Let's go through the problem step by step, with explanations and calculations. --- ### **1. (a) Use a plot of the harvesting function \( H(N) \) to argue why policy based on this harvesting function may be an improvement over constant quota harvesting.** **Harvesting Function:** \[ H(N) = \frac{BN}{A + N} \] - For small \( N \), \( H(N) \approx \frac{B}{A} N \) (harvesting is proportional to population, i.e., effort-based). - For very large \( N \), \( H(N) \approx B \) (saturates to a maximum quota, i.e., quota-based). **Plot:** - \( H(N) \) increases quickly for small \( N \), then levels off as \( N \) grows. - This is **better than constant quota harvesting**, which is \( H(N) = h \) (horizontal line), because: - For small \( N \), constant quota can over-harvest, risking collapse. - This function naturally reduces harvest when \( N \) is low, helping to prevent extinction. **Conclusion:** This function is a blend of proportional and quota harvesting, providing a built-in safeguard against over-exploitation at low population levels. --- ### **1. (b) Non-dimensionalization** Original model: \[ \frac{dN}{dt} = RN \left(1 - \frac{N}{K}\right) - H(N) \] with \( H(N) = \frac{BN}{A + N} \). Let - \( x = \frac{N}{K} \) (non-dimensional population, \( 0 \leq x \leq 1 \)), - \( \tau = Rt \) (non-dimensional time). Then, \[ N = xK \] \[ \frac{dN}{dt} = K \frac{dx}{dt} \] Substitute into the equation: \[ K \frac{dx}{dt} = R x K (1 - x) - \frac{B x K}{A + x K} \] \[ \frac{dx}{dt} = R x (1 - x) - \frac{B x}{A + x K} \] Change to non-dimensional time (\( \tau \)): \[ \frac{dx}{d\tau} = x(1 - x) - \frac{B x}{R (A + x K)} \] Let - \( a = \frac{A}{K} \) - \( b = \frac{B}{R K} \) So, \[ A + x K = K(a + x) \] \[ \frac{B x}{R (A + x K)} = \frac{B x}{R K (a + x)} = \frac{b x}{a + x} \] Therefore, \[ \boxed{ \frac{dx}{d\tau} = x(1 - x) - \frac{b x}{a + x} } \] **Non-dimensional variables and parameters:** - \( x = \frac{N}{K} \) - \( \tau = R t \) - \( a = \frac{A}{K} \) - \( b = \frac{B}{R K} \) --- ### **2. (a) Fixed Point at \( x^* = 0 \) and its stability** The equation: \[ \frac{dx}{d\tau} = x(1 - x) - \frac{b x}{a + x} \] Fixed points: Set RHS to zero: \[ x(1 - x) - \frac{b x}{a + x} = 0 \] \[ x \left[ (1 - x) - \frac{b}{a + x} \right] = 0 \] So, \( x^* = 0 \) is always a fixed point. **Linear stability at \( x = 0 \):** Let \( f(x) = x(1-x) - \frac{b x}{a + x} \). Calculate the derivative at \( x = 0 \): \[ f(x) = x(1-x) - \frac{b x}{a + x} \] \[ f'(x) = (1-x) - x - \frac{b (a + x) - b x}{(a + x)^2} \] At \( x = 0 \): \[ f'(0) = 1 - \frac{b}{a} \] - If \( 1 - \frac{b}{a} < 0 \), i.e., \( b > a \): \( x^* = 0 \) is **stable**. - If \( 1 - \frac{b}{a} > 0 \), i.e., \( b < a \): \( x^* = 0 \) is **unstable**. --- ### **2. (b) Other Fixed Points** Other fixed points are found by: \[ (1-x) - \frac{b}{a + x} = 0 \] \[ (1-x)(a + x) - b = 0 \] \[ a + x - a x - x^2 - b = 0 \] \[ -x^2 - a x + x + a - b = 0 \] \[ x^2 + (a - 1)x + (b - a) = 0 \] This is a quadratic in \( x \): \[ x^2 + (a - 1)x + (b - a) = 0 \] Discriminant: \[ \Delta = (a-1)^2 - 4(b-a) \] Number of (biologically relevant) fixed points depends on: - The discriminant (\( \Delta \)) - Whether the roots are non-negative (\( x \ge 0 \)) --- ### **3. (a) Bifurcation Diagrams** For each \( a = 1/2, 1, 2 \), the bifurcation diagram plots fixed points as \( b \) varies. \[ x^2 + (a - 1)x + (b - a) = 0 \] Roots: \[ x = \frac{1-a \pm \sqrt{(a-1)^2 - 4(b-a)}}{2} \] The number of