CUBO A Mathematical Journal
Vol.15, No¯ 01, (33–47). March 2013

On Fokker-Planck and linearized Korteweg-de Vries type
equations with complex spatial variables 1

Ciprian G. Gal

Florida International University,

Department of Mathematics, DM 435B,

Miami, Florida 33199

cgal@fiu.edu

Sorin G. Gal

Department of Mathematics and Computer

Science, University of Oradea,

Str. Universitatii No. 1 410087, Romania

galso@uoradea.ro

ABSTRACT

In the present work, we construct solutions to a Fokker-Planck type equation with real

time variable and complex spatial variable, and prove some properties. The equations

are obtained from the complexification of the spatial variable by two different methods.

Firstly, one complexifies the spatial variable in the corresponding convolution integral

in the solution, by replacing the usual sum of variables (translation) by an exponen-

tial product (rotation). Secondly, one complexifies the spatial variable directly in the

corresponding evolution equation and then one searches for analytic solutions. These

methods are also applied to a linear evolution equation related to the Korteweg-de Vries

equation.

RESUMEN

En este trabajo construimos soluciones de una ecuación tipo Fokker-Planck con variable

de tiempo real y variable espacial compleja. Las ecuaciones se obtienen de la comple-

jización de la variable espacial por dos métodos diferentes. Primero, se complejiza la

variable espacial en la integral de convolución respectiva en la solución reemplazando la

suma usual de las variables (traslaciones) por un producto de exponenciales (rotación).

Luego, se complejiza la variable espacial directamente en la respectiva la ecuación de

evolución y se busca por las soluciones anaĺıticas. Estos métodos también se aplican a

la ecuación de evolución lineal relacionada a la ecuación Korteweg-de Vries.

Keywords and Phrases: Fokker-Planck equation, Korteweg-de Vries equation, complex convo-

lution integrals, complex spatial variables.

2010 AMS Mathematics Subject Classification: 47D03, 47D06, 47D60.

1Dedicated to Professor Gaston N’Guerekata on the occasion of his 60th birthday.



34 Ciprian G. Gal and Sorin G. Gal CUBO
15, 1 (2013)

1 Introduction

Let us consider the following initial value problem:

{
∂u
∂t

(t, x) = α(t)∂
2
u

∂x2
(t, x) + β(t)xu(t, x), (t, x) ∈ R+ × R,

u(0, x) = f(x), x ∈ R,
(1)

where α ∈ C ([0, +∞), R+) and β ∈ C ([0, +∞), R) . Using the exponential operator method, it is
shown in [6] that

u(t, x) =
ea(t)+d(t)[c(t)d(t)+b(t)+x]

2
√

πc(t)

∫∞

−∞

e−[u+b(t)+2c(t)d(t)]
2
/(4c(t))f(x − u)du, (2)

is a solution of (1) provided that the integral in (2) converges, with a(t), b(t), c(t), d(t) depending

on α(t), β(t), and given in [6, (54)] as follows:

c(t) =

∫t

0

α(u)du, d(t) =

∫t

0

β(u)du, (3)

b(t) = −2

∫t

0

[
β(u)

∫u

0

α(s)ds

]
du,

a(t) = 2

∫t

0

β(u)

{∫u

0

[
β(s)

∫s

0

α(v)dv

]
ds

}
du.

Note that by the assumptions on α, β, the functions a, b, c and d are differentiable for every t > 0,

and that c(t) > 0, for all t > 0. For β(t) ≡ 0 and α(t) ≡ C (C =constant), we recapture the initial
value problem for the classical heat equation. For β (t) 6= 0, α (t) 6= 0, the main equation in (1) is
known as a Fokker-Planck type equation.

In the second part of this article, we’ll devote our attention to the ”linearized” Korteweg-de

Vries equation with real time variable and complex spatial variable. Indeed, let us consider the

well-known Korteweg-de Vries equation

∂u

∂t
+ αu

∂u

∂x
=

∂3u

∂x3
, (4)

where α ∈ R (see, e.g., Widder [7]), and the related linear problem
{

∂u
∂t

(t, x) = ∂
3
u

∂x3
(t, x), (t, x) ∈ R+ × R,

limtց0 u(t, x) = f(x), x ∈ R.
(5)

For problem (5), the following is known to hold.

