# Solve Laplace equation on unit disk

• MHB
• evinda
In summary, the conversation discusses the method of solving the Laplace equation on the unit disk with given boundary data of $u(\theta)=\cos\theta$ on the unit circle. The conversation also delves into the proof that small oscillations of the boundary data will result in small oscillations of the corresponding solution of the Dirichlet problem. The conversation covers the separation of variables, the characteristic equation, and the general form of the solution for $\Theta(\theta)$. It also addresses the meaning of "little oscillations of the boundary data" and suggests using a basis of $\cos(m\theta)$ and $\sin(m\theta)$ to represent all real functions with period $2\pi$.
evinda
Gold Member
MHB
Hello! (Wave)

I want to solve the Laplace equation on the unit disk, with boundary data $u(\theta)=\cos{\theta}$ on the unit circle $\{ r=1, 0 \leq \theta<2 \pi\}$. I also want to prove that little oscillations of the above boundary data give little oscillations of the corresponding solution of the Dirichlet problem after first stating strictly the statemtent.

How do we solve the Laplace equation on the unit disk with the given boundary data? Could you give me a hint? (Thinking)

The Laplace equation is the following, right?

$$u_{rr}+\frac{1}{r} u_r+\frac{1}{r^2}u_{\theta \theta}=0$$

Don't we look for a solution of the form $u(r,\theta)=R(r) \Theta(\theta)$ ? (Thinking)

Hey evinda!

Suppose we write Laplace's equation in the intermediate form:
$$\frac 1r \pd {}r(ru_r)+\frac 1{r^2}u_{\theta\theta}=0$$
Can we separate the variables as you already suggested? (Wondering)

I like Serena said:
Hey evinda!

Suppose we write Laplace's equation in the intermediate form:
$$\frac 1r \pd {}r(ru_r)+\frac 1{r^2}u_{\theta\theta}=0$$
Can we separate the variables as you already suggested? (Wondering)

Supposing that $u(r, \theta)=R(r) \Theta(\theta)$, we get the following, right?

$$r \frac{R'(r)}{R(r)}+r^2 \frac{R''(r)}{R(r)}+\frac{\Theta''(\theta)}{\Theta(\theta)}=0$$

So,

$$r \frac{R'(r)}{R(r)}+r^2 \frac{R''(r)}{R(r)}=-\frac{\Theta''(\theta)}{\Theta(\theta)}=-\lambda$$

Thus, we have the following system:

$$\left\{\begin{matrix} r^2 R''(r)+rR'(r)+\lambda R(r)=0\\ \Theta''(\theta)-\lambda \Theta(\theta)=0 \end{matrix}\right.$$

$\Theta''(\theta)-\lambda \Theta(\theta)=0$

The characteristic equation is $\mu^2-\lambda=0$.

$\mu^2=\lambda \Rightarrow \mu=\pm \sqrt{\lambda}$.

So, $\Theta(\theta)=c_1 e^{\sqrt{\lambda}\theta}+c_2 e^{-\sqrt{\lambda}\theta}$.

Is it right so far? (Thinking)

All correct as far as I can tell. (Nod)

I like Serena said:
All correct as far as I can tell. (Nod)

Nice... And how can we solve the equation $r^2 R''(r)+rR'(r)+\lambda R(r)=0$ ? (Thinking)

evinda said:
Nice... And how can we solve the equation $r^2 R''(r)+rR'(r)+\lambda R(r)=0$ ? (Thinking)

WolframAlpha tells us that it's a Cauchy-Euler equation.
Alternatively we can write it as a Sturm-Liouville equation.

Either way, WolframAlpha also tells us that the solution is:
$$R(r) = c_1\cos(\sqrt \lambda \ln r) + c_2\sin(\sqrt\lambda\ln r)$$

evinda said:
Thus, we have the following system:

$$\left\{\begin{matrix} r^2 R''(r)+rR'(r)+\lambda R(r)=0\\ \Theta''(\theta)-\lambda \Theta(\theta)=0 \end{matrix}\right.$$

$\Theta''(\theta)-\lambda \Theta(\theta)=0$

The characteristic equation is $\mu^2-\lambda=0$.

$\mu^2=\lambda \Rightarrow \mu=\pm \sqrt{\lambda}$.

So, $\Theta(\theta)=c_1 e^{\sqrt{\lambda}\theta}+c_2 e^{-\sqrt{\lambda}\theta}$.

