This thesis examines the fundamental solution to a generalized Laplace equation in a three-dimensional sub-Riemannian space called Grushin space. The author presents the main result as a theorem stating that the generalized Laplace operator applied to a particular function f equals zero, with f containing vector fields and functions that define the Grushin space. The proof of the theorem involves directly computing the generalized Laplace operator by taking derivatives of f with respect to the vector fields. This generalizes previous work that solved the two-dimensional case.

1. University of South Florida
Honors Thesis
The Fundamental Solution of
an Extension to a
Generalized Laplace
Equation
Johnathan Gray
directed by
Thomas Bieske, Ph.D.
November 18th, 2015

2. 1
1. Introduction and Motivation
In basic mathematics and classical physics, as well as many other
fields, models rely on Euclidean spaces and the geometry within. An
important property of Euclidean space is that every direction is iden-
tical. This idea is unsuitable for many situations. One example is
driving in a parking garage. One can not move straight in the vertical
direction, and thus movement in the vertical direction is dependent
on movement in the horizontal directions. Because of this, Euclidean
space isn’t an accurate representation of this. The correct spaces for
modeling such situations are sub-Riemannian spaces. Sub-Riemannian
spaces generalize Euclidean spaces by restricting certain tangent vec-
tors to model the difficulty in moving in specific directions.
The study of partial differential equations in sub-Riemannian spaces
is an area of interest in modern mathematics, and is an area which is
still under investigation. The p-Laplace equation is considered foun-
dational in the study of partial differential equations, and thus is of
particular interest in this area.
In [1], the fundamental solution to the 2-Laplace equation is found
in a wide class of sub-Riemannian spaces. This solution is found the-
oretically, and is expressed using integrals which may or may not be
evaluated. In this thesis, we look at this solution in the three di-
mensional Grushin space, and we write this equation as an algebraic
expression. Using a direct technique, we evaluate this closed form to
verify the solution found in [1].
The equation we present is a generalization of computations done in

3. 2
[2]. The difference between the work presented in our thesis and the
computations in [2] is that [2] considers the fundamental solution in
the two dimensional Grushin plane, while we extend this to the three
dimensional Grushin space. This is a non-trivial higher dimensional
generalization and is therefore of interest to experts in this area. Our
result sets up future exploration in this area, as in [2]. This thesis will
be part of a research paper submitted to a refereed journal and is the
first step in achieving results similar to that of [2].

4. 3
2. The Environment
We are in R3
and, for 0 = c ∈ R, a1, a2 ∈ R and 0 ≤ k ∈ R, have the
following vector fields:
X1 =
∂
∂x1
X2 =
∂
∂x2
and X3 = c (x1 − a1)2
+ (x2 − a2)2
k
2 ∂
∂x3
.
Because the vector X3 vanishes at a point (a1, a2, b), we have a sub-
Riemannian system of vector fields. That is, we have 3 vectors at any
point not of the form (a1, a2, b) but only 2 at points (a1, a2, b).
Our main function of interest is
f(x1, x2, x3) = c (x1 − a1)2
+ (x2 − a2)2
k+1
2
− i(k + 1)(x3 − b)
−1
2
(1−L)
× c (x1 − a1)2
+ (x2 − a2)2
k+1
2
+ i(k + 1)(x3 − b)
−1
2
(1+L)
for b ∈ R and L ∈ R.

5. 4
For convenience we define the following functions:
d = (x1 − a1)2
+ (x2 − a2)2
g = cd
k+1
2 − i(k + 1)(x3 − b)
h = cd
k+1
2 + i(k + 1)(x3 − b)
α = −
1
2
(1 − L)
and β = −
1
2
(1 + L)
so that we have f = gα
hβ
.
Note that X3 reduces to cd
k
2
∂
∂x3
.
3. The Main Result
Theorem 1. Given the vector fields and functions above, we have,
∆2f + iLc(k + 1) (x1 − a1)2
+ (x2 − a2)2
k−1
2 ∂
∂x3
f = 0
in R3
{(a1, a2, b)}.
Proof.
Recall that
∆2f = X1X1f + X2X2f + X3X3f.

6. 5
We begin by computing X1X1f:
X1f = αgα−1
hβ
X1g + βgα
hβ−1
X1h
X1X1f = α(α − 1)gα−2
hβ
(X1g)2
+ αβgα−1
hβ−1
(X1g)(X1h)
+ αgα−1
hβ
X1X1g + β(β − 1)gα
hβ−2
(X1h)2
+ αβgα−1
hβ−1
(X1g)(X1h) + βgα
hβ−1
X1X1h.
Simplifying, we obtain
X1g = c(k + 1)(x1 − a1)d
k−1
2
X1h = c(k + 1)(x1 − a1)d
k−1
2 = X1g
X1X1g = c(k + 1)d
k−1
2 + c(k − 1)(k + 1)(x1 − a1)2
d
k−3
2
X1X1h = X1X1g

