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 References [1] Logan, J. David, Applied Mathematics: A Contemporary Approach, New York: J. Wiley, 1987. [2] Sachdev, P.L., Self-similarity and Beyond: Exact Solutions of Nonlinear Problems, Boca Raton, FL: Chapman and Hall, 2000. [3] Olver, Peter J., Applications of Lie Groups to Differential Equations, New York: Springer-Verlag, 1986. [4] Kudryavtsev, A.G., Exactly solvable two-dimensional stationary Schrodinger operators obtained by the nonlocal Darboux transformation, Physics Letters A 377 (2013) 2477-2479. [5] Zakeri, G.-A., Similarity solutions for two-dimensional weak shock waves, Journal of Engineering Mathematics (2010) 67:275-288. [6] Ibragimov, N.H. (ed), Handbook of Lie group analysis of differential equations. Volume 1: Symmetries, exact solutions and conservation laws. (1994) CRC Press, Boca Raton. [7] Ibragimov, N.H. (ed), Handbook of Lie group analysis of differential equations. Volume 2: Applications in engineering and physical sciences. (1995) CRC Press, Boca Raton. [8] Ibragimov, N.H. (ed), Handbook of Lie group analysis of differential equations. Volume 3: New trends in theoretical developments and computational methods. (1996) CRC Press, Boca Raton. [9] Engui Fan and Manwai Yuen, Similarity reductions and new nonlinear exact solutions for the 2D incompressible Euler equations, Physics Letters A 378 (2014) 623-626. 72