real, non-negative roots changes as \( b \) varies. **Bifurcation points:** Values of \( b \) where the number of real positive roots changes, i.e., discriminant \( = 0 \): \[ (b-a) = \frac{1}{4}(a-1)^2 \implies b = a + \frac{1}{4}(a-1)^2 \] At \( b = a \), the fixed point at \( x = 0 \) changes stability (from 2a). --- ### **3. (b) Bifurcation at \( a = b \)** At \( b = a \): - Discriminant: \( (a-1)^2 \), always non-negative. - Fixed point: \( x = 0 \) is always a root. - Stability at \( x = 0 \) changes as shown in 2(a): this is a **transcritical bifurcation**. --- ### **3. (c) Bifurcation at \( b = \frac{1}{4}(a+1)^2 \) and its classification** Using the quadratic equation, set discriminant to zero: \[ (a-1)^2 - 4(b-a) = 0 \implies 4(b-a) = (a-1)^2 \implies b = a + \frac{1}{4}(a-1)^2 \] This is a **saddle-node bifurcation**. It is biologically relevant when \( a \leq 1 \), as it ensures non-negative fixed points. --- ### **4. (a) Region of Hysteresis in Parameter Space \((a, b)\)** Hysteresis occurs when the bifurcation diagram has a region where three fixed points exist (two stable, one unstable). This happens for values of \( (a, b) \) such that the quadratic for \( x \) has two positive roots (i.e., discriminant \( > 0 \), both roots \( x > 0 \)), and the lower root is not at \( x = 0 \). --- ### **4. (b) Values of \( a \) and \( b \) to Avoid Catastrophe** To avoid catastrophic collapse (no positive fixed point), policy should avoid: - Large \( b \) (overharvesting). - Values of \( b \) near where the only fixed point is \( x = 0 \) (i.e., \( b > b_{sn} \)), where \( b_{sn} \) is the saddle-node bifurcation value. --- ### **4. (c) Which values of \( a \), \( b \) ensure sustainable harvesting?** Sustainable harvesting requires at least one positive, stable fixed point. This occurs for: - \( b < a + \frac{1}{4}(a-1)^2 \) (before saddle-node bifurcation). - Also, avoid \( b > a \), where \( x = 0 \) is stable (population collapse). --- ## **Summary Table of Key Points** | Condition | Meaning | Formula | |----------------------------|------------------------------|-----------------------------------------| | \( a, b > 0 \) | Physical requirement | | | \( b < a \) | \( x = 0 \) is unstable | | | \( b > a \) | \( x = 0 \) is stable | (collapse) | | \( b < a + \frac{1}{4}(a-1)^2 \) | 2 positive fixed points (hysteresis region) | | | \( b = a \) | Transcritical bifurcation | | | \( b = a + \frac{1}{4}(a-1)^2 \) | Saddle-node bifurcation | | --- ### **Final Answers (in brief):** **1. (a):** The function \( H(N) \) prevents overharvesting at low population, unlike constant quota harvesting. **1. (b):** Non-dimensionalized model: \[ \frac{dx}{d\tau} = x(1-x) - \frac{b x}{a + x} \] with \( x = \frac{N}{K}, \tau = R t, a = \frac{A}{K}, b = \frac{B}{RK} \). **2. (a):** \( x^* = 0 \) is stable if \( b > a \), unstable if \( b < a \). **2. (b):** Other fixed points from \( x^2 + (a-1)x + (b-a) = 0 \), depend on discriminant. **3. (a-c):** - Draw bifurcation diagrams for each \( a \). - Bifurcation at \( b = a \) is transcritical. - Saddle-node at \( b = a + \frac{1}{4}(a-1)^2 \). **4. (a):** Hysteresis for \( a < 1 \) and \( a < b < a + \frac{1}{4}(a-1)^2 \). **4. (b):** Avoid \( b > a + \frac{1}{4}(a-1)^2 \) and \( b > a \). **4. (c):** Sustainable harvesting when \( b < a + \frac{1}{4}(a-1)^2 \) and \( a, b > 0 \). --- **If you need detailed sketches, bifurcation diagrams, or specific algebraic steps for each part, please let me know!**