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Two Dimensions in Accordance to Dirichlet Problem Brandon Allen Winona State University May 16, 2015 1
Two Dimensions in Accordance to Dirichlet Problem
Brandon AllenWinona State University
May 16, 2015
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1 Acknowledgments
I would like to thank Dr. Debnath of the Mathematics department at Winona State Univer-sity for taking me under her wing to do research. I would also like to thank other professorssuch as Dr. Leonhardi for giving me guidance of what to do in Mathematics. I would liketo thank a few cohorts of mine that have made me learn that Mathematics is just not asubject that one can learn by themselves. I thank Tiaxia Jia at Winona State University forhelping me through this project. I would especially like to thank the professors that havebeen mentors to me such as Dr. Guentchev and Dr. Driscoll of Bemidji State University forinspiring me always to do my best no matter what challenge lies ahead.
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Abstract
One requirement of this paper for reading it is to have a understanding of at leasta Calculus I level of Mathematical background. I see very few papers ever givinga prelude in mathematics, but I think they are important. Over my time of tryingto understand the Dirichlet Problem it has occurred to me that this problem is verydifficult to understand and solve. We shall try and accomplish some basic ideas ofthe Dirichlet problem and take a Feminist perspective onto the Dirichlet problem tounderstand it in a greater manner. I haven’t seen papers done so far that go into thefeminist perspective of the Dirichlet problem. However, with the help of Mathematicawe can go into a Feminist prospective of the Dirichlet Problem.
It has been argued in the Philosophy of Science that Feminist perspective intothe Philosophy of Science in Mathematics seems absurd because of how can we setMathematics to be subjective. Isn’t it just more objective if we include other forms ofdata? Sure, but a lot of people are turned off by looking at numerous equations thatare long and hard to compute. Dirichlet problems though have a way to introduce in acolorful picture with the assistance of Mathmatica that can aide us understanding theproblem.
What the reader should get from this paper is what exactly is the Dirichlet problemis. Another aspect in which the reader should gain from this paper is the solving oftwo dimensional problems of the Dirichlet Problem.
The Dirichlet problem is important to the field of applied Mathematics. The Dirich-let Problem can be found in the fields of Thermodynamics concerning how heat flows,Hydrodynamics concerning how water flows, Electrostatics how electricity flows, andAreonatutics by how gravitation works in a body’s domain.
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Contents
1 Acknowledgments 2
2 What Is the Dirichlet Problem? 52.1 Dirichlet Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Characteristics of the Laplace Equation . . . . . . . . . . . . . . . . . . . . . 6
3 Why Feminist Philosophy of Science in Mathematics? 83.1 Incorporation of Positivist Philosophy into the Dirichlet Problem . . . . . . . 9
4 Infinite Lines 104.1 Harmonic Functions and Analytic Mapping . . . . . . . . . . . . . . . . . . . 134.2 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
5 Problem of the Square 16
6 The Problem of the Annulus 186.1 Revisiting the Problem of the Annulus . . . . . . . . . . . . . . . . . . . . . 216.2 Interior and Exterior Dirichlet Problems . . . . . . . . . . . . . . . . . . . . 24
7 Problem of the Disk 257.1 Removing the Singularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
8 Future Projects 30
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2 What Is the Dirichlet Problem?
2.1 Dirichlet Problem
When looking up the Dirichlet problem we see even in famous websites the complexity ofit’s definition.
If f on the boundary of a unit open disk ∂D(0, 1), then define:
u(z) =
{1
2π
∫ 2π
0f(eiψ) 1−|z|2
|z−eiψ |2dψ if zε∂D(0, 1)
f(z) if zε∂D(0, 1)
However, the definition of the open unit disk seems to be very limited for the DirichletProblem. There is a way to define a Dirichlet problem however for something that is not aunit disk, but rather all space itself! Without further ado we shall state the DirichletProblem.
Definition 2.1. Suppose that D is a domain of a Cartesian plane and that g is a functionthat is defined on the boundary C of D. the problem of finding a function φ(x, y), whichsatisfies Laplace equation in D and which equals gon the boundary C of D is called a Dirichletproblem.Laplace’s Equation:
∂2φ
∂x2+∂2φ
∂y2= ∇2φ = 0 (1)
[Figure 1]
Proof. Suppose that there is a large heat plate such that there is a flow in g(x) and g(y)when heat flows out in the x and y direction of the heat plate then we get g(y) and g(x) .So we obtain the following:
g(x)∆y∆z∆t+ g(y)∆x∆z∆t = g(x+ ∆x)∆y∆z∆t+ g(y + ∆y)∆x∆z∆t
We’ll next multiply by ∆x∆x
and ∆y∆y
respectively which corrosponds to:
g(x+ ∆x)− g(x)
∆x∆x∆y +
g(y + ∆y)− g(y)
∆y∆x∆y = 0
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This should look familiar to any calculus I student. We see that the definition of the derivativepops up such that if we take the limit as x goes to zero and the limit of y goes to zero weobtain the following:
−dgdx− dg
dy= 0
We shall use a result found out in partial differential equations to arrive at the derivation ofthe Laplace equation. The following result is:
gi = kρc∂φ
∂i
Where ρ is just the pressure and k and c are constants and i is a dependant on what thevariable we’re looking at. So, with plugging this function in we shall arive at the lapaceEqaution being the following:
∂2φ
∂x2+∂2φ
∂y2= 0
How I like to think about the Dirichlet Problem is by picturing my own body as thedomain and then the boundary is just my skin. There must be some function that mapsfrom the domain of my own body to my skin.
2.2 Characteristics of the Laplace Equation
There are many characteristics that the laplace equation holds. I will first go over the proofof the laplace equation and I will also go over some of the characteristics of the Laplaceequation.
Proposition 2.1. Let u1 and u2 be harmonic functions (i.e.) solution to the laplaceequation. We shall show that c1u1 + c2u2 is harmonic for constants c1 and c2 .
