Let's analyze the **phase diagram** of the system: \[ \begin{cases} \dot{x} = 20x - x^2 - x y \\ \dot{y} = 20y - 2y^2 - x y \end{cases} \] where \( x \) is the population of smaller squirrels, \( y \) is the population of larger squirrels. --- ## (a) Nullclines **Nullclines** are where one of the derivatives is zero. ### **x-nullcline** (\(\dot{x} = 0\)): \[ 20x - x^2 - x y = 0 \implies x(20 - x - y) = 0 \] So, either \( x = 0 \) or \( y = 20 - x \). ### **y-nullcline** (\(\dot{y} = 0\)): \[ 20y - 2y^2 - x y = 0 \implies y(20 - 2y - x) = 0 \] So, either \( y = 0 \) or \( x = 20 - 2y \). **Summary:** - \( x = 0 \): y-axis - \( y = 0 \): x-axis - \( y = 20 - x \): straight line, negative slope - \( x = 20 - 2y \): straight line, negative slope, twice as steep --- ## (b) Critical Points Critical points are intersections of the nullclines: solve simultaneously. ### **Case 1:** \( x = 0 \) - From \( x = 0 \), y-nullcline: \( y = 0 \) or \( 0 = 20 - 2y \implies y = 10 \) - \( (0,0) \) - \( (0,10) \) ### **Case 2:** \( y = 0 \) - From \( y = 0 \), x-nullcline: \( x = 0 \) or \( 0 = 20 - x \implies x = 20 \) - \( (0,0) \) (already found) - \( (20,0) \) ### **Case 3:** \( y = 20 - x \) and \( x = 20 - 2y \) Substitute \( y = 20 - x \) into \( x = 20 - 2y \): \[ x = 20 - 2y = 20 - 2(20 - x) = 20 - 40 + 2x = -20 + 2x \implies x - 2x = -20 \implies -x = -20 \implies x = 20 \] Then \( y = 20 - x = 0 \). (Already found: \( (20,0) \)) Try the other way: \( x = 20 - 2y \) into \( y = 20 - x \): \[ y = 20 - (20 - 2y) = 2y \implies y = 0 \] So all found already. ### **Case 4:** Both \( x \neq 0, y \neq 0 \): Set \( y = 20 - x \) and \( x = 20 - 2y \): As above, this just gives us \( (20,0) \). But, let's check if there's another intersection: Set \( y = 20 - x \) into \( x = 20 - 2y \): \[ x = 20 - 2(20 - x) = 20 - 40 + 2x = -20 + 2x \implies x - 2x = -20 \implies -x = -20 \implies x = 20, y = 0 \] But try \( x \neq 0 \), \( y \neq 0 \) and try to solve: From x-nullcline: \( y = 20 - x \) Plug into y-nullcline: \( x = 20 - 2y \) \[ x = 20 - 2(20 - x) = 20 - 40 + 2x = -20 + 2x \implies x - 2x = -20 \implies -x = -20 \implies x = 20, y = 0 \] Try another approach: set \( x = 20 - 2y \) into \( y = 20 - x \): \[ y = 20 - (20 - 2y) = 2y \implies y = 0 \] Alternatively, let's try equating the two nullcline lines: From x-nullcline: \( y = 20 - x \) From y-nullcline: \( x = 20 - 2y \) Plug \( x \) from y-nullcline into x-nullcline: \[ y = 20 - (20 - 2y) = 2y \implies y = 0 \] So, again, \( y = 0 \) Alternatively, set \( x = a \), \( y = b \), and solve the system: \[ 20x - x^2 - x y = 0 \\ 20y - 2y^2 - x y = 0 \] Assume \( x \neq 0, y \neq 0 \): From first: \( 20 - x - y = 0 \implies x + y = 20 \implies y = 20 - x \) Plug into second: \[ 20y - 2y^2 - x y = 0 \implies 20y - 2y^2 - x y = 0 \implies y(20 - 2y - x) = 0 \] Since \( y \neq 0 \), \( 20 - 2y - x = 0 \implies x = 20 - 2y \) But also \( x = 20 - y \), so \[ 20 - y = 20 - 2y \implies y = 0 \] No new critical point. So, let's try the other way: first, \( x = 0 \); second, \( y = 0 \); third, can both nullclines be satisfied for \( x \neq 0, y \neq 0 \)? From the above, **no**. **Therefore, the only critical points are:** - \( (0,0) \) - \( (0,10) \) - \( (20,0) \) --- ## (c) Linearization (Jacobian) at each Critical Point The **Jacobian matrix** is: \[ J = \begin{bmatrix} \frac{\partial \dot{x}}{\partial x} & \frac{\partial \dot{x}}{\partial y} \\ \frac{\partial \dot{y}}{\partial x} & \frac{\partial \dot{y}}{\partial y} \end{bmatrix} \] Compute the partial derivatives: \[ \frac{\partial \dot{x}}{\partial x} = 20 - 2x - y \\ \frac{\partial \dot{x}}{\partial y} = -x \\ \frac{\partial \dot{y}}{\partial x} = -y \\ \frac{\partial \dot{y}}{\partial y} = 20 - 4y - x \] Now, compute at each point: --- ### **At (0, 0):** \[ J_{(0,0)} = \begin{bmatrix} 20 & 0 \\ 0 & 20 \end{bmatrix} \] Eigenvalues are \( 20, 20 \) (both positive). --- ### **At (20, 0):** \[ \frac{\partial \dot{x}}{\partial x} = 20 - 2*20 - 0 = 20 - 40 = -20 \\ \frac{\partial \dot{x}}{\partial y} = -20 \\ \frac{\partial \dot{y}}{\partial x} = -0 = 0 \\ \frac{\partial \dot{y}}{\partial y} = 20 - 4*0 - 20 = 0 \] \[ J_{(20,0)} = \begin{bmatrix} -20 & -20 \\ 0 & 0 \end{bmatrix} \] Eigenvalues: The matrix is triangular, so eigenvalues are the diagonal: \(-20, 0\). --- ### **At (0, 10):** \[ \frac{\partial \dot{x}}{\partial x} = 20 - 0 - 10 = 10 \\ \frac{\partial \dot{x}}{\partial y} = 0 \\ \frac{\partial \dot{y}}{\partial x} = -10 \\ \frac{\partial \dot{y}}{\partial y} = 20 - 4*10 - 0 = 20 - 40 = -20 \] \[ J_{(0,10)} = \begin{bmatrix} 10 & 0 \\ -10 & -20 \end{bmatrix} \] Eigenvalues: The matrix is lower triangular, so eigenvalues are \(10, -20\). --- ## (d) Approximate flow lines near each critical point (using eigenvalues/vectors) ### **(0,0):** - Eigenvalues: \(20, 20\) (both positive) - Both directions are unstable (repelling node), trajectories move away from the origin along both axes. ### **(20,0):** - Eigenvalues: \(0, -20\) - One zero eigenvalue: center manifold possible, need to check stability by nonlinear terms. - The negative eigenvalue along x-direction (\(x\)), zero along y-direction (\(y\)). - Trajectories approach along x, but are "neutral" along y. ### **(0,10):** - Eigenvalues: \(10, -20\) - Saddle point: unstable along x-direction (positive eigenvalue), stable along y-direction (negative eigenvalue). - Trajectories approach along y-direction, escape along x-direction. --- ## (e) Stability of each critical point - **(0,0):** Unstable node (source) - **(20,0):** Non-hyperbolic (one zero eigenvalue), but locally stable along x-direction, neutral along y. - **(0,10):** Saddle (unstable in x, stable in y) --- ## (f) Flow lines near nullclines - **On \( x = 0 \) (y-axis):** \(\dot{x} = 0\), only \( y \) changes: - \( \dot{y} = 20y - 2y^2 \implies \) logistic-type growth for \( y \), max at \( y = 10 \). - **On \( y = 0 \) (x-axis):** \(\dot{y} = 0\), only \( x \) changes: - \( \dot{x} = 20x - x^2 \implies \) logistic-type growth for \( x \), max at \( x = 20 \). - **On \( y = 20 - x \):** \(\dot{x} = 0\), so movement is parallel to x-nullcline. - **On \( x = 20 - 2y \):** \(\dot{y} = 0\), so movement is parallel to y-nullcline. --- ## (g) Qualitative sketch of several solutions in the phase plane - The axes (\( x = 0 \), \( y = 0 \)) are nullclines. - The lines \( y = 20 - x \) and \( x = 20 - 2y \) cross at \((20,0)\). - Solutions starting near the origin move away from it (unstable source). - Near \( (0,10) \), solutions approach along y-direction and move away along x-direction (saddle). - Near \( (20,0) \), solutions approach along x-direction, and along y-direction, flow is slow/neutral (must check higher order terms to determine if locally stable or not). --- ## **Summary Table** | Critical Point | Type | Stability | Eigenvalues | |----------------|------------------|-------------|-------------| | (0, 0) | Unstable node | Unstable | 20, 20 | | (20, 0) | Non-hyperbolic | Saddle-like | 0, -20 | | (0, 10) | Saddle point | Saddle | 10, -20 | --- ## **Overall Conclusion** - The system has three critical points: the origin (both extinct), a pure small-squirrel population \((20, 0)\), and a pure large-squirrel population \((0, 10)\). - The origin is an unstable source (both populations increase from there). - The pure large-squirrel population is a saddle (unstable in x, stable in y), so it's not robustly stable. - The pure small-squirrel population is non-hyperbolic (one zero, one negative eigenvalue): it's stable along x, but neutral along y; nonlinear terms determine global stability. - The nullclines divide the phase plane into regions with different flow behavior; trajectories generally tend toward the small-squirrel-only equilibrium, with large squirrels going extinct unless started with a large population and few small squirrels. - **Ecological interpretation:** Small squirrels are competitively superior; unless the initial large-squirrel population is very high, the system tends toward dominance by small squirrels. --- **If you need a hand-drawn phase plane sketch or further step-by-step on nonlinear stability near (20,0),