i ii
Abstract
Motivated by the Navier-Stokes equations, which are a set of unsolved equations related to fluid motion in R, we explored the incompressibility condition and the Neumann boundary problem. After exploring, we noticed that using iterated Riesz transforms of the boundary data could be used to get information about the velocity field directly from the boundary data. We then used this same logic to look at the same problem in a spherical space, and found no definition for the Riesz transforms. By using the incompressibility condition and solving the Neumann boundary problem on a sphere, using both separation of variables and potential theory, we can define the Riesz transforms of a function on the sphere. This allows us to use boundary data on a sphere to describe the divergence-free field, or in the case of the incompressible Navier-Stokes equation, allows us to solve for the velocity field given initial data.
1/29 1 Introduction
1.1 Motivation and Objective
One of the great, unanswered questions in physics and mathematics is related to solving the incompressible
Navier-Stokes equations in R3. The Clay Mathematics Institute has put up a one-million dollar prize for a correct, verified solution [5]. As a Millennium Prize Problem, it is an important problem that is currently being used to design
a number of practical things by using ”approximations” that idealize the systems they’re modeling. This includes
many sorts of fluid dynamics problems such as designing boat hulls, airplanes, aerodynamic cars, and other systems.
The motivation to define the Riesz transforms on the sphere comes from the motion of fluids, which are described
by the 3-dimensional incompressible Navier-Stokes equations. We will be using the Riesz transform, which we will
define on both the interior and exterior of the sphere, so that we can use the boundary data we measure around a
sphere to solve for the velocity field in the incompressible Navier-Stokes equations. By defining the Riesz transforms
on the sphere, we can use the result of the transforms to explore systems using measurable physical data such as the velocity of water passing over a surface. So we will begin by discussing the Navier-Stokes equations, what they mean,
and then how they relate to Riesz transforms. From there, we discuss the definition of the Riesz transform and how it
is related to finding solutions to the equation. Then we will solve our Neumann boundary problem on the sphere, and we will find a result for the Riesz transform, in spherical coordinates, related to Neumann boundary data.
Our goal was to find a definition for Riesz transforms on the sphere in both infinite series and integral forms using
analysis and potential theory. Although Riesz transforms are well understood and have applications in mathematics
and physics, we were unable to find any explicit representations for the Riesz transforms on the sphere. In the
following sections, we show you the background we used to provide the basis of our definition, the steps we followed
to get there, and then show the result and its potential uses.
1.2 Navier-Stokes Equations
3 The Navier-Stokes equations are used to describe fluid velocities u(x, t) = (u1(x, t), u2(x, t), u3(x, t)), x ∈ R , t ≥ 0,
and will include some initial data u(x, 0) = u0(x). These equations are based on Newton’s laws of motion and conservation of mass for fluids of constant density. The Navier-Stokes equation can be expressed as:
2/29 ∂uu ρ + (u · ∇)u = ν∇2u − ∇p + ff, ∇ · u = 0, u(x, 0) = u (x). (1) ∂t 0
P3 ∂ The definitions of the operators are: ∇ = (∂/∂x1, ∂/∂x2, ∂/∂x3), u · ∇ = uj , and j=1 ∂xj 2 P3 2 2 ∇ = ∇ · ∇ = j=1 ∂ /∂xj , where the last two are evaluated component-wise for the vector functions. The term ρ is the density, p(x, t) is a scalar function representing pressure, and f (x, t) represents external force, and ν > 0 is the
kinetic viscosity. The incompressibility condition is the requirement that ∇ · u = 0.
These equations make up a non-linear system of four partial differential equations that govern four unknown
scalars, three coordinates of velocity (u1, u2, u3) and a scalar quantity for pressure. The equations are used to model the motion of a viscous, incompressible, homogeneous fluid in free space and are derived from conservation of mass
and momentum of packets of fluid. We place these under the assumption that stress and strain rates have a localized,
linear relationship among these packets of fluid. The left hand side of (1) represents the acceleration in a reference
frame where one views the motion as a parade these fluid packets where ρ is the mass density and u(x, t) is the velocity of the fluid packet at that location and time. The right hand side represents the forces acting on the fluid
packet. The first term ν∇2u represents the viscous friction forces, which is a result of the shear stresses from
interactions between the molecules. The term f ≡ f (x, y, z, t) represents body forces, such as gravity or other
external fields. The −∇p term defines the pressure force on a fluid packet, and is the gradient of the pressure, which
is a force that maintains the homogeneous fluid density under our incompressibility condition [2].
1.3 Riesz Transforms on R3
Riesz transforms are a multi-dimensional generalizations of the Hilbert transform, which means that Riesz transforms
are singular integral operators that consist of a convolution of one function with another that has a singularity at the
origin. These transforms are important in the studying of harmonic potentials in harmonic analysis and potential
theory. This makes them useful for many different systems, and are connected to many areas of study, but it has not
been characterized for many spaces and effectively has not been explored or investigated much beyond Marcel Riesz’s
original descriptions.
