California State University, Northridge Similarity

CALIFORNIA STATE UNIVERSITY, NORTHRIDGE

SIMILARITY ANALYSIS OF THE ONE-DIMENSIONAL FOKKER-PLANCK EQUATION

A thesis submitted in partial fulﬁllment of the requirements For the degree of Master of Science in Mathematics

Areej Alsafri

May 2014 The thesis of Areej Alsafri is approved:

Dr. Rabia Djellouli Date

Dr. Emmanuel Yomba Date

Dr. Ali Zakeri, Chair Date

California State University, Northridge

ii Acknowledgments

I would like to express my special appreciation and thanks to my advisor Professor Ali Zaker, you have been a tremendous mentor for me. I would like to thank you for encouraging my research and for allowing me to grow as a research scientist. Your advice on both research as well as on my career have been priceless. I would also like to thank my committee members, professor Rabia Djellouli and professor Emmanuel Yomba, for serving as my committee members even at hardship. I also want to thank you for letting my defense be an enjoyable moment, thanks to you. A special thanks to my family and to my family-in-law. Words cannot express how grateful I am to my mother, father, mother-in-law, fater-in-law, and uncl Hamed for all of the sacriﬁces that you’ve made on my behalf. Your prayer for me was what sustained me thus far. I would also like to thank all of my brothers and sisters who supported and motivated me to strive towards my goal. At the end I would like express appreciation to my husband Essam who spent sleepless nights with and was always my support in the moments when there was no one to answer my queries.

iii Table of Contents

Signature Page ii Acknowledgments iii Abstract v Introduction 1

1 Local Lie Group 3

2 Stretching Group 5

3 The Lie Variables 19

4 The General Similarity Method 28

5 The Fokker-Planck Equation 43

6 Similarity Solution Under The Stretching Transformation 44

7 A General Similarity Solution For The Fokker-Planck Equation 54

8 The Clarkson-Kruskal Reduction Method 63 Conclusion 73 Bibliography 74

iv ABSTRACT

SIMILARITY ANALYSIS OF THE ONE-DIMENSIONAL FOKKER-PLANCK EQUATION

Areej Alsafri

Master of Science in Mathematics

The goal of this study is to investigate the conditions under which the Fokker- Planck equation assumes closed form solutions. The Fokker-Planck equation arises in the study of ﬂuctuations in physical and biological systems. This equation has been subject of several recent studies. We have analyzed the existence of solutions of this equation analytically and have established results using various invariant approaches. For example, we have employed a group of stretching transformations that keeps the original equation invariant under this group actions. This resulted in obtaining a class of closed form solutions. Then we analyzed the conditions under which a Clarkson-Kruskal type similarity functional form would result in closed form solutions. The importance of this type of solutions relies on the fact that they can be used as a test functional tool in checking the accuracy of numerical algorithms. General form of such class of solutions is obtained and several speciﬁc examples of this class is discussed. In addition, a general group invariant approach is investigated.

v Introduction

There are many important nonlinear partial differential equations that we are unable to solve analytically. However, under certain conditions, for some of these nonlinear partial differential equations we can transform them into other equations that are invariant or unchanged due the symmetries properties in the partial differential equations. Basically, the method of similarity is determining the set of transformations under which the partial differential equations is invariant and then using the transformations to find special conditions where the partial differential equations can be simplified. These conditions drive a new single independent variable that is a function of the independent variables in the partial differential equations; where the new independent variable reduces the partial differential equation into an ordinary differential equation. The solution to the ordinary differential equation pro- vides a class of self-similar solutions to the partial differential equation. Such solutions are very impotent closed form solutions that can be used to study the physics of the problem, and even if the obtained solutions have no physical values, they can be used as the check solutions for complex numerical method for accuracy of the method. In this thesis, we give an elementary brief review of the general Lie group method with ample examples and explain key defini- tions and main results that are applicable in the problem that we are going to study.We like to point out that G.W. Bluman has already obtained a classs of similarity solutions for the one-dimensional Fokker-Planck equation in the

∂2P ∂ ∂P form of ∂x2 + ∂x (f(x)P ) = ∂t for the initial value P (x, 0/x0) = δ(x − x0). α He assumed that f(x) is an odd function, given by f(x) = x + βx where β > 0 and −∞ < α < 1. In this thesis, we consider a general case of Fokker-

∂W ∂ (1) ∂2 (2) Planck equation in the form of ∂t = − ∂x [D (x, t)W ] + ∂x2 [D (x, t)W ]. So, in Bluman assumed D(1)(x, t) = −f(x) for that particular form and D(2)(x, t) = 1. The main goal of this thesis is to analyze and study the

1 closed form solution of one-dimensional Fokker-Plank equation that is used to study fluctuations in physical and biological models. To this end, we first study a group of stretching infinitesimal transformations. We obtain such solution and the required specific condition on the drift and diffusion coefficients for which such solutions exist. Then, we consider the general Lie group of transformations. Finally, we use the Clarkson-Kruskal type solutions by selecting a specific closed form expression and then we obtained the neces- sary requirements for the existence of this class of similarity solutions. A few special cases are analyzed and closed form solutions are given.

2 1 Local Lie Group

We consider a transformation T of the plane that is mapping from R2 to R2, i.e. T (x, t) = (¯x, t¯). This means that T : R2 → R2 which is deﬁned as T : t¯= φ(x, t), x¯ = ψ(x, t) where φ and ψ are given functions. Geometrically, this means T transforms the point (x, t) to another point (¯x, t¯) in the plane R2. We could obtain the inverse transformation T −1 deﬁned by equations: T −1 : t = φ(¯x, t¯), x = ψ(¯x, t¯) Clearly, TT −1 and T −1T are identity transformation I given by the equations t¯= t, x¯ = x. The types of symmetry transformations under the invariance properties of an equation are transformations of the plane with the additional stipulation that the transformation also depends on a real parameter which we denote by

, where varies over an open interval || < 0 and we consider the collection

T of all transformations included in the family deﬁned by

T : t¯= φ(x, t, ), x¯ = ψ(x, t, ) (1)

2 where φ and ψ are given functions on R × (−0, 0), and (1) is said to be a one parameter family of transformation on R2.

A one parameter family of transformations T is called a local Lie group of transformation if it satisﬁes the local closer property, contains the identity, and inverses exist for small .

Therefore, we want (1) to satisfy the local closure property, namely if 1 and

2 are suﬃciently small, then there exists 3 in the interval || < 0 for which

T1 T2 = T3 .

3 Also, we assume that the family T contains the identity transformation I when = 0 t¯= φ(x, t, 0) = t, x¯ = ψ(x, t, 0) = x

Finally, for each suﬃciently small we want the transformation T1 to have an inverse T2 . Examples of local Lie groups: Translation Group t¯= t + , x¯ = x + Rotation Group t¯= t cos − x sin ,x ¯ = t sin + x cos Stretching Group t¯= eat,x ¯ = ebx where a, b are constants.

4 2 Stretching Group

Similarity solutions establish by the invariance properties of the partial differential equations under local Lie groups of transformations. In this section, we discuss an example of local Lie groups of transformations which is the Stretching Group. The idea is to ﬁnd new independent and dependent variables that are multiples of the old ones where the partial diﬀerential equation in the new coordinate system is the same up to some multiple as in the old coordinate system. Example 1 Consider the heat equation

ut − uxx = 0 (2)

ﬁrst we make change of variables of the form x¯ = ax t¯= bt u¯ = cu where is a real parameter in the open interval containing 1, and a,b, and c are ﬁxed real numbers. so, we have

c−b c−2a u¯t¯ − u¯x¯x¯ = ut − uxx (3)

if we choose a, b and c such that c − b = c − 2a so b = 2a then we get

5 c−b u¯t¯ − u¯x¯x¯ = (ut − uxx) (4)

therefore, we can say the original equation is stretched by a multiple constant c−b. Consider a single second order partial diﬀerential equation with one dependent variable u, and independent variables x and t, u = u(x, t)

F (x, t, u, ux.ut, uxx, uxt, utt) = 0 (5)

3 3 deﬁne the family of transformation T : R → R such that

x¯ = ax

t¯= bt (6)

u¯ = cu where a, b and c are ﬁxed real constants and is real parameter that varies over an open interval I that contains 1.

