1. No category

Integrating Factors - (Brief) Review
c 2015 Prof. Philip Pennance1 -Version: March 30, 2015
1. Let U ⊆ IR2 be open. By a vector field
on U we mean a function V : U → IR2 .
(a) α1 (t) is invertible.
(b) The trajectory (image) of α is the
graph of the function φ given by
2. Let V ((x, y) = (v1 (x, y), v2 (x, y)) be a
vector field. The expression:
dx
= v1 (x, y)
dt
dy
= v2 (x, y)
dt
φ(x) = α2 ◦ α1−1 (x)
(c) φ is a solution of the nonautonomous ODE determined by
V viz:
(1)
v2 (x, y)
dy
=
dx
v1 (x, y
is called the autonomous system of
ODE’s determined by V .
3. A solution of (1) is a curve α : I → IR2
defined on an interval I such that
α10 (t) = v1 (α1 (t), α2 (t))
α20 (t)
I
α2
(2)
φ
α10 (t) = v1 (α1 (t), α2 (t))
Since v1 is continuous and nonvanishing α10 (t) must be everywhere
positive or everywhere negative on I.
In either case, we deduce that α1 is a
bijection from I onto its image. Let
φ(x) = α2 ◦ α1−1 (x) Then using (2) and
(4)
(3)
or in matrix form
x˙
0 −1
x
=
y˙
1 0
y
φ0 (x) = α20 (α1−1 (x))(α1 −1 )0 (x)
v2 (x, φ(x))
(6)
=
v1 (x, φ(x))
By inspection α(t) = (cos t, sin t) is a
solution of (3).
Thus φ is a solution of (5).
Let Gφ be the graph of φ. We claim
Im α = Gφ . Let (x, y) ∈ Im α. Then
there exists τ ∈ I such that (x, y) =
(α1 (τ ), α2 (τ )). Therefore τ = α1−1 (x)
and so y = α2 ◦ α1 −1 = φ(x). Hence
(x, y) ⊆ Gφ . The proof of the converse
inclusion is similar.
5. Claim: Let V = (v1 , v2 ) be a vector field
with v1 and v2 continuous, v1 6= 0. If
t∈I
is a solution of the autonomous system
(1) then
1
/ Im α1
Proof:
Since α is a solution of (1),
4. Example:
Let V (x, y) = (−y, x).
The autonomous system determined by V is:
α(t) = (α1 (t), α2 (t)),
|
α1
(5)
IR
= v2 (α1 (t), α2 (t))
dx
= −y
dt
dy
= x
dt
(4)
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6. Corollary
Near (x0 , y0 ), this system of two autonomous equations in one dimension
has solution α(t) = (α1 (t), α2 (t) given
by:
Z α1 (t)
Z α2 (t)
ξdξ =
−ηdη
(10)
x0
y0
Let q : D ⊆ IR2 → IR be continuous and
V ((x, y) = (v1 (x, y), v2 (x, y)) a vector
field on D. If qv1 6= 0 on D and
α(t) = (α1 (t), α2 (t))
is a solution of
= t − t0
dx
= q(x, y)v1 (x, y)
dt
dy
= q(x, y)v2 (x, y)
dt
which yields:
(7)
α1−1 (x)
then φ = α2 ◦α1−1 is a solution of the non
autonomous ODE determined by V .
α2 (t) = ±
φ(x) = α2 ◦ α1−1 (x)
p
= − x0 2 + y 0 2 − x 2
(11)
The maximal solution of the IVP is a
semicircle in the lower half plane, which
is in agreement with the fact that trajectories of the autonomous system (3)
have the form α(t) = (r cos t, r sin t).
7. Example:
Let V (x, y) = (−y, x). The nonautonomous ODE determined by V
takes the separable form:
8. Let D ⊆ IR2 and
V (x, y) = (v1 (x, y), v2 (x, y))
(8)
be a vector field on D. A function
Φ : D → IR such that ∇Φ = V is called
a potential function for V . A level curve
of V is called an equipotential. A vector
field is called exact if it admits a potential function.
Suppose we have an the initial condition, y(x0 ) = y0 with y0 < 0. Take as
integrating factor
q(x, y) =
p
y0 2 − 2(t − t0 )
Since y0 < 0 the negative sign applies.
By (4) above a solution of the non autonomous ODE (8) is
Proof: Notice that the presence of the
integrating factor q changes the speed
of the trajectories of the autonomous
system (1). By claim (6b), the solution φ is a solution of the non autonomous equation associated with vector field qV = (qv1 , qv2 ), which (since
q cancels) coincides with the non autonomous ODE determined by V .
x
dy
=
dx
−y
x2 − x0 2
= t0 +
2
−1
xy
9. If V ((x, y) = (v1 (x, y), v2 (x, y)) is a vector field on an open set U , the vector
field determined by:
The autonomous system determined by
the vector field qV is:
V ⊥ (x, y) = (−v2 (x, y), v1 (x, y))
dx
1
=
dt
x
dy
1
=−
dt
y
will be called the vector field perpendicular to V .
(9)
2
10. Claim. Let V is vector field on D ⊆ IR2
with potential function Φ. Let I be an
interval and α : I → D be a solution of
the autonomous system determined by
V ⊥:
dx
= −v2 (x, y)
dt
(12)
dy
= v1 (x, y)
dt
then Φ(α(t)) is constant on I.
12. Remark: If V is continuously differentiable and D2 V 6= 0 the implicit
function theorem guarantees the (local) existence of a function φ such that
V (x, φ(x)) = c where c is constant.
13. Example. Consider the vector field V =
(x, y). Notice V is exact with potential
function
Φ(x, y) =
Proof
Φ(α(t))
= h∇Φ(α(t), α0 (t)i
dt
= h(v1 , v2 ), (−v2 , v1 )i
x2 y 2
+
2
2
The non autonomous ODE determined
by V ⊥ = (−y, x) is:
(13)
dy
x
=
dx
−y
=0
11. Corollary
(b) Im α is an equipotential curve.
Level curves of Φ (equipotentials) are
circles:
x2 + y 2 = C
(c) If v2 6= 0 the solutions of the non
autonomous equation
and maximal solutions (e.g., (11)) are
semicircles.
0
(a) α is perpendicular to V .
v1 (x, y)
dy
=
dx
−v2 (x, y
(14)
14. The reason for the term potential derives from physics. If F is a force, and
F = −∇Φ then the energy
are equipotentials of Φ.
(d) If v2 6= 0, then any equipotential φ
is a solution of (14)
1
m|x(t)0 |2 + Φ(x(t))2
2
Proof. (a) and (b) are clear from the
proof. (c) is an instance of (5). Finally,
let φ is an equipotential, viz,
is constant along solutions of the Newton/Euler equation F = mx00
d
Φ(x, φ(x) = 0.
dx
Then, by the chain rule
15. Suppose that a vector field V = (v1 , v2 )
is not exact but it happens that there
exists a function q such that Vˆ =
(qv1 , qv2 ) is exact. Since the solutions of
the non-autonomous ODE corresponding to a vector field are unchanged by
the introduction of an integrating factor, (12) can be solved (implicitly) by
finding a potential function Φ such that
∇Φ = Vˆ and finding its level curves.
h∇Φ(x, φ(x)), (1, φ0 (x)i = 0
Since ∇Φ = V , it follows that
dφ
v1 (x, φ)
=
dx
−v2 (x, φ(x)
and so φ is a solution of (14).
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