CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
SIMILARITY SOLUTIONS OF THE TWO-DIMENSIONAL
STATIONARY SCHRÖDINGER EQUATION
A thesis submitted in partial fulfillment of the requirements
For the degree of Master of Science
in Mathematics
By
Heather Pielaet
May 2014
The thesis of Heather Pielaet is approved:
Dr. Werner Horn
Date
Dr. Emmanuel Yomba
Date
Dr. Ali Zakeri, Chair
Date
California State University, Northridge
ii
Contents
Signature page . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ii
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iv
1 Introduction
1
2 Lie groups and Similarity Solutions
2
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2.2
Lie Group Transformations . . . . . . . . . . . . . . . . . . . .
4
2.3
Stretching Transformation . . . . . . . . . . . . . . . . . . . .
7
2.4
Infinitesimal Transformation . . . . . . . . . . . . . . . . . . . 18
2.5
General Similarity Method . . . . . . . . . . . . . . . . . . . . 21
3 The Schrödinger Equation
37
3.1
The two-dimensional stationary Schrödinger equation . . . . . 37
3.2
Similarity Method applied to the Two-dimensional Stationary
Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . . 39
3.3
Examples of Solutions to the Schrödinger Equation (45) . . . . 64
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
iii
ABSTRACT
SIMILARITY SOLUTIONS OF THE TWO-DIMENSIONAL
STATIONARY SCHRÖDINGER EQUATION
By
Heather Pielaet
Master of Science in Mathematics
We have investigated a similarity analysis for a two-dimensional stationary
Schrödinger equation that is based on the group theoretic method introduced
by Lie. Working in the complex plane, and using a family of one-parameter
transformations, forming a group, we have found a requirement on the ”energy function” for the model problem to have Lie group solutions. This requirement reduces the parameter coefficients for the matrix in the invariance
condition to be a diagonal matrix. Furthermore, analyzing its characteristic
system, we have shown there are subclasses that one can obtain solutions in
closed form. Several of such solutions obtained are illustrated graphically.
In addition, we used a predetermined form in the extended similarity form
in the real plane, we obtained several closed form solutions from a reduced
ODE. Some of these solutions are presented graphically.
iv
1
Introduction
The majority of fundamental equations of mathematical and theoretical physics
admit wide symmetry groups. It is the symmetries that give rise to transformations that leave the system invariant, or unchanged. This enables us to
develop efficient methods, such as the similarity method, for mathematical
analysis of differential equations. The similarity method allows us to take a
partial differential equation and reduce it to an ordinary differential equation
giving us the same solutions called similarity solutions or group-invariant solutions.
If a given system possesses these types of symmetries, it seems reasonable
that special properties can be derived. In chapter 2, we will first look at the
properties and simple examples of Lie groups and invariant functions. Then
we will examine the infinitesimal transformation and see how we can derive
a transformation for which our function is invariant under.
Then, in chapter 3 we will look at the two-dimensional stationary Schrödinger
equation and apply the methods discussed in chapter 2 to attempt to derive
similarity solutions. We obtained several cases that lead to the existence of
closed form solutions. Some of these solutions are illustrated graphically.
1
2
2.1
Lie groups and Similarity Solutions
Introduction
Partial differential equations are often used to model mathematical and theoretical physics. The majority of the fundamental equations admit wide
symmetry groups. Often, partial differential equations are difficult to solve
directly which makes the similarity method so helpful. In order to apply the
similarity method we must first determine the transformation under which
the problem is invariant, or unchanged, and then simplify the problem. A
few different types of transformations are rotation, translation and dilation.
A simple example of this are angles and ratios of distance. They are invariant
under scalings, rotations, translations, and reflections. Our goal is to find a
transformation in which our PDE is invariant under and use this to reduce it
to an ODE. To motivate this discussion let us start with a simple example.
Example 2.1.1 Consider the nonlinear one-dimensional wave equation:
utt − uuxx = 0
(1)
under a stretching transformation:
x̄ = αa x
t̄ = αb t
ū = αc u
We have :
ūt̄t̄ − ūūx̄x̄ =
∂ 2 ū
dt̄∂ t̄
2
− ū ∂∂x̄∂ūx̄ =
∂ 2 αc u
∂αb t∂αb t
2 c
2
2
α u
c−2b ∂ u
− αc u ∂α∂a x∂α
− α2c−2a u ∂∂xu2
ax = α
∂t∂t
2
Then c-2b must equal 2c-2a in order for it to be invariant. We can let c = 0
and a=b=1. Now we have
ūt̄t̄ − ūūx̄x̄ =
1
(utt − uuxx )
α2
(2)
Therefore, the differential operator that defines the nonlinear one-dimensional
wave equation is unchanged or invariant under the following stretching transformation:
x̄ = αx
t̄ = αt
ū = u
up to a constant factor
defined by s =
x
t
1
.
α2
We can now introduce an independent variable s
and assume that
u(x, t) = f (s)
(3)
Also, note that s is invariant under this transformation since
s̄ =
x̄
t̄
=
αx
αt
=
x
t
=s
We can substitute (3) into (1) to reduce our partial differential equation to an
ordinary differential equation and solve for a solution known as a similarity
solution. Our variable s is known as the similarity variable. This procedure
is known as the similarity method for partial differential equations, and it
applies to a wide variety of important problems in science and engineering
[1]. Invariance properties are not only used to simplify PDE’s, but also in
the calculus of variations.
3
2.2
Lie Group Transformations
2.2.1 Definition Let α vary continuously over an open interval |α| ≤ α0 .
Let φ(t, x, α) and ψ(t, x, α) be given analytic functions on <2 × (−α0 , α0 ). A
one parameter family of transformations on <2 defined by
Tα :
t̄ = φ(t, x, α),
x̄ = ψ(t, x, α)
(4)
is a local Lie group if the following conditions hold:
• Tα closed, i.e. If Tα1 , Tα2 in Tα then Tα1 Tα2 = Tα3 in Tα
• Tα Contains the Identity, i.e. there exists a T0 such that t̄ = φ(t, x, 0) =
t,
x̄ = ψ(t, x, 0) = x
• Tα has an Inverse for small α, i.e. For every α1 in (−α0 , α0 ), there
exists an α2 in (−α0 , α0 ) such that Tα1 Tα2 = Tα2 Tα1 = I
Example 2.2.1
Consider
Tα =



t̄ = tcosα − xsinα


x̄ = tsinα + xcosα
This is the one-parameter family of rotations about the origin in the tx-plane
which takes a point (t,x) to another point ( t̄, x̄) through an angle α, where
0 ≤ α < 2π. In order to check the closure property we must first define the
following
4
Tα1 =



t̄ = tcosα1 − xsinα1


x̄ = tsinα1 + xcosα1
Tα2 =



t̄¯ = t̄cosα2 − x̄sinα2


x̄¯ = t̄sinα2 + x̄cosα2
Calculations show that Tα2 Tα1 is
t̄¯ = φ(t̄, x̄, α2 ) = (tcosα1 − xsinα1 )cosα2 − (tsinα1 + xcosα1 )sinα2
= t(cosα1 cosα2 − sinα1 sinα2 ) − x(cosα1 sinα2 + sinα1 cosα2 )
= tcos(α1 + α2 ) − xsin(α1 + α2 )
Similarly,
x̄¯ = ψ(t̄, x̄, α2 ) = (tcosα1 − xsinα1 )sinα2 + (tsinα1 + xcosα1 )cosα2
= t(cosα1 sinα2 + sinα1 cosα2 ) + x(cosα1 cosα2 − sinα1 sinα2 )
= tsin(α1 + α2 ) + xcos(α1 + α2 )
Therefore, Tα2 Tα1 = Tα3 belongs to the group of rotations with α3 = α1 + α2 .
Next, we check that an Identity exists. Consider α = 0, then we have that
T0 =



t̄ = tcos(0) − xsin(0) = t


x̄ = tsin(0) + xcos(0) = x
5
Lastly, the inverse transformation for Tα is T−α . Clearly, if α ∈ (−α0 , α0 )
then −α ∈ (−α0 , α0 ).
Other examples of local Lie groups are the following.
• Translation Group:
• Stretching Group:
x̄ = x + α,
x̄ = αa x,
t̄ = t
t̄ = αb t,
a, b constants
Let us take a further look at the stretching transformation.
6
2.3
Stretching Transformation
Let us consider a single second order partial differential equation for u =
u(x, t) given by
F (x, t, u, ux , ut , uxx , uxt , utt ) = 0
(5)
Let the transformation Tα : <3 → <3 be defined as follows




x̄ = αa x




Tα = t̄ = αb t






ū = αc u
(6)
where a, b, and c are fixed real constants and α is a real parameter in an
open interval I containing α = 1.
Let us check that this stretching group is indeed a local Lie group. First,
note that Tα1 Tα2 = Tα1 α2 since
Tα1 Tα2




x̄¯ = α2a (α1a x) = (α2 α1 )a x




= t̄¯ = α2b (α1b t) = (α2 α1 )b t






ū¯ = α2c (α1c u) = (α2 α1 )c u
Also, if we let α = 1 we get the identity and for every α ∈ I the inverse of
Tα is Tα−1 . Therefore the stretching group is a local Lie group.
The stretching group takes our surface in <3 defined by u = φ(x, t) to a
surface in x̄t̄ū space defined by
ū = φ̄(x̄, t̄),
7
(x̄, t̄) ∈ D̄
(7)
where D̄ = {(x̄, t̄)|x̄ = αa x, t̄ = αb t, (x, t) ∈ D} and φ̄ is defined by
φ̄(x̄, t̄) = αc φ(α−a x, α−b t)
(8)
Our transformed partial differential equation is written as
F (x̄, t̄, ū, ūx̄ , ūt̄ , ūx̄x̄ , ūx̄t̄ , ūt̄t̄ ) = 0
(9)
In general φ̄(x̄, t̄) is not the same as φ(x̄, t̄). Here is an example to illustrate
this notion.
Example 2.3.1
Consider a stretching transformation
x̄ = αx,
t̄ = αt,
ū = α2 u
and let φ(x, t) = x − t2 . Then, φ(x̄, t̄) = x̄ − t̄2 , whereas φ̄(x̄, t̄) = α2 (α−1 x̄ −
α−2 t̄2 ) = αx̄ − t̄2 .
Now, let us define what it means for a partial differential equation to be
constant conformally invariant.
Definition 2.3.1 The partial differential equation (5) is constant conformally
invariant under the one parameter family of stretching Tα defined by (6) if,
and only if,
F (x̄, t̄, ū, ūx̄ , ūt̄ , ūx̄x̄ , ūx̄t̄ , ūt̄t̄ ) = A(α)F (x, t, u, ux , ut , uxx , uxt , utt )
(10)
for all α in I, for some function A with A(1) = 1. If A(α) ≡ 1, then we say
that (5) is absolutely invariant. A(α) is known as the conformal factor.
8
Example 2.3.2
Find a stretching transformation under which the equation
ut = vuxx
is invariant. Let
Tα =