If u1 and u2 are harmonic functions then we have that: ∂2u1∂x2
+ ∂2u1∂y2
= 0 and ∂2u2∂x2
+ ∂2u2∂y2
= 0.
c1(∂2u1
∂x2+∂2u1
∂y2) + c2(
∂2u2
∂x2+∂2u2
∂y2) = c1(0) + c2(0) = 0
Proposition 2.2. If u(x, y) is harmonic, deduce that xu(x, y) is harmonic only ifu(x, y) = ay + b for some constants a and b.
Proof. If u(x, y) = ay + b then,
f(x, y) = xu(x, y) = x(ay + b)
fx(x, y) = ay + b
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fxx(x, y) = 0
fy(x, y) = (0)(ay + b) + ax
fyy(x, y) = 0
fxx(x, y) + fyy(x, y) = 0
Proposition 2.3. Not all functions such as u1 and u2 such that u1u2 are harmonic.
Proof. Let u1 = x and u2 = 2x. Then,
u(x, y) = u1u2 = x ∗ 2x = 2x2
ux(x, y) = 4x
uxx(x, y) = 4
uyy(x, y) = uy(x, y) = 0
uxx(x, y) + uyy(x, y) = 4 6= 0
Proposition 2.4. Find all functions that are harmonic in the form of U(x,y) =X(x,y)Y(x,y).
Proof.uxx = X ′′(x)Y (y)
uyy = Y ′′(y)X(x)
X ′′(x)
X(x)=−Y ′′(y)
Y (y)= −λ
Case 1: Let λ = 0; X ′′ = 0 we do the double integration to receive the following for X(x).
X(x) = ax+ b, a, bεZ
The same also goes for Y(y).Y (y) = cy + d, c, dεZ
U(x, y) = (ax+ b)(cy + d)
Case 2: Let λ < 0;X ′′ + λX = 0r2erx + rerxλ = 0X(x) = c1 cos(bx) + c2sin(by)Y (y) = d1e
by + d2e−by
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U(x, y) = (c1Cos(bx) + c2Sin(bx))(d1eby + d2e
−by)
Case 3: Let λ > 0;do the same derivation as in case two and we get the following:
U(x, y) = (c1ebx + c2e
−bx)(d1Cos(by) + d2Sin(by))
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3 Why Feminist Philosophy of Science in
Mathematics?
Feminist Philosophy of Science concerning mathematics as I have stated in my abstractstatement is something that mathematicians are not used too. This is because A lot ofMathematicians don’t think that there functions feel feelings. Numbers are abstractconcepts that may exist out in the world but when we don’t ever think that if I poke thenumber 2 it will giggle. However, mathematics has made methods of which to tell thecertain properties of what the mathematician is observing. A great example of thisconcerns the derivative in which tells the observer of abstract mathematical entities of thefunction’s velocity in it’s first continuous derivative and it’s acceleration for it’s secondderivative. The mathematician knows that these properties are part of the certain functionthat is evaluated and shall stay eterenal as long as the function is percieved under the samemathematical axiomatic system.To really understand where I am getting from one should take into perspective myperspective about how I perceive how mathematics works and operates. I am first off not arealist. I do not believe that numbers exists independently of the human consciousness. Istem from a school of thought called Idealism which started back with George Barkley.2 Ibelieve numbers do not exist independently from the human consciousness and are themanifestations of the human mind itself. They stem from each of our minds but theyappear to be the same numbers across the variations of the human consciouses. This isvery similar to what is seen as the Absolute spirit that Hegel created with his work ”ThePhenomenology of Spirit”. Because I stem from the Absolute Spirit and others do too weshare the same informational knowledge that has been gained through time. When I thinkabout a mathematical function it stems from my inner being that was created by Absolute.The mathematical function that comes out of my being then mathematicians say it hascertain ”properties” to the function. However, I say that it has a deterministic set of
1one can use simple ODE with a guess of ert to figure out what X(x) and Y(y) are. These equations arevery interesting due to seeing these in your first semester of ODE in which you use very fequently. It’s veryinteresting for seeing the derivation of this. Please take note of these three equations as they’ll be highlyused later.
2George Barkley isn’t the first Idealist, Plato is usually considered the first Idealist, but the school ofthought kicked off with Barkley to Kant to German Idealists
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feeling to itself created by the previous human conciousness’s and stems from myself due tomy will’s interest into the function that is observed. When I plug in numbers into thefunction it reacts in a following way telling the user of the function how did it react. Thisreaction is transcribed into my inner being from the observed effects of what has been doneto the function when I interact with it. The property of reflection which is needed in orderfor my theory to work is used to compare and contrast against the other forms of data usedwith that function.When looking forward on we shall see that when using the perspective of the Dirichletproblem we will see the function’s deterministic will in accordance to the laws of nature weperceive outside of ourselves. These functions will be represented as contour and densityplots such that anyone who reads this will see that functions under the Dirichlet problemhave special properties to them. However, for the first section of what I am going to speakabout brings up a very important issue in Philosophy and has even concernedmathematicians such as Poincare about even fathoming the size of infinite. How can I evenextend my domain to my boundary even if it’s infinite and also how can I understand aninfinite function with its properties? I attempt to make this leap by putting into aconversion of polar coordinates with the help of mathematica in order to get anunderstanding of the full function that we are looking at.I admit that there are many mathematicians who do philosophy. The mathematician mayreply back to what I am doing ”Well this is just a representational form of the equationthat you are looking at it’s not subjective but objective still then.” I agree that theexperience that we see is objective but to the mathematical function it isn’t. (Yes, I amsaying equations have feelings.) The mathematical function follows certain laws of it’s ownnature of what is bestowed upon itself. For example, take the function f(x) = x+ 1 I forcea condition on the function making x=1. f(1) = 1 + 1 in which arguably makes f(1) = 2.Somehow the function f(x) knows that there is some x=1 put into itself. It behaves in asubjective experience based on the condition of addition with the number one. However,this subjective experience is determined by the conditions in which f(x) = x+ 1 is putunder. This leads to the conclusion that mathematics becomes more subjective than whatmany mathematicians believe to be so. We shall use a subjective experience ofmathematics when imploring into the next section which deals with infinite lines.
