Consider the heat flow problem of Section 3. Solve this by Laplace transforms (with respect to t) by starting as in Example 1. You should get $\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{U}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{p}}{{\mathbf{\alpha }}^{\mathbf{2}}}\mathit{U}\mathbf{=}\mathbf{-}\frac{\mathbf{100}}{{\mathbf{\alpha }}^{\mathbf{2}}\mathbf{l}}\mathit{x}$and $\mathit{U}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{p}\mathbf{)}\mathbf{=}\mathit{U}\mathbf{(}\mathbf{l}\mathbf{,}\mathbf{p}\mathbf{)}\mathbf{=}\mathbf{0}$.

Solve this differential equation to get$\mathit{U}\mathbf{(}\mathit{x}\mathbf{,}\mathit{p}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{100}\mathbf{sinh}\mathbf{(}{\mathbf{p}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{/}\mathbf{\alpha }\mathbf{)}\mathbf{x}}{\mathbf{p}\mathbf{sinh}\mathbf{(}{\mathbf{p}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{/}\mathbf{\alpha }\mathbf{)}\mathbf{l}}\mathbf{+}\frac{\mathbf{100}}{\mathbf{p}\mathbf{l}}\mathit{x}$

Assume the following expansion, and find u by looking up the inverse Laplace transforms of the individual terms of U:

$\frac{\mathbf{sinh}\mathbf{(}{\mathbf{p}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{/}\mathbf{\alpha }\mathbf{)}\mathbf{x}}{\mathbf{p}\mathbf{sinh}\mathbf{(}{\mathbf{p}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{/}\mathbf{\alpha }\mathbf{)}\mathbf{l}}\mathbf{=}\frac{\mathbf{x}}{\mathbf{p}\mathbf{l}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{\pi }}\mathbf{[}\frac{\mathbf{sin}\mathbf{(}\mathbf{\pi }\mathbf{x}\mathbf{/}\mathbf{l}\mathbf{)}}{\mathbf{p}\mathbf{+}\mathbf{(}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\alpha }}^{\mathbf{2}}\mathbf{/}{\mathbf{l}}^{\mathbf{2}}\mathbf{)}}\mathbf{-}\frac{\mathbf{sin}\mathbf{(}\mathbf{2}\mathbf{\pi }\mathbf{x}\mathbf{/}\mathbf{l}\mathbf{)}}{\mathbf{2}\mathbf{[}\mathbf{p}\mathbf{+}\mathbf{(}\mathbf{4}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\alpha }}^{\mathbf{2}}\mathbf{/}{\mathbf{l}}^{\mathbf{2}}\mathbf{)}\mathbf{]}}\mathbf{+}\frac{\mathbf{sin}\mathbf{(}\mathbf{3}\mathbf{\pi }\mathbf{x}\mathbf{/}\mathbf{l}\mathbf{)}}{\mathbf{3}\mathbf{[}\mathbf{p}\mathbf{+}\mathbf{(}\mathbf{9}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\alpha }}^{\mathbf{2}}\mathbf{/}{\mathbf{l}}^{\mathbf{2}}\mathbf{)}\mathbf{]}}\mathbf{\cdots }\mathbf{]}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathit{\theta }$

The given equation is mentioned below.

$u=\frac{100}{\pi }\arctan \left(\frac{\sin 2\theta }{r^2-\cos 2\theta }\right)$

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.

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

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.

$U(x,p)={\int }_0^{\infty }u(x,t)e^{-pt}dt$

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

$\begin{array}{c}L\left(\frac{\partial u}{\partial t}\right)=\frac{\partial }{\partial t}{\int }_0^{\infty }u(x,t)e^{-pt}dt\\ ={\int }_0^{\infty }[\frac{\partial u}{\partial t}{e}^{-pt}+u(-p{e}^{-pt})]dt\\ =\frac{\partial U}{\partial t}-p{u}_0\\ =\frac{\partial U}{\partial t}-p(U-100x/l)\end{array}$

Laplace transform of the boundary conditions.

$\begin{array}{c}u(0,p)={\int }_0^{\infty }u(0,0)e^{-pt}dt\\ ={\int }_0^{\infty }0dt\\ =0\end{array}$

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

$\frac{{\partial }^{2}U}{\partial x^2}-\frac{p}{{\alpha }^2}U=-\frac{100}{{\alpha }^{2}l}x$and $U(0,p)=U(l,p)=0$.

First solve the homogeneous part.

$\frac{{\partial }^{2}U}{\partial x^2}-\frac{p}{{\alpha }^2}U=0$

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

$U_h(x,p)=A_1{e}^{x\sqrt{p}/\alpha }+A_2{e}^{-x\sqrt{p}/\alpha }$

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

$U_p=100x/lp$

Put everything together.

$U(x,p)=(A_1{e}^{x\sqrt{p}/\alpha }+A_2{e}^{-x\sqrt{p}/\alpha })+\frac{100x}{lp}$

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

$U(x,p)=-\frac{100\sinh (p^{1/2}/\alpha )x}{p\sinh (p^{1/2}/\alpha )l}+\frac{100}{pl}x$

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

$\frac{\sinh (p^{1/2}/\alpha )x}{p\sinh (p^{1/2}/\alpha )l}=\frac{x}{pl}-\frac{2}{\pi }[\frac{\sin (\pi x/l)}{p+({\pi }^2{\alpha }^2/l^2)}-\frac{\sin (2\pi x/l)}{2[p+(4{\pi }^2{\alpha }^2/l^2)]}+\frac{\sin (3\pi x/l)}{3[p+(9{\pi }^2{\alpha }^2/l^2)]}\cdots ]$

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.

$u=\frac{200}{\pi }\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n}{e}^{-{(n\pi \alpha /l)}^{2}t}\sin \left(\frac{n\pi x}{l}\right)$