The most famous form of the Riesz transform on Rn is the singular integral: [3]
3/29 Z yj Rj(f)(x) = lim cn n+1 f(x − y)dy, (j = 1, ..., n) (2) ε→∞ |y|≥ε |y|
n+1 Γ( 2 ) yj where the normalization constant is cn = π(n+1)/2 , the non-constant part of the Riesz kernel is yn+1 , and f(x) is the function to be transformed. However, there are other ways of describing the definition of the Riesz transform, and we will discuss those later.
4/29 2 Background and Methods
2.1 Incompressibility
Following all the information so far, one may ask why we can justify using the incompressibility condition we
stipulated, that ∇ · u = 0. The discussion for this begins with the idea of mass conservation. In our system, we are
not creating or destroying mass, we are simply looking at a fixed subspace of some larger region we care about.
Let W be a fixed subspace of D, which is a region in R3 that is filled with some fluid. The rate of change of mass in W with respect to time is
d Z ∂ρ m(W, t) = (x, t)dV, (3) dt W ∂t
where dV is the volume element of W , ρ is the mass density of the fluid, and we can take the time derivative
inside the integral because our space W is fixed and does not change with time. It’s easy to see that the rate of
change of mass in W is equal to the rate at which mass is crossing the boundaries of W or ∂W . Using equation (3)
and the divergence theorem, we can show that
Z ∂ρ Z Z (x, t)dV = − ρu · ηdA = − ∇ · (ρu)dV (4) W ∂t ∂W W
and finally
Z ∂ρ + ∇ · (ρu)dV = 0. (5) W ∂t
Since this is to hold for all W, we can generate the continuity equation for this flux, resulting in
∂ρ + ∇ · (ρu) = 0 (6) ∂t
If we then decide that our fluid is homogeneous, which is to say that our density (ρ) is a constant, then it must follow that ∇ · u = 0 is true for equation (6) to be true. Because of this, we can justify using ∇ · u = 0 to solve this problem for incompressible fluids.
5/29 2.2 Fourier Transform on Rn
It’s useful to look at Fourier transforms now because we will use them in further definitions and because they are
helpful for understanding what transforms can do and are used for. The Fourier transform is a transform that takes a
”time-based” function and decomposes it into the frequencies that describe the function. This method of describing
the same function using different variables and form means that you can more easily identify different parts of the
equation or perform different operations that are easier.
The Fourier transform is typically defined as
Z ∞ F(f)(ξ) = ce−iξxf(x)dx. (7) −∞ where c is a normalization constant. Using this transform allows us to take a function in ”x space” and convert it to
”ξ space.” As a result, we can see that
∂f F = −iξjF(f) (8) ∂xj
is true as well. We will use these equations and their properties for helping show how the Riesz transform is defined.
2.3 One Way to Define Riesz Transform
For us to move forward with developing a definition for the Riesz transform on the sphere, it is important for us to
show how we came up with the justification for our form for the transform. Starting with the definition of the Fourier
transform and the Fourier transform of a derivative, equations (7) and (8), we can then show that
∂u ∂u ∂u F(∇ · u) = F 1 + 2 + 3 , (9) ∂x1 ∂x2 ∂x3
∂u ∂u ∂u F(∇ · u) = F 1 + F 2 + F 3 , (10) ∂x1 ∂x2 ∂x3
where u =< u1, u2, u3 > and x1, x2, x3 are the generalized coordinates. The step from equation (9) to (10) is
6/29 permissible due to the linearity of the Fourier transform. Now, if we use our incompressibility condition for ∇ · u = 0
and our definition for Fourier transforms of partial derivatives, equation (8), we get
F(∇ · u) = −iξ1F(u1) − iξ2F(u2) − iξ3F(u3) = 0, (11)
which leads to the conclusion that
ξ ·F(u) = 0, (12)
where ξ =< ξ1, ξ2, ξ3 > and F(u) =< F(u1), F(u2), F(u3) >.
Now we are going to use a projection operator, called the Leray projection operator, denoted P. The operator has a lot of history and its usage is very specialized, so I won’t go into huge detail here, but the informal description is
that it is the orthogonal projection onto the divergence-free subspace of the original vector field. This means that whatever our field is, we’re going to project it onto the divergence-free portion of the field.
After applying this projection to our field, u, we can show that ξ ·F(P(u))(ξ) = 0 (from equation (12)) and then we can notice in Figure 1 that F(u) − F(P(u)) ⊥ F(P(u)).
Fig 1. Leray projection. A representation of the projection from the Fourier transform of the field, F(u), onto the divergence-free subspace of the field, and how that compares to the Fourier transform of that projected field, F(P(u)).
