Exercise 2.3.5 Each ODE below depends on a parameter h. Assume the parameter h can be a real number of any sign. A bifurcation occurs in the ODE for a single value of the parameter, call it h = h*. Determine this value, sketch representative phase portraits for each of hh*, and h=h*. Use this to make a bifurcation diagram in the spirit of Figure 2.15. (a) u' = hu-u².

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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2.35 just do a)

 

**Bifurcation Diagram for the Harvested Logistic Equation**

This diagram illustrates the bifurcation behavior of a harvested logistic equation. It shows how the system dynamics change as the harvesting level \( h \) varies relative to the intrinsic growth rate \( r \).

- **Axes:**
  - The vertical axis represents the population level \( u \).
  - The horizontal axis segments are defined by:
    - \( h < r \)
    - \( h = r \)
    - \( h > r \)

- **Critical Points:**
  - When \( h < r \), there is a stable equilibrium point at \( u^* \) (indicated by a filled blue circle) above \( u = 0 \).
  - At \( h = r \), the stable equilibrium is at \( u = 0 \) (indicated by a transition line) and \( u^* \) continues as another equilibrium point (blue circle).
  - For \( h > r \), \( u = 0 \) remains the stable equilibrium, with \( u^* \) (dashed line and open circle) below becoming unstable or non-existent.

**Figure 2.15: Bifurcation diagram for the harvested logistic equation (2.61).**
Transcribed Image Text:**Bifurcation Diagram for the Harvested Logistic Equation** This diagram illustrates the bifurcation behavior of a harvested logistic equation. It shows how the system dynamics change as the harvesting level \( h \) varies relative to the intrinsic growth rate \( r \). - **Axes:** - The vertical axis represents the population level \( u \). - The horizontal axis segments are defined by: - \( h < r \) - \( h = r \) - \( h > r \) - **Critical Points:** - When \( h < r \), there is a stable equilibrium point at \( u^* \) (indicated by a filled blue circle) above \( u = 0 \). - At \( h = r \), the stable equilibrium is at \( u = 0 \) (indicated by a transition line) and \( u^* \) continues as another equilibrium point (blue circle). - For \( h > r \), \( u = 0 \) remains the stable equilibrium, with \( u^* \) (dashed line and open circle) below becoming unstable or non-existent. **Figure 2.15: Bifurcation diagram for the harvested logistic equation (2.61).**
**Exercise 2.3.5**

Each ODE below depends on a parameter \( h \). Assume the parameter \( h \) can be a real number of any sign. A bifurcation occurs in the ODE for a single value of the parameter, call it \( h = h^* \). Determine this value, sketch representative phase portraits for each of \( h < h^* \), \( h > h^* \), and \( h = h^* \). Use this to make a bifurcation diagram in the spirit of Figure 2.15.

(a) \( u' = hu - u^2 \).

(b) \( u' = hu - u^3 \). The resulting bifurcation diagram illustrates a *pitchfork bifurcation*.

(c) \( u' = ru(1 - u/K) - h \), with \( r = 1 \) and \( K = 1 \). This is similar to the harvested logistic equation (2.61), but here the harvest rate is a constant \( h \), instead of \( hu \). Assume \( h > 0 \), and confine your attention to the region \( u \geq 0 \).

---

**The Existence and Uniqueness of Solutions**

**Some Inspiration from Calculus 1**

Let's forget about differential equations for a moment and instead go back to Calculus 1, by considering the equation

\[ 2x - \cos(x) = 0. \; \; \; \; (2.63) \]

The goal is to find a real number \( x \) that satisfies (2.63). Consider trying to solve (2.63) by using the elementary algebraic operations \( +, -, \times, \div \), as well as square roots, inverse cosines, etc. You will not succeed, yet a plot of the function \( f(x) = 2x - \cos(x) \) clearly reveals that (2.63) has a real root between \( x = 0 \) and \( x = 1 \), as illustrated in Figure 2.16. It’s pretty clear this root is unique (there can be only one) since \( f \) appears to be strictly increasing; once \( f \) exceeds zero it can never decrease again.
Transcribed Image Text:**Exercise 2.3.5** Each ODE below depends on a parameter \( h \). Assume the parameter \( h \) can be a real number of any sign. A bifurcation occurs in the ODE for a single value of the parameter, call it \( h = h^* \). Determine this value, sketch representative phase portraits for each of \( h < h^* \), \( h > h^* \), and \( h = h^* \). Use this to make a bifurcation diagram in the spirit of Figure 2.15. (a) \( u' = hu - u^2 \). (b) \( u' = hu - u^3 \). The resulting bifurcation diagram illustrates a *pitchfork bifurcation*. (c) \( u' = ru(1 - u/K) - h \), with \( r = 1 \) and \( K = 1 \). This is similar to the harvested logistic equation (2.61), but here the harvest rate is a constant \( h \), instead of \( hu \). Assume \( h > 0 \), and confine your attention to the region \( u \geq 0 \). --- **The Existence and Uniqueness of Solutions** **Some Inspiration from Calculus 1** Let's forget about differential equations for a moment and instead go back to Calculus 1, by considering the equation \[ 2x - \cos(x) = 0. \; \; \; \; (2.63) \] The goal is to find a real number \( x \) that satisfies (2.63). Consider trying to solve (2.63) by using the elementary algebraic operations \( +, -, \times, \div \), as well as square roots, inverse cosines, etc. You will not succeed, yet a plot of the function \( f(x) = 2x - \cos(x) \) clearly reveals that (2.63) has a real root between \( x = 0 \) and \( x = 1 \), as illustrated in Figure 2.16. It’s pretty clear this root is unique (there can be only one) since \( f \) appears to be strictly increasing; once \( f \) exceeds zero it can never decrease again.
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