ON THE STATIONARY SOLUTIONS
OF VAN DER POLtS EQUATION WITH A FORCING TERM
by
WARREN SIMMS LOUD
S.B., Massachusetts Institute of Technology
(1942)
SUBMITTED IN PARTIAL FULFILLMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
* **.. *
(1946)
Signature
of
*-
Author..
.*
Department of Mathematics, Sept.9, 1946
Certified by . .
4
·.
*
0·
·
·
*
Thesis Supervisor
*
*
*
*
*
0
*
*
*
*
*
* 0 *
**
* * * * *
*
*
*
*0
0
Chairman, Department Committee on Graduate Students
II
ate
IF
Acknowledgement
The author wishes to thank Professor Norman Levinson for
suggesting the topic of this thesis, and for his many helpful
suggestions.
§
9
XJ
Xs
Abstract
In this paper van der Pol'sequation
uAith
a forcing term is
studied with the purpose of determining the nature of its stationary solutions.
The equation is
_a
dxd
_
d
dx
Em
- 1~x;
lift
+ x=
e(t).
The equation with right member zero arose in van der Pol's
study of relaxation oscillations.
The solutions of the equation
tend to a single periodic solution having a period approximately
2rr if
f
is small, but proportional to /4
if
/A
is large.
In this paper the effect of a periodic forcing term is investigated.
It develops that if
is large enough, all solutions but
one tend to a quasi-periodic behavior with a "period" proportional
roughly to /
.
There is a single unstable solution having the
same period as the forcing term.
The results obtained are collected in two theorems, in which
e(t) is a general periodic function of amplitude less than one, and
a sinusoid respectively.
I.
Let max
t
let E
e(t)t
= E, and let e(t) be of period L.
Furthermore
5
1, />
,
___
,
> 3L, i
+ /U~x''
>200.
Then the equation
dx
d X + /4A(x2 - l)-t
+ x = e(t)
dt
dt
has solutions with the following properties;
1.
Moreover
2.
There is a unique unstable solution x (t) of period L.
x o (t)
l
<1 for all t.
The solutions
of the related
system
I I.
dt
=
f
-
=x
-
dv
d=
with the
-x + e(t),
) eventuallY
exception of x = x (t), y = x (t) + (x t
The time
into a region of zero area encircling the oriin.
fal
r
/
3 + 21
4to encircle the origin lies between
3.
and6
and5
(1
+
22) .
In the original equation the solutions assume a cyclic
behavior with the time between successive ascending nodes lying
between P + 21 and 15 + 3 / + 22) 5
Let
TI.
be different from zero, and let
/4 ->
/, 10(2,)
>20
10 /(n>s
)
2w
E
> 21.
Then the equation
dax+
dt
( 2(
- 1)d + x = E os t
has solutions with the following properties.
1.
2iT
(0
*
There is a unique unstable solution x
o (t) with period
2.
< 1 for all t.
xo(t)l
Moreover
The solutions of the related system
dx
dtd= y
dl$=
with
( 33 - x) = Y - F(x)
-x + EcosOt
the exception of x = xo (t), y = xl(t) + Fx
(t)] eventually
fall into a region of zero area encircling the origin.
encircle the origin lies between . 8
3.
0
k
+ 236
and 6.2 /
The time to
+ 117
In the original equation the solutions assumae a cyclic
behavior with the time between successive ascending nodes lying
between .0/
+
238 and6.2r
+ 117
Table of Contents
I
Introduction
II
Van der Pol's Equation with a Forcing Term which is
Periodic and'Less than One in Magnitude.
Page
1
4
III
Van der Pol's Equation with a Sinusoidal Forcing Term. 17
IV
Conclusion
26
Bibliography
27
1.
Introduct. on
In the study of relaxation oscillations, Van der Pol was led
to the study of the equation
d x + A(X - 1)X- + x
dt
o.
is a constant.
where f
In the paper referred to he studies the solutions of the above
equation by means of the transformation
dx
dt =v
dv
dt
_
_t
-x- f(x' -1)V.
a single limit cycle if
The solutions are shown to approach
different
from zero.
/
is
When translated back into the x - t plane,
the solutions are periodic, with a period proportional to j
if
is large.
The problem was attacked
Lienard
studied
more generally
the more general equation
da
dx +
dt2
d
dt
f(x)At
where f(x) is an even function.
y =' "'
where
F(x)
by Lienard in 1928.
dt+
+ m x
=O
He made the transformation
F(x)
f(x)
f
dx is an odd function.
The equation becomes
the system
dt =y - F(x)
&dt
-X
W
A)
lVan der Pol, Relaxation Oscillations. Phil.Mag.,Ser. VII,Vol.2 (1926)
p.978
2 Lienard
- Etude des Oscillations entretenues Rev.Gen.de
Vol.23 (1928) p. 901 and 946.
lElec.
2.
or when t is ellminated
dx
This
is
equation
Is
y - F (x)
shown to have a unique closed solution, which
approached by all other solutions if F(x) is positive and in-
creasing for all x greater than a certain x.
For Van der Polls equation
(x') r (Xa 1)
For largevaluesof /
Fig. 1.
the limit
3
F(x)= ( x).
cycle has the appearance shown in
The direction of increasing t is shown by the arrow.
large values of
(3 - ln4)=
the period is asymptotic to
For
1.61/'.
This will be the period of x as a function of t in the x - t plane.
