Lie Symmetries of Fundamental Solutions to the Leutwiler-Weinstein Equation

In this article, we study Lie symmetries to fundamental solutions to the Leutwiler-Weinstein equation Lu:=Δu+kxn∂u∂xn+ℓ(xn)2u=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ Lu:={\Delta} u+\frac{k}{x^{n}}\frac{\partial u}{\partial x^{n}}+\frac{\ell}{(x^{n})^{2}}u=0 $$\end{document} in the upper half-space ℝ+n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathbb {R}^{n}_{+}$\end{document}. Starting from the infinitesimal generators of the equation Lu = 0, we deduce symmetries of the equation Lu = δ(x − x0), and using its invariant solutions, we construct a fundamental solution. As an application, we study a Green functions of the operator in the hyperbolic unit ball.


Introduction
In 1953, Alexander Weinstein published his paper on axially symmetric potentials [25]. He started to study the problem u + k x n ∂u ∂x n = 0 in the upper half-space R n + = {(x 1 , ..., x n ) ∈ R n : x n > 0}; which is now known as the Weinstein equation. The parameter k can be real or complex. The problem has mathematical significance. Indeed, it is a generalization of the Laplace equation, and it is maybe a one of the most simple partial differential equations with non-constant coefficients.
In 1987, Heinz Leutwiler published the first paper in which he started to study the extension of the Weinstein equation with two parameters (see [21] and also [2]) This equation is called the Leutwiler-Weinstein equation. The parameters k and are considered as real numbers. Finding a fundamental solution to the equation is an interesting and complicated task. It has already been studied in some special cases by Sirkka-Liisa Eriksson and the second author, for example in [11][12][13][14][15]. In these papers, the approach is based on differential equations and hyperbolic geometry.
The constructive approach to find a fundamental solution based on local Lie symmetries has been introduced by the first author in [5,6]. The method itself is applicable, if a partial differential equation has enough symmetry. The method can be applied in different interesting cases; see for example [4,7,20].
In this paper, we give a general description of the method and the detailed construction of a fundamental solution to the Leutwiler-Weinstein equation. We hope that this gives the reader a good picture of the method itself and motivates the application of the algorithm in different cases.

Symmetries of Fundamental Solutions to Linear Partial Differential Equations
In [5] and [6], the first author introduced a constructive method to find fundamental solutions to linear partial equations of the form P u := |α|≤m a α (x)D α u = 0, (2) defined in an open set ⊂ R n . We assume that a α ∈ C ∞ ( ). In Eq. 2, we use the standard multi-index notation α = (α 1 , ..., α n ), |α| = α 1 + ... + α n and Let us now describe the method of how to find fundamental solutions. We assume that the reader knows basics of the local symmetry theory of partial differential equations, what is represented, for example, in [19,22,24]. Let g be a symmetry Lie algebra generated by the infinitesimal operators admitted by Eq. 2.
Proposition 1 [10] The symmetry Lie algebra g may be represented as the direct sum where g f is a finite dimensional Lie subalgebra generated by the infinitesimal generators of the form and g ∞ is an infinite dimensional Lie algebra generated by where β is an arbitrary smooth solution of Eq. 2.
The explicit description of the symmetry Lie algebra of fundamental solutions is based on the existence of the canonical function θ(x) described in the next proposition.
Proposition 2 [5] Let X ∈ g be an infinitesimal generator. Then X ∈ g f if and only if there exists a function θ ∈ C ∞ ( ) satisfying the equation Using the preceding function, we may give the following description for a symmetry Lie algebra of fundamental solutions.
Theorem 3 [5] The symmetry Lie algebra h of is a subalgebra of g, which may be represented as a direct sum The finite dimensional subalgebra h f is a subalgebra of g f where the coefficients of infinitesimal generators (3) satisfy the system ξ j (x 0 ) = 0, j = 1, ..., n, We observe, that if the Lie algebra h f is wide enough, we may try to construct a fundamental solution of operator P using invariants of the Lie algebra h f . These observations allow us to formulate the following algorithm. The preceding algorithm works in principle in every case, when the Lie subalgebra h f is wide enough, i.e., it allows us to construct invariant solutions. Examples of the use of the algorithm may be found in [3,5,20].
Step (d) is demonstrated for example in [8]. In this article, we will represented a comprehensive illustration of steps (a), (b), and (c) in the case of the Leutwiler-Weinstein equation.
In this section, we compute the infinitesimal generators of the Lie symmetry subalgebra g f for Eq. 1. The second prolongation of the infinitesimal generator (3) is of the form where the coeffients are given as where the total derivative is of the form see all details in [19,22,24]. Infinitesimal generators of g f may be obtained by solving the equation X 2 (Lu)| Lu=0 = 0. The equivalent system is described in the following lemma.
Proof The prolongation (6) acting on Eq. 1 gives Using the total derivative (9), we obtain that the coefficients (7) and (8) take the form Using these formulas, the prolongation (15) takes the form We make the restriction on Lu = 0 by substituting and we obtain Assuming u and its partial derivatives are linearly independent, we obtain the result.
The solution of the system expressed in the preceding lemmas is the following.

