Do Problem 5 if the $\mathit{x}\mathbf{=}\mathbf{0}$end is insulated and the $\mathit{x}\mathbf{=}\mathit{l}$end held at $\mathbf{20}\mathbf{°}$for $\mathit{t}\mathbf{>}\mathbf{0}$. (See Problem 3.9)

It has been asked to use problem 5 and make certain changes and solve further.

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 $({\frac{\partial u}{\partial x}|}_{x=0}=0\text{and}T=20°\text{for}x=l)$

$u_0=20°$ ….. (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 to 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.

$u=e^{-{(n\pi \alpha /l)}^{2}t}\cos \left(\frac{kx}{l}\right)$

The solution of our problem will be the series

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

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

$\begin{array}{c}u=\stackrel{\infty }{}{b}_n\cos \frac{\pi }{2l}(2n+1)\\ =u_0\\ =20°\end{array}$

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

$b_n=\underset{0}{\overset{l}{\int }}20\cos \left[\frac{\pi }{2l}(2n+1)x\right]dx$

$\begin{array}{c}{b}_n=20{\left[\frac{sin\left[\frac{\pi }{2l}(2n+1)x\right]}{\left[\frac{\pi }{2}(2n+1)\right]}\right]}_0^l\\ =20[\frac{sin\left(\frac{\pi }{2l}(2n+1)l\right)}{\left[\frac{\pi }{2}(2n+1)\right]}-\frac{sin\left(0\right)}{\left[\frac{\pi }{2}(2n+1)\right]}]\\ =20\left[\frac{sin(\pi n-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.)}{\left[\frac{\pi }{2}(2n+1)\right]}\right]\end{array}$

$\begin{array}{c}{b}_n=40\left[\frac{sin\left(\pi n\right)-\sin \left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\left[\pi (2n+1)\right]}\right]\\ =40\left[\frac{0-1}{\left[\pi (2n+1)\right]}\right]\\ =\frac{40{(-1)}^n}{\pi (2n+1)}\end{array}$

Now, replace $b_n$and $u_0$and add the subtracted temperature.

$u=20+\frac{40}{\pi }\sum _{n=0}^{\infty }\frac{{(-1)}^n}{2n+1}{e}^{-{\left[\right(2n+1)\pi \alpha /(2l\left)\right]}^{2}t}\cos \left(\frac{2n+1}{2l}\pi x\right)$.

Hence, the solution is found as below.

$\begin{array}{c}u=20+\frac{40}{\pi }\sum _{\text{odd}n}\frac{1}{n}{e}^{-{(n\pi \alpha /l)}^{2}t}\sin \frac{n\pi x}{l}u\\ =20+\frac{40}{\pi }\sum _{n=0}^{\infty }\frac{{(-1)}^n}{2n+1}{e}^{-{\left[\right(2n+1)\pi \alpha /(2l\left)\right]}^{2}t}\cos \left(\frac{2n+1}{2l}\pi x\right)\end{array}$