More general conditions under which a periodic solution
exists
have been found by Levlnson and Smith.3
Vander Pol's equation with a sinusoidal forcing term is
considered for small values of
The original
by Friedrichs
4
and Stoker.
equation can be interpreted
as an equation of
damped harmonic motion with the character
ing on the amplitude.
of the damping depend-
If x is greater than 1 in absolute
value,
the damping is positive,
so that amplitudes can not increase in-
definitely.
Howeverif x is less than 1 in absolute value, the
damping becomes negative so that
atively
damped.
Thus qualitatively
very small oscillations
are neg-
it can be seen that the oscill-
ations will reach a stable finite amplitude.
If
A
is large,
the
3 Levinson
and Smith, A General Equation for Relaxation Oscillations,
Duke M4ath.Journal, Vol. 9 (1942) p.382
4 Friedrichs and
1943, Chap.V.
Stokes - Intro. to Non-Linear
Mechanics,
Brown Univ.
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final mode of oscillation is assumed very quickly.
As can be
seen from figure 1, x is greater than 1 in absolute value for most
of the period, so that the negative damping is not effective very
much of the time.
It should be noted that there is one singular
solution, x = O, y = 0 which is unstable because of the negative
damping.
In this paper Van der Pol's equation is studied with a periodic
forcing term e(t)
dta
d+xd +x=
e(t).
The equation is studied for large values of the parameter
.
The results obtained can be described qualitatively as follows.
Corresponding
of
to the
singular solution there is a unique solution
the same period as e(t) which is unstable if e(t) is not too large.
All solutions, when the x - y plane is used, tend to a region of
zero area which roughly follows the limit cycle of the non-forced
case.
The solutions thus do not tend to strictly periodic behavior.
There is no limit cycle.
It is well to keep a limit cycle in mind
visualizing the behavior of the solutions.
Two functions e(t) are considered.
The first is any periodic
function which is less than one in absolute value.
a sinusoid Ecosat
where w
is different from zero.
The second is
-a
II
Van der Poi'sEquationwith a Forcing Term which is Periodic
and Less than One in Magnitude.
The equation is
(x
dX +
l)d + x - e(t).
We make the transformation of Lienard to the variables x,y, and t.
_x -x + e3(t).
dt
The result is stated in the following Theorem:
Let max e(t) = E, and let the period of e(t) be L. Further-
t
more let
EC 1,
Thaenthe equation d
3L
5E)
1 $A)
(x3 -
+
1A
> 200.
)d + x - e(t) has solutions
dt
with the following properties:
1.
L.
There is a unique, unstable solution xo(t) which has period
Moreover
2.
x o0 (t)l
¥ 1
for all t.
The solutions of the related system
aX =yY -
x) = y - F(x)
-
dt-x +e(t)
with the exception of x = xo(t), y = x'(t) + F(xo(t)) eventually
fall into a region of zero area encircling the origin.
to encircle the origin lies between
3.
+ 81
The time
and 16 + (16
In terms of the original equation the solutions assume a
cyclic behavior with the time between successive ascending nodes
lying between -,+ 8LP and 56 + (16 + 22)
Tr
.
.
5,
The condition E
If e(t)
I is necessary for the fol Lowing reason.
E, the singular solution x = E, y = F(E)
is stable if
E >1 because the damping is positive.
The existence of a periodic solution of period L is guaranteed
by a result of Levinson.5
To prove the statements made about the behavior of the nonperiodic solutions, we proceed as follows.
Step 1. Weestablish- the fact that all solutions passing
through points- outside the rectangle
Ixl< 1
must cross
the line
< 3Yr
y = 0 for
xl
The various cases are illustrated
1.
by figure 2.
A. The solution starts at PA, in the first quadrant, to the right
of x = 1, and belol the curve y = F(x).
y<F(x), both dX and dt are negative.
At PA since x>E and
The solution can not
cross y = F(x) from below to above if x >+E, since dy is always
negative.
Since
is negative, the solution can not increase
in x.
Therefore it must cut the line y = 0.
B. The solution starts at PB'in the first quadrant, to the right
of x = 1, but above
but y is decreasing.
to the right of PB'
J.
the curve y = F(x).
The solution
At PB' x is increasing,
must cut the curve y = F(x)
To do this it must either
pass into case A,
or cut the line y =
0 before it intersects y = F(x).
The solution
at P
between x = 5 Levinson
starts
and x = 1.
which lies above y =
and lies
Let x C and yC be the coordinates of
- On the Existence of Periodic Solutions
for Second Order
Differential Equations with a Forcing Term, Jour. Math. and Phys.
Vol. XXII (1943) p.41.
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d=
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I//
J-6
-
---II --- . ......
,I
1
y/
..........
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6#
P.,
x(l
For
>-i - E.
Y >YY - (
thus
Therefore for
+ E)t
> YC- (1 + E)t-
F(x)> y
-
(1 + E)t- F(-E)
(l+ E)t + -(E3 - 3E)
>2
But by hypothesis
the solution
(E- 1) 2 >5
(l+ E)t
_-(E-l)(2)
E>
EŽO
'., dtX> 10
3 - 2t.
Therefore
o
Therefore for t
dx is positive for t =
55
x> 3 t - t + x >¥3 t - t - 1
5
At t
= x>
50
T
25 -
-
16
9
Y > yc - (1 + E)t )>3/ Therefore at t = 5
y >2
1 0
t.
since
/>5*
Therefore the solution passes into case B, and y
0 when it
crosses x = 1.
