SME
Differential Equations
Systems of ODEs
Use this for “Classify the equilibrium of x′ = x + 2y, y′ = 2x + y.”
Diagram
Trace T = 2, determinant D = −3, T² − 4D = 16
λ = 3 along ⟨1, 1⟩ and λ = −1 along ⟨1, −1⟩ (the dashed lines)
The origin is a saddle
This is an example. Type your own values to start a new problem.
x′
=
a
x
+
b
y
y′
=
c
x
+
d
y
T
Trace of A
worked out (not typed)
D
Determinant of A
worked out (not typed)
Δ
Discriminant T² − 4D
worked out (not typed)
type
Type of equilibrium
worked out (not typed)
α
Midpoint of the eigenvalues
worked out (not typed)
β
Offset from the midpoint
worked out (not typed)
λ₁
Larger eigenvalue
worked out (not typed)
λ₂
Smaller eigenvalue
worked out (not typed)
Formulas
T
=
a
+
d
2 = 1 + 1
D
=
a
×
d
−
b
×
c
−3 = 1 × 1 − 2 × 2
Δ
=
T
2
−
4
×
D
16
=
2
2
−
4
×
(−3)
type
=
saddle
if
D
<
0;
else
node
if
Δ
≥
0,
spiral
if
Δ
<
0
(center
if
T
=
0);
stable
if
T
<
0
−3 < 0, so saddle
α
=
T
2
1
=
2
2
β
=
(
1
2
)
×
√
|
Δ
|
2
=
(
1
2
)
×
√
|16|
λ₁
=
T
+ √
Δ
2
3
=
2 +
√
16
2
λ₂
=
T
− √
Δ
2
−1
=
2 −
√
16
2
Assumptions
•
The system is x′ = Ax with A = [[a, b], [c, d]]. Its eigenvalues solve λ² − Tλ + D = 0, with T = a + d and D = ad − bc.
•
D < 0 is a saddle. With D > 0, T² − 4D ≥ 0 gives a node and T² − 4D < 0 a spiral, or a center when T = 0.
•
T < 0 is stable (arrows in) and T > 0 unstable. Straight-line solutions run along the eigenvectors.
Step-by-step
WE KNOW
a = 1, b = 2, c = 2, d = 1
FIND
trace of A (T), determinant of A (D), discriminant T² − 4D (Δ), type of equilibrium (type), midpoint of the eigenvalues (α), offset from the midpoint (β), larger eigenvalue (λ₁), smaller eigenvalue (λ₂)
Step 1 · Find trace of A (T)
Use T = a + d
The
trace
adds
the
diagonal
of
A,
and
it
equals
λ₁
+
λ₂
,
the
sum
of
the
eigenvalues.
T
=
1
+
1
T
=
2
Step 2 · Find determinant of A (D)
Use D = a × d − b × c
A
=
[[1,
2],
[2,
1]].
Its
determinant
equals
λ₁
λ₂
,
the
product
of
the
eigenvalues:
negative
when
they
have
opposite
signs.
D
=
1
×
1
−
2
×
2
D
=
1
−
4
D
=
−3
Step 3 · Find discriminant T² − 4D (Δ)
Use Δ = T² − 4 × D
det(A
−
λI)
=
0
is
λ
2
−
Tλ
+
D
=
0.
Its
discriminant
says
whether
the
eigenvalues
are
real
(
Δ
≥
0)
or
complex
(
Δ
<
0).
Δ
=
2
2
−
4
×
(−3)
Δ
=
4
−
4
×
(−3)
Δ
=
4
−
(−12)
Δ
=
16
Step 4 · Find type of equilibrium (type)
Use type = saddle if D < 0; else node if Δ ≥ 0, spiral if Δ < 0 (center if T = 0); stable if T < 0
D
=
λ₁
λ₂
<
0:
one
eigenvalue
is
positive
and
one
negative.
Trajectories
come
in
along
one
eigenvector
and
leave
along
the
other.
−3 < 0: saddle
type = saddle
Step 5 · Find midpoint of the eigenvalues (α)
Use α = T ÷ 2
λ
2
−
Tλ
+
D
=
0
gives
λ
=
T
± √
Δ
2
.
Its
first
part,
T
2,
is
halfway
between
the
eigenvalues,
or
their
real
part
when
they
are
complex.
α
=
2
2
α
=
1
Step 6 · Find offset from the midpoint (β)
Use β = (1 ÷ 2) × √(|Δ|)
With
Δ
≥
0
the
eigenvalues
are
real:
α
+
β
and
α
−
β
.
β
=
(
1
2
)
×
√
Δ
β
=
(
1
2
)
×
√
16
β
=
1
2
×
4
β
=
0.5
×
4
β
=
2
Step 7 · Find larger eigenvalue (λ₁)
Use λ₁ = (T + √Δ) ÷ 2
The
larger
eigenvalue:
a
solution
starting
on
its
eigenvector
line
stays
on
it,
scaled
by
e
λ₁t
.
λ₁
=
2 +
√
16
2
λ₁
=
2 + 4
2
=
3
(A − 3I)v = 0: (1 − 3)x + 2y = 0, so v₁ = ⟨1, 1⟩
λ₁
=
3
Step 8 · Find smaller eigenvalue (λ₂)
Use λ₂ = (T − √Δ) ÷ 2
The smaller eigenvalue: its eigenvector line is the second straight-line direction.
λ₂
=
2 −
√
16
2
λ₂
=
2 − 4
2
=
−1
(A − (−1)I)v = 0: (1 − (−1))x + 2y = 0, so v₂ = ⟨1, −1⟩
λ₂
=
−1
Check
Put the numbers back in. Both sides should match.
2 = 1 + 1 ✓
−3 = 1 × 1 − 2 × 2 ✓
16
=
2
2
−
4
×
(−3)
✓
−3 < 0, so saddle ✓
1
=
2
2
✓
2
=
(
1
2
)
×
√
|16|
✓
3
=
2 +
√
16
2
✓
−1
=
2 −
√
16
2
✓
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