Proposition 6
The coefficients of the infinitesimal generators (3) are where e i j = −e Proof We give a detailed proof in the Appendix A.
we may write it in the form Hence, we obtain the compete list of infinitesimal generators.

Remark 8
In the Appendix A we give a detailed proof to find the preceding infinitesimal generators. There is also a another way to find them. If We make the substitution u(x) = (x n ) − k 2 v(x), it transform Eq. 1 to the Helmholz equation with a singular potential Hence, if k(2−k)+4 = 0 is just the Laplace equation and in other cases symmetry algebra is should be a subalgebra of that of Laplace equation (see, e.g., [19,22,24] or in above put k = 0). Because the lack of the explicit dependence of x n , most of the infinitesimal generators of the invariance algebra of the Laplace equation and and the preceding equation will remain the same. The only difference is the infinitesimal generator X i . With these, we can proceed as follows. We start from the corresponding infinitesimal generator of the Laplace equation, and define for α ∈ R as a parameter. Its second prolongation gives

Function θ(x) for the Leutwiler-Weinstein Equation
To compute infinitesimal generators of the Lie subalgebra h f , we need to compute the function θ(x) described in Proposition 2. We need to substitute the coefficients given in Proposition 6 into the formula (16). We first observe, that Eq. 11 gives Then using Eq. 12, we obtain x n u x n and these together gives On the other hand, using (A.6), (A.7) and the information, that e j i = −e i j , we compute Substituting these into (16), we obtain where we use the information η x n = η x i x i = 0 for i = 1, ..., n − 1. Substituting ξ n (x), we obtain This gives us the following result.

Proposition 9 The function θ(x) for the Weinstein-Leutwiler equation is
a j x j .

Symmetry of the Equation Lu = δ(x − x 0 )
Using the function θ(x), we can find the infinitesimal generators of the symmetry Lie algebra h f . We observe, that since Eq. 1 is translation invariant with respect to the variables x = (x 1 , ..., x n−1 ), it is enough to consider the symmetry for the equation In Theorem 3, we deduce, that the Lie algebra h f is generated by the infinitesimal generators (3) and they should satisfy the system . In our case, the first equation gives The general form of an infinitesimal generator of h f is where infinitesimal generators X j and Y ij are given in Theorem 7. We obtain the following theorem.

Fundamental Invariants of the Equation Lu = 0
The notion "fundamental invariant" means an invariant solution of the equation Lu = 0 where the solution is invariant with respect to the Lie subalgebra h f , depending on a point x 0 . These invariants are natural candidates to build fundamental solutions. We start from the equation and we obtain the corresponding Lagrange-Charpit equations The first and last terms gives us Since this does not depend on x i , for i = 1, ..., n − 1, and we obtain Then if and only if Using the classical method of characteristics for the first-order partial differential equations, we obtain the solution

Proposition 12
The fundamental invariants of Eq. 1 are

Finding Invariant Solutions
Using fundamental invariants, we may construct an invariant solution for the equation Lu = 0 depending on point x 0 . To obtain an invariant (fundamental) solution for Lu = δ(x − x 0 ), we use the form for the weak invariants of Berest, expressed in [9]. We make the Anzats where we denote Then Let us next prove the following proposition.

