Answer to Question #219622 in Differential Equations for Rehema

Question

Consider the boundary value problem(BVP) defined by laplaces equation.

u_xx+u_yy=0 on the square o<x,y<2

subjected to the boundary conditions:

u(0,y)=0 ,u(x,2)=0 ,u(2,y)=0 ,u(x,0)=100sin(πx/2)

Solve the BVP using method of separation of variable.

Answer 1

The given equation is -

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

Let $u=XY.............................................2)$

where $X$ is a function of $x$ and $Y$ is a function of $y$ , only then ,

$\dfrac{\partial^{2}u}{\partial x^{2}}=\dfrac{\partial^{2}(XY)}{\partial x^{2}}=Y\dfrac{d^{2}X}{dx^{2}}$ and

$\dfrac{\partial^{2}u}{\partial y^{2}}=\dfrac{\partial^{2}(XY)}{\partial y^{2}}=X\dfrac{d^{2}Y}{dy^{2}}$

$\therefore$ From (1), $YX^{''}+XY^{''}=0$

$\implies$ $\dfrac{X^{''}}{X}+\dfrac{Y^{''}}{Y}=0$

Now , we will make some cases -

Case I. When $\dfrac{X^{''}}{X}=-\dfrac{Y^{''}}{Y}$ $=p^{2}(say)$

$i) \ \dfrac{X^{''}}{X}=p^{2}$

$X^{''}-P^{2}X=0$

Auxiliary equation is-

$m^{2}-p^{2}=0$

$m={\pm}p$

$\therefore$ $CF=c_1e^{px}+c_2e^{-px}$

$PI=0$

$\therefore$ $X=$ $c_1e^{px}+c_2e^{-px}$

$ii)$ $\ \dfrac{Y^{''}}{Y}=p^{2}\implies$ $Y^{''}+P^{2}Y=0$

Auxiliary equation is $-$

$m^{2}+p^{2}=0\implies$ $m={\pm}pi$

$\therefore$ $CF=c_3cospy+c_4sinpy$

$PI=0$

$\therefore$ $y=c_3cospy+c_4sinpy$

Now , we have given some boundary conditions making use of them now -

$X(0)=0$

$\implies$ $c_{1}+c_{2}=0$ $\implies c_2=-c_1$

$X(x)=0$

$\implies$ $c_{1}e^{px}+c_2e^{-px}=0$ $\implies$ $c_1(e^{px}+e^{-px})=0$

$\implies$ $c_1=0$ (Since $e^{px}-e^{-px}$ $\neq0(as\ p\neq0\neq l)$

$\therefore$ $c_2=0$

$\therefore$ $X=0\implies$ $u=XY=0$ which is impossible .

Hence Case 1 is rejected .

$Case \ 2)$ when $\dfrac{X^{''}}{X}=-\dfrac{Y^{''}}{Y}=0$

$i)$ $\dfrac{X^{''}}{X}=0$

$\implies$ $X^{''}=0\implies$ $X=c_5x+c_6$

$ii)$ $-\dfrac{Y^{''}}{Y}=0$

$\implies$ $Y^{''}=0\implies$ $Y=c_7y+c_8$

Now , $X(0)=0\implies$ $c_6$ $=$ 0

$X(x)=0$

$\implies$ $c_5+c_6=0\implies$$c_5l=0$

$\implies$ $c_5=0$

$\therefore X=0$

$\therefore u=XY=0$ which is impossible .

Hence we reject case 2 also .

$Case \ 3)$ when

$\dfrac{X^{''}}{X}=-\dfrac{Y^{''}}{Y}=-p^{2}(say)$

$(i)$ $\dfrac{X^{''}}{X}=-p^{2}$ $\implies \dfrac {d^{2}X}{dx^{2}}+p^{2}=0$

$\implies$ $X^{''}+p^{2}X=0$

Auxiliary equation is given by -

$=m^{2}+p^{2}=0\implies$ $m={\pm}pi$

$CF=c_9cospx+c_{10}sinpx$

$PI=0$

$X=c_9cospx+c_{10}sinpx$

$(ii)$ $-\dfrac{Y^{''}}{Y}=-p^{2}$ $\implies \dfrac{d^{2}y}{dy^{2}}-p^{2}y=0$

Auxiliary equation is -

$m^{2}-p^{2}=0$

$m={\pm}p$

$\therefore CF=c_{11}e^{py}+c_{12}e^{-py}$

$PI=0$

hence , $Y=c_{11}e^{py}+c_{12}e^{-py}$

Now , $X(0)=0 \implies c_{9}=0$

$\therefore X=c_{10}sinpx$

$X(l)=0$

$=c_{10}sinpl=0$

$\therefore sinpl=0=sinn{\pi}, n\in I$

$\therefore$ $p=\dfrac{n{\pi}}l$

$\therefore$ $X=c_{10}sin\dfrac{n{\pi}x}{l}$ .............................................................3)

Again $Y(0)=0$

$\therefore c_{11}+c_{12}=0 \therefore c_{11}=-c_{12}$

$Y=c_{11}(e^{py}-e^{-py})=c_{11}(e^\dfrac{n{\pi}}{l}-e^\dfrac{-n{\pi}}{l})$

$\therefore u=XY=c_{10}c_{11}sin\dfrac{n{\pi}x}{l}[(e^\dfrac{n{\pi}}{l}-e^\dfrac{-n{\pi}}{l})]$

or $u(x,y)=b_nsin\dfrac{n{\pi}x}{l}[(e^\dfrac{n{\pi}}{l}-e^\dfrac{-n{\pi}}{l})]$

Now $u(x,0)=100sin\dfrac{{\pi}x}{2}=b_nsin\dfrac{n{\pi}x}{l}[(e^\dfrac{n{\pi}}{l}-e^\dfrac{-n{\pi}}{l})]$

$=b_{n}=\dfrac{100sin\dfrac{{\pi}x}{2}}{sin\dfrac{n{\pi}x}{l}[(e^\dfrac{n{\pi}}{l}-e^\dfrac{-n{\pi}}{l})]}$

$u(x,y)=$ $[(e^\dfrac{n{\pi}}{l}-e^\dfrac{-n{\pi}}{l})]\dfrac{100sin\dfrac{{\pi}x}{2}}{sin\dfrac{n{\pi}x}{l}[(e^\dfrac{n{\pi}}{l}-e^\dfrac{-n{\pi}}{l})]}$ $sin\dfrac{n{\pi}x}{l}$

$u(x,y)=100sin\dfrac{{\pi}x}{2}$

which is given solution .