CALIFORNIA STATE UNIVERSITY, NORTHRIDGE SIMILARITY SOLUTIONS OF THE TWO-DIMENSIONAL STATIONARY SCHRÖDINGER EQUATION

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1 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

2 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

3 Contents Signature page ii Abstract iv 1 Introduction 1 2 Lie groups and Similarity Solutions Introduction Lie Group Transformations Stretching Transformation Infinitesimal Transformation General Similarity Method The Schrödinger Equation The two-dimensional stationary Schrödinger equation Similarity Method applied to the Two-dimensional Stationary Schrödinger Equation Examples of Solutions to the Schrödinger Equation (45) References iii

4 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

5 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

6 2 Lie groups and Similarity Solutions 2.1 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 Consider the nonlinear one-dimensional wave equation: u tt uu xx = 0 (1) under a stretching transformation: x = α a x t = α b t ū = α c u We have : ū t t ūū x x = 2 ū d t t ū 2 ū x x = 2 α c u α b t α b t αc u 2 α c u = α a x α a x αc 2b 2 u t t α2c 2a u 2 u x 2 2

7 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 ū t t ūū x x = 1 α 2 (u tt uu xx ) (2) Therefore, the differential operator that defines the nonlinear one-dimensional wave equation is unchanged or invariant under the following stretching transformation: x = αx t = αt ū = u up to a constant factor 1 α 2. We can now introduce an independent variable s defined by s = x t and assume that u(x, t) = f(s) (3) Also, note that s is invariant under this transformation since s = x t = 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

8 2.2 Lie Group Transformations Definition Let α vary continuously over an open interval α α 0. Let φ(t, x, α) and ψ(t, x, α) be given analytic functions on R 2 ( α 0, α 0 ). A one parameter family of transformations on R 2 defined by T α : t = φ(t, x, α), x = ψ(t, x, α) (4) is a local Lie group if the following conditions hold: T α closed, i.e. If T α1, T α2 in T α then T α1 T α2 = T α3 in T α T α Contains the Identity, i.e. there exists a T 0 such that t = φ(t, x, 0) = t, x = ψ(t, x, 0) = x T α has an Inverse for small α, i.e. For every α 1 in ( α 0, α 0 ), there exists an α 2 in ( α 0, α 0 ) such that T α1 T α2 = T α2 T α1 = I Example Consider t = tcosα xsinα T α = x = 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 ( t, x) through an angle α, where 0 α < 2π. In order to check the closure property we must first define the following 4

9 t = tcosα 1 xsinα 1 T α1 = x = tsinα 1 + xcosα 1 t = tcosα 2 xsinα 2 T α2 = x = tsinα 2 + xcosα 2 Calculations show that T α2 T α1 is t = φ( t, x, α 2 ) = (tcosα 1 xsinα 1 )cosα 2 (tsinα 1 + xcosα 1 )sinα 2 = t(cosα 1 cosα 2 sinα 1 sinα 2 ) x(cosα 1 sinα 2 + sinα 1 cosα 2 ) = tcos(α 1 + α 2 ) xsin(α 1 + α 2 ) Similarly, x = ψ( t, x, α 2 ) = (tcosα 1 xsinα 1 )sinα 2 + (tsinα 1 + xcosα 1 )cosα 2 = t(cosα 1 sinα 2 + sinα 1 cosα 2 ) + x(cosα 1 cosα 2 sinα 1 sinα 2 ) = tsin(α 1 + α 2 ) + xcos(α 1 + α 2 ) Therefore, T α2 T α1 = T α3 belongs to the group of rotations with α 3 = α 1 + α 2. Next, we check that an Identity exists. Consider α = 0, then we have that t = tcos(0) xsin(0) = t T 0 = x = tsin(0) + xcos(0) = x 5

10 Lastly, the inverse transformation for T α is T α. Clearly, if α ( α 0, α 0 ) then α ( α 0, α 0 ). Other examples of local Lie groups are the following. Translation Group: x = x + α, t = t Stretching Group: x = α a x, t = α b t, a, b constants Let us take a further look at the stretching transformation. 6

11 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, u x, u t, u xx, u xt, u tt ) = 0 (5) Let the transformation T α : R 3 R 3 be defined as follows x = α a x T α = t = α b t (6) ū = α c u 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 T α1 T α2 = T α1 α 2 since x = α2(α a 1x) a = (α 2 α 1 ) a x T α1 T α2 = t = α2(α b 1t) b = (α 2 α 1 ) b t ū = α2(α c 1u) c = (α 2 α 1 ) c u Also, if we let α = 1 we get the identity and for every α I the inverse of T α is T α 1. Therefore the stretching group is a local Lie group. The stretching group takes our surface in R 3 defined by u = φ(x, t) to a surface in x tū space defined by ū = φ( x, t), ( x, t) D (7) 7

12 where D = {( x, t) x = α a x, t = α b t, (x, t) D} and φ is defined by φ( x, t) = α c φ(α a x, α b t) (8) Our transformed partial differential equation is written as F ( x, t, ū, ū x, ū t, ū x x, ū x t, ū t t) = 0 (9) In general φ( x, t) is not the same as φ( x, t). Here is an example to illustrate this notion. Example Consider a stretching transformation x = αx, t = αt, ū = α 2 u and let φ(x, t) = x t 2. Then, φ( x, t) = x t 2, whereas φ( x, t) = α 2 (α 1 x α 2 t 2 ) = α x t 2. Now, let us define what it means for a partial differential equation to be constant conformally invariant. Definition The partial differential equation (5) is constant conformally invariant under the one parameter family of stretching T α defined by (6) if, and only if, F ( x, t, ū, ū x, ū t, ū x x, ū x t, ū t t) = A(α)F (x, t, u, u x, u t, u xx, u xt, u tt ) (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

13 Example Find a stretching transformation under which the equation u t = vu xx is invariant. Let x = α a x T α = t = α b t ū = α c u Then, ū t vū x x = αc u α b t v 2 α c u α a x α a x = αc b u t vα c 2a u xx If we let c b = c 2a, then we have ū t vū x x = α c b (u t vu xx ) Let a = 1, b=2, and c=0. By definition we have that our partial differential equation is constant conformally invariant under the following transformation x = αx T α = t = α 2 t ū = u with a conformal factor of A(α) = α 2. Theorem If the partial differential equation (5) is constant conformally invariant under the one parameter family of stretchings T α given by 9

14 (6) and if φ(x, t) is a solution of (5), then φ( x, t) defined by (8) is a solution of the partial differential equation (9). Definition A solution u = φ(x, t) of (5) is an invariant solution if, and only if, φ( x, t) = φ( x, t) (11) under the transformation T α. Note that (11) is equivalent to φ(α a x, α b t) = α c 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 axφ x + btφ t = cφ (13) It s characteristic equations are dx ax = dt bt = dφ cφ Integrating the first and second pair of equations gives x b t a = constant and φt c b = constant respectively. Therefore, the general solution of (13) is ψ(φt c b, x b t a ) = 0 for some function ψ. The invariant surfaces are φ(x, t) = t c b f(s) (14) where f is an arbitrary function and s is the similarity variable defined by 10