x̄ = αa x




t̄ = αb t






ū = αc u
Then,
ūt̄ − vūx̄x̄
∂αc u
∂ 2 αc u
=
−v a
= αc−b ut − vαc−2a uxx
b
a
∂α t
∂α x∂α x
If we let c − b = c − 2a, then we have
ūt̄ − vūx̄x̄ = αc−b (ut − vuxx )
Let a = 1, b=2, and c=0. By definition 2.3.1 we have that our partial
differential equation is constant conformally invariant under the following
transformation
Tα =



x̄ = αx




2
t̄ = α t






ū = u
with a conformal factor of A(α) = α−2 .
Theorem 2.3.1 If the partial differential equation (5) is constant conformally invariant under the one parameter family of stretchings Tα given by
9
(6) and if φ(x, t) is a solution of (5), then φ̄(x̄, t̄) defined by (8) is a solution
of the partial differential equation (9).
Definition 2.3.2 A solution u = φ(x, t) of (5) is an invariant solution if, and
only if,
φ(x̄, t̄) = φ̄(x̄, t̄)
(11)
under the transformation Tα .
Note that (11) is equivalent to
φ(αa x, αb t) = αc φ(x, t)
(12)
If we differentiate (12) with respect to α and then set α = 1 we obtain the
following first order partial differential equation, which is called the invariant
surface condition
axφx + btφt = cφ
(13)
It’s characteristic equations are
dx
dt
dφ
=
=
ax
bt
cφ
Integrating the first and second pair of equations gives xb t−a = constant
and φt
ψ(φt
−c
b
−c
b
= constant respectively. Therefore, the general solution of (13) is
, xb t−a ) = 0 for some function ψ. The invariant surfaces are
c
φ(x, t) = t b f (s)
(14)
where f is an arbitrary function and s is the similarity variable defined by
10
s=
xb
ta
(15)
Theorem 2.3.2. If the partial differential equation (5) is constant conformally invariant under the one parameter family of stretching transformations
defined by (6) then substitution of the expression
c
u = t b f (s)
(16)
where s is defined by (15), into (5) yeilds an equation of the form
H(s, f, f 0 , f 00 ) = 0
which is an ordinary differential equation for f.
Proof
Recall that Tα being invariant means we have
F (x̄, t̄, ū, ūx̄ , ūt̄ , ūx̄x̄ , ūx̄t̄ , ūt̄t̄ ) = A(α)F (x, t, u, ux , ut , uxx , uxt , utt )
where A is some function. Observe that ūx̄ = αc−a ux ,
αc−2a uxx ,
ūx̄t̄ = αc−a−b uxt ,
ūt̄ = αc−b ut ,
(17)
ūx̄x̄ =
ūt̄t̄ = αc−2b utt .
Differentiate (17) with respect to α and then setting α = 1 we get
axFx + btFt + cuFu + (c − a)ux Fux + (c − b)ut Fut + (c − 2a)uxx Fuxx
+ (c − a − b)uxt Fuxt + (c − 2b)utt Futt = A0 (1)F
Our characteristic system is
11
dx
dt
du
dux
dut
duxx
=
=
=
=
=
ax
bt
cu
(c − a)ux
(c − b)ut
(c − 2a)uxx
dutt
dF
duxt
=
= 0
=
(c − a − b)uxt
(c − 2b)utt
A (1)F
Eight independent first integrals are
xb
,
ta
ut
−c
b
,
ux t
a−c
b
,
ut t
b−c
b
,
uxx t
2a−c
b
,
uxt t
b+a−c
b
,
utt t
2b−c
b
,Ft
−A0 (1)
b
(18)
Consequently,
F =t
A0 (1)
b
G(s, ut
−c
b
, ux t
a−c
b
, ut t
b−c
b
, uxx t
2a−c
b
, uxt t
b+a−c
b
, utt t
2b−c
b
)
For some function G. Now, u is given by (16) and it follows that
c
ux = bt b−a xb−1 f 0 (s)
c
c c
ut = t b−1 f (s) − axb t b−a−1 f 0 (s)
b
uxx = b(b − 1)xb−2 t
c−ab
b
uxt = (c − ab)xb−1 t
c−b(a+1)
b
utt =
f 0 (s) + b2 x2b−1 t
c−2ab
b
f 0 (s) − abx2b−1 t
f 00 (s)
c−b(2a+1)
b
f 00 (s)
c−2b(a+1)
c(c − b) c−2b
a(2c − ab − b) b c−b(a+2) 0
t b f (s) −
x t b f (s) + a2 x2b t b f 00 (s)
2
b
b
When we substitute these values into G we obtain an ordinary differential
equation of the form
H(s, f, f 0 , f 00 ) = 0
Let us now illustrate this procedure with an example.
12
Example 2.3.3
Consider the nonlinear partial differential equation
uut + u2x = 0
We assume a stretching transfomation
Tα : x̄ = αa x,
t̄ = αb t,
ū = αc u
Then
ūūt̄ + ū2x̄ = α2c−b uut + α2c−2a u2x = α2c−b (uut + u2x )
provided 2c-b = 2c-2a or b = 2a. Thus the partial differential equation is
invariant under the following transformation
Tα : x̄ = αa x,
t̄ = α2a t,
ū = αc u
The invariant surface condition is
axux + 2atut = cu
having characteristic system
dx
dt
du
=
=
ax
2at
cu
and first integrals
x
√
t
−c
and ut 2a . Therefore, letting s =
(19)
x
√
,
t
the invariant
surfaces are given by
c
u(x, t) = t 2a f (s)
Let c = 0 and we have that u(x,t) = f(s). After substituting u(x,t) = f(s) into
13
our partial differential equation we obtain an ordinary differential equation
for the function f,
−x
1
f (s)f 0 (s) + (f 0 (s))2 = 0
t
2t
3
2
(20)
We can simplify (20) to get
1
0
− sf (s) + f (s) f 0 (s) = 0
2
Example 2.3.4 Consider the nonlinear one-dimensional wave equation from
example 2.1.1. We had that
ūt̄t̄ − ūūx̄x̄ =
∂ 2 ū
∂ 2 αc u
∂ 2 αc u
∂ 2 ū
c
− ū
=
−
α
u
dt̄∂ t̄
∂ x̄∂ x̄
∂αb t∂αb t
∂αa x∂αa x
= αc−2b
∂ 2u
∂ 2u
− α2c−2a u 2 = αc−2b (utt − uuxx )
∂t∂t
∂x
provided that c-2b = 2c-2a or c = 2a-2b. Thus the partial differential equation is invariant under
Tα : x̄ = αa x,
t̄ = αb t,
ū = α2a−2b u
The invariant surface condition is
axux + btut = (2a − 2b)u
having characteristic system
dt
du
dx
=
=
ax
bt
(2a − 2b)u
14
and first integrals
xb
ta
and ut
2b−2a
b
. Therefore, letting s =
xb
ta
the invariant
surfaces are given by
u(x, t) = t
2a−2b
b
f (s)
Let a = b = 1 and c = 0. Then we have u(x, t) = f (s). now differentiate u(x, t) and substitute the values into our partial differential equation to
obtain a second order ordinary differential equation for the function f,
x
x2
2 f 0 (s) + 2 f 00 (s) − f (s)f 00 (s) = 0
t
t
(21)
We can simplify (21) to get
2sf 0 (s) + (s2 − f (s))f 00 (s) = 0
Example 2.3.5 Consider the following acoustic approximation equations
ut − vx = 0
vt + uux = 0
We assume a stretching transformation
Tα : x̄ = αa x,
t̄ = α2a t,
ū = αc u,
Then,
ūt̄ − v̄x̄ = αc−b ut − αe−a vx
and
15
v̄ = αe v
v̄t̄ + ūūx̄ = αe−b vt + α2c−a uux
provided c-b = e-a and e-b = 2c-a, or a = b and c = e = 0. Thus, the system
of partial differential equations are invariant under
x̄ = αa x,
t̄ = αa t,
ū = u,
v̄ = v
Let a =1 and our stretching transformation becomes
x̄ = αx,
t̄ = αt,
ū = u,
v̄ = v
Then, by theorem 2.3.2, the system of partial differential equations are of
the form
u = f (s),
where the similarity variable s =
x
.
t
v = g(s)
(22)
Now, after differentiating (22) and
substituting the results into our original partial differential equation we get
x
1
− 2 f 0 (s) − g 0 (s) = 0
t
t
x 0
1
− 2 g (s) + f (s)f 0 (s) = 0
t
t
multiply both equations by x so they can be written as
s2 f 0 (s) + sg 0 (s) = 0
−s2 g 0 (s) + sf (s)f 0 (s) = 0
Now, divide the second equation by s and add them together to get
16
f 0 (s)(s2 + f (s)) = 0
Case 1:
f 0 (s) = 0, which implies that f (s) = constant giving us that g(s) = constant
as well.
Case 2:
s2 + f (s) = 0 implies that f (s) = −s2 . If we substitute this into −s2 g 0 (s) +
sf (s)f 0 (s) = 0 we find that g(s) =
2s2
.
3
Therefore, the similarity solution for the acoustic approximation equations
are given by
x2
t2
3
2s
2x3
v(x, t) =
= 3
3
3t
u(x, t) = −s2 = −
17
2.4
Infinitesimal Transformation
For many purposes it suffices to study only transformations close to the
identity. We assume that the functions φ and ψ in (4) are differentiable a
sufficient number of times with respect to α. If Tα defined by (4) is a local Lie