Proposition 13 Function (19) is a solution for (1) if and only if
for j = 1, ..., n − 1 and Then we compute x j x n w (z), Using the preceding observations, we have Then We see, that the unknown function w(z) may be found by the associated Legendre functions P μ ν and Q μ ν , which solves the associated Legendre equation where ν and μ are real or complex numbers.

Proposition 14
Invariant solutions of Eq. 1 with respect to the Lie algebra h f are of the form Proof Let us prove the equation given in the preceding proposition. First, we define new variables z = yz n 0 and the function g(y) := w(yx n 0 ). Then After the substitution, the equation in Proposition 13 takes the form . We make a substitution g(y) = (y 2 − 1) α v(y), and the equation in the above takes the form assuming that |y| = 1. We compare this equation with Eq. 20 and obtain Putting δ = 2α + γ and = 4α(α − 1) + 2nα, we have where the coefficient of v(y) is simplified using Comparing now Eq. 21 with the Legendre equation (20), we obtain We see, that the general solution of the system is a linear combination of Legendre functions with the preceding coefficients ν and μ, i.e., We first observe, that using formulas 8.
Taking into account (19), we obtain the result.
The associated Legendre functions in the above may be represented using the hypergeometric functions (see [1,16] where (q) 0 = 1 and (q) k = q(q + 1) · · · (q + k − 1) for k ≥ 1. Then, we obtain (see [1,16]) converging in |1 − z/x n 0 | < 2 and converging in |x n 0 /z| < 1. In general, parameters ν and μ are arbitrary complex numbers. The behaviour of functions depends on their numerical values and relations. Gelfand and Shilov's method of analytic continuation allows us to study this dependence systematically, see e.g. [16] and their references.

Proposition 15
The preceding function satisfies the equation Using Eq. 18, we obtain |x − (0, x n 0 )| 2 < 4x n 0 x n . It is well known that P(x, x n 0 ) is bounded when z → x n 0 .

Computing the Fundamental Solution
In the preceding section, we infer, that the fundamental solution may be founded by using the function Q(x; x n 0 ). Since it is an invariant solution of Eq. 17, we infer LQ(x; x n 0 ) = cδ(x − x 0 ). In this section, we compute the constant c and obtain a fundamental solution.
First, we prove the following technical lemma.

Lemma 16 If we define
Proof We first compute derivatives Then we have Since the Dirac delta satisfies for all smooth functions f , we obtain the following corollary.
There preceding lemma shows, that it is enough to find the fundamental solution of L. We will denote in this section λ = z/x n 0 . The preceding corollary motivates us to study the function The preceding function multiplied with a constant give us the fundamental solution for L.
We obtain Then we compute We can define the function f (λ) = f 0 (λ) f 0 (1) and we find completing the proof.
The preceding function F (x; x n 0 ) is a candidate for the fundamental solution. Next we extend L to distributions by LF, ϕ = F, Lϕ , where ϕ ∈ D(R n + ) is a test function.

Proposition 19 (Green's formula) Assume ⊂ R n is a bounded set with a smooth enough boundary. If u and v are twice differentiable real-valued functions on an open set including , we have
where dS is the Euclidean surface measure, n the outward unit normal on the boundary ∂ and ∂u ∂n = ∇u · n.
Proof The proposition follows from the classical Green's formula Let us define the r-ball with the centre ( 0, x n 0 ) by B r (x n 0 ) = {x ∈ R n : | x| 2 + (x n − x n 0 ) 2 < r 2 }.
We will always assume, that r > 0 is defined such, that B r (x n 0 ) ⊂ R n + . Let us compute the following crucial formula. Proof We compute ∂λ ∂x j = x j x n x n 0 for j = 1, ..., n − 1 and The unit normal on ∂B r (x n 0 ) is completing the proof.
Next, we recall the classical localization theorem.
Theorem 21 [17] If u : → R is a continuous function and B r (x n 0 ) ⊂ for some r > 0, then where ω n−1 is the surface area of the unit sphere S n−1 ⊂ R n .
Using the localization theorem, we can compute the following limits.

Lemma 22
If F is the function defined above and ϕ ∈ D(R n + ) a test function, then Proof Using the representation given in Proposition 18, we compute Since λ = r 2 2x n x n 0 + 1, we have 1 (x n 0 ) 2 for λ → 1, or equivalently r → 0 and especially then x n → x n 0 . Since μ = n−2 2 we have (λ 2 − 1) μ = r n−2 (λ) μ , Hence, using the localization Theorem 21 and Lemma 20, we obtain Similarly, we compute . We see that the first integral formula is true. To prove the second integral, we compute Let us now define Hence we obtain the following corollary.