Then we can say that T contains the identity transformation T1. Also, for

every 1 and 2 we have T1 T2 = T12 which means the family T satisﬁes the local closure property. Finally, the inverse of T is T−1 . Therefore, T is a local Lie group of transformation on R3. Let φ(x, t) be a solution for (5) so, u = φ(x, t)

The surface φ(x, t) gets mapped under the transformation T to the surface u¯ = φ¯(¯x, t¯). where φ¯ is deﬁned by

6 φ¯(¯x, t¯) = cφ(−ax, −bt) (7)

Furthermore, we could transform (5) as

F (¯x, t,¯ u,¯ u¯x¯, u¯t¯, u¯x¯x¯, u¯x¯t¯, u¯t¯t¯) = 0 (8)

since u = φ(x, t) is solution for (5), thenu ¯ = φ(¯x, t¯) is solution to (8), but u¯ = φ¯(¯x, t¯) is not necessarily a solution to (8) because φ(¯x, t¯) 6= φ¯(¯x, t¯). Deﬁnition1

The partial diﬀerential equation F (x, t, u, ux.ut, uxx, uxt, utt) = 0 is constant conformally invariant under the one parameter family of stretching T de- ﬁned by x¯ = ax t¯= bt u¯ = cu if and only if

F (¯x, t,¯ u,¯ u¯x¯, u¯t¯, u¯x¯x¯, u¯x¯t¯, u¯t¯t¯) = A()F (x, t, u, ux.ut, uxx, uxt, utt) for all in I, and for some function A whereA(1) = 1. If A() = 1 then we say that F (x, t, u, ux.ut, uxx, uxt, utt) = 0 is absolutely invariant. If we consider our ﬁrst example, we could say from (4) that A() = c−b; so the heat equation is constant conformally invariant under the given stretching transformation with b = 2a. Example 2

In this example we would like to ﬁnd the conformal factorA() under the T deﬁned as in (5)

utt − uxx = 0 (9)

7 we would have

∂2u¯ ∂2u¯ u¯¯¯ − u¯ = − tt x¯x¯ ∂t¯2 ∂x¯2 ∂2cu ∂2cu = − ∂2bt2 ∂2ax2 c−2b c−2a = utt − uxx

Leta = b,

c−2a u¯t¯t¯ − u¯x¯x¯ = (utt − uxx) then we have showed that A() = c−2a, which means that the wave equation is constant conformally invariant under T provided a = b. Theorem 1 If the partial differential equation (5) is constant conformally invariant under the one parameter family of stretchings T given by (6) and if φ(x, t) is a solution of (5), then φ¯(¯x, t¯) defined by (7) is a solution of the partial differential equation (8).

proof It follows from Deﬁnition 1.

Deﬁnition 2 A solution u = φ(x, t) of is an invariant solution ( or invariant surface) if and only if

φ(¯x, t¯) = φ¯(¯x, t¯) (10)

under the transformation T. Similarity solutions for partial diﬀerential equations are obtained by the in-

8 variant solutions. We may use (10) to obtain the invariant solutions, so we can say (10) is equivalent to

φ(ax, bt) = cφ(x, t) (11)

where in I, by taking the derivative with the respect to and then fix = 1, we obtain the first order partial differential equation

axφx + btφt = cφ (12)

which called the invariant surface condition, we are going to use the methods of characteristics to ﬁnd the general form of φ. We set the characteristic system

dx dt dφ = = ax bt cφ

if we integrate the ﬁrst two, we get

xb = constant ta

and if we integrate the second two, we would get

−c φt b = constant

so the general solution of (12) is

9 −c b −a ψ(φt b , x t ) = 0

for some functionψ. Then we get the invariant surfaces are

b c x φ(x, t) = t b f (13) ta

where f is an arbitrary function. Equation (13) shows the possible form of self-similar solution for partial differential equation under stretching transformations.

xb We set s = ta , where s called The similarity variable then equation(13) becomes

c u = φ(x, t) = t b f(s). The following theorem shows if we substituted (13) in the partial differential equation, it reduces to an ordinary differential equation. Theorem 2 If the partial differential equation (5) is constant conformally invariant under the one parameter family of stretching transformations defined by

x¯ = ax

t¯= bt

u¯ = cu the substitution of the expression

c u = t b f(s) (14)

10 xb s = (15) ta

into (5) yields an equation of the form

H(s, f, f 0, f 00) = 0

which is an ordinary diﬀerential equation for f.

proof Consider a single second order partial equation

F (x, t, u, ux, ut, uxx, uxt, utt) = 0

by Deﬁnition 1, since the partial diﬀerential equation is constant conformally invariant under T the we have

F (¯x, t,¯ u,¯ u¯x¯, u¯t¯, u¯x¯x¯, u¯x¯t¯, u¯t¯t¯) = A()F (x, t, u, ux.ut, uxx, uxt, utt) (16)

where A is some function. We note that

11 c−a u¯x¯ = ux

c−b u¯t¯ = ut

c−2a u¯x¯x¯ = uxx

c−a−b u¯x¯t¯ = uxt

c−2b u¯t¯t¯ = utt

By diﬀerentiation equation (16) with respect to , we get

a−1 b−1 c−1 c−a−1 a xFx + b tFt + c uFu + (c − a) uxFux

c−b−1 c−2a−1 + (c − b) utFut + (c − 2a) uxxFuxx

c−a−b−1 c−2b−1 0 + (c − a − b) uxtFuxt + (c − 2b) uttFutt = A ()F by setting = 1, we have

axFx+btFt + cuFu + (c − a)uxFux + (c − b)utFut

+ (c − 2a)uxxFuxx + (c − a − b)uxtFuxt

0 + (c − 2b)uttFutt = A (1)F we can solve the above invariant surface condition by the methods of characteristics. The characteristic system is

dx dt du du du = = = x = t ax bt cu (c − a)ux (c − b)ut duxx duxt dutt dF = = = = 0 (c − 2a)uxx (c − a − b)uxt (c − 2b)utt A (1)F

Integrating the above system give us eight independent ﬁrst integrals

b 0 x −c a−c 1− c 2a−c 1+ a−c 2− c −A (1) , ut b , u t b , u t b , u t b , u t b , u t b , F t b ta x t xx xt tt

12 Therefore

0 b A (1) x −c a−c 1− c 2a−c 1+ a−c 2− c F = t b G , ut b , u t b , u t b , u t b , u t b , u t b ta x t xx xt tt

for some G. From the above relation, we observe that if we use

c u = t b f(s)

where

xb s = ta

Then,

c − ab b−1 0 ux = bx t b f (s) c − b c − ab − b c u = t b f(s) − axbt b f 0(s) t b c − ab c − 2ab b−2 0 2 2(b−1) 00 uxx = b(b − 1)x t b f (s) + b x t b f (s) c − b(a + 1) c − b(2a + 1) b−1 0 2b−1 00 uxt = (c − ab)x t b f (s) − abx t b f (s) c − 2b c − b(a + 2) c(c − b) a(2c − ab − b) u = t b f(s) − xbt b f 0(s) tt b2 b c − 2b(a + 1) + a2x2bt b f 00(s)

by substituting in G the value of x, t, u, ux, ut, uxx, uxt, and utt we obtain an

13 ordinary diﬀerential equation of the form H(s, f, f 0, f 00) = 0

Example 3 Lets consider the nonlinear heat equation

uut − uxx = 0

we would like to reduce the nonlinear heat equation to an ordinary diﬀeren- tial equation. First, we consider the following stretching transformation

x¯ = ax

t¯= bt

u¯ = cu then

c c c ∂( u) ∂(∂ u) u¯u¯¯ − u¯ = u − t x¯x¯ ∂bt (∂ax)(∂ax)

2c−b c−2a = uut − uxx if we choose 2c − b = c − 2a or c = b − 2a, then the heat equation is invariant under the following transformation

x¯ = ax

t¯= bt

u¯ = b−2au then the invariant surface condition is

14 axux + btut = (b − 2a)u

which means the characteristic system is

dx dt du = = ax bt (b − 2a)u

xb 2a b −1 The frist integrals are ta and ut xb Letting s = ta the invariant surface are given by

b 1− 2a x u(x, t) = t b f ta

1 since a, b, and c are arbitrary values, we could choose a = 2 , b = 1, so c = 0 then we have s = √x and u(x, t) = f(s) t so we get

x 0 ut = − 3 f (s) 2t 2 −1 00 uxx = t f (s)

Thus, the nonlinear heat equation reduces to

x √ f(s)f 0(s) + f 00(s) = 0 2 t

or equivalent,

s f(s)f 0(s) + f 00(s) = 0 2

15 Example 4 In this example we are going to reduce the nonlinear wave equation directly by using theorem 2

utt − uuxx = 0

by using the following transformation

x¯ = ax

t¯= bt

u¯ = cu we get

c c ∂(∂ u) c ∂(∂ u) u¯¯¯ − u¯u¯ = − u tt x¯x¯ (∂bt)(∂bt) (∂ax)(∂ax)

c−2b 2c−2a = utt − uuxx if we choose c − 2b = 2c − 2a, then the nonlinear wave equation is invariant under the above transformation. Lets choose a = 1, b = 1 and c = 0 so we have by Theorem 2

x u(x, t) = f(s) where s = t Diﬀerentiating u = f(s) ﬁrst with respect to x twice, then with respect to t twice. we obtain

1 u = f 00(s) xx t2 2x x2 u = f 0(s) + f 00(s) tt t3 t4

16 Therefore, the nonlinear wave equation reduces to

2x x2 f 0(s) + f 00(s) − f(s)f 00(s) = 0 t t2

Equivalently,

(s2 − f(s))f 00(s) + sf 0(s) = 0

We can conclude from Example 3 and Example 4 that both the nonlinear hate equation and the nonlinear wave equation has been transformed into a second order ordinary equation. Example 5 In this example we consider The Korteweg-de Vries equation

ut + 6uux + uxxx = 0

applying the following stretching transformation

x¯ = ax

t¯= bt

u¯ = cu we get

c c c ∂ u c ∂ u ∂(∂(∂ u)) u¯¯ + 6¯uu¯ +u ¯ = + 6 u + t x¯ x¯x¯x¯ ∂bt ∂ax (∂ax)(∂ax)(∂ax)

c−b 2c−a c−3a = ut + 6uux + uxxx the above gives us the invariant condition is

17 c − b = 2c − a = c − 3a

so we have b = 3a and c = −2a choosing a = 1 and by using theorem 2 we say

−2 u(x, t) = t 3 f(s)