Is it right so far? (Thinking)

First, don't assume that $\lambda$ is positive. If it is negative, then the $\Theta$ equation will be a simple harmonic motion equation, with solutions of the form $\Theta(\theta) = c_1\cos(\sqrt{-\lambda}\theta) + c_2\sin(\sqrt{-\lambda}\theta)$. In particular, if $\lambda = -1$ then one of the solutions will be $\Theta(\theta) = \cos\theta$, which fits in rather well with the boundary data $u(\theta) = \cos\theta$.

Next, if indeed $\lambda=-1$ then the $R$ equation becomes $rR'(r) + r^2R''(r) = R(r)$, which has a very simple solution $R(r) = r$.

That still leaves you with the problem of how to handle "little oscillations of the boundary data". That seems much more challenging to me.

Opalg said:

First, don't assume that $\lambda$ is positive. If it is negative, then the $\Theta$ equation will be a simple harmonic motion equation, with solutions of the form $\Theta(\theta) = c_1\cos(\sqrt{-\lambda}\theta) + c_2\sin(\sqrt{-\lambda}\theta)$. In particular, if $\lambda = -1$ then one of the solutions will be $\Theta(\theta) = \cos\theta$, which fits in rather well with the boundary data $u(\theta) = \cos\theta$.

So we consider firstly $\lambda$ to be negative. Then we find the general form of $\Theta(\theta)$.
Since one possible solution is $\Theta(\theta)=\cos{\theta}$, i.e. for $\lambda=-1$, do we deduce that this is the desired value of $\lambda$ ? Or have I understood it wrong? (Thinking)

evinda said:
So we consider firstly $\lambda$ to be negative. Then we find the general form of $\Theta(\theta)$.
Since one possible solution is $\Theta(\theta)=\cos{\theta}$, i.e. for $\lambda=-1$, do we deduce that this is the desired value of $\lambda$ ? Or have I understood it wrong? (Thinking)

Yes.
We can write the boundary condition as:
$$u(1,\theta)=R(1)\Theta(\theta)=\cos\theta$$
And as Opalg said, we can write the solution for $\Theta(\theta)$ also as:
$$\Theta(\theta) = c_3 \cos(\sqrt{-\lambda}\theta) + c_4 \sin(\sqrt{-\lambda}\theta)$$
So we can solve for the boundary condition first, and then look at $R(r)$. (Thinking)

I like Serena said:
Yes.
We can write the boundary condition as:
$$u(1,\theta)=R(1)\Theta(\theta)=\cos\theta$$
And as Opalg said, we can write the solution for $\Theta(\theta)$ also as:
$$\Theta(\theta) = c_3 \cos(\sqrt{-\lambda}\theta) + c_4 \sin(\sqrt{-\lambda}\theta)$$
So we can solve for the boundary condition first, and then look at $R(r)$. (Thinking)

Ok... And do we know that only for $\lambda=-1$ the boundary data $u(\theta)=\cos{\theta}$ is satisfied? (Thinking)

- - - Updated - - -

Opalg said:
That still leaves you with the problem of how to handle "little oscillations of the boundary data". That seems much more challenging to me.
What is meant with "little oscillations of the boundary data" ? Do we pick for example $\cos{(m \theta)}$ ? Or do we show it somehow else? (Thinking)

evinda said:
Ok... And do we know that only for $\lambda=-1$ the boundary data $u(\theta)=\cos{\theta}$ is satisfied?

We have the restriction that $\sqrt{-\lambda}$ must be an integer, since otherwise we wouldn't have that $\Theta(0)=\Theta(2\pi)$.

The set $\{\cos(m \theta),\sin(m \theta)\}$ is a basis for the space of all real functions with period $2\pi$.
It means that each real function with period $2\pi$ can be uniquely written as a linear sum of those functions.
Since with $\lambda=-1$ we have a solution, it must be the only solution for $\lambda$. (Thinking)

evinda said:
What is meant with "little oscillations of the boundary data" ? Do we pick for example $\cos{(m \theta)}$ ? Or do we show it somehow else?

I think that it indeed means that we have $u(1,\theta)=\cos\theta + \varepsilon\cos(m(\theta -\theta_0))$.
In that case we can split the problem into 2 problems and add the respective solutions together, can't we? (Wondering)

I like Serena said:
We have the restriction that $\sqrt{-\lambda}$ must be an integer, since otherwise we wouldn't have that $\Theta(0)=\Theta(2\pi)$.

