An introduction to partial differential equations 527
from which,X=Acospx+Bsinpx
and Y=Cepy+De−py
or Y=Ccoshpy+Dsinhpy
or Y=Esinhp(y+φ)
Hence u(x,y)=XY
={Acospx+Bsinpx}{Esinhp(y+φ)}or u(x,y)
={Pcospx+Qsinpx}{sinhp(y+φ)}whereP=AEandQ=BE.
The first boundary condition is: u( 0 ,y)= 0 , hence
0 =Psinhp(y+φ)from which,P=0.
Hence,u(x,y)=Qsinpxsinhp(y+φ).
The second boundary condition is:u( 1 ,y)= 0 ,hence
0 =Qsinp( 1 )sinhp(y+φ)from which,
sinp= 0 ,hence,p=nπ forn= 1 , 2 , 3 ,...
The third boundary condition is:u(x, 0 )= 0 ,hence,
0 =Qsinpxsinhp(φ)from which,
sinhp(φ)=0andφ=0.
Hence,u(x,y)=Qsinpxsinhpy.
Since there are many solutions for integer values ofn,
u(x,y)=∑∞n= 1Qnsinpxsinhpy=∑∞n= 1Qnsinnπxsinhnπy (a)The fourth boundary condition is:u(x, 1 )= 4 =f(x),
hence,f(x)=∑∞n= 1Qnsinnπxsinhnπ( 1 ).From Fourier series coefficients,
Qnsinhnπ= 2 ×the mean value of
f(x)sinnπxfromx=0tox= 1i.e. =
2
1∫ 104sinnπxdx= 8[
−cosnπx
nπ] 10=−
8
nπ(cosnπ−cos0)=8
nπ( 1 −cosnπ)= 0 (for even values ofn),=16
nπ(for odd values ofn)Hence, Qn=16
nπ(sinhnπ)=16
nπcosechnπHence, from equation (a),u(x,y)=∑∞n= 1Qnsinnπxsinhnπy=16
π∑∞n(odd)= 11
n(cosechnπsinnπxsinhnπy)Now try the following exerciseExercise 203 Further problemson the
Laplace equation- A rectangular plate is bounded by the
linesx= 0 ,y= 0 ,x=1andy= 3 .Apply the
Laplace equation∂^2 u
∂x^2+∂^2 u
∂y^2=0 to determine
thepotentialdistributionu(x,y)overtheplate,
subject to the following boundary conditions:
u=0whenx= 00 ≤y≤2,
u=0whenx= 10 ≤y≤ 2 ,
u=0wheny= 20 ≤x≤ 1 ,
u=5wheny= 30 ≤x≤1.
⎡
⎣u(x,y)=^20
π∑∞
n(odd)= 11
n
cosechnπsinnπxsinhnπ(y− 2 )⎤
⎦- A rectangular plate is bounded by the
linesx= 0 ,y= 0 ,x= 3 ,y= 2 .Determine the
potential distributionu(x,y)over the rec-
tangle using the Laplace equation
∂^2 u
∂x^2
+∂^2 u
∂y^2=0, subject to the following
boundary conditions:
u( 0 ,y)= 00 ≤y≤2,
u( 3 ,y)= 00 ≤y≤2,
u(x, 2 )= 00 ≤x≤3,
u(x, 0 )=x( 3 −x) 0 ≤x≤3.
⎡
⎣u(x,y)=^216
π^3∑∞
n(odd)= 11
n^3 cosech2 nπ
3 sinnπx
3 sinhnπ
3 (^2 −y)⎤
⎦