7. 6
X1X1f = α(α − 1)gα−2
hβ
c2
(k + 1)2
(x1 − a1)2
dk−1
+ αβgα−1
hβ−1
c2
(k + 1)2
(x1 − a1)2
dk−1
+ αgα−1
hβ
c(k + 1)d
k−1
2
+ c(k − 1)(k + 1)(x1 − a1)2
d
k−3
2
+ β(β − 1)gα
hβ−2
c2
(k + 1)2
(x1 − a1)2
dk−1
+ αβgα−1
hβ−1
c2
(k + 1)2
(x1 − a1)2
dk−1
+ βgα
hβ−1
c(k + 1)d
k−1
2
+ c(k − 1)(k + 1)(x1 − a1)2
d
k−3
2
X1X1f = gα−2
hβ−2
c(k + 1)d
k−3
2 α(α − 1)h2
c(k + 1)(x1 − a1)2
d
k+1
2
+ (αgh2
+ βg2
h) d + (k − 1)(x1 − a1)2
+ β(β − 1)g2
c(k + 1)(x1 − a1)2
d
k+1
2 + 2αβghc(k + 1)(x1 − a1)2
d
k+1
2 .

8. 7
Similarly we compute X2X2f:
X2f = αgα−1
hβ
X2g + βgα
hβ−1
X2h
X2X2f = α(α − 1)gα−2
hβ
(X2g)2
+ αβgα−1
hβ−1
(X2g)(X2h)
+ αgα−1
hβ
X2X2g + β(β − 1)gα
hβ−2
(X2h)2
+ αβgα−1
hβ−1
(X2g)(X2h) + βgα
hβ−1
X2X2h.
Simplifying, we have
X2g = c(k + 1)(x2 − a2)d
k−1
2
X2h = X2g
X2X2g = c(k + 1)d
k−1
2 + c(k − 1)(k + 1)(x2 − a2)2
d
k−3
2
X2X2h = X2X2g

9. 8
X2X2f = α(α − 1)gα−2
hβ
c2
(k + 1)2
(x2 − a2)2
dk−1
+ αβgα−1
hβ−1
c2
(k + 1)2
(x2 − a2)2
dk−1
+ αgα−1
hβ
c(k + 1)d
k−1
2
+ c(k − 1)(k + 1)(x2 − a2)2
d
k−3
2
+ β(β − 1)gα
hβ−2
c2
(k + 1)2
(x2 − a2)2
dk−1
+ αβgα−1
hβ−1
c2
(k + 1)2
(x2 − a2)2
dk−1
+ βgα
hβ−1
c(k + 1)d
k−1
2
+ c(k − 1)(k + 1)(x2 − a2)2
d
k−3
2
X2X2f = gα−2
hβ−2
c(k + 1)d
k−3
2 α(α − 1)h2
c(k + 1)(x2 − a2)2
d
k+1
2
+ (αgh2
+ βg2
h) d + (k − 1)(x2 − a2)2
+ β(β − 1)g2
c(k + 1)(x2 − a2)2
d
k+1
2 + 2αβghc(k + 1)(x2 − a2)2
d
k+1
2 .
Noting that (x1 − a1)2
+ (x2 − a2)2
= d, we combine X1X1f and X2X2f
X1X1f + X2X2f = gα−2
hβ−2
c(k + 1)d
k−3
2 α(α − 1)h2
c(k + 1)d
k+3
2
+ (αgh2
+ βg2
h)(k + 1)d
+ β(β − 1)g2
c(k + 1)d
k+3
2 + 2αβghc(k + 1)d
k+3
2
= gα−2
hβ−2
c(k + 1)2
d
k−1
2 α(α − 1)h2
cd
k+1
2
+ αgh2
+ βg2
h + β(β − 1)g2
cd
k+1
2 + 2αβghcd
k+1
2 .

10. 9
We now compute X3X3f, beginning with ∂f
∂x3
.
∂f
∂x3
= αgα−1
hβ ∂g
∂x3
+ βgα
hβ−1 ∂h
∂x3
∂g
∂x3
= −i(k + 1)
∂h
∂x3
= i(k + 1)
∂f
∂x3
= −i(k + 1)αgα−1
hβ
+ i(k + 1)βgα
hβ−1
∂2
f
∂x3
2
= − i(k + 1)
2
α(α − 1)gα−2
hβ
− i(k + 1)
2
αβgα−1
hβ−1
− i(k + 1)
2
αβgα−1
hβ−1
+ i(k + 1)
2
β(β − 1)gα
hβ−2
.
Simplifying, we get
∂2
f
∂x3
2
= −(k + 1)2
α(α − 1)gα−2
hβ
+ 2(k + 1)2
αβgα−1
hβ−1
− (k + 1)2
β(β − 1)gα
hβ−2
.

