91Ó°ÊÓ

The given equation is mentioned below.

$u=\frac{100}{\mathrm{π}}\arctan \left(\frac{\sin 2\mathrm{θ}}{r^2−\cos 2\mathrm{θ}}\right)$

The Laplace transform of an ordinary differential equation converts it into an algebraic equation. Taking the Laplace transform of a partial differential equation reduces the number of independent variables by one, and so converts a two-variable partial differential equation into an ordinary differential equation.

The initial steady-state temperaturesatisfies Laplace's equation, which in this one-dimensional case is,

$\frac{d^2{u}_0}{d{x}^2}=0$

The solution of this equation is $u_0=ax+b,$where a and are constants which must be found to fit the given conditions. Since $u_0=0$at x = 0 and $u_0=100$at x = l.

$u_0=\frac{100}{l}x$

The differential equation satisfied by u.

$\frac{{∂}^{2}u}{∂x^2}=\frac{1}{{\mathrm{Î\pm }}^2}\frac{∂u}{∂t}$

Take the t Laplace transform of the differential equation. The variable x will just be a parameter in this process. Let U be the Laplace transform of u.

$U(x,p)={∫}_0^{\mathrm{∞}}u(x,t)e^{−pt}dt$

So, for further solution it can be written as mentioned below.

$\begin{array}{c}L\left(\frac{∂u}{∂t}\right)=\frac{∂}{∂t}{∫}_0^{\mathrm{∞}}u(x,t)e^{−pt}dt\\ ={∫}_0^{\mathrm{∞}}[\frac{∂u}{∂t}{e}^{−pt}+u(−p{e}^{−pt})]dt\\ =\frac{∂U}{∂t}−p{u}_0\\ =\frac{∂U}{∂t}−p(U−100x/l)\end{array}$

Laplace transform of the boundary conditions.

$\begin{array}{c}u(0,p)={∫}_0^{\mathrm{∞}}u(0,0)e^{−pt}dt\\ ={∫}_0^{\mathrm{∞}}0dt\\ =0\end{array}$

Thus, the following differential equation and boundary conditions is obtained.

$\frac{{∂}^{2}U}{∂x^2}−\frac{p}{{\mathrm{Î\pm }}^2}U=−\frac{100}{{\mathrm{Î\pm }}^{2}l}x$and $U(0,p)=U(l,p)=0$.

First solve the homogeneous part.

$\frac{{∂}^{2}U}{∂x^2}−\frac{p}{{\mathrm{Î\pm }}^2}U=0$

It can be solved by proposing a solution like $U(x,p)=A{e}^{−xr}$where A and r are constant values.

$U_h(x,p)=A_1{e}^{x\sqrt{p}/\mathrm{Î\pm }}+A_2{e}^{−x\sqrt{p}/\mathrm{Î\pm }}$

Now, solve the particular part by proposing a solution like$U_p=Bx+D{x}^2$.

$U_p=100x/lp$

Put everything together.

$U(x,p)=(A_1{e}^{x\sqrt{p}/\mathrm{Î\pm }}+A_2{e}^{−x\sqrt{p}/\mathrm{Î\pm }})+\frac{100x}{lp}$

Now apply the boundary conditions and use the identity$\sinh x=\frac{e^x−e^{−x}}{2}$

$U(x,p)=−\frac{100\sinh (p^{1/2}/\mathrm{Î\pm })x}{p\sinh (p^{1/2}/\mathrm{Î\pm })l}+\frac{100}{pl}x$

Now find the inverse Laplace transform, look for the inverse transform of the individual terms of U.

$\frac{\sinh (p^{1/2}/\mathrm{Î\pm })x}{p\sinh (p^{1/2}/\mathrm{Î\pm })l}=\frac{x}{pl}−\frac{2}{\mathrm{Ï€}}[\frac{\sin (\mathrm{Ï€}x/l)}{p+({\mathrm{Ï€}}^2{\mathrm{Î\pm }}^2/l^2)}−\frac{\sin (2\mathrm{Ï€}x/l)}{2[p+(4{\mathrm{Ï€}}^2{\mathrm{Î\pm }}^2/l^2)]}+\frac{\sin (3\mathrm{Ï€}x/l)}{3[p+(9{\mathrm{Ï€}}^2{\mathrm{Î\pm }}^2/l^2)]}⋯]$

The inverse Laplace transform of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$(p+a)$}\right.$is which can be identified with our case but as an infinite series. Therefore, the final solution is as given below.

$u=\frac{200}{\mathrm{Ï€}}\underset{n=1}{\overset{\mathrm{∞}}{∑}}\frac{{(−1)}^{n−1}}{n}{e}^{−{(n\mathrm{Ï€}\mathrm{Î\pm }/l)}^{2}t}\sin \left(\frac{n\mathrm{Ï€}x}{l}\right)$