Following this geometry, we are able to conclude that
7/29 F(u) − F(P(u)) = cξ, (13)
and note that c is a scalar that has not yet been determined. From this, we observe that
ξ ·F(u) − ξ ·F(P(u))) = cξ · ξ (14)
results in the ξ ·F(P(u)) term going to zero and the right hand side becoming c|ξ|2, we can then rearrange and solve for c.
−ξ ·F(u) c = (15) |ξ|2
Now, if we go back to equation (13), plug in for c and rearrange, we get
ξ ξ F( (u)) = F(u) − ·F(u) , (16) P |ξ| |ξ|
ξ ⊗ ξ F( (u)) = I − F(u), (17) P |ξ|2
where I is the identity matrix in R3, and
ξ1ξ1 ξ1ξ2 ξ1ξ3 ξ ⊗ ξ = ξ ξ ξ ξ ξ ξ . (18) 2 1 2 2 2 3 ξ3ξ1 ξ3ξ2 ξ3ξ3
ξiξj The matrix elements of |ξ|2 are Fourier symbols of iterated Riesz transforms (Ri,Rj), meaning that
ξj F(Rj(f)(ξ)) = − |ξ| F(f). We can conclude that the Leray projection operator is thus P = I + R where
8/29 R1R1 R1R2 R1R3 R = R R R R R R . (19) 2 1 2 2 2 3 R3R1 R3R2 R3R3
This is getting us there! Now, we can use this information to define the Riesz transform. One of the ways this can
n+1 n be done is to solve the Neumann boundary value problem on the half-space (defined as R+ = [(x, y)|x ∈ R , y > 0])
2 n+1 ∂w n by using ∇ w = 0 on R+ and ∂η = f(x) on R (y = 0), where η represents the outward normal unit vector y=0 ∂w ∂w (such that in this instance, ∂η = − ∂y ). By taking the Fourier transform of the Laplacian (∇2w), we get that
∂2F(w) F(∇2w) = −|ξ|2F(w) + = 0 (20) ∂y2
2 ∂ F(w) 2 since ∂y2 = |ξ| F(w). We also get that
∂w ∂ F − = − F(w) = F(f). (21) ∂y y=0 ∂y y=0
2 ∂ F(w) 2 Using ∂y2 = |ξ| F(w) again, we can also see that
F(w) = ce−|ξ|y, (22)
where we got rid of the e|ξ|y term since we expect for F(w) → 0 as |w| → ∞. Using equation (22) with equation
(21), we get
∂ − (F(w)) = c|ξ|e−|ξ|y = F(f), (23) ∂y y=0 y=0 which we can then rearrange to see that
F(f) c = , (24) |ξ|
9/29 and thus can plug c back into equation (22) and get
F(f) F(w) = e−|ξ|y. (25) |ξ|
Noting that
∂w −iξ F = j F(f), (26) ∂xj y=0 |ξ|
we can show that
∂w Rj(f) = . (27) ∂xj y=0
This is it! This means that we’ve got a possible way of defining the Riesz transform of a function f on the sphere!
From here, we just need to take our Neumann boundary problem, apply it to our spherical space, and the result from
these calculations will yield our result.
2.4 Separation of Variables Method
In the study of partial differential equations, many different methods of solving these difficult equations have been
used. One particularly powerful way, that is common for solving Laplace equations such as ours, is to use separation
of variables. Separation of variables, in the realm of partial differential equations, is a way of solving a multi-variable
function by ”decomposing” the multi-variable function into a convolution of two single-variable functions. A simple
example is to take the one-dimensional heat equation,
∂u ∂2u − α = 0, (28) ∂t ∂x2
where u in this case represents temperature. With the boundary condition u = u = 0, where we assume that x=0 x=L the function u is not zero and that both X and T are independent functions that depend only on x and t, respectively.
This result is that
10/29 u(x, t) = X(x)T (t). (29)
We can then substitute equation (29) back into equation (28), and use the product rule to get
0 00 T (t) X (x) = . (30) αT (t) X(x)
Since the left hand side only depends on t and the right hand side only depends on x, then they must each be equal
to some constant value. We’ll call this constant −λ. As a result,
0 T (t) = −λαT (t) (31)
and
00 X (x) = −λX(x), (32) which results in solutions of
T (t) = Ae (33)
and
√ X(x) = Bsin( λx), (34)
√ π where λ = n L . This solves the heat equation for the special case that it is defined as the convolution of two functions that only depend on x or t, respectively. We will apply this method for our infinite series solution of the
Riesz transform on the sphere.
11/29 2.5 Potential Theory Method
After we had progress with the separation of values method, we realized there are some limitations to an infinite
series solution and wanted to find if we could come up with an integral form for the same thing.
Potential theory has classically been used in mathematical physics for all manner of problems, the most famous of which is Newton’s law of universal gravitation. However, potential theory has been expanded to be used for
electromagnetism, heat, and can generally be thought of as the study and application of the Laplace equation and its
generalizations. Laplace’s equation is a differential equation which characterizes the steady flow of ideal fluids and is
important in solving many other physics and math problems as well.