Theorem 1.1. ([7, Theorem 4]) Let f : R → R be continuous and of bounded variation, such
that it satisfies the conditions:

(i) The integral

F(s) =

∫∞

−∞

e−sx f (x)dx



CUBO
15, 1 (2013)

On Fokker-Planck and linearized Korteweg-de Vries ... 35

converges absolutely for s ∈ C with Re(s) = c;

(ii) Let ∫∞

−∞

|f (c + iτ)|dτ < ∞

Setting

u(t, x) :=

∫∞

−∞

K(x − y, t)f(y)dy =

∫∞

−∞

K(u, t)f(x − u)du, (6)

where

K(u, t) :=
1

π

∫∞

0

cos(uτ − tτ3)dτ =
1

(3t)1/3
Ai

(
−u

(3t)1/3

)
, (7)

the function u (t) satisfies (5).

Above,

Ai(u) :=
1

π

∫∞

0

cos(uτ + τ3/3)dτ

is also called the Airy function. Moreover, it is well-known (see, e.g., Widder [7, (5.1)]) that

∂K

∂t
(x, t) =

∂3K

∂x3
(x, t), t > 0, x ∈ R. (8)

It is natural to ask what happens if in the above equations we complexify the spatial variable

and keep the time variable real? We shall proceed as follows. The complexification of the spatial

variable in the above mentioned equations is made by two different methods which produce different

equations: first, one complexifies the spatial variable in the corresponding formula for the solutions

in (2) and (5), respectively, by replacing in the integral the usual sum of variables (translation)

by an exponential product(rotation) and looking for solutions in a disk DR of radius R > 1. This

method yields solutions that satisfy differential equations similar to (2) and (5). Secondly, one

directly complexifies the spatial variable in the corresponding evolution equations, and then one

searches for analytic and non-analytic solutions for the resulting equation. The topic was already

developed in detail for complex heat and Laplace equations in [1, 2], for complex wave and telegraph

equations in [3] and for complex Schrödinger type equations in [4].

2 Generalized heat type equations with complex spatial vari-

able

Let R ≥ 1 and let us now consider the open disk DR = {z ∈ C; |z| < R} and A(DR) = {f : DR → C;
f is analytic on DR, continuous on DR}, endowed with the uniform norm ‖f‖R = sup{|f(z)|; z ∈ DR}.
Is well-known that (A(DR), ‖ · ‖R) is a Banach space. If f ∈ A(DR), then we have f(z) =

∞∑
k=0

akz
k,

for all z ∈ DR. Finally, ω1(f; δ)DR denotes the modulus of continuity,

ω1 (f ; δ)DR = sup{|f (u) − f (v)|; |u − v | ≤ δ, u, v ∈ DR}.



36 Ciprian G. Gal and Sorin G. Gal CUBO
15, 1 (2013)

Concerning the system (1), we will first complexify the solution in (2) as follows. For f ∈ A(DR)
and t > 0, let us replace x and the translation x−u in (2) by z and the rotation ze−iu, respectively,

and consider the complex integral

Gt(f)(z) =
ea(t)+d(t)[c(t)d(t)+b(t)+z]

2
√

πc(t)

∫∞

−∞

e−[u+b(t)+2c(t)d(t)]
2
/(4c(t))f(ze−iu)du, z ∈ DR. (9)

The first goal of this section is to prove some properties of the complex integral (9).

Theorem 2.1. Let R > 1 and f ∈ A(DR).

(i) For all t > 0, Gt(f) ∈ A(DR), namely Gt(f) is continuous on DR, is analytic in DR, and
the following holds:

Gt(f)(z) = e
Φ(t,z)

∞∑

k=0

akdk(t)z
k, z ∈ DR, (10)

where

Φ(t, z) = a(t) + d(t)[c(t)d(t) + b(t) + z]

and, for all k ≥ 0,
dk(t) := e

−k
2
c(t)+ikg(t), g(t) := b(t) + 2c(t)d(t).

(ii) Setting

Wt(f)(z) := e
−Φ(t,z)Gt(f)(z) =

1

2
√

πc(t)

∫∞

−∞

f(ze−iu)e−[u+g(t)]
2
/(4c(t))du,

the following estimate holds:

|Wt(f)(z) − f(z)| ≤ (R + 1)
[
1 +

2√
π
+

|g(t)|

2
√

c(t)

]
ω1(f;

√
c(t))

DR
,

for all z ∈ DR, t > 0.

(iii) The operator Wt is contractive, that is, ‖Wt(f )‖DR ≤ ‖f ‖DR , for all t > 0, f ∈ A(DR).

