Systems of differential equations. 0 x (t) x(t) a b Consider the system of differential equations (S) : = y 0 (t) y(t) c d | {z } A Step 1 : The determinant of A is det(A) = ad − cb. — If det(A) = 0 thenA is not invertible and there are infinitely many equilibrium solutions, i.e. solutions x(t) 0 to the equation A = . y(t) 0 x(t) 0 — If det(A) 6= 0 then A is invertible and there is a unique equilibrium solution = . y(t) 0 Step 2 : The characteristic polynomial of A is by definition PA (λ) = det(A − λI), where I is the identity matrix, i.e. the matrix with 1’s on the diagonal and 0’s elsewhere. For a 2 × 2 matrix a − λ b = λ2 − Tr(A)λ + det(A), PA (λ) = c d − λ where Tr(A) is the trace of the matrix, i.e. the sum of the diagonal coefficients. The eigenvalues of A are the roots of the characteristic polynomial, i.e. λ is an eigenvalue of A if and only if PA (λ) = 0. A vector V is called an eigenvector for the eigenvalue λ if : 1. V satisfies the equation AV = λV , 2. V is a non-zero vector. Step 3 : Let A be a 2 × 2 matrix. 1. If A has two distinct real eigenvalues λ1 < λ2 , then the general solution of (S) is given by : x(t) = Aeλ1 t V1 + Beλ2 t V2 , y(t) where V1 is an eigenvector for λ1 and V2 is an eigenvector for λ2 . x(t) 0 a. If λ1 6= 0 and λ2 = 6 0, = is the unique equilibrium solution of (S), the origin of the phase y(t) 0 portrait is : — a saddle point if λ1 < 0 < λ2 , — a stable node, also called a nodal sink, if λ1 < λ2 < 0, — an unstable node, also called a nodal source, if 0 < λ1 < λ2 . x(t) = AV1 y(t) is an unstable equilibrium solution for every constant A. The other trajectories are lines parallel to the vector V2 . b. If 0 = λ1 < λ2 , there are infinitely many unstable equilibrium solutions of (S), indeed x(t) = BV2 is a y(t) stable equilibrium solution for every constant B. The other trajectories are lines parallel to the vector V1 . c. If λ1 < λ2 = 0, there are infinitely many stable equilibrium solutions of (S), indeed 2. If A has two complex eigenvalues thenthe two are conjugated and if one writes eigenvalues λ1 = α +iβ u1 u2 u1 u2 then λ2 = α − iβ. Moreover if V1 = +i is an eigenvector for λ1 then V2 = −i is v1 v2 v1 v2 an eigenvector for λ2 . The general solution of (S) is given by : x(t) u u u u = Aeαt cos(βt) 1 − sin(βt) 2 + Beαt cos(βt) 2 + sin(βt) 1 . y(t) v1 v2 v2 v1 1 If β 6= 0, x(t) 0 = is the unique equilibrium solution of (S), the origin of the phase portrait is : y(t) 0 — a stable spiral point, also called a spiral sink, if α < 0 and β 6= 0, — an unstable spiral point, also called a spiral source, if α > 0 and β 6= 0, — a center if α = 0 and β 6= 0. In this case, the trajectories are ellipses. 3. If A has one repeated eigenvalue, denoted λ, one says that λ has algebraic multiplicity 2. a. If one can find two linearly independent eigenvectors V and W then one says that λ has geometric multiplicity 2, and the general solution is given by : x(t) = Aeλt V + Beλt W. y(t) If λ 6= 0, x(t) 0 is the unique equilibrium solution of (S), the origin of the phase portrait is : = y(t) 0 — a stable star point, also called a star sink or a proper stable node, if λ < 0, — an unstable star point, also called a star source or a proper unstable node, if λ > 0. b. If one cannot find two linearly independent eigenvectors, then one says that λ has geometric multiplicity 1, and the general solution is given by : x(t) = Aeλt V + B(teλt V + eλt W ), y(t) where V is an eigenvector for λ and W a generalized eigenvector satisfying (A − λI)W = V . x(t) 0 If λ 6= 0, = is the unique equilibrium solution of (S), the origin of the phase portrait is : y(t) 0 — a stable improper node, also called an improper sink or a degenerate stable node, if λ < 0 — an unstable improper node, also called an improper source or a degenerate unstable node, if λ > 0. In the following diagram the various situations are summarized (q = det(A), p = Tr(A), the red parabola represents the discriminant of the characteristic polynomial) 2