3.1 Incorporation of Positivist Philosophy into the DirichletProblem
One aspect the Positivists were worried about was the sense that the instruments that weuse for measuring specific gradients of substance could be wrong. Simply our perceptions ofwhat we record may not actually be that temperature but some other temperature that isout there in the world. For example, if I took the thermometer outside and read thetemperature outside with the thermometer and it said 20 Degrees Celsius versus theoutside temperature being 27 degrees Celsius. There must be something wrong with theinstrument that we are using. This is where the Dirichlet problem comes into play. Due to
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math not being mainly focused on a-posteriori (The individual acquires knowledge ofmathematics through experiences rather than mental processes) methods of interpretationwe can get rid of the troublesome worry that the Positivist school of thought hadconcerning the boundary temperatures of objects. However, this can be debated uponfurther in on the text due to Fourier series only being approximations (Unless we take theseries to infinity. However, we still get the Poincare problem of trying to fathom theinfinite.) of what we are trying to find. However, in simple Complex Analysis examplesthis works out just fine since we get a general solution out of what we are inquiring about.
4 Infinite Lines
Since we are in the Cartesian Plane, it is best if we start with a simple example first. If thereader has taken some Ordinary Differential Equations the reader would have calculatedhow heat flows out of two poles that are infinite in length and are parallel to one another.Being stuck to a Cartesian Plane though we are limited to two dimensional space in whichwe’ll picture two parallel infinite lines crossing the x-axis at x0 and x1 .
The Dirichlet problem that we have is the following:
Solve :∂2φ
∂x2+∂2φ
∂y2= 0, x0 < x < x1, −∞ < y <∞ (2)
Subject to : φ(x0, y) = k0, φ(x1, y) = k1, −∞ < y <∞ (3)
Looking at the picture we see that φ is in fact independent of y if this is the case, then wehave that ∂
2φ∂2x2
= 0. Carrying through with integration, we receive the following:φ(x) = ax+ b. We know that this is a linear equation. When plugging both x-coordinatesin we receive back k0 and k1Through simple point-slope form of Algebra I we see thefollowing:
φ(x) =k1 − k0
x1 − x0
(x− x0) + k0 (4)
Because this equation satisfies (1) it’s a harmonic function. A harmonic function is definedas follows:
Definition 4.1. A real valued function φ of two real variables x and y that has continuousfirst and second-order partial derivatives in a domain D and satisfies Laplace’s equation issaid to be Harmonic in D.
Proof. Let ux = vy and uy = −vy such that u and v are C2 functions which satisfy theCauchy-Riemann equation.
∂2u
∂x2= vyx
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∂2u
∂y2= −vxy
vyx − vxy = 0
since vxy = vxy. Therefore, u and v are harmonic.
Proving a real valued function is harmonic shows that there is a correlation between theconditions that we have used previously with Definition 4.1 and the Laplace Equation.
Definition 4.2. The Cauchy-Riemann Equations Suppose f(z) = φ(x, y) + iψ(x, y) isdifferentiable at a point z = x + iy. Then at z the first order partial derivatives of u and vexist and satisfy the Cauchy-Riemann Equations
Proof. Note that ∂2
∂x2+ ∂2
∂y2= ( ∂
∂x− i ∂
∂y)( ∂∂x
+ i ∂∂y
) we get then fx + ify = 0 and fx− ify = 0.
Let u(x, y) and v(x, y) satisfy the system ux = vy and uy = −vx. Let f(x, y) = u(x, y) +iv(x, y). Then,
fx(x, y) = ux(x, y) + ivx(x, y)fy(x, y) = uy(x, y) + ivy(x, y)
fx + ify = 0
ux(x, y) + ivx(x, y) + i(uy(x, y) + ivy(x, y))
ux(x, y) + ivx(x, y) + iuy(x, y) +−vy(x, y)
ux(x, y)− iuy(x, y) + iuy(x, y) +−ux(x, y) = 0
Let f(x, y) = u(x, y) + iv(x, y) solve fx + ify = 0
ux(x, y) + ivx(x, y) + i(uy(x, y) + ivy(x, y)) = 0
ux(x, y) + ivx(x, y) + iuy(x, y) +−vy(x, y) = 0
ux − vy + ivx + iuy = 0
ux + iuy = vy − ivxThis means ux = vy and uy = −vx in order to be harmonic.
∂φ
∂x=∂ψ
∂yand
∂φ
∂y= −∂ψ
∂x(5)
The result ends for us getting a ψ(y) = k1−k0x1−x0y . We obviously hold the constant to be zero
for convenience purposes. We then combine both φ(x) and ψ(y) into a complex potentialfunction as the following:
ω(z) =k1 − k0
x1 − x0
(x− x0) + k0 + i(k1 − k0
x1 − x0
y) (6)
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With the form of this equation we can solve any problem concerning two infinitely longplates. Take for example two infinite plates that are parallel with each other withboundary conditions such as φ(2, y) = 3 and φ(7, y) = −2. By using equation (6) we obtainthe following result:
ω(z) = −x+ 5− yi
0.0 0.5 1.0 1.5 2.0 2.5 3.0
2.0
2.5
3.0
3.5
4.0
4.5
5.0
[Figure 2]
This result seems to be very boring to us. The result obtains that it’s contour and densityplot are the same as well. We shall switch to polar coordinates in order to hopefully obtaina better result for a more subjective result with the Dirichlet problem.