D.
The solution
x=
-
since
starts
and x = -E.
d
at PD which lies
At PD d
and dt
Z-X + ,!(t) remains finite,
above y = 2k,
and between
are both positive, and
the solution must pass
into case C.
E.
The solution starts at PE which is in the second quadrant to
the left of x = -1, and above the curve y = F(x).
Both x and
y are increasing at PE, and for the reasons in case A the solution can not cut y F(x) from above to below. Also since
ddx remains bounded for fixed x, the solution can not become
infinite in y.
Therefore it must cross x
and pass into case D.
-1 above y
F.
The solution starts at PF which is
in
the second quadrant to
At PF x is
the left of x = -1, but below y = F(x).
decreasing
Therefore the solution must cut the curve
but y is increasing.
y = F(x) to the left of PF and pass into case E.
If the solution starts in the third or fourth quadrant it is
handled in an exactly analogous manner. From the above we see that
the conclusion
of step 1 is Justified.
If the solution starts within
the rectangle Ixl( <1
yl
2j,
and will be dealt with later.
the reasoning is more difficult
Havingestablished the fact that all solutions starting outside
the rectangle Ixl
1
IYIL
t
must cross the line y = 0 outside
the fact that all sol-
jxj = 1, we proceed to Step2. Weestablish
utions crossing the line y = 0 for x
A. Cross the line x = -1 for 3
B. Cross the line x
-
-1 must
+ 21
<
y<
for 3 Y'2f~
C. Cross the line x +l for 23 / -
10
4/,.
1.5
Y
4
+ 5
D. Cross the curve y = F(x) for x< 2.5
x 2.5.
J /7(:
E. Cross the line y = for
The steps are illustrated in figure 3.
A.
The solution
must
cross
x = -1 above y =
as we saw
in cases E and F of step 1.
The maximum for y(-l) is obtained
<)
dY = - + e}
if
4-2
y>O
dx
- F
-F(x)
as follows:
ix
X
3
Therefore for the solution
.zf;
Yfif
6 axx aoth
aJx
1
3
/ X- 3
6
r
1
3
Y
....
i
~~~i
....
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-
3
,3-F('x)
At
... A - 4 e
tI
.
.
Y
.
.. - .
I..
For x = -2, y
cot
2
_ 4.57
To pass from -2 to -1, we use the relationship
xdx + ydy = F(x)dy + e(t)dx
or
xo + ya
=2
[F(x)dy + e(t)dx
P1
P1
where the integral is a line integral along the solution.
the values
of y at -2 and -1
be yl and
Y2 respectively.
rP2
Then
Y
Y2 < y
Nowif2
y1
3
2
=
+ 5
2
J
2
(F(x)dy + e(t)dx) 2 [M(y
2 - Sg Yl(
+
6
Let
+ 5 +
kY2
-Yl1 ) + 1]
2
3/Y
2
+ 6
+
2
If the point on y = 0 lies between -2 and -1, the same line integral
holds except that Y
= 0, and
e(t)dx must be less than 2 instead
of 1 because x can both increase and decrease.
Thus
4-- 33<< 2/IAY2
/A
a<4
Y2 <
+
222]
Y2 + 7
!/,
Y2(3
r+ 421
In either case Y2
B.
5c1+
2
The solution crosses x = -E for y2
negative for -1 - x -
, and y
at x
since dt
3/Y since
-1.
is non-
Theupper limit is obtained from consideration of the solutions of the equation
~d ,-x
=
y
+ E
F(
.
Solutions of our equation
cross solutions of this equation, and have when crossing a smaller
slope since e(t)
E, and y - F(x) is positive in the region
I4
considered.
But since t does not aooear in
solutions have the further
another.
y
x
ta
property that they do not cross one
Now on the solution o-
+ 21 d 1
=
=
since x-
+ E
starting at
x
=
-1
y - F(x)>/F~ Thereforeif
the change in x is 1 - E, the change in y is no more than
2
< 2/
Now no solution of our equation can cross the curve from below to
above.
4
y =
4 +
Therefore every solution of our equation starting below
21
+
must cross x
4
+
-E below
=2
2A
<
/
+
1,5 since
> 5.
C. We saw in Step 1, Case C that the solution
x = 1 for y
2
will cross
1 .
We saw also in case C, step 1 that the time from x = -E to
x = 1 was less than 3
the most y can increase
Y<43 j + 1.5 + 103 43
t
dt
is
at most 2 for
is 10
10 .
Therefore
-E
we cross
x
1 so that
x = 1 for
+ 5.
D, Since y must decrease for x>l,
the solution can not inter-
sect y = F(x) for y
/ + 5.
4
Therefore the maximumvalue of x is less than
+ 5 since A> 5.
NowF(2.5) = 2.7/p which exceeds
2.5.
E. Since x decreases for y<F(x), the solution must cross
y = 0 for
x<2.5.
To show that the solution crosses y = 0 for x>J3,
we pro-
ceed as follows:
dt- = -x + e(t)
therefore y)
-3. 5
- 3.5t where t is measured from the tme of
1-13
'10.
orossin x = 1.
F(x)0 so dy
Nowif xj,
dx 210 i0
_d 2
1 - 3.5t>
20
t3
up to t = 21
This is positive
all the time, x> 1 +
1•~l)
2>
t
3
At that time, if x<
- 1,75ta -t
:
=
1
3 21°t 4 4
+
1
The conclusion is that x does reach
*3inceince Y>
=k'
-- 3.5t since 4> 10.