Corollary 23
If G is the function defined above and ϕ ∈ D(R n + ) a test function, then Now we are ready to prove the following proposition.

Proposition 24
For the preceding G, we have Proof Assume ϕ ∈ D(R n + ) and supp(ϕ) ⊂ ⊂ R n + , where is a set with a smooth enough boundary. Assume ( 0, x n 0 ) ∈ supp(ϕ) and take B r (x n 0 ) ⊂ and r (x n 0 ) := \B r (x n 0 ). Since ∂ r (x n 0 ) = ∂ ∪ (−∂B r (x n 0 )), we compute by Green's formula In the last part, we use the information, that ϕ and ∂ϕ ∂n vanish in the boundary ∂ . Since LG = 0 in r (x n 0 ), we have We observe, that since G is continuous outside of λ = 1 and Lϕ is smooth with compact support, then G Lϕ is locally integrable, and we can compute the limit r → 0. Using the preceding corollary, we have LG, ϕ = G, Lϕ We may give the following crucial result.

Theorem 25 The function
In the formula, ω n−1 is the surface area of the unit sphere Proof By virtue of Corollary 17 and the preceding proposition, we obtain that which satisfies the equation LH (x; x n 0 ) = δ( x)δ(x n − x n 0 ). Using Proposition 18, we obtain We define

The function f (λ) is given in Proposition 18, and we have
Using the formula 6.1.15 of [1], we have μ (μ) = (μ + 1), that is, We complete the paper by making the following remarks.

Remark 26 Since the Weinstein-Leutwiler equation is translation invariant with respect to transformations
x → x + x 0 , we obtain a fundamental solution H (x; x 0 ) at any point x 0 ∈ R n + just making the substitution.

Remark 27
The preceding fundamental solution H (x; x n 0 ) is not unique, since we can always add an arbitrary solution. We can say that all invariant fundamental solutions with respect to the Lie algebra h f are of the form H (x; Proposition 14), where c ∈ R.
By representing the formula given in Theorem 25 using the Legendre function Q −μ ν , we obtain the main result of the study.

Application: Mean Value Principle for the Hyperbolic Unit Ball at ( 0, x n 0 )
The classical Dirichlet problem in Euclidean space is usually formulated as follows: Given a function f that has values everywhere on the boundary of a region in R n , is there a unique continuous function u twice continuously differentiable in the interior and continuous on the boundary, such that u is harmonic in the interior and u = f on the boundary?
A solution to the problem depends on the geometry of the domain. For example, in unit ball B(0, 1), a solution is given by the so-called Poisson integral formula where P (x, y) is the so-called Poisson kernel and S n−1 = ∂B(0, 1) is the unit sphere.
In this section, we consider a Dirichlet problem of the Leutwiler-Weinstein operator L.
Recently, there has been a growing interest in such problems, see for example, [18,23]. The Dirichlet problem is then As in the Euclidean case k = = 0, to obtain an explicit representation formula, we need to restrict a geometrically suitable case. In this section, we consider the case where is the so-called hyperbolic unit ball in the upper half-space, defined in the next section. Using the fundamental solution, we can find a Poisson-type kernel and general representation formula for the preceding Dirichlet problem at the origin of the ball. Unfortunately, a general formula is still an open question.

Poincaré Upper Half-Space
We assume, that in the upper half-space, is endowed with the non-Euclidean metric The Riemannian manifold (R n + , ds 2 ) is called the hyperbolic Poincaré half-space. The straight lines in the preceding hyperbolic space are represented by circular arcs crossing perpendicular to the x n = 0 plane. The distance between two points x, x 0 ∈ R n + with respect to the preceding metric is computed by where the auxiliary function λ is These observations allow us to define balls in the upper half-space. We consider the r-ball, with the centre ( 0, x n 0 ), and we denote B h (x n 0 , r) = {x ∈ R n + : 0 ≤ d(x, x n 0 ) < r} = {x ∈ R n + : 1 ≤ λ(x, x n 0 ) < R} where R = cosh(r). For the unit ball B h (x n 0 ) := B h (x n 0 , 1), we denote R 1 = cosh(1) ≈ 1.543... Geometrically, the preceding r-ball is just the Euclidean ball B e (z e , r e ) = {x ∈ R n : |x − z e | < r e }, with the centre z e = ( 0, x n 0 cosh(r)) and the radius r e = x n 0 sinh(r). See all details of the preceding discussion and more, e.g., in [11][12][13][14][15].

