A slab of thickness 10 cm has its two faces at ${\mathbf{10}}^{\mathbf{°}}$and $\mathbf{20}\mathbf{°}$. At t = 0 , the face temperatures are interchanged. Find $\mathit{u}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}$for t > 0.

It has been given that a slab of thickness 10cm has two faces at ${10}^{°}$and${20}^{°}$.

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 initial steady-state temperature $u_0$satisfies Laplace's equation, which in this one-dimensional case is$\raisebox{1ex}{$d^2{u}_0$}\!\left/ \!\raisebox{-1ex}{$d{x}^2$}\right.=0$.

The solution of this equation is $u_0=ax+b,$where a and b are constants that must be found to fit the given conditions.

It has been given that $u_0=10$at x = 0 and $u_0=20$at x = l.

${\text{u}}_{\text{0}}\text{=x+10}$ ….. (1)

The flow equation is mentioned below.

${\nabla }^{2}u=\frac{1}{{\alpha }^2}\frac{\partial u}{\partial t}$

Here, u is the temperature and ${\alpha }^2$is a constant characteristic of the material through which heat is flowing. Assume a solution of the form mentioned below.

$u=F\left(x\right)T\left(t\right)$

Replacing this into the differential equation.

$\frac{1}{F}{\nabla }^{2}F=\frac{1}{{\alpha }^2}\frac{1}{T}\frac{dT}{dt}$

The left side of this identity is a function only of the space variable x, and the right side is a function only of time. Therefore, both sides are the same constant.

$\frac{dT}{dt}=-k^2{\alpha }^{2}T$

The time equation can be integrated to give the equation mentioned below.

$T=e^{-k^2{\alpha }^{2}t}$

For the space equation.

$\frac{d^{2}F}{d{x}^2}+k^{2}F=0$

The above equation is the solutions of the equation mentioned below.

$F\left(x\right)=\left\{\begin{array}{l}\sin \left(kx\right)\\ \cos \left(kx\right)\end{array}\right.$

Given the boundary conditions of the problem. Discard the $\cos \left(kx\right)$solution. Thus eigenfunctions.

$u=e^{-{(n\pi \alpha /10)}^{2}t}\sin \left(\frac{n\pi x}{10}\right)$

The solution of our problem will be the series as given below.

$u=u_f+\sum _{n=1}^{\infty }{b}_n{e}^{-{(n\pi \alpha /10)}^{2}t}\sin \frac{n\pi x}{10}$

At t = 0 there is a need of $u=u_0$.

$\begin{array}{c}u=\sum _{n=1}^{\infty }{b}_n\sin \frac{n\pi x}{l}\\ =u_0\\ =x+10-u_f\end{array}$

This means finding the Fourier sine series for $x+10$on $(0,\text{10})$.

$\begin{array}{c}{b}_n=\frac{2}{10}\underset{0}{\overset{10}{\int }}(x+10)\sin (n\pi x/10)dx\\ =\frac{2}{10}[(x+10)\underset{0}{\overset{10}{\int }}\sin (n\pi x/10)dx-\underset{0}{\overset{10}{\int }}\left[\frac{d}{dx}(x+10)\underset{0}{\overset{10}{\int }}\sin (n\pi x/10)dx\right]dx]\\ =\frac{2}{10}{[-(x+10)\left[\frac{\cos (n\pi x/10)}{n\pi }\right]-\left[\frac{sin(n\pi x/10)}{{\left(n\pi \right)}^2}\right]]}_0^{10}\\ =-\frac{40}{\left(n\pi \right)}\text{;forneven}\end{array}$

Now, replace $b_n$and $u_0$.

$u=20-x-\frac{40}{\pi }\sum _{\begin{array}{c}2\\ \text{ven}n\end{array}}^{\infty }\frac{1}{n}{e}^{-{(n\pi \alpha /10)}^{2}t}\sin \frac{n\pi x}{10}$

Hence, the solution is found to be$u=20-x-\frac{40}{\pi }\sum _{\begin{array}{c}2\\ \text{ven}n\end{array}}^{\infty }\frac{1}{n}{e}^{-{(n\pi \alpha /10)}^{2}t}\sin \frac{n\pi x}{10}$.