unit – i fourier series problem 1 - … – i fourier series ... expand f(x) = cosx in a fourier...
TRANSCRIPT
UNIT – I
FOURIER SERIES
PROBLEM 1:
The turning moment T lb feet on the crankshaft of steam engine is given for a series of values of the crank angle degrees
0 30 60 90 120 150 180T 0 5224 8097 7850 5499 2626 0
Expand T in a series of sines.
Solution:
Let T = b1 sin + b2 sin2 + b3 sin3 + ………,Since the first and last values of T are repeated neglect last one
T T sin T sin2 T sin3
0 0 0 0 030 5224 2612 4524.11 522460 8097 7012.2 7012.2 090 7850 7850 0 -7850120 5499 4762.27 - 4762.27 0150 2624 1313 - 2274.18 2626
Total 23549.47 4499.86 0
b1 = 2 [mean value of T sin ]
= 2[23549.47
6] = 7849.8
b2 = 2 [mean value of T sin2 ]
= 2[6
86.4499] = 1499.95
b3 = 2 [mean value of T sin2 ] = 0
f ( x ) = 7849.8 sin + 1499.95 sin2
PROBLEM 2:
Analyze harmonically the given below and express y in Fourier series upto the third harmonic.
x 0
3
3
2
3
43
5 2
y 1.0 1.4 1.9 1.7 1.5 1.2 1Solution:
Since the last value of y is a repetition of the first, only the six values will be used. The length of the interval is 2.
Let y = 2
a0+ (a1 cosx + b1 sinx) + (a2 cos2x + b2 sin2x) + (a3 cos3x + b3 sin3x) + … (1)
x y cosx sinx cos2x sin2x cos3x sin3x
0 1.0 1 0 1 0 1 0
3
1.4 0.5 0.866 - 0.5 0.866 -1 0
3
2 1.9 - 0.5 0.866 - 0.5 - 0.866 1 0
1.7 -1 0 1 0 -1 0
3
4 1.5 - 0.5 - 0.866 - 0.5 0.866 1 0
3
5 1.2 0.5 - 0.866 - 0.5 - 0.866 -1 0
a0 = 2 [mean value of y]
= 6
2(8.7) = 2.9
a1 = 2 [mean value of y cosx]
= 6
2[1 (1.0) + 0.5 (1.4) – 0.5 (1.9) – 1 (1.7) – 0.5 (1.5) + 0.5 (1.2)]
= - 0.37b1 = 2 [mean value of y sinx]
= 6
2[0.866 (1.4 + 1.9 -1.5 – 1.2)]
= 0.17a2 = 2 [mean value of y cos2x]
= 6
2[1(1.0 + 1.7) – 0.5 (1.4 + 1.9 + 1.5 + 1.2)]
= - 0.1b2 = 2 [mean value of y sin2x]
= 6
2[0.866(1.4 – 1.9 + 1.5 – 1.2)]
= - 0.06a3 = 2 [mean value of y cos3x]
= 6
2[1.0 – 1.4 + 1.9 – 1.7 + 1.5 – 1.2]
= 0.03b3 = 2 [mean value of y sin3x] = 0
Hence y = 1.45 + (-0.37 cosx + 0.17 sinx) – (0.1 cos2x + 0.06 sin2x) + 0.03 cos3x.
PROBLEM 3:
Find the Fourier series expansion for the function f(x) = x sinx in 0 < x < 2 and deduce
that 4
2.........