−1 where s = xt 3 note that

−2 −5 1 x 0 u = t 3 f(s) − f (s) t 3 3 t2 −1 0 ux = t f (s)

−5 000 uxxx = t 3 f (s)

Then the Korteweg-deVries equation reduces to

−5 000 0 h si 2 t 3 f (s) + f (s) 6f(s) − − f(s) = 0 3 3

h si 2 f 000(s) + f 0(s) 6f(s) − − f(s) = 0 3 3

18 3 The Lie Variables

In the previous section we showed how we reduce a partial differential equation to an ordinary differential equation under the one parameter family of transformations in order to find a self similar solution; however, it is not always easy to solve analytically a given ordinary differential equation of the second order or higher. A solution to this problem is to reduce the second order ordinary differential equation to a first order ordinary differential equation, and we do that by taking advance of its invariance properties as the same way we did with the invariance properties of the partial differential equation.

Defnation 3 An ordinary diﬀerential equation of second order written in the form

f 00 − G(s, f, f 0) = 0 (17)

is said to be constant conformally invariant under the following stretching transformation

s¯ = s (18) f¯ = bf where ∈ I if

d2f¯ df¯ − G(¯s, f,¯ ) = A(s)(f 00 − G(s.f.f 0)) (19) ds¯2 ds¯

for some function A, for all ∈ I.

19 Theorem 3 If the second order ordinary differential equation (17) is constant conformally invariant under the stretching transformation defined by (18), then (17) can be reduced to a first order ordinary differential equation of the form

dv H(u, v) − (b − 1)v = (20) du v − bu

where u = φ(s, f) and v = ψ(s, f, f 0) are ﬁrst integrals of the characteristic system

ds df = = df 0(b − 1)f 0 (21) s bf

and H is some ﬁxed function depending on G. The quantities u and v are called the Lie Variables.

proof Consider the second order ordinary diﬀerential equation(17)

f 00 − G(s, f, f 0) = 0

by invariance under the transformation (18), we have

d2f¯ d(dbf) 2 = ds¯ (ds)(ds) (22) = b−2f 00 similarly,

20 df¯ dbf = ds¯ ds (23) = b−1f 0

d2f¯ ¯ df¯ by substituting (20) and(21) to ds¯2 − G(¯s, f, ds¯ ), we obtain

d2f¯ df¯ − G(¯s, f,¯ ) = b−2f 00 − G(s, bf, b−1f 0) ds¯2 ds¯

from (17) we have f 00 = G(s, f, f 0), and since (17) is constant conformally invariant under the given transformation, then we get

G(s, bf, b−1f 0) = b−2G(s, f, f 0)

for all ∈ I. By taking the derivative with the respect to , we have

b−1 b−2 0 b−3 sGs + b fGf + (b − 1) f Gf 0 = (b − 2) G

Set = 1, we obtain

0 sGs + bfGf + (b − 1)f Gf 0 = (b − 2)G

It is a partial diﬀerential equation for G. It can be solved by the method of characteristics. The characteristics system is

ds df df 0 dG = = = s bf (b − 1)f 0 (b − 2)G

21 and the ﬁrst integrals are

f s f 0 sb−1 G sb−2 Let

f u = φ(s, f) = s f 0 v = ψ(s, f, f 0) = sb−1

Thus

G = sb−2H(u, v)

For some function H. and u, v are Lie variables to ﬁnd du and dv, we have

du f 0 bf = − ds sb sb+1 v − bu = s and we also have

dv f 00 bf 0 f 0 = − + ds sb−1 sb sb H(u, v) − bv + v = s

Thus, dv H(u, v) − (b − 1)v = du v − bu

22 Example 7 In this example, we are going to resolve the nonlinear heat equation

uut − uxx = 0, x > 0, t > 0 and we are going to consider the following initial condition and boundary conditions:

u(x, 0) = 0, x > 0

lim u(x, t) = 0, t > 0 (24) x→∞

ux(0, t) = −C, t > 0 where C > 0 is a ﬁxed constant. we consider the same stretching transformation

x¯ = ax

t¯= bt

u¯ = cu we got that b 1− 2a x u(x, t) = t b f ta

we will not select particular values of a, b, and c since we have to satisfy the initial and boundary conditions

First consider the boundary condition ux(0, t) = −C , note that

1−(2a/b)−a b−1 0 b a ux = t bx f (x /t )

23 since we do not want the left side to vanish, so we have to choose b = 1 to get t1−3af 0(x/ta) = −C atx = 0 (25) so from the ﬁrst boundary condition we determined the value of b; however,

1 the left side of the of eq (24) cannot depend on t, so we must have a = 3 Therefore, we can say that we found the values of a, b, and c by the ﬂux condition at x = 0, and we have

1 x¯ = 3 x

t¯= t

1 u¯ = 3 u and we obtain

1 u(x, t) = t 3 f(s) (26)

where

x s = 1 t 3

note that

1 −2 −1 0 ut = (t 3 f(s) − t xf (s)) 3 (27) −1 00 uxx = t 3 f (s)

by substituting the values of u, ut, and uxx into the nonlinear heat equation, we obtain the following ordinary diﬀerential equation

24 1 f(f − sf 0) − f 00 = 0 0 < s < ∞ (28) 3

and the ﬂux condition ux(0, t) = −C t > 0 becomes

f 0(0) = −C (29)

Now we want to study the initial condition u(x, 0) = 0 x > 0 Note that as t → 0 we get s → ∞ for ﬁxed x so the initial condition becomes

lim f(s) = 0 (30) s→∞

Finally the boundary condition limx→∞ u(x, t) = 0, t > 0 becomes

1 lim t 3 f(s) = 0, t > 0 (31) s→∞ which is the same as eq (30) we conclude the nonlinear heat equation with the initial and boundary conditions (24) transformed under the family of transformation into

1 f(f − sf 0) − f 00 = 0 0 < s < ∞ (32) 3

f 0(0) = −C where C is constant (33) lim f(s) = 0 s→∞

However, the (32) is a formidable equation, which cannot be solve analytically. Therefore, we need to reduce it to a ﬁrst order ordinary diﬀerential

25 equation. Consider the following transformation

s¯ = s

f¯ = bf then

df¯ dbf = ds¯ ds = b−1f 0

d2f¯ d2bf = ds¯2 d2s = b−2f 00 and we have

1 df¯ d2f¯ 1 f¯(f¯− s¯ ) − = 2bf(f − sf 0) − b−2f 00 3 ds¯ ds¯2 3

For invariance b − 2 = 2b, so we have b = −2. From theorem 3 we get the values u and v

u = s2f

v = s3f 0

Diﬀerentiating with respect to s

du v 2u = + ds s s dv u2 uv 3v = − + ds 2 s s

26 therefore, the ﬁrst order equation is

dv u2 − uv + 3v = (34) du v + 2u

We have reduced the second order ordinary diﬀerential equation (32) to a ﬁrst order ordinary equation in the uv plane which called the Lie Plane in terms of the variables u and v.

27 4 The General Similarity Method

In Section 2, we studied the reducing of the partial diﬀerential equation to ordinary diﬀerential equation under stretching transformations. Also, we saw in Section 1 that stretching transformation is a kind of the local Lie groups of transformations on R3. In this section, we are going to develop more general local Lie groups.

3 3 Consider a general one parameter family of transformation T : R → R deﬁned as follow: x¯ = Φ(x, t, )

t¯= Ψ(x, t, ) (35)

u¯ = Ω(x, t, u, )

where is a real parameter that varies over an open interval || < 0 containing 0, and Φ, Ψ, and Ω are arbitrary functions. When = 0 assume that the transformation (35) is the identity transformation I given by:

x¯ = Φ(x, t, 0) = x

t¯= Ψ(x, t, 0) = t

u¯ = Ω(x, t, u, o) = u

Furthermore, for any 1 and 2 in || < 0 there is an 3 in || < 0 such

that T1 T2 = T3 , which means the local closure property satisﬁes. Finally,

1 for each in || < 0 there exists such that TT−1 = T−1 T = I; hence,

−1 T = T−1 .

