Analytical Solution of Mathieu Equation
Dmitri Yerchuck (a), Alla Dovlatova (b), Yauhen Yerchak(c), Felix Borovik (a)
(a) - Heat-Mass Transfer Institute of National Academy of Sciences of RB, Brovka Str., 15, Minsk, 220072, RB
(b) - M.V.Lomonosov Moscow State University, Moscow, 119899
(c) Belarusian State University, Nezavisimosti Avenue 4, Minsk, 220030, RB
(Dated: January 22, 2014)
The general solution of the homogeneous damped Mathieu equation in the analytical form, allowing its practical using in many applications, including superconductivity studies, without numerical
calculations has been found.
PACS numbers:
Keywords: Mathieu equation, superconductivity
arXiv:1401.5348v1 [quant-ph] 19 Jan 2014
I.
INTRODUCTION AND BACKGROUND
It is known, that a number of physical phenomena can
be described mathematically by Mathieu equation. For
instance, the equation of motion for the flux lattice in
the supeconductor theory is the following [1]
dy(t)
BJ0
d2 y(t)
+η
+ (K0 + k cos ωt)y(t) =
cos Ωt,
m
dt2
dt
c
(1)
where B is the magnetic induction in the sample, m is
the total mass (per unit length) of the flux lattice, η is
the viscosity coefficient, ω is the frequency of the modulating magnetic field, c is the light velocity, J0 and Ω are
the amplitude and the frequency of the microwave current, respectively, K0 and k are amplitudes of constant
component and an alternating component in the function
K(t) in the relationship
F (t) = −K(t)y(t)
(2)
between the force F (t) and small displacement y(t) of
the flux lattice from its equilibrium position, that is
K(t) = K0 + k cos ωt.
(3)
The equation (1) is the inhomogeneous damped Mathieu equation.
Mathieu equation is well knowm in the theory of differential equations, see for example [2], [3], [4], [5]. At
that,the so-called general Mathieu equation is the following equation
d2 y(t)
+ [h − 2θ cos(2t)]y(t) = 0,
dt2
(4)
where h and θ are real or complex constants. The
known solution of the general Mathieu equation (4) is
built in the form
y(t) = exp(µt)P (t),
(5)
where P (t) is a periodical function with the period,
equal to π, µ is so-called characteristic index, depending
on the values of h and θ. The function
y(t) = exp(−µt)P (−t),
(6)
represents itself the second solution. The solutions (5)
and (6) are linearly independent, they produce the fundamental system of the solutions, excluding the case, when
iµ ∈ Z, that is, to the set of whole numbers [4]. Further,
the solution (5) is written formally in the form of the
following infinite series
y(t) =
∞
X
cn exp (µ + 2ni)t,
(7)
n=−∞
in which for the set of coefficients {cn } the following
recurrent relations
−θcn−1 + [h + (µ + 2ni)2 ]cn − θcn+1 = 0, n ∈ Z
(8)
were obtained. It is the only algorithm for the numerical solution. In fact, the analytical solution of the general
Mathieu equation (4) was not found. The existence the
only algorithn for the numerical solution is inconveniently
for the practical usage of Mathieu equation, especially in
the cases of physical applications when analytical dependences are required to understand the physical processes.
For instance, the authors of the work [1] have preferred
instead of trying to solve the equation (1) numerically to
use the linearized equation
m
d2 δy
+ ηdδydt + K0 δy = −k cos ωty0 ,
dt2
(9)
in which y0 is the particular solution of (1) for the
case of k = 0. Let us remark, that an (in)homogeneous
damped Mathieu equation (4) can be reduced to an
(in)homogeneous general Mathieu equation by the following transformation of the function y(t): y(t) =
w(t) exp(− 12 ηt).