15 s = xb t a (15) Theorem 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 u = t c b f(s) (16) where s is defined by (15), into (5) yeilds an equation of the form H(s, f, f, f ) = 0 which is an ordinary differential equation for f. Proof Recall that T α being invariant means we have F ( x, t, ū, ū x, ū t, ū x x, ū x t, ū t t) = A(α)F (x, t, u, u x, u t, u xx, u xt, u tt ) (17) where A is some function. Observe that ū x = α c a u x, ū t = α c b u t, ū x x = α c 2a u xx, ū x t = α c a b u xt, ū t t = α c 2b u tt. Differentiate (17) with respect to α and then setting α = 1 we get axf x + btf t + cuf u + (c a)u x F ux + (c b)u t F ut + (c 2a)u xx F uxx + (c a b)u xt F uxt + (c 2b)u tt F utt = A (1)F Our characteristic system is 11

16 dx ax = dt bt = du cu = = du xt (c a b)u xt = du x (c a)u x = Eight independent first integrals are du t (c b)u t = du tt = df (c 2b)u tt A (1)F du xx (c 2a)u xx x b t a, c ut b, ux t a c b, u t t b c b, uxx t 2a c b, u xt t b+a c b, u tt t 2b c b, F t A (1) b (18) Consequently, F = t A (1) b G(s, ut c b, ux t a c b, u t t b c b, uxx t 2a c b, u xt t b+a c b, u tt t 2b c b ) For some function G. Now, u is given by (16) and it follows that u x = bt c b a x b 1 f (s) u t = c b t c b 1 f(s) ax b c t b a 1 f (s) u xx = b(b 1)x b 2 t c ab b u xt = (c ab)x b 1 t c b(a+1) b u tt = c(c b) t c 2b b 2 b f(s) f (s) + b 2 x 2b 1 t c 2ab b f (s) f (s) abx 2b 1 t c b(2a+1) b f (s) a(2c ab b) b x b t c b(a+2) b f (s) + a 2 x 2b t c 2b(a+1) b f (s) When we substitute these values into G we obtain an ordinary differential equation of the form H(s, f, f, f ) = 0 Let us now illustrate this procedure with an example. 12

17 Example Consider the nonlinear partial differential equation uu t + u 2 x = 0 We assume a stretching transfomation T α : x = α a x, t = α b t, ū = α c u Then ūū t + ū 2 x = α 2c b uu t + α 2c 2a u 2 x = α 2c b (uu t + u 2 x) provided 2c-b = 2c-2a or b = 2a. Thus the partial differential equation is invariant under the following transformation T α : x = α a x, t = α 2a t, ū = α c u The invariant surface condition is axu x + 2atu t = cu having characteristic system dx ax = dt 2at = du cu (19) and first integrals x t and ut c 2a. Therefore, letting s = x t, the invariant surfaces are given by u(x, t) = t c 2a f(s) Let c = 0 and we have that u(x,t) = f(s). After substituting u(x,t) = f(s) into 13

18 our partial differential equation we obtain an ordinary differential equation for the function f, x f(s)f (s) + 1 t (f (s)) 2 = 0 (20) 2t 3 2 We can simplify (20) to get ( 1 ) 2 sf(s) + f (s) f (s) = 0 Example Consider the nonlinear one-dimensional wave equation from example We had that ū t t ūū x x = 2 ū d t t ū 2 ū x x = 2 α c u α b t α b t αc u 2 α c u α a x α a x = α c 2b 2 u t t α2c 2a u 2 u x 2 = αc 2b (u tt uu xx ) provided that c-2b = 2c-2a or c = 2a-2b. Thus the partial differential equation is invariant under T α : x = α a x, t = α b t, ū = α 2a 2b u The invariant surface condition is axu x + btu t = (2a 2b)u having characteristic system dx ax = dt bt = du (2a 2b)u 14

19 and first integrals xb t a surfaces are given by and ut 2b 2a b. Therefore, letting s = xb t a the invariant 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, We can simplify (21) to get 2 x t f (s) + x2 t 2 f (s) f(s)f (s) = 0 (21) 2sf (s) + (s 2 f(s))f (s) = 0 Example Consider the following acoustic approximation equations u t v x = 0 v t + uu x = 0 We assume a stretching transformation T α : x = α a x, t = α 2a t, ū = α c u, v = α e v Then, ū t v x = α c b u t α e a v x and 15

20 v t + ūū x = α e b v t + α 2c a uu x 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 x = α a x, t = α a t, ū = u, v = v Let a =1 and our stretching transformation becomes x = αx, t = αt, ū = u, v = v Then, by theorem 2.3.2, the system of partial differential equations are of the form u = f(s), v = g(s) (22) where the similarity variable s = x t. Now, after differentiating (22) and substituting the results into our original partial differential equation we get x t 2 f (s) 1 t g (s) = 0 x t 2 g (s) + 1 t f(s)f (s) = 0 multiply both equations by x so they can be written as s 2 f (s) + sg (s) = 0 s 2 g (s) + sf(s)f (s) = 0 Now, divide the second equation by s and add them together to get 16

21 f (s)(s 2 + f(s)) = 0 Case 1: f (s) = 0, which implies that f(s) = constant giving us that g(s) = constant as well. Case 2: s 2 + f(s) = 0 implies that f(s) = s 2. If we substitute this into s 2 g (s) + sf(s)f (s) = 0 we find that g(s) = 2s2 3. Therefore, the similarity solution for the acoustic approximation equations are given by u(x, t) = s 2 = x2 t 2 v(x, t) = 2s3 3 = 2x3 3t 3 17

22 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 T α 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 t = φ(t, x, 0) + φ α (t, x, 0)α + 1 2! φ αα(t, x, 0)α = t + τ(t, x)α + o(α) x = ψ(t, x, 0) + ψ α (t, x, 0)α + 1 2! ψ αα(t, x, 0)α = x + ξ(t, x)α + o(α) where τ and ξ are defined by τ(t, x) φ α (t, x, 0), ξ(t, x) ψ α (t, x, 0) (23) For small α the infinitesimal transformation t = t + τ(t, x)α + o(α), x = x + ξ(t, x)α + o(α) (24) approximates T α. The quantities τ and ξ are called the generators of T α 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

23 Example Referring to the transformation in Example we have τ(t, x) = α φ(t, x, α) α=0 = ( tsinα xcosα) α=0 = x ξ(t, x) = α ψ(t, x, α) α=0 = (tcosα xsinα) α=0 = t Therefore the infinitesimal transformation is t = t αx + o(α), x = x + αt + o(α) which approximates T α 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 subject to the initial conditions d t τ( t, x) = d x ξ( t, x) = dα t = t, x = x, at α = 0 Example We calculate the group whose generators are given by τ = t 2 and ξ = tx. The dynamical system is d t t = d x 2 t x = dα 19

24 with t = t and x = x at α = 0. Integrating d t t 2 = d x t x gives t x = c 1 where c 1 is some constant. Applying the initial conditions gives t x = c 1 = tx. Integrating d t t 2 = dα and applying the initial conditions gives t = 1 αt. Thus t x = t2 x. The group is therefore given by 1 αt t = 1 αt, x = t2 x t 1 αt 20

25 2.5 General Similarity Method Let us extend the similarity method to more general local LIe groups of transformations on R 3. Such transformations include translations, rotations, and other kinds of geometrical mappings. Consider a general one parameter family of transformations T α defined by the equations T α : x = Φ(x, t, α), t = Ψ(x, t, α), ū = Ω(x, t, u, α) (25) where α is a real parameter that varies over some open interval α < α 0 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 x = Φ(x, t, 0) = x t = Ψ(x, t, 0) = t ū = Ω(x, t, u, 0) = u Furthermore, let us assume that T α 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 x = x + αξ(x, t) + o(α) t = t + ατ(x, t) + o(α) (26) ū = u + αω(x, t, u) + o(α) where 21