3t
fort
3.5t, yy )0 for
-
t
63
-
=
63
/J3 before
so tt
that weros
we cross
,
y = 0 for x >.
From the result of step 2 we conclude that once a solution
nrorsses v = 0 niutild.
Il
= 1 it
area and cross y = 0 on the
other
miit
side
tr-vel
thrtlgh
j/
between
Then it enters a smaller region, and crosses y
slightly narrower band on the original side.
a l mittd
and 2.5.
0=in perhaps a
We shall eventually
see that with enough circuits of the origin, the allowable area
does indeed shrink.
To continue we proceed to
Step 3.
r3
21
+81"
The time taken in the circuit of the origin lies between
a
and
5 + ( 3+Do 22)^
The time is divided as follows.
A. From x =
least
B.
2
Over a half-cycle
to x =-E the time is at most4/-
+
and at
5
From x = -E to x = 1 the
time is
at most s and at least
2l+E+
C. Outsidex = 1 the time is at most (8/ + 7)
and at least
21
The time for the cycle is obtained by adding and multiplying by two.
'1ii
A.
The mnnimnumtimes
inr A and B are both
+ 5 for (xll<
, F (-)) -
that y(
dt
-V+
<
5 for Ix l' .
+__+EEfor the two parts
> 2
for Ixl <
and- is more than 2
fact
so that
I-E+
to
nd
separately.
to x = -E is more difficult to
The maximum time from x = -
If
from the
Therefore the time from x = -l
2+
x =3 +l is more than 2
fix.
computed
+(_l)
+
(1 - E
dx is greaterthan 1 - E for
x) -1, so that the time is at most L from -1 tc -E.
If y(Al)
+
<1+-
2, -
since the
= -x
E passing through x = - +FX)
solution
solutionof dv
dy
has
( - E); y(-E)< 2 t+
y = 2/k
2
2+
This curve crosses x = -E below -. /t+
ahas
E .
+
(1
(l-E)
- E
+
+E=
3/"2 +2.
Thusif - = x 4 -E
Now from -1
to x,
..
Fromx to -E,
y(Xo)< t
dt)-E0'dt
-
x
t
3 -F(x
ax = -X
a y<2.
>
)
2 time <
the
2..E
-X0
E
-
2
B.
2
.,
o
+
(1 - E) I
2
(
E E =
-E
...
+ Xo )a
E
2
4
+
4,,/:-+
3
--E
/A(3 7-V
-1-
(4 5/+1since
/>45.
3
4
In Step 1, part C we found the time from x = -E to x = 1 was
at most
C.
+
4
The time
45
outside x =
a1X,)
-x -E
X+
-1 - E
-- - E- =
2
'
2
j+
2?'(1 + xo )
o'
t <
2
time( 2
- E,
The total time is less than
fx
andifX
+ 2.
has a minimum value determined from the
fact that the change in y is at least !
an
at
t
most
3.5 in absolute value.
'. The time is more than 2
3 5
2
The maximum time is found from the fact that the change in y
at most 44/A +5
i
4
isatinm niue +§+
os
i - E in magnitude.
I
8
+ 21 </f+
7 and
+~+dt t
is at least
Therefore the time is at most (3/
+ 7).
1 - E< '3 /A+ 7.)/
' 5 '
Some conclusions can be drawn now.
The total time is at least
3 +
and since /03L, this is greater than L. This proves
#
21
that the periodic solution must have x(t)< 1, y(t) < /
for
all t, since we have seen that any solution for which Ixl
-
1 must
fall into the cycle about the origin and can not have a period as
small as L.
We can now proceed
Step 4
to
x = x(t)
Y = Yo(t) be the periodic solution
= xl (t)
y = Y l (t) be any other solution distinct from it
dto
dt
d()
To show this, let
The periodic solution is unique.
dto
=
Y - F(xl)
dt
=
-xi + e(t)
= (Y1 - Yo) - [F(x
1 ) - F(xo)]
- xo)(Xl +XxO+ o -3)
=(y1- o)- (Xl
dt(Y-
Yo) =
X - xo).
x+tx_ 3
--(- )2
Therefore (x - x, )d -C
x( l - xo ) + (Y1
-
"
Yo)t (Y1 - Y)
oi
"0"
=
"
"0
o -
13,
+ x
Now if lxI< 1, (xo 1, (x + xx,
di-
+ (y1 - yo ) 2
(X - Xoj
Now if (x
1 -
were to approach a limit, this limit
since this is different
Therefore
would have to be 0.
and can not decrease,
the start,
(Xl - Xo
is positive or zero all the time.
o )a
+ (Yl -
o)
]
)
yo,)
+ (yl
so x(t)Yl(t)
so
- 3) is negatie
there can be no limit.
must increase to beyond
from 0 at
Therefore
a + 4 and
-96-
must go outside the rectangle.
Now finally we take
Step 5.
The solutions come to a region of zero area.
we consider the solutions which start in an area
A A as it varies
To do this
A. Wefollow
with time, and see what happens to it as time
increases.
dX Y
=
AA(O)
o
4A
d(J ) d
4A(t) =
dy
AA
What we show is that the Jacobian
=
J=
approaches zero as t-
.