2
The authors of the work [1] have found, to the first
order in k, η, and ω, that the value of an electric field
E(t) induced is
E(t) =
B 2 J0 Ω
[1 − ǫ cos(ωt − φ)] sin(Ωt − α),
|K0 |c2
(10)
ηΩ
where ǫ = Kk0 , tan φ = 2ηω
K0 , and tan α = K0 . So,
the equation (10) indicates that the induced E(t) field
is amplitude modulated. In other words, the concrete
physical mechanism and its analytical description were
established, alhough for rather restricted ranges of the
parameters used. Really, the equation (10) is valid for
k ≪ |K0 |, ω ≪ Ω, and |K0 | ≫ mΩ2 . At the same time,
the authors of [1] remark, that when static magnetic field
H becomes too large [larger than 800 G], the flux structure becomes probably too complex for the simple model
proposed to remain valid.
It is clear on the given example, that an analytical solution of Mathieu equation remains to be very actual for
its applications in physical sciences and in engineering.
On the other hand, an analytical solution of Mathieu
equation has also the mathematical theoretical aspect.
It is determined by the fact that the solution of a number of differential equations is reduced to the solution of
Mathieu equation. They, for example, are
1)
dy(t)
d2 y(t)
+t
+ (2at2 + b)y(t) = 0,
2
dt
dt
(1 − t2 )
(11)
The transformation of variable t = cosz leads to the
Mathieu equation
d2 y(z)
+ (a + b + acos2z)y(z) = 0,
dz 2
(12)
2)
2t(t − 1)
d2 y(t)
dy(t)
+ (2t − 1)
+ (at + b)y(t) = 0,
dt2
dt
(13)
The transformation of variable t = cos2 z leads to the
Mathieu equation
2
(14)
d2 y(t)
+ (a sin λt + b)y(t) = 0,
dt2
(15)
3)
π
2
(16)
4)
The equations
d2 y(t)
+ (a sin2 t + b)y(t) = 0,
dt2
d2 y(t)
+ (a cos2 t + b)y(t) = 0
dt2
(17)
are transformed to Mathieu equations by using of
trigonometric formulae
2 sin2 t = 1 − cos 2t,
2 cos2 t = 1 + cos 2t
(18)
correspondingly.
The aim of the given work is to find the analytical
solution of the general Mathieu equation.
II.
RESULTS
Theorem. The general solution of the general Mathieu equation can be represented analytically to be the
superposition of Bessel functions of the first and the second kinds.
Proof.
It is evident, that the equation (1) can be represented
in the form
1 d2 y 1 dy
1
1
BJ0
+ η
m
+ K0 y + k exp(iωt)y −
cos Ωt+
2 dt2
2 dt
2
2
2c
1
1
1 d2 y 1 dy
+ K0 y + k exp(−iωt)y−
[ m 2 + η
2 dt
2 dt
2
2
BJ0
cos Ωt] = (L̂1 + L̂2 )y = 0,
2c
(19)
where L̂1 and L̂2 are differential operators
1 d
1
1
BJ0
1 d2
+ η + K0 + k exp(iωt) −
m
cos Ωt
2 dt2
2 dt 2
2
2c
1 d2
1 d
1
1
L̂2 = [ m 2 + η + K0 + k exp(−iωt)−
2 dt
2 dt 2
2
BJ0
cos Ωt].
2c
(20)
L̂1 =
d y(z)
− (a + 2b + acos2z)y(z) = 0,
dz 2
The transformation of variable λt = 2z +
the Mathieu equation
d2 y(z)
4b
4a
− ( 2 + 2 cos2z)y(z) = 0,
dz 2
λ
λ
leads to
It is also evident that the partial solution of the starting equation will correspond to the intersection of sets of
the solutions satysfying simultaneously to the equations
3
L̂1 y = 0,
L̂2 y = 0.