Therefore, we can say the one parameter family of transformation T deﬁned by (35) is a local Lie group. By using Taylor’s expanding we can expand the right sides of (35) as follow:

28 ∂ x¯ = Φ(x, t, ) = Φ(x, t, 0) + Φ(x, t, 0) + o(2) ∂ ∂ t¯= Ψ(x, t, ) = Ψ(x, t, 0) + Ψ(x, t, 0) + o(2) (36) ∂ ∂ u¯ = Ω(x, t, u, ) = Ω(x, t, u, 0) + Ω(x, t, u, 0) + o(2) ∂

Then the Inﬁnitesimal Transformation

x¯ = x + ξ(x, t) + o(2)

t¯= t + τ(x, t) + o(2) (37)

u¯ = u + ω(x, t, u) + o(2) where

∂ ξ(x, t) = Φ(x, t, 0) ∂ ∂ τ(x, t) = Ψ(x, t, 0) (38) ∂ ∂ ω(x, t, u) = Ω(x, t, u, 0) ∂

ξ, τ, and ω are called the generators of transformation (35). Moreover, these generators are determined uniquely by the local Lie group T deﬁned by (35). On the other hand, we can determine the local Lie group by the generators by solving the initial value problem:

dx¯ dt¯ du¯ = = = d ξ(¯x, t¯) τ(¯x, t¯) ω(¯x, t,¯ u¯) where

x¯ = x

t¯= t at = 0

u¯ = u

29 Example 8 Find the local Lie group by the following generators:

ξ = 2x + t

τ = t

ω = u + 1

by substituting into the system, we get

dx¯ dt¯ du¯ = = = d 2¯x + t¯ t¯ u¯ + 1

The three ﬁrst integrals are

x¯ + t¯ = c t¯2 1 t¯ = c u¯ + 1 2

ln(¯u + 1) = + c3

to ﬁnd the values of c1, c2, and c3 we apply the initial conditionx ¯ = x, t¯= t, andu ¯ = u at = 0 we obtine

x + t c = 1 t2 t c = 2 u + 1

c3 = ln(u + 1) so the three equations become

30 x¯ + t¯ x + t = t¯2 t2 t¯ t = u¯ + 1 u + 1 ln(¯u + 1) = + ln(u + 1) yields

x¯ = (x + 1)e2 − te

t¯= te

u¯ = (u + 1)e − 1 which deﬁne the local Lie group.

The group of transformation T deﬁned by (35) maps for each ﬁxed a surface in xtu space having equation

u = g(x, t)(x, t) ∈ D

to a surface inx ¯t¯u¯ having equation

u¯ =g ¯(¯x, t¯) ∈ D¯

where

D¯ = {(¯x, t¯)s.t.x¯ = Φ(x, t, ), t¯= Ψ(x, t, ), (x, t) ∈ D}

31 To ﬁndg ¯ we invert the transformation

x¯ = Φ(x, t, )

t¯= Ψ(x, t, )

on D to get

x = x(¯x, t¯)

t = t(¯x, t¯)

Theng ¯ is

g¯(¯x, t¯) = Ω(x(¯x, t¯), t(¯x, t¯), g(x(¯x, t¯), t(¯x, t¯)), ) (39)

Definition 4 The first order partial differential equation

F (x, t, u, ux, ut) = 0, (x, t) ∈ D (40)

is constant conformally invariant under the local Lie group (35) if and only if ∂ F (¯x, t,¯ u,¯ u¯ , u¯¯) | = kF (x, t, u, u , u ) (41) ∂ x¯ t =0 x t

For some constant k. If k = 0, the partial diﬀerential equation is said to be absolutely invariant.

If k = k(x, t, u, ux, ut), then the partial diﬀerential equation is conformally invariant.

32 By expanding the left side of (40) we get

∂ ∂F ∂ ∂F ∂ F (¯x, t,¯ u,¯ u¯ , u¯¯) | = Φ(x, t, 0) + Ψ(x, t, 0) ∂ x¯ t =0 ∂x ∂ ∂t ∂ (42) ∂F ∂ ∂F ∂u¯x¯ ∂F ∂u¯t¯ + Ω(x, t, u, 0) + |=0 + |=0= kF ∂u ∂ ∂u¯x¯ ∂ ∂u¯t¯ ∂

Let

∂u¯ π = π(x, t, u, u , u ) = x¯ | x t ∂ =0 (43) ∂u¯¯ χ = χ(x, t, u, u , u ) = t | x t ∂ =0 substituting the values of ξ, τ, ω, π, and χ from (38) and (43) into (42) we get

Fxξ + Ftτ + Fuω + Fu¯x¯ π + Fu¯t¯χ = kF (44)

which is the condition for invariance in terms of the generators of local Lie group. We can write invariant condition of the partial equation equation F = 0 as

VFˆ = kF (45)

where Vˆ is called the Lie derivative operator express as

∂ ∂ ∂ ∂ ∂ Vˆ = ξ + τ + ω + π + χ ∂x ∂t ∂u ∂u¯x¯ ∂u¯t¯

We know if u = g(x, t) is a solution to (40) thenu ¯ = g(¯x, t¯) is solution to

F (¯x, t,¯ u,¯ u¯x¯, u¯t¯) = 0 (46)

33 It can be shown thatu ¯ =g ¯(¯x, t¯) is also a solution to (46); however, it is not always true that g(¯x, t¯) =g ¯(¯x, t¯). In order to ﬁnd a similarity solutions among the invariant surfaces, where u = g(x, t) is a solution to (40), we must have

g(¯x, t¯) =g ¯(¯x, t¯) (47)

by substituting from (39) into (47) we get

g(¯x, t¯) = Ω(x(¯x, t¯), t(¯x, t¯), g(x(¯x, t¯), t(¯x, t¯)), ) by taking the derivative with respect to then set = 0 we get

ξgx(x, t) + τgt(x, t) = ω(x, t, g) (48)

where

∂ ξ(x, t) = Φ(x, t, 0) ∂ ∂ τ(x, t) = Ψ(x, t, 0) ∂ ∂ ω(x, t, u) = Ω(x, t, g, 0) ∂

Equation (48) called the invariant surface condition, which is the ﬁrst order partial diﬀerential equation in g and we can solve it by the characteristics method. The characteristic system is

dx dt dg = = (49) ξ(x, t) τ(x, t) ω(x, t, g)

34 If we integrate the ﬁrst pair we get

s = s(x, t) = constant

which is the similarity variable. Let x = X(s, t), then the second pair becomes

dt dg = τ(X(s, t), t) ω(X(s, t), t, g)

by integration we get

w(s, t, g) = constant then the general solution to (48) is

F (s, w(s, t, g)) = 0 (50)

then we can get the value of g where it is the self-similar solutions for (40). Now consider the general second order partial diﬀerential equation in the dependent variable u and independent variables x and t

F (x, t, u, ux, ut, uxx, utt) = 0 (51)

Now let consider the one parameter family of transformation T which takes the xtu space to itself deﬁned as

35 x¯ = x + X

t¯= t + T (52)

u¯ = u + U which transform takes the solution u = g(x, t) intou ¯ =g ¯(¯x, t¯), and (51) is invariant under the transformations and we have

F (¯x, t,¯ u,¯ u¯x¯, u¯t¯, u¯x¯x¯, u¯t¯t¯) = 0 (53)

we also haveu ¯ = g(¯x, t¯) so

u¯ = u + U = g(¯x, t¯) = g(x + X, t + T ) (54) ∂g ∂g = g(x, t) + X + T + o(2) ∂x ∂t collecting the terms of we get

∂u ∂u X + T = U (55) ∂x ∂t where it is called the invariant surface condition as before we solve it by the method of characteristic, and the characteristic system is

dx dt du = = X T U

by integration the ﬁrst pair of (55) we get

s(x, t) = constant

36 X where T is a function in x and t , so s is the similarity variable. now if we integrate the second pair of (55) we have

h(x, t, u) = constant

The general solution of (55) is

h(x, t, u) = f(s)

where f is an arbitrary function. solving for u we get

u = F (x, t, f(s)) (56)

In order to determined f(s) we substitute from (56) into (51), and we get an ordinary differential equation in f(s), where each solution for the ordinary differential equation is a similarity solution for the partial differential equation (51). Now we need to determine the generators X(x, t, u),T (x, t, u), and U(x, t, u) of the group of transformation T. Therefore, we have to show how the derivatives transform under the infinitesimal transformation T defined by (52).