However, we won’t be using separation of variables. We, instead, use the method of images to meet our boundary
conditions to solve for the Neumann function that allows us to define the Riesz transform.
The method of images is another tool for solving differential equations, wherein we allow symmetry to assist us in
satisfying our boundary conditions and removing any singularities from the domain of our function. We employ this
method for our integral form solution of the Riesz transform on the sphere.
From this, we are able to determine a very important function that we need later when we solve the Neumann
boundary problem. This is called the Neumann function, but it is also known as Green’s function of the second kind.
This important function is used to solve boundary value problems, such as ours. Fortunately for us, we have our
Neumann function defined already from past work, done at the Thapar College of Engineering back in the late
1970’s [6].
12/29 3 Results
3.1 Infinite Series Form
For our definition for the Riesz transform, we will spend some time solving the Neumann boundary value problem.
This is done using the spherical coordinate system described by Figure 2.
Fig 2. Spherical coordinate system. Spherical coordinate system by which r is the radial distance outward from the origin, θ is the polar angle, and φ is the azimuthal angle.
Section 2.3 describes one possible way to get our Riesz transform. We then take that concept, and apply it to our
spherical space by solving the spherical Neumann boundary problem.
Then we begin our work with the Laplacian:
2 1 1 ∇2u = u + u + (u sin(θ)) + u , (35) rr r r r2sin(θ) θ θ r2sin2(θ) φφ
∂u(r, θ, φ) W ith boundary condition : = f(θ, φ) at r = 1 ∂η
∂u(r,φ,θ) where ∂η is the normal derivative. We then use separation of variables, where we assume that u(r, θ, φ) can be reduced to the product of three functions which only depend on one variable each. Thus, we are looking for solutions
of ∇2u = 0 where u has the form u = R(r)Θ(θ)Φ(φ). This results in the following equation:
13/29 R00 2R0 (Θ0sin(θ))0 Φ00 r2sin2(θ) + + sin(θ) = − . (36) R rR Θ Φ
Since both sides are independent of their variables, it means that both sides must equal a constant, we’ll call it m2. Dealing with the right side of the equation first, Φ00 + m2Φ = 0, which has solutions of Φ(φ) = Aeimφ + Be−imφ.
Because of how φ is defined, Φ must be 2π periodic and m must be an integer, and we’ll decide m ≥ 0.
Now that we’ve dealt with the right side, we’ll set the left side equal to m2 and work on it, rearranging to get
r2R00 + 2rR0 m2 (Θ0sin(θ))0 = − . (37) R sin2(θ) Θsin(θ)
This means, again, that both sides are equal to a constant. This time, we’ll call it λ. The result is that the remaining functions can be written as:
(Θ0sin(θ))0 m2Θ − + λΘ = 0, (38) sin(θ) sin2(θ)
r2R00 + 2rR0 − λR = 0. (39)
Equation (38) can be transformed by using the substitution s = cos(θ). This will turn the equation into an associated Legendre equation. This substitution results in
m2S [(1 − s2)S0]0 − + λS = 0. (40) 1 − s2
We then notice that if m = 0, we get the ordinary Legendre equation. If we then suppose that w is a solution to the ordinary Legendre equation, we get
[(1 − s2)w0]0 + λw = 0 (41)
14/29 Now, we use properties of the Legendre polynomials and the associated Legendre polynomials to find that if w
satisfies equation (40), then S = (1 − s2)m/2w(m) satisfies equation (41). Further, if λ = n(n + 1) and w is the
ordinary Legendre polynomial, then we get the associated Legendre function:
(1 − s2)m/2 dn+m P m(s) = (s2 − 1)n. (42) n 2nn! dsn+m
From here, we are going to want to return to equation (37) since we have shown that Θ(θ) has a form of y(cos(θ)) where y is the solution to the associated Legendre equation (40). Now we plug in λ = (n(n + 1)) into equation (39) to
get
r2R00 = 2rR0 − n(n + 1)R = 0. (43)
This equation has a general solution of the form R(r) = arn + br−n−1, but we want the solution to be continuous
at the origin, so we’ll take b = 0. Using this, and the solutions we found for our other independent functions, we can
then say that we have the following family of solutions to our Laplace equation:
n imφ |m| um,n(r, θ, φ) = r e Pn (cos(θ)), (n = 0, 1, 2, ...; |m| ≤ n). (44)
Thus, after separating and using the appropriate constants, associated Legendre polynomials, and spherical
harmonics, we come to find we can solve the incompressible Neumann boundary value problem using the separation
of variables method for both the interior and exterior of the sphere [4]. The superposition of our family of solutions, where the 0th index of our associated Legendre polynomial is taken to be the ordinary Legendre polynomial, and the
index is done such that we can combine our eimφ terms into one, is:
∞ n X X n ımφ |m| u(r, θ, φ) = cm,nr e Pn (cos(θ)), (interior) (45) n=0 m=−n
15/29 ∞ n X X 1 u(r, θ, φ) = d eımφP |m|(cos(θ)), (exterior). (46) m,n rn+1 n n=0 m=−n
As a reminder, the m index comes from the constant we get through the separation of variables method and the n index comes from the superposition of associated Legendre equation. Now we take this solution and solve for the
∂u(r,φ,θ) boundary condition ∂η = f(θ, φ). r=1 This results in interior and exterior functions:
∞ n X X ımφ |m| f(θ, φ) = am,ne Pn (cos(θ)), (interior) (47) n=0 m=−n
∞ n X X ımφ |m| f(θ, φ) = bm,ne Pn (cos(θ)), (exterior) (48) n=0 m=−n
on the sphere. The interior solution’s constant cm,n = am,n/n and the exterior’s constant dm,n = bm,n/(−n − 1).
