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