Green Function of Unit Ball
A Green function on a domain ⊂ R n + is a function G(x; x 0 ) satisfying for all x, x 0 ∈ . In general, such a function is not easy to find, since it depends on the shape of the . Usually, to compute a Green's function, the set should have enough symmetry.
One of the cases with enough symmetry is a hyperbolic unit ball, which we consider next. Moreover, we have the following useful transformation formula.

Proposition 30 (Symmetry property) A fundamental solution satisfies
Using the preceding information, we can compute the Green function at the origin ( 0, x n 0 ) as follows.
Theorem 31 (Green function at the origin of the hyperbolic unit ball) The function The P −μ ν part of the fundamental solution exists, if |λ−1| < 2 or equivalently −1 < λ < 3. Especially, R 1 < 3, that is, the construction exists on the unit ball.
Unfortunately, the preceding formula is not valid at every point of the unit ball, only at the origin. The usual technique used in the Euclidean case seems to be hard to apply directly. We leave this question open and just give the following conjecture.
Conjecture 32 (Green function of the hyperbolic unit ball) There exists a Green function G(x; x 0 ) with the symmetry property (x n ) k G(x; y) = (y n ) k G(y; x) (maybe up to a constant) satisfying LG(x; Using the classical methods of partial differential equations, one can prove that the preceding Green function exists. The symmetry property must also be true, since all Green functions are fundamental solutions.

Representation Formula for Solutions to the Dirichlet Problem
In this section, we derive an integral representation of solutions to the Dirichlet problem assuming that the Green function, given in Conjecture 32 exists. This motivates us to find an explicit expression for the Green function in future studies. Our problem is to study The necessary condition for integral representations is the existence of the Green's type integral formula. Replacing u by (x n ) k u, we have Changing the role of u and v, we have Subtracting the preceding integrals from the upper on, we obtain We add and subtract the term (x n ) k (x n ) 2 uv in the volume integral and we have completing the proof.
Using Green's formula, we can prove an integral representation formula for the Dirichlet problem. We will write L x , if we want to emphasize the variable.

Theorem 34 A solution to the Dirichlet problem
in ∂B h (x n 0 ). can be given by ∂G(x; y) ∂n x n y n k dS(x).
Proof By Green's formula Taking v = G(x; y), i.e., LG(x; y) = δ(x − y), we have We have G(x; y) ∂u ∂n x n y n k dS(x).
Since G(x; y) = 0 in x ∈ ∂B(x n 0 ), we have x n y n k dS(x).
If Lu = f in the interior and u = g on a boundary, we have ∂G(x; y) ∂n x n y n k dS(x).
Using (x n ) k G(x; y) = (y n ) k G(y; x), we have ∂G(x; y) ∂n x n y n k dS(x).
If f ≡ 0, then we obtain the following Poisson-type representation formula.

Conclusions
In this paper, we study the symmetries of fundamental solutions of the Leutwiler-Weinstein equation. The method is described by the first author in [5]. As a result, we compute detailed the fundamental solution and study, how to use it to find Green's function for the problem. In the future, we will complete this task and construct give a detailed construction for it. Also some other interesting linear partial differential equations with non-constant coefficients should be studied. We hope, that our text motivates researchers to apply the method in their studies.
Assume that the root α = 1. Then ξ n (x) = (x n ) α h( x) and by Eq. 14 ξ n x n (x) = ξ j x j (x) = α(x n ) α−1 h( x) and by Eq. 13 ξ n x j x j + ξ j x j x n = (x n ) α h x j x j ( x) + α(α − 1)(x n ) α−2 h( x) = 0, that is, h = 0, and we see that these solutions do not give us a nontrivial symmetry.
Then we have by (13) and (14) that g j x j + c.
Let us now compute the coefficients ξ i . By Eq. 14, we obtain Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.