37 ∂x ∂ = [¯x − X(x, t, u) + o(2)] ∂x¯ ∂x¯ ∂ = 1 − (X(x, t, u)) + o(2) ∂x¯ (57) ∂X ∂X ∂u ∂x = 1 − + + o(2) ∂x ∂u ∂x ∂x¯

2 = 1 − (Xx + Xuux) + o( )

Now we take the derivative of x with respect to t¯

∂x ∂ = [¯x − X(x, t, u) + o(2)] ∂t¯ ∂t¯ ∂ = − (X(x, t, u)) + o(2) (58) ∂t¯ 2 = −(Xt + Xuut) + o( )

Similarly, we take the derivative of t with respect to t¯ andx ¯

∂t = 1 − (T + T u ) + o(2) (59) ∂t¯ t u t

∂t = −(T + T u ) + o(2) (60) ∂x¯ x u x

We Know

∂u ∂x ∂t = u + u (61) ∂x¯ ∂x¯ x ∂x¯ t

by substitute from (57) and (58) into (61) we get

∂u = [1 − (X + X u )] u + [−(X + X u )] + o(2) ∂x¯ x u x x t u t

38 Now we get the following operator

∂ ∂ ∂ = [1 − (Xx + Xuux)] − (Tx + TuUx) ∂x¯ ∂x ∂t (62) ∂ + [(1 − (X + X u ))u − (T + T u )u ] x u x x x u x t ∂u we are going to keep terms up to order only, so the derivative ofu ¯ with respect tox ¯ is

∂u¯ ∂ = [u + U(x, t, u)] ∂x¯ ∂x¯

= [1 − (Xx + Xuux)] ux − (Tx + Tuux)ut (63)

2 + [Ux + Uuux] + o( ) or equivalently,

∂u¯ = u + V˜ + o(2) (64) ∂x¯ x x

where

˜ 2 Vx = Ux + Uuux − Xxux − Xuux − Txut − Tuuxut (65)

Similarly, we obtain

∂u¯ = u + V˜ + o(2) (66) ∂t¯ t t

where

˜ 2 Vt = Ut + UuUt − Xtux − Xuutux − Ttut − Tuut (67)

39 From (64) we have

∂2u¯ ∂ = [u + V˜ ] ∂x¯2 ∂x¯ x x

If we use (62) we obtain the second derivative ofu ¯ with respect tox ¯

∂2u¯ = u + V˜ + o(2) (68) ∂x¯2 xx xx

where

˜ 2 Vxx =Uxx + (2Uxu − Xxx)ux − Txxut + (Uuu − 2Xxu)ux

3 2 − 2Txuuxut − Xuuux − Tuuux ut + (Uu − 2Xx)uxx (69)

− 2Txuxt − 3Xuuxxux − Tuuxxut − 2Tuuxtux

Similarly, we get

∂2u¯ = u + V˜ + o(2) (70) ∂x∂¯ t¯ xt xt

where

˜ Vxt =Uxt + (Utu − Xtx)ux + (Uxu − Ttx)ut

2 2 2 + (Uuu − Xxu − Ttu)uxut − Xtuux − Xuuux ut − Txuut (71) 2 − Tuuuxut + (Uu − Xx − Tt)uxt − Xtuxx − 2Xuuxuxt

− Xuutuxx − Txutt − Tuuxutt − 2Tuutuxt

Also we have

40 ∂2u¯ = u + V˜ + O(2) (72) ∂t¯2 tt tt

˜ ˜ where Vtt is obtained from Vxx by interchanging x and t as well as X and T as the following

˜ 2 Vtt =Utt + (2Utu − Ttt)ut − Xttux + (Uuu − 2Ttu)ut

3 2 − 2Xtuutux − Tuuut − Xuuut ux + (Uu − 2Tt)utt (73)

− 2Xtuxt − 3Tuuttut − Xuuttux − 2Xuuxtut

The partial differential equation (51) is invariant under the group of transformation T defined as (52) if (53) is satisfied for any solution u = g(x, t) of (51) The condition for this is

VFˆ = KF when F = 0 (74)

For some constant k. If k = 0, the partial diﬀerential equation is said to be absolutely invariant.

If k = k(x, t, u, ux, ut), then the partial diﬀerential equation is conformally invariant. And

∂ ∂ ∂ ∂ ∂ Vˆ =X + T + U + V˜ + V˜ ∂x ∂t ∂u x ∂u t ∂u x t (75) ˜ ∂ ˜ ∂ ˜ ∂ + Vxx + Vxt + Vtt ∂uxx ∂uxt ∂utt

By substituting (65) - (73) into (74), by using (51), then equating the terms not involving derivatives of u and the coeﬃcients of like derivative terms in

41 u to zero, we obtain a system of linear partial differential equations to find the generators X,T and U. We use the invariant surface condition (55) after we get X,T and U in order to obtain the similarity form of solution. These results can be generalized to systems of partial differential equations with arbitrary number of dependent and independent variables.

42 5 The Fokker-Plank Equation

The Fokker-Planck equation was named after Adriaan Fokker (1887 - 1972) who was a Dutch physicist and musician, and Max Planck (1858 - 1947) who was a German theoretical physicist, originated quantum theory which won him the Nobel Prize in Physics in 1918. They used it to describe the Brownian motion of particles. The Fokker-Planck equation is a powerful tool in statistical physics and in the study of fluctuations some in physical and biological systems. It is con- nected with the theory of stochastic differential equations. A solution to the Fokker-Planck equation represents the probability density for a particle posi- tion or velocity which motion is described by a corresponding It¯o stochastic differential equation. In one spatial dimension x, for an It¯o process given by the stochastic differential equation

q (1) (2) dXt = D (Xt, t)dt + 2D (Xt, t)dWt

(1) (2) where D (Xt, t) is the drift coefficient and D (Xt, t) is the diffusion coefficient, then the Fokker-Planck equation for the probability density distribution function W (x, t) of the random variable Xt is :

∂ ∂ ∂2 W (x, t) = − [D(1)(x, t)W (x, t)] + [D(2)(x, t)W (x, t)] (76) ∂t ∂x ∂x2

where W (x, t) is the probability density distribution function of the random

(1) (2) variable Xt, D (x, t) is the drift coefficient, and D (x, t) is the diffusion coefficient. The drift coefficient represents the external force acting on the particle, while the diffusion coefficient accounts for the effect of fluctuation. W (x, t) is a probability normalized distribution function, i.e. R W (x, t)dx = 1 for t ≥ 0.

43 6 Similarity Solution Under A Group of Stretching Transformation

In this section, we are going to solve the Fokker-Planck equation (76) by similarity method as introduced in Section 2. Consider the Fokker-Planck equation

(1) (2) Wt = −(D W )x + (D W )xx (77)

and T an inﬁnitesimal transformation deﬁned by T(x, t) = (¯x, t¯), where

x¯ = ax

t¯= bt

Thus, under this group of transformations ( varies), we have the following invariant relation:

W¯ (¯x, t¯) = cW (x, t)

D¯ (1)(¯x, t¯) = dD(1)(x, t)

D¯ (2)(¯x, t¯) = eD(2)(x, t) where , a, b, c, d, and e are real parameters. We want to determine if the transformed diﬀerential equation (77) in terms of new variables has the same form as the original equation (77). First note

44 ¯ c−b Wt¯ = Wt ∂ [D¯ (1)(¯x, t¯)W¯ (¯x, t¯)] = D¯ (1)W¯ + D¯ (1)W¯ ∂x¯ x¯ x¯ d+c−a (1) (1) = [D Wx + D xW ] ∂ = d+c−a [D(1)W ] (78) ∂x ∂2 ∂ ∂ [D¯ (2)(¯x, t¯)W¯ (¯x, t¯)] = ( [D¯ (2)(x, t)W¯ (x, t)]) ∂x¯2 ∂x¯ ∂x¯ ∂ = [e+c−aD¯ (2)W¯ ] ∂x¯ ∂2 = e+c−2a [D(2)W ] ∂x2

By substituting from (78) into (77) we have

2 b−c a−c−d ∂ (1) 2a−c−e ∂ (2) W¯ ¯ = − [D¯ (¯x, t¯)W¯ (¯x, t¯)] + [D¯ (¯x, t¯)W¯ (¯x, t¯)] (79) t ∂x¯ ∂x¯2

so if b = a − d = 2a − e then (79) has the same form as (77). Then by using Theorem 2 form Section 2 the similarity variable is

xb x a s = = ; α = , a, b 6= 0 (80) ta tα b

The general scaling form of the probability density function W (x, t) is W (x, t) =

tδ ( xλ )y(s) where δ and λ are real parameters and y(s) is a scale-invariant function under T. Thus, we have

tδ W¯ (¯x, t¯) = cW (x, t) = c( )y(s) xλ −bδt¯δ = c( )y(s) −aλx¯λ −δ −δ −c α t¯ α = a ( )y(s) λx¯λ

−c δ then a = λ − α .

45 αc Let λ = 0, we get δ = a . Then

αc W (x, t) = t a y(s). (81)

We have from (80) that

x s = tα dx = tαds

The normalization of the distribution function by using the above relation is

Z Z α(1+ c ) W (x, t)dx = [t a y(s)]ds = 1

For the above relation to hold for all t ≥ 0, the power of t should vanish, and we must have c = −a then, we have

W (x, t) = t−αy(s). (82)

Now let D(1)(x, t) = tδ1 ρ (s) where ρ (s) is a scale-invariant function, then xλ1 1 1

¯δ1 (1) −d ¯ (1) −d t D (x, t) = D (¯x, t¯) = ( )ρ1(s) x¯λ1 −bδ1 δ1 −d t = ( )ρ1(s) −aλ1 xλ1

and we must have d = bδ1 − aλ1.

a Setλ1 = 0 then we have δ1 = α − 1 where b = a − d and α = b . So, we obtain

(1) α−1 D (x, t) = t ρ1(s). (83)

46 Similarly, we set D(2) = ( tδ2 )ρ (s), where ρ (s) is a scale-invariant function, xλ2 2 2 then

¯δ2 (2) −e ¯ (2) −e t D (x, t) = D (¯x, t¯) = ( )ρ1(s) x¯λ2 −bδ2 δ2 −e t = ( )ρ2(s) −aλ2 xλ2

similarly we have e = bδ2 − aλ2.

a Setλ2 = 0 then we have δ2 = 2α − 1 where e = 2a − d and α = b and we get

(2) 2α−1 D (x, t) = t ρ2(s) (84)

From (81) we have

αc αc −1 −1 0 W = t a y(s) − αt sy (s) t a αc −α 0 Wx = t a y (s) (85)

αc −2α 00 Wxx = t a y (s).