imφ |m| Since e evaluated from −∞ to ∞ is a complete orthogonal set on (−π, π) and since [Pn (cos(θ))] evaluated from 0 to ∞ where |m| ≤ n is a complete orthogonal set on (0, π), we can define the spherical harmonics of our unit sphere as
imφ |m| Ym,n(θ, φ) = e Pn (cos(θ)). (49)
Since equations (47) and (48) are an expansion of f(θ, φ) with respect to these spherical harmonics, we can show that the coefficients am,n and bm,n are given by:
< f, Ym,n > am,n = 2 , (interior) (50) ||Ym,n||
< f, Ym,n > bm,n = 2 . (exterior) (51) ||Ym,n||
Now we can solve for the constants am,n and bm,n:
16/29 Z π Z π (2n + 1)(n − |m|)! −ımφ |m| am,n = f(θ, φ)e Pn (θ)(sin(θ))dφdθ, (interior) (52) 4π(n + |m|)! 0 −π
Z π Z π (2n + 1)(n − |m|)! −ımφ |m| bm,n = f(θ, φ)e Pn (θ)(sin(θ))dφdθ. (exterior) (53) 4π(n + |m|)! 0 −π
Now we take our solution to the boundary data from our Neumann boundary value problem, and we’re going to use it to come up with the result for our Riesz transform. Because of how the Riesz transform is defined, we know that the Riesz transform of a coordinate is going to be the derivative of the boundary data with respect to either coordinate. This will result in a definition of the Riesz transform for both the θ and φ directions. In addition, we will refine our definition of the associated Legendre functions based on the symmetry of the superposition and the orthogonality of the Legendre functions, because it allows us to take a much easier derivative. The associated
|m| Legendre function Pn (cos(θ)), using the same definitions for n and m, is then defined as
dm P |m|(cos(θ)) = (1 − cos2(θ))m/2 (P (cos(θ))) (54) n dcosm(θ) n
Now we go back to equations (45) and (46), remembering (27), take the tangential derivatives, resulting in
∞ n X X 1 R (f) = a eımφ(mcot(θ)P |m|(cos(θ)) − P |m+1|(cos(θ)), (interior) (55) θ n m,n n n n=1 m=−n
∞ n X X m R (f) = a eımφP |m|cos(θ), (interior) (56) φ n m,n n n=1 m=−n
∞ n X X 1 R (f) = b eımφ(mcot(θ)P |m|(cos(θ)) − P |m+1|(cos(θ)), (exterior) (57) θ (−n − 1) m,n n n n=1 m=−n
∞ n X X m R (f) = b eımφP |m|cos(θ). (exterior) (58) φ (−n − 1) m,n n n=1 m=−n
17/29 And so now we have successfully defined the Riesz transforms of a function f on the unit sphere using separation
of variables for an infinite series form.
3.2 Integral Form
For the integral form, we return to our Neumann boundary problem. However, we are going to use a different
approach to solve it.