group, then the right sides of (4) can be expanded in a Taylor series about
α = 0 to obtain
t̄ = φ(t, x, 0) + φα (t, x, 0)α +
1
φαα (t, x, 0)α2 + ...
2!
= t + τ (t, x)α + o(α)
x̄ = ψ(t, x, 0) + ψα (t, x, 0)α +
1
ψαα (t, x, 0)α2 + ...
2!
= x + ξ(t, x)α + o(α)
where τ and ξ are defined by
τ (t, x) ≡ φα (t, x, 0),
ξ(t, x) ≡ ψα (t, x, 0)
(23)
For small α the infinitesimal transformation
t̄ = t + τ (t, x)α + o(α),
x̄ = x + ξ(t, x)α + o(α)
(24)
approximates Tα . The quantities τ and ξ are called the generators of Tα and
they define the principal linear part of the transformation in α; that is, the
generators are the coefficients of the lowest order terms in α in the Taylor
expansion [1].
18
Example 2.4.1 Referring to the transformation in Example 2.2.1 we have
τ (t, x) =
∂
φ(t, x, α)|α=0 = (−tsinα − xcosα)α=0 = −x
∂α
ξ(t, x) =
∂
ψ(t, x, α)|α=0 = (tcosα − xsinα)α=0 = t
∂α
Therefore the infinitesimal transformation is
t̄ = t − αx + o(α),
x̄ = x + αt + o(α)
which approximates Tα for small α.
We can see that the local Lie group uniquely determines the generators by
equation (23 ). The converse statement is also true, we can determine the
local Lie group if given the generators. The global representation of the
group can be determined by solving the system of differential equations
dt̄
dx̄
=
= dα
τ (t̄, x̄)
ξ(t̄, x̄)
subject to the initial conditions
t̄ = t,
x̄ = x,
at α = 0
Example 2.4.2 We calculate the group whose generators are given by τ = t2
and ξ = −tx. The dynamical system is
dt̄
dx̄
=
= dα
2
t̄
−t̄x̄
19
with t̄ = t and x̄ = x at α = 0. Integrating
dt̄
t̄2
=
dx̄
−t̄x̄
gives
t̄x̄ = c1
where c1 is some constant. Applying the initial conditions gives t̄x̄ = c1 = tx.
Integrating
x̄ =
t2 x
.
1−αt
dt̄
t̄2
= dα and applying the initial conditions gives t̄ =
The group is therefore given by
t̄ =
1 − αt
,
t
x̄ =
20
t2 x
1 − αt
1−αt
.
t
Thus
2.5
General Similarity Method
Let us extend the similarity method to more general local LIe groups of
transformations on <3 . Such transformations include translations, rotations,
and other kinds of geometrical mappings. Consider a general one parameter
family of transformations Tα defined by the equations
Tα :
x̄ = Φ(x, t, α),
t̄ = Ψ(x, t, α),
ū = Ω(x, t, u, α)
(25)
where α is a real parameter that varies over some open interval |α| < α0
containing zero, and Φ, Ψ and Ω are functions analytic on their respective
domains. When α = 0 we assume the transformation (25) is the identity
transformation I given by
x̄ = Φ(x, t, 0) = x
t̄ = Ψ(x, t, 0) = t
ū = Ω(x, t, u, 0) = u
Furthermore, let us assume that Tα defined by (25) is a local Lie group of
transformations.
Using Taylor’s Theorem we can expand the right sides of (25) in a power
series in α to obtain the infinitesimal transformation
x̄ = x + αξ(x, t) + o(α)
t̄ = t + ατ (x, t) + o(α)
ū = u + αω(x, t, u) + o(α)
where
21
(26)
∂
Φ(x, t, 0)
∂α
∂
Ψ(x, t, 0)
τ (x, t) ≡
∂α
∂
Ω(x, t, u, 0)
ω(x, t, u) ≡
∂α
ξ(x, t) ≡
(27)
The quantities ξ, τ and ω are called the generators of transformation (25).
The local Lie group Tα defined by (25) determines by definition a unique set of
generators ξ, τ and ω by formula (27). Conversely, the generators determine
the group. The group (25) can be determined by solving the initial value
problem
dx̄
dt̄
dū
=
=
= dα
ξ(x̄, t̄)
τ (x̄, t̄)
ω(x̄, t̄, ū)
x̄ = x,
t̄ = t,
ū = u,
at α = 0
Example 2.5.1 Determine the group with generators
ξ = 2x + t,
τ = t,
ω =u+1
We form the system
dx̄
dt̄
dū
=
=
= dα
t̄
2x̄ + t̄
ū + 1
subject to the initial condition x̄ = x,
first integrals are found to be
22
t̄ = t,
ū = u,
at α = 0. Three
x̄ + t̄
= c1 ,
t̄2
t̄
= c2 ,
ū + 1
ln(ū + 1) = α + c3
Applying the initial conditions determines c1 , c2 and c3 and we obtain the
three equations
x̄ + t̄
x+t
= 2 ,
2
t
t̄
t
t̄
=
,
ū + 1
u+1
ln(ū + 1) = α + ln(u + 1)
The third yeilds
ū = (u + 1)eα − 1
The second, therefore gives
t̄ = teα
Finally, from the first
x̄ = (x + t)e2α − teα
and the equations for the group are determined.
The transformation Tα defined by (25) maps for each fixed α a surface having
equation
u = g(x, t),
(x, t) ∈ D
in xtu space to a surface having equation
ū = ḡ(x̄, t̄),
23
(x̄, t̄) ∈ D̄
in x̄t̄ū space, where
D̄ = {(x̄, t̄)|x̄ = Φ(x, t, α),
t̄ = Ψ(x, t, α),
(x, t) ∈ D}
To determine ḡ we invert the transformation
x̄ = Φ(x, tα),
t̄ = Ψ(x, t, α)
on D to obtain
x = x(x̄, t̄),
t = t(x̄, t̄)
Then ḡ is defined by
ḡ(x̄, t̄) = Ω(x(x̄, t̄), t(x̄, t̄), g(x(x̄, t̄), t(x̄, t̄)), α)
Now we will take a look at the definition of invariance of a first order partial
differential equation in the general sense.
Consider the first order partial differential equation
F (x, t, u, ux , ut ) = 0
(28)
Definition 2.5.1 The partial differential equation (28) is constant conformally invariant under the local Lie group (25) if, and only if,
∂
F (x̄, t̄, ū, ūx̄ , ūt̄ )|α=0 = kF (x, t, u, ux , ut )
∂α
(29)
for some constant k. If k = 0, the partial differential equation is said to be
24
absolutely invariant. In (29) if it occurs that k = k(x, t, u, ux , ut ), then the
partial differential equation is conformally invariant.
Example 2.5.2 Consider the diffusion operator ut − uxx and the group of
stretchings
x̄ = (1 + α)x
t̄ = (1 + α)2 t
(30)
ū = u
We have that
ūt̄ − ūx̄x̄ = (1 + α)−2 (ut − uxx )
The right side of the last equation can be expanded about α = 0 to obtain
ūt̄ − ūx̄x̄ = (1 − 2α + ...)(ut − uxx ) = (ut − uxx ) − 2α(ut − uxx ) + o(α2 )
Taking the derivative with respect to α at α = 0 gives
∂
(ūt̄ − ūx̄x̄ )α=0 = −2(ut − uxx )
∂α
Therefore, the diffusion operator is constant confromally invariant under the
local Lie group (30).
Now, returning to definition 2.5.1 let us expand the left side of (29) to obtain
a condition for invariance in terms of the generators of the local Lie group.
Let us define p ≡ ux and q ≡ ut . Differentiating the left side of (29) gives
25
Fx ξ + Ft τ + Fu ω + Fp
∂ p̄
∂α
+ Fq
α=0
∂ q̄
∂α
= kF
(31)
α=0
Now, we must compute the following two expressions
π = π(x, t, u, p, q) ≡ Fp
∂ p̄
∂α
∂ q̄
∂α
α=0
(32)
χ = χ(x, t, u, p, q) ≡ Fq
α=0
Theorem 2.5.1 If u = g(x, t), then under the transformation Tα given by
(25) the derivatives of ū = ḡ(x̄, t̄) are
t̄t Dx Ω − t̄x Dt ω
detJ
x̄t Dx Ω − x̄x Dt ω
q̄ ≡ ḡt̄ (x̄, t̄) = −
detJ
p̄ ≡ ḡx̄ (x̄, t̄) =
(33)
(34)
where
Dx Ω ≡ Ωx + gx Ωu ,
Dt Ω ≡ Ωt + gt Ωu
and J is the Jacobi matrix of the transformation x̄ = Φ(x, t, α),
Ψ(x, t, α) given by


 x̄x x̄t 
J ≡

t̄x t̄t
26
t̄ =
The right sides of (33) and (34) are evaluated at x = x(x̄, t̄) and t = t(x̄, t̄).
Proof
Recall that ḡ is defined by
ḡ(x̄, t̄) = Ω(x(x̄, t̄), t(x̄, t̄), g(x(x̄, t̄), t(x̄, t̄)), α)
Thus,
ḡx̄ = Ωx xx̄ + Ωt tx̄ + Ωu (gx xx̄ + gt tx̄ )
= xx̄ Dx Ω + tx̄ + Dt Ω
Similarly ḡt̄ = x̄t̄ Dx Ω + tt̄ Dt Ω. In matrix form, this is


(ḡx̄
 xx̄ xt̄ 
ḡt̄ ) = (Dx Ω Dt Ω) 

tx̄ tt̄
The matrix on the right is the Jacobi matrix of the inverse transformation
x = x(x̄, t̄), t = t(x̄, t̄) and hence is the inverse of J. Therefore

(ḡx̄
ḡt̄ ) = (Dx Ω Dt Ω)J −1 = (Dx Ω Dt Ω)

