Find the steady-state temperature distribution in a plate with the boundary temperatures$\mathbf{T}\mathbf{=}\mathbf{30}\mathbf{°}$for $\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{y}\mathbf{=}\mathbf{3}$;$\mathbf{T}\mathbf{=}\mathbf{20}\mathbf{°}$for $\mathit{y}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{5}$. Hint: Subtract$\mathbf{20}\mathbf{°}$from all temperatures and solve the problem; then add $\mathbf{20}\mathbf{°}$. (Also see Problem 2.)

It has been given that the rectangular plate is covering the area $T=30°$for $x=0$and $y=3$; $T=20°$for $y=0$and$x=5$.

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.

Subtract $20°$so you end up with the sum of two cases. Set the first case where $T=0$but for $x=0$, and another case where $T=0$but for$y=3$.

Solution 1:

Thus, the possible solution for Y(x).

$Y\left(y\right)=\sinh \left(ky\right)$

Thus, the possible solutions for X(x).

$X\left(x\right)=\sin \left(kx\right)$

The boundary condition $T=10°$for $y=3$gives us the equation mentioned below.

$\begin{array}{c}T(x,y)=\sum _{n=1}^{\infty }{c}_n\sin (n\pi x/5)\sin (n\pi 3/5)\\ =10\end{array}$

The Fourier coefficient.

$c_n=\frac{40}{\pi }\frac{1}{\sinh (3\pi n/5)}$

Thus, the solution is as mentioned below.

$T_1(x,y)=\frac{40}{\pi }\sum _{\text{oddn}}^{\infty }\frac{1}{\sinh (3\pi n/5)}\sin (n\pi y/5)\cos (n\pi x/5)$

Solution 2:

Thus, the possible solution for Y(x).

$Y\left(y\right)=\sin \left(ky\right)$

Thus, the possible solutions for X(x).

$X\left(x\right)=\sinh k(5-x)$

The boundary condition $T=10°$for $x=0$gives us the equation mentioned below.

$T(x,y)=\sum _{n=1}^{\infty }{c}_n\sinh (n\pi 5/3)\sin (n\pi y/3)=10$

The Fourier coefficient.

$c_n=\frac{40}{\pi }\frac{1}{\sinh (5\pi n/3)}$

Thus, the solution is as mentioned below.

$T_2(x,y)=\frac{40}{\pi }\sum _{oddn}^{\infty }\frac{1}{\sinh (\pi n5/3)}\sinh (\pi n(5-x)/3)\sin (n\pi y/3)$

Combine the result and the solution becomes as mentioned below.

$\begin{array}{c}T(x,y)=20+\frac{40}{\pi }\sum _{oddn}^{\infty }\frac{1}{\sinh (3\pi n/5)}\sin (n\pi y/5)\cos (n\pi x/5)\\ +\frac{40}{\pi }\sum _{\text{oddn}}^{\infty }\frac{1}{\sinh (\pi n5/3)}\sinh (\pi n(5-x)/3)\sin (n\pi y/3)\end{array}$

Hence, this is the steady-state temperature distribution.