(Nod)

I like Serena said:
The set $\{\cos(m \theta),\sin(m \theta)\}$ is a basis for the space of all real functions with period $2\pi$.
It means that each real function with period $2\pi$ can be uniquely written as a linear sum of those functions.
Since with $\lambda=-1$ we have a solution, it must be the only solution for $\lambda$. (Thinking)

So if we are given any $2\pi$-periodic function and we find some $m$ so that the function is written as a linear combination of $\cos(m \theta)$, $\sin(m \theta)$, then this $m$ is unique? (Thinking)

I like Serena said:
I think that it indeed means that we have $u(1,\theta)=\cos\theta + \varepsilon\cos(m(\theta -\theta_0))$.
In that case we can split the problem into 2 problems and add the respective solutions together, can't we? (Wondering)

Why do we pick $u(1,\theta)=\cos\theta + \varepsilon\cos(m(\theta -\theta_0))$? (Thinking)

Picking this, we get that $u(1, \theta)=\cos{\theta}+ \epsilon [\cos{(m \theta)} \cos{(m \theta_0)}+\sin{(m \theta)} \sin{(m \theta_0)}]$. $\Theta(\theta)$ doesn't get this value for any $\lambda$, does it? (Thinking)

evinda said:
So if we are given any $2\pi$-periodic function and we find some $m$ so that the function is written as a linear combination of $\cos(m \theta)$, $\sin(m \theta)$, then this $m$ is unique? (Thinking)

Any $2\pi$-periodic function can be written as the Fourier series $A_0 + A_1\cos(\theta) + B_1\sin(\theta) + A_2\cos(2\theta) + B_2\sin(2\theta) + ...$.
Those $A_i$ and $B_i$ are unique.

In our case we have the function $\cos\theta$, which means that $A_1=1$ and all other $A_i$ and $B_i$ must be $0$.

evinda said:
Why do we pick $u(1,\theta)=\cos\theta + \varepsilon\cos(m(\theta -\theta_0))$?

Isn't this the original boundary equation with a small oscillation on top of it?

evinda said:
Picking this, we get that $u(1, \theta)=\cos{\theta}+ \epsilon [\cos{(m \theta)} \cos{(m \theta_0)}+\sin{(m \theta)} \sin{(m \theta_0)}]$. $\Theta(\theta)$ doesn't get this value for any $\lambda$, does it? (Thinking)

Suppose we solve the problem separately for $u(1, \theta)=\cos{\theta}$ and $u(1, \theta)=\epsilon [\cos{(m \theta)} \cos{(m \theta_0)}+\sin{(m \theta)} \sin{(m \theta_0)}]$.
Then the first problem has $\lambda=-1$ and the second problem has $\lambda=-m^2$ doesn't it? (Wondering)
After all, for a given $m$ the expression $\epsilon\cos{(m \theta_0)}$ is a constant, and $\epsilon\sin{(m \theta_0)}$ is a constant as well.

I like Serena said:
Any $2\pi$-periodic function can be written as the Fourier series $A_0 + A_1\cos(\theta) + B_1\sin(\theta) + A_2\cos(2\theta) + B_2\sin(2\theta) + ...$.
Those $A_i$ and $B_i$ are unique.

In our case we have the function $\cos\theta$, which means that $A_1=1$ and all other $A_i$ and $B_i$ must be $0$.

I see... (Smile)

I like Serena said:
Isn't this the original boundary equation with a small oscillation on top of it?

Ok...
I like Serena said:
Suppose we solve the problem separately for $u(1, \theta)=\cos{\theta}$ and $u(1, \theta)=\epsilon [\cos{(m \theta)} \cos{(m \theta_0)}+\sin{(m \theta)} \sin{(m \theta_0)}]$.
Then the first problem has $\lambda=-1$ and the second problem has $\lambda=-m^2$ doesn't it? (Wondering)
After all, for a given $m$ the expression $\epsilon\cos{(m \theta_0)}$ is a constant, and $\epsilon\sin{(m \theta_0)}$ is a constant as well.

Yes. So is then the solution of the $\Theta$-problem this one?

$\Theta(\theta)=\cos{\theta}+ \epsilon \cos{(m \theta_0)}\cos{(m \theta)}+\epsilon \sin{(m \theta_0)}\sin{(m \theta)}$

If so, then do we find for this $\Theta$ the corresponding $R(r)$ in order to show that we have little oscillations of the solution of the problem? (Thinking)

evinda said:
I see... (Smile)

Ok...

Yes. So is then the solution of the $\Theta$-problem this one?