7.5
1
5.3
1
3.1
1
SOLUTION:
The Fourier series is
0n n
n 1
af (x) (a cosnx b sin nx)..................(1)
2
Given f(x) = x sinx
a0 = 12
0
dx)x(f
= 12
0
xdxsinx
= 1 2
0 sinx)](-(1)- cosx)(-[x
= 1
(-2) = -2
an = 12
0
nxdxcos)x(f
= 12
0
nxdxcosxsinx
= 2
1
2
0
dx]x)n1sin(x)n1sin(x[
)]BA(Sin)BA(SinSinACosB2[ 2
2 20
1 cos(1 n)x cos(1 n)x sin(1 n)x sin(1 n)xx( ) ( )
2 1 n 1 n ( 1 n) (1 n)
= 1 cos 2(1 n) cos 2(1 n)[ 2 ( )]
2 1 n 1 n
(
1 1[ ]1 n 1 n
21 n 1 n
[ ]1-n
= n 22
a where n 1n 1
a1 = 12
0
xdxcos)x(f
= 12
0
xdxcosxsinx
= 2
12
0
xdx2sinx
= 2
1
2
0
)4
x2sin)(1()
2
x2cos(x
= 2
1[2 (-
2
4cos ) = -
2
1
)2
4cos(2[
2
1
bn = 2
0
1f (x)sin nxdx
= 2
0
1x sin xsin nxdx
= 2
0
1x[cos(1 n)x cos(1 n)x]dx
2
)]BA(Cos)BA(CosSinASinB2[
=
2
2 20
1 sin(1 n)x sin(1 n)x cos(1 n)x cos(1 n)xx( ) ( )
2 1 n 1 n (1 n) (1 n)
= 2 2 2 2
1 cos2(1 n) cos2(1 n) cos0 cos0
2 (1 n) (1 n) (1 n) (1 n)
= 2 2 2 2
1 1 1 1 1
2 (1 n) (1 n) (1 n) (1 n)
= 0 where n 1
b1 = 2
0
1f (x)sin xdx
= 2
0
1x sin x sin xdx
= 2
2
0
1x sin xdx
= 2
0
1 1 cos2xx( )dx
2
= 2
0
1x(1 cos2x)dx
2
22
0
1 sin 2x x cos2xx(x ) (1)( )
2 2 2 4
= 21 1 1[2 ]
2 4 4
=
Substitute in (1), we get
f (x) = 2
n 2
1 21 cos x cosnx sin x .......(2)
2 n 1
+ ……………….(2)
Deduction Part:
Put x [x isa point of continuity]2
[x=
2
is a point of continuity]
f (2
) = f ( ) sin
2 2 2 2
=
2
( sin
2
= 1)
(2) 2
n 2
2 n(2) 1 0 cos
2 2n 1
= -1-0+
2n 2
1 n1 2 cos
2 2n 1
- +1 =
-n 2
1 n1 2 cos
2 (n 1)(n 1) 2
+1 =
-n 2
1 1 ncos
4 2 (n 1)(n 1) 2
=
= n 2
1 n 2cos
(n 1)(n 1) 2 4
4
2........)1(
5.3
1)0(
4.2
1)1(
3.1
1
]4
2[........]
7.5
1
5.3
1
3.1
1)[1(
]4
2[........
7.5
1
5.3
1
3.1
1
Problem:4
Find the Fourier series for f(x) = xsin in - < x < Solution:
Given f(x) = xsin
This is an even function. bn = 0
f(x) = )1.......(..............................Cosnxa2
a
1nn
0
a0 =
0
dx)x(f2
=
0
dxSinx2
=
0
xdxsin2
= 0]xcos[2
= -2
(cos cos0)
= -2
[ 1 1]
[-1-1]
= 4
an =
0
nxdxcos)x(f2
=
0
nxdxcosSinx2
=
0
nxdxcosxsin2
=
0
dx]x)1nsin(x)1n[sin(2
12
=
0]1n
x)1ncos(
1n
x)1ncos([
1
= ]1n
1
1n
1
1n
)1ncos(
1n
)1ncos([
1
= ]1n
1
n1
1
1n
)1(
1n
)1([
1 1n1n
= ]1n
1
1n
1
1n
)1(
1n
)1([
1 nn
= ]1n
1)1(
1n
1)1([
1 nn
= n( 1) 1 1 1
[ ]n 1 n 1
= n
2
( 1) 1 n 1 n 1[ ]
n 1
= n
2
( 1) 1 2[ ]n 1
=
)1n(
]1)1[(22
n
if n 1.
a1 =
0
xdxcos)x(f2
=
0
xdxcosSinx2
=
0
xdxcosxsin2
=
0
xdx2sin2
12
= 0]2
x2cos[
1
= ]2
1
2
1[
1
= 0
Substitute in equation (1), we get
f(x) = nxcos)1n(
]1)1[(20
2
2n2
n
Hence f(x) = nxcos)1n(
]1)1[(22
2n2
n
Problem:5
Expand f(x) = xcos in a Fourier series in the interval (-,)
Solution:
Given f(x) = xcos .
