MHB Why should the solution be periodic?

[mathmari]

Hey!

The Laplace's equation with polar coordinates is:
$$u_{rr}+\frac{1}{r} u_r +\frac{1}{r^2} u_{\theta \theta}=0$$

We suppose the boundary value problem with Dirichlet boundary conditions for the Laplace's equation on a disc with center the origin of axis and radius $a$:

$$u_{rr}+\frac{1}{r} u_r +\frac{1}{r^2} u_{\theta \theta}=0, \ \ \ \ 0 \leq r \leq a, \ \ \ 0 \leq \theta \leq 2 \pi$$
$$u(a,\theta)=h(\theta), \ \ \ \ 0 \leq \theta \leq 2 \pi$$

The boundary condition should be continuous everywhere, so $h(0)=h(2 \pi)$.

Solving the problem with the method separation of variables, we have the following:

The solution is of the form $u(r , \theta)= R(r) \Theta(\theta)$, which are continuous at $[0,a] \times [0 , 2 \pi]$ and periodic as for $\theta$, with period $2 \pi$.

So we get the problems:

$$(1):\Theta''+\lambda\Theta=0$$
$$\Theta(0)=\Theta(2 \pi)$$

and

$$(2):r^2 R'+rR'-\lambda R=0$$

For the problem $(1)$:

So that there is a non-trivial periodic solution it should be $\lambda \geq 0$.

$\lambda_0=0: \Theta_0(\theta)=1$

$\lambda=n^2: \Theta(\theta)=A_n \cos{(n \theta)}+B_n \sin{(n \theta)}$
Why should the solution of $\Theta$ be periodic?? (Wondering)

And the case $\lambda<0$, where the solution would be of the form: $\Theta=c_1e^{n \theta}+c_2e^{-n \theta}$, is rejected, because the exponential is not periodic?? (Wondering)

And also, why at the case $\lambda>0$ do we take: $\lambda=n^2$?? (Wondering)

 

[I like Serena]

Hi! ;)

Why should the solution of $\Theta$ be periodic?? (Wondering)

$\Theta$ is a function of the angle $\theta$.
Since there are $2\pi$ radians in a full circle that means that $\Theta(0)= \Theta(2\pi)$.
Moreover, any value must be identical to a value $2\pi$ radians along on the circle.
That is because it is the same point. (Nerd)

And the case $\lambda<0$, where the solution would be of the form: $\Theta=c_1e^{n \theta}+c_2e^{-n \theta}$, is rejected, because the exponential is not periodic?? (Wondering)

Correct.
It will not hold that $\Theta(\theta) = \Theta(\theta + 2\pi)$.

And also, why at the case $\lambda>0$ do we take: $\lambda=n^2$?? (Wondering)

Perhaps you can solve it with just $\lambda>0$?
What does your solution look like? (Wondering)

 

[mathmari]

$\Theta$ is a function of the angle $\theta$.
Since there are $2\pi$ radians in a full circle that means that $\Theta(0)= \Theta(2\pi)$.
Moreover, any value must be identical to a value $2\pi$ radians along on the circle.
That is because it is the same point. (Nerd)

Ok! I understand! (Nod)

Correct.
It will not hold that $\Theta(\theta) = \Theta(\theta + 2\pi)$.

Besides noticing that the exponential is not periodic, if we would solve this problem taking cases for $\lambda$ would the case $\lambda<0$ be rejected when applying the boundary conditions?

I mean the following:

The characteristic equation is $m^2+\lambda =0 \Rightarrow m^2=-\lambda$

$\lambda<0: \lambda=-k, k>0$
$m^2=k \Rightarrow m= \pm \sqrt{k}$
$\Theta(\theta)=c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta}$
$\Theta(0)=c_1+c_2 $
$\Theta(2 \pi)=c_1e^{\sqrt{k} 2 \pi}+c_2 e^{-2 \sqrt{k} \pi }$
$\Theta(0)=\Theta(2 \pi) \Rightarrow c_1+c_2=c_1e^{\sqrt{k} 2 \pi}+c_2 e^{-2 \sqrt{k} \pi } \Rightarrow e^{\sqrt{k} 2 \pi}c_1=c_2$
Is this correct?? How could I continue to find out that we cannot have a non-trivial solution at this case?? Or can we not do this in that way?? (Wondering)

Perhaps you can solve it with just $\lambda>0$?
What does your solution look like? (Wondering)