$\Theta(\theta)=\cos{\theta}+ \epsilon \cos{(m \theta_0)}\cos{(m \theta)}+\epsilon \sin{(m \theta_0)}\sin{(m \theta)}$

If so, then do we find for this $\Theta$ the corresponding $R(r)$ in order to show that we have little oscillations of the solution of the problem? (Thinking)

Yes...

I like Serena said:
Yes...

For $\lambda=-1$, the $R$-equation had the solution $R(r)=r$.

For $\lambda=-m^2$, the $R$-equation gets the form $r^2R''(r)+rR'(r)=m^2R(r)$, the solution of which is $R(r)=m^2 r$, right?

Having the boundary data $u(1,\theta)=\cos{\theta}+ \epsilon \cos{(m(\theta-\theta_0))}$, do we pick as $R$ the sum of the above two solutions? If so, how is this justified? (Thinking)

For reference, we have:
[box=blue]
The polar Laplace equation:
$r^2u_{rr}+ru_r+u_{\theta\theta} = 0 \tag 1$
$\Theta(\theta) = c_1 e^{\sqrt\lambda\theta}+c_2 e^{-\sqrt\lambda\theta} =c_3 e^{\sqrt{-\lambda}\theta}+c_4 e^{-\sqrt{-\lambda}\theta}$
$=c_5 \cos(\sqrt{-\lambda}\theta)+c_6 \sin(-\sqrt{-\lambda}\theta) =c_7 \cos(\sqrt{-\lambda}(\theta-c_8)) \tag 2$
The $R$-equation:
$r^2R''(r)+rR'(r)+\lambda R(r) = 0 \tag 3$
[/box]

evinda said:
For $\lambda=-1$, the $R$-equation had the solution $R(r)=r$.

With the help of W|A I found that the general solution of the 2nd order $R$-equation (3) for $\lambda=-1$ is:
$$R(r)=c_1r+\frac{c_2}{r}$$
(Thinking)

evinda said:
For $\lambda=-m^2$, the $R$-equation gets the form $r^2R''(r)+rR'(r)=m^2R(r)$, the solution of which is $R(r)=m^2 r$, right?

Not quite.
$R(r)=m^2 r$ is actually the solution for $\lambda=-1$, just with the constant $c_1=m^2$, isn't it? (Wondering)

evinda said:
Having the boundary data $u(1,\theta)=\cos{\theta}+ \epsilon \cos{(m(\theta-\theta_0))}$, do we pick as $R$ the sum of the above two solutions? If so, how is this justified?

Let's try!

Let $\Theta_1(\theta)=\cos(\theta)$ with corresponding solution $R_1(r)$ for $\lambda=-1$.
Let $\Theta_m(\theta)=\epsilon \cos{(m(\theta-\theta_0))}$ with corresponding solution $R_m(r)$ for $\lambda=-m^2$.

Suppose we try the solution $u(r,\theta)=(R_1(r)+R_m(r))(\Theta_1(\theta)+\Theta_m(\theta))$.
Does it satisfy the Laplace equation (1)? (Wondering)

I like Serena said:
With the help of W|A I found that the general solution of the 2nd order $R$-equation (3) for $\lambda=-1$ is:
$$R(r)=c_1r+\frac{c_2}{r}$$
(Thinking)

(Nod)

I like Serena said:
Not quite.
$R(r)=m^2 r$ is actually the solution for $\lambda=-1$, just with the constant $c_1=m^2$, isn't it? (Wondering)

I found in W|A that for $\lambda=-m^2$ we get that $R(r)=c_1 \cosh{(m \ln{r})}+c_2 \sinh{(m \ln{r})}$, right?

I like Serena said:
Let's try!

Let $\Theta_1(\theta)=\cos(\theta)$ with corresponding solution $R_1(r)$ for $\lambda=-1$.
Let $\Theta_m(\theta)=\epsilon \cos{(m(\theta-\theta_0))}$ with corresponding solution $R_m(r)$ for $\lambda=-m^2$.

Suppose we try the solution $u(r,\theta)=(R_1(r)+R_m(r))(\Theta_1(\theta)+\Theta_m(\theta))$.
Does it satisfy the Laplace equation (1)? (Wondering)

Do we substitute the quantities that we found? (Thinking)

evinda said:
I found in W|A that for $\lambda=-m^2$ we get that $R(r)=c_1 \cosh{(m \ln{r})}+c_2 \sinh{(m \ln{r})}$, right?

Yes.
But after a bit of twiddling I also found the nicer solution:
$$R(r)=c_1r^m + \frac{c_2}{r^m}$$
(Thinking)

evinda said:
Do we substitute the quantities that we found?

