A metal plate covering the first quadrant has the edge which is along the y axis insulated and the edge which is along the x-axis held at

$\mathit{u}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\{\begin{array}{l}\begin{array}{cc}100(2-x),& \text{for}0<x<2\end{array}\\ \begin{array}{cc}0,& \text{for}x>2\end{array}\end{array}$

Find the steady-state temperature distribution as a function of x and y. Hint: Follow the procedure of Example 2, but use a cosine transform (because $\mathbf{\partial }\mathit{u}\mathbf{/}\mathbf{\partial }\mathit{x}\mathbf{=}\mathbf{0}$for x = 0 ). Leave your answer as an integral like (9.13).

The given equation is mentioned below.

$u(x,0)=\left\{\begin{array}{l}\begin{array}{cc}100(2-x),& \text{for}0<x<2\end{array}\\ \begin{array}{cc}0,& \text{for}x>2\end{array}\end{array}\right.$

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.

Propose a solution to the form mentioned below.

$u(x,y)={\int }_0^{\infty }B\left(k\right)e^{ky}\cos \left(ky\right)dk$

Put y = 0.

$\begin{array}{c}u(x,0)={\int }_0^{\infty }B\left(k\right)\cos \left(kx\right)dk\\ =100(2-x)\text{if}0<x<2\\ =0\text{if}x>2\end{array}$

Define B(k) as the inverse transform.

$\begin{array}{c}B\left(k\right)=\frac{2}{\pi }{\int }_0^{\infty }u(x,0)\cos \left(kx\right)dx\\ =\frac{2}{\pi }[200{\int }_0^2\cos \left(kx\right)dx-100{\int }_0^{2}x\cos \left(kx\right)dx]\end{array}$

First part of the integral can be solved directly and the second part is easily solved using integration by parts.

$\int x\cos \left(kx\right)dx=x\sin \left(kx\right)/k+\cos \left(kx\right)/k^2$

Therefore, write the expression for B(k).

$\begin{array}{r}B\left(k\right)=\frac{2}{\pi }[200\sin \left(2k\right)/k-100(\sin \left(2k\right)+\cos \left(2k\right)-1)/k\\ \begin{array}{l}=\frac{2}{\pi k}(100-100\cos \left(2k\right))\\ =\frac{200}{\pi k^2}(1-\cos \left(2k\right))\end{array}\end{array}$

Now, by replacing B(k) in equation (1) the final solution is obtained.

$u(x,y)=\frac{200}{\pi }{\int }_0^{\infty }{k}^{-2}(1-\cos \left(2k\right))e^{-ky}\cos \left(kx\right)dk$