$\lambda>0:$
$\Theta(\theta)=c_1 \cos{(\sqrt{\lambda} \theta)}+c_2 \sin{(\sqrt{\lambda} \theta)}$
$\Theta(0)=c_1 $
$\Theta(2 \pi)=c_1 \cos{(\sqrt{\lambda} 2 \pi)}+c_2 \sin{(\sqrt{\lambda} 2 \pi)}$
$\Theta(0)=\Theta(2 \pi) \Rightarrow c_1=c_1 \cos{(\sqrt{\lambda} 2 \pi)}+c_2 \sin{(\sqrt{\lambda} 2 \pi)} \Rightarrow c_1(1-\cos{(\sqrt{\lambda} 2 \pi)})=c_2 \sin{(\sqrt{\lambda} 2 \pi)}$

How could I continue?? (Wondering)

 

[I like Serena]

Besides noticing that the exponential is not periodic, if we would solve this problem taking cases for $\lambda$ would the case $\lambda<0$ be rejected when applying the boundary conditions?

I mean the following:

The characteristic equation is $m^2+\lambda =0 \Rightarrow m^2=-\lambda$

$\lambda<0: \lambda=-k, k>0$
$m^2=k \Rightarrow m= \pm \sqrt{k}$
$\Theta(\theta)=c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta}$
$\Theta(0)=c_1+c_2 $
$\Theta(2 \pi)=c_1e^{\sqrt{k} 2 \pi}+c_2 e^{-2 \sqrt{k} \pi }$
$\Theta(0)=\Theta(2 \pi) \Rightarrow c_1+c_2=c_1e^{\sqrt{k} 2 \pi}+c_2 e^{-2 \sqrt{k} \pi } \Rightarrow e^{\sqrt{k} 2 \pi}c_1=c_2$
Is this correct?? How could I continue to find out that we cannot have a non-trivial solution at this case?? Or can we not do this in that way?? (Wondering)

The boundary condition is a bit stronger.
We need that:
$$\Theta(\theta) = \Theta(\theta + 2\pi n)$$
for any $\theta$ and for any integer $n$.

What would happen if we substitute that? (Wondering)

$\lambda>0:$
$\Theta(\theta)=c_1 \cos{(\sqrt{\lambda} \theta)}+c_2 \sin{(\sqrt{\lambda} \theta)}$
$\Theta(0)=c_1 $
$\Theta(2 \pi)=c_1 \cos{(\sqrt{\lambda} 2 \pi)}+c_2 \sin{(\sqrt{\lambda} 2 \pi)}$
$\Theta(0)=\Theta(2 \pi) \Rightarrow c_1=c_1 \cos{(\sqrt{\lambda} 2 \pi)}+c_2 \sin{(\sqrt{\lambda} 2 \pi)} \Rightarrow c_1(1-\cos{(\sqrt{\lambda} 2 \pi)})=c_2 \sin{(\sqrt{\lambda} 2 \pi)}$

How could I continue?? (Wondering)

Same thing. (Sweating)

 

[mathmari]

The boundary condition is a bit stronger.
We need that:
$$\Theta(\theta) = \Theta(\theta + 2\pi n)$$
for any $\theta$ and for any integer $n$.

What would happen if we substitute that? (Wondering)

$\lambda<0: \lambda=-k, k>0$
$m^2=k \Rightarrow m= \pm \sqrt{k}$
$\Theta(\theta)=c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta}$
$\Theta(\theta +2 \pi n)=c_1e^{\sqrt{k} (\theta +2 \pi n)}+c_2 e^{-\sqrt{k}( \theta +2 \pi n)}$

$\Theta(\theta) = \Theta(\theta + 2\pi n) \Rightarrow c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta}=c_1e^{\sqrt{k} (\theta +2 \pi n)}+c_2 e^{-\sqrt{k}( \theta +2 \pi n)} \Rightarrow c_1(e^{\sqrt{k} \theta}-e^{\sqrt{k}( \theta +2 \pi n)})=c_2(e^{-\sqrt{k} (\theta+ 2 \pi n)} -e^{-\sqrt{k} \theta}) $

How could I continue?? (Wondering) I got stuck right now.. (Worried)

 

[I like Serena]

$\lambda<0: \lambda=-k, k>0$
$m^2=k \Rightarrow m= \pm \sqrt{k}$
$\Theta(\theta)=c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta}$
$\Theta(\theta +2 \pi n)=c_1e^{\sqrt{k} (\theta +2 \pi n)}+c_2 e^{-\sqrt{k}( \theta +2 \pi n)}$