I don't think that's really necessary, although we could.
We can use that $u_1=R_1(r)\Theta_1(\theta)$ and $u_m=R_m(r)\Theta_m(\theta)$ are solutions. (Thinking)

I like Serena said:
Yes.
But after a bit of twiddling I also found the nicer solution:
$$R(r)=c_1r^m + \frac{c_2}{r^m}$$
(Thinking)

Ok... (Nod)

I like Serena said:
I don't think that's really necessary, although we could.
We can use that $u_1=R_1(r)\Theta_1(\theta)$ and $u_m=R_m(r)\Theta_m(\theta)$ are solutions. (Thinking)

So we use the fact that $u_1$, $u_2$ are solutions, and so is also their sum?

$u(r, \theta)=\left( c_1 r+ \frac{c_2}{r}\right) \cos{\theta}+\epsilon \left( c_3 r^m+\frac{c_4}{r^m}\right) \cos{(m(\theta-\theta_0))}$

satisfies the boundary condition if $\left( c_1 r+ \frac{c_2}{r}\right)=1$ and $\left( c_3 r^m+\frac{c_4}{r^m}\right)$=1, right? (Thinking)

evinda said:
So we use the fact that $u_1$, $u_2$ are solutions, and so is also their sum?

$u(r, \theta)=\left( c_1 r+ \frac{c_2}{r}\right) \cos{\theta}+\epsilon \left( c_3 r^m+\frac{c_4}{r^m}\right) \cos{(m(\theta-\theta_0))}$

satisfies the boundary condition if $\left( c_1 r+ \frac{c_2}{r}\right)=1$ and $\left( c_3 r^m+\frac{c_4}{r^m}\right)$=1, right?

Yep.
That is, for $r=1$. (Nod)

I have looked again at it... Isn't it

$$u(r, \theta)= \left( c_1 r+\frac{c_2}{r}\right) \cos{\theta}+ \epsilon \left( c_3 r^m+\frac{c_4}{r^m}\right)[ \cos{(m \theta)} \cos{(m \theta_0)}+ \sin{(m \theta)} \sin{(m \theta_0)}]$$

? Or am I wrong? (Thinking)

evinda said:
I have looked again at it... Isn't it

$$u(r, \theta)= \left( c_1 r+\frac{c_2}{r}\right) \cos{\theta}+ \epsilon \left( c_3 r^m+\frac{c_4}{r^m}\right)[ \cos{(m \theta)} \cos{(m \theta_0)}+ \sin{(m \theta)} \sin{(m \theta_0)}]$$

? Or am I wrong?

Isn't that the same? (Wondering)On a different note, what is $u(0,\theta)$ and what should it be?

I like Serena said:
Isn't that the same? (Wondering)On a different note, what is $u(0,\theta)$ and what should it be?

Oh yes, right... $u(1, \theta)=\cos{\theta}+ \epsilon \cos{(m(\theta-\theta_0))}$.

It has to hold that $c_1 r+\frac{c_2}{r}=1$ and $c_3 r^m+\frac{c_4}{r^m}=1$.

Do we solve the first equation as for $r$ and substitute at the second one? (Thinking)
Or do we do something else?

evinda said:
Oh yes, right... $u(1, \theta)=\cos{\theta}+ \epsilon \cos{(m(\theta-\theta_0))}$.

It has to hold that $c_1 r+\frac{c_2}{r}=1$ and $c_3 r^m+\frac{c_4}{r^m}=1$.

Do we solve the first equation as for $r$ and substitute at the second one?
Or do we do something else?

Isn't it a boundary condition for $r=1$, so shouldn't we subtitute that? (Wondering)

And what happens if we substitute $r=0$? (Wondering)

I like Serena said:
Isn't it a boundary condition for $r=1$, so shouldn't we subtitute that? (Wondering)

And what happens if we substitute $r=0$? (Wondering)

Ah yes! So it has to hold that $c_1+c_2=1$ and $c_3+c_4=1$. Can we get also an other information?

Can we substitute $r=0$ although we have $\frac{1}{r}$ and thus $r \neq 0$ ? (Thinking)

evinda said:
Ah yes! So it has to hold that $c_1+c_2=1$ and $c_3+c_4=1$.

(Nod)

evinda said:
Can we get also an other information?

Can we substitute $r=0$ although we have $\frac{1}{r}$ and thus $r \neq 0$ ?