From (83) we have

(1) −1 0 D x = t ρ1(s) (86)

and we get from (84) (2) α−1 0 D x = t ρ2(s) (87) (2) −1 00 D xx = t ρ2(s)

From (77) we have

47 (2) (1) (2) (1) (2) Wt = (2D x − D )Wx + (D xx − D x)W + D Wxx (88) substituting from (81), (83), (84), (85), (86), and (87) into the above equation (88), we get

αc αc −1 −1 0 α−1 0 α−1 αc −α 0 t a y(s) − αt sy (s) = [2t ρ (s) − t ρ (s)]t a y (s) a 2 1 αc αc −1 00 −1 0 a 2α−1 a −2α 00 +[t ρ2(s) − t ρ1(s)]t y(s) + t ρ2(s)t y (s).

So, the Fokker-Planck equation is reduced to the ordinary diﬀerential equation

00 0 0 00 0 ρ2(s)y (s) + [2ρ2(s) − ρ1(s) + αs]y (s) + [ρ2(s) − ρ1(s) + α]y(s) = 0 (89) and by integrating (89) we get

0 0 ρ2(s)y (s) + [ρ2(s) − ρ1(s) + αs]y(s) = c (90) where c is a constant. We know ∂ ∂ W (x, t) = − J(x, t) ∂t t ∂x where J(x, t) is the associated probability current density. By substituting in (77), we have

∂ J(x, t) = D(1)(x, t)W (x, t) − [D(2)(x, t)W (x, t)]. ∂x

By using (80), (82), (86), and (87), we have

−1 0 0 J(x, t) = t [(ρ1(s) − ρ2(s))y(s) − ρ2(s)y (s)]. (91)

By substituting from (90) into (91) we get

48 J(x, t) = t−1[αsy(s) − c] = t−1[αxW (x, t) − c] where we have used (80) and (82), to get the above result. Consider the natural boundary conditions

J(x, t)|boundary = W (x, t)|boundary = 0 which implies c = 0 and J(x, t) is proportional to W (x, t) and x, i.e.,

J(x, t) = t−1αsy(s) = t−1αxW (x, t) since c = 0 equation (90) becomes

0 0 ρ2(s)y (s) + [ρ2(s) − ρ1(s) + αs]y(s) = 0.

We solve for y(s), we get

y0(s) ρ0 (s) − ρ (s) + αs = 2 1 . y(s) ρ2(s)

Then we obtain

ρ0 (s)−ρ (s)+αs R 2 1 ds y(s) = e ρ2(s)

where ρ2(s) 6= 0. So the probability density function W (x, t) is given by

R W (x, t) = t−αe f(s)ds

0 ρ2(s)−ρ1(s)+αs x where f(s) = , ρ2(s) 6= 0 and s = α . ρ2(s) t We can obtain an exact solution to the Fokker-Planck equation if for any

49 integrable function f(s), we can ﬁnd ρ1(s) and ρ2(s), from equations (83) and (84), such as

1−α (1) ρ1(s) = t D (x, t)

1−2α (2) ρ2(s) = t D (x, t)

Let’s consider some cases:

Let us take ρ1(s) and ρ2(s)

ρ1(s) = a1s + a2

ρ2(s) = a3 where a1, a2, and a3 are real constants. From this choice of ρ1(s) and ρ2(s), we have

x D(1)(x, t) = a + a tα−1 1 t 2 (2) 2α−1 D (x, t) = a3t

0 ρ2(s)−ρ1(s)+αs x by substituting into f(s) = , ρ2(s) 6= 0 and s = α ρ2(s) t we get y(s) is

1 s2 y(s) = exp{ [(a1 − α) + a2s]}, a1 6= α a3 2 a2 y(s) = exp{ s}, a1 = α. a3

When a1 6= α, we get the following normalized solution,

r α 2 α − a1 α − a1 a2t W (x, t) = 2α exp − 2α (x − ) (92) 2πa3t 2a3t α − a1

where either (a3 > 0, a1 < α) or (a3 < 0, a1 > α) must be satisﬁed.

The Figure 1 shows the evolutions of the probability density in solution (92)

50 Figure 1: W (x, t) versus x and t for solution (92) with α = 1, a1 = .5, a2 = 1, and a3 = 1

for a set of the parameters, where a1 6= α. We can see that the peak of the probability distribution W (x, t) moves to the right as time increases. This is due to the presence of the drift force.

Figure 2 shows the probability current J(x, t) corresponding to the solution (92) with the same set of the parameters as in Figure 1.

Figure 2: J(x, t) versus x and t for solution (92) with α = 1, a1 = .5, a2 = 1, and a3 = 1

51 When a1 = α, in this case, the normalized solution is

a2 a2 W (x, t) = | α | exp{ α x} (93) a3t a3t

where it is valid in x ≥ 0 for ( a2 ) < 0; x ≤ 0 for ( a2 ) > 0. a3 a3

Figure 3: W (x, t) versus x and t for solution (93) with α = 1, a2 = −6, and a3 = 2

Figure 3 displays the evolutions of the probability density in solution (93) for a set of parameters, where a1 = α. The time evolution of W (x, t) in Figure 3 shows the decrease of the concentration near the origin and the spread to the area away from the origin with time.

1 Let’s take a2 = 0 and consider α = 2 so we have

x D(1)(x, t) = a 1 t (2) D (x, t) = a3 with the probability distribution

52 r1 − 2a 1 − 2a W (x, t) = 1 exp{−( 2 )x2} (94) 4πt 4a3t

As a2 = 0, the peak of the probability distribution will not move with time, as shown in Figure 4.

Figure 4: W (x, t) versus x and t for solution (94) with a1 = .25, a2 = 0, and a3 = 1

53 7 A General Similarity Solution For The Fokker-Planck Equation

We want to discuss a general similarity method for solving the Fokker-Planck equation with time-dependent drift and diﬀusion coeﬃcients. We have the Fokker-Planck equation as following

∂ ∂ ∂2 W (x, t) = − [D(1)(x, t)W (x, t)] + [D(2)(x, t)W (x, t)] (95) ∂t ∂x ∂x2

This can be rewritten as

(1) (1) (2) (2) (2) Wt = −(D Wx + D xW ) + (D Wxx + 2D xWx + D xxW )

(2) (1) (2) (1) (2) Wt = (2D x − D )Wx + (D xx − D x)W + D Wxx (96)

Let the function F be deﬁned by above equation (96) as follows

F (x, t, W, Wx,Wt,Wxx,Wxt,Wtt) (97) (2) (1) (2) (1) (2) = Wt − (2D x − D )Wx − (D xx − D x)W − D Wxx = 0.

The invariance condition that needs to be satisfied is VFˆ = KF where K is a constant, and Vˆ is the infinitesimal operator defined by

ˆ ˜ ˜ ˜ VF =ζFx + τFt + γFW + VxFWx + VtFWt + VxxFWxx (98) ˜ ˜ + VxtFWxt + VttFWtt = KF

By taking the derivative of (97) with respect to x, t, Wxt,Wtt,Wt,Wx,W and

Wxx we get

54 (2) (1) (2) (1) (2) Fx = [−(2D xx − D x)Wx − (D xxx − D xx)W − D xWxx]

(2) (1) (2) (1) (2) Ft = [−(2D xt − D t)Wx − (D xxt − D xt)W − D tWxx]

FWxt = 0

FWtt = 0

FWt = 1

(2) (1) FWx = −[2D x − D ]

(2) (1) FW = −[D xx − D x]

(2) FWxx = −D

˜ ˜ ˜ by substituting these and the values of Vx, Vt and Vxx as deﬁned in Section 4

55 into (98) we get

ˆ (2) (1) (2) (1) (2) VF =ζ[−(2D xx − D x)Wx − (D xxx − D xx)W − D xWxx]

(2) (1) (2) (1) (2) + τ[−(2D xt − D t)Wx − (D xxt − D xt)W − D tWxx]

(2) (1) (2) (1) − γ[D xx − D x] − [2D x − D ][γx + γW Wx − ζxWx

2 − τxWt − τwWxWt − ζW Wx ] + γt − ζtWx + γW Wt − τtWt

2 (2) − ζW WxWt − τW Wt − D [γxx + (2γxW − ζxx)Wx − τxxWt

2 + (γWW − 2ζxW )Wx − 2τxW WxWt + (γW − 2ζx)Wxx − 2τxWxt

3 2 − 3ζW WxxWx − ζWW Wx − τWW Wx Wt − τW WxxWt − 2τW WxtWx]

(2) (1) (2) (1) (2) = K[Wt − (2D x − D )Wx − (D xx − D x)W − D Wxx] (99) then equating the terms not involving derivatives of W and the coeﬃcients

2 3 2 of W , Wx, Wt, WxWt, Wx , Wx , Wxt, WxtWx, and Wx Wt to zero. We get

(2) (1) (2) (1) (2) − γ[D xx − D x] − [2D x − D ]γx + γt − D γxx = 0 (100)

(2) (1) (2) (1) (2) (1) − ζ(D xxx − D xx) − τ(D xxt − D xt) = −K(D xx − D x) (101)

(2) (1) (2) (1) (2) (1) −ζ(2D xx − D x) − τ(2D xt − D t) − (2D x − D )(γW − ζx)

(2) (2) (1) − ζt − D (2γxW − ζxx) = −K(2D x − D ) (102)

(2) (1) (2) [2D x − D ]τx + γW − τt + D τxx = K (103)

56 (2) (2) (2) (2) − ζD x − τD t − D (γW − 2ζx) = −KD (104)

(2) (1) (2) [2D x − D ]τW − ζW + 2τxW D = 0 (105)

(2) (1) (2) [2D x − D ]ζW − D [γWW − 2ζxW ] = 0 (106)

− τW = 0 (107)

(2) D τWW = 0 (108)

(2) D ζWW = 0 (109)

(2) 2D τx = 0 (110)

(2) 3D ζW = 0 (111)

57 (2) D τW = 0 (112)

From (107) and (110) τx = τW = 0 then τ is function of t, so τ = τ(t).