(21)
So, we have to solve the equations
1 d2 y 1 dy 1
+ η
m
+ K0 y +
2 dt2
2 dt
2
BJ0
cos Ωt,
2c
1 d2 y 1 dy 1
+ η
m
+ K0 y +
2 dt2
2 dt
2
BJ0
cos Ωt
2c
d2 y
η dy K0
+
+
y+
2
dt
m dt
m
η dy K0
d2 y
+
+
y+
dt2
m dt
m
1
k exp(iωt)y =
2
1
k exp(−iωt)y =
2
(22)
k
exp(iωt)y =
m
η
=a
can be solved strictly. Let us designate m
k
′
=
c,
iω
=
λ,
−iω
=
λ
.
m
Then, according to [6] the solutions are
k
exp(−iωt)y =
m
K0
m
iω
exp(
iωt
))+
2
= b,
(28)
q
K0
2
η
−iωt
m
[2]
y(t)[2] = exp(−
t)[C1 Jν ′ (
exp(
))+
2m
−iω
2
(29)
q
K0
2
−iωt
m
[2]
exp(
))],
C2 Yν ′ (
−iω
2
can be also studied. There are a number of variants to
simplify the solution of (24) using the symmetry of y(t).
d2 y
In the case, if y(t) is even, dy
dt is uneven, dt2 is even from
(24) we obtain by t′ → −t
for the second equation in (27). Here
Jν (
Yν (
2
q
2
K0
m
iω
q
K0
m
iω
2
d y
1
1
+ K0 y + k exp(iωt′ )y + k exp(−iωt′ )y =
dt′2
2
2
BJ0
cos Ωt′ .
c
(25)
Given case is equivalent to the task above formulated
by the relations (19) - (21). Further, if y(t) is even, dy
dt is
is even from (24) we obtain by t′ → −t
dy BJ0
−
cos Ωt = 0,
dt
2c
(26)
K0
(30)
are Bessel functions of the first kind and of the second
kind of index
ν=
m
q
iωt
−iωt
m
)), Jν ′ (
exp(
)),
2
−iω
2
q
2 Km0
iωt
−iωt
exp(
)), Yν ′ (
exp(
))
2
−iω
2
exp(
2
η
q
K0
m
for the first equation in (27) and
1 dy
1
1
1 d2 y
+ η
m
+ K0 y + k exp(iωt)y−
2 dt2
2 dt
2
2
BJ0
cos Ωt =
2c
(24)
1
1 d2 y 1 dy 1
+ K0 y + k exp(−iωt)y−
−[ m 2 + η
2 dt
2 dt
2
2
BJ0
cos Ωt]
2c
d2 y
dt2
2
(27)
(23)
The case
uneven,
k
exp(iωt)y = 0
m
k
exp(−iωt)y = 0
m
η
[1]
y(t)[1] = exp(−
t)[C1 Jν (
2m
q
K0
2
iωt
m
[1]
C2 Yν (
exp(
))]
iω
2
or in the equivalent form
d2 y
η dy K0
+
+
y+
2
dt
m dt
m
BJ0
cos Ωt,
mc
η dy K0
d2 y
+
+
y+
dt2
m dt
m
BJ0
cos Ωt,
mc
that is the simple differential equation of the first order
, the solution of which is evident.
The homogeneous equations, corresponding to inhomogeneous equations (1) that is the equations
q
η 2
k
) − 4m
(m
(31)
iω
for the equation (30) and of index
ν′ =
q
η 2
k
) − 4m
(m
(32)
−iω
for the equation (31).
Then the set of functions, satisfying to the relation
y(t)[1] ∩ y(t)[2]
(33)
4
will be the solution of the homogeneous equation, corresponding to the starting inhomogeneous equation (1).
