Solve Problem 1 if $\mathbf{T}=\mathbf{0}$for $\mathit{x}\mathbf{=}\mathbf{0}$, $\mathit{x}\mathbf{=}\mathbf{1}$, $\mathit{y}\mathbf{=}\mathbf{0}$, and $\mathit{T}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{x}$for $\mathit{y}\mathbf{=}\mathbf{2}$. Hint: Use $\mathbf{sinh}\mathbf{ky}$as the y solution; then $\mathbf{T}=\mathbf{0}$when $\mathit{y}\mathbf{=}\mathbf{0}$as required.

It has been given that the rectangular plate is covering the area $T=0$for $x=0$, $x=1$, $y=0$, and $T=1-x$for$y=2$.

Laplace’s equation in cylindrical coordinates is

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

The steady state temperature distribution.

$T(x,y)=(c_1{e}^{-ky}+c_2{e}^{ky})(c_3\cos kx+c_4\sin kx)$

First set the value of the constants $c_1,c_2,c_3$and $c_4$. So, if $T=0$for $x=0$.

$\begin{array}{c}T(0,y)=(c_1{e}^{-ky}+c_2{e}^{ky})\left(c_3\right)\\ =0\end{array}$

Since an exponential can't be zero, $c_3=0.$If $T=0$for$x=1$.

$\begin{array}{c}T(1,y)=(c_1{e}^{-ky}+c_2{e}^{ky})\left(c_4\sin k\right)\\ =0\end{array}$

It is known that $\sin kx=0$if $kx=n\pi ,$where $n=0,1,2..,$therefore for $x=1$there is $k=n\pi .$Finally, there is that for $y=2T=0$then need $c_1$and$c_2$.

$\begin{array}{c}(c_1{e}^{-ky}+c_2{e}^{ky})=\frac{1}{2}(e^{ky}-e^{-ky})\\ =\sinh \left(ky\right)\end{array}$

Thus$c_1=\frac{1}{2}{e}^k$and$c_2=-\frac{1}{2}{e}^{-k}.$Thus for any integral n, the solution is mentioned below.

$T=\sinh \left(n\pi y\right)\sin \left(n\pi x\right)$

It satisfies the given boundary conditions on the three $T=0$sides.

The $y=0$condition is not satisfied by any value of n. But a linear combination of solutions is a solution. Thus, write an infinite series of T.

$T(x,y)=\sum _{n=1}^{\infty }{B}_n\sinh n\pi (2-y)\sin \left(n\pi x\right)$

Find the expression for$B_n$.

$\begin{array}{c}T(x,2)=1-x\\ =\sum _{n=1}^{\infty }{B}_n\sinh \left(2n\pi \right)\sin \left(n\pi x\right)\\ =\sum _{n=1}^{\infty }{b}_n\sin \left(n\pi x\right)(b_n=B_n\sinh \left(2n\pi \right))\end{array}$

Find the value of$B_n$.

$\begin{array}{c}{b}_n=2{\int }_0^1(1-x)\sin \left(n\pi x\right)dx\\ =2{[(1-x)\left(\frac{-\cos \left(n\pi x\right)}{n\pi }\right)-\frac{\sin \left(n\pi x\right)}{n^2{\pi }^2}]}_0^1\\ =2[(1-1)\left(\frac{-\cos n\pi }{n\pi }\right)-\frac{\sin \left(n\pi \right)}{n^2{\pi }^2}-(1-0)\left(\frac{-\cos 0}{n\pi }\right)+\frac{\sin \left(0\right)}{n^2{\pi }^2}]\\ =2[0-0-\left(\frac{-1}{n\pi }\right)+0]\end{array}$

$b_n=\frac{2}{n\pi }$

Hence $B_n$is derived as mentioned below.

$\begin{array}{c}{B}_n=\frac{b_n}{\sinh \left(2n\pi \right)}\\ =\frac{2}{n\pi \sinh \left(2n\pi \right)}\end{array}$

Finally, substitute $B_n$into the distribution$T(x,y)$.

$T(x,y)=\sum _{n=1}^{\infty }\frac{2}{n\pi \sinh \left(2n\pi \right)}\sinh \left(n\pi y\right)\sin \left(n\pi x\right)$

Hence, this is the steady-state temperature distribution.