ω(z) = −rcos(x) + 5− rsin(y)I
4.2
4.2
4.4
4.4
4.6
4.6
4.8
4.8
5 5
5.2
5.45.6
5.8
0 1 2 3 4 5 6
0.0
0.2
0.4
0.6
0.8
1.0
[Figure 3] 0 1 2 3 4 5 6
0.0
0.2
0.4
0.6
0.8
1.0
[Figure 4]
0
1
2
3
0
1
2
3
2
4
6
8
[Figure 5]
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We see here for the equation put above for its Real part takes a form of a high temperaturewhen reaching to the stage of about π and also the most concentrated amount of flowoccurs right at the start of the function. We know this by taking the contour and densityplot’s of these functions in mathematica as seen above. One could computationally checkthis as well. However, a picture provides us with a more laid out approach of how tounderstand the problem in all cases in coordination with how much concentration there isat a particular time to what is the temperature of what we are observing. 3
There are other interesting features of the Dirichlet problem that are just infinite plateproblems that are transfixed into one position in the Cartesian coordinate plane. Onewould certainly want the structure of a domain to preserve it’s structure over time. It does!
4.1 Harmonic Functions and Analytic Mapping
Previously, we have stated what a Harmonic function is. However, what is an analyticmapping? A function f that is analytic in a domain D and that maps D onto a domain D’is called an analytic mapping of D onto D’. This means we can map a function on ourCartesian plane to another plane. Let’s call this plane the uv-plane. This is shown as aconventional theorem.
Theorem 1 (Harmonic function under Analytic Mapping). Let f(z) = u(x, y) + iv(x, y) ananalytic mapping of a domain D the z-plane onto a domain D′ in the w-plane. If the functionφ(x, y) is harmonic in D′, then the function φ(x, y) = φ(u(x, y), v(x, y)) is harmonic in D.
This Theorem will prove to be useful because it takes every point in the domain of ourspace we are considering and transmorphs the domain into a new space with a differentlooking boundary.Assume we have a Domain D such that it’s bounded by lines y =
√3x and y =
√3x+ 4,
and the boundary conditions are φ(x,√
3x) = 10 and φ(x,√
3x+ 4) = 5.we can rotate these plates by π
6counter-clockwise. So our new space will look like what
we’ve dealt with parallel plates beforehand. Thus our equation turns out to be thefollowing:
f(z) = eiπ6 z (7)
f(z) = (
√3x
2+i
2)z (8)
u(x, y) =
√3x
2− y
2(9)
v(x, y) =x
2−√
3y
2(10)
3One could certainly take first and second derivatives and find out the the maxima and minima temper-atures that a function produces. This is just an alternative method that I find more aesthetically pleasingand more telling of the function presented before us.
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Boundary Conditions: φ(−2, y) = 5 and φ(0, y) = 10
φ(u, v) = 5/2u+ 10 (11)
φ(x, y) =5√
3
4x− 5
4y + 10 (12)
Since this is an equation of only φ we will show ψ(x, y) and find the following solution to
be the following: ω(z) = 5√
34x− 5
4y + 10 + i5
√3
4y.
01
23
0 1 2 3
5
10
15
[Figure 6]
88.5 9
9.5
10 10
10.5
10.5
11
11
11.5
11.5
1212
0 1 2 3 4 5 6
0.0
0.2
0.4
0.6
0.8
1.0
[Figure 7]
0 1 2 3 4 5 6
0.0
0.2
0.4
0.6
0.8
1.0
[Figure 8]
We have taken the real part of ω(z) again to show how the temperature works out and thedensity of heat flow through polar coordinates. There are other properties that a functioncan have that are very important. 4
4.2 Characteristics
There are two properties of functions we can recognize with the Dirichlet problem. We canthink of u(x, y) and v(x, y) in the Cartesian plane that are velocity components in both thehorizontal and vertical directions respectively. We say that the flow is irrotational ifωz = 1
2( dvdx− du
dy) = 0. We say that the flow of a function is incompressible if du
dx+ dv
dy= 0.
4Note that we have only used the aspect of rotation with our transformation of the domain. We canalso move the domain left or right through subtraction and addition and can expand or contract the domainthrough multiplication.
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Incompressibility is important because it satisfies the Conservation of Mass. This meansthe mass doesn’t change, then volume of the function doesn’t change therefore density ofthe function we are looking at doesn’t change. Now that the concepts of irrotational andincompressibility have been gone over.We will imagine a liquid substance in which the liquid in our domain is flowing parallellayers. This is known in Hydrodynamics as Laminar flow. Steady laminar flow in thex-direction between two parallel plates: Let’s take the following equation where µ standsfor the momentum in the x-direction, ρ for the pressure function, and h is the distanceaway from the layer of fluid to the boundary. One can see where the first equation comesfrom by the following source listed.: [4].
0 = −∂p∂x
+ µ∂2u
∂y2(13)
u(x, y) =1
2µ
∂p
∂xy2 + C1y + C2 (14)
Boundary Conditions: y = +h, u = 0. Let’s also say ω(z) is irrotational and incompressible.
0 =1
2µ
∂p
∂xh2 + C1h+ C2 (15)
0 =1
2µ
∂p
∂xh2 − C1h+ C2 (16)
0 =1
µ
∂p
∂xh2 + 2C2 (17)
C2 = − 1
2µ
∂p
∂xh2, C1 = 0 (18)
u(x, y) =1
2µ
∂p
∂x(y2 − h2) (19)
Assuming that p is a continuous function with respect of taking derivatives we obtain thefollowing due to the Cauchy-Riemann equation:
v(x, y) =1
2µ
∂2p
∂x2y(y2
3− h2) (20)
This results into the following complex potential function for all laminar flow:
1
2µ
∂p
∂x(y2 − h2) + i(
1
2µ
∂2p
∂x2y)(
y2
3− h2) (21)
We see that the first boundary Condition cancels out and gives us a result to work with sothat C2 and C1 can be found with ease. With knowing how to solve for the momentum wecan find the velocity5 of a laminar fluid flow.
5The derivative of function.
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5 Problem of the Square
Theorem 2 (Uniqueness Theorem for Dirichlet problem). There must be at most onesolution of the Dirichlet problem for a bounded open set D, with a continuous function Fgiven the boundary D.