1  t̄t −x̄t 


detJ
−t̄x x̄x
Theorem 2.5.2 The quantities π and χ defined by (32) are given by
π = ωx + pωu − pξx − qτx
(35)
χ = ωt + qωu − pξt − qτt
(36)
27
where ξ, τ and ω are the generators of (25).
The invariance condition (31) can now be expressed as
Fx ξ + Ft τ + Fu ω + Fp π + Fq χ = kF
(37)
where π and χ are given by (35) and (36), respectively. The left side of (37)
is denoted by Ũ F where Ũ is the Lie derivative operator, or infinitesimal
operator [2], defined by
Ũ ≡ ξ
∂
∂
∂
∂
∂
+τ +ω
+π
+χ
∂x
∂t
∂u
∂p
∂q
The partial differential equation (28) will be invariant under the group of
transformations (26) if
Ũ F = kF
(38)
Substituting (35) and (36) into (38), using (28) and equating to zero the
coefficients of like derivative terms in u, we get an over determined system
of linear PDEs to obtain the generators ξ, τ and ω. After finding ξ, τ and
ω, the invariant surface condition is used to find the similarity form of the
solution. Here is an example to illustrate this idea.
Example 2.5.3 Consider the equation ut + uux = 0 or
q + up = 0
We have that F(x,t,u,p,q) = q + u p and Fx = Ft = 0, Fu = p, Fp = u and
Fq = 1.
Substituting these values, (35) and (36) into (37) gives
pω + u(ωx + ωu p − ξx p − τx q) + (ωt + qωu − pξt − qτt ) = k(q + up)
28
Rearranging terms we can write
p(ω + uωu − uξx − ξt − ku) + q(ωu − uτx + τt − k) + (uωx + ωt ) = 0
Therefore, we have
uωx + ωt = 0
ω + uωu − uξx − ξt − ku = 0
ωu − uτx + τt − k = 0
which is a system of three coupled first order linear partial differential equations for the generators ξ, τ and ω that can be solved.
We find that
ξ = (2b − k)x + ct + h
τ = (b − k)t + g
ω = bu + c
where b, c, g, h and k are arbitrary constants.
Now, consider
F (m) (x, u, p) = 0,
m = 1, ..., n
This is a system of n partial differential equations, where x = (x1 , x2 ), u =
j
(u1 , ..., un ), and p = (pji ) = ∂u
. Consider the extended local Lie group T̃α
∂xi
written in infinitesimal form
29




x̄i = xi + αXi (x) + o(α)




T̃α = ūj = uj + αU j (x, u) + o(α)






p̄ji = pji + αPij (x, u, p) + o(α)
where i = 1, 2 and j = 1, ..., n and the generators Pij are given in terms of
the generators Xi and U j by
n
Pij
∂U j X
=
+
∂xi
k=1
∂U j k
p
∂uk i
−
2 X
∂Xl
l=1
∂xi
pjl
If the generators Xi depend on u as well, then there is an additional term
n X
2 X
∂Xl
k=1 l=1
pj pk
∂uk l i
subtracted on the right side of the last equation. The invariance condition
takes the form
Ũ F
(m)
≡
2
X
i=1
Fx(m)
Xi
i
+
n
X
(m)
Fuj U j
+
2 X
n
X
i=1 j=1
j=1
(m)
Fpj Pij
i
=
n
X
kmr F (r)
r=1
for m = 1, ..., n.
Now we have seen what it means for a partial differential equation defined by
(28) to be invariant and how to solve for generators that create an invariant
local Lie group, but what do we know about its solutions? Well, if u = g(x, t)
is a solution of (28), then ū = g(x̄, t̄) is a solution of
F (x̄, t̄, ū, p̄, q̄) = 0
It is also true that ū = ḡ(x̄, t̄) is a solution as well, but it is not always true
that g(x̄, t̄) = ḡ(x̄, t̄), recall example 2.3.1.
Similarity solutions are found among the invariant surfaces, that is, solutions
30
u = g(x, t) for which
g(x̄, t̄) = ḡ(x̄, t̄)
(39)
And similar to how we arrived to the invariant surface condition (13), differentiating (39) with respect to α at α = 0 we get
ξ(x, t)gx + τ (x, t)gt = ω(x, t, g)
(40)
Condition (40) is the invariant surface condition, which can be solved by the
method of characteristics. The characteristic system is
dt
du
dx
=
=
ξ(x, t)
τ (x, t)
ω(x, t, g)
(41)
Now, let us expand our previous notions to account for several independent
and dependent variables. Consider the following partial differential equation
j
∂ 2 uj
j ∂u
Ha xi , u ,
,
, ... = 0,
∂xi ∂xi ∂xj
where i = 1, ..., n and j = 1, ..., m with n independent variables, xi , and m
dependent variables, uj (xi ).
Consider the following one-parameter group of transformations
x̄i = Φi (xk , ul ; α)
ūj = Ωj (xk , ul ; α)
The corresponding infinitesimal transformations are
31
x̄i = xi + αXi (xk , ul ) + o(α)
ūj = uj + αU j (xk , ul ) + o(α)
where
∂Φi
(xk , ul ; 0)
∂α
∂Ωj
U j (xk , ul ) =
(xk , ul ; 0)
∂α
Xi (xk , ul ) =
For convenience, we denote
pji ≡
∂uj
,
∂xi
pjih =
∂ 2 uj
,
∂xi ∂xh
p̄ji ≡
∂ ūj
,
∂ x̄i
p̄jih =
∂ 2 ūj
∂ x̄i ∂ x̄h
We introduce the total derivative operator
m
m
n
X
XX
∂
∂
∂
D
pai a +
paih a + ...
=
+
Dxi
∂xi a=1 ∂u
∂ph
a=1 h=1
Now, taking the derivative of x̄i , we obtain
dx̄i = d (xi + αXi + o(α))
= dxi + αdXi + d(o(α))
dxi
=
+α
dxh
∂Xi ∂Xi ∂ua
+
∂xh
∂ua ∂xh
32
dxh + o(α)
= δih + α
∂Xi ∂Xi ∂ua
+
∂xh
∂ua ∂xh
dxh + o(α)
Where a = 1, ..., m and δih is the Kronecker delta. Using the total derivative
operator, we rewrite dx̄i as
DXi
dxh + o(α)
dx̄i = δih + α
Dxh
Similarly, for ūj ,
dūj = d uj + αU j + o(α)
= duj + αdU j + d(o(α))
duj
=
+α
dxb
∂U j ∂U j ∂ua
+
∂xb
∂ua ∂xb
dxb + o(α)
∂uj
DU j
=
+α
dxb + o(α)
∂xb
Dxb
Then,
h
i
DU j
+
α
dxb + o(α)
Dxb
∂ ū
i
p̄ji =
=h
∂Xi ∂ua
∂ x̄i
i
δih + α ∂X
+
dxh + o(α)
∂xh
∂ua ∂xh
j
∂uj
∂xb
∂uj
∂xb
j
+ α DU
+ o(α)
dxb
Dxb
=
a
i
i ∂u
+ ∂X
δih + α ∂X
+ o(α) dxh
∂xh
∂ua ∂xh
33
∂uj
DU j
δbh
+α
+ o(α)
DXi
∂xb
Dxb
δih + α Dxh + o(α)
∂uj
DU j
DXb
+α
+ o(α)
δib − α
+ o(α)
∂xb
Dxb
Dxi
=
=
∂uj
DU j
∂uj DXb
=
+α
−
+ o(α)
∂xi
Dxi
∂xb Dxi
= pji + αPij + o(α)
where
Pij =
DU j
∂uj DXb
−
Dxi
∂xb Dxi
(42)
=
∂U j
+
∂xi
If we substitute
m X
a=1
∂uj
∂xi
∂U j a
p −
∂ua i
for
∂ 2 uj
,
∂xi ∂xh
n X
b=1
∂Xb j
p
∂xi b
−
m X
n X
a=1 b=1
∂Xb j a
pp
∂ua b i
U j for Uij , and xi for xh in the equation
DU j
∂uj DXb
∂uj
+α
−
+ o(α)
∂xi
Dxi
∂xb Dxi
we obtain p̄jih , in other words
34
∂ 2 ūj
∂ x̄i ∂ x̄h
p̄jih =
"
#
DUij
∂ 2 uj DXb
∂ 2 uj
+α
−
+ o(α)
=
∂xi ∂xh
Dxh
∂xi ∂xb Dxh
(43)
j
+ o(α)
= pjih + αPih
where
j
Pih
DUij
∂ 2 uj DXb
=
−
Dxh
∂xi ∂xb Dxh
m
m
n
2 j
X
∂ 2 U j a X ∂ 2 U j a X ∂ 2 Xb j
∂ U
+
p +
p −
p
=
∂xi ∂xh a=1 ∂xi ∂ua h a=1 ∂xh ∂ua i
∂xi ∂xh b
b=1
+
−
m
X
∂U j
paih
a
∂u
a=1
m X
n
X
a=1 b=1
−
−
∂xi
b=1
pjhb
−
n
X
∂Xb
b=1
m
∂xh
pjib
m X
m
X
∂ 2U j c a
+
pp
∂uc ∂ua i h
c=1 a=1
(44)
n
∂ 2 X b a j X X ∂ 2 Xb a j
p p −
p p
∂xi ∂ua h b a=1 b=1 ∂xh ∂ua i b
m X
n
X
∂Xb
a=1 b=1
n
X
∂Xb
[pj pa
∂ua b ih
+
pai pjbh
+
pah pjib ]
m X
n X
m
X
∂ 2 Xb c a j
−
p p p
c ∂ua h i b
∂u
a=1 b=1 c=1
The last term
m X
n X
m
X
∂ 2 Xb c a j
p p p
c ∂ua h i b
∂u
a=1 b=1 c=1
is only added on when the generators Xi depend on ui as well.
The rth derivative, Pij1 i2 ...ir , can be generated by applying the total derivative
operator to the (r − 1)th derivative and adds the term
35
−
DXb j
p
Dxir i1 i2 ...ir1 b
where iρ = 1, 2, ..., n for ρ = 1, 2, ..., r.
Define our infinitesimal operator,Ũ , as
Ũ = Xi
∂
∂
∂
∂
∂
+ ...
+ U j j + Pij1 j + Pij1 i2 j + Pij1 i2 i3 j
∂xi
∂u
∂pi1
∂pi1 i2
∂pi1 i2 i3
Our invariance condition becomes
Ũ Ha =
r
X
kab Hb (xi , uj , pji , pjih , ...)
b=1
We have generalized our result in finding the group of transformations that
will leave the partial differential equation(s) invariant.
36
3
3.1
The Schrödinger Equation
The two-dimensional stationary Schrödinger equation
Exact solutions of differential equations of mathematical physics, linear and
nonlinear, are very important for the understanding of various physical phenomena. In this chapter, we examine the two-dimensional stationary Schrödinger
equation, which, in quantum mechanics, is a partial differential equation that
describes how the quantum state of some physical system changes over time.
In classical mechanics, the Schrödinger equation plays the role of Newton’s
method and conservation of energy. Erwin Schrödinger, an Austrian physicist, formulated this equation in late 1925.
Consider the two-dimensional stationary Schrödinger equation
∂ 2 Φ(x, y) ∂ 2 Φ(x, y)
+
− v(x, y)Φ(x, y) = 0
∂x2
∂y 2
(45)
where Φ(x, y) and v(x, y) are complex valued functions, thus
Φ(x, y) = f (x, y) + ig(x, y)
(46)
v(x, y) = v1 (x, y) + iv2 (x, y)
where Re Φ(x, y) = f (x, y) is the real part of Φ(x, y) and Im Φ(x, y) = g(x, y)
is the imaginary part of Φ(x, y) and similarly for v(x, y).
Equation (45) can be split into real and imaginary parts
fxx + fyy − (v1 f − v2 g) = 0
(47)
gxx + gyy − (v1 g + v2 f ) = 0
37
So, we have a system of two equations for two unknown functions f and g. It
is assumed that v1 and v2 are known functions. In the case v = −E +U (x, y),
equation (45) describes a nonrelativistic quantum system with energy E.
This form uses the principle of the conservation of energy. The terms of the
nonrelativistic Schrödinger equation can be interpreted as total energy of the
system, equal to the system kinetic energy plus the system potential energy.
In this respect, it is just the same as in classical physics.
In the case v =
ω2
c(x,y)2
equation (45) describes an acoustic pressure field with
temporal frequency ω in inhomogeneous media with sound velocity c. The
case of fixed energy E for the two-dimensional equation is of interest for
the multidimensional inverse scattering theory due to connections with twodimensional integrable nonlinear systems. The case of fixed frequency ω is
of interest for modeling in acoustic tomography [4].
In this next section we will employ the methods from Chapter 2 to attempt
to find similarity solutions to the two-dimensional stationary Schrödinger
equation.
38
3.2
Similarity Method applied to the Two-dimensional Stationary
Schrödinger Equation
Recall system of equations (47). Let us define the following infinitesimal
transformation
Tα =




