Find the steady-state temperature distribution in a rectangular plate covering the area $\mathbf{0}<\mathbf{x}<\mathbf{1}$, $\mathbf{0}<\mathbf{y}<\mathbf{2}$, if $\mathit{T}\mathbf{=}\mathbf{0}$for $\mathit{x}\mathbf{=}\mathbf{0}$, $\mathit{x}\mathbf{=}\mathbf{1}$, $\mathit{y}\mathbf{=}\mathbf{2}$, and $\mathit{T}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{x}$for$\mathit{y}\mathbf{=}\mathbf{0}$.

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

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$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Assume that the plate is so long compared to its width that the mathematical approximation can be made that it extends to infinity in the y direction, the temperature T satisfies Laplace's equation inside the plate that is${\nabla }^{2}T=0$.

To solve this equation, try a solution of the form mentioned below.

$\begin{array}{l}T(x,y)=X\left(x\right)Y\left(y\right)\\ Y\frac{d^{2}X}{d{x}^2}+X\frac{d^{2}Y}{d{y}^2}=0\end{array}$

Divide the above equation by XY.

$\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}=0$

Now, following the process of separation of variables it can be written

$\begin{array}{c}\frac{1}{X}\frac{d^{2}X}{d{x}^2}=-\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}\\ =\text{constant}\\ =-k^2,k\ge 0\end{array}$

$X^{\text{'}\text{'}}=-k^{2}X$and $Y^{\text{'}\text{'}}=k^{2}Y$

Here, the constant $k^2$is called the separation constant. The solutions for these equations.

$\begin{array}{l}X=\left\{\begin{array}{l}\sin \left(kx\right)\\ \cos \left(kx\right)\end{array}\right.\\ Y=\left\{\begin{array}{l}{e}^{ky}\\ e^{-ky}\end{array}\right.\end{array}$

Since none of the four solutions satisfies the given boundary temperatures, take a combination of the previous solutions.

Write the steady state temperature distribution.

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

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$x=1$.

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

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 for $y=2T=0$ then $c_1$and $c_2$are needed.

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

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

$c_2=-\frac{1}{2}{e}^{-2k}\cdot T=\sinh n\pi (2-y)\sin n\pi x$.

The above equation 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.

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

Find the expression for$B_n$.

$\begin{array}{c}T(x,0)=1-x=\sum _{n=1}^{\infty }{B}_n\sinh \left(2n\pi \right)\sin \left(n\pi x\right)\\ =\sum _{n=1}^{\infty }{b}_n\sin \left(n\pi x\right)(b_n=B_n\sinh 2n\pi )\end{array}$

Find the value of$B_n$.

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

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 $T(x,y)$.

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

Hence, the steady-state temperature distribution is obtained as below.

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