Also, from (103) γW − τt = K which means γW is function of t, then γ has the form γ = (τt + K)W + γ1(x, t).

From (111) ζW = 0, then ζ is a function of x and t, i.e. ζ = ζ(x, t). We note that equation (100) can be integrated w.r.t x to get

Z (2) (1) (2) γ[D x − D ] + D γx = γtdx (113)

So, we can summaries the result as follows:

Theorem 4 For the Fokker-Planck equation (96), the generators of the local Lie group are given by τ = τ(t)

ζ = ζ(x, t)

γ = (τt + K)W + γ1(x, t) where τ, and ζ are satisfying the following PDEs

(2) (1) (2) (1) (2) (1) ζ(D xx − D x)x + τ(D xx − D x)t = K(D xx − D x) (114)

(2) (2) 2 (ζD )x + (τD )t − 3ζxD = 0 (115)

(2) (1) (2) (2) (1) (ζ(2D x − D ) − D ζx)x + (τ(2D x − D ))t (116) (2) (1) − (3D x − 2D )ζx + ζt = 0

58 and γ(x, t, W ) is satisfying the following equation

(2) (1) (2) (1) (2) [D xx − D x]γ + [2D x − D ]γx − γt + D γxx = 0 (117)

The over determined equations in (114), (115), and (116) give the speciﬁc forms for D(1) and D(2). Equation (114) be solved for τ and substituted into (115) to get a PDE involving only ζ. Once ζ is solved from that equation, equation (114) gives the functional form of τ. Now, these functional forms for τ and ζ are substituted into equation (116) that gives the requirement of

(1) (2) 0 D and D . Let us assume γ is a function of s only, then γt = stγ (s) and

γ = s γ0(s). Dividing these, we get γ = st γ . Now, assumaing st = f(t), x x t sx x sx R R we have γtdx = f(t)γxdx = f(t)γ + C1(t). Then we can integrate (117) to get

1 Z D(1) − f(t) Z Z f(t) − D(1) γ = Exp dx C (t, W )D(2)Exp dx dx D(2) D(2) 1 D(2) (118) The characteristic equation is then given by

dt dx dW = = (119) τ(t) ζ(x, t) γ(x, t, W )

If we select K = 0 then from (114) we get

(2) (1) −τ(D xx − D x)t ζ = (2) (1) (120) (D xx − D x)x and the similarity variable is obtained by integrating

(2) (1) (2) (1) (D xx − D x)x dx + (D xx − D x)t dt = 0 (121)

59 For a diﬀerentiable function F (x, t), its diﬀerential is

∂F ∂F dF = dx + dt ∂x ∂t

Comparing this with (121) we get dF = 0 or F = constant i.e.

(2) (1) D xx − D x = constant

Therefore, we can choose our similarity variable as

(2) (1) s(x, t) = D xx − D x (122) so (120) becomes −τs ζ = t (123) sx

Then, we have ζ, τ, and γ function of s, from (115) we get

3 ζ0D(2) (D(2) − D(1) ) − + ζD(2) xx x x (D(2) − D(1) ) x xx x x (124) (2) (1) (2) 0 (2) + (D xx − D x)tD τ + D tτ = 0 and from (116)

0 (2) (1) (2) (1) (2) (1) (2) 00 τ (D xx − D x)t(2D x − D ) + τ(2D x − D )t − D ζ (2) (1) 0 (2) (1) (2) (1) 2(2D x − D ) + ζ (D xx − D x)x(2D x − D ) − (2) (1) (D xx − D x) x (125) 1 (D(2) − D(1) ) + + D(2) xx x xx (D(2) − D(1) ) (2) (1) 2 xx x t [(D xx − D x)x] (2) (1) + ζ(2D x − D )x = 0

It is a system of 2 ODEs for ζ(s) and τ(s) if the coeﬃcients of ζ, τ and their derivatives are functions of s .

60 If ζ and τ are functions of s then equation (123) shows that st/sx is a function of s. Now we can rewrite (124) using (123) to obtain

0 (2) 3 (2) 0 (2) 0 (2)0 ζ D (s) sx − − τstD (s) + stD (s)τ + stD τ = 0 (126) sx

This equation reduces to

s 3 τ 0 + ζ0 x − (127) st sxst which shows that

sxst must be a function of s. Now let us reconsider (125) and following the same manipulation as above, we get D(2) s (2D(2) − D(1))τ 0 − ζ00 + ζ0 x (2D(2) − D(1)) x s s x t t (128) (2) (1) (2) 2(2D x − D ) 1 D sxx − + 2 + 2 = 0 stsx (st) st (sx) Analyzing the above equation we note that the following expressions must be functions of s:

(2) (2) (1) D sxx 2D x − D , , 2 st (sx) where s is given by (122). We can summarize our result as follows.

Theorem 5 The one-dimensional Fokker-Planck equation (96) has similarity solutions with the similarity variable given by (122) provided that the following expressions are functions of s:

(2) (2) (1) (2) (1) D (D xx − D x)xx 2D x − D , (2) (1) , (2) (1) 2 (D xx − D x)t ((D xx − D x)x)

61 and

(2) (1) (2) (1) (D xx − D x)t(D xx − D x)x.

This shows that we have an over-determined system on D(1) and D(2), thus only for certain choices of D(1) and D(2) we have similarity solutions. In this case, the generators of the local Lie group are solved by using (123), and (127) to get Z s3 − 3s d s τ = Exp x x t ds (129) 3st ds sx s Z s3 − 3s d s ζ = − t Exp x x t ds (130) sx 3st ds sx and γ is given by (118). Then, we can use the characteristic equation (119) to ﬁnd W . This conclude our investigation of general Lie group method on the one-dimensional Fokker-Planck equation. In the next section, we consider a direct approach to ﬁnd certain classes of similarity solutions.

62 8 The Clarkson- Kruskal Reduction Method

The idea of the Clarkson- Kruskal reduction method is to find a reduction of a given partial differential equation, which in this case is the Fokker-planck equation, by substituting W (x, t) = α(x, t) + β(x, t)γ(s(x, t)) into the partial differential equation, where s is the similarity variable. α, β, and γ are arbitrary functions to be determined. Then by demanding the result to be an ordinary differential equation of γ(s). We have

(1) (2) Wt = −(D W )x + (D W )xx (131)

(1) (1) (2) (2) (2) = −(D Wx + D xW ) + (D Wxx + 2D xWx + D xxW ),

(2) (1) (2) (1) (2) Wt = (2D x − D )Wx + (D xx − D x)W + D Wxx. (132)

Let

W = α(x, t) + β(x, t)γ(s(x, t)). (133)

Note that

0 Wt = αt + βtγ(s) + βγ (s)st

0 Wx = αx + βxγ(s) + βγ (s)sx

0 00 2 0 Wxx = αxx + βxxγ(s) + sβxγ (s)sx + βγ (s)sx + βγ (s)sxx

63 by substituting the values of W, Wt,Wx, and Wxx into (132) we get,

0 (2) (1) 0 αt + βtγ(s) + βγ (s)st = (2Dx − D )αx + βxγ(s) + βγ (s)sx

(2) (1) (2) 0 + (Dxx − Dx )α + βγ(s) + D αxx + βxxγ(s) + sβxγ (s)sx

00 2 0 + βγ (s)sx + βγ (s)sxx or equivalently,

(2) 2 00 (2) (1) (2) 0 D βsx γ (s) + −βst + (2D x − D )βsx + D (2βxsx + βsxx) γ (s)

(2) (1) (2) (1) (2) + −βt + (2D x − D )βx + (D xx − D x)β + D βxx γ(s)

(2) (1) (2) (1) (2) − αt + (2D x − D )αx + (D xx − D x)α + D αxx = 0. (134)

We want to choose α, β, and s such that the above equation is an ordinary diﬀerential equation in terms of γ(s).