This is an even function, bn = 0
f(x) = )1.......(..............................Cosnxa2
a
1nn
0
a0 =
0
dx)x(f2
=
0
dxxcos2
= /2
0 /2
2[ cos xdx (cosx)dx]
= /2
0 /2
2[ cos xdx cos xdx]
= /20 /2
2[(sin x) (sin x) ]
= 2
[(1 0) (0 1)]
a0 = 4
an =
0
nxdxcos)x(f2
=
0
nxdxcosxcos2
= /2
0 /2
2[ cos x cosnxdx (cos x)cosnxdx]
= /2
0 /2
2 2cos x cosnxdx cos x cosnxdx
= /2
0 /2
1 1[cos(n 1)x cos(n 1)x]dx [cos(n 1)x cos(n 1)x]dx
-
= /20 /2
1 sin(n 1)x sin(n 1)x 1 sin(n 1)x sin(n 1)x[ ] [ ]
n 1 n 1 n 1 n 1
= 1 sin(n 1) / 2 sin(n 1) / 2 1 sin(n 1) / 2 sin(n 1) / 2
[ ] [ ]n 1 n 1 n 1 n 1
= ]1n
2/)1nsin(
1n
2/)1nsin([
2
= ]1n
)2/2/sin(
1n
)2/n2/sin([
2
= ]1n2
ncos
1n2
ncos
[2
= ]1n
1
1n
1[
2
ncos
2
= ]1n
1n1n[
2
ncos
22
= ]1n
2[
2
ncos
22
= ]1n
1[
2
ncos
42
= ]1n
1[
42
if n is even
= 0 if n is odd [n 1]Hence
f(x) = 2
n 2
2 4cosnx
(n 1)
Complex Form of a Fourier series:
Problem 6:
Find the complex form of the Fourier series of the function f(x) = ex when - < x < and f(x+2) = f(x).
Solution:
We know that f(x) =
n
inxneC -----------------------------(1)
Cn =
inxe)x(f
2
1dx
=
inxxee
2
1dx
=
dx(e
2
1 x)in1(
=
in1
e
2
1 x)in1(
=
x)in1(e
)in1(2
1
= ]ee[)in1(2
1 )in1()in1(
= ]eeee[)in1(2
1 inin
nnin )1(0i)1(nsinincose
nnin )1(0i)1(nsinincose
= ])1(e)1(e[)in1(2
1 nn
= ]2
ee[
)in1(2
)1( n
Cn =
sinh)n1(
)in1()1(22
n
(1) f(x) =
n
sinh)n1(
)in1()1(22
n
einx
i.e., ex =
n
inx22
n en1
in1)1(
sinh
Problem7:Find the complex form of the Fourier series of f(x) = Cosax in ( -, ) where ‘a’ is neither zero nor an integer.
Solution:
Here 2c = 2 or c = .
We know that f(x) =
n
inxneC -----------------------------(1)
Cn =
inxe)x(f
2
1dx
= dxCosaxe2
1 inx
=
)]axsinaaxcosin(na
e[
2
122
inx
= )asinaacosin(e)asinaacosin(e[)na(2
1 inin22
= )]ee(asina)ee(acosin[)na(2
1 inininin22
= )]ncos2(asina)nsini2(acosin[)na(2
122
Cn =
asina2)1()na(2
1 n22
Hence (1) becomes
Cosax = inx
n22
n
e)na(
)1(
2
asina
Problem8:
Find the complex form of the Fourier series of f(x) = e-x in -1 ≤ x ≤ 1
Solution:
We know that f(x) =
n
xinneC -----------------------------(1)
Cn = dxe)x(f2
1 1
1
xin
= dxee2
1 1
1
xinx
= dxe2
1 1
1
x)in1(
=
]
)in1(
e[
2
1 x)in1(
= )in1(2
ee )in1(in1
= )in1(2
)nsinin(cose)nsinin(cose 1
= )in1(2
)1(e)1(e n1n
=
= 2
)1)(ee( n1
22n1
)in1(
= 1sinhn1
)in1()1(22
n
Hence (1) becomes e-x = xin22
n
n
e1sinhn1
)in1()1(
UNIT- 2
FOURIER TRANSFORMS
1. Find the Fourier transform of xae , if 0a . Deduce that
0
3222 4)( aax
dx
Sol:
The Fourier transform of the function )(xf is
dxexfxfF isx)(2
1)(
.
)(
2
cos22
1
0cos22
1
sincos2
1
)sin(cos2
1
2
1)(
22
0
0
as
a
sxdxe
sxdxe
sxdxeisxdxe
dxsxisxe
dxeexfF
ax
xa
xaxa
xa
isxxa
By Parseval’s identity
dxxfdssF22
)()(
22
a x
2 2
2 2ax2ax
2 2 20 0
2 ads e dx
(s a )
2a 1 eds 2 e dx 2
(s a ) 2a
10 1
a
2 2 2 30
2 2 2 30
12 ds
(s a ) 2a
dx
(x a ) 4a
2.Show that the Fourier transform of
0;0
;)(
22
ax
axxaxf
is 3
2 sinsa sacossa2
s
. Hence deduce that4
cossin
03
dtt
ttt.
