Solving series with complex exponentials.

[yungman]

$$[Acos(θ)+Bsin(θ)]$$ is less than one ?
the entire term being less than 1 is what matters

There is no saying that $A_n$ and $B_n$ are smaller than one. Besides, your assumption will limit the solution to be less than 1, that is not right.

 

[Ray Vickson]

There is no saying that $A_n$ and $B_n$ are smaller than one. Besides, your assumption will limit the solution to be less than 1, that is not right.

You seem to think that a solution of Laplace's equation in the open disk $r < a$ must necessarily be continuous on the closed disk $r \leq a$. There is no reason for anything like that to be true; we often face problems of solving a PDE in the interior of a set $S$, but with a boundary condition on the boundary $\partial S$ that is not everywhere continuous.

Take the current example: the book gives you a function
$$U(r,\theta-\phi) = \frac{a^2 - r^2}{a^2 + r^2 - 2 a r \cos(\theta - \phi)}.$$
Below, let $t = \theta - \phi$ for ease of writing.

It is easy to verify directly that $U$ solves Laplace's equation in the open region $r < a$. Furthermore, for every point $(a,t_0)$ with $t_0 \neq 0$ the function is continuous at $(a,t_0)$; that is,
$$\lim_{(r,t) \to (a,t_0)} = 0 = U(a,t_0).$$
However, $U$ is not continuous at $(a,0)$ (or at $(a, \pm 2 \pi n), n = 0, 1, 2, \ldots$). For example, for $r = a$ we have
$$U(a,t) = 0\; \forall\: t \neq 0,$$
so $\lim_{t \to 0} U(a,t) = 0.$ However, for $r \neq a$ we have
$$U(r,0) = \frac{a^2 - r^2}{a^2+r^2 - 2 a r} = \frac{a^2 - r^2}{(a-r)^2} = <br /> \frac{a+r}{a-r} \to +\infty\text{ as } r \to a-.$$
So, the function given to you in the book does satisfy Laplace's equation in the open disc and is continuous except at a single point on the boundary of the disc.

Exactly the same thing happens in the infinite-series representation of $U$. The fact that the series fails to converge at a single point with $r = a$ is not problem; the function $U$ is just not continuous there, and that has absolutely nothing to do with the infinite series (or, rather, is reflected properly in the infinite series, as it should be).

 

[yungman]

Yes, I agree that there is no guaranty the solution is continuous. BUT if you look at the original equation:
$$u=A_0+\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n[A_n\cos(n\theta)+B_n\sin(n\theta)]$$

It is continuous at $r=a$.

But if convert the above formula into this form:

$$\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}$$

You have undefined solution whenever $x=1$. That is not right!

You convert a continuous solution into one that has undefined solution.

 

[Ray Vickson]

Yes, I agree that there is no guaranty the solution is continuous. BUT if you look at the original equation:
$$u=A_0+\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n[A_n\cos(n\theta)+B_n\sin(n\theta)]$$

It is continuous at $r=a$.

But if convert the above formula into this form:

$$\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}$$

You have undefined solution whenever $x=1$. That is not right!

You convert a continuous solution into one that has undefined solution.

Stop blaming ME; all I did was expand on what the BOOK says. If you have a problem, take it up with the authors of the book!

Anyway, you cannot just say that your original Fourier series is continuous at r = a; you have to PROVE it. You will have trouble doing this, because such arguments are a lot harder than you might think. Anyway, I suspect you are wrong; I bet the function is NOT continuous on the whole circle r = a, and just saying so or wishing it to be true will not work.

 

[yungman]

Stop blaming ME; all I did was expand on what the BOOK says. If you have a problem, take it up with the authors of the book!

Anyway, you cannot just say that your original Fourier series is continuous at r = a; you have to PROVE it. You will have trouble doing this, because such arguments are a lot harder than you might think. Anyway, I suspect you are wrong; I bet the function is NOT continuous on the whole circle r = a, and just saying so or wishing it to be true will not work.

I am not blaming you or disrespect you at all.

All I am saying is from the original formula where the exponential formula supposing derived from is continuous unless $A_n$ or $B_n$ is infinite.

$$u=A_0+\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n[A_n\cos(n\theta)+B_n\sin(n\theta)]$$

But the derived formula goes to infinite at $r=a$ no matter what.

This is really a question than disrespecting you. I am an engineer that trying to learn math, I am not in the same league to disrespect you or anyone that is homework helper here! I am desperately trying to derive and making sense of this.

Far as I know, using $\sum_0^{\infty} x^n=\frac{1}{1-x}$ require $x\neq 1$ no matter what. That's the limitation that has to be imposed on this series expansion.