26 ξ(x, t) Φ(x, t, 0) α τ(x, t) Ψ(x, t, 0) α ω(x, t, u) Ω(x, t, u, 0) α (27) The quantities ξ, τ and ω are called the generators of transformation (25). The local Lie group T α 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 d x ξ( x, t) = d t τ( x, t) = dū ω( x, t, ū) = dα x = x, t = t, ū = u, at α = 0 Example Determine the group with generators ξ = 2x + t, τ = t, ω = u + 1 We form the system d x 2 x + t = d t t = dū ū + 1 = dα subject to the initial condition x = x, t = t, ū = u, at α = 0. Three first integrals are found to be 22

27 x + t t 2 = c 1, t ū + 1 = c 2, ln(ū + 1) = α + c 3 Applying the initial conditions determines c 1, c 2 and c 3 and we obtain the three equations x + t t 2 = x + t t 2, t ū + 1 = t, ln(ū + 1) = α + ln(u + 1) u + 1 The third yeilds ū = (u + 1)e α 1 The second, therefore gives t = te α Finally, from the first x = (x + t)e 2α te α and the equations for the group are determined. The transformation T α 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( x, t), ( x, t) D 23

28 in x tū space, where D = {( x, t) x = Φ(x, t, α), t = Ψ(x, t, α), (x, t) D} To determine ḡ we invert the transformation x = Φ(x, tα), t = Ψ(x, t, α) on D to obtain x = x( x, t), t = t( x, t) Then ḡ is defined by ḡ( x, t) = Ω(x( x, t), t( x, t), g(x( x, t), t( x, t)), 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, u x, u t ) = 0 (28) Definition The partial differential equation (28) is constant conformally invariant under the local Lie group (25) if, and only if, α F ( x, t, ū, ū x, ū t) α=0 = kf (x, t, u, u x, u t ) (29) for some constant k. If k = 0, the partial differential equation is said to be 24

29 absolutely invariant. In (29) if it occurs that k = k(x, t, u, u x, u t ), then the partial differential equation is conformally invariant. Example Consider the diffusion operator u t u xx and the group of stretchings x = (1 + α)x t = (1 + α) 2 t (30) ū = u We have that ū t ū x x = (1 + α) 2 (u t u xx ) The right side of the last equation can be expanded about α = 0 to obtain ū t ū x x = (1 2α +...)(u t u xx ) = (u t u xx ) 2α(u t u xx ) + o(α 2 ) Taking the derivative with respect to α at α = 0 gives α (ū t ū x x ) α=0 = 2(u t u xx ) Therefore, the diffusion operator is constant confromally invariant under the local Lie group (30). Now, returning to definition 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 u x and q u t. Differentiating the left side of (29) gives 25

30 ( ) ( ) p q F x ξ + F t τ + F u ω + F p + F q = kf (31) α α=0 α α=0 Now, we must compute the following two expressions ( ) p π = π(x, t, u, p, q) F p α α=0 ( ) q χ = χ(x, t, u, p, q) F q α α=0 (32) Theorem If u = g(x, t), then under the transformation T α given by (25) the derivatives of ū = ḡ( x, t) are p ḡ x ( x, t) = t t D x Ω t x D t ω detj q ḡ t( x, t) = x td x Ω x x D t ω detj (33) (34) where D x Ω Ω x + g x Ω u, D t Ω Ω t + g t Ω u and J is the Jacobi matrix of the transformation x = Φ(x, t, α), t = Ψ(x, t, α) given by J x x t x x t t t 26

31 The right sides of (33) and (34) are evaluated at x = x( x, t) and t = t( x, t). Proof Recall that ḡ is defined by ḡ( x, t) = Ω(x( x, t), t( x, t), g(x( x, t), t( x, t)), α) Thus, ḡ x = Ω x x x + Ω t t x + Ω u (g x x x + g t t x ) = x x D x Ω + t x + D t Ω Similarly ḡ t = x td x Ω + t td t Ω. In matrix form, this is (ḡ x ḡ t) = (D x Ω D t Ω) x x t x x t t t The matrix on the right is the Jacobi matrix of the inverse transformation x = x( x, t), t = t( x, t) and hence is the inverse of J. Therefore (ḡ x ḡ t) = (D x Ω D t Ω)J 1 1 = (D x Ω D t Ω) detj t t t x x t x x Theorem The quantities π and χ defined by (32) are given by π = ω x + pω u pξ x qτ x (35) χ = ω t + qω u pξ t qτ t (36) 27

32 where ξ, τ and ω are the generators of (25). The invariance condition (31) can now be expressed as F x ξ + F t τ + F u ω + F p π + F q χ = kf (37) where π and χ are given by (35) and (36), respectively. The left side of (37) is denoted by ŨF where Ũ is the Lie derivative operator, or infinitesimal operator [2], defined by Ũ ξ x + τ t + ω u + π p + χ q The partial differential equation (28) will be invariant under the group of transformations (26) if ŨF = 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 Consider the equation u t + uu x = 0 or q + up = 0 We have that F(x,t,u,p,q) = q + u p and F x = F t = 0, F u = p, F p = u and F q = 1. Substituting these values, (35) and (36) into (37) gives pω + u(ω x + ω u p ξ x p τ x q) + (ω t + qω u pξ t qτ t ) = k(q + up) 28

33 Rearranging terms we can write p(ω + uω u uξ x ξ t ku) + q(ω u uτ x + τ t k) + (uω x + ω t ) = 0 Therefore, we have uω x + ω t = 0 ω + uω u uξ x ξ t ku = 0 ω u uτ x + τ t 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 = (x 1, x 2 ), u = ( (u 1,..., u n ), and p = (p j i ) = u j x i ). Consider the extended local Lie group T α written in infinitesimal form 29

34 x i = x i + αx i (x) + o(α) T α = ū j = u j + αu j (x, u) + o(α) p j i = pj i + αp j i (x, u, p) + o(α) where i = 1, 2 and j = 1,..., n and the generators P j i the generators X i and U j by are given in terms of P j i = U j x i + n ( U j k=1 u k pk i ) 2 l=1 ( ) Xl p j l x i If the generators X i depend on u as well, then there is an additional term n k=1 2 l=1 ( ) Xl u k pj l pk i subtracted on the right side of the last equation. The invariance condition takes the form ŨF (m) 2 i=1 F (m) x i X i + n j=1 F (m) u j U j + 2 i=1 n j=1 F (m) P j p j i = i n k mr 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( x, t) is a solution of F ( x, t, ū, p, q) = 0 It is also true that ū = ḡ( x, t) is a solution as well, but it is not always true that g( x, t) = ḡ( x, t), recall example Similarity solutions are found among the invariant surfaces, that is, solutions 30

35 u = g(x, t) for which g( x, t) = ḡ( x, t) (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)g x + τ(x, t)g t = 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 dx ξ(x, t) = dt τ(x, t) = du 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 ( ) H a x i, u j, uj 2 u j,,... = 0, x i x i x j where i = 1,..., n and j = 1,..., m with n independent variables, x i, and m dependent variables, u j (x i ). Consider the following one-parameter group of transformations x i = Φ i (x k, u l ; α) ū j = Ω j (x k, u l ; α) The corresponding infinitesimal transformations are 31

