A bar of length l is initially at $\mathbf{0}\mathbf{°}$.From $\mathit{t}\mathbf{=}\mathbf{0}$on, the ends are held at $\mathbf{20}\mathbf{°}$. Find $\mathit{u}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}$for$\mathit{t}\mathbf{>}\mathbf{0}$.

It has been given that a bar of length l is initially at$0°$.

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

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 $u_0=20$at $x=0$and $x=l$.

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.

$\begin{array}{l}\frac{dT}{dt}=-k^2{\alpha }^{2}T\\ \frac{dT}{T}=-k^2{\alpha }^{2}dt\end{array}$

Take an integration both the side.

$\begin{array}{l}\int \frac{dT}{T}=\int -k^2{\alpha }^{2}dt\\ \ln \left(T\right)=-k^2{\alpha }^{2}t\\ T=\exp (-k^2{\alpha }^{2}t)\end{array}$

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.

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

The solution of our problem will be the series

$u=u_0\sum _{n=1}^{\infty }{b}_n{e}^{-{(n\pi \alpha /l)}^{2}t}\sin \frac{n\pi x}{l}$

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

$\begin{array}{c}u=u_{0}u\\ =\sum _{n=1}^{\infty }{b}_n\sin \frac{n\pi x}{l}\\ =u_0\\ =20°\end{array}$

This means finding the Fourier sine series for $20$on$(0,l).$

$\begin{array}{c}{b}_n=\underset{0}{\overset{l}{\int }}20\sin (n\pi x/l)dx\\ =20{\left[\frac{-\cos (n\pi x/l)}{n\pi }\right]}_0^l\\ =20\left[\frac{-\cos (n\pi l/l)+\cos 0}{n\pi }\right]\\ =\frac{20}{n\pi }[-(-1)+1]\end{array}$

$b_n=\frac{40}{n\pi }\text{;fornodd}$

Now, replace $b_n$and $u_0$.

$u=20+\frac{40}{\pi }\sum _{\text{odd}n}\frac{1}{n}{e}^{-{(n\pi \alpha /l)}^{2}t}\sin \frac{n\pi x}{l}$

Hence, the solution is found to be $u=20+\frac{40}{\pi }\sum _{\text{odd}n}\frac{1}{n}{e}^{-{(n\pi \alpha /l)}^{2}t}\sin \frac{n\pi x}{l}$.