x̄ = x + αX + O(α2 )







ȳ = y + αY + O(α2 )
(48)



f¯ = f + αF + O(α2 )







ḡ = g + αG + O(α2 )
where
x = (x1 , x2 ) = (x, y)
u = (u1 , u2 ) = (f, g)
(49)
X = (X1 , X2 ) = (X, Y )
U = (U 1 , U 2 ) = (F, G)
We need to solve for f¯x̄x̄ . From equation (43) and (44) we have that
1
f¯x̄x̄ = fxx + αP11
+ O(α2 )
where
39
(50)
1
P11
= Fxx + Fxf fx + Fxg gx + Fxf fx + Fxg gx − Xxx fx − Yxx fy
+ Ff fxx + Fg gxx − Xx fxx − Yx fxy − Xx fxx − Yx fxy + Ff f fx fx
+ Ff g fx gx + Fgf gx fx + Fgg gx gx − Xxf fx fx − Yxf fx fy − Xxg gx fx
− Yxg gx fy − Xxf fx fx − Yxf fx fy − Xxg gx fx − Yxg gx fy − Xf [3fx fxx ]
− Yf [fy fxx + fx fxy + fx fxy ] − Xg [fx gxx + gx fxx + gx fxx ] − Yg [fy gxx
+ gx fxy + gx fxy ] − Xf f fx fx fx − Yf f fx fx fy − Xf g fx gx fx
− Yf g fx gx fy − Xf g gx fx fx − Yf g gx fx fy − Xgg gx gx fx − Ygg gx gx fy
(51)
Similarly,
1
f¯ȳȳ = fyy + αP22
+ O(α2 )
(52)
where,
1
P22
= Fyy + Fyf fy + Fyg gy + Fyf fy + Fyg gy − Xyy fx − Yyy fy
+ Ff fyy + Fg gyy − Xy fyx − Yy fyy − Xy fxy − Yy fyy + Ff f fy fy
+ Ff g fy gy + Fgf gy fy + Fgg gy gy − Xyf fy fx − Yyf fy fy − Xyg gy fx
− Yyg gy fy − Xyf fy fx − Yyf fy fy − Xyg gy fx − Yyg gy fy − Xf [fx fyy
+ fy fxy + fy fxy ] − Yf [3fy fyy ] − Xg [fx gyy + gy fxy + gy fxy ] − Yg [fy gyy
+ gy fyy + gy fyy ] − Xf f fy fy fx − Yf f fy fy fy − Xf g fy gy fx
− Yf g fy fy gy − Xf g gy fy fx − Yf g gy fy fy − Xgg gy gy fx − Ygg gy gy fy
(53)
Also, we have
40
2
+ O(α2 )
ḡx̄x̄ = gxx + αP11
(54)
where,
2
P11
= Gxx + Gxf fx + Gxg gx + Gxf fx + Gxg gx − Xxx gx − Yxx gy
+ Gf fxx + Gg gxx − Xx gxx − Yx gxy − Xx gxx − Yx gxy + Gf f fx fx
+ Gf g fx gx + Ggf gx fx + Ggg gx gx − Xxf fx gx − Yxf fx gy − Xxg gx gx
− Yxg gx gy − Xxf fx gx − Yxf fx gy − Xxg gx gx − Yxg gx gy − Xf [gx fxx + fx gxx + fx gxx ]
− Yf [gy fxx + fx gxy + fx gxy ] − Xg [gx gxx + gx gxx + gx gxx ] − Yg [gy gxx
+ gx gxy + gx gxy ] − Xf f fx fx gx − Yf f fx fx gy − Xf g fx gx gx
− Yf g fx gx gy − Xf g gx fx gx − Yf g gx fx gy − Xgg gx gx gx − Ygg gx gx gy
(55)
Finally, we have
2
ḡȳȳ = gyy + αP22
+ O(α2 )
where,
41
(56)
2
P22
= Gyy + Gyf fy + Gyg gy + Gyf fy + Gyg gy − Xyy gx − Yyy gy
+ Gf fyy + Gg gyy − Xy gyx − Yy gyy − Xy gxy − Yy gyy + Gf f fy fy
+ Gf g fy gy + Ggf gy fy + Ggg gy gy − Xyf fy gx − Yyf fy gy − Xyg gy gx
− Yyg gy gy − Xyf fy gx − Yyf fy gy − Xyg gy gx − Yyg gy gy − Xf [gx fyy
+ fy gxy + fy gxy ] − Yf [gy fyy + fy gyy + fy gyy ] − Xg [gx gyy + gy gxy + gy gxy ] − Yg [gy gyy
+ gy gyy + gy gyy ] − Xf f fy fy gx − Yf f fy fy gy − Xf g fy gy gx
− Yf g fy gy gy − Xf g gy fy gx − Yf g gy fy gy − Xgg gy gy gx − Ygg gy gy gy
(57)
Now, our invariance condition is
(m)
Ũ H (m) = Hx(m)
X1 + Hx(m)
X2 + Hu(m)
U1 + Hu(m)
U2 + Hp1 P11
1
2
1
2
1
(m)
(m)
(m)
(m)
(m)
11
12
1
1
+ Hp2 P12 + Hp1 P21 + Hp2 P22 + Hp1 P11
+ Hp1 P12
1
+
2
(m) 1
Hp1 P22
22
+
(m) 2
Hp2 P11
11
2
+
(m) 2
Hp2 P12
12
+
(58)
(m) 2
Hp2 P22
22
= km1 H (1) + km2 H (2)
where Ũ is the Lie derivative operator.
For our particular problem we have that
H 1 = fxx + fyy − (v1 f − v2 g)
(59)
H 2 = gxx + gyy − (v1 g + v2 f )
(60)
42
Therefore, our invariance conditions become
1
1
Ũ H 1 = −v1 F + v2 G + P11
+ P22
= k11 H 1 + k12 H 2
(61)
2
2
+ P22
= k21 H 1 + k22 H 2
Ũ H 2 = −v2 F − v1 G + P11
(62)
Substituting (51), (53), (59) and (60) into (61) we get
− v1 F + v2 G + Fxx + Fxf fx + Fxg gx + Fxf fx + Fxg gx − Xxx fx − Yxx fy
+ Ff fxx + Fg gxx − Xx fxx − Yx fxy − Xx fxx − Yx fxy + Ff f fx fx
+ Ff g fx gx + Fgf gx fx + Fgg gx gx − Xxf fx fx − Yxf fx fy − Xxg gx fx
− Yxg gx fy − Xxf fx fx − Yxf fx fy − Xxg gx fx − Yxg gx fy − Xf [3fx fxx ]
− Yf [fy fxx + fx fxy + fx fxy ] − Xg [fx gxx + gx fxx + gx fxx ] − Yg [fy gxx
+ gx fxy + gx fxy ] − Xf f fx fx fx − Yf f fx fx fy − Xf g fx gx fx
− Yf g fx gx fy − Xf g gx fx fx − Yf g gx fx fy − Xgg gx gx fx − Ygg gx gx fy
+ Fyy + Fyf fy + Fyg gy + Fyf fy + Fyg gy − Xyy fx − Yyy fy
+ Ff fyy + Fg gyy − Xy fyx − Yy fyy − Xy fxy − Yy fyy + Ff f fy fy
+ Ff g fy gy + Fgf gy fy + Fgg gy gy − Xyf fy fx − Yyf fy fy − Xyg gy fx
− Yyg gy fy − Xyf fy fx − Yyf fy fy − Xyg gy fx − Yyg gy fy − Xf [fx fyy
+ fy fxy + fy fxy ] − Yf [3fy fyy ] − Xg [fx gyy + gy fxy + gy fxy ] − Yg [fy gyy
+ gy fyy + gy fyy ] − Xf f fy fy fx − Yf f fy fy fy − Xf g fy gy fx
− Yf g fy fy gy − Xf g gy fy fx − Yf g gy fy fy − Xgg gy gy fx − Ygg gy gy fy
= k11 [fxx + fyy − (v1 f − v2 g)] + k12 [gxx + gyy − (v1 g + v2 f )]
(63)
43
Now, we equate like term coefficients to get the following system of equations
1:
−v1 F + v2 G + Fxx + Fyy = k11 [v2 g − v1 f ] − k12 [v1 g + v2 f ]
fx :
2Fxf − Xxx − Xyy = 0
(65)
fy :
−Yxx + 2Fyf − Yyy = 0
(66)
gx :
2Fxg = 0
(67)
gy :
2Fyg = 0
(68)
fx2 :
Ff f − 2Xxf = 0
(69)
gx2 :
Fgg = 0
(70)
2Ff g − 2Xxg = 0
fx gx :
0
(64)
(71)
fy2 :
Ff f − 2Yyf = 0
(72)
gy2 :
Fgg = 0
(73)
fy gy :
2Ff g − 2Yyg = 0
(74)
fx gy :
−2Xyg = 0
(75)
fx fy :
−2Yxf − 2Xyf = 0
(76)
gx fy :
−2Yxg = 0
(77)
fxx :
Ff − 2Xx = k11
(78)
fyy :
Ff − 2Yy = k11
(79)
fxy :
−2Yx − 2Xy = 0
(80)
gxx :
Fg = k12
(81)
gyy :
Fg = k12
(82)
fx fxx :
−3Xf = 0
(83)
fy fxx :
−Yf = 0
(84)
fx fxy :
−2Yf = 0
(85)
fx gxx :
−Xg = 0
(86)
44
gx fxx :
−2Xg = 0
(87)
fy gxx :
−Yg = 0
(88)
gx fxy :
−2Yg = 0
(89)
fx fyy :
−Xf = 0
(90)
fy fxy :
−2Xf = 0
(91)
fy fyy :
−3Yf = 0
(92)
fx gyy :
−Xg = 0
(93)
gy fxy :
−2Xg = 0
(94)
fy gyy :
−Yg = 0
(95)
gy fyy :
−2Yg = 0
(96)
Equations (83)-(96) tell us that
Xf = Xg = Yf = Yg = 0
In other words, X = X(x, y) and Y = Y (x, y).
