264 Chapter 12 Applications in Circuit Theory
Proof: Any solution starting on v
+
immediately enters the region A since
x
(0) > 0. In A we have y
< 0, so this solution must decrease in the y direction.
Since the solution cannot tend to the source, it follows that this solution must
eventually meet g
+
.Ong
+
we have x
= 0 and y
< 0. Consequently, the
solution crosses g
+
, then enters the region B. Once inside B, the solution
heads southwest. Note that the solution cannot reenter A since the vector field
points straight downward on g
+
. There are thus two possibilities: Either the
solution crosses v
−
, or else the solution tends to −∞ in the y direction and
never crosses v
−
.
We claim that the latter cannot happen. Suppose that it does. Let (x
0
, y
0
)
be a point on this solution in region B and consider φ
t
(x
0
, y
0
) = (x(t ), y(t )).
Since x(t) is never 0, it follows that this solution curve lies for all time in the
strip S given by 0 <x≤ x
0
, y ≤ y
0
, and we have y(t ) →−∞as t → t
0
for
some t
0
. We first observe that, in fact, t
0
=∞. To see this, note that
y(t) − y
0
=
t
0
y
(s) ds =
t
0
−x(s) ds.
But 0 <x(s) ≤ x
0
, so we may only have y(t) →−∞if t →∞.
Now consider x(t) for 0 ≤ t<∞. We have x
= y − x
3
+ x. Since the
quantity −x
3
+ x is bounded in the strip S and y(t) →−∞as t →∞,it
follows that
x(t ) − x
0
=
t
0
x
(s) ds →−∞
as t →∞as well. But this contradicts our assumption that x(t ) > 0.
Hence this solution must cross v
−
. Now the vector field is skew-symmetric
about the origin. That is, if G(x, y) is the van der Pol vector field, then
G(−x, −y) =−G(x, y). Exploiting this symmetry, it follows that solutions
must then pass through regions C and D in similar fashion.
As a consequence of this result, we may define a Poincaré map P on the
half-line v
+
. Given (0, y
0
) ∈ v
+
, we define P(y
0
)tobethey coordinate of the
first return of φ
t
(0, y
0
)tov
+
with t>0. See Figure 12.4. As in Section 10.3,
P is a one to one C
∞
function. The Poincaré map is also onto. To see this,
simply follow solutions starting on v
+
backward in time until they reintersect
v
+
, as they must by the proposition. Let P
n
= P ◦ P
n−1
denote the n-fold
composition of P with itself.
Our goal now is to prove the following theorem:
Theorem. The Poincaré map has a unique fixed point in v
+
. Furthermore,
the sequence P
n
(y
0
) tends to this fixed point as n →∞for any nonzero
y
0
∈ v
+
.