Sol:
The Fourier transform of the function )(xf is
dxexfxfF isx)(2
1)(
.
Given 22)( xaxf in axa and 0 otherwise.
3
32
032
22
0
22
2222
22
22
cossin22)(
000sin2cos2
02
sin)2(
cos)2(
sin)(
2
0cos)(22
1
sin)(cos)(2
1
)sin)(cos(2
1
)(2
1)(
s
sasasasF
s
sa
s
saa
s
sx
s
sxx
s
sxxa
sxdxxa
sxdxxaisxdxxa
dxsxisxxa
dxexaxfF
a
a
a
a
a
a
a
a
a
a
isx
By inverse Fourier transform
dsesFxf isx)(2
1)(
03
33
3
3
0coscossin
22
)(
sincossin
coscossin2
)sin(coscossin2
cossin22
2
1)(
sxdss
sasasaxf
sxdss
sasasaisxds
s
sasasa
dssxisxs
sasasa
dses
sasasaxf isx
Put 1&0 ax , we get
4
cossin.
cossin41
03
03
dss
sssei
dss
sss
Replace s by t , we get4
cossin
03
dtt
ttt
3.Find the Fourier transform of
1;0
1;1)(
xif
xifxxf . Hence deduce that
3
sin4
0
dtt
tand dt
t
t2
0
sin
.
Sol:
The Fourier transform of the function )(xf is
dxexfxfF isx)(2
1)(
.
Given xxf 1)( in 11 x and 0 otherwise.
2
22
1
02
1
0
1
1
1
1
1
1
1
1
cos12)(
10
cos0
2
cos)1(
sin)1(
2
0cos)1(22
1
sin)1(cos)1(2
1
)sin)(cos1(2
1
)1(2
1)(
s
ssF
ss
s
s
sx
s
sxx
sxdxx
sxdxxisxdxx
dxsxisxx
dxexxfF isx
By inverse Fourier transform
dsesFxf isx)(2
1)(
02
22
2
2
0coscos1
21
)(
sincos1
coscos11
)sin(coscos11
cos12
2
1)(
sxdss
sxf
sxdss
sisxds
s
s
dssxisxs
s
dses
sxf isx
Put 0x , we get
1.
22sin2
.
cos121
02
2
02
dss
sei
dss
s
Let ,2
ts then dtds 2
ts
ts 00
2
sin
22
2
sin2
02
2
02
2
dtt
t
dtt
t
By Parseval’s identity
dxxfdssF22
)()(
2 12
21
2 12
20
2 12
20
2 1 cossds 1 x dx
s
2 1 cossds 2 1 x dx
s
2 1 cossds 2 1 x dx
s
2 1
22
0 0
2 1 coss2 ds 2 1 x 2x dx
s
12 3 2
20 0
2 1 coss x 2xds x
s 3 2
0
2
2
2
0
2
2
32sin2
2
00013
11
cos12
dss
s
dss
s
Let ,2
ts then dtds 2
ts
ts 00
3
sin
3
sin
32
2
sin22
0
4
0
2
2
2
0
2
2
2
dtt
t
dtt
t
dtt
t
4.Find the Fourier transform of22xae . Hence prove that 2
2x
e
is self reciprocal.
Sol:
The Fourier transform of the function )(xf is
dxexfxfF isx)(2
1)(
.
dxee
dxe
dxe
dxe
dxeeeF
a
isax
a
s
a
s
a
isax
isxxa
isxxa
isxxaxa
2
2
2
2
22
22
22
2222
24
42
)(
)(
2
1
2
1
2
1
2
1
2
1
Put ya
isax
2, then
a
dydx
yx
yx
2
2
22
2
2
22
2
4
4
4
2
1
2
1
2
1
a
sxa
a
s
ya
s
ea
eF
ea
a
dyee
Put2
1a , then
22
2
2
22
22
2
14
2
1
22
11
sx
s
x
eeF
eeF
2
2x
e
is self reciprocal with respect to Fourier transform.