Since the generators X and Y do not depend on f and g, we can drop the
additional term
−
m
m X
n X
X
∂ 2 Xb c a j
p p p
c ∂ua h i b
∂u
a=1 b=1 c=1
on the right side of equation (44).
45
Similarly, substituting (55), (57), (59), and (60) into (62) we have
− v2 F − v1 G + Gxx + Gxf fx + Gxg gx + Gxf fx + Gxg gx − Xxx gx − Yxx gy
+ Gf fxx + Gg gxx − Xx gxx − Yx gxy − Xx gxx − Yx gxy + Gf f fx fx
+ Gf g fx gx + Ggf gx fx + Ggg gx gx − Xxf fx gx − Yxf fx gy − Xxg gx gx
− Yxg gx gy − Xxf fx gx − Yxf fx gy − Xxg gx gx − Yxg gx gy − Xf [gx fxx + fx gxx + fx gxx ]
− Yf [gy fxx + fx gxy + fx gxy ] − Xg [gx gxx + gx gxx + gx gxx ] − Yg [gy gxx
+ gx gxy + gx gxy ] − Xf f fx fx gx − Yf f fx fx gy − Xf g fx gx gx
− Yf g fx gx gy − Xf g gx fx gx − Yf g gx fx gy − Xgg gx gx gx − Ygg gx gx gy
+ Gyy + Gyf fy + Gyg gy + Gyf fy + Gyg gy − Xyy gx − Yyy gy
+ Gf fyy + Gg gyy − Xy gyx − Yy gyy − Xy gxy − Yy gyy + Gf f fy fy
+ Gf g fy gy + Ggf gy fy + Ggg gy gy − Xyf fy gx − Yyf fy gy − Xyg gy gx
− Yyg gy gy − Xyf fy gx − Yyf fy gy − Xyg gy gx − Yyg gy gy − Xf [gx fyy
+ fy gxy + fy gxy ] − Yf [gy fyy + fy gyy + fy gyy ] − Xg [gx gyy + gy gxy + gy gxy ] − Yg [gy gyy
+ gy gyy + gy gyy ] − Xf f fy fy gx − Yf f fy fy gy − Xf g fy gy gx
− Yf g fy gy gy − Xf g gy fy gx − Yf g gy fy gy − Xgg gy gy gx − Ygg gy gy gy
= k21 [fxx + fyy − (v1 f − v2 g] + k22 [gxx + gyy − (v1 g + v2 f ]
(97)
Again, we equate like term coefficients to get the following system of equations
1:
fx :
−v2 F − v1 G + Gxx + Gyy = k21 [v2 g − v1 f ] − k22 [v1 g + v2 f ]
2Gxf = 0
(98)
(99)
46
fy :
2Gyf = 0
(100)
gx :
2Gxg − Xxx − Xyy = 0
(101)
gy :
2Gyg − Yxx − Yyy = 0
(102)
fx2 :
Gf f = 0
(103)
gx2 :
Ggg − 2Xxg = 0
(104)
2Gf g − 2Xxf = 0
fx gx :
0
(105)
fy2 :
Gf f = 0
(106)
gy2 :
Ggg − 2Yyg = 0
(107)
fy gy :
2Gf g − 2Yyf = 0
(108)
fx gy :
−2Yxf = 0
(109)
gx fy :
−2Xyf = 0
(110)
gx gy :
−2Yxg − 2Xyg = 0
(111)
fxx :
Gf = k21
(112)
fyy :
Gf = k21
(113)
gxx :
Gg − 2Xx = k22
(114)
gyy :
Gg − 2Yy = k22
(115)
gxy :
−2Yx − 2Xy = 0
(116)
gx fxx :
Xf = 0
(117)
fx gxx :
−2Xf = 0
(118)
gy fxx :
Yf = 0
(119)
fx gxy :
−2Yf = 0
(120)
gx gxx :
−3Xg = 0
(121)
gy gxx :
−Yg = 0
(122)
gx gxy :
−2Yg = 0
(123)
fx fyy :
−Xf = 0
(124)
47
fy gxy :
−2Xf = 0
(125)
gy fyy :
−Yf = 0
(126)
fy gyy :
−2Yf = 0
(127)
gx gyy :
−Xg = 0
(128)
gy gxy :
−Xg = 0
(129)
gy gyy :
−3Yg = 0
(130)
Equations (117)-(130) tell us that
Xf = Xg = Yf = Yg = 0
In other words, X = X(x, y) and Y = Y (x, y).
So once again, since the generators X and Y do not depend on f and g, we
can drop the additional term
m
m X
n X
X
∂ 2 Xb c a j
−
p p p
∂uc ∂ua h i b
a=1 b=1 c=1
on the right side of equation (44).
After simplifying equations (64) - (130) we have
Fxg = Fyg = Ff f = Fgg = Ff g = 0
1
Fxf = (Xxx + Xyy )
2
1
Fyf = (Yxx + Yyy )
2
(131)
(132)
(133)
Fg = k12
(134)
Ff = 2Xx + k11
(135)
− v1 F + v2 G + Fxx + Fyy = k11 [v2 g − v1 f ] − k12 [v1 g + v2 f ]
(136)
Gxf = Gyg = Ggg = Gf f = Gf g = 0
(137)
48
1
Gxg = (Xxx + Xyy )
2
1
Gyg = (Yxx + Yyy )
2
(138)
(139)
Gf = k12
(140)
Gg = 2Xx + k11
(141)
− v2 F − v1 G + Gxx + Gyy = k21 [v2 g − v1 f ] − k22 [v1 g + v2 f ]
(142)
Xx = Yy
(143)
Xy = −Yx
(144)
Now, we need to solve for the generators X, Y, F, and G.
Proposition 1: The two-dimensional stationary state Schrödinger equation,
∂ 2 Φ(x, y) ∂ 2 Φ(x, y)
+
− v(x, y)Φ(x, y) = 0,
∂x2
∂y 2
has the following general Lie group generators X, Y, F, and G, satisfying the
following relations:
Y = ηy − ζx
X = ηx + ζy,
where η and ζ are some constants, and F and G satisfy
− v1 F + v2 G + Fxx + Fyy = k11 [v2 g − v1 f ] − k12 [v1 g + v2 f ]
− v2 F − v1 G + Gxx + Gyy = k21 [v2 g − v1 f ] − k22 [v1 g + v2 f ]
49
Let us consider the special case when Fxx + Fyy = c1 and Gxx + Gyy = c2 ,
where c1 and c2 are constants. Equations, (136) and (142) become
− v2 F − v1 G = k21 [v2 g − v1 f ] − k22 [v1 g + v2 f ] − c2
(145)
− v1 F + v2 G = k11 [v2 g − v1 f ] − k12 [v1 g + v2 f ] − c1
(146)
From these two equations we find that
F =
λv2 + µv1
v12 − v22
(147)
G=
λv1 − µv2
v12 − v22
(148)
where
λ = k21 [v2 g − v1 f ] − k22 [v1 g + v2 f ] − c2
µ = k11 [v2 g − v1 f ] − k12 [v1 g + v2 f ] − c1
and, v1 and v2 are functions of x, y.
Corollary 1: The two-dimensional stationary state Schrödinger equation,
Φ(x, y)xx + Φ(x, y)yy − v(x, y)Φ(x, y) = 0, has the following Lie group gener-
50
ators X, Y, F, and G satisfying the following relations:
Y = ηy − ζx
X = ηx + ζy,
F =
λv2 + µv1
,
v12 − v22
G=
λv1 − µv2
v12 − v22
where η and ζ are some constants, v1 and v2 are functions of x, y and
λ = k21 [v2 g − v1 f ] − k22 [v1 g + v2 f ] − c2
µ = k11 [v2 g − v1 f ] − k12 [v1 g + v2 f ] − c1
subject to Fxx + Fyy = c1 and Gxx + Gyy = c2 .
Recall, similarity solutions are among the invariant surface conditions. The
invariant surface conditions give the following characteristic system
dx
dy
=
ηx + ζy
ηy − ζx
=
=
df
(k21 [v2 g−v1 f ]−k22 [v1 g+v2 f ]−c2 )v2 +(k11 [v2 g−v1 f ]−k12 [v1 g+v2 f ]−c1 )v1
v12 −v22
(149)
dg
(k21 [v2 g−v1 f ]−k22 [v1 g+v2 f ]−c2 )v1 −(k11 [v2 g−v1 f ]−k12 [v1 g+v2 f ]−c1 )v2
v12 −v22
Our first pair of equations are
dx
dy
=
ηx + ζy
ηy − ζx
rearranging gives
dy
ηy − ζx
=
dx
ηx + ζy
51
(150)
Integrating both sides gives the similarity variable as
s ≡ ζln(x2 + y 2 ) + ηarctan
y
x
(151)
The conditions Fxx + Fyy = c1 and Gxx + Gyy = c2 give a nonlinear coupled
second order system of PDE for v1 and v2 . On the other hand, this system
can be considered as a system of two linear equations for f and g.
   
