Chapter 14, Problem 1RE

In Problems 1-20 solve the given boundary-value problem by an appropriate integral transform. Make assumptions about boundedness where necessary.

1. $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$, x > 0, 0 < y < π

${\frac{\partial u}{\partial x}|}_{x=0}=0$, 0 < y < π

u(x, 0) = 0, ${\frac{\partial u}{\partial y}|}_{y=\pi }=e^{-x}$, x > 0

To determine

The solution of the given boundary value problem under given boundary conditions.

Answer to Problem 1RE

The solution of boundary value problem is $u\left(x,y\right)=\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sinh \alpha y}{\alpha \left(1+{\alpha }^2\right)\cosh \alpha \pi }\cos \alpha x}d\alpha$.

Explanation of Solution

Given:

The given boundary value problem is $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0,x>0,0<y<\pi$ and boundary conditions are $u_x\left(0,y\right)=0,u\left(x,0\right)=0\text{and}{u}_y\left(x,\pi \right)=e^{-x}$.

Calculation:

The given boundary value problem is,

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$.                                                                                                           (1)

Take Fourier transform on both sides of the above equation,

$\begin{array}{c}F\left\{\frac{{\partial }^{2}u}{\partial x^2}\right\}+F\left\{\frac{{\partial }^{2}u}{\partial y^2}\right\}=0\\ \frac{d^{2}U}{d{x}^2}-{\alpha }^{2}U\left(\alpha ,y\right)=0\end{array}$

Therefore, the equation is,

$\frac{d^{2}U}{d{x}^2}-{\alpha }^{2}U\left(\alpha ,y\right)=0$

Apply Fourier cosine transform then the particular solution of the above equation is,

$U\left(\alpha ,y\right)=c_1\cosh \alpha y+c_2\sinh \alpha y$.                                                                        (2)

At the given boundary condition $u\left(x,0\right)=0$, $c_1=0$

Substitute the value of $c_1$ in equation (2),

$U\left(\alpha ,y\right)=c_2\sinh \alpha y$.                                                                                             (3)

At boundary condition,

$u_y\left(x,\pi \right)=e^{-x}$

Take Fourier transform of the above equation,

$U^{\prime }\left(\alpha ,\pi \right)=\frac{1}{\alpha \left(1+{\alpha }^2\right)}$.                                                                                              (4)

Partially differentiate the equation (3) with respect to $y$ and substitute $y=\pi$,

$U^{\prime }\left(\alpha ,\pi \right)=c_2$.                                                                                                           (5)

Equate the equations (4) and (5),

$c_2=\frac{1}{\alpha \left(1+{\alpha }^2\right)}$

Substitute the value of $c_2$ in equation (3),

$U\left(\alpha ,y\right)=\frac{\sinh \alpha y}{\alpha \left(1+{\alpha }^2\right)}$

Take inverse Fourier transform of the above equation and apply Fourier cosine transform,

$\begin{array}{l}{F}^{-1}\left\{U\left(\alpha ,y\right)\right\}=F^{-1}\left\{\frac{\sinh \alpha y}{\alpha \left(1+{\alpha }^2\right)}\right\}\\ u\left(x,y\right)=\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sinh \alpha y}{\alpha \left(1+{\alpha }^2\right)\cosh \alpha \pi }\cos \alpha x}d\alpha \end{array}$

Thus, the solution of boundary value problem is $u\left(x,y\right)=\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sinh \alpha y}{\alpha \left(1+{\alpha }^2\right)\cosh \alpha \pi }\cos \alpha x}d\alpha$.