$\Theta(\theta) = \Theta(\theta + 2\pi n) \Rightarrow c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta}=c_1e^{\sqrt{k} (\theta +2 \pi n)}+c_2 e^{-\sqrt{k}( \theta +2 \pi n)} \Rightarrow c_1(e^{\sqrt{k} \theta}-e^{\sqrt{k}( \theta +2 \pi n)})=c_2(e^{-\sqrt{k} (\theta+ 2 \pi n)} -e^{-\sqrt{k} \theta}) $

How could I continue?? (Wondering) I got stuck right now.. (Worried)

It is not possible. (Worried)

We have:
$$c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta} = c_1e^{\sqrt{k} (\theta +2 \pi n)}+c_2 e^{-\sqrt{k}( \theta +2 \pi n)}$$
$$c_1e^{\sqrt{k} (\theta +2 \pi n)} - c_1e^{\sqrt{k} \theta} = c_2 e^{-\sqrt{k} \theta} - c_2 e^{-\sqrt{k}( \theta +2 \pi n)}$$
Suppose $c_1 > 0$ and $n > 0$, then both sides are positive.
And for large enough $\theta$ the right hand side becomes as small as we want it to be.
Let's say the right hand side is smaller than $\epsilon > 0$.

Then we have:
$$|c_1 e^{\sqrt k (\theta + 2\pi n)} - c_1 e^{\sqrt k \theta}| < \epsilon$$
$$c_1 e^{\sqrt k \theta} | e^{\sqrt k \cdot 2\pi n} - 1| < \epsilon$$
But the left hand side can become as big as we want it to be, since we can pick $n$ as big as we want.
This is a contradiction.

So there is no solution for $\lambda<0$. (Wasntme)

 

[mathmari]

It is not possible. (Worried)

We have:
$$c_1e^{\sqrt{k} \theta}+c_2 e^{-\sqrt{k} \theta} = c_1e^{\sqrt{k} (\theta +2 \pi n)}+c_2 e^{-\sqrt{k}( \theta +2 \pi n)}$$
$$c_1e^{\sqrt{k} (\theta +2 \pi n)} - c_1e^{\sqrt{k} \theta} = c_2 e^{-\sqrt{k} \theta} - c_2 e^{-\sqrt{k}( \theta +2 \pi n)}$$
Suppose $c_1 > 0$ and $n > 0$, then both sides are positive.
And for large enough $\theta$ the right hand side becomes as small as we want it to be.
Let's say the right hand side is smaller than $\epsilon > 0$.

Then we have:
$$|c_1 e^{\sqrt k (\theta + 2\pi n)} - c_1 e^{\sqrt k \theta}| < \epsilon$$
$$c_1 e^{\sqrt k \theta} | e^{\sqrt k \cdot 2\pi n} - 1| < \epsilon$$
But the left hand side can become as big as we want it to be, since we can pick $n$ as big as we want.
This is a contradiction.

So there is no solution for $\lambda<0$. (Wasntme)

Ahaa..Ok! (Thinking)(Sweating)Is this the same at the case $\lambda>0$?

$\Theta(\theta)=c_1 \cos{(\sqrt{\lambda} \theta)}+c_2 \sin{(\sqrt{\lambda} \theta)}$
$\Theta(\theta+2 \pi n)=c_1 \cos{(\sqrt{\lambda} (\theta+2 \pi n))}+c_2 \sin{(\sqrt{\lambda} (\theta+2 \pi n))}$

$\Theta(\theta)=\Theta(\theta+2 \pi n) \Rightarrow c_1 \cos{(\sqrt{\lambda} \theta)}+c_2 \sin{(\sqrt{\lambda} \theta)}=c_1 \cos{(\sqrt{\lambda} (\theta+2 \pi n))}+c_2 \sin{(\sqrt{\lambda} (\theta+2 \pi n))} \Rightarrow c_1(\cos{(\sqrt{\lambda} \theta)}-\cos{(\sqrt{\lambda} (\theta+2 \pi n))})=c_2(\sin{(\sqrt{\lambda} (\theta+2 \pi n))}- \sin{(\sqrt{\lambda} \theta)}) $

Could I continue from here? (Wondering)
Or is there an other way to check this case?
How can I conclude that $\lambda = n^2$ ? (Wondering)

 

[I like Serena]

Ahaa..Ok! (Thinking)(Sweating)Is this the same at the case $\lambda>0$?

$\Theta(\theta)=c_1 \cos{(\sqrt{\lambda} \theta)}+c_2 \sin{(\sqrt{\lambda} \theta)}$
$\Theta(\theta+2 \pi n)=c_1 \cos{(\sqrt{\lambda} (\theta+2 \pi n))}+c_2 \sin{(\sqrt{\lambda} (\theta+2 \pi n))}$

$\Theta(\theta)=\Theta(\theta+2 \pi n) \Rightarrow c_1 \cos{(\sqrt{\lambda} \theta)}+c_2 \sin{(\sqrt{\lambda} \theta)}=c_1 \cos{(\sqrt{\lambda} (\theta+2 \pi n))}+c_2 \sin{(\sqrt{\lambda} (\theta+2 \pi n))} \Rightarrow c_1(\cos{(\sqrt{\lambda} \theta)}-\cos{(\sqrt{\lambda} (\theta+2 \pi n))})=c_2(\sin{(\sqrt{\lambda} (\theta+2 \pi n))}- \sin{(\sqrt{\lambda} \theta)}) $

Could I continue from here? (Wondering)
Or is there an other way to check this case?
How can I conclude that $\lambda = n^2$ ? (Wondering)

This has to be true for any $\theta$.
That can only happen if the terms cancel.

That is, if we assume that $c_1 \ne 0$, we get that:
$$\cos(\sqrt{\lambda} \theta)-\cos(\sqrt{\lambda} (\theta+2 \pi n)) = 0$$
This can only be true if $\sqrt \lambda$ is an integer. (Thinking)

 

[mathmari]

This has to be true for any $\theta$.
That can only happen if the terms cancel.

That is, if we assume that $c_1 \ne 0$, we get that:
$$\cos(\sqrt{\lambda} \theta)-\cos(\sqrt{\lambda} (\theta+2 \pi n)) = 0$$
This can only be true if $\sqrt \lambda$ is an integer. (Thinking)

I haven't understood why that happens only if the terms cancel.. (Doh)

 

[I like Serena]

I haven't understood why that happens only if the terms cancel.. (Doh)

Can you simplify:
$$c_1(\cos{(\sqrt{\lambda} \theta)}-\cos{(\sqrt{\lambda} (\theta+2 \pi n))})=c_2(\sin{(\sqrt{\lambda} (\theta+2 \pi n))}- \sin{(\sqrt{\lambda} \theta)})$$

I suggest to use the sum-to-product identities:
$$\cos p - \cos q = -2\sin\frac 1 2(p+q) \sin\frac 1 2(p-q)$$
$$\sin p - \sin q = 2\cos\frac 1 2(p+q) \sin\frac 1 2(p-q)$$
(Thinking)

 

[mathmari]

Can you simplify:
$$c_1(\cos{(\sqrt{\lambda} \theta)}-\cos{(\sqrt{\lambda} (\theta+2 \pi n))})=c_2(\sin{(\sqrt{\lambda} (\theta+2 \pi n))}- \sin{(\sqrt{\lambda} \theta)})$$

I suggest to use the sum-to-product identities:
$$\cos p - \cos q = -2\sin\frac 1 2(p+q) \sin\frac 1 2(p-q)$$
$$\sin p - \sin q = 2\cos\frac 1 2(p+q) \sin\frac 1 2(p-q)$$
(Thinking)

$$c_1(\cos{(\sqrt{\lambda} \theta)}-\cos{(\sqrt{\lambda} (\theta+2 \pi n))})=c_2(\sin{(\sqrt{\lambda} (\theta+2 \pi n))}- \sin{(\sqrt{\lambda} \theta)}) \Rightarrow \\

-2 c_1 \sin{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} =2 c_2 \cos{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} \Rightarrow \\

-c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))} = c_2 \cos{(\sqrt{\lambda} (\theta+ \pi n))} $$

Is this correct so far??

How can I continue?? (Wondering)

 

[I like Serena]

$$c_1(\cos{(\sqrt{\lambda} \theta)}-\cos{(\sqrt{\lambda} (\theta+2 \pi n))})=c_2(\sin{(\sqrt{\lambda} (\theta+2 \pi n))}- \sin{(\sqrt{\lambda} \theta)}) \Rightarrow \\

-2 c_1 \sin{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} =2 c_2 \cos{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} \Rightarrow \\

-c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))} = c_2 \cos{(\sqrt{\lambda} (\theta+ \pi n))} $$

Is this correct so far??

You seem to have dropped a $-$ sign.
Note that:
$$\sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))}
= \sin{(- \sqrt{\lambda} \pi n)}
= - \sin{(\sqrt{\lambda} \pi n)}$$

How can I continue?? (Wondering)

This should be true for any $\theta$ and $n$.
Suppose we pick $\theta = 0$ and $n=0$... (Thinking)

 

[mathmari]

You seem to have dropped a $-$ sign.
Note that:
$$\sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))}
= \sin{(- \sqrt{\lambda} \pi n)}
= - \sin{(\sqrt{\lambda} \pi n)}$$
This should be true for any $\theta$ and $n$.
Suppose we pick $\theta = 0$ and $n=0$... (Thinking)

$$c_1(\cos{(\sqrt{\lambda} \theta)}-\cos{(\sqrt{\lambda} (\theta+2 \pi n))})=c_2(\sin{(\sqrt{\lambda} (\theta+2 \pi n))}- \sin{(\sqrt{\lambda} \theta)}) \Rightarrow \\

-2 c_1 \sin{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} =-2 c_2 \cos{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} \Rightarrow \\

c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))} = c_2 \cos{(\sqrt{\lambda} (\theta+ \pi n))} \Rightarrow \\

c_1 \sin{( \sqrt{\lambda} \theta+ \sqrt{\lambda}\pi n)} = c_2 \cos{(\sqrt{\lambda} \theta+ \sqrt{\lambda}\pi n)} \Rightarrow \\

c_1(\sin{(\sqrt{\lambda} \theta)} \cos{(\sqrt{\lambda} \pi n)}+\cos{(\sqrt{\lambda} \theta)} \sin{(\sqrt{\lambda} \pi n)})=c_2(\cos{(\sqrt{\lambda} \theta)} \cos{(\sqrt{\lambda} \pi n)}-\sin{(\sqrt{\lambda} \theta)} \sin{(\sqrt{\lambda} \pi n)})

$$

Is it better now? Can I get from the last relation something usefull? Or didn't I have to write in that way? (Wondering)

 

[I like Serena]

Yep! All better! (Nod)

Let's stick with:
$$c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))} = c_2 \cos{(\sqrt{\lambda} (\theta+ \pi n))}$$
What do you get if you substitute $\theta=0$ and simultaneously $n=0$? (Thinking)

 

[mathmari]

Yep! All better! (Nod)

Let's stick with:
$$c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))} = c_2 \cos{(\sqrt{\lambda} (\theta+ \pi n))}$$
What do you get if you substitute $\theta=0$ and simultaneously $n=0$? (Thinking)

For $\theta=0$ and $n=0$:
$$0= c_2 $$

So do we take $c_2=0$ and replace it at $c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))} = c_2 \cos{(\sqrt{\lambda} (\theta+ \pi n))}$?

Then we get:
$$c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))}=0 \Rightarrow \\ \sin{( \sqrt{\lambda} (\theta+ \pi n))}=0 \Rightarrow \\ \sqrt{\lambda} (\theta+ \pi n)=n \pi \Rightarrow \\ \sqrt{\lambda}=\frac{n \pi}{\theta+n \pi} \Rightarrow \\ \lambda=(\frac{n \pi}{\theta+n \pi})^2 $$Is $\displaystyle{\frac{n \pi}{\theta+n \pi}}$ an integer?? (Thinking) It isn't, right?

 

[I like Serena]

$$\lambda=(\frac{n \pi}{\theta+n \pi})^2 $$

Is $\displaystyle{\frac{n \pi}{\theta+n \pi}}$ an integer?? (Thinking) It isn't, right?

Hold on!
$\lambda$ is a constant, while $(\frac{n \pi}{\theta+n \pi})^2$ is not a constant.
This is supposed to be true for any $\theta$ and $n$.
I don't think it is, is it?Let's back up a step.

$$-2 c_1 \sin{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} =-2 c_2 \cos{(\frac{1}{2} (\sqrt{\lambda} \theta+\sqrt{\lambda} (\theta+2 \pi n)))} \sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} \Rightarrow \\

c_1 \sin{( \sqrt{\lambda} (\theta+ \pi n))} = c_2 \cos{(\sqrt{\lambda} (\theta+ \pi n))}
$$

Actually, didn't you drop a possible solution here?

What if $\sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} = 0$? (Thinking)

See my avatar for what could be going wrong. (Evilgrin)

 

[mathmari]

Actually, didn't you drop a possible solution here?

What if $\sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} = 0$? (Thinking)

See my avatar for what could be going wrong. (Evilgrin)

(Blush)

$\sin{(\frac{1}{2} (\sqrt{\lambda} \theta-\sqrt{\lambda} (\theta+2 \pi n)))} = 0 \Rightarrow \sin{(\sqrt{\lambda} \pi n)}=0 \Rightarrow \sqrt{\lambda} \pi n= k \pi \Rightarrow \sqrt{\lambda}=\frac{k}{n}$

Is this correct? (Wondering)

But we do not know if $\displaystyle{\frac{k}{n}}$ is an integer, do we? (Wondering)