Solve Laplace equation on unit disk

[evinda]

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)

 

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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

 

[evinda]

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)

 

[I like Serena]

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:
$$R(r) = c_1\cos(\sqrt \lambda \ln r) + c_2\sin(\sqrt\lambda\ln r)$$

 

[Opalg]

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)

Two comments about this.

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.

 

[evinda]

Two comments about this.

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)

 

[I like Serena]

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)

 

[evinda]

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 - - -

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)

 

[I like Serena]

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)

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)

 

[evinda]

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

(Nod)

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 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)

 

[I like Serena]

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 $0$.

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?

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.

 

[evinda]

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)

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

Ok...

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)

 

[I like Serena]

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...

 

[evinda]

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)

 

[I like Serena]

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]

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)

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)

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)

 

[evinda]

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)

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?

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)

 

[I like Serena]

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)

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)

 

[evinda]

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 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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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?

 

[evinda]

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?

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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

(Nod)

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)

 

[evinda]

(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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

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)

 

[I like Serena]

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)

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)