36 x i = x i + αx i (x k, u l ) + o(α) ū j = u j + αu j (x k, u l ) + o(α) where X i (x k, u l ) = Φ i α (x k, u l ; 0) U j (x k, u l ) = Ωj α (x k, u l ; 0) For convenience, we denote p j i uj x i, p j ih = 2 u j x i x h, p j i ūj x i, p j ih = 2 ū j x i x h We introduce the total derivative operator D Dx i = x i + m a=1 p a i u + m a n a=1 h=1 p a ih p a h +... Now, taking the derivative of x i, we obtain d x i = d (x i + αx i + o(α)) = dx i + αdx i + d(o(α)) = [ ( dxi Xi + α + X )] i u a dx dx h x h u a h + o(α) x h 32

37 [ ( Xi = δ ih + α + X )] i u a dx x h u a h + o(α) x h Where a = 1,..., m and δ ih is the Kronecker delta. Using the total derivative operator, we rewrite d x i as Similarly, for ū j, d x i = [ δ ih + α DX ] i dx h + o(α) Dx h dū j = d ( u j + αu j + o(α) ) = du j + αdu j + d(o(α)) [ du j = + α dx b ( U j + U )] j u a dx x b u a b + o(α) x b = [ u j + α DU ] j dx b + o(α) x b Dx b Then, p j i = ūj x i = [ u j [ δ ih + α ] x b + α DU j Dx b dx b + o(α) ( )] X i x h + X i u a u a x h dx h + o(α) = u j x b + α DU j Dx b + o(α) dx ( ) b X δ ih + α i x h + X i u a u a x h + o(α) dx h 33

38 ( u j = + α DU ) j + o(α) x b Dx b δ bh δ ih + α DX i Dx h + o(α) = ( u j + α DU ) ( j + o(α) δ ib α DX ) b + o(α) x b Dx b Dx i [ = uj DU j + α x i Dx i ] uj DX b + o(α) x b Dx i = p j i + αp j i + o(α) where P j i = DU j Dx i uj x b DX b Dx i = U j x i + m ( U j a=1 u a pa i ) n b=1 ( ) Xb p j b x i m n a=1 b=1 ( ) Xb u a pj b pa i (42) If we substitute uj x i for 2 u j x i x h, U j for U j i, and x i for x h in the equation [ u j DU j + α x i Dx i we obtain p j ih, in other words ] uj DX b + o(α) x b Dx i 34

39 2 ū j p j ih = x i x h = 2 u j x i x h + α [ DU j i Dx h ] 2 u j DX b + o(α) x i x b Dx h (43) = p j ih + αp j ih + o(α) where P j ih = DU j i Dx h = 2 U j x i x h u j DX b x i x b Dx h m 2 U j x i u a pa h + a=1 m U j u a pa ih m n a=1 a=1 b=1 m n a=1 b=1 n b=1 m a=1 X b n p j hb x i 2 m X b x i u a pa hp j b n a=1 b=1 b=1 2 U j x h u a pa i n 2 X b p j b x i x h m b=1 X b p j ib x + m h 2 X b x h u a pa i p j b m X b u a [pj b pa ih + p a i p j bh + pa hp j ib ] c=1 a=1 n m a=1 b=1 c=1 2 U j u c u a pc ip a h 2 X b u c u a pc hp a i p j b (44) The last term m n m a=1 b=1 c=1 2 X b u c u a pc hp a i p j b is only added on when the generators X i depend on u i as well. The r th derivative, P j i 1 i 2...i r, can be generated by applying the total derivative operator to the (r 1) th derivative and adds the term 35

40 DX b Dx ir p j i 1 i 2...i r1 b where i ρ = 1, 2,..., n for ρ = 1, 2,..., r. Define our infinitesimal operator,ũ, as Ũ = X i + U j x i u + P j j i 1 p j i 1 + P j i 1 i 2 p j i 1 i 2 + P j i 1 i 2 i 3 p j +... i 1 i 2 i 3 Our invariance condition becomes ŨH a = r k ab H b (x i, u j, p j i, pj ih,...) b=1 We have generalized our result in finding the group of transformations that will leave the partial differential equation(s) invariant. 36

41 3 The Schrödinger Equation 3.1 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 Consider the two-dimensional stationary Schrödinger equation 2 Φ(x, y) x Φ(x, y) y 2 v(x, y)φ(x, y) = 0 (45) where Φ(x, y) and v(x, y) are complex valued functions, thus Φ(x, y) = f(x, y) + ig(x, y) v(x, y) = v 1 (x, y) + iv 2 (x, y) (46) 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 f xx + f yy (v 1 f v 2 g) = 0 g xx + g yy (v 1 g + v 2 f) = 0 (47) 37

42 So, we have a system of two equations for two unknown functions f and g. It is assumed that v 1 and v 2 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

43 3.2 Similarity Method applied to the Two-dimensional Stationary Schrödinger Equation Recall system of equations (47). Let us define the following infinitesimal transformation x = x + αx + O(α 2 ) ȳ = y + αy + O(α 2 ) T α = f = f + αf + O(α 2 ) (48) ḡ = g + αg + O(α 2 ) where x = (x 1, x 2 ) = (x, y) u = (u 1, u 2 ) = (f, g) X = (X 1, X 2 ) = (X, Y ) (49) U = (U 1, U 2 ) = (F, G) We need to solve for f x x. From equation (43) and (44) we have that f x x = f xx + αp O(α 2 ) (50) where 39

44 P 1 11 = F xx + F xf f x + F xg g x + F xf f x + F xg g x X xx f x Y xx f y + F f f xx + F g g xx X x f xx Y x f xy X x f xx Y x f xy + F ff f x f x + F fg f x g x + F gf g x f x + F gg g x g x X xf f x f x Y xf f x f y X xg g x f x Y xg g x f y X xf f x f x Y xf f x f y X xg g x f x Y xg g x f y X f [3f x f xx ] Y f [f y f xx + f x f xy + f x f xy ] X g [f x g xx + g x f xx + g x f xx ] Y g [f y g xx + g x f xy + g x f xy ] X ff f x f x f x Y ff f x f x f y X fg f x g x f x Y fg f x g x f y X fg g x f x f x Y fg g x f x f y X gg g x g x f x Y gg g x g x f y (51) Similarly, fȳȳ = f yy + αp O(α 2 ) (52) where, P 1 22 = F yy + F yf f y + F yg g y + F yf f y + F yg g y X yy f x Y yy f y + F f f yy + F g g yy X y f yx Y y f yy X y f xy Y y f yy + F ff f y f y + F fg f y g y + F gf g y f y + F gg g y g y X yf f y f x Y yf f y f y X yg g y f x Y yg g y f y X yf f y f x Y yf f y f y X yg g y f x Y yg g y f y X f [f x f yy + f y f xy + f y f xy ] Y f [3f y f yy ] X g [f x g yy + g y f xy + g y f xy ] Y g [f y g yy + g y f yy + g y f yy ] X ff f y f y f x Y ff f y f y f y X fg f y g y f x Y fg f y f y g y X fg g y f y f x Y fg g y f y f y X gg g y g y f x Y gg g y g y f y (53) Also, we have 40