5. Find Fourier sine and cosine transform of 1nx and hence prove that
C
1 1F
x s
Sol:
We know that
0
1n
nnax
adxxe
Put isa ,
0
1n
nnisx
isdxxe
nn 1n n
0
nn 1 n 1n n
0 0
nn 1 n 1n
0 0 n
cossx isin sx x dxi s
x cossxdx x isin sxdxi s
x cossxdx i x sin sxdx
cos i sin s2 2
n
nn 1 n 1
n0 0
cos i sin2 2
x cossxdx i x sin sxdxs
nn 1 n 1
n0 0
n nn 1 n 1
n n0 0
n ncos isin
2 2x cossxdx i x sin sxdx
s
n ncos i sin
2 2x cossxdx i x sin sxdx
s s
Equating Real and imaginary parts, we get
n
nn
n
nn
s
n
sxdxx
s
n
sxdxx
2sin
sin
2cos
cos
0
1
0
1
n
nn
S
n
nn
C
n
nn
n
nn
s
n
xF
s
n
xF
s
n
sxdxx
s
n
sxdxx
2sin
2
2cos
2
2sin
2sin
2
2cos
2cos
2
1
1
0
1
0
1
Put2
1n in the above results, we get
1
211
2C 1
2
12c o s2
2F x
s
1
2C
cos42
F xs
C
C
11 2 2Fx s
1 1F
x s
Thereforex
1is self reciprocal with respect to Fourier cosine transform.
5. Find Fourier sine transform and Fourier cosine transform of 0, ae xa . Hence
evaluate
0222
2
)(dx
xa
xand
02222 )()( bxax
dx.
Sol:
The Fourier sine transform of the function )(xf is
0
sin)(2
)( sxdxxfxfFS .
22
0
2
sin2
as
s
sxdxeeF axaxS
The Fourier cosine transform of the function )(xf is
0
cos)(2
)( sxdxxfxfFC .
22
0
2
cos2
as
a
sxdxeeF axaxC
By parseval’s identity
0
2
0
2dxxfdsxfFS
0
2
0
2
22
0
2
0
2
22
2
2
dxedsas
s
dxedsas
s
ax
ax
ads
as
s
ads
as
s
a
eeds
as
s
a
eds
as
s
dxedsas
s
aa
ax
ax
4
2
12
2
2
2
2
2
0222
2
0222
2
022
0222
2
0
2
0222
2
0
2
0222
2
Replace s by x , we get
adx
ax
x
40222
2
Let axexf and bxexg
By parseval’s identity
00
dxxgxfdsxgFxfF CS
002222
002222
2..
22
dxedsasas
abei
dxeedsas
b
as
a
xba
bxax
baabasas
ds
baasas
dsab
ba
ee
asas
dsab
ba
e
asas
dsab
baba
xba
2
12
2
2
02222
02222
0
02222
002222
Replace s by x , we get
baabaxax
dx
202222
6. Verify Parseval’s theorem of Fourier transform for the function
0;
0;0)(
xe
xxf
x.
Sol:
By Parseval’s theorem
dssFdsxf22
…………(1)
Given
0;
0;0)(
xe
xxf
x
0
1
0
1
0
0
12
1
2
1
2
1
2
1
is
e
dxe
dxe
dxeesF
xis
xis
isxx
isxx
1
1
2
1
11
1
2
1
1
1
2
1
1
1
1
0
2
1
112
1
2
011
s
issF
isis
is
is
isis
is
e
is
esF
isis
L. H. S. of (1)
22 x
0
2x
0
f x dx e dx
e dx
2x
0
2 0
e
2
e e
2
1
2
R. H. S. of (1)
s
s
ds
dss
s
dss
is
dss
is
dss
isdssF
1
2
22
2
22
2
2
2
2
2
2
tan2
1
12
1
1
1
2
1
1
1
2
1
1
1
2
1
1
1
2
1
2
1
2
1
222
1
tantan2
1
tantan2
1
11
112
dssF
L. H. S of (1) = R. H. S of (1)Hence Parseval’s theorem verified.
7. Solve for )(xf from the integral equation
0
cos)( exdxxf .
