddle point. Draw a phase-plane diagram of this system.

Short Answer

Expert verified
Simplified to dimensionless form, for m=1 the steady state \(E=M=0\) is stable if \(\alpha \beta > 1\), with phase diagrams indicating stability. For m=2, two steady states exist if \(2 \alpha \beta < 1\), demonstrated by the phase-plane diagram.

Step by step solution

01

Define Dimensionless Variables

Introduce dimensionless variables \( M' = M \tilde{M} \) and \( E' = E \tilde{E}\). Redefine parameters such that \( \tilde{M} \) and \( \tilde{E} \) make the equations dimensionless.
02

Non-dimensionalize the Equations

Rewrite the original equations using the dimensionless variables \( M' \) and \( E' \). Set constants \( \tilde{M} = \frac{1}{\beta} \) and \( \tilde{E} = \frac{1}{c} \) to balance units. As a result, \( \alpha = b \beta \) and \( \beta = \frac{d}{c} \).
03

Show Steady State Eqs for Part (b)

Set \( \dot{M} = \dot{E} = 0 \) in the dimensionless equations. Solve for \( M \) and \( E \). For \( M = E = 0 \), the steady state is straightforward. For other steady states, set up \( E^{m-1} = \alpha \beta (1 + E^m) \).
04

Analyze for m=1

For \( m=1 \), simplify the steady state equation to \( E = \alpha \beta (1 + E) \). Solve for \( E \), showing it exists if and only if \( \alpha \beta \leq 1 \).
05

Stability Analysis for m=1

Show that for \( m=1 \) and \( \alpha \beta > 1 \), the only steady state \( E = M = 0 \) is stable by linearizing around \( E=M=0 \) and finding eigenvalues of the Jacobian.
06

Phase-plane Diagram for Case 1

Draw the phase-plane diagram illustrating trajectories that converge to the stable point \( E = M = 0 \).
07

Steady State Eqs for m=2

Solve for steady states by setting \( \frac{dE}{dt} = 0 \) and \( \frac{dM}{dt} = 0 \) for \( m=2 \). Solve quadratic equation \( E = \frac{1 \,\pm\, \sqrt{1 + 4 \alpha^2 \beta^2}}{2 \alpha \beta} \). Identify conditions when solutions exist: \( 2 \alpha \beta < 1 \).
08

Stability Analysis for m=2

Analyze the stability by finding eigenvalues of the Jacobian matrix at steady states. Show that \( E=M=0 \) is stable and identify the second steady state as a saddle if \( 2 \alpha \beta < 1 \).
09

Phase-plane Diagram for m=2

Draw the phase-plane diagram for \( m=2 \). Show two stable nodes (one at \( E=M=0 \)) and a saddle point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dimensionless Variables
Dimensionless variables are used in mathematical modeling to simplify equations and reduce the number of parameters. They make equations universal, meaning they can apply to different systems with similar behavior without being tied to a specific unit system. For example, let’s introduce dimensionless variables for mRNA (M) and protein (E) concentrations as follows:
Assume new variables as \ M' = \frac{M}{M^0} \ and \ E' = \frac{E}{E^0}, where \( M^0 \) and \( E^0 \) are characteristic or typical values of M and E, respectively. These values normalize the concentrations.
The next step involves transforming the original equations to use these dimensionless variables, setting constants \( \tilde{M} \) and \ \tilde{E}\ such that the transformed equations govern the normalized variables. In our context, rewrite the parameters \( \alpha = b \beta \ and \beta = \frac{d}{c} \). By doing this, the complexity of solving the system is reduced, making it easier to analyze and interpret biological interactions.
Steady State Analysis
Steady state analysis in mathematical models helps identify equilibrium points where the system variables do not change over time. For our given model:
\ \dot{M}=\frac{E^{m}}{1+E^{m}} - \alpha M, \quad \dot{E}=M - \beta E.
Set the derivatives to zero, \(\dot{M} = 0\) and \(\dot{E} = 0\), to solve for steady states:
For \(M = 0\) and \(E = 0\), the equations read \(0 = \frac{0}{1+0} - \alpha 0\) and \(0 = 0 - \beta 0\), yielding the trivial steady state at \(M = E = 0\).
Non-trivial solutions satisfy \(E^{m-1} = \alpha \beta (1 + E^m)\).
For \(m = 1\), it simplifies to \(E = \alpha \beta (1 + E)\). Solving for E shows that an additional steady state exists only if \(\alpha \beta \ \leq 1\).
Phase-plane Diagrams
Phase-plane diagrams graphically represent the trajectories of a dynamic system in the state space. In our system, the state space is defined by the variables M and E. These diagrams help in visualizing the stability and behavior of solutions over time.
For \(m=1\) and \(\alpha \beta > 1\), draw a phase-plane diagram illustrating trajectories converging to the stable equilibrium point \(M=E=0\).
Steps to draw:
  • Plot Nullclines: Set \(\dot{M}=0\) and \(\dot{E}=0\) to get M and E nullclines.
  • Draw Arrows: Indicate the direction of movement in the M-E plane based on the sign of \(\dot{M}\) and \(\dot{E}\).
  • Stable Points: Identify and mark steady states.

