I have the equation $$\bigtriangleup u=\frac{1}{r} \frac{\partial}{\partial r}(r\frac{\partial u}{\partial r}) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta}=0$$ where $0<r<1$ , $-\pi< \theta \leq\pi$ and $$u(1,\theta)=g(\theta)$$ This is my solution. To solve this I assume the solution is of the form $$u(r,\theta)=R(r)\Theta(\theta)$$ I then work out $u_{r}$,$u_{rr}$, $u_\theta$ and $u_{\theta \theta}$ and I plug these into $\bigtriangleup u$ to get $$\frac{1}{r} R' \Theta +\frac{1}{r^2}R\Theta''+R''\Theta=0$$
However my lecture notes get $$\frac{-r}{R}(rR')'=\frac{\Theta''}{\Theta}=k$$ How do you get this result?
My lecture notes then look at 3 cases. Case 1: K=0. The lecture notes then say from this we get $$\Theta=a\theta+b$$ and $$R=c_1 ln r+c_2$$ I now understand how to get these expressions. But the notes then say: With boundary/consistency conditions, $\Theta=b$, $=c_2$ $\rightarrow$$u(r,\theta)=1$ What does this mean?
We then look at the case for when $k=-p^2<0$ and $k=p^2>0$.
Looking at $k=p^2>0$ we then I get $$T'=-p^2T$$ and $$x''=-p^2x$$Where do I go from here?
You have
$$\begin{align} \frac{1}{r} R' \Theta + \frac{1}{r^{2}} R \Theta '' + R'' \Theta &= 0 \\ \implies r R' \Theta + R \Theta '' + r^{2} R'' \Theta &= 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(multiplying through by $r^{2}$)}\\ \implies r R' \Theta + r^{2} R'' \Theta &= -R \Theta '' \\ \implies r R' + r^{2} R'' &= -R \frac{\Theta ''}{\Theta} \ \ \ \ \ \ \ \text{(dividing both sides by $\Theta$)}\\ \implies r (R' + r R'') &= -R \frac{\Theta ''}{\Theta} \\ \implies r (rR')' &= -R \frac{\Theta ''}{\Theta} \ \ \ \ \ \ \ \ \text{as $(rR')' = R' + r R''$} \\ \implies \frac{-r}{R} (rR')' &= \frac{\Theta ''}{\Theta} \ \ (*) \\ \end{align}$$
Now at $(*)$, we have functions of different variables, but the functions are equal to each other. The only way this can occur is if the LHS and RHS are equal to a constant, the same constant, which your lecturer has called $k$, which is also known as the separation constant.
So we should rewrite $(*)$ as
$$\frac{-r}{R} (rR')' = \frac{\Theta ''}{\Theta} = k \ \ (**)$$
(Note: Some books use $k^{2}$ for the separation constant, some use $-\lambda$, the main thing to remember is that it is just a constant so it doesn't matter what you call it).
Now to your expressions. From $(**)$ and what I explained earlier, we know that
$$\frac{-r}{R} (rR')' = k \ \ (1)$$
and
$$\frac{\Theta ''}{\Theta} = k \ \ (2)$$
If $k = 0$ then $(1)$ implies
$$\begin{align} \frac{-r}{R} (r R')' &= 0 \\ \implies (r R')' &= 0 \\ \implies r R' &= C \\ \implies R' &= \frac{C}{r} \\ \implies R &= C \ln r + D \\ \end{align}$$
where $C$ and $D$ are constants. If $k = 0$ then $(2)$ implies
$$\begin{align} \frac{\Theta ''}{\Theta} &= 0 \\ \implies \Theta '' &= 0 \\ \implies \Theta ' &= A \\ \implies \Theta &= A \theta + B \\ \end{align}$$
where $A$ and $B$ are constants.