Proposition 5.1. Solving for any square.D.E. uxx + uyy = 0, 0 < x < L, 0 < y < M
B.C.
{u(x, 0) = f(x) u(x,M) = 0, 0 ≤ x ≤ L
u(0, y) = 0 u(L, y) = 0, 0 ≤ y ≤M
Proof. Let U(x, y) = (c1Cos(bx) + c2Sin(bx))(d1eby + d2e
−by
assume U(0, y) = c2Sin(bx)(d1eby + d2e
−by)Assume c2 6= 0, then sin(bL) = 0, so bL = nπ, b = nπ
L
Un(x, y) = sin(nπx
L)(d1e
by + d2e−by)
Apply U(x,M) = 0,
d2 = −d1enπML
e−nπML
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Un(x, y) = sin(nπx
L)
2d1
e−nπML
enπ(y−M)
L − e−nπ(y−M)
L
2
notice the resemblance to sinh(z) = ez−e−z2
un(x, y) =N∑n=1
sin(nπx
L)sinh(
nπ(M − y)
L), n = 1, 2, ...
u(x, 0) =N∑n=1
d1sin(nπx
L)sinh(
nπy
L), n = 1, 2, ... (22)
proof of the other three sides are left to the reader, however they are very much similarwith what we have just done. We now achieve our big theorem due to superposition.
Theorem 3. Let an, bn, cn, dn be the fourier sine coefficents of f(x), g(x), h(y), andk(y) befinite fourier series. Then the solution for the Dirichlet Problem is
U(x, y) =∑Nn=1(An sin(nπx
L)sinh(nπ(M−y)
L) + (Bn sin(nπx
L)sinh(nπx
L)) + (Cn sin(nπy
M)sinh(nπ(M−x)
M)
+ (Dn sin(nπyM
)sinh(nπxM
))
Where as;
An =bn
sinh(nπML
)
Bn =bn
sinh(nπML
)
Cn =cn
sinh(nπLM
)
Dn =cn
sinh(nπLM
)
However, there needs to be a way to check if a contradiction arises out of the boundaries ofthe edges. We would not like 1=0 in our problems. This seems such a childish way oftrying to prove a contradiction. There must be a more formal mathematical way we can dothis. Luckily there is a way we can check this formally we should check this lemma beforedoing anything to the square.
Lemma 5.2. The Dirichlet problem of the square has no solution unless the compatabilitycondition holds for what is called the Nuemann Problem as the following:∫ L
0
g(x)dx−∫ L
0
f(x)dx+
∫ M
0
k(y)dy −∫ M
0
h(y)dy = 0
Proof.∫M
0
∫ L0
(uxx + uyy)dxdy =∫M
0
∫ L0uxxdx+
∫M0
∫ L0uyydy =∫M
0(ux(L, y)− ux(0, y)dy +
∫ L0
(uy(x,M)− uy(x, 0))dx =∫M0k(y)dy −
∫M0h(y)dy +
∫ L0g(x)dx−
∫ L0f(x)dx
17
6 The Problem of the Annulus
The steady-temperature φ(r) between the two concentric circular conducting cylinderssatisfies Laplace equation in polar coordinates form
r2d2φ
dr2+ r
dφ
dr= 0 (23)
We can use Cauchy-Euler’s equation in which the form of the following second orderDifferential Equation:
Definition 6.1 (Cauchy-Euler Equation (Homogeneous)). ax2y′′ + bxy′ + cy = g(x) Weuse the axillary equation with respect to: am(m− 1) + bm+ c = 0. We then find the zero’sof the equation taking the form of (m− a)(m− b).Take g(x) = 0.Then the roots will take the form of any of the following where c’s areconstants:If m1 and m2 are two distinct real roots then:
yc = c1xm1 + c2x
m2
If m1 = m2 = m are two equal real roots then:
yc = c1xm1 + c2x
m2Ln|x|
If m1 and m2 are complex roots taking the form α± iβ then:
yc = xα(c1cos(βLn|x|) + c2sin(βx)
a = 1, b = 1, c = 0 Then,m2 = 0,m = 0
φ(r) = Aln|r|+B, (24)
φ′(r) =A
|r|, (25)
φ′′(r) = − A
|r|2, (26)
We plug in for equation (9) and we see that this equation is a solution to our annulusproblem.
Now we assume that φ(a) = k0 and φ(b) = k1
ALn|a|+B = k0 (27)
ALn|b|+B = k1 (28)
18
We preform simple algebra to achieve the following results:
A =k0 − k1
Ln|ab|, B = −−k0Ln|b|+ k1Ln|a|
Ln|ab|
(29)
Now that these conditions have been satisfied. Let us find the Complex potential functionof this problem as we have done in the past for other problems. However, we will be doingthis in polar coordinates. The Cauchy-Riemann equation in polar coordinates is thefollowing:
∂φ
∂r=
1
r
∂ψ
∂θand
∂ψ
∂r= −1
r
∂φ
∂θ(30)
ω(z) = eiθ(∂φ
∂r+ i
∂ψ
∂r) =
1
reiθ(
∂φ
∂θ− i∂ψ
∂θ) (31)
LaplaceEquation(PolarCoordinates) :∂2φ
∂r2+
1
r
∂φ
∂r+
1
r2
∂2φ
∂θ= 0 (32)
Proof. Suppose f is a complex valued function differentiable at some point z0 in the complexplane. Suppose that z0 6= 0. We can write z = reiθ and z0 = r0e
iθ such that
f(reiθ) = φ(r, θ) + iψ(r, θ)
Let z− > z0 along the ray of θ = θ0. Then the following limits exists:
f ′(z0) = limr→r0
f(reiθ0)− f(r0eiθ0)
reiθ0 − r0eiθ0
= limr→r0
1
eiθ0u(r, θ0)− v(r0, θ0) + i[v(r, θ0 − v(r0, θ0)]
r − r0
= e−iθ0 [ limr→r0
u(r, θ0)− u(r0, θ0)
r − r0
+ i limr→r0
v(r, θ0)− v(r0, θ0)
r − r0
]
The real and the imaginary parts both have a limit thus the complex function has a limit atthe point z0. Therefore, we arrive at the result:
f ′(z0) = e−iθ0 [∂u
∂r(r0, θ0) +
∂v
∂r(r0, θ0)]
Now we let z− > z0 along the circle r = r0. Then,
f ′(z0) = limθ→θ0
f(reiθ0)− f(r0eiθ0)
reiθ0 − r0eiθ0
=1
r0
limθ→θ0
[u(r0, θ)− u(r0, θ0)
eiθ − eiθ0+ i
v(r0, θ)− v(r0, θ0)
eiθ − eiθ0]
= 1r0
limθ→θ0{[u(r0,θ)−u(r0,θ0)
θ−θ0 + iv(r0,θ)−v(r0,θ0)θ−θ0 ] x θ−θ0
eiθ−eiθ0 }
19
The differences in theta start to converse when taking the limit of the following functionabove to the partial derivatives of φ and ψ in respect to θ.