Indeed. But shouldn't $u(r,\theta)$ be defined for $r=0$ as well? (Wondering)

I like Serena said:
(Nod)
Indeed. But shouldn't $u(r,\theta)$ be defined for $r=0$ as well? (Wondering)

In order $u(r, \theta)$ to be defined for $r=0$, it has to hold $c_2=c_4=0$.
Thus $c_1=c_3=1$. Or not? (Thinking)

So $u(r, \theta)=r \cos{\theta}+ \epsilon r^m \cos{(m(\theta-\theta_0))}$, right? (Thinking)

- - - Updated - - -

But if it is like that, shouldn't it also be $c_2=0$ ? (Thinking)

evinda said:
In order $u(r, \theta)$ to be defined for $r=0$, it has to hold $c_2=c_4=0$.
Thus $c_1=c_3=1$. Or not? (Thinking)

So $u(r, \theta)=r \cos{\theta}+ \epsilon r^m \cos{(m(\theta-\theta_0))}$, right? (Thinking)

- - - Updated - - -

But if it is like that, shouldn't it also be $c_2=0$ ? (Thinking)

Indeed.
Why should $c_2=0$? (Wondering)

I like Serena said:
Indeed.
Why should $c_2=0$? (Wondering)

Don't we have that $u(r, \theta)=\left( c_1 r+\frac{c_2}{r}\right) \cos{\theta}$ ?

And this $u(r, \theta)$ is only defined for $r=0$ if $c_2=0$... Or am I wrong? (Thinking)

evinda said:
Don't we have that $u(r, \theta)=\left( c_1 r+\frac{c_2}{r}\right) \cos{\theta}$ ?

And this $u(r, \theta)$ is only defined for $r=0$ if $c_2=0$... Or am I wrong?

Yes.

To be fair, now that I read the OP again, it doesn't say that $u$ has to be defined for $r=0$. (Thinking)

I like Serena said:
Yes.

To be fair, now that I read the OP again, it doesn't say that $u$ has to be defined for $r=0$. (Thinking)

So we pick this solution:

$u(r, \theta)=\left( c_1 r+ \frac{c_2}{r}\right) \cos{\theta}+\epsilon \left( c_3 r^m+\frac{c_4}{r^m}\right) \cos{(m(\theta-\theta_0))}$

Right? (Thinking)

And can we just say that this is a little oscillation of the solution of the first problem? Or do we prove it somehow? (Wondering)

evinda said:
So we pick this solution:

$u(r, \theta)=\left( c_1 r+ \frac{c_2}{r}\right) \cos{\theta}+\epsilon \left( c_3 r^m+\frac{c_4}{r^m}\right) \cos{(m(\theta-\theta_0))}$

Right?

Shall we make that:

$u(r, \theta)=\left( c_1 r+ \frac{1-c_1}{r}\right) \cos{\theta}+\epsilon \left( c_3 r^m+\frac{1-c_3}{r^m}\right) \cos{(m(\theta-\theta_0))}$

(Wondering)

evinda said:
And can we just say that this is a little oscillation of the solution of the first problem? Or do we prove it somehow?

Yes, we can just say that for a small oscillation, its amplitude $\epsilon$ is small, and then the contribution to the solution is also small. (Emo)

## 1. What is the Laplace equation?

The Laplace equation is a partial differential equation that describes how a scalar field varies over a given region. It is named after the French mathematician Pierre-Simon Laplace and is commonly used in physics and engineering to model various physical phenomena.

## 2. What is the unit disk?

The unit disk is a mathematical term that refers to a disk with a radius of 1 unit. In other words, it is a circle with a diameter of 2 units and a circumference of approximately 6.28 units.

## 3. Why is it important to solve the Laplace equation on a unit disk?

Solving the Laplace equation on a unit disk allows us to understand and predict the behavior of scalar fields within a circular region with a radius of 1 unit. This can be useful in many applications, such as determining the electric potential or temperature distribution within a circular object.

## 4. How is the Laplace equation solved on a unit disk?

The Laplace equation on a unit disk can be solved using various methods, including separation of variables, conformal mapping, and complex analysis. These methods involve breaking down the equation into simpler parts and using mathematical techniques to find a solution.

## 5. What are some real-world applications of solving the Laplace equation on a unit disk?

The Laplace equation on a unit disk has a wide range of applications, including modeling the flow of heat in a circular object, predicting the electric potential in a circular circuit, and understanding the behavior of fluid flow in a circular pipe. It is also commonly used in image processing and computer graphics to smooth out edges and create realistic images.

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