(iv) Let

U(t, ϕ) := ea(t)d(t)[c(t)d(t)+b(t)+ϕ]
∞∑

k=0

akdk(t)z
k,

for every z 6= 0, z = re−iϕ such that r ∈ (0, R) , ϕ ∈ (0, 2π]. Then, U(t, ϕ) satisfies

∂U

∂t
(t, ϕ) = α(t)

∂2U

∂ϕ2
(t, ϕ) + β(t)ϕU(t, ϕ), (11)

for (t, z) ∈ R+ × DR\ {0} , z = reiϕ, r ∈ (0, R), and

U(0, ϕ) = f(reiϕ), ϕ ∈ (0, 2π], 0 < r ≤ R, f ∈ A(DR). (12)



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On Fokker-Planck and linearized Korteweg-de Vries ... 37

Proof. (i) For fixed z ∈ DR, we have

f(ze−iu) =

∞∑

k=0

ake
−ikuzk. (13)

Since |ake
−iku| = |ak|, for all u ∈ R, and since

∞∑
k=0

akz
k is absolutely convergent, it follows that

∞∑
k=0

ake
−ikuzk is uniformly convergent with respect to u ∈ R. Thus, on account of (13), we can

integrate in (9) term by term. This yields

Gt(f)(z) = e
Φ(z)

∞∑

k=0

ak

[
1

2
√

πc(t)

∫+∞

−∞

e−iku · e−[u+g(t)]
2
/(4c(t))du

]
zk

== eΦ(z)
∞∑

k=0

ak

[
1

2
√

πc(t)

∫+∞

−∞

(cos[k(v − g(t)] − i sin[k(v − g(t)])e−v
2
/(4c(t))dv

]
zk

:= eΦ(z)[I1 − iI2].

Since sin(kv)e−v
2

/(4c(t)) is odd as function of v, we have

I1 =

∞∑

k=0

ak

[
cos(kg(t))

1

2
√

πc(t)

∫+∞

−∞

cos(kv)e−v
2
/(4c(t))dv

]
zk

=

∞∑

k=0

ak[cos(kg(t)) · e−k
2
c(t)]zk

and

I2 = −

∞∑

k=0

ak[sin(kg(t))e
−k

2
c(t)]zk.

Hence, these calculations give the formula (10) and prove the analyticity of Gt(f)(z), as function

of z ∈ DR. To prove the continuity in DR, it suffices to prove the continuity in z, of the function

Ht(f)(z) :=
1

2
√

πc(t)

∫∞

−∞

e−[u+g(t)]
2
/(4c(t))f(ze−iu)du.

To this end, let z0, zn ∈ DR be such that limn→∞ zn = z0. We get

|Ht(f)(zn) − Ht(f)(z0)| ≤
1

2
√

πc(t)

∫+∞

−∞

|f(zne
−iu) − f(z0e

−iu)|e−[u+g(t)]
2
/(4c(t)) du (14)

≤
1

2
√

πc(t)

∫+∞

−∞

ω1(f; |zne
−iu − z0e

−iu|)
DR

e−[u+g(t)]
2
/(4c(t)) du

=
1

2
√

πc(t)

∫+∞

−∞

ω1(f; |zn − z0|)DRe
−[u+g(t)]

2
/(4c(t)) du

= ω1(f; |zn − z0|)DR.



38 Ciprian G. Gal and Sorin G. Gal CUBO
15, 1 (2013)

Passing to the limit as n → ∞, the continuity of Ht(f)(z) at z0 ∈ DR is a consequence of (14)
since f is continuous on DR.

(ii) First, note that we can also write

Wt(f) (z) =
1

2
√
πc(t)

∫+∞

−∞

e−v
2
/(4c(t))f(ze−i(v−g(t)))dv.

A simple calculation gives

|Wt(f)(z) − f(z)| ≤
1

2
√
πc

∫+∞

−∞

|f(ze−i(u−g)) − f(z)|e−u
2
/(4c) du (15)

≤
R + 1

2
√
πc

∫∞

−∞

ω1(f; |1 − e
−i(u−g)|)

DR
e−u

2
/(4c) du

=
R + 1

2
√
πT

∫+∞

−∞

ω1

(
f; 2

∣∣∣∣sin
u − g

2

∣∣∣∣
)

DR

e−u
2
/(4c) du

≤
R + 1

2
√
πT

∫+∞

−∞

ω1(f; |u − g|)DRe
−u

2
/(4c) du

≤
R + 1

2
√
πT

∫+∞

−∞

ω1(f;
√
c)

DR

(
|u − g|√

c(t)
+ 1

)
e−u

2
/(4c) du

≤ (R + 1)
[
ω1(f;

√
c)

DR
+

ω1(f;
√
c)

DR√
c2
√
πc

∫∞

0

2ue−u
2
/(4c) du

]

+ (R + 1)

[
|A| ·

ω1(f;
√
c)

DR√
c2
√
πc

∫∞

0

e−u
2
/(4c) du

]
.