eiθ − eiθ0θ − θ0
=cosθ − cosθ0
θ − θ0
+ isinθ − sinθ0
θ − θ0
taking θ− > θ0 we achieve the following result:
[d
dθcosθ] + i[
d
dθsinθ] = −sinθ0 + icosθ0 = ieiθ0
f ′(z) =1
r0eiθ0[∂u
∂θ(r0, θ0) + i
∂v
∂θ(r0, θ0)]
=e−iθ0
r0
[∂v
∂θ(r0, θ0)− i∂u
∂θ(r0, θ0)]
Combining two expressions of the two expressions for f ′(z) creates the following by equalingthe real and imaginary parts of both equations used to get Cauchy-Riemann Equations.
∂u
∂r=
1
r
∂v
∂θand
∂v
∂r=−1
r
∂u
∂θ
For the problem of the annulus, we obtain the following while taking into consideration ofthe three equations listed above through some simple calculus and algebra manipulation:
ω(z) = (ALn|r|)(cos θ − i sin θ) (33)
20
2.02.53.03.54.0
0.00.51.01.5
0.0
0.5
1.0
[Figure 9]
-2 -2
-1.5-1.5
-1-1
-0.5-0.5
0 0
0.5
1
1.5
2
0 1 2 3 4 5 6
0.0
0.2
0.4
0.6
0.8
1.0
[Figure 10]
0 1 2 3 4 5 6
0.0
0.2
0.4
0.6
0.8
1.0
[Figure 11]
Once again, we see the temperature and how the heat flows within the domain of theannulus to its boundary through contour and density plotting with mathematica. We arenoticing a real relationship that is happening in the domain of bodies. Throughmathematics and looking at data representations we can bring up a conjecture about whatwe have seen. It seems as if cold states want to flow faster away while warmer states of adomain don’t. We notice this may seem obvious, but remember that cold particles moveslower than heated particles in their atomic forms. However, in space, they are reactingquite differently and telling us a different story of what our common notions of Physicstells us.
6.1 Revisiting the Problem of the Annulus
Proposition 6.1. uxx + uyy = Urr + 1rUr + 1
r2Uθθ(r > 0)
Proof. see [8].
What if no differential equation was given to us and our boundaries wouldn’t be defined asbefore? Reluctantly there is a method to do this.
21
we see that r0 > r1 for the annulus giving us conditions in which the outer and innerboundaries have functions ascribed to them. We get the following Dirichlet Problem then:
Definition 6.2 (Dirichlet Problem for Annulus Generalization). The differential equationthat must satisfy the annulus is:
Urr +1
rUr +
1
r2Uθθ = 0, r1 < r < r0
Boundary Conditions as: {U(r0, θ) = f(θ), f(θ + 2π) = f(θ)
U(r1, θ) = g(θ), g(θ + 2π) = g(θ)
and Periodicity as:
U(r, θ + 2π) = U(r, θ), r1 < r < r0
Proof. Formalization of Dirichlet Problem for the Annulus Guess the functionU(r, θ) = R(r)T (θ) and input it into the following to get:
R′′(r)T (θ) +1
rR′(r)T (θ) +
1
r2R(r)T ′′(θ) = 0
Simple rearranging acquires the following:
r2R”(r) + rR′(r)
R(r)=−T”(θ)
T (θ)= ±b2(b ≥ 0)
Using Propositon 2.4 we can see the following solution comes out:
Tn(θ) = an cos(nθ) + bn sin(nθ), an, bnεconstants, n = 0, 1, 2, ...
6 Having r2R′′(r) + rR′(r)− b2R(r) = 0, we can just use the definition of Euler’s equationas set above to receive a solution for when:
6Notice here that T0(θ) is not a solution because it doesn’t fall under the form of periodicity and justforms into constant.