To restate for this perspective, our Neumann boundary problem is
∇2u = 0, on G (59)
∂u = g, on H, (60) ∂η where G is a ball in R3 and H is a sphere such that ∂G = H. In this problem, our boundary condition g cannot be arbitrarily designated. We instead have to apply a necessary condition that
ZZ g dA = 0. (61) H
This is necessary because of the conditions of the question, where
ZZZ ZZZ ∇2u dV = ∇ · (∇u) dV, (62) G G which can be converted, by the divergence theorem, to
ZZ ∂u ZZ dA = g dA = 0. (63) H ∂η H
From here, we know we have the Neumann function, which is also called Green’s function of the second kind, on
the sphere. The Neumann function came from the method of images, thus is it necessary to describe that it based on
two points, P (ρ, θ, φ) and Q(ρ0, θ0, φ0), and the angle between them, γ. Now we can begin using Green’s identity to
18/29 work on solving our boundary problem. The Green identity is given as
ZZZ ZZ ∂v ∂u (u∇2v − v∇2u) dV = u − v dA (64) G H ∂η ∂η with u = u(Q) and v = N(P,Q) such that we can obtain
ZZZ 2 2 u(P ) = (u(Q)∇3(P,Q) − N(P,Q)∇Qu) dVQ, (65) G
because the Neumann function acts like a delta function when the Laplace operator is acted on it, which allows us to
reduce to
ZZ ∂N ∂u u(P ) = u(Q) − N(P,Q) dAQ. (66) H ∂ηQ ∂ηQ
RR The first term is imposed to be zero as a choice of normalization, by choosing H u dAQ = 0, and we are left with
ZZ u(P ) = − g(Q)N(P,Q) dAQ. (67) H
This is exactly what we were looking for - a representation that allows us to take the tangential derivatives which
is based on the boundary data. Now we need to address the Neumann function, which was defined by previous work
done by the Department of Applied Mathematics at the Thapar College of Engineering [6].The Neumann function is
defined as
1 1 1 c N(P,Q) = − + + log (68) 2 2 1/2 4π rPQ rP Q¯ 1 − ρrcos(γ) + (r ρ + 1 − 2rρcos(γ)) which can be simplified after limiting our Neumann function to the boundary, meaning that ρ, and ρ0 are all equal to
1, the boundary of the sphere. This simplification is:
19/29 1 2 c N(P,Q) = − + log (69) 2 1/2 ρ=1 4π rPQ 1 − rcos(γ) + (r + 1 − 2rcos(γ))
2 0 0 0 where rPQ = r + 1 − 2rcos(γ) and cos(γ) = cos(θ)cos(θ ) + sin(θ)sin(θ )cos(φ − φ ). Now that we have our Neumann function defined, we can then take out tangential derivatives after setting r = 1.
For calculation details, please see the Appendix. The resulting Riesz transforms of a function on the sphere have the
form
ZZ ∂N(P,Q) Rθ(g) = − lim g(Q) dAQ, (70) γ→0 π π ∂θ − 2 ≤γ≤ 2
ZZ ∂N(P,Q) Rφ(g) = − lim g(Q) dAQ. (71) γ→0 π π ∂φ − 2 ≤γ≤ 2
Where we have imposed a limit and bounds on γ to indicate our representation of the Cauchy principal value for
the singular integral. As seen below, if γ = 0, our integral does not converge, and so we use the symmetry of γ and
the limit to get a value when we evaluate the integral. The respective partial derivatives of the Neumann function are
∂N(P,Q) 1 −2sin(θ)cos(θ0) + 2cos(θ)sin(θ0)cos(φ − φ0) = − − ∂θ r=1 4π (2 − 2cos(γ))3/2 0 0 0 sin(θ)cos(θ )−cos(θ)sin(θ )cos(φ−φ ) 0 0 0 ( 1/2 + sin(θ)cos(θ ) − cos(θ)sin(θ )cos(φ − φ ) (2−2cos(γ)) . (72) 1 − cos(γ) + (2 − 2cos(γ))1/2
sin(θ)sin(θ0)sin(φ−φ0) 0 0 0 0 ∂N(P,Q) 1 2sin(θ)sin(θ )sin(φ − φ ) ( 1/2 + sin(θ)sin(θ )sin(φ − φ )) = − + (2−2cos(γ)) , (73) ∂φ r=1 4π (2 − 2cos(γ))3/2 1 − cos(γ) + (2 − 2cos(γ))1/2
This result gives us our Riesz transforms of a function g in integral form for the interior of the sphere only, having
used potential theory and the method of images.
20/29 4 Discussion and Future Work
This result is exciting because we now have an explicit form for defining the Riesz transforms on the sphere. This
means that if we are given the boundary data for a sphere, we are able to use these Riesz transforms to find the
divergence-free solution to u, our velocity field that describes the interior or exterior velocity of our fluid packets,
using the boundary data. In the context of our incompressible Navier-Stokes equation, this helps us solve for u! This
gives us understanding of the incompressible fluid on a sphere.
Looking at equations (56) and (58), we can also see that for the φ Riesz transforms in either the interior or
exterior cases is just a scaling factor of the boundary data function f.
We think that the result for this application is already spectacular, but we also think that this definition and
process has application outside of the incompressible Navier-Stokes equation. One example of this would be any other
system which has a divergence free requirement on the field. Another, more extensive use, however, would be through
Helmholtz decomposition.
Simply put, Helmholtz decomposition is the concept that a vector field that decays quickly enough and is smooth
enough can be described as the sum of the divergence-free vector field and curl-free vector field. The method we
could use to use this would be to find the divergence-free field using the Riesz transform, and then finding the
curl-free field by subtracting our divergence-free field from the total field. This is exciting because Helmholtz
decomposition is used effectively in other physics analysis, including electromagnetism.