Since
∫∞
0

2ue−u
2
/(4c(t))du = 4c (t), we infer

|Wt(f)(z) − f(z)| ≤ ω1(f;
√
c)

DR

[
1 +

2√
π
+

|g|

2
√
c

]
(R + 1).

This proves (ii).

(iii) Since 1
2
√
πc

∫+∞
−∞

e−u
2
/(4c)du = 1, we also have

|Wt(f)(z)| ≤
1

2
√
πc

∫+∞

−∞

|f(ze−i(u−g))|e−u
2
/(4c)du ≤ ‖f‖

DR
, z ∈ DR,

which proves the claim.

(iv) Let f ∈ A(DR), and z ∈ DR, z = reiϕ, 0 < r < R. Set

B(t, ϕ) := a(t) + d(t)[c(t)d(t) + b(t) + ϕ], Ak(t) := −k
2c(t) + ikg(t);

by (i), we have dk(t) = e
Ak(t), and we can write

U(t, ϕ) = eB(t,ϕ)
∞∑

k=0

ake
Ak(t)rkekiϕ. (16)



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On Fokker-Planck and linearized Korteweg-de Vries ... 39

Consequently, since the series representation (10) for Gt(f)(z) is uniformly convergent in any

compact disk included in DR, it follows that U(t, ϕ) can be differentiated term by term with

respect to t and ϕ. Therefore, simple calculations show that U, given by (16), satisfies. We

emphasize again that in (11) we must take z 6= 0 simply because z = 0 has no polar representation,
that is, z = 0 cannot be represented as function of ϕ. Finally, it is also easy to check that U(t, ϕ)

satisfies (12) since a(0) = b(0) = c(0) = d(0) = 0 on account of (3). This completes the proof of

the theorem.

Remark 2.2. In Theorem 2.1-(ii), we have limt→0 c(t) = 0. Therefore, there exists a sufficiently

small δ0 > 0 such that 0 < c(t) < 1, for all t ∈ [0, δ0). This implies that

|g (t)| /
√

c (t) ≤ |b (t)| /c (t) + 2 |d (t)| , (17)

for all t ∈ (0, δ0) . Exploiting (3) once more again it is easy to show that |b (t)| /c (t) ≤ 2β0t, for all
t ∈ (0, δ0), where β0 = ‖β‖[0,δ0]. This together with the inequality (17) yields limt→0 |g(t)|/

√
c(t) =

0. As a consequence, cf. the estimate of Theorem 2.1-(ii), it also follows that limtց0 Wt(f)(z) =

f(z), for all z ∈ DR.

In what follows, the system (1) is complexified, by replacing x ∈ R with z ∈ C directly in the
equations. More precisely, our goal is to study the following initial value problem

{
∂u
∂t

(t, z) = α(t)∂
2
u

∂z2
(t, z) + β(t)zu(t, z),

u(0, z) = f(z).
(18)

We will first consider the case when f is analytic. Our first goal is to search for analytic solutions

u(t, z), as functions of z, for any t > 0. First, we need some basic notations. For r > 0, define the

strip

Sr = {z = x + iy ∈ C; x ∈ R, |y| ≤ r}

and

A(Sr) = {f : Sr → C; f is analytic in Sr},

( i.e., f is analytic in a domain that contains Sr). Next, let Mr be the set of all f ∈ A (Sr) such that
there exists g ∈ L1 (R+) ∩ L∞ (R+) with the property that |f′ (z)| ≤ g (|z|), as |z| goes to infinity.
Finally, let

u(t, z) :=
eΦ(t,z)

2
√

πc(t)

∫∞

−∞

f(z − ξ)e−[ξ+g(t)]
2
/(4c(t))dξ, (t, z) ∈ R+ × Sr,

where c(t), Φ(t, z), and g(t) are defined in the statement of Theorem 2.1-(i).

Theorem 2.3. For each f ∈ Mr we have u = u(t, ·) ∈ A (Sr) , for any t > 0. Moreover, u(t, z)
solves (18) for (t, z) ∈ R+ × Sr.



