Answer to Question #194761 in Differential Equations for Arpan Chakraborty

Question #194761

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


1
Expert's answer
2021-07-05T17:56:25-0400

Given equation-


d2udx2+d2udy2=0\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)=sinnmxlu(x,0)=0,u(x,a)=sin\dfrac{nmx}{l}



Let u=XYu=XY


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


=2ux2=(XY)x2=Yd2Xdx2=\dfrac{\partial^{2}u}{\partial x^{2}}=\dfrac{\partial(XY)}{\partial x^{2}}=Y\dfrac{d^{2}X}{dx^{2}}


== 2uy2=(XY)y2=Xd2Ydy2\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=0YX''+XY''=0


    \implies XX+YY=0\dfrac{X''}{X}+\dfrac{Y''}{Y}=0


Case 1, we have


XX=\dfrac{X''}{X}= YY=p2-\dfrac{Y''}{Y}=p^{2}


(i)(i) XX=p2\dfrac{X''}{X}=p^{2}


XP2X=0X''-P^{2}X=0


auxiliary equation is given by -


m2p2=0m^{2}-p^{2}=0


m=±pm={\pm}p


CF=C1epx+C2epxCF=C_1e^{px}+C_2e^{-px}


PI=0PI=0


X=C1epx+C2epxX=C_1e^{px}+C_2e^{-px}


(ii)(ii)

YY=p2    \dfrac{-Y''}{Y}=p^{2}\implies Y+p2Y=0Y''+p^{2}Y=0



Auxiliary equation is m2+p2=0    m^{2}+p^{2}=0\implies m=±pim={\pm}pi


CF=C3cospy+C4sinpyCF=C_3cos py+C_4sinpy


PI=0PI=0


y=y= C3cospy+C4sinpyC_3cos py+C_4sinpy


Now X(0)=0


c1+c2=0    c_1+c_2=0\implies c2=c1c_2=-c_1


X(l)=0X(l)=0


    \implies C1epl++C2epl=0C_1e^{pl}++C_2e^{-pl}=0


    \implies C1(eplepl)=0C_1(e^{pl}-e^{-pl})=0


c1=0c_1=0

c2=0c_2=0


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


hence we reject case 1 .


case 2) , when



XX=YY=0\dfrac{X''}{X} =-\dfrac{Y''}{Y}=0


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


X=0    X''=0\implies X=c5x+c6X=c_5x+c_6


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


    \implies Y=0Y''=0     \implies Y=c7y+c8Y=c_7y+c_8



Now , X(0)=0    X(0)=0\implies c6=0c_6=0


X(l)=0X(l)=0


    \implies c5l+c6=0    c_5l+c_6=0\implies c5l=0c_5l=0


    \implies c5=0c_5=0 (since 0){\neq}0)


\therefore X=0X=0


Hence , we reject case 2.



Case 3)


XX=YY=p2\dfrac{X''}{X}=-\dfrac{Y''}{Y}=p^{2}


(i)XX=p2(i)-\dfrac{X''}{X}=p^{2}


    \implies X+p2X=0    X''+p^{2}X=0\implies d2Xdx2+p2X=0\dfrac{d^{2}X}{dx^{2}}+p^{2}X=0


Auxiliary equation is m2+p2=0m^{2}+p^{2}=0 =0 mpim{\neq}pi


CF=c9cospx+c10sinpxCF=c_9cos px+c_{10}sinpx



PI=0PI=0


X=X= c9cospx+c10sinpxc_9cos px+c_{10}sinpx



(ii)(ii) YY=P2\dfrac{-Y''}{Y}=-P^{2}



    \implies d2Ydy2p2Y=0\dfrac{d^{2}Y}{dy^{2}}-p^{2}Y=0



auxiliary equation is m2p2=0m^{2}-p^{2}=0



\therefore CF=C11ePY+C12ePYC_{11}e^{PY}+C_{12}e^{-PY}


PI=0PI=0


\therefore CF=CF= C11ePY+C12ePYC_{11}e^{PY}+C_{12}e^{-PY}


PI=0PI=0


Hence,Hence , Y=C11epy+C12epyY=C_{11}e^{py}+C_{12}e^{-py}


Now, X(0)=0    X(0)=0\implies c9=0c_9=0


X=c10sinpx\therefore X=c_{10}sinpx


X(l)=0X(l)=0


c10sinpl=0c_{10}sinpl=0


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


\therefore p=nπlp=\dfrac{n{\pi}}{l}

\therefore X=c10sinnπxlX=c_{10}sin\dfrac{n{\pi}x}{l}


Again,Y(0)=0Again ,Y(0)=0


    c11+c12=0    c11=c12\implies c_{11}+c_{12}=0\implies c_{11}=-c_{12}


Y=c11(epyepy)=c11(enπylenπyl)Y=c_{11}(e^{py}-e^{-py})=c_{11}(e^\dfrac{n{\pi}y}{l}-e^\dfrac{-n{\pi}y}{l})



\therefore u=XYu=XY =c11c10sinnπxl(enπylenπyl)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)=bnsinnπxl(enπylenπyl)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)=sinnπxl=bnsinnπxlnow, u(x,a)=sin\dfrac{n{\pi}x}{l}=b_nsin\dfrac{n{\pi}x}{l} (enπalenπal)(e^\dfrac{n{\pi}a}{l}-e^\dfrac{-n{\pi}a}{l})



    \implies bn=1(enπalenπal)b_n=\dfrac{1}{(e^\dfrac{n{\pi}a}{l}-e^\dfrac{-n{\pi}a}{l})} =12sinh(nπa)l=\dfrac{1}{2sinh \dfrac{(n{\pi}a)}{l}}




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