f  m
A  =  
g
n
A closer examination of this system, we note that the determinant of its
coefficients is zero, and thus the system has no solutions for f , and g unless
m = n = 0. That is,
− c1 (v12 − v22 )3 − (c1 vxx + c2 v2xx )(v12 − v22 )2 − (c1 vyy + c2 v2yy )(v12 − v22 )2
2
+ 4(c1 vx + c2 v2x )(v1 vx − v2 v2x )(v12 − v22 ) + 2(c1 v1 + c2 v2 )(vx2 − v2x
+ v1 vxx − v2 v2xx )(v12 − v22 )
2
+ 4(c1 vy + c2 v2y )(v1 vy − v2 v2y )(v12 − v22 ) + 2(c1 v1 + c2 v2 )(vy2 − v2y
+ v1 vyy − v2 v2yy )(v12 − v22 )
− 8(c1 v1 + c2 v2 )(v1 vx − v2 v2x )2 − 8(c1 v1 + c2 v2 )(v1 vy − v2 v2y )2 = 0
and
c2 (v12 − v22 )3 − (c1 vxx + c2 v2xx )(v12 − v22 )2 − (c1 vyy + c2 v2yy )(v12 − v22 )2
2
+ 4(c1 vx + c2 v2x )(v1 vx − v2 v2x )(v12 − v22 ) + 2(c1 v1 + c2 v2 )(vx2 − v2x
+ v1 vxx − v2 v2xx )(v12 − v22 )
2
+ 4(c1 vy + c2 v2y )(v1 vy − v2 v2y )(v12 − v22 ) + 2(c1 v1 + c2 v2 )(vy2 − v2y
+ v1 vyy − v2 v2yy )(v12 − v22 )
− 8(c1 v1 + c2 v2 )(v1 vx − v2 v2x )2 − 8(c1 v1 + c2 v2 )(v1 vy − v2 v2y )2 = 0
Comparing the above two equations for consistency we must have c2 = −c1 .
Now, letting L = v1 +v2 , we get the necessary requirement to have non-trivial
52
solutions for f , and g as follows:
L3 + 2|∇L|2 − L∇2 L = 0
(152)
Theorem 1: The 2D stationary state Schrödinger equation, Φ(x, y)xx +
Φ(x, y)yy − v(x, y)Φ(x, y) = 0, has a class of similarity solutions provided
that v = v1 + i v2 satisfies
(v1 + v2 )3 + 2|∇(v1 + v2 )|2 − (v1 + v2 )∇2 (v1 + v2 ) = 0
and Φ = f + i g, satisfies the invariant surface conditions
dx
dy
df
dg
=
=
=
X
Y
F
G
with generators X, Y, F and G, given earlier, and the similarity variable given
by
s ≡ ζln(x2 + y 2 ) + ηarctan
y
x
.
Going back to equation (151), we note that if we choose
−k12 v12 + (k11 − k22 ) v2 v1 + k21 v22 = 0,
(153)
−k21 v12 + (k11 − k22 ) v2 v1 + k12 v22 = 0
(154)
then selecting the Lie group transformations free parameters as k11 = k22
53
and k21 = k12 = 0, the above equations are valid for all v1 and v2 , and our
invariance condition becomes