To do this we note that
d
c!i "
ddtt
x
0
ad (x)
=
dt dy
dt
dYo
{(io
do
+
ax
ax
o
dt
dt dyO
14.
No~,_i
dX
d
No\dx
-F (x)ax
+
8x
dtd
I
3
x
o
8t
=
t~
= o xt
'
.z° =
+ (dt( x) ax
o
a__ax
xx
_
;
fi
dJ
x
o-Ftx)
Il
X
0x
d~s
dt
-Ft (x )
o
y
+]L~dxAdr
0
-Q
7X
+r
t0
ao
= -F' (x) J
- 1)
ThusJ = J
e
I t (xa_i)dt
.
(x2- )dt = + .
Thus to show that J approaches 0, we must show that
To do this we consider one cycle.
Ixl( 1 is no more than 25+ 8
cycle the time for
Now for
IxI<
x
We have seen that in one
-
>-,
in A of step 3.
so
2
f(xz-i3dt>
Ixl<1
Let Y2 be the value of
Now we consider the time when x1.4.
t when x = 1.4 on the solution,
x
1
In the rectangle
Z is negative
Y = Y2 since
1
for x> 1, y
and let yl be the value of y when
1.4,
x
F(x)),
Y2
Yl
Y1 since
since
Y=(Y2 <yl
a(l is maximumat x = 1.4,
dyI increases with x but decreases with y.
for the rectangle
I-•l Y2 -F(I.[4)
1.4
+
4
<
2.4
Y2 +
485r
Therefore
15.
n>+
Since x 2
x
=
k4,
-
> 0 X
gin33
L
h Y2 >-242
FromPo
thi.s
>
.96
- .2436
> -. 79
4
,
'V
6
_
.45
>O*
965
)
242
+
..
242
ince
t
Ow that
we
,
.96
242/ )
-
.96
.96
-
> 5.
Now the solution crosses x = 1.4 above y = -.790, and can not
re-cross it until y<-.485
y >-.485
least
since dt is positive at x = 1.4 if
Therefore the y-distance traveled for x)1.4
.
485P
-
.790.
|
Since
at least .485 357.790 - 138
is at
+ E (3.5, the time taken is
<+2.5
- .226.
Thus in a cycle, the time
forx}x> 1.4 is greater than .276/ - 4:2
f
(x
- l)dt ) (.27 6
452)(.96)
-
.265
.434
-
Ixl >1.4
Now for our cycle the integral gives at least
.265p - .434
- .400
This last is positive if
Therefore
j
.265/
- 3.57
>)187.
- .834.
But by hypothesis, /
(x2 _ 1)dt for a cycle is at least 2.5 so that
around a cycle is at least 500.
>200.
/J(xa
- 1)dt
Thus as t becomes infinite the
Jacobian approaches zero rapidly.
There is no way of determining the precise location of the
region of zero area.
An area b A will become and remain very small,
but as t increases it will traverse an extremely complicated path
which merely is contained in the region found in Step 2.
To find the solution of the original equation, x is plotted
as a function of t.
Points where the x - y locus crosses the y-axis
for x increasing are ascending nodes.
There is precisely one per
16.
cycle.
Ixl remains greater than 1 for the major part of the
as we have seen the time
to the total period.
of traverse
of
xl < 1 is
short
ycle,
compared
Vanider Pol s quat.on with a inusoidal Fc-inf
Term.
The equation is
dax + l/(x 2 - 1)
+ x
E os w t.
dtd
'1e make the same transformation as in II to obtain
3
dx
t
= y -/(
x) = y - F(x)
dt = -x + E coswt.
at
The results
are
stated in the following
Theorem:
Let w be different
from 0, and let
> 10(2 .f(2',)
Then the equation dx +
dt
> 20E E
(x - 1)d
dt
-with the following properties.
1.
> 2
21.
t>
+ x = E cosw t has solutions
There is a unique unstable solution x o ( t)
which has
period .* Moreover
xo (t) l< 1 forall t.
2.
The solutions of the related system
Y - r(
-
) = y-
F(x)
dy= -x + Eosit
dt
with the exception of x = xo (t),
y = x(t) + F(x
o
(t)),
fall into a region of zero area encircling the origin.
to encircle the origin lies between .80/4 + 236
3.
eventually
The time
and 6./
+
17
In terms of the original equation, the solutions assume
a cyclic behavior with the time between successive ascending nodes
lying between
.80so + 2t3
and 6./ ,2/
+ 117
.6
Ay
is~~~~~
Tle existence of a periodic solution of period 2 is
guaranteed by Levinson's result (5) as in
II.
First, however
The method of proof is similar to that of II.
we derive some consequences of the sinusoidal nature of e(t).
Sincedt
-x + Ecoswt, it followsthat
2dv
t
t1
2E
for tl
t = t
Y(tl +
2
Y(tl) - y(t2 )
t
-
(t
2
-
tl)
Max Ix +(1)
t2
(t
Also
1
dt
From this, if x)O0 for t
and if x O
+ Eint
dtt
-
(t2) - (tl ) (t2- t ) MaxIx+ E
1
) _ y(t) = _
1t
'W~
(2)
(3)
x(t) dt
Now as before we proceed with
Step 1..
We establish the fact that all solutions passing through
points outside the rectangle
Ix
must cross the
2
3l
line y = 0 for Ixl
2 1.
The various cases are illustrated by figure 4.
A.