11. 10
Using ∂2f
∂x3
2 , we compute X3X3f
X3X3f = c2
dk ∂2
f
∂x3
2
X3X3f = c2
dk
2(k + 1)2
αβgα−1
hβ−1
− (k + 1)2
α(α − 1)gα−2
hβ
− (k + 1)2
β(β − 1)gα
hβ−2
= gα−2
hβ−2
c2
(k + 1)2
dk
2αβgh − α(α − 1)h2
− β(β − 1)g2
.
Combining our results thus far, we get ∆2f:
∆2f = X1X1f + X2X2f + X3X3f
= gα−2
hβ−2
c(k + 1)2
d
k−1
2 α(α − 1)h2
cd
k+1
2
+ αgh2
+ βg2
h + β(β − 1)g2
cd
k+1
2 + 2αβghcd
k+1
2
+ 2αβghcd
k+1
2 − α(α − 1)h2
cd
k+1
2 − β(β − 1)g2
cd
k+1
2
= gα−2
hβ−2
c(k + 1)2
d
k−1
2 αgh2
+ βg2
h + 4αβghcd
k+1
2
= gα−1
hβ−1
c(k + 1)2
d
k−1
2 αh + βg + 4αβcd
k+1
2 .

12. 11
To simplify this, we compute the terms in the last set of parentheses.
αh + βg = −
1
2
(1 − L) cd
k+1
2 + i(k + 1)(x3 − b)
−
1
2
(1 + L) cd
k+1
2 − i(k + 1)(x3 − b)
= −
1
2
cd
k+1
2 +
1
2
Lcd
k+1
2 −
1
2
i(k + 1)(x3 − b)
+
1
2
Li(k + 1)(x3 − b) −
1
2
cd
k+1
2 −
1
2
Lcd
k+1
2
+
1
2
i(k + 1)(x3 − b) +
1
2
Li(k + 1)(x3 − b)
= Li(k + 1)(x3 − b) − cd
k+1
2
αβ =
1
4
(1 − L2
)
and 4αβcd
k+1
2 = cd
k+1
2 − L2
cd
k+1
2 .
Substituting into ∆2f, we get
∆2f = gα−1
hβ−1
c(k + 1)2
d
k−1
2
× Li(k + 1)(x3 − b) − cd
k+1
2 + cd
k+1
2 − L2
cd
k+1
2
∆2f = Lgα−1
hβ−1
c(k + 1)2
d
k−1
2 i(k + 1)(x3 − b) − Lcd
k+1
2 .

13. 12
We now turn our attention to the other half of our equation, iLc(k + 1)d
k−1
2
∂
∂x3
f.
We start by simplifying ∂
∂x3
f.
∂f
∂x3
= −i(k + 1)αgα−1
hβ
+ i(k + 1)βgα
hβ−1
= i(k + 1)gα−1
hβ−1
(βg − αh)
βg − αh = −
1
2
(1 + L) cd
k+1
2 − i(k + 1)(x3 − b)
+
1
2
(1 − L) cd
k+1
2 + i(k + 1)(x3 − b)
= −
1
2
cd
k+1
2 −
1
2
Lcd
k+1
2 +
1
2
i(k + 1)(x3 − b)
+
1
2
Li(k + 1)(x3 − b) +
1
2
cd
k+1
2 −
1
2
Lcd
k+1
2
+
1
2
i(k + 1)(x3 − b) −
1
2
Li(k + 1)(x3 − b)
= i(k + 1)(x3 − b) − Lcd
k+1
2
∂f
∂x3
= i(k + 1)gα−1
hβ−1
i(k + 1)(x3 − b) − Lcd
k+1
2
and so
iLc(k + 1)d
k−1
2
∂
∂x3
f = iLc(k + 1)d
k−1
2
× i(k + 1)gα−1
hβ−1
i(k + 1)(x3 − b) − Lcd
k+1
2
= −Lgα−1
hβ−1
c(k + 1)2
d
k−1
2 i(k + 1)(x3 − b) − Lcd
k+1
2 .

14. 13
This is exactly −∆2f, and thus we have our result:
∆2f + iLc(k + 1) (x1 − a1)2
+ (x2 − a2)2
k−1
2 ∂
∂x3
f = 0.
The next step in our investigation is to find a generalized equation as
in [2]. The approach used in [2] will require significant changes. That
equation does not work for us and so we will have to use a different
method. This next step is what we are doing now. We hope for a
submitted journal article in the next few months.

15. 14
References
[1] Beals, Richard.; Gaveau, Bernard.; Greiner, Peter. On a Geometric Formula
for the Fundamental Solution of Subelliptic Laplacians. Math. Nachr. 1996,
181, 81–163.
[2] Bieske, Thomas; Childers, Kristen. Generalizations of a Laplacian-type equa-
tion in the Heisenberg group and a class of Grushin-type spaces. Proc. Amer.
Math. Soc. 2014, 142:3, 989–1003.