40 Ciprian G. Gal and Sorin G. Gal CUBO
15, 1 (2013)

Proof. If f ∈Mr, there exists M > 0 such that

sup{|f′(z)|; z ∈ Sr} = ‖f′‖Sr ≤ M.

By the Mean Value Theorem in complex analysis, f is uniformly continuous in Sr, and that

|f(z − ξ)| ≤ |f(0)| + ‖f′‖Sr · |z − ξ| ≤ |f(0)| + M|z − ξ|.

The latter implies that the integral I(t, z) := 1
2
√

πc(t)

∫∞
−∞

f(z − ξ)e−[ξ+g(t)]
2
/(4c(t))dξ exists and

is absolutely convergent in C, and that I(t, z) is differentiable with respect to any z ∈ Sr, with

∂zI(t, z) =

∫∞

−∞

f′(z − ξ)e−[ξ+g(t)]
2
/(4c(t))dξ.

The analicity of Φ(t, z) with respect to z ∈ Sr implies that u = u(t, ·) also belongs to A (Sr), for
any t > 0. Analogous calculations to those performed in (15) yield

|I(t, z) − f(z)|

≤ ω1(f; |g(t)|)Sr + ω1(f;
√

c (t))Sr
1

2
√

πc (t)

∫∞

−∞

(
|u| (c (t))

−1/2
+ 1
)
e−u

2
/(4c(t))du

= ω1(f; |g(t)|)Sr +

(
1 +

2√
π

)
ω1(f;

√
c(t))Sr.

Taking now into account the Remark 2.2, and the uniform continuity of f on Sr, it easily follows

that limtց0 I(t, z) = f(z), for all z ∈ Sr. This together with the fact that Φ(0, z) = 1 yields
u(0, z) = limtց0 u(t, z) = f(z), for all z ∈ Sr, i.e., u satisfies the initial condition of (18). It
remains to show that u also solves the main equation of (18). To this end, define

F(t, z) :=
∂u

∂t
(t, z) − α(t)

∂2u

∂z2
(t, z) − β(t)zu(t, z),

for all (t, z) ∈ R+ × Sr, and recall that u(0, z) = f(z), z ∈ Sr. For each t > 0, F (t, ·) is analytic in
Sr. Taking now z = x ∈ R in all the equations of (18), we can now apply known theory to deduce
that u (t, z) = u (t, x) also solves (1). Hence, F(t, x) = 0, for all (t, x) ∈ R+ × R. The identity
theorem for holomorphic functions (in a domain that contains Sr) implies that we must also have

F(t, z) = 0, (t, z) ∈ R+ × Sr. This finishes the proof of the theorem.

3 Linearized Korteweg-de Vries type equations

For R > 1, let us define the open disk

DR := {z ∈ C : |z| < R}.

Next we endow the local convex space

A∗(DR) := {f : DR → C : f is analytic on DR},



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On Fokker-Planck and linearized Korteweg-de Vries ... 41

with the countable family of seminorms

‖f‖n := sup{|f(z)| : z ∈ DRn}, Rn ր R, Rn ≥ 1,

and metric

d(f, g) :=

∞∑

n=0

1

2n
‖f − g‖n

1 + ‖f − g‖n
.

Then A∗(DR) is a Fréchet space.

Let f ∈ A∗(DR) such that

f(z) =

∞∑

k=0

akz
k,

for all z ∈ DR. We consider the integral operator

TK (t) (f) (z) :=
∫∞

−∞

K(u, t)f(ze−iu)du, z ∈ DR, (19)

with K(u, t) given by (7). Evidently, since K(−u, t) 6= K(u, t), we can naturally introduce another
complex integral by

T̃K (t) (f) (z) :=
∫∞

−∞

K(−u, t)f(zeiu)du, z ∈ DR. (20)

The first goal of this section is to prove some properties for (19) and (20).

Theorem 3.1. Let R > 1, f ∈ A∗(DR), that is, f(z) =
∞∑

k=0

akz
k, for all z ∈ DR. Let

TK (·) (f) and T̃K (·) (f) be as in (19) and (20), respectively.

(i) For all t ≥ 0, as functions of z, we have TK (·) (f) (z) ∈ A∗(DR), T̃K (·) (f) (z) ∈ A∗(DR),
and there hold

TK (t) (f) (z) =
∞∑

k=0

akdk(t)z
k and T̃K (t) (f) (z) =

∞∑

k=0

akbk(t)z
k, z ∈ DR, (21)

where for all k ≥ 0,
dk(t) = e

itk
3

and bk(t) = e
−itk

3

.