45 ḡ x x = g xx + αp O(α 2 ) (54) where, P 2 11 = G xx + G xf f x + G xg g x + G xf f x + G xg g x X xx g x Y xx g y + G f f xx + G g g xx X x g xx Y x g xy X x g xx Y x g xy + G ff f x f x + G fg f x g x + G gf g x f x + G gg g x g x X xf f x g x Y xf f x g y X xg g x g x Y xg g x g y X xf f x g x Y xf f x g y X xg g x g x Y xg g x g y X f [g x f xx + f x g xx + f x g xx ] Y f [g y f xx + f x g xy + f x g xy ] X g [g x g xx + g x g xx + g x g xx ] Y g [g y g xx + g x g xy + g x g xy ] X ff f x f x g x Y ff f x f x g y X fg f x g x g x Y fg f x g x g y X fg g x f x g x Y fg g x f x g y X gg g x g x g x Y gg g x g x g y (55) Finally, we have ḡȳȳ = g yy + αp O(α 2 ) (56) where, 41

46 P 2 22 = G yy + G yf f y + G yg g y + G yf f y + G yg g y X yy g x Y yy g y + G f f yy + G g g yy X y g yx Y y g yy X y g xy Y y g yy + G ff f y f y + G fg f y g y + G gf g y f y + G gg g y g y X yf f y g x Y yf f y g y X yg g y g x Y yg g y g y X yf f y g x Y yf f y g y X yg g y g x Y yg g y g y X f [g x f yy + f y g xy + f y g xy ] Y f [g y f yy + f y g yy + f y g yy ] X g [g x g yy + g y g xy + g y g xy ] Y g [g y g yy + g y g yy + g y g yy ] X ff f y f y g x Y ff f y f y g y X fg f y g y g x Y fg f y g y g y X fg g y f y g x Y fg g y f y g y X gg g y g y g x Y gg g y g y g y (57) Now, our invariance condition is ŨH (m) = H (m) x 1 X 1 + H (m) X 2 + H (m) x 2 u 1 U 1 + H (m) u 2 U 2 + H (m) p 1 1 P H (m) p H (m) p 1 22 P H (m) p 1 2 P H (m) p 2 11 P2 1 + H (m) P2 2 + H (m) P H (m) p 2 2 P H (m) p 2 12 p 1 11 P H (m) P p P p (58) = k m1 H (1) + k m2 H (2) where Ũ is the Lie derivative operator. For our particular problem we have that H 1 = f xx + f yy (v 1 f v 2 g) (59) H 2 = g xx + g yy (v 1 g + v 2 f) (60) 42

47 Therefore, our invariance conditions become ŨH 1 = v 1 F + v 2 G + P P 1 22 = k 11 H 1 + k 12 H 2 (61) ŨH 2 = v 2 F v 1 G + P P 2 22 = k 21 H 1 + k 22 H 2 (62) Substituting (51), (53), (59) and (60) into (61) we get v 1 F + v 2 G + F xx + F xf f x + F xg g x + F xf f x + F xg g x X xx f x Y xx f y + F f f xx + F g g xx X x f xx Y x f xy X x f xx Y x f xy + F ff f x f x + F fg f x g x + F gf g x f x + F gg g x g x X xf f x f x Y xf f x f y X xg g x f x Y xg g x f y X xf f x f x Y xf f x f y X xg g x f x Y xg g x f y X f [3f x f xx ] Y f [f y f xx + f x f xy + f x f xy ] X g [f x g xx + g x f xx + g x f xx ] Y g [f y g xx + g x f xy + g x f xy ] X ff f x f x f x Y ff f x f x f y X fg f x g x f x Y fg f x g x f y X fg g x f x f x Y fg g x f x f y X gg g x g x f x Y gg g x g x f y + F yy + F yf f y + F yg g y + F yf f y + F yg g y X yy f x Y yy f y + F f f yy + F g g yy X y f yx Y y f yy X y f xy Y y f yy + F ff f y f y + F fg f y g y + F gf g y f y + F gg g y g y X yf f y f x Y yf f y f y X yg g y f x Y yg g y f y X yf f y f x Y yf f y f y X yg g y f x Y yg g y f y X f [f x f yy + f y f xy + f y f xy ] Y f [3f y f yy ] X g [f x g yy + g y f xy + g y f xy ] Y g [f y g yy + g y f yy + g y f yy ] X ff f y f y f x Y ff f y f y f y X fg f y g y f x Y fg f y f y g y X fg g y f y f x Y fg g y f y f y X gg g y g y f x Y gg g y g y f y = k 11 [f xx + f yy (v 1 f v 2 g)] + k 12 [g xx + g yy (v 1 g + v 2 f)] (63) 43

48 Now, we equate like term coefficients to get the following system of equations 1 : v 1 F + v 2 G + F xx + F yy = k 11 [v 2 g v 1 f] k 12 [v 1 g + v 2 f] (64) f x : 2F xf X xx X yy = 0 (65) f y : Y xx + 2F yf Y yy = 0 (66) g x : 2F xg = 0 (67) g y : 2F yg = 0 (68) fx 2 : F ff 2X xf = 0 (69) gx 2 : F gg = 0 (70) f x g x : 2F fg 2X xg = 0 (71) fy 2 : F ff 2Y yf = 0 (72) gy 2 : F gg = 0 (73) f y g y : 2F fg 2Y yg = 0 (74) f x g y : 2X yg = 0 (75) f x f y : 2Y xf 2X yf = 0 (76) g x f y : 2Y xg = 0 (77) f xx : F f 2X x = k 11 (78) f yy : F f 2Y y = k 11 (79) f xy : 2Y x 2X y = 0 (80) g xx : F g = k 12 (81) g yy : F g = k 12 (82) f x f xx : 3X f = 0 (83) f y f xx : Y f = 0 (84) f x f xy : 2Y f = 0 (85) f x g xx : X g = 0 (86) 44

49 g x f xx : 2X g = 0 (87) f y g xx : Y g = 0 (88) g x f xy : 2Y g = 0 (89) f x f yy : X f = 0 (90) f y f xy : 2X f = 0 (91) f y f yy : 3Y f = 0 (92) f x g yy : X g = 0 (93) g y f xy : 2X g = 0 (94) f y g yy : Y g = 0 (95) g y f yy : 2Y g = 0 (96) Equations (83)-(96) tell us that X f = X g = Y f = Y g = 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 n m a=1 b=1 c=1 on the right side of equation (44). 2 X b u c u a pc hp a i p j b 45

50 Similarly, substituting (55), (57), (59), and (60) into (62) we have v 2 F v 1 G + G xx + G xf f x + G xg g x + G xf f x + G xg g x X xx g x Y xx g y + G f f xx + G g g xx X x g xx Y x g xy X x g xx Y x g xy + G ff f x f x + G fg f x g x + G gf g x f x + G gg g x g x X xf f x g x Y xf f x g y X xg g x g x Y xg g x g y X xf f x g x Y xf f x g y X xg g x g x Y xg g x g y X f [g x f xx + f x g xx + f x g xx ] Y f [g y f xx + f x g xy + f x g xy ] X g [g x g xx + g x g xx + g x g xx ] Y g [g y g xx + g x g xy + g x g xy ] X ff f x f x g x Y ff f x f x g y X fg f x g x g x Y fg f x g x g y X fg g x f x g x Y fg g x f x g y X gg g x g x g x Y gg g x g x g y + G yy + G yf f y + G yg g y + G yf f y + G yg g y X yy g x Y yy g y + G f f yy + G g g yy X y g yx Y y g yy X y g xy Y y g yy + G ff f y f y + G fg f y g y + G gf g y f y + G gg g y g y X yf f y g x Y yf f y g y X yg g y g x Y yg g y g y X yf f y g x Y yf f y g y X yg g y g x Y yg g y g y X f [g x f yy + f y g xy + f y g xy ] Y f [g y f yy + f y g yy + f y g yy ] X g [g x g yy + g y g xy + g y g xy ] Y g [g y g yy + g y g yy + g y g yy ] X ff f y f y g x Y ff f y f y g y X fg f y g y g x Y fg f y g y g y X fg g y f y g x Y fg g y f y g y X gg g y g y g x Y gg g y g y g y = k 21 [f xx + f yy (v 1 f v 2 g] + k 22 [g xx + g yy (v 1 g + v 2 f] (97) Again, we equate like term coefficients to get the following system of equations 1 : v 2 F v 1 G + G xx + G yy = k 21 [v 2 g v 1 f] k 22 [v 1 g + v 2 f] (98) f x : 2G xf = 0 (99) 46