21/29 5 Conclusion
Inspired by the incompressible Navier-Stokes equations and noting a lack of explicit spherical definition for the Riesz
transform, we set out to find a solution to the Riesz transforms on the sphere. Using Fourier transforms, we were able
to define one method of defining the Riesz transform. Then we decided we would look at solutions by first
considering the incompressibility condition, and looking for solutions that would involve that. Once we were done
addressing the incompressibility condition, we then solved our Neumann boundary problem by using either separation
of variables or potential theory to reduce that down to the functions that meet the boundary condition. From here, we were able to define our Riesz transforms of a function on the sphere for both θ and φ directions. These explicit
definitions allow us to take boundary data, which is attainable from our system, and apply it to solve for u in our
incompressible Navier-Stokes equations, with initial data, or apply it to solve for the divergence-free portion of a
Helmholtz decomposition.
22/29 Appendix
The following shows the the steps taken to take the tangential derivatives of our Neumann function from the integral
result section. We begin with the θ derivative and state it:
∂N(P,Q) ∂ 1 2 c = − + log , (A1) 2 1/2 ∂θ ∂θ 4π rPQ 1 − rcos(γ) + (r + 1 − 2rcos(γ))
2 0 0 0 where we remind you that rPQ = r + 1 − 2rcos(γ) and cos(γ) = cos(θ)cos(θ ) + sin(θ)sin(θ )cos(φ − φ ).
−1 We then set 4π aside and then take the derivatives term-by-term, beginning with:
∂ 2 ∂ 2 ∂γ = , (A2) ∂θ rPQ ∂γ rPQ ∂θ
by the chain rule. Dealing with ∂ 2 first, we have: ∂γ rPQ
∂ 2 −2rsin(γ) = . (A3) ∂γ r2 + 1 − 2rcos(γ) (r2 + 1 − 2rcos(γ))3/2
∂γ Now to tackle ∂θ , we use the relationship that
∂cos(γ) ∂γ = −sin(γ) , (A4) ∂θ ∂θ
resulting in:
∂γ −sin(γ) = −sin(θ)cos(θ0) + cos(θ)sin(θ0)cos(φ − φ0), (A5) ∂θ
thus
∂γ sin(θ)cos(θ0) − cos(θ)sin(θ0)cos(φ − φ0) = . (A6) ∂θ sin(γ)
23/29 We can then return to (A2) and plug in our results:
∂ 2 −2rsin(γ) sin(θ)cos(θ0) − cos(θ)sin(θ0)cos(φ − φ0) = , (A7) 2 3/2 ∂θ rPQ (r + 1 − 2rcos(γ)) sin(γ) where the sin(γ) cancels, and we distribute to get
∂ 2 −2rsin(θ)cos(θ0) + 2rcos(θ)sin(θ0)cos(φ − φ0) = . (A8) 2 3/2 ∂θ rPQ (r + 1 − 2rcos(γ))
Now we move onto the second term:
∂ c log = ∂θ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 ∂ c ∂γ log . (A9) ∂γ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 ∂θ
∂γ We’ve already solved for ∂θ , so we only need to work on the first part of the right hand side. We begin by using the property of logarithms that allows us to separate the numerator from the denominator:
∂ c log = ∂γ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 ∂ log(c) − log(1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 . (A10) ∂γ
0 0 u (x) where it’s easy to see the first term, log(c) goes to 0 when the derivative is taken. Then, since (log(u(x))) = u(x) , we have
∂ 2 1/2 ∂ − 1 − rcos(γ) + (r + 1 − 2rcos(γ)) −log(1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 = ∂γ , (A11) ∂γ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 where the derivative can be done term-by-term, resulting in
24/29 ∂ (1) = 0, (A12) ∂γ ∂ (rcos(γ)) = −rsin(γ), (A13) ∂γ ∂ 1 ∂ (r2 + 1 − 2rcos(γ))1/2 = (r2 + 1 − 2rcos(γ))−1/2 (r2 + 1 − 2rcos(γ)). (A14) ∂γ 2 ∂γ
In (A14) both the product and chain rule have been applied, and again the derivative in it can be evaluated
term-by-term, resulting in
∂ −2(−rsin(γ)) (r2 + 1 − 2rcos(γ))1/2 = , (A15) ∂γ 2(r2 + 1 − 2rcos(γ))1/2 which simplifies to
∂ rsin(γ) (r2 + 1 − 2rcos(γ))1/2 = . (A16) ∂γ (r2 + 1 − 2rcos(γ))1/2
We can then plug (A12), (A13), and (A16) into (A11) to get
rsin(γ) ∂ − (r2+1−2rcos(γ))1/2 + rsin(γ) −log(1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 = , (A17) ∂γ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 which we factor the numerator to get
1 ∂ −rsin(γ) (r2+1−2rcos(γ))1/2 + 1 −log(1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 = . (A18) ∂γ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2
Going back to (A9) and plugging in (A18) and (A6) for the respective factors, we now have the evaluated
derivative of the second term:
25/29 ∂ c log = ∂θ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 1 0 0 0 −rsin(γ) (r2+1−2rcos(γ))1/2 + 1 sin(θ)cos(θ ) − cos(θ)sin(θ )cos(φ − φ ) . (A19) 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 sin(γ)
We then cancel sin(γ) from the numerator and denominator and distribute, resulting in:
∂ c log = ∂θ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 0 0 0 sin(θ)cos(θ )−cos(θ)sin(θ )cos(φ−φ ) 0 0 0 −r (r2+1−2rcos(γ))1/2 + sin(θ)cos(θ ) − cos(θ)sin(θ )cos(φ − φ ) . (A20) 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2
We then replace the terms in (A1) with (A8) and (A20), yielding our result:
∂N(P,Q) 1 −2rsin(θ)cos(θ0) + 2rcos(θ)sin(θ0)cos(φ − φ0) = − − ∂θ 4π (r2 + 1 − 2rcos(γ))3/2 0 0 0 sin(θ)cos(θ )−cos(θ)sin(θ )cos(φ−φ ) 0 0 0 r (r2+1−2rcos(γ))1/2 + sin(θ)cos(θ ) − cos(θ)sin(θ )cos(φ − φ ) . (A21) 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2
We can see that when we set r = 1, (A21) = (72), giving us our solution to the θ partial derivative of the
Neumann function.