For m = 2 and \( \2 \alpha \beta < 1\), there are three steady states. The diagram will show two stable nodes (one is at \(M=E=0\)) and one saddle point. This representation clarifies the dynamics and helps to visualize different possible behaviors of the system depending on initial conditions.
Stability Analysis
Stability analysis determines if a system returns to equilibrium after a small disturbance. It's crucial for understanding the long-term behavior of biological models.
For steady states:\(\E=M=0\) and \(E^{m-1}=\alpha \beta (1 + E^m)\)
  • Linearize Equations: Approximate the system near the equilibrium points.
  • Jacobian Matrix: For the system \{\(\dot{M}, \)\dot{E}\}, calculate the Jacobian matrix at the steady state.
  • Eigenvalues: Determine the eigenvalues of the Jacobian matrix. The sign of the real part of the eigenvalues reveals stability:

For \(m=1\) and \(\alpha \beta > 1\), eigenvalues at \(M=E=0\) are both negative (stable).
For \(m=2\) and \(\2 \alpha \beta < 1\), one eigenvalue may be positive (unstable) while the other is negative (saddle point). This helps classify the type of steady states (nodes, saddles, spirals) and their stability properties, thereby indicating whether the system will return to equilibrium or diverge after perturbation.

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Most popular questions from this chapter

In this problem we examine a continuous plant-herbivore model. We shall define \(q\) as the chemical state of the plant. Low values of \(q\) mean that the plant is toxic; higher values mean that the herbivores derive some nutritious value from it. Consider a situation in which plant quality is enhanced when the vegetation is subjected to a low to moderate level of herbivory, and declines when herbivory is extensive. Assume that herbivores whose density is \(I\) are small insects (such as scale bugs) that attach themselves to one plant for long periods of time. Further assume that their growth rate depends on the quality of the vegetation they consume. Typical equations that have been suggested for such a system are $$ \begin{aligned} \frac{d q}{d t} &=K_{1}-K_{2} q I\left(t-I_{0}\right) \\ \frac{d l}{d t} &=K_{3} I\left(1-\frac{K_{4} I}{q}\right) \end{aligned} $$ (a) Explain the equations, and suggest possible meanings for \(K_{1}, K_{2}, I_{0}, K_{3}\), and \(K_{4}\). (b) Show that the equations can be written in the following dimensionless form: $$ \begin{aligned} &\frac{d q}{d t}=1-K q I(I-1), \\ &\frac{d I}{d t}=\alpha I\left(1-\frac{I}{q}\right) . \end{aligned} $$ (c) Find qualitative solutions using phase-plane methods. Is there a steady state? What are its stability properties? (d) Interpret your solutions in part (c).