Moreover,

TK (0) (f) = T̃K (0) (f) = f.

(ii) For all z ∈ Dr with 1 ≤ r < R and t ∈ R+, the following estimate holds:

|TK (t) (f) (z) − f(z)| ≤
t2

2

∞∑

k=0

|ak|k
6rk + |t|

∞∑

k=0

|ak|k
3rk,

where
∞∑

k=0

|ak|k
6rk < ∞, since f(6) ∈ A∗(DR).



42 Ciprian G. Gal and Sorin G. Gal CUBO
15, 1 (2013)

(iii) For all z ∈ Dr with 1 ≤ r < R and t, s ∈ R+, there holds:

|TK (t) (f) (z) − TK (s) (f) (z) | ≤ 2
∞∑

k=0

|ak|k
3rk|t − s|

and

|T̃K (t) (f) (z) − T̃K (s) (f) (z) | ≤ 2
∞∑

k=0

|ak|k
3rk|t − s|.

(iv) The families {TK (t)}t≥0 and
{
T̃K (t)

}
t≥0

are (C0)-semigroups of linear operators, locally

equicontinuous (that is, equicontinuous for t ∈ [0, a], for some a > 0) on A∗(DR). For each
f ∈ A∗(DR), the corresponding Cauchy problems

{
∂u
∂t

+ ∂
3
u

∂ϕ3
= 0, (t, z) ∈ R+ × DR\ {0} ,

u(0, z) = f(z), z ∈ DR
(22)

and {
∂w
∂t

= ∂
3
w

∂ϕ3
, (t, z) ∈ R+ × DR\ {0} ,

u(0, z) = f(z), z ∈ DR
(23)

are well-posed, with solutions given by

u (t) = TK (t) (f) ∈ C∞ (R+; A∗(DR)) ,
w (t) = T̃K (t) (f) ∈ C∞ (R+; A∗(DR)) ,

respectively.

Proof. We will prove the above statements only for the family {TK (t)}t≥0 (the proof for
{
T̃K (t)

}
t≥0

is the same).

(i) Let f(z) =
∞∑

k=0

akz
k, for all z ∈ DR. For fixed z ∈ DR, we get

f(ze−iu) =

∞∑

k=0

ake
−ikuzk.

Since |ake
−iku| = |ak|, for all u ∈ R, and since

∞∑
k=0

akz
k is absolutely convergent, the series

∞∑
k=0

ake
−ikuzk is also uniformly convergent with respect to u ∈ R. Therefore, the latter can be

integrated term by term. Using (19), we deduce

TK (t) (f) (z) =
∞∑

k=0

ak

{∫∞

−∞

[
1

π

∫+∞

0

cos(uτ − tτ3)dτ

]
e−ikudu

}
zk

=

∞∑

k=0

akdk(t)z
k,



CUBO
15, 1 (2013)

On Fokker-Planck and linearized Korteweg-de Vries ... 43

where

dk(t) =

∫∞

−∞

[
1

π

∫+∞

0

cos(uτ − tτ3)dτ

]
cos(ku)du

− i

∫∞

−∞

[
1

π

∫+∞

0

cos(uτ − tτ3)dτ

]
sin(ku)du

=

∫0

−∞

[
1

π

∫+∞

0

cos(uτ − tτ3)dτ

]
cos(ku)du

+

∫∞

0

[
1

π

∫+∞

0

cos(uτ − tτ3)dτ

]
cos(ku)du

− i

∫0

−∞

[
1

π

∫+∞

0

cos(uτ − tτ3)dτ

]
sin(ku)du

+

∫∞

0

[
1

π

∫+∞

0

cos(uτ − tτ3)dτ

]
sin(ku)du

=
1

π

∫∞

0

[∫∞

0

(
cos(uτ + tτ3) + cos(uτ − tτ3)

)
dτ

]
cos(ku)du

+
i

π

{∫∞

0

[∫∞

0

(
cos(uτ + tτ3) − cos(uτ − tτ3)

)
dτ

]
sin(ku)du

}
.