Sol:
0
cos)( exdxxf
edxxe2
cos22
0
Inverse Fourier cosine transform is
2
2
2
02
0
1
2
1
12
)1(1
10
2
)sincos.1(1
2
cos22
)(
x
x
x
xxxx
e
dxexf
UNIT-3
PARTIAL DIFFERENTIAL EQUATIONS
Problems on Lagrange’s equations
1. Solve p x q y z
Solution:
p x q y z ------ ( 1 )
This is of the form Pp + Qq = R
Here , , P x Q y R z
Subsidiary equations are
dx dy dz
P Q R
dx dy dz
x y z
Grouping the first two members
dx dy
x y
Integrating
1 1
2 2
1 1
2 2
1 1
2 2
2( )
2
2
dx dx
x y
x dx y dy
x y a
x y a
ax y
ax y c c
u x y
|||ly Grouping the another two members
dy dz
y z
Integrating
2 2
2 2
2( )
2
2
y z b
y z b
by z
by z c c
v y z
The general solution of (1) is (u,v) = 0
, 0 x y y z
2. Obtain the general solution of pzx + qzy = xy.
Solution:
Given pzx + qzy = xy
This is of the form Pp + Qq = R
Here P = zx, Q= zy, R = xy
Subsidiary equations are
dx dy dz
P Q R
dx dy dz
zx zy xy
Grouping the first two members
dx dy
zx zy
Integrating
(1)
dx dy
x y
og x og y og a
xog og a
y
xa
y
xu
y
Grouping the another two members
( (1) )
dy dz
zy xy
dy dz
z xdy dz
from x ayz ay
ay dy z dz
Integrating
2 2
2 2
2 2
2 2
2
2
2 2
2
( 2 )
.
ay zc
ay z c
ay z b b c
xy z b
y
xy z b
v xy z
The general solution is (u,v) = 0
2, 0
xxy z
y
3. Solve 2 2 2 2 2 2( ) ( ) ( ) 0 x y z p y z x q z x y
Solution:
Given 2 2 2 2 2 2( ) ( ) ( ) 0 x y z p y z x q z x y
This is of the form Pp + Qq = R
Here 2 2( ) P x y z , 2 2( ) Q y z x , 2 2( ) R z x y
Subsidiary equations are
2 2 2 2 2 2(1)
( ) ( ) ( )
dx dy dz
P Q R
dx dy dz
x y z y z x z x y
Choose the set of multipliers x,y,z
0dzzdyydxx
0
dzzdyydxx
)yx(z)xz(y)zy(x
dzzdyydxx
)yx(z
dz
)xz(y
dy
)zy(x
dx222222222222
Integrating we get
2 2 2
2 2 2
2 2 2
2 2 2
( 2 )
x y zc
x y z a et c a
u x y z
|||ly consider another set of multipliers z
1,
y
1,
x
1
Each member of (1)
0z
dz
y
dy
x
dx0
z
dz
y
dy
x
dx
yxxzzy
z
dz
y
dy
x
dx
222222
Integrating we get
log x + log y + log z = log b
log(xyz) = log b
xyz = b
v = xyz
The general solution is (u,v) = 0
2 2 2, 0 x y z x y z
4. Solve 2 2 2(x y z ) p ( y z x ) q z x y
Solution:
Given 2 2 2(x y z ) p ( y z x ) q z x y
This is of the form Pp + Qq = R
Here 2 P x y z , 2 Q y z x , 2 R z x y
Subsidiary equations are
2 2 2(1)
) )
dx dy dz
P Q R
dx dy dz
x y z y z x z x y
Each of (1)
2 2 2 2 2 2
d x d y d y d z d x d z
x y z (x y ) y z x (y z ) x z y (x z )
d ( x y ) d ( y z ) d ( x z )i.e.,
( x y) ( x y z ) ( y z ) ( x y z ) ( x z) ( x y z )
d ( x y ) d ( y z ) d ( x z )i.e.,
x y y z x z
Grouping the members
d ( x y ) d ( y z )
( x y) ( y z )
Integrating we get
1
1
1
log ( x y ) log ( y z) log c
x ylog log c
y z
x yc
y z
x yu
y z
Choose multipliers x , y , z
3 3 3
2 2 2
x dx y dy z dzEach of 1 ( 2 )
x y z 3x y z
Choose another set of multipliers 1,1,1
x dx ydy z dzEach of 1 (3)
x y z y z z x x y
3 3 3 2 2 2
2 2 2 2 2 2
x dx ydy z dz x dx ydy z dz
x y z 3x y z x y z y z z x x y
x dx ydy z dz x dx ydy z dzi.e.,
(x y z ) ( x y z y z z x x y) x y z y z z x x y
x dx ydy z dzdx dy dz
(x y z )
(x y z )d( x y z ) x dx y dy z dz
Integrating weget
( x y z
2 2 2 2
2 2 2 2
2
) x y zb
2 2 2 2
( x y z ) ( x y z )b
2x y y z zx 2b
x y y z zx c
v x y y z zx
x yThe generalsolution is , x y y z zx 0.