The solution starts at PA in the first quadrant to the right
The solution
of x = 1 and below the curve y = F(x).
x = 1 from right to left for y =-3
.
can not
cross
Therefore by (1) the de-
crease in y will be at least equal to the time taken
2E-, so that
the solution must cross y = 0.
B.
The solution starts at PB in the first quadrant, to the right
of x = 1, but above the curve y = F(x).
crease an amount at least eual
Again by (1) y must de2E
and so the solution
to the time
-··---·-·------··--
Y....
.....'
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i
i
.i...
I
.... ,....:.;...i.
-.. i- ....
:_.
. : : _
j·
I··--T`:.
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.;.-··-.
--';-l.-.--jii
-- L .....
·
i.._
._.._,..
~~~~~~~~~~~
' ':
-- ·-- -··-·I---·-·-;--:-· -- r---i
; ·
i-
-i'·:'·
3/:
....-. ..........
i
:.-......
'-.....-
-
.-".-
.....--
'.
.......;
i 1
I·· · - --
·- ·-··-·-·
-··
,-
·- · ·-··---· · ·i -·-.···:
r
1 :r
i ..-. ,
:
:·'
---- -- :
:":
· 1 ;-·
;rsF--··:
.:·
-·-·-------·
i
..- ;.1........
i
-.
;;
.... i
i..
1
: .-..-.
C ·!
I
]·ji·
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i.
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i.
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I
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I"-
i
I
r
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i
i
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:i
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PA~~~~
: '.:
:t
I
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:~~~~~~~~~
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i
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·
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:'; : . "
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i
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1
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i
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j
i··
i
1
· T·--·-·- i.'
j---.:-F.:-:-..
r
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i
i
i
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341 "
i.
i-_...·.
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i
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1.-..
?...-
4
:
/-Fx
-- ·- -· ----- ---- ·C--·-
i5c
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--.--..
i!
i
i.r
-1·
,.
··
o''":':.
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- i
......,
ii
i~~~~~~~~~~~-:i..'~
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'i
i
::i
i~~~~~~~~~~
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.-'"'
.....-- ,- .....
ri
'S3
i'
I
.~~[ .....·.........
V-T-
r
'~'EtJw
dt~~~~~~~~~~~~~~~~~~
19.
It crosses y = 0 before crossing
must cross the line y = 0.
y = F(x) or it crosses y = F(x) first in which case we pass
into
case A.
which lies between x = 0 and x = 1
The solution starts at P
0.
and above y = 2rA.
In one period y can descend no more than
than
2v which is less
2+
.
Thus y is at least 15
1, so that x advances at least
for 0 _ x
O0
for this period since F(x)
dt is at least
first period.
for the
0/5A
*1
=-
2
281
24 )1.65. Thus in this one period x advances to at least 1.5 and
so passes into case B.
D.
The solution starts at PD which lies between x = -1 and x = 0
and above y =
2E
3
3.
So long as xO,
the most y can descend is
2
1
<2
d2
23
(2), so that y
Elo
by
for x<O.
But this means
>°0
so that we must pass into case 0.
for x,0,
E.
23
The solution starts at PE which lies between x = -1 and x
2
and also between y = 2A
23
and y = 23A
-7~~~~O
Here the possible descent
of
may permit the solution to intersect y = F(x) and so be
carried
back.
Since
17
y> tA3
always for x <0,
the solution if
carried back would have to intersect y = F(x) for x-.6
for x
-1.4.
and again
This means that while we are being carried back, y is
ascending at least .6x
23 x and we
O
21
- every period, so eventually y will exceed
ass into case D.
The other alternative is
that the
solution is not carried back and reaches x = O without falling into
case D.
17
Then y 230 P when it crosses x = 0 and by reasoning as in
in one period.
case C y can descend to no less than
dx
-3.-ufor one period, so the increase in x exceeds
dt
30f.;a>
*
2
==
Then
) 1.5, so that even here x reaches +1.5.
20.
F, The solution starts at PF in the second quadrant to the left
off x = -1 and above y = F(x).
will
cut x = -1 above y
2
x will decrease for a time.
or it
will
the solution
ncreasing
Since x is
intersect
y = F(x),
and
However, since y increases at least
2- every period, this will not continue indefinitely and the solution will ultimately cross x = -1 above y = 2/
and pass into
case D or E.
G.
The solution
starts at PG in the
second quadrant
to the left
of x = -1
and below the curve y = F(x). Since x is decreasing, and
2E
y can not decrease more than , while x is negative, the solution
will cross y = F(x) and fall into case F.
The solution is not prevented from temporarily entering the
third quadrant in cases F and G.
As before if the solution starts
in the third or fourth quadrants, it is handled in an analogous
manner.
From case A to G the conclusion of step 1 is Justified.
If the solution starts within the rectangle Ixl< l,
<22y
the reasoning is more difficult and will be dealt with later.
Having established the fact that all
the rectangle
xi(
xl <l,
lykl 2r
starting
solutions
outside
must cross the line y = 0 outside
i, we proceed with
Step 2.
We establish the fact that all solutions crossing the line
y = 0 for x
-1
must:
<y
<
A.
B.
Cross the line x = -1 for 2
Cross the line x = 0 for lo/
C.
Cross the line x = 1 for
D.
Cross the curve y = F(x) for x( 2.3
E.
Cross the line y = 0 for 1.51
1.2/
y< 1.3/
-1/k yz 1.
x
2.3
+ ,
+
+
Y
! :....