51 f y : 2G yf = 0 (100) g x : 2G xg X xx X yy = 0 (101) g y : 2G yg Y xx Y yy = 0 (102) fx 2 : G ff = 0 (103) gx 2 : G gg 2X xg = 0 (104) f x g x : 2G fg 2X xf = 0 (105) fy 2 : G ff = 0 (106) gy 2 : G gg 2Y yg = 0 (107) f y g y : 2G fg 2Y yf = 0 (108) f x g y : 2Y xf = 0 (109) g x f y : 2X yf = 0 (110) g x g y : 2Y xg 2X yg = 0 (111) f xx : G f = k 21 (112) f yy : G f = k 21 (113) g xx : G g 2X x = k 22 (114) g yy : G g 2Y y = k 22 (115) g xy : 2Y x 2X y = 0 (116) g x f xx : X f = 0 (117) f x g xx : 2X f = 0 (118) g y f xx : Y f = 0 (119) f x g xy : 2Y f = 0 (120) g x g xx : 3X g = 0 (121) g y g xx : Y g = 0 (122) g x g xy : 2Y g = 0 (123) f x f yy : X f = 0 (124) 47

52 f y g xy : 2X f = 0 (125) g y f yy : Y f = 0 (126) f y g yy : 2Y f = 0 (127) g x g yy : X g = 0 (128) g y g xy : X g = 0 (129) g y g yy : 3Y g = 0 (130) Equations (117)-(130) tell us that X f = X g = Y f = Y g = 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 n m a=1 b=1 c=1 on the right side of equation (44). 2 X b u c u a pc hp a i p j b After simplifying equations (64) - (130) we have F xg = F yg = F ff = F gg = F fg = 0 (131) F xf = 1 2 (X xx + X yy ) (132) F yf = 1 2 (Y xx + Y yy ) (133) F g = k 12 (134) F f = 2X x + k 11 (135) v 1 F + v 2 G + F xx + F yy = k 11 [v 2 g v 1 f] k 12 [v 1 g + v 2 f] (136) G xf = G yg = G gg = G ff = G fg = 0 (137) 48

53 G xg = 1 2 (X xx + X yy ) (138) G yg = 1 2 (Y xx + Y yy ) (139) G f = k 12 (140) G g = 2X x + k 11 (141) v 2 F v 1 G + G xx + G yy = k 21 [v 2 g v 1 f] k 22 [v 1 g + v 2 f] (142) X x = Y y (143) X y = Y x (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) x Φ(x, y) y 2 v(x, y)φ(x, y) = 0, has the following general Lie group generators X, Y, F, and G, satisfying the following relations: X = ηx + ζy, Y = ηy ζx where η and ζ are some constants, and F and G satisfy v 1 F + v 2 G + F xx + F yy = k 11 [v 2 g v 1 f] k 12 [v 1 g + v 2 f] v 2 F v 1 G + G xx + G yy = k 21 [v 2 g v 1 f] k 22 [v 1 g + v 2 f] 49

54 Let us consider the special case when F xx + F yy = c 1 and G xx + G yy = c 2, where c 1 and c 2 are constants. Equations, (136) and (142) become v 2 F v 1 G = k 21 [v 2 g v 1 f] k 22 [v 1 g + v 2 f] c 2 (145) v 1 F + v 2 G = k 11 [v 2 g v 1 f] k 12 [v 1 g + v 2 f] c 1 (146) From these two equations we find that F = λv 2 + µv 1 v 2 1 v 2 2 (147) G = λv 1 µv 2 v 2 1 v 2 2 (148) where λ = k 21 [v 2 g v 1 f] k 22 [v 1 g + v 2 f] c 2 µ = k 11 [v 2 g v 1 f] k 12 [v 1 g + v 2 f] c 1 and, v 1 and v 2 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

55 ators X, Y, F, and G satisfying the following relations: X = ηx + ζy, Y = ηy ζx F = λv 2 + µv 1, G = λv 1 µv 2 v1 2 v2 2 v1 2 v2 2 where η and ζ are some constants, v 1 and v 2 are functions of x, y and λ = k 21 [v 2 g v 1 f] k 22 [v 1 g + v 2 f] c 2 µ = k 11 [v 2 g v 1 f] k 12 [v 1 g + v 2 f] c 1 subject to F xx + F yy = c 1 and G xx + G yy = c 2. Recall, similarity solutions are among the invariant surface conditions. The invariant surface conditions give the following characteristic system dx ηx + ζy = = = dy ηy ζx df (k 21 [v 2 g v 1 f] k 22 [v 1 g+v 2 f] c 2 )v 2 +(k 11 [v 2 g v 1 f] k 12 [v 1 g+v 2 f] c 1 )v 1 v 2 1 v2 2 dg (k 21 [v 2 g v 1 f] k 22 [v 1 g+v 2 f] c 2 )v 1 (k 11 [v 2 g v 1 f] k 12 [v 1 g+v 2 f] c 1 )v 2 v 2 1 v2 2 (149) Our first pair of equations are rearranging gives dx ηx + ζy = dy ηy ζx dy dx = ηy ζx ηx + ζy (150) 51

56 Integrating both sides gives the similarity variable as ( y s ζln(x 2 + y 2 ) + ηarctan x) (151) The conditions F xx + F yy = c 1 and G xx + G yy = c 2 give a nonlinear coupled second order system of PDE for v 1 and v 2. On the other hand, this system can be considered as a system of two linear equations for f and g. A f = m 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, c 1 (v1 2 v2) 2 3 (c 1 v xx + c 2 v 2xx )(v1 2 v2) 2 2 (c 1 v yy + c 2 v 2yy )(v1 2 v2) (c 1 v x + c 2 v 2x )(v 1 v x v 2 v 2x )(v1 2 v2) 2 + 2(c 1 v 1 + c 2 v 2 )(vx 2 v2x 2 + v 1 v xx v 2 v 2xx )(v1 2 v2) 2 + 4(c 1 v y + c 2 v 2y )(v 1 v y v 2 v 2y )(v1 2 v2) 2 + 2(c 1 v 1 + c 2 v 2 )(vy 2 v2y 2 + v 1 v yy v 2 v 2yy )(v1 2 v2) 2 8(c 1 v 1 + c 2 v 2 )(v 1 v x v 2 v 2x ) 2 8(c 1 v 1 + c 2 v 2 )(v 1 v y v 2 v 2y ) 2 = 0 and c 2 (v1 2 v2) 2 3 (c 1 v xx + c 2 v 2xx )(v1 2 v2) 2 2 (c 1 v yy + c 2 v 2yy )(v1 2 v2) (c 1 v x + c 2 v 2x )(v 1 v x v 2 v 2x )(v1 2 v2) 2 + 2(c 1 v 1 + c 2 v 2 )(vx 2 v2x 2 + v 1 v xx v 2 v 2xx )(v1 2 v2) 2 + 4(c 1 v y + c 2 v 2y )(v 1 v y v 2 v 2y )(v1 2 v2) 2 + 2(c 1 v 1 + c 2 v 2 )(vy 2 v2y 2 + v 1 v yy v 2 v 2yy )(v1 2 v2) 2 8(c 1 v 1 + c 2 v 2 )(v 1 v x v 2 v 2x ) 2 8(c 1 v 1 + c 2 v 2 )(v 1 v y v 2 v 2y ) 2 = 0 Comparing the above two equations for consistency we must have c 2 = c 1. Now, letting L = v 1 +v 2, we get the necessary requirement to have non-trivial 52