Given conclusion is correct, if y(t)[1] ∩ y(t)[2] 6= ∅. In
[1]
[2]
particular, it is satisfied for the case C1 = 0, C1 = 0
and by
2
q
K0
m
iωt
exp(
)) =
iω
2
q
2 Km0
−iωt
[2]
C2 Y−ν (
exp(
)),
−iω
2
[1]
C2 Yν (
(34)
q
q
K0
K0
2
2
m
m
iωt
−iωt
exp(
))
cos
πν
−
J
(
J
(
−ν
ν
[1]
iω
2
−iω exp( 2 ))
=
C2
sin πν
q
q
K0
K0
2
2
m
m
−iωt
−J
(
exp(
))
cos
πν
+
J
(
exp( iωt
−ν
ν
[2]
−iω
2
iω
2 ))
C2
,
(35)
Hence we obtain the conditions, by which the relation
[1]
C2 exp(−
η
t)Yν (
2m
2
q
iω
exp(
iωt
))
2
(36)
[2]
C2 cos πν − C2
= 0,
sin πν
[1]
[1]
C2 cos πν − C2
=0
sin πν
(37)
cos2 πν − 1
= 0,
sin πν
[1]
C2 6= 0.
(38)
that is, by
Consequently, ν ∈ Z, that is ν has to belong to the set
of whole numbers.
Therefore, the relation (36) is the solution of homoge[1]
neous damped Mathieu equation by ∀C2 6= 0 ∈ C and
by ν ∈ Z. It means, that there are restrictions on possible values of parameters in the starting homogeneous
damped Mathieu equation. They are the following
ν=(
q
that is
ν=(
η 2
k
) − 4m
(m
iω
q
η 2
k
− (m
)
4m
ω
q
K0
m
for ∀ ν ∈ Z.
Since the replacement −iω → iω is equivalent to
2
q
K0
m
−iωt
J (
exp(
)) → J−ν (
−iω
2
) ∈ Z,
)∈Z
(39)
(40)
2
q
ν′
[1]
K0
m
will be the solution of the homogeneous damped Mathieu equation. They are
[1]
q
2 Km0
−iωt
iωt
exp(
)) = J−ν (
exp(
)) =
Jν ′ (
−iω
2
iω
2
q
0
2 K
iωt
m
(−1)ν Jν (
exp(
))
iω
2
(41)
2
that is, in the case
sin πν
It is evident that by η = 0 we obtain the solution of
the homogeneous undamped Mathieu equation, that is
the solution of the general Mathieu equation.
To obtain the fundamental system of the solutions of
the homogeneous damped Mathieu equation, we have to
find the second linearly independent solution. It can be
easily done, if to take into account, that
K0
m
iω
exp(
iωt
)), (42)
2
[2]
we obtain, that by C2 = 0, C2 = 0 the expression (33)
that is
[1]
C1
η
t)Jν (
exp(−
2m
2
q
K0
m
iω
exp(
iωt
))
2
(43)
will be the second linearly independent solution of the
[1]
homogeneous damped Mathieu equation ∀C1 6= 0 ∈ C
[2]
[1]
and by ∀ν ∈ Z at C1 = C1 for even ν ∈ Z and
[1]
[2]
[1]
at C1 = −C1 for uneven ν ∈ Z, that is at C1 =
[2]
C1 sgn(−1)ν . Its linear independence from the first solution follows from linear independence of Bessel func[1]
tions on the first and the second kinds. At fixed C1
[1]
and C2 the relations (36) and (43) represent themselves
the fundamental system of the solution of homogeneous
damped Mathieu equation. So, the general solution of
homogeneous damped Mathieu equation is
exp(−
[1]
η
[1]
t)[C1 Jν (
2m
q
C2 Yν (
2
K0
m
iω
exp(
2
q
K0
m
iω
exp(
iωt
))+
2
(44)
iωt
))]
2
The theorem is proved.
Thus, we have found the general solution of the homogeneous damped Mathieu equation in the essentially
more simple form, allowing its practical using in many
applications without numerical calculations, since Bessel
functions are well known.
5
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887-893
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[3] Meixner J and Schäfke F W, Mathieusche Funktionen und
Sphäroidfunktionen, Springer, Berlin,1954
[4] Bateman H,Erdelyi A, Higher Transcendental Functions,
V.3, New York, Toronto, London, MC Graw-Hill Book
Company, Inc., 1955, M., Nauka, 1967, 300 pp
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