22
n ≥ 1 : Rn(r) = cnrn + dnr
−n
n = 0 : R0(r) = c0 + d0 log(r)
In summary now we have:
U0(r, θ) = R0(r) = c0 + d0 log(r)
Un(r, θ) = Tn(r)Rn(r) = (ancnrn + andnr
−n) cos(nθ) + (bncnrn + bndnr
−n) sinnθ, n ≥ 1
By the superposition principle we receive the following equation for the annulus:
U(r, θ) = U0(r, θ) +N∑n=1
Un(r, θ)
Great, we have an equation for the Laplace equation to satisfy the conditions. How aboutf(θ) and g(θ) though. We assume f(θ) and g(θ) are finite Fourier series taking the form ofwhat we usually see when doing finite Fourier series:
f(θ) =1
2A0 +
N∑n=1
An cosnθ +Bn sinnθ
g(θ) =1
2C0 +
N∑n=1
Cn cosnθ +Dn sinnθ
a0 + α0 log(r0) =1
2A0 =
1
2π
∫ 2π
0
f(θ)dθ
a0 + α0 log(r1) =1
2C0 =
1
2π
∫ 2π
0
g(θ)dθ
anrn0 + αnr
−n0 = An =
1
π
∫ 2π
0
f(θ) cos(nθ)dθ
bnrn0 + βnr
−n0 = Bn =
1
π
∫ 2π
0
g(θ) cos(nθ)dθ
anrn1 + αnr
−n1 = Cn =
1
π
∫ 2π
0
f(θ) sin(nθ)dθ
bnrn1 + βnr
−n1 = Dn =
1
π
∫ 2π
0
g(θ) sin(nθ)dθ
one can easily prove that the next 6 statements are true by algebra.Let M ∼= r0
r1;
23
a0 = C0 log(r0)−A0 log(r1)2 log(M)
an =Anr
−n1 −Cnr
−n0
Mn−M−n
bn =Bnr
−n1 −Dnr
−n0
Mn−M−n
α0 = A0−C0
2 logM
αn =Cnrn0−Anrn1Mn−M−n
βn =Dnrn0−Anrn1Mn−M−n
Now we get the full solution to any annulus in our system which is the following:
U(r, θ) = a0 + α0 log(r) +N∑n=1
((anrn + αr−n)) cos(nθ) + (bnr
n + βnr−n) sin(nθ))
Example 6.1. Consider the following problem with constant boundaries on the annulus:{uxx+ uyy = 0, 1 < r < 2, 0 ≤ θ < 2π
u(1, θ) = 2, u(2, θ) = 1, 0 ≤ θ < 2π{a0 + b0 log(1) = 2
a0 + b0 log(2) = 1
a0 = 2, b0 = − log2(e)
∴ u(r) = 2− ln(r)ln2
6.2 Interior and Exterior Dirichlet Problems
Consider the case in which the annulus becomes the exterior of the disk. The interiorDirichlet problem (R0 = 0, R2 = R){r2urr + rur + uθθ = 0, 0 ≤ r < R, 0 ≤ θ < 2π
u(R, θ) = f(θ), 0 ≤ θ ≤ 2π
We can use the same methods as we solved for the annulus. However, we must disregardfunctions that are not bounded as r goes to zero. (i.e)ln(r), r−n cos(nθ), r−n sin(nθ), n = 1, 2, 3, ... Solving for what we are given we are leftwith the following:
u(r, θ) =∞∑n=0
(r
R)n(an cos(nθ) + bn sin(nθ)
coefficients are calculated by the following methods:a0 = 1
2π
∫ 2π
0f(θ)dθ
an = 1π
∫ 2π
0f(θ) cos(nθ)dθ, n > 0
bn = 1π
∫ 2π
0f(θ) sin(nθ)dθ, n > 0
24
Therefore, we are expanding the function f(θ) such that
f(θ) =∞∑n=0
an cos(nθ + bn sin(nθ)
with multiplication of each term of the series with ( rR
)n.The exterior Dirichlet problem (R1 = R,R2 =∞):{r2urr + rur + uθθ = 0, R ≤ r <∞, 0 ≤ θ < 2π
u(R, θ) = f(θ), 0 ≤ θ ≤ 2π
We would like bounded solutions for when r goes to infinity. So, just like last derivation weleave out the following cases again. The solution then takes the form of the following:
u(r, θ) =∞∑n=0
(r
R)−n(an cos(nθ) + bnsin(nθ))
In which coefficients are solved the same way as previously stated for interior Dirichletproblems.
7 Problem of the Disk
Okay, Remember what we said long ago on the first page? well we are finally at that point!Suppose that we have a disk that satisfies the laplace equation Urr + 1
rUr + 1
r2Uθθ = 0 with
the boundary condition u(a, θ) = f(θ).where u is periodic in θ with period 2π and u isbounded for r ≤ a.
Proof. Let u(r, θ) = θ(θ)
R′′(r)θ(θ) +1
rR′(r)θ(θ) +
1
r2R(r)θ′′(θ)
r2R′′(r)
R(r)+ r
R′(r)
R(r)=−θ′′(θ)θ
= λ
r2R′′(r) + rR′(r)− λR(r) = 0
θ′′(θ) + λθ(θ) = 0
Case 1: λ < 0;Remember proposition 2.4? We’ll use that here.
θ(θ) = c1ebθ + c2e
−bθ
Since u is periodic in respect with θ because e±bθ is never periodic. Therefore, c1 = c2 = 0.Case 2: λ = 0;
25
θ(θ) = d1θ(θ) + d2
Since θ is periodic then d1 = 0. so,
θ(θ) = d2
r2R′′(r) + rR′(r)− λR(r) = 0
Taking from the definition of the Cauchy-Euler equation we find the following:m(m− 1) +m = 0, s2 = 0R(r) = e1r
0 + e2r0 log(r)
As r gets closer to 0 then the function will trend to go into infinity which would not makethe problem bounded anymore. 7 However, when e2 = 0then R(r) is bounded.R(r) = e1
Case 3: λ > 0;access the useful proposition again to obtain the following equation for θ(θ).
θ(θ) = f1 cos(bθ) + f2 sin(bθ)
since θ is periodic with period 2π. So b = n, n = N
r2R′′(r) + rR′(r)− n2R(r) = 0
s(s− 1) + s− n2 = 0
s = ±n
R(r) = g1rn + c2r
−n
This function is bounded when c2 = 0 because again we have another radius problem.so R(r) = g1r
n
U(r, θ) = g1rnf1 cos(bθ) + g1r
nf2 sin(bθ)
Therefore, we have three solutions.
u(r, θ) =
λ < 0 0
λ = 0 d2e1 = 1
λ > 0 g1f1rn cos(bθ) + g1f2r
n sin(bθ) = cnrn cos(bθ) + knr
n sin(bθ)
7This is a huge problem because the function will turn into an infinite the closer we get to 0 hopefully Iprovide some methodology to fix this problem
26
We can now set up and equation:
u(r, θ) =c0
2+∞∑n=1
cnrn cos(bθ) + knr
n sin(bθ)
u(a, θ) = f(θ) =c0
2+∞∑n=1
cnan cos(bθ) + kna
n sin(bθ)
We can take the Fourier Series to find cnan and kna
n and c0.
c0 =1
π
∫ π
−πf(θ)dθ
cnan =
1
π
∫ 2π
0
f(θ) cos(nθ)dθ, nε0,N
knan =
1
π
∫ π
0
f(θ) sin(nθ)dθ, nε0,N
7.1 Removing the Singularity
We explicitly see a problem with doing the last problem in polar coordinates. It seems as ifthere is a singularity at (0,0) when it comes to Polar-coordinates. We would like thissingularity to be removed.