57 solutions for f, and g as follows: L 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 = v 1 + i v 2 satisfies (v 1 + v 2 ) (v 1 + v 2 ) 2 (v 1 + v 2 ) 2 (v 1 + v 2 ) = 0 and Φ = f + i g, satisfies the invariant surface conditions dx X = dy Y = df F = dg G with generators X, Y, F and G, given earlier, and the similarity variable given by ( y s ζln(x 2 + y 2 ) + ηarctan. x) Going back to equation (151), we note that if we choose k 12 v (k 11 k 22 ) v 2 v 1 + k 21 v 2 2 = 0, (153) k 21 v (k 11 k 22 ) v 2 v 1 + k 12 v 2 2 = 0 (154) then selecting the Lie group transformations free parameters as k 11 = k 22 53

58 and k 21 = k 12 = 0, the above equations are valid for all v 1 and v 2, and our invariance condition becomes α H 1 H 2 α=0 = k H 1 H 2 Therefore, our system of PDEs are constantly conformally invariant and our characteristic system reduces to dx ηx + ζy = dy ηy ζx = df c 1 (v 1 v 2 ) k 11 (v 2 1 +v2 2 )f v 2 1 v2 2 = dg c 1 (v 1 v 2 ) k 11(v 2 1 +v2 2)g v 2 1 v2 2 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 ηx + ζy = dy ηy ζx = η(aydx + bxdy) ζ(axdx bydy) 2ab(ηx + ζy)(ηy ζx) Now, we choose the free parameters a, b, η, ζ so that the above fraction can be written as ds p(s) for some function p(s), where s ζln(x2 +y 2 )+ηarctan ( y x). Case 1: We select s = y, η = 1, ζ = 0, and a = b = 1 then (158) x reduces to ds 2s = df c 1 (v 1 v 2 ) k 11 (v 2 1 +v2 2 )f v 2 1 v2 2 = dg c 1 (v 1 v 2 ) k 11(v1 2+v2 2)g v1 2 v2 2. (155) If 54

59 c 1 v 1 + v 2, and, k 11 (v v 2 2) v 2 1 v 2 2 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 c 1 = 0. Assuming v2 1 +v2 2 v 2 1 v2 2 is a function of s then ( ) k11 (v1 2 + v 2 f(s) = g(s) = Exp 2) 2s(v1 2 v2) ds 2 therefore, ( ) k11 (v1 2 + v 2 Φ(x, y) = Φ(s) = Exp 2) 2s(v1 2 v2) ds 2 case 1.2: Let k 11 = 0. Assuming v 1 + v 2 is a function of s then Φ(x, y) = Φ(s) = c 1 2s(v 1 + v 2 ) ds Case 2: We select s = x 2 + y 2, η = 0, ζ = 1, and a = b = 2 then we get ds = 8xy(v 2 1 v 2 2)df c 1 (v 1 v 2 ) k 11 (v v 2 2)f = 8xy(v 2 1 v 2 2)dg c 1 (v 1 v 2 ) k 11 (v v 2 2) g (156) If c 1 xy(v 1 + v 2 ), and, k 11 (v1 2 + v2) 2 xy(v1 2 v2) 2 are functions of s then equation (156) can be integrated to obtain f and g in terms of s = x 2 + y 2. Let us now consider two special cases. 55

60 case 2.1: Let c 1 = 0. Assuming v1 2+v2 2 xy(v1 2 v2 2 ) is a function of s then ( ) k11 (v1 2 + v 2 Φ(x, y) = Φ(s) = Exp 2) 8xy(v1 2 v2) ds 2 case 2.2: Let k 11 = 0. Assuming v 1 + v 2 is a function of s then Φ(x, y) = Φ(s) = c 1 8xy(v 1 + v 2 ) ds 56

61 Some special cases: Let h(s) and k(s) be arbitrary functions of s, s is the similarity variable, and c 3 and c 4 are some constants. We consider the following cases. (1) ζ = 0, c 1 = 0 and k 11 = 1 then integrating above system, provided that v 2 1 v2 2 v 2 1 +v2 2 is a constant, and s = y/x, we get f = x 1 c 3η h(s), g = x 1 c 4η k(s) (157) (2) ζ = 0, c 1 = 1 and k 11 = 0 then integrating above system, provided that v 1 + v 2 is a constant, and s = y/x, we get ) ) f = ln (h(s)x 1 c 3 η, g = ln (k(s)x 1 c 4 η (158) (3) ζ = 0, c 1 = 0 and k 11 = 1 then integrating above system, provided that v 2 1 +v2 2 v 2 1 v2 2 = x is a constant, and s = y/x, we get f = e c 3 η x+h(s), g = e c 4 η x+k(s) (159) (4) η = 0, c 1 = 0 and k 11 = 1 then integrating above system, provided that v 2 1 +v2 2 v 2 1 v2 2 = y is a constant, and s = x 2 + y 2, we get f = e c 3 ζ x+h(s), g = e c 4 ζ x+k(s) (160) 57

62 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 (βs 2 x+βs 2 y)w +(2β x s x +2β y s y +βs xx +βs yy )W +(βxx+β yy vβ)w +(α xx +α yy vα) = 0 (162) where := d. In order to make these equations be a system of ordinary differ- ds 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 2β x s x + 2β y s y + βs xx + βs yy = (βs 2 x + βs 2 y)γ 1 (s) (163) β xx + β yy vβ = (βs 2 x + βs 2 y)γ 2 (s) (164) α xx + α yy vα = (βs 2 x + βs 2 y)γ 3 (s) (165) where Γ 1 (s), Γ 2 (s), and Γ 3 (s) are some arbitrary functions of s. We will consider the case where α = 0. Therefore, equation (162) becomes 58

63 (βs 2 x + βs 2 y)w + (2β x s x + 2β y s y + βs xx + βs yy )W + (β xx + β yy βv)w = 0 (166) (166) is an ODE provided that equations (167) and (168) are satisfied for some Γ 1 (s) and Γ 2 (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 0, then equations (167) and (168) reduce to s xx + s yy = (s 2 x + s 2 y)γ 1 (s) (167) v = (s 2 x + s 2 y)γ 2 (s) (168) When s = x 2 + y 2 equations (171) and (172) hold and (166) reduces to 4s 2 W + 4sW svw = 0 (169) Notice that if we let v = 4c s the following solutions: then (169) is a Cauchy-Euler equation and has W = c 1 s c + c 2 s c ; if c > 0 W = c 1 cos( c lns) + c 2 sin( c lns); if c < 0 where c, c 1, and c 2 are some constants. Therefore, ( Φ(x, y) = k c 1 s c ) + c 2 s c ; if c > 0 59