On the other hand, it is well-known that the Fourier transform of the Airy’s function is given by

(see, e.g., [5, p. 87, Table 4.2])

∫∞

−∞

[
1

π

∫∞

0

cos(uτ + τ3/3)dτ

]
e−ikωudu = ei(ωk)

3
/3 =: I1 + I2,

where

I1 :=
1

π

∫∞

0

[∫∞

0

(
cos(τ3/3 + uτ) + cos(τ3/3 − uτ)

)
dτ

]
cos(kωu)du,

I2 :=
i

π

{∫∞

0

[∫∞

0

(
cos(τ3/3 + uτ) − cos(τ3/3 − uτ)

)
dτ

]
sin(kωu)du

}
.

Now, by a change of variable τ = (3t)1/3η, and then another u = v
(3t)1/3

, (t > 0 is a fixed

parameter), simple calculations yield

I1 + I2 =
1

π

∫∞

0

[∫∞

0

(cos(tη3 + vη) + cos(tη3 − vη))dη

]
cos(kωv/(3t)1/3)dv

+
i

π

∫∞

0

[∫∞

0

(cos(tη3 + vη) − cos(tη3 − vη))dη

]
sin(kωv/(3t)1/3)dv

= ei(ωk)
3
/3.

Choosing ω = (3t)1/3, and taking into account that I1 + I2 = dk(t), we easily arrive at

dk(t) = e
itk

3

.



44 Ciprian G. Gal and Sorin G. Gal CUBO
15, 1 (2013)

This proves the analyticity of TK (·) (f) (z) , as function of z ∈ DR. Finally, the relation TK (0) = I
is immediate in view of (21).

(ii) Let |z| ≤ r. We obtain

|TK (t) (f) (z) − f(z)| =
∣∣∣∣∣

∞∑

k=0

akz
k[eik

3
t − 1]

∣∣∣∣∣

=

∣∣∣∣∣

∞∑

k=0

akz
k[−2 sin2(k3t/2) + i sin(k3t/2) cos(k3t/2)]

∣∣∣∣∣

≤
∞∑

k=0

|ak|r
k|2 sin2(k3t/2)| +

∞∑

k=0

|ak|r
k|2 sin(k3t/2)|

≤
t2

2

∞∑

k=0

|ak|k
6rk + |t|

∞∑

k=0

|ak|k
3rk,

since | sin(x)| ≤ |x|, for all x ∈ R.

(iii) We have

|TK (t) (f) (z) − TK (s) (f) (z)|

=

∣∣∣∣∣

∞∑

k=0

akz
k[eik

3
t − eik

3
s]

∣∣∣∣∣

=

∞∑

k=0

[cos(k3t) − cos(k3s) + i(sin(k3t) − sin(k3s))]

≤ 4
∞∑

k=0

|ak|r
k

∣∣∣∣sin
(
k3(t − s)

2

)∣∣∣∣

≤ 2
∞∑

k=0

|ak|k
3rk|t − s|.

(iv) From (i), it is immediate that TK (t + s) = TK (t) TK (s) , for all t, s ∈ R+. The strong
continuity of TK (t) follows from (iii). We can argue as in the proof of Theorem 6.2.1 to deduce
the first part of the statement in (iv). Let f ∈ A∗(DR). We can compute the generators of the
semigroups TK (t) and T̃K (t), t ∈ R+, respectively, as follows:

(
d

dt
TK (t) (f) (z)

)

|t=0

= i

∞∑

k=0

k3ake
ik

3
tzk = −

∂3

∂ϕ3
TK (t) (f) (z)

and (
d

dt
T̃K (t) (f) (z)

)

|t=0

= −i

∞∑

k=0

k3ake
−ik

3
tzk =

∂3

∂ϕ3
d

dt
T̃K (t) f (z) ,

for all z = reiϕ ∈ DR\ {0} . The proof of the theorem is complete.



CUBO
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On Fokker-Planck and linearized Korteweg-de Vries ... 45

In what follows, the linearized Korteweg-de Vries equation from (5) is complexified by replacing

x ∈ R with z ∈ Ω ⊂ C directly in the equations. More precisely, we aim to study the following
initial value problem 




∂u
∂t

= ∂
3
u

∂z3
, (t, z) ∈ R+ × Ω,

lim
tց0

u(t, z) = f(z), z ∈ Ω. (24)

We look for classical solutions of (24) which belong to the class:

u ∈ C1 (R+; A◦ (Sr)) , (25)

where

A◦(Sr) := {f : Sr → C : f is analytic in Sr},

i.e., f is analytic in a domain that contains the closure Sr of

Sr := {z = x + iy ∈ C : x ∈ R, |y| < r}.