Curves in the plane. The locus of points for which \(x=y^{3}\) can be written in the form \((x(t), y(t))\) by choosing some parameter \(t\). For example, $$ x(t)=t, \quad y(t)=t^{3} . $$ Other choices are possible, for example, $$ x=s^{1 / 3}, \quad y=s \text {. } $$ These would depict the same curve but a different rate of motion along the curve. Then, using this parameterized form we can depict any position on the curve by the vector $$ \mathbf{x}(t)=\left(t, t^{3}\right) $$ and any tangent vector to the curve by the vector $$ \mathbf{v}(t)=\frac{d x}{d t}=\left(\frac{d x(t)}{d t}, \frac{d y(t)}{d t}\right)=\left(1,3 t^{2}\right) \text {. } $$ For example, at \(t=1, \mathbf{x}=(1,1)\) and \(\mathbf{v}=(1,3)\). (a) Using the parameterized form given here, sketch the curve, and compute the tangent vectors at points \((0,0),(2,8)\), and \((-1,-1)\). (b) Find a way of parameterizing the following curves, and determine the form of the tangent vector to the curve: (1) \(y=x(x-1)\). (4) \(x^{2}+y^{2}=1\). (2) \(y^{2}=\sin x\). (5) \(y=a x+b\). (3) \(x=1 / y\). (6) \(y=4 x^{2}\).

For the following first-order ordinary differential equations, sketch solution curves \(y(t)\) by first plotting the tangent vectors specified by the differential equations: (a) \(\frac{d y}{d t}=y^{2}\). (d) \(\frac{d y}{d t}=y e^{(r-1)}\) (b) \(\frac{d y}{d t}=1-\frac{y}{1+y}\). (e) \(\frac{d y}{d t}=\sin y \cos y\). (c) \(\frac{d y}{d t}=y(y-2)\).

In modeling the effect of spruce budworm on forest, Ludwig et al. (1978) defined the following set of variables for the condition of the forest: \(S(t)=\) total surface area of trees \(E(t)=\) energy reserve of trees. They considered the following set of equations for these variables in the presence of a constant budworm population \(B\) : $$ \begin{aligned} &\frac{d S}{d t}=r_{S} S\left(1-\frac{S}{K_{S}} \frac{K_{E}}{E}\right), \\ &\frac{d E}{d t}=r_{E} E\left(1-\frac{E}{K_{E}}\right)-P \frac{B}{S} . \end{aligned} $$ The factors \(r, K\), and \(P\) are to be considered constant. *(a) Interpret possible meanings of these equations. (b) Sketch nullclines and determine how many steady states exist. (c) Draw a phase-plane portrait of the system. Show that the outcomes differ qualitatively depending on whether \(B\) is small or large. *(d) Interpret what this might imply biologically. [Note: You may wish to consult Ludwig et al. (1978) or to return to parts (a) and (d) after reading Chapter 6.]

Write a system of linear first-order ODEs whose solutions have the following qualitative behaviors: (a) \((0,0)\) is a stable node with eigenvalues \(\lambda_{1}=-1\) and \(\lambda_{2}=-2\). (b) \((0,0)\) is a saddle point with eigenvalues \(\lambda_{1}=-1\) and \(\lambda_{2}=3\). (c) \((0,0)\) is a center with eigenvalues \(\lambda=\pm 2 i\). (d) \((0,0)\) is an unstable node with eigenvalues \(\lambda_{1}=2\) and \(\lambda_{2}=3\). Hint: Use the fact that \(\lambda_{1}\) and \(\lambda_{2}\) are eigenvalues of a matrix \(A\), then $$ \begin{aligned} &\lambda_{1}+\lambda_{2}=\operatorname{Tr} \mathrm{A}=a_{11}+a_{22} \\ &\lambda_{1} \lambda_{2}=\operatorname{det} \mathrm{A}=a_{11} a_{22}-a_{12} a_{21} . \end{aligned} $$ Note that there will be many possible choices for each of the above.
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