y z
5. Solve ( mz – ny ) p + ( nx – lz ) q = ly – mx
Solution:
Given ( mz – ny ) p + ( nx – lz ) q = ly – mx
This is of the form Pp + Qq = R
Here P = mz – ny, Q = nx – lz, R = ly – mx
The Subsidiary equations are
(1)
dx dy dz
P Q R
dx dy dz
m z n y n x l z l y m x
Choose a set of multipliers x, y ,z
x dx ydy z dzEach of 1
x ( m z n y) y ( n x z ) z ( y m x )
x dx ydy z dz
0x dx ydy z dz 0
l l
Integrating we get
x 2 + y 2 + z 2 = a
v = x 2 + y 2 + z 2
Choose a set of multipliers l, m ,n
dx m dy n dzEach of 1
( m z n y) m ( n x z ) n ( y m x )
dx m dy n dz
0dx m dy n dz 0
Integrating we get
x m y n z b
v x m y n z
l
l l l
l
l
l
l
The general solution is (u,v) = 0
2 2 2, 0 x y z l x m y n z
Equations reducible to standard types:
Type V:
Equations of the form
f ( x m p , y n q ) = 0 or f (x m p, y n q , z ) = 0 ------------(1)
where m and n are constants
This type of equations can be reduced to Type I or Type III by the following
transformations.
Case ( i ):
If m 1 , n 1
Put x 1 – m = X, y 1– n = Y
m
m
n
n
z z X zp (1 m) x P , where P
x X x X
i.e., x p ( 1 m ) P
z z Y zq (1 n ) y Q, where Q
y Y y Y
i.e., y q ( 1 n )Q
m n
m n
The equation f x p, y q 0 reduces to f (1 m) P , (1 n )Q 0 which is a Type I eqn.
The equation f x p, y q , z 0 reduces to f (1 m) P ,(1 n )Q ,z 0 which is aType III eqn.
Case ( ii ):
If m = 1, n = 1 then (1) f (x p ,y q ) = 0 or f (x p ,y q, z ) = 0
Put log x = X, log y = Y.
z z X z 1 1 zp P, where P
x X x X x x Xi.e., x p P
z z Y z 1 1 zq Q, where Q
y Y y Y y y Y
i.e., y q Q
The equation f ( px , qy) = 0 reduces to f ( P, Q) = 0 which Type I eqn.
The equation f ( px , qy , z ) = 0 reduces to f ( P, Q , z ) = 0 which Type III eqn.
Type VI :
Equations of the form f ( z k p, z n q) = 0 or f (z k p , y n q , x , y ) = 0 ------------(1)
Where k is constant,
This can be reduced to Type I or Type IV equations by the following
substitutions.
Case ( i ):
If k –1
Put Z = z k + 1
k
k
k
k
k k
k k
Z Z zP ( k 1) z p
x z xP
z pk 1
Z Z zQ ( k 1) z q
y z y
Qz q
k 1
P QThe equation f z p, z q 0 reduces to f , 0 which is a Type I eqn.
k 1 k 1
PThe equation f z p, z q , x , y 0 reduces to f ,
k 1
Q, x , y 0 which is a Type IV eqn..
k 1
Case ( ii ):
If k = –1
( 1 )
p q p qf , 0 or f , , x , y 0
z z z z
put z log z
z Z z 1P p.
x z x zp
Pz
z Z z 1Q q.
y z y z
zp q
The eqn f , 0 reduces to Type Ieqn.z z
p qThe eqn f , , x , y 0 reduces to Type IV eqn.
z z
k
k
k
k
k k
k k
Z Z zP ( k 1) z p
x z xP
z pk 1
Z Z zQ ( k 1) z q
y z y
Qz q
k 1
P QThe equation f z p, z q 0 reduces to f , 0 which is a Type I eqn.
k 1 k 1
PThe equation f z p, z q , x , y 0 reduces to f ,
k 1
Q, x , y 0 which is a Type IV eqn..
k 1
1– m
2
1
H ere m – 1 , n 1.
Pu t X x
X = x Y log y
X Y 12 x
x y y
z zp q
x y
z X z Y
X x Y y
1p P 2 x q Q
y
x p 2 P y q Q
2 2 2 1 Reduces to 2 P Q z 2
This is of the form f p, q , z 0 which is TypeIII eqn.
