Question

Solve two dimensional Laplace equation d²u/dx²+d²u/dy²=0 ,subject to the condition u(0,y)=u(l,y)=u(x,0)=0, u(x,a)=sin nmx/l

Answer 1

Given equation-

$\dfrac{d^{2}u}{dx^{2}}+\dfrac{d^{2}u}{dy^{2}}=0$.........1)

, subject to condition u(0,y)=u(l,y)=$u(x,0)=0,u(x,a)=sin\dfrac{nmx}{l}$

Let $u=XY$

where X is a function of x only and Y is a function of y only .

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

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

From 1) ,we can say that ,

$YX''+XY''=0$

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

Case 1, we have

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

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

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

auxiliary equation is given by -

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

$m={\pm}p$

$CF=C_1e^{px}+C_2e^{-px}$

$PI=0$

$X=C_1e^{px}+C_2e^{-px}$

$(ii)$

$\dfrac{-Y''}{Y}=p^{2}\implies$ $Y''+p^{2}Y=0$

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

$CF=C_3cos py+C_4sinpy$

$PI=0$

$y=$ $C_3cos py+C_4sinpy$

Now X(0)=0

$c_1+c_2=0\implies$ $c_2=-c_1$

$X(l)=0$

$\implies$ $C_1e^{pl}++C_2e^{-pl}=0$

$\implies$ $C_1(e^{pl}-e^{-pl})=0$

$c_1=0$

$c_2=0$

X=0$\implies$ U=XY =0 which is impossible.

hence we reject case 1 .

case 2) , when

$\dfrac{X''}{X} =-\dfrac{Y''}{Y}=0$

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

$X''=0\implies$ $X=c_5x+c_6$

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

$\implies$ $Y''=0$ $\implies$ $Y=c_7y+c_8$

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

$X(l)=0$

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

$\implies$ $c_5=0$ (since ${\neq}0)$

$\therefore$ $X=0$

Hence , we reject case 2.

Case 3)

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

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

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

Auxiliary equation is $m^{2}+p^{2}=0$ =0 $m{\neq}pi$

$CF=c_9cos px+c_{10}sinpx$

$PI=0$

$X=$ $c_9cos px+c_{10}sinpx$

$(ii)$ $\dfrac{-Y''}{Y}=-P^{2}$

$\implies$ $\dfrac{d^{2}Y}{dy^{2}}-p^{2}Y=0$

auxiliary equation is $m^{2}-p^{2}=0$

$\therefore$ CF=$C_{11}e^{PY}+C_{12}e^{-PY}$

$PI=0$

$\therefore$ $CF=$ $C_{11}e^{PY}+C_{12}e^{-PY}$

$PI=0$

$Hence ,$ $Y=C_{11}e^{py}+C_{12}e^{-py}$

Now, $X(0)=0\implies$ $c_9=0$

$\therefore X=c_{10}sinpx$

$X(l)=0$

$c_{10}sinpl=0$

$\implies$ $sinpl=0=sinn{\pi},n\isin$ I

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

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

$Again ,Y(0)=0$

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

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

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

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

$now, u(x,a)=sin\dfrac{n{\pi}x}{l}=b_nsin\dfrac{n{\pi}x}{l}$ $(e^\dfrac{n{\pi}a}{l}-e^\dfrac{-n{\pi}a}{l})$

$\implies$ $b_n=\dfrac{1}{(e^\dfrac{n{\pi}a}{l}-e^\dfrac{-n{\pi}a}{l})}$ $=\dfrac{1}{2sinh \dfrac{(n{\pi}a)}{l}}$