Given equation-
dx2d2u+dy2d2u=0.........1)
, subject to condition u(0,y)=u(l,y)=u(x,0)=0,u(x,a)=sinlnmx
Let u=XY
where X is a function of x only and Y is a function of y only .
=∂x2∂2u=∂x2∂(XY)=Ydx2d2X
= ∂y2∂2u=∂y2∂(XY)=Xdy2d2Y
From 1) ,we can say that ,
YX′′+XY′′=0
⟹ XX′′+YY′′=0
Case 1, we have
XX′′= −YY′′=p2
(i) XX′′=p2
X′′−P2X=0
auxiliary equation is given by -
m2−p2=0
m=±p
CF=C1epx+C2e−px
PI=0
X=C1epx+C2e−px
(ii)
Y−Y′′=p2⟹ Y′′+p2Y=0
Auxiliary equation is m2+p2=0⟹ m=±pi
CF=C3cospy+C4sinpy
PI=0
y= C3cospy+C4sinpy
Now X(0)=0
c1+c2=0⟹ c2=−c1
X(l)=0
⟹ C1epl++C2e−pl=0
⟹ C1(epl−e−pl)=0
c1=0
c2=0
X=0⟹ U=XY =0 which is impossible.
hence we reject case 1 .
case 2) , when
XX′′=−YY′′=0
(i) XX′′=0
X′′=0⟹ X=c5x+c6
(ii) −YY′′=0
⟹ Y′′=0 ⟹ Y=c7y+c8
Now , X(0)=0⟹ c6=0
X(l)=0
⟹ c5l+c6=0⟹ c5l=0
⟹ c5=0 (since =0)
∴ X=0
Hence , we reject case 2.
Case 3)
XX′′=−YY′′=p2
(i)−XX′′=p2
⟹ X′′+p2X=0⟹ dx2d2X+p2X=0
Auxiliary equation is m2+p2=0 =0 m=pi
CF=c9cospx+c10sinpx
PI=0
X= c9cospx+c10sinpx
(ii) Y−Y′′=−P2
⟹ dy2d2Y−p2Y=0
auxiliary equation is m2−p2=0
∴ CF=C11ePY+C12e−PY
PI=0
∴ CF= C11ePY+C12e−PY
PI=0
Hence, Y=C11epy+C12e−py
Now, X(0)=0⟹ c9=0
∴X=c10sinpx
X(l)=0
c10sinpl=0
⟹ sinpl=0=sinnπ,n∈ I
∴ p=lnπ
∴ X=c10sinlnπx
Again,Y(0)=0
⟹c11+c12=0⟹c11=−c12
Y=c11(epy−e−py)=c11(elnπy−el−nπy)
∴ u=XY =c11c10sinlnπx(elnπy−el−nπy)
or u(x,y)=bnsinlnπx(elnπy−el−nπy)
now,u(x,a)=sinlnπx=bnsinlnπx (elnπa−el−nπa)
⟹ bn=(elnπa−el−nπa)1 =2sinhl(nπa)1
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