I have this exercise: Find the bounded steady temperatures $T (x, y)$ in the semi-infinite solid $y \geq 0$ if $T = 0$ on the part $x < −1$ $(y = 0)$ of the boundary, if $T = 1$ on the part $x > 1$ $(y = 0)$, and if the strip $−1 < x < 1$ $(y = 0)$ of the boundary is insulated.
then $$T(x,0)=0, \;x<-1; \\ \frac{\partial T(x,0)}{\partial y}=0, \; -1<x<1; \\ T(x,0)=1, \; x>1.$$
How can I solve this problem, I have to demonstrate all that I say, then I can't do nothing by inspection. Thank you for your help.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ It's convenient to use polar coordinates $\ds{\pars{r,\theta}}$: $\quad\ds{x \equiv r\cos\pars{\theta}\quad\mbox{and}\quad y = r\sin\pars{\theta}}$. By symmetry reasons we'll consider $\ds{\varphi\pars{r,\theta} = {\rm T}\pars{x,y} - \half}$ and confine our attention to the upper half $\ds{xy}$-plane: \begin{align} &\nabla^{2}\varphi\pars{r,\theta}=0\,,\qquad \left\lbrace\begin{array}{rclrcl} \left.\partiald{\varphi\pars{r,0}}{\theta}\right\vert_{\theta\ =\ 0} & = & 0\,,\quad & r & < & 1 \\[1mm] \varphi\pars{r,0} & = & \half\,,\quad & r & > & 1 \end{array}\right.\tag{1} \\[3mm]&\mbox{with}\quad \varphi\pars{r,\theta} = -\varphi\pars{r,\pi - \theta} \tag{2} \end{align} $\pars{2}$ reduces the problem to the first quadrant $\ds{\pars{~0 \leq \theta \leq {\pi \over 2}~}}$.
$\pars{2}$ reduces this expression to: \begin{align} \left.\varphi\pars{r,\theta}\right\vert_{r\ <\ 1}&= a\pars{\theta - {\pi \over 2}} + \sum_{n = 1}^{\infty}\bracks{ A_{2n}\sin\pars{2n\theta}r^{2n} +B_{2n + 1}\cos\pars{\bracks{2n + 1}\theta}r^{2n + 1}} \\[3mm] \left.\varphi\pars{r,\theta}\right\vert_{r\ >\ 1}&= c\pars{\theta - {\pi \over 2}} + \sum_{n = 1}^{\infty}\bracks{% C_{2n}\sin\pars{2n\theta}r^{-2n} +D_{2n + 1}\cos\pars{\bracks{2n + 1}\theta}r^{-2n - 1}} \end{align}
The condition $\ds{\lim_{r \to 1^{-}}\varphi\pars{r,\theta} =\lim_{r \to 1^{+}}\varphi\pars{r,\theta}}$ leads to: $$\color{#66f}{\large% \sum_{n = 1}^{\infty}B_{2n + 1}\cos\pars{\bracks{2n + 1}\theta} =\half - {\theta \over \pi} + \sum_{n = 1}^{\infty}C_{2n}\sin\pars{2n\theta}} \tag{3} $$ With $$ \left\lbrace\begin{array}{rcl} \int_{0}^{\pi/2}\cos\pars{\bracks{2m + 1}\theta}\cos\pars{\bracks{2n + 1}\theta} &=&\pi\,\delta_{mn} \\[3mm] \int_{0}^{\pi/2}\cos\pars{\bracks{2m + 1}\theta}\pars{\half - {\theta \over \pi}} &=&{1 \over \pars{2m + 1}^{2}\pi} \\[3mm] \int_{0}^{\pi/2}\cos\pars{\bracks{2m + 1}\theta}\sin\pars{2n + 1\theta} &=&-\,{2n \over 4n + 1}\,\delta_{mn} \end{array}\right. $$ and condition $\pars{3}$ we find $$\color{#66f}{\large% \pi B_{2n + 1} + {2n \over 4n + 1}\,C_{2n} = {1 \over \pars{2n + 1}^{2}\pi}} \tag{4} $$ Similarly, $$ \left\lbrace\begin{array}{rcl} \int_{0}^{\pi/2}\sin\pars{2m\theta}\cos\pars{\bracks{2n + 1}\theta} &=&-\,{2n \over 4n + 1}\,\delta_{mn} \\[3mm] \int_{0}^{\pi/2}\sin\pars{2m\theta}\pars{\half - {\theta \over \pi}} &=&{1 \over 4m} \\[3mm] \int_{0}^{\pi/2}\sin\pars{2m\theta}\sin\pars{2n\theta} &=&{\pi \over 4}\,\delta_{mn} \end{array}\right. $$ these expressions and condition $\pars{3}$ leads to $$ \color{#66f}{\large -\,{2n \over 4n + 1}\,B_{2n + 1} - {\pi \over 4}\,C_{2n}= {1 \over 4n}}\tag{5} $$