_..(. :.- ..... -)~I
~~~~~~~~~
I...
...
,~
.. :i. 'i·
;~~~~~~~~~~~~~~
.
.
~:.
'__
!..
'"'
i-:""i:...
...
~"......
.
:x_
.........
' .,l f-
,
:
....
:
I~ .
I
·
i.i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
:':.:':i-!!i
.....!-......·-r-····-·
...
t i,
:·,
-. £
. .
.
..
~.."~
i
';:i
..
.
.
.i
.S
.
'
.
~
... /
.
:·-.·
;·~~~~~~~~~~~~~
~~~~~I'
''! :
~.
.:....... .~-
",',:
'
i
.
-
1--...
·
-- · · · ·----
:I
........
·
P':~~~~~~~~~~~~~~~~~
I~~~~~~~~~~~
. ·
,
,
-
i~~~'...;
i:.:,
11
1~.--.- j .·
-
~~~~~~~~~~~~~~~.
I ..-,.-.
'
-'-- i"
.L
i
~~~
i..
.!.: ~
...
.~.r
....-....
~..~-
,'/
~~~~~~~~~~~----L.
i~~~~
',t
"~...'..I...........[...---
·
F
IP
x
!".-
~~~~~~~~~~~~~~~~~~~~~~~:
I
i~......
· · · r~~~~~~~~~
k
~ -·1
· 1·
'i
.'~
:i
:
i..~~~~~~~~~~~~~~~~~~~~~~
... .
.I
ix
I~~~~~~~~~~~~~~~~~~~~~~~~~~
I~~~~~
i~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
.'
L
i
:
!
..
.
-
.
7/I
'
~
.
'
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~..........
i
~~
i
f
.-1.
dr
,~-F-
-
Fcx)
X'l IEc,~' ,t
i
'7..........
21.
The steps are illustrated
A. The solution
First,
in figure 5.
must cross x = -1 for 2/
for x< -2,
ds
= -
y +- Ecost<
(x
y
.2
E -3 x
3
L (1 +Ex
x
/
J"
3
20
c 029
08/
Now
so y2) .1
Therefore every solution crossing y =
x = -2 below y = .2/,.
for -2
x
-1.
y will exceed
by (2).
Thus
-24
xZT-1 will
be less
--<1f(-1~
++
.
+.l
f
+
.2
outside x
-2 crosses
cross y = 0
Now if y becomes greater than 1.1/a in so long
E<
+
All others under consideration
1.0/
is
x
x
/
X
2
A.
+
than
as x
O0, since the most
will exceed
7.
x z-1,
xL
y can decrease
and the time in
Now y can increase no more than
so
+
The onlyother
posso that
thaty(-l)c 1.2/^
1 ~ +.1/
ibility is that y never got as large as 1.1.
The solution
for y
B.
crosses
x = -1 for y >
/A since dt is negative
+-
The solution crosses x = 0 for 30
If in -1
x
14
_ 0, y ever exceeds 1.1/4, y
l.3/
+
.
1.0/i always so
least L3 and the time is no more than .
y can increase
3
?,-than 3 + 2E in this time, so we cross x = 0 below
1.2 +
+
17
no more
2E<1.
As we saw in E step 1, the solution must cross x
=
is at
0
O above
22.
C.
The solution crosses x =
for L-~A
2E !-1~
1
Since x 0, the most y can increase is
y G1.3f+
9
+ .. t
-
1.4p + A9
Y 1.4
+ .
80
This incidentally
is the
maximum y attains for x >O.
In E of step 1 it was shown that y(!) indeed y(l.5)
was
11
greater than 30r.
D.
The solution crosses y = F(x)
for
x
2.3.
The very maximum of y has been found to be 1.4/4
y> 10.
x
E.
But F(2.3) = 1.74f,
+
,
1.5f
so we must cross y = F(x)
since
for
2.3.
The solution crosses the line y = 0 for 1.5 4x C2.3.
Wehave seenthaty(1.5)
-/,
so the solution
must cross y = 0
to the right of x = 1.5.
Since x reaches its maximum when we cross y = F(x), the maximum value of x is 2.3, so we must cross y = 0 to the left of
x = 2.3.
From the result
of step 2 we conclude that once a solution
crosses y = 0 outside x = 1 it must travel in a limited area
and cross y = 0 on the other side between 1.5 and 2.3.
It then
enters a smaller region and comes back to perhaps a slightly
narrower interval on y = 0 on the original side.
We shall see as
in II that there is no way to make the interval shrink to zero, as
it probably does not.
We now proceed to
Step 3. The time taken in the circuit
.80 + 2
and 62/ + 117
tk
/A maIP.+
of the origin lies between
Over a half-cycle:
The time is divided as follows.
From x = -1 to x - 0 the time is at least 7
A
+ t558
and at most
7/
and at most
B.
From x
C.
The time outside x = 1 is at least .40/k and at most 2.9/A.
O to x
0
1 the time is at least ~
27
The times for the cycle are obtained by adding and multiplying by
two.
A.
The minimum time in A is found from the fact that since
y(l.3A +- (<1.4k for-1 = xS , dt <,4 for -1=x- 0,
1
so the time is at least 1.4
5
=
-
If the solutions are turned
The maximum time is more difficult.