Therefore, we select Γ1(s), Γ2(s), and Γ3(s) so that

(2) (1) (2) (2) 2 − βst + (2D x − D )βsx + D (2βxsx + βsxx) = D βsx Γ1(s) (135)

(2) (1) (2) (1) (2) (2) 2 −βt +(2D x −D )βx +(D xx −D x)β +D βxx = D βsx Γ2(s) (136)

(2) (1) (2) (1) (2) (2) 2 −αt+(2D x−D )αx+(D xx−D x)α+D αxx = D βsx Γ3(s). (137)

(2) We divide (135) by D βsx to get

64 (2) (1) −st 2D x D 2βx sxx (2) + (2) − (2) + + = sxΓ1(s) D sx D D β sx

If we integrate with respect to x, we get

−st (2) (1) Z + 2D x − D sx dx + ln s + 2 ln β = Γ (s) D(2) x 1

by dividing the above equation by 2 and solve for β to get

s +D(1)s R t x 1 1 (2) dx Γ (s) 2D sx 2 1 β = (2)√ e e . (138) D sx

1 Γ1(s) Since β has the form β(x, t) = β0(x, t)Ω(s), where Ω(s) = e 2 , we can absorb Ω(s) into γ(s) by taking Ω(s) = 1 and have β(x, t) = β0(x, t).

1 Γ1(s) Letting Ω(s) = e 2 = 1, so Γ1(s) = 0 and we get (138) reduces to

(1) R st+D sx 1 (2) dx 2D sx β = (2)√ e (139) D sx

i.e. the right hand side of (135) is zero, then the coeﬃcient of γ0(s) is zero. Hence, (134) reduces to

(2) 2 00 (2) (1) (2) (1) (2) D βsx γ (s) + −βt + (2D x − D )βx + (D xx − D x)β + D βxx γ(s)

(2) (1) (2) (1) (2) − αt + (2D x − D )αx + (D xx − D x)α + D αxx = 0. (140)

On the other hand, substituting for β from (139) and letting Γ2(s) = 1, we get an equation that gives similarity variable, s given by

65 Z (2) (1) (2) (1) − At + A( Bdx) + (2D x − D )(Ax + AB) + (D xx − D x)A t (2) 2 3 + D (Axx + 2AxB + ABx + AB ) = sx 2 (141) where

1 A = (2)√ D sx (1) st + sxD B = (2) . 2sxD Let’s consider some special case. First case If β = constant and α = constant, then (140) becomes

(2) 2 00 (2) (1) (2) (1) D βsx γ (s) + (D xx − D x)βγ(s) + (D xx − D x)α = 0 (142)

Without loss of generality, we can set α = 0 and β = 1, so (133) becomes W (x, t) = γ(s), and the above equation becomes

(2) 2 00 (2) (1) D sx γ (s) + (D xx − D x)γ(s) = 0 (143)

(2) (1) D xx−D x under the requirement that (2) 2 is a function of s only, then (143) D sx reduces to an ordinary diﬀerential equation. Let

(2) (1) D xx − D x 0 2 (2) 2 = l(f (s)) (144) D sx

where l is a constant, and f(s) is an invertible and diﬀerentiable function.

66 By integrating (144) with respect to x we get

r Z D(2) − D(1) Z xx x dx = s f 0(s)dx = f(s) lD(2) x

then our similarity variable is given by

r ! Z D(2) − D(1) s = f −1 xx x dx . (145) lD(2)

Second case Now let’s look to a more general case by choosing α = 0 and β = g(t), so we have W (x, t) = g(t)γ(s), where g(t) is an arbitrary diﬀerentiable function. Then (140) becomes

(2) 2 00 (2) (1) D βsx γ (s) + −βt + (D xx − D x)β γ(s) = 0. (146)

(2) 2 By dividing (146) by D βsx we get

(D(2) − D(1) ) β 00 xx x t (147) γ (s) + (2) 2 − (2) 2 γ(s) = 0. D sx D βsx

(2) (1) (D xx−D x) βt We must have (2) 2 − (2) 2 to be a function of s so (147) is an D sx D βsx ordinary diﬀerential equation. Let

(2) (1) 0 (D xx − D x) g (t) 0 2 (2) 2 − (2) 2 = l(f (s)) (148) D sx D g(t)sx

where l is a constant, and f(s) is an invertible and diﬀerentiable function.

67 By integrating (148) with respect to x we get

s Z (D(2) − D(1) ) g0(t) Z √ √ xx x − dx = ls f 0(s)dx = lf(s) D(2) D(2)g(t) x

then the similarity variable s is

s ! 1 Z (D(2) − D(1) ) g0(t) s = f −1 √ xx x − dx . (149) l D(2) D(2)g(t)

Now let’s consider some example. Example 9

(2) (1) (D xx−D x) βt If (2) 2 − (2) 2 = k where k is a constant, then (147) reduces to D sx D βsx

γ00(s) + kγ(s) = 0. (150)

Where the similarity variable is given by

s Z (D(2) − D(1) ) g0(t) s = xx x − dx D(2)k D(2)g(t)k

here we set l = 1. If k < 0 then

√ √ s −k −s −k γ(s) = c1e + c2e . (151)

If k ≥ 0 then √ √ γ(s) = c1 cos(s k) + c2 sin(s k), (152)

where c1 and c2 are arbitrary constants, and s is given by (149).

68 So the general solution is

√ √ s −k −s −k W (x, t) = g(t)γ(s) = g(t)[c1e + c2e ] where k < 0 (153)

√ √ W (x, t) = g(t)γ(s) = g(t)[c1 cos(s k) + c2 sin(s k)] where k ≥ 0. (154)

Example 10 r (2) (1) 1 R D xx−D x dx 0 k k lD(2) If f (s) = s then f(s) = k ln s, where s = e . By substituting above relations into (147), we get

s2γ00(s) + lk2γ(s) = 0.

which is a Cauchy-Euler equation and has the following characteristic equation obtained by letting γ(s) = sm where m is a constant.

m(m − 1) + lk2 = 0 (155)

or equivalently,

m2 − m + lk2 = 0 (156)

√ 1± 1−4lk2 which has two possible roots m1,2 such that m1,2 = 2 , by letting l = 1 1 If |k| < 2 , then m1,2 are real and m1 6= m2, and we have

m1 m2 γ(s) = c1s + c2s (157)

69 1 If |k| > 2 , then m1,2 = p ± iq, and we get

√ √ 2 2 1 4k − 1 4k − 1 γ(s) = s 2 c cos( ln s) + c sin( ln s) (158) 1 2 2 2

1 1 If k = ± 2 , then m1 = m2 = 2 , and we have

1 1 γ(s) = c1s 2 + c2s 2 ln s (159)

where c1 and c2 are arbitrary constant. Note that (157), (158) and (159) are three possible diﬀerent solutions to (147)

0 k when f (s) = s and s is deﬁned by (149). So the general solutions for the above cases are given, respectively, as

m1 m2 W (x, t) = g(t)γ(s) = g(t)[c1s + c2s ] (160)

√ 1± 1−4k2 1 where m1,2 = 2 with |k| < 2 .

√ √ 2 2 1 4k − 1 1 4k − 1 W (x, t) = g(t)γ(s) = g(t)[c s 2 cos( ln s) + c s 2 sin( ln s)] 1 2 2 2 (161)

1 where |k| > 2 .

1 1 W (x, t) = g(t)γ(s) = g(t)[c1s 2 + c2s 2 ln s]. (162)

1 where k = ± 2

70 r (2) (1) 1 R D xx−D x dx where s = e k D(2) .

1 Figure 5: W (x, t) versus x and t for solution (162) when k = 2

Figure 5 shows evolutions of the probability distribution for initial Gaussian proﬁle, we used g(t) = t1/2, γ(s) = e−3(x−t)2 , D(1) = xt−1 − 2t−2, D(2) =

−tx2 1 e ,k = 2 . We conclude our result as the following: Theorem 6 The Fokker-Planck equation given by (131) has a general similarity solution given by

W (x, t) = g(t)γ(s)

where γ(s) is a solution of the ordinary diﬀerential equation given by

γ00(s) + l(f 0(s))2γ(s) = 0

q (D(2) −D(1) ) g0(t) √1 R xx x where f(s) = (2) 2 − (2) 2 dx l D sx D g(t)sx provided that f(s) is diﬀerentiable and has an inverse, under the requirement

71 that s is a solution of

(2) (2) (1) D sxx + (2D x − D )sx − st = 0

and g(t) is an arbitrary diﬀerential function.

(1) R st+D sx 1 (2) dx 2sxD g(t) = (2)√ e D sx

72 Conclusion

We analyze Fokker-Planck equation which is essential tool is study of ﬂuc- tuations in physical and biological systems. We took three approaches in our analysis. First, we found a class of similarly solution under a group of transformation that describe stretching transformations. A natural boundary condition is considered. Then we considered the general Lie group method and we obtained the speciﬁc form of similarity variable in terms of D(1) and D(2) and the requirement for existence of such solution. The general form of group generators are also given. Finally, we considered a functional approach due to Clarkson-Kruskal reduction method and we obtained the condition under which such solution exist. We give several example of such solutions.

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