Question

Solve Problem 2 if the sides $\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{1}$are insulated.

Short answer

The solution is found to be$T(x,y)=y+\frac{4}{{\pi }^2}\sum _{n=1}^{\infty }\frac{1}{n^2\sinh \left(2n\pi \right)}\sinh \left(n\pi y\right)\cos \left(n\pi x\right)$.

Step-by-step solution

Given Information:

It has been asked to solve problem 2 with the given condition.

Definition of Laplace’s equation.

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

Apply Boundary condition:

The steady-state temperature distribution.

$T(x,y)=c_{0}y+(c_1{e}^{-ky}+c_2{e}^{ky})(c_3\cos kx+c_4\sin kx)$

First set the value of the constants $c_1,c_2,c_3$and $c_4$. So, if $\frac{\partial T}{\partial x}=0$for $x=0$and $x=1$.

$\begin{array}{c}\frac{\partial T(0,y)}{\partial x}=(c_1{e}^{-ky}+c_2{e}^{ky})\left(c_4\right)\\ =0\end{array}$

Since an exponential can't be zero, $c_4=0.$If for $x=1$:

$\begin{array}{c}T(1,y)=(c_1{e}^{-ky}+c_2{e}^{ky})\left(c_3\sin k\right)\\ =0\end{array}$

It is known that $\sin \left(kx\right)=0$if $kx=n\pi ,$where $n=0,1,2..,$therefore for $x=1$ there is $k=n\pi$. Finally, there is $y=2T=0$then $c_1$and $c_2$are needed.

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

Thus,

$c_1=\frac{1}{2}{e}^k$and$c_2=-\frac{1}{2}{e}^{-k}$

Therefore, for any integral n, the solution is as follow.

$T=c_{0}y+\sinh \left(n\pi y\right)\cos \left(n\pi x\right)$

The above equation satisfies the given boundary conditions on the three T=0 sides.

Find the series solution:

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, namely

$T(x,y)=b_{0}y+\sum _{n=1}^{\infty }{B}_n\sinh \left(n\pi y\right)\cos \left(n\pi x\right)$

Find the expression for$B_n$.

$\begin{array}{c}T(x,2)=1-x\\ =2c_0+\sum _{n=1}^{\infty }{B}_n\sinh \left(2n\pi \right)\cos \left(n\pi x\right)\\ =b_0+\sum _{n=1}^{\infty }{b}_n\cos \left(n\pi x\right)(b_n=B_n\sinh \left(2n\pi \right),b_0=\raisebox{1ex}{$c_0$}\!\left/ \!\raisebox{-1ex}{$4$}\right.)\end{array}$

Solve further and you have,

$\begin{array}{c}{b}_0=2{\int }_0^1(1-x)dx\\ =2{[x-\frac{x^2}{2}]}_0^1\\ =2[1-\frac{1}{2}]\\ =1\end{array}$

$\begin{array}{c}{b}_n=2\underset{0}{\overset{1}{\int }}(1-x)\cos \left(n\pi x\right)dx\\ =2\underset{0}{\overset{1}{\int }}(1-x)\cos \left(n\pi x\right)dx\\ =2[\underset{0}{\overset{1}{\int }}\cos \left(n\pi x\right)dx-\underset{0}{\overset{1}{\int }}x\cos \left(n\pi x\right)dx]\\ =2{[\frac{\sin \left(n\pi x\right)}{n\pi }-\frac{x\sin \left(n\pi x\right)}{n\pi }+\frac{\cos \left(n\pi x\right)}{{\left(n\pi \right)}^2}]}_0^1\end{array}$

$\begin{array}{c}{b}_n=2[(\frac{\sin \left(n\pi \right)}{n\pi }-\frac{1\times \sin \left(n\pi \right)}{n\pi }+\frac{\cos \left(n\pi \right)}{{\left(n\pi \right)}^2})-(\frac{\sin \left(0\right)}{n\pi }-\frac{0\times \sin \left(0\right)}{n\pi }+\frac{\cos \left(0\right)}{{\left(n\pi \right)}^2})]\\ =2\left[\frac{1-1\cos n\pi }{{\left(n\pi \right)}^2}\right]\\ =2\left[\frac{1-1(-1)}{{\left(n\pi \right)}^2}\right]\\ =\frac{4}{{\left(n\pi \right)}^2}\end{array}$

Write the value of $B_n$.

$\begin{array}{c}{B}_n=\frac{b_n}{\sinh \left(2n\pi \right)}\\ =\frac{4}{{\left(n\pi \right)}^2\sinh \left(2n\pi \right)}\end{array}$

Finally, substitute $B_n$and $b_0$into the distribution $T(x,y)$.

$T(x,y)=y+\frac{4}{{\pi }^2}\sum _{n=1}^{\infty }\frac{1}{n^2\sinh \left(2n\pi \right)}\sinh \left(n\pi y\right)\cos \left(n\pi x\right)$

Hence, this is the required solution.