64 Φ(x, y) = k ( c 1 cos( c lns) + c 2 sin( c lns) ) ; if c < 0 When s = y x equations (171) and (172) hold and (166) reduces to s W + 2s x 2 x W vw = 0 (170) 2 If we let v = 2 x 2 then we have the following solution to (170) W (s) = c 1 (sarctan(s) + 1) + c 2 s where c 1 and c 2 are some constants. Therefore, we have Φ(x, y) = k (c 1 (sarctan(s) + 1) + c 2 s) Case 2: Let β = g(x, y) constant. When s = x y equation (166) reduces to ( ) [ ( ) ( ) β s2 + 1 y 1 W + 2β x 2 x + 2β x 2 y + β x ( 2y x 3 )] W +(β xx + β yy βv) W = 0 (171) (2.1) Let β = x n, then (171) reduces to (s 2 + 1)W + [2s(1 n)] W + (n(n 1) x 2 v)w = 0 (172) (2.2) Let β = ax + by, then (171) reduces to (s 2 + 1)W + 2axs + 2bx + 2sβ W x 2 vw = 0 (173) β 60

65 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 x 2 then (172) has the following solution W (s) = c 1 ( sarctan(s) 1) + c 2 s where c 1 and c 2 are some constants. Therefore, we have Φ(x, y) = x 3 (c 1 ( sarctan(s) 1) + c 2 s) Case 3: Let β = g(x, y) constant. When s = x 2 + y 2 equation (166) reduces to β(4s)w + (4xβ x + 4yβ y + 4β)W + (β xx + β yy βv)w = 0 (174) (3.1) Let β = (xy) n, then (174) reduces to ( n(n 1)s 4sW + (8n + 4)W + (xy) 2 ) v W = 0 (175) If we let n = 1 and v = 4c s then (175) becomes s 2 W + 3sW cw = 0 and has the following solutions: 61

66 W = c 1 s 1+ 1+c + c 2 s 1 1+c ; if c > 1 W = c 1 s cos( 1 + c lns) + c 2 s sin( 1 + c lns); if c < 1 where c 1 and c 2 are some constants. Therefore, we have ( ) Φ(x, y) = (xy) c 1 s 1+ 1+c + c 2 s 1 1+c ; if c > 1 ( c1 Φ(x, y) = (xy) s cos( 1 + c lns) + c ) 2 s sin( 1 + c lns) ; if c < 1 (3.2) Let β = y n, then (174) reduces to sw + (n + 1)W + ( n(n 1) v ) W = 0 (176) 4y 2 4 If we let n = 1 and v = 4c s then (176) becomes s 2 W + sw cw = 0 (177) and has the following solutions: W = c 1 s c + c 2 s c ; if c > 0 W = c 1 cos( c lns) + c 2 sin( c lns); if c < 0 62

67 ( Φ(x, y) = (y) c 1 s c ) + c 2 s c ; if c > 0 Φ(x, y) = (y) ( c 1 cos( c lns) + c 2 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

68 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 = a 2 (x 2 + y 2 ), where a is some constant. We get Φ(x, y) = e axy + ie axy Case 2: Let v = 4atanh(2a(x 2 +y 2 ))+4a 2 (x 2 +y 2 ), where a is some constant. We find Case 3: Let v = (x+y)2 e 2x y +(x y) 2 e 2x y Φ(x, y) = e a(x2 +y 2) + ie a(x2 +y 2 ) ( y 4 e 2xy +e 2x ) 2ix y ( sech y 3 ) 2x. We find that y Φ(x, y) = e x y x + ie y Case 4: Let v = (4a2 +b 2 )(x 2 +y 2 )+8abxy+4a)e a(x2 +y 2 )+bxy +(4a 2 +b 2 )(x 2 +y 2 )+8abxy 4a)e a(x2 +y 2 ) bxy e 2a(x2 +y 2 )+bxy +e 2(a(x2 +y 2 )+bxy) were a and b are some constants. We find that Φ(x, y) = e a(x2 +y 2 )+bxy + ie a(x2 +y 2 ) bxy Figure A1: Plot of Modulus square of Φ for case 1 using a = 2 64

69 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

70 Group B: Example of choices of v functional that leads to a real-valued Φ(x, y). Case 1: Let v = 4 N j=1 j2 a j (x 2 +y 2 ) j 1 N j=0 a j(x 2 +y 2 ) j. We find Φ(x, y) = N a j (x 2 + y 2 ) j j=0 Case 2: Let v = 2(x2 +y 2 ) b 2 We find that Case 3: Let v = that sech ( ) 2 xy a b, where a and b are some constants. ( ) xy a Φ(x, y) = tanh b 2a 2 (x 2 +y 2 ) a 0 +a 1 (xy)+a 2 (xy) 2, where a j s are some constants. We find Φ(x, y) = a 0 + a 1 (xy) + a 2 (xy) 2 Case 4: Let v = ( 2ay 2ax)sech 2 (axy). We find that Φ(x, y) = tanh(axy) Case 5: Let v = a 0b 2 0 (x+y)2 e b 0 xy +a 1 b 2 1 (x+y)2 e b 1 xy a 0 e b 0 xy +a 1, where a e b 1 xy j s and b j s are some constants. We find that Φ(x, y) = a 0 e b 0xy + a 1 e b 1xy Case 6: Let v = 4c, where c is some constant. We find that x 2 +y 2 ( Φ(x, y) = k c 1 (x 2 + y 2 ) c ) + c 2 (x 2 + y 2 ) c ; if c > 0 Φ(x, y) = k ( c 1 cos( c ln(x 2 + y 2 )) + c 2 sin( c ln(x 2 + y 2 )) ) ; if c < 0 66

71 Case 7: Let v = 4c, where c is some constant. We find that x 2 +y 2 ( ) Φ(x, y) = (xy) c 1 s 1+ 1+c + c 2 s 1 1+c ; if c > 1 ( c1 Φ(x, y) = (xy) s cos( 1 + c lns) + c ) 2 s sin( 1 + c lns) ; if c < 1 Figure B1(a): Plot of Φ (left) and v (right) for case 1, using N = 2, a 0 = 1, a 1 = 1, and a 2 = 0 Figure B1(b): Plot of Φ (left) and v (right) for case 1, using N = 2, a 0 = 1, a 1 = 1, and a 2 = 1 67

72 Figure B1(c): Plot of Φ (left) and v (right) for case 1, using N = 2, a 0 = 1, a 1 = 1, and a 2 = 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 a 0 = 1, a 1 = 0, and a 2 = 1 68

73 Figure B3(b): Plot of Φ (left) and v (right) for case 3, using a 0 = 1, a 1 = 1, and a 2 = 1 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 a 0 = 1, a 1 = 1, b 0 = 1 and b 1 = 1 69

74 Figure B5(b): Plot of Φ (left) and v (right) for case 5, using a 0 = 1, a 1 = 1, b 0 = 1 and b 1 = 2 Figure B6: Plot of Φ (left) and v (right) for case 6, using c = 4, β = 2, and c 1 = c 2 = 1 Figure B7: Plot of Φ (left) and v (right) for case 7, using c = 3, β = xy, and c 1 = c 2 = 1 70