Let z f X a Y be a trial Solution.
z f u
u X a Y
u u1 , a
x y
z zP Q
X Yz u z u
u X u Yz z
P Q au u
2 22
22 2
2 2
2
2
2
( 2 ) r e d u c e s t o
d z d z2 a z
d u d u
d z( 4 a ) z
d u
d z z
d u ( 4 a )
d z z.
d u 4 a
d z d u
z 4 a
Integrating
2
2
2
22
2
dz du
z 4 a
1log z u c
4 a
X a Ylogz c
4 a
x a log ylogz c ( X x , Y log y)
4 a
which is the complete int egral
2 2 2 2 2
2 2 2 2 2
2 2 2 2
k k
2 . Solve z p q x y
Solution :
Given z p q x y
z p z q x y 1
This is of the form f z p, z q , x , y 0 Type VI
1 1
2
Here k 1
Put Z z
Z z
Z zP Q
x y
Z z Z z
z x z y
P 2 z p Q 2 z q
P Qz p z q
2 2
2 2 2 2
2 2 2 2
1 2
Eqn 1 reduces to P Q 4 x y
i.e., P – 4 x 4 y – Q
This is of Standard Type IV f x , p f y , q
2 2 2 2
2 2 2 2
2 2 2 2
L e t P – 4 x 4 y – Q 4 c s a y
P – 4 x 4 c 4 y – Q 4 c
P 4 c 4 x Q 4 y – 4 c
2 2
2 2
2 2
P 4 ( c x ) Q 4( y c)
P 2 ( c x ) Q 2 ( y c)
W e know
Z ZdZ dx dy
x y
Z ZdZ P dx Q dy P , Q
x x
dZ 2 ( c x ) dx 2 ( y c ) dy
2 2
2 1 2 1
2 2 1 2 1 2
dZ 2 (c x ) dx 2 ( y c) dy
x c x y c xZ 2 ( c x ) sinh ( y c) cosh a
2 2 2 2C C
x xz x (c x ) c sinh y ( y c) ccosh a ( Z z )
C C
Examples:
1.Form the p d e by eliminating the arbitrary
function f
from the relation z = f(x2 + y2)
Solution:
Given z = f(x2 + y2)--------------(1)
Differentiating (1) partially w. r.t. x and y,
' 2 2
' 2 2
zp f ( x y ) .2 x, ( 2 )
x
zq f (x y ) .2 y ( 3 )
y
( 2 ) p x
(3) q y
y p x q 0
y p x q 0 is the required partial differential equation.
2.Form the p d e by eliminating the arbitrary functions f and g
from the relation z = f ( x +ct ) + g (x -ct) .
Solution :
Given z = f ( x +ct ) + g (x -ct) -------- ( 1 )
2
2
D i f f e r e n t i a t i n g 1 p a r t i a l ly w i t h r e s p e c t t o x
zf '( x c t ) g '( x c t ) ,
x
A g a i n d i f f e r e n t i a t i n g p a r t i a l ly w i t h r e s p e c t t o x
zf " ( x c t ) g " ( x c t ) ( 2 )
x
22 2
2
22
2
2 2
2 2
D i f f e r e n t i a t i n g 1 p a r t i a l l y w i t h r e s p e c t t o t
zc f '( x i t ) c g '( x i t ) ,
t
A g a i n d i f f e r e n t i a t i n g p a r t i a l l y w i t h r e s p e c t t o t
zc f " ( x i t ) c g " ( x i t )
t
zc ( 3 )
x
z z( 2 ) ( 3 ) 0
t x
UNIT-IV
Applications of Partial Differential Equations
1. A tightly stretched string with fixed end points 0x and lx is initially in a position
given by
l
xyxy
30 sin)0,( . It is released from rest from this position. Find the
displacement at any time ''t .
Sol:
One dimensional wave equation 2
22
2
2
x
ya
t
y
The conditions are
(i) 0),0( ty (ii) 0),( tly (iii) 0)0,(
xt
y
(iv)
l
xyxy
30 sin)0,(
The correct solution which satisfying the given boundary conditions is
)sincos)(sincos(),( 4321 patcpatcpxcpxctxy -------------(1)
Apply the boundary condition (i) in (1)
),0( ty 0)sincos( 431 patcpatcc
weget 01 c , then equation (1) becomes
)sincos(sin),( 432 patcpatcpxctxy -------------(2)
Apply the boundary condition (ii) in (2)
0)sincos(sin),( 432 patcpatcplctