MHB Solve Laplace equation on unit disk

[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)