Theorem 3.2. Let f ∈ A◦(Sr) such that the following are satisfied:

(i) f′ is bounded on Sr;

(ii) for all |y| ≤ r, the integral F(s, y) =
∫+∞
−∞

e−sxf(x + iy)dx is absolutely convergent for

s = c1 + iτ;

(iii) for all |y| ≤ r, the integral G(s, y) =
∫+∞
−∞

e−sxf′(x + iy)dx is absolutely convergent for

s = c1 + iτ;

(iv) for all |y| ≤ r,
∫+∞
−∞

|F(c1 + iτ, y)|dτ < +∞.
Setting

TKV (t) (f) (z) :=

∫+∞

−∞

K(u, t)f(z − u)du, (t, z) ∈ R+ × Sr,

there holds u (t) = TKV (t) (f) ∈ C∞ (R+; A◦(Sr)) , and TKV (t) (f) solves the initial value problem
(24).

Proof. Let f ∈ A◦(Sr) and decompose f(z) = U(x, y) + iV(x, y), with U and V having continuous
partial derivatives of first order. Moreover, U, V satisfy the Cauchy-Riemann conditions at any

(x, y) ∈ Sr. We can write

TKV (t) (f) (z) =

∫+∞

−∞

K(u, t)U(x − u, y)du + i

∫+∞

−∞

K(u, t)V(x − u, y)du

:= T1(t, x, y) + iT2(t, x, y).

Step 1. Let z = x + iy ∈ Sr. Since

f′(z) =
∂U

∂x
(x, y) + i

∂V

∂x
(x, y), (26)



46 Ciprian G. Gal and Sorin G. Gal CUBO
15, 1 (2013)

we note that

f′′′(z) =
∂3U

∂x3
(x, y) + i

∂3V

∂x3
(x, y).

We claim that for any fixed |y| ≤ r, conditions (i)-(ii) in the statement of Theorem 7.1.1, are
fulfilled for T1 and T2, with respect to (t, x) ∈ R+ × R. As a consequence,





∂T1
∂t

(t, x, y) = ∂
3
T1

∂x3
(t, x, y), (t, x, y) ∈ R+ × Sr,

lim
tց0

T1(t, x, y) = U(x, y), (x, y) ∈ Sr.





∂T2
∂t

(t, x, y) = ∂
3
T2

∂x3
(t, x, y), (t, x, y) ∈ R+ × Sr,

lim
tց0

T2(t, x, y) = V(x, y), (x, y) ∈ Sr,

and, therefore, TKV (t) (f) solves (24). Indeed, by the first hypothesis above (see (i)), there exists

M > 0 such that

sup{|f′(z)| : z ∈ Sr} = ‖f′‖Sr ≤ M.

Now, from (26), ∂U
∂x

(x, y) and ∂V
∂x

(x, y) are bounded on Sr. Therefore, for any fixed |y| ≤ r, U(xy)
and V(x, y) are continuous and of bounded variation with respect to x ∈ R. Setting

F1(s, y) =

∫+∞

−∞

e−sxU(x, y)dx, F2(s, y) =

∫+∞

−∞

e−sxV(x, y)dx,

clearly, F(s, y) = F1(s, y) + iF2(s, y), and by virtue of (ii), both F1(s, y) and F2(s, y) are absolutely

convergent, for a fixed (but otherwise arbitrary) |y| ≤ r. Furthermore, in view of (iv), we deduce
∫+∞

−∞

|F1(c1 + iτ, y)|dτ < +∞,
∫+∞

−∞

|F2(c1 + iτ, y)|dτ < +∞.

Therefore, we can apply Theorem 1.1 to the functions T1 and T2, respectively. This yields the

above claim.

Step 2. Clearly, TKV (t) (f) belongs to the class (25) for any f ∈ A◦(Sr). Indeed, the fact that
∂xT1(·, x, y) and ∂xT2(·, x, y) are continuous on Sr was already proved in Step 1. The existence and
continuity of the partial derivatives ∂yT1(·, x, y), ∂yT2(·, x, y) follow from condition (iii). Finally,
the functions Ti (·, x, y) , i = 1, 2, also satisfy the Cauchy-Riemann equations since U, V do. The
proof is finished. �

Remark 3.1. A simple example of boundary data in (24) that satisfies all the hypotheses of

Theorem 3.2 is

f(z) = e−z
2

.

In this case, one can prove that F(s, y) =
√
πes

2
/4 cos(sy).

Received: October 2012. Revised: February 2013.



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On Fokker-Planck and linearized Korteweg-de Vries ... 47

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