Find all critical points of the given system and investigate the stability of each.
dx/dt = y - 1
dy/dt = x^2 - y
To find the critical points of the given system, we need to find the values of x and y where dx/dt = 0 and dy/dt = 0.
1. Set dx/dt = y - 1 equal to 0:
y - 1 = 0
y = 1
2. Set dy/dt = x^2 - y equal to 0:
x^2 - y = 0
x^2 - 1 = 0
x^2 = 1
x = ±1
So, we have two critical points: (1, 1) and (-1, 1).
To investigate the stability of each critical point, we need to find the eigenvalues of the Jacobian matrix J at each critical point. The Jacobian matrix is defined as:
J = [∂f/∂x, ∂f/∂y]
[∂g/∂x, ∂g/∂y]
where f and g are the functions dx/dt and dy/dt, respectively.
1. For the critical point (1, 1):
∂f/∂x = 0 (∂y/∂x is constant, so its derivative is 0)
∂f/∂y = 1
∂g/∂x = 2x
∂g/∂y = -1
Substitute the critical point values:
∂f/∂x = 0
∂f/∂y = 1
∂g/∂x = 2
∂g/∂y = -1
Jacobian matrix J = [0, 1]
[2, -1]
The eigenvalues of J can be found by solving the characteristic equation:
det(J - λI) = 0, where I is the identity matrix and λ is an eigenvalue.
det([0-λ, 1], [2, -1-λ]) = 0
(0-λ)(-1-λ) - (1)(2) = 0
λ^2 + λ + 2 = 0
Solve for λ using the quadratic formula:
λ = (-1 ± √(1^2 - 4(1)(2))) / 2
λ = (-1 ± √(1 - 8)) / 2
λ = (-1 ± √(-7)) / 2
λ = (-1 ± i√7) / 2
The eigenvalues are complex with negative real parts, since Re(λ) = -1/2. Therefore, the critical point (1, 1) is a stable spiral.
2. For the critical point (-1, 1):
Perform the same steps as above, substituting the critical point values into the Jacobian matrix.
∂f/∂x = 0
∂f/∂y = 1
∂g/∂x = -2
∂g/∂y = -1
Jacobian matrix J = [0, 1]
[-2, -1]
Solve the characteristic equation:
det(J - λI) = 0
(0-λ)(-1-λ) - (1)(-2) = 0
λ^2 + λ - 2 = 0
Solve for λ:
λ = (-1 ± √(1^2 - 4(1)(-2))) / 2
λ = (-1 ± √(1 + 8)) / 2
λ = (-1 ± √9) / 2
λ = (-1 ± 3) / 2
The eigenvalues are real and have opposite signs, with λ = -2 and λ = 1/2. Therefore, the critical point (-1, 1) is a saddle point.
In summary, the critical point (1, 1) is a stable spiral, while the critical point (-1, 1) is a saddle point.