-. 6.
back, they must be turned back for x
, x
Now after a time
will have increased to more than -.6 or else y will have increased
to
A
.26-)
2
/, so thatthereis no chanceof
' 0-A + /.A
is up, x-1.4,
Now when the time
turning back.
the time over to -.6 does not exceed
y > .57
so dx
dt
.566/ - .528/A= .038 /,
B.
6
40 . .38/&
Now for
x)-.6,
so the time from -. 6 to 0
5
+ 55.8
A
The minimum time in B is found from the fact that since y
for -t -= x =
.45
1
2. 214
t
Since y
.36t
C.
- 40
41.5
+ 2 /k
1
2.2,
l.5A
so the time is greater than
d
31/ = . 3 6 7/ * dt ).36/A
and the time is lessthan
f ·
Since x
time t.
,At
so
time is less than
. Thusthe total
is less than 06
-+
+5
.8
d> .02/
2.3 always, y will go down no more than 2.3t + .1^
Now the least y has to decrease is .36/A + 2
Therefore 2.3t + .
= 1.03/
gives the minimum time t
in
= 1.03/.
2.3
=
01Ap
24.
Sincey<1.5/A for x>,
y)-1.3,/for x 1, the most y has to
travel is 2.8/4, and since the change in y is at least t .
-1
k
t maximum will be 2.9/.
We are now able to draw some conclusions.
circuit of the origin is at least
+
80r
36 ,>
-
:- The time of
/
-.
proves that the periodic solution X 0 (t) must have IX(t)l
all t, by the same reasoning
This
1 for
as in II.
The same argument as in step 4 of II proves that x (t)is a
unique periodic solution.
Finally as in II we showthat /
(x - l)dt = + whichwill
prove the zero area result.
x!( 1, x
For
-i1
and t<
-1
f(x2 - l)dt> -(.4/A
Thus
.4
+ 117
+ 17)
ih a single cycle.
Ixf(1
At x = 1.5, y2 .36/4 and x can not return to less than 1.5 until
y< F($.5) = -. 37p
, Thus y moves a distance at least .73/A
The time taken is at least
-73-
.54/A for a cycle.
Thussincex2 - 1) 1.25 for
(x -
.27
outside.
on one side and
x/)>1.5,
l)dt>(l.25)(.54k)- .675/
Ixt > 1.5
thereforejx
This function is positive if r>
so
Jx
17 = .275-
- l)dt > .675f- .4/r
21.
But by hypothesis,
- l)dt > .88 - 4.68 = 2.20, so
J(x2
Thus as we take more and more cycles the Jacobian
Again
as in
II there
- 1)dt
/f21,
)46.2.
approaches
zero.
is no way of determining the precise loc-
ation of the region of zero area.
It can travel all over the
25.
region found in Step 2.
Finally to find the solutions forthe original equation we
plot x as a function of t.
xl remains greater than for the
major part of the cycle, the time of transit of the region
lx
1( being only a small part of the period.
26.
The rs ults
obtained
are by no means the best
They can be sharpened to a ezree by more attention
detail.
possible.
to cose
However, it is to be doubted that much improvement is
possble.
a-re a.
The region of zeroAis called
Invariant Domain.0
In
by Levinson
a Maximum Finite
he paper referred to it is shrownthat such
domains, although of zero area, can be extremely complicated. As
we have seen, the domains discussed in this paper are not prevented from having a complicated form.
The conditions
under which
the complications are excluded are not known.
Possible improvementslie in the direction of relaxing the
requirements on the size of / , narrowing the differences between
minimumand maximumperiods,
solutions
cross
lines
like
and narrowing the intervals
x = -1,
at which
x = 1, y = 0.
Levinson - Transformation Theory of Non-T ne ar Differential
Equations of the Second Orier. Annals of IMathematics. Vol. 45,
No. 4, Oct. 1944, p.723.
27.
Bibliography
Bieberbach, L. - Theorie der Differentialgelechungen, Sprineer,
Berlin, 1923
Brown
Friedrichs and Stoker - Introduction to Non-Linear iltechanics,
Univ., 1943, Chap. V.
Levinson and Smith - A General Equation for Relaxation Oscillations,
Duke Mathematical Journal, Vol. 9, 1942,
p. 38 2.
Levinson - On the Existence of Periodic Solutions for Second Order
Differential Equations with a Forcing Term,
Journal of Math.
and Physics, Vol. XXII (1943),
p.41.
Levinson - Transformation Theory of Non-Linear Differential Equations of the Second Order.
Annals of Math-
ematics, Vol. 45, No.4, Oct. 1944,p. 723.
van der Pol - Relaxation Oscillations. Philosophical Magazine,
Ser. VII, Vol.2 (1926), p.978.
ioE
oa-hi
ca1 Note
arren S. Loid was born in Boston,
the son of Roger P. Loud and Esther
He received his
schools
elementary and
Mass. on September
Ni. Loud, of ~Weymouth,Mass.
secondary education in the public
of %{eymouth, Mass. He was graduated from
School in 1938.
3, 1921,
Jeymouth High
His advanced training, both undergraduate and
graduate, has been at the Massachusetts Institute of Technology.
In 1942 he was awarded the degree of Bachelor of Science in
iathematics.
Since that time he has pursued a program of graduate
study in pure and applied mathematics.
In addition to graduate
study he has been engaged in teaching as an Instructor of Mathematics,
and
in research in the department of Electrical Engineering
and the Servomechanisms Laboratory.
At the Servomechanisms Labor-
atory he was the author of several unpublished reports.