Now we move on to the φ partial derivative of our Neumann, stated as:
∂N(P,Q) ∂ 1 2 c = − + log , (A22) 2 1/2 ∂φ ∂φ 4π rPQ 1 − rcos(γ) + (r + 1 − 2rcos(γ))
−1 We, again, set 4π aside and tackle the derivatives term-by-term, starting with:
∂ 2 ∂ 2 ∂γ = , (A23) ∂φ rPQ ∂γ rPQ ∂φ
∂γ where we can notice that (A3) gives us the first factor, leaving just the ∂φ factor. Similar to (A4), we use the
26/29 relationship:
∂cos(γ) ∂γ = −sin(γ) , (A24) ∂φ ∂φ
giving us
∂γ −sin(θ)sin(θ0)sin(φ − φ0) = . (A25) ∂φ sin(γ)
Replacing (A3) and (A25) into (A23) gives our first term:
∂ 2 −2rsin(γ) −sin(θ)sin(θ0)sin(φ − φ0) = , (A26) 2 3/2 ∂φ rPQ (r + 1 − 2rcos(γ)) sin(γ) which we can distribute, cancel the sin(γ), and end with:
∂ 2 2rsin(θ)sin(θ0)sin(φ − φ0) = . (A27) 2 3/2 ∂φ rPQ (r + 1 − 2rcos(γ))
Now we move to the next term:
∂ c log = ∂φ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 ∂ c ∂γ log . (A28) ∂γ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 ∂φ
∂γ Where we can see that the first factor’s derivative is (A18) and ∂φ is (A25), giving us:
∂ c log = ∂φ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 1 0 0 −rsin(γ) (r2+1−2rcos(γ))1/2 + 1 −sin(θ)sin(θ )sin(φ − φ ) , (A29) 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 sin(γ)
27/29 which, after distribution, simplifies to:
∂ c log = ∂φ 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2 0 0 sin(θ)sin(θ )sin(φ−φ ) 0 0 r (r2+1−2rcos(γ))1/2 + sin(θ)sin(θ )sin(φ − φ ) . (A30) 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2
We can then take the results from (A27) and (A30), plug them into (A22) to get:
0 0 sin(θ)sin(θ )sin(φ−φ ) 0 0 0 0 ∂N(P,Q) 1 2rsin(θ)sin(θ )sin(φ − φ ) r (r2+1−2rcos(γ))1/2 + sin(θ)sin(θ )sin(φ − φ ) = − + . (A31) ∂φ 4π (r2 + 1 − 2rcos(γ))3/2 1 − rcos(γ) + (r2 + 1 − 2rcos(γ))1/2
When we set r = 1, we can see that (A31) = (73), giving us the solution to the the φ partial derivative of the
Neumann function.
28/29 References
1. J.P. Ward, K.N. Chaudhury, and M. Unser. Decay Properties of Riesz Transforms and Steerable Wavelets.
SIAM Journal on Imaging Sciences 6, 984-1005 (2013).
2. E.C. Waymire. Probability & Incompressible Navier-Stokes Equations: An Overview of Some Recent
Developments. Probability Surveys 2, 1-32 (2005).
3. E.M. Stein. Singular Integrals and Differentiability Properties of Functions. (Princeton University Press,
Princeton, NJ, 1970).
4. G.B. Folland. Fourier Analysis and its Applications. (Wadsworth & Brooks/Cole, Pacific Grove, CA, 1992).
5. C.L. Fefferman. Existence and Smoothness of the Navier-Stokes Equation. (Clay Mathematics Institute, 2000).
6. B.M Nayar. Neumann Function for the Sphere. Indian J. pure appl. Math., 12(10): 1266-1282 (1981).
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