Definition 7.1. (Dirichlet problem of the Disk) The Dirichlet Problem for the disk ofradium r0 centered at (0, 0) can be expressed as the following:
uxx + uyy = 0√x2 + y2 < r0
U(r0, θ) = f(θ) f(θ + 2π) = f(θ)
Where U(r, θ) = u(x, y) = u(rcosθ, rsinθ) and f(θ) is a given periodic, continuous functionof period 2π.
We shall now prove a useful corollary.
Corollary. (Poisson’s Kernel) The solution to the Dirichlet problem using the PoissonKernel is
u(r, θ) =1
2π
∫ π
−π
1− r2
1− 2r cos(θ − α) + r2g(α)dα
27
Proof. our definitions of Fourier series under the annulus we prove to be useful here as well.
u(r, θ) =a0
2+∞∑n=1
anrncos(nθ) + bnr
nsin(nθ)
=1
2π
∫ π
−πg(α)dα +
∞∑n=1
(1
π
∫ π
−πg(α)cos(nα)dα)rncos(nθ) + (
1
π
∫ π
−πg(α)sin(nα)dα)rnsin(nθ)
=1
2π
∫ π
−π(1 + 2
∞∑n=1
rn(cos(nα)cos(nθ) + sin(nα)sin(nθ)))g(α)gα
Let there exist a function P (r, θ, α) such that
P (r, θ, α) = 1 + 2∞∑n=1
rn(cos(nα)cos(nθ) + sin(nα)sin(nθ))
Through using trigonometric identities we achieve the following:
= 1 + 2∞∑n=1
rncos(n(θ − α))
Please recognize this is now in a form that we can now devolve back into Demoivre’sformula.
= 1 + 2∞∑n=1
(rei(θ−α))n +∞∑n=1
(re−i(θ−α))n
Recall the definition of a geometric series from Calculus
∞∑n=1
zn =z
1− z, z < 1
We achieve the following consequence from the geometric series:
= 1 + (rei(θ−α)
1− rei(θ−α)) +
re−i(θ−α)
1− re−i(θ−α)
Through simplification we achieve the following:
=1− r2
1− rei(θ−α) − rei(θ−α) + r2
P (r, θ, α) =1− r2
1− 2rcos(θ − α) + r2
Therefore, finally achieving the following:
u(r, θ) =1
2π
∫ π
−π
1− r2
1− 2r cos(θ − α) + r2g(α)dα
.
28
This proof was for the outer boundary to be 1. However, this problem can be extendedinfinitely, not just to 1. The proof is about the same and I encourage the reader to do thegeneralization proof. One should get the result of
P (r, r0, θ − t) =r2
0 − r2
r20 − 2rr0cos(θ − t) + r2
, (r < r0)
Achieving the following solution:{U(r, θ) = 1
2π
∫ π−π P (r, r0, θ − t)f(t)dt (r < r0)
U(r0, θ) = f(θ), (r = r0)
One can easily plug in their favorite number to check the trivial second condition to see ifthis holds true. The singularity is now removed from the problem we have been dreadingabout.
29
8 Future Projects
To move up into three space. I have done research over the winter of transitioningrectangular functions into gradients and then using vector analysis about how to decide if afunction satisfies the Dirichlet problem conditions. It seems as if the mathematicalcommunity hasn’t developed a method in which one can find a satisfaction withCauchy-Riemann in three dimensions. However, shifting the coordinate functions does.To develop more into Naiver-Stokes with applications. I find this subject to be interestingconcerning fluid flow. its a bit out of bounds for this paper concerning how Dirichletproblem figures out how fluid flows from the domain to it’s boundary. But none the lessit’s a characteristic of fluids so I choose to put it in my paper.I have recently been in the phase of nostalgia and have looked into a few previous booksthat I own in the field of Mathematics. I have found the Dirichlet problem highly relatiblein the field of Partial Differential Equations and has a whole chapter dedicated to it. Thisseems like the next logical step forward into understanding the Dirichlet problem once I goover three dimensional space.
30
References
[1] Dr. Blecker, Dr. Csordas Basic Partial Differential Equations. pg. 339-414. Honolulu,Hawaii
[2] Dr. Mathews, Dr. Howell http://mathfaculty.fullerton.edu/mathews//complex.htmlComplex Analysis Project for Undergradate Students. Fullerton, California
[3] Shanahan Patrick, Zill, Dennis Complex Analysis: A first Course with Applications.pg.130-134,137-138,141-146, 190-196 3rd edition, 2014
[4] Weber-Shirk, Monroe Basics Governing of Differential Equations. Cornell University.PPT.
[5] Shapiro http://www.mth.msu.edu/ shapiro/teaching/classes/425/crpolar.pdf MichiganState University 1998
[6] Slowik http://course1.winona.edu/eslowik/Philsciencenotes.htm Winona, MN 2001
[7] Department of Chemical and Biological Engineering. Irrotational and Incompressabil-ity. https://www.youtube.com/watch?v=CwO8a5mP13A University of Colorado Boulder2011.
[8] University of California Davis https://www.math.ucdavis.edu/ saito/courses/21C.w11/polar-lap.pdf.
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