∂  H1 


∂α
H2

α=0


 1 0   H1 
= k11 


0 1
H2
Therefore, our system of PDEs are constantly conformally invariant and our
characteristic system reduces to
dy
dx
=
=
ηx + ζy
ηy − ζx
df
−c1 (v1 −v2 )−k11 (v12 +v22 )f
v12 −v22
=
dg
c1 (v1 −v2 )−k11 (v12 +v22 )g
v12 −v22
and can be integrated under suitable conditions.
We examine the conditions under which f and g can be obtained in terms of
s only from integrating the characteristic system. The idea is to express each
fraction in terms of s. We note that for some arbitrary non-zero constants,
a and b, we have
dx
dy
η(aydx + bxdy) − ζ(axdx − bydy)
=
=
ηx + ζy
ηy − ζx
2ab(ηx + ζy)(ηy − ζx)
Now, we choose the free parameters a, b, η, ζ so that the above fraction can
ds
for some function p(s), where s ≡ ζln(x2 +y 2 )+ηarctan xy .
be written as p(s)
Case 1: We select s =
y
,
x
η = −1, ζ = 0, and a = −b = −1 then (158)
reduces to
ds
=
2s
df
−c1 (v1 −v2 )−k11 (v12 +v22 )f
v12 −v22
If
54
=
dg
c1 (v1 −v2 )−k11 (v12 +v22 )g
v12 −v22
.
(155)
k11 (v12 + v22 )
v12 − v22
c1
, and,
v1 + v2
are functions of s then equation (155) can be integrated to obtain f and g
in terms of s. Let us now consider two special cases.
case 1.1: Let c1 = 0. Assuming
v12 +v22
v12 −v22
is a function of s then
−k11 (v12 + v22 )
ds
2s(v12 − v22 )
Z
f (s) = g(s) = Exp
therefore,
Z
Φ(x, y) = Φ(s) = Exp
−k11 (v12 + v22 )
ds
2s(v12 − v22 )
case 1.2: Let k11 = 0. Assuming v1 + v2 is a function of s then
Z
Φ(x, y) = Φ(s) =
−c1
ds
2s(v1 + v2 )
Case 2: We select s = x2 + y 2 , η = 0, ζ = −1, and a = −b = 2 then we get
ds =
8xy(v12 − v22 )df
8xy(v12 − v22 )dg
=
−c1 (v1 − v2 ) − k11 (v12 + v22 )f
c1 (v1 − v2 ) − k11 (v12 + v22 ) g
(156)
If
c1
, and,
xy(v1 + v2 )
k11 (v12 + v22 )
xy(v12 − v22 )
are functions of s then equation (156) can be integrated to obtain f and g
in terms of s = x2 + y 2 . Let us now consider two special cases.
55
case 2.1: Let c1 = 0. Assuming
v12 +v22
xy(v12 −v22 )
Z
Φ(x, y) = Φ(s) = Exp
is a function of s then
−k11 (v12 + v22 )
ds
8xy(v12 − v22 )
case 2.2: Let k11 = 0. Assuming v1 + v2 is a function of s then
Z
Φ(x, y) = Φ(s) =
56
−c1
ds
8xy(v1 + v2 )
Some special cases:
Let h(s) and k(s) be arbitrary functions of s, s is the similarity variable, and
c3 and c4 are some constants.
We consider the following cases.
(1) ζ = 0, c1 = 0 and k11 = −1 then integrating above system, provided that
v12 −v22
v12 +v22
is a constant, and s = y/x, we get
1
1
f = x c3 η h(s),
g = x c4 η k(s)
(157)
(2) ζ = 0, c1 = 1 and k11 = 0 then integrating above system, provided that
v1 + v2 is a constant, and s = y/x, we get
f = ln h(s)x
1
c3 η
,
g = ln k(s)x
1
c4 η
(158)
(3) ζ = 0, c1 = 0 and k11 = −1 then integrating above system, provided that
v12 +v22
v12 −v22
= x is a constant, and s = y/x, we get
c3
c4
f = e η x+h(s) ,
g = e η x+k(s)
(159)
(4) η = 0, c1 = 0 and k11 = −1 then integrating above system, provided that
v12 +v22
v12 −v22
= y is a constant, and s = x2 + y 2 , we get
c3
c4
f = e ζ x+h(s) ,
g = e ζ x+k(s)
57
(160)
Another approach: The Clarkson-Kruskal reduction method is one that
seeks a reduction of a given partial differential equation in the form
Φ(x, y) = α(x, y) + β(x, y)W (s(x, y))
(161)
We substitute (161) into our original partial differential equation, (45), and
demand that the result be an ordinary differential equation for Φ of s.The unusual characteristic of this method is that it does not use Lie group theory [7].
Substituting (161) into (45), and collecting coefficients of monomials of W
and its derivatives yields
(βs2x +βs2y )W 00 +(2βx sx +2βy sy +βsxx +βsyy )W 0 +(βxx+βyy −vβ)W +(αxx +αyy −vα) = 0
(162)
where 0 :=
d
.
ds
In order to make these equations be a system of ordinary differ-
ential equations for W, the ratios of their coefiicients of different derivatives
and powers have to be functions of s only. That is, the following equations
should be satisfied
2βx sx + 2βy sy + βsxx + βsyy = (βs2x + βs2y )Γ1 (s)
(163)
βxx + βyy − vβ = (βs2x + βs2y )Γ2 (s)
(164)
αxx + αyy − vα = (βs2x + βs2y )Γ3 (s)
(165)
where Γ1 (s), Γ2 (s), and Γ3 (s) are some arbitrary functions of s.
We will consider the case where α = 0. Therefore, equation (162) becomes
58
(βs2x + βs2y )W 00 + (2βx sx + 2βy sy + βsxx + βsyy )W 0 + (βxx + βyy − βv)W = 0
(166)
(166) is an ODE provided that equations (167) and (168) are satisfied for
some Γ1 (s) and Γ2 (s). Depending on the different values of β we can find
values for s and v that will satisfy equation (167) and (168).
Case 1: Let β = k = constant 6= 0, then equations (167) and (168) reduce
to
sxx + syy = (s2x + s2y )Γ1 (s)
(167)
− v = (s2x + s2y )Γ2 (s)
(168)
When s = x2 + y 2 equations (171) and (172) hold and (166) reduces to
4s2 W 00 + 4sW 0 − svW = 0
Notice that if we let v =
4c
s
(169)
then (169) is a Cauchy-Euler equation and has
the following solutions:
√
√
+ c2 s− c ; if c > 0
√
√
W = c1 cos( c lns) + c2 sin( c lns);
W = c1 s
c
where c, c1 , and c2 are some constants. Therefore,
Φ(x, y) = k c1 s
√
c
+ c2 s
√ − c
;
59
if c > 0
if c < 0
√
√
Φ(x, y) = k c1 cos( c lns) + c2 sin( c lns) ;
When s =
y
x
if c < 0
equations (171) and (172) hold and (166) reduces to
s2 + 1 00 2s 0
W + 2 W − vW = 0
x2
x
If we let v =
2
x2
(170)
then we have the following solution to (170)
W (s) = c1 (sArctan(s) + 1) + c2 s
where c1 and c2 are some constants. Therefore, we have
Φ(x, y) = k (c1 (sArctan(s) + 1) + c2 s)
Case 2: Let β = g(x, y) 6= constant. When s =
x
y
equation (166) reduces to
2
s +1
2y
−y
1
00
β
+β
W + 2βx
+ 2βy
W 0 +(βxx + βyy − βv) W = 0
2
2
x
x
x
x3
(171)
(2.1) Let β = xn , then (171) reduces to
(s2 + 1)W 00 + [2s(1 − n)] W 0 + (n(n − 1) − x2 v)W = 0
(172)
(2.2) Let β = ax + by, then (171) reduces to
(s2 + 1)W 00 +
−2axs + 2bx + 2sβ 0
W − x2 vW = 0
β
60
(173)
Note that
−2ay+2bx
ax+by
is a function of s only if a = 0 or b = 0.
If we let n = 3 and v =
2
x2
then (172) has the following solution
W (s) = c1 (−sArctan(s) − 1) + c2 s
where c1 and c2 are some constants. Therefore, we have
Φ(x, y) = x3 (c1 (−sArctan(s) − 1) + c2 s)
Case 3: Let β = g(x, y) 6= constant. When s = x2 + y 2 equation (166)
reduces to
β(4s)W 00 + (4xβx + 4yβy + 4β)W 0 + (βxx + βyy − βv)W = 0
(174)
(3.1) Let β = (xy)n , then (174) reduces to
00
0
4sW + (8n + 4)W +
If we let n = 1 and v =
−4c
s
n(n − 1)s
−v W =0
(xy)2
then (175) becomes
s2 W 00 + 3sW 0 − cW = 0
and has the following solutions:
61
(175)
√
√
W = c1 s−1+ 1+c + c2 s−1− 1+c ; if c > 1
√
√
c2
c1
W = cos( 1 + c lns) + sin( 1 + c lns);
s
s
if c < 1
where c1 and c2 are some constants. Therefore, we have
√
−1+ 1+c
√
−1− 1+c
Φ(x, y) = (xy) c1 s
+ c2 s
; if c > 1
c
√
√
c2
1
Φ(x, y) = (xy)
cos( 1 + c lns) + sin( 1 + c lns) ;
s
s
if c < 1
(3.2) Let β = y n , then (174) reduces to
00
0
sW + (n + 1)W +
If we let n = 1 and v =
−4c
s
n(n − 1) v
−
4y 2
4
W =0
then (176) becomes
s2 W 00 + sW 0 − cW = 0
(177)
and has the following solutions:
√
√
+ c2 s− c ; if c > 0
√
√
W = c1 cos( c lns) + c2 sin( c lns);
W = c1 s
c
(176)
62
if c < 0
√
√ Φ(x, y) = (y) c1 s c + c2 s− c ;
if c > 0
√
√
Φ(x, y) = (y) c1 cos( c lns) + c2 sin( c lns) ;
if c < 0
Through the Clarkson-Kruskal reduction method we were able to find solutions to the two dimensional stationary Schrödinger equation without having
to use Lie group theory. Let us look further into solutions to the Schrödinger
equation by plotting the energy functions and corresponding solutions.
63
3.3
Examples of Solutions to the Schrödinger Equation (45)
Now we will consider various choices of different functional values for v, the
energy function, and give the corresponding solution of (45).
Group A: Example of choices of v functional that leads to a complex-valued
Φ(x, y).
Case 1: Let v = a2 (x2 + y 2 ), where a is some constant. We get
Φ(x, y) = eaxy + ie−axy
Case 2: Let v = 4atanh(2a(x2 +y 2 ))+4a2 (x2 +y 2 ), where a is some constant.
We find
Φ(x, y) = ea(x
Case 3: Let v =
(x+y)2
e
2 +y 2 )
−2x
2x
y +(x−y)2 e y
y4 e
2x
y +e
−2x
−
y
+ ie−a(x
2ix
sech
y3
x
Φ(x, y) = e y + ie
2
Case 4: Let v =
2 +y 2 )
2x
. We find that
y
−x
y
2
(4a2 +b2 )(x2 +y 2 )+8abxy+4a)ea(x +y )+bxy +(4a2 +b2 )(x2 +y 2 )+8abxy−4a)e−a(x
2
2
2
2
e2a(x +y )+bxy +e−2(a(x +y )+bxy)
were a and b are some constants. We find that
2 +y 2 )+bxy
Φ(x, y) = ea(x
+ ie−a(x
2 +y 2 )−bxy
Figure A1: Plot of Modulus square of Φ for case 1 using a = 2
64
2 +y 2 )−bxy
Figure A2: Plot of Modulus square of Φ for case 2 using a = 2
Figure A3: Plot of Modulus square of Φ for case 3
Figure A4: Plot of Modulus square of Φ for case 4 using a = 1 and b = 1
65
Group B: Example of choices of v functional that leads to a real-valued
Φ(x, y).
Case 1: Let v =
4
PN
j=1
j 2 aj (x2 +y 2 )j−1
PN
j=0
aj (x2 +y 2 )j
Φ(x, y) =
. We find
N
X
aj (x2 + y 2 )j
j=0
2
2
)
Case 2: Let v = − 2(x b+y
sech2
2
xy−a
b
, where a and b are some constants.
We find that
Φ(x, y) = tanh
2a2 (x2 +y 2 )
,
a0 +a1 (xy)+a2 (xy)2
Case 3: Let v =
xy − a
b
where aj ’s are some constants. We find
that
Φ(x, y) = a0 + a1 (xy) + a2 (xy)2
Case 4: Let v = (−2ay − 2ax)sech2 (axy). We find that
Φ(x, y) = tanh(axy)
Case 5: Let v =
a0 b20 (x+y)2 eb0 xy +a1 b21 (x+y)2 eb1 xy
,
a0 eb0 xy +a1 eb1 xy
where aj ’s and bj ’s are some
constants. We find that
Φ(x, y) = a0 eb0 xy + a1 eb1 xy
Case 6: Let v =
4c
,
x2 +y 2
2
where c is some constant. We find that
2
√
Φ(x, y) = k c1 (x + y )
c
√ 2 − c
2
+ c2 (x + y )
;
if c > 0
√
√
Φ(x, y) = k c1 cos( c ln(x2 + y 2 )) + c2 sin( c ln(x2 + y 2 )) ;
66
if c < 0
Case 7: Let v =
−4c
,
x2 +y 2
where c is some constant. We find that
√
√
Φ(x, y) = (xy) c1 s−1+ 1+c + c2 s−1− 1+c ; if c > 1
c
√
√
c2
1
Φ(x, y) = (xy)
cos( 1 + c lns) + sin( 1 + c lns) ;
s
s
if c < 1
Figure B1(a): Plot of Φ (left) and v (right) for case 1, using N = 2, a0 =
1, a1 = 1, and a2 = 0
Figure B1(b): Plot of Φ (left) and v (right) for case 1, using N = 2, a0 =
1, a1 = 1, and a2 = 1
67
Figure B1(c): Plot of Φ (left) and v (right) for case 1, using N = 2, a0 =
1, a1 = 1, and a2 = −1
Figure B2: Plot of Φ (left) and v (right) for case 2, using a = 0 and b = 2
Figure B3(a): Plot of Φ (left) and v (right) for case 3, using a0 = 1,
0, and a2 = 1
68
a1 =
Figure B3(b): Plot of Φ (left) and v (right) for case 3, using a0 = 1,
1, and a2 = 1
a1 =
Figure B4: Plot of Φ (left) and v (right) for case 4, using a = 1
Figure B5(a): Plot of Φ (left) and v (right) for case 5, using a0 = 1,
1, b0 = 1 and b1 = 1
69
a1 =
Figure B5(b): Plot of Φ (left) and v (right) for case 5, using a0 = 1,
1, b0 = 1 and b1 = −2
a1 =
Figure B6: Plot of Φ (left) and v (right) for case 6, using c = 4,
2, and c1 = c2 = 1
β =
Figure B7: Plot of Φ (left) and v (right) for case 7, using c = 3,
xy, and c1 = c2 = 1
β =
70
To Conclude
Through the use of Lie group theory we were able to form the group that
would lead us to similarity solutions of the two dimensional stationary Schrödinger
equation. We had to set requirements on the energy function and the transformation free parameters in order to find these solutions. We then investigated the Clarkson-Kruskal reduction method and we’re able to find solutions
without the use of any Lie group theory. The plots in Chapter 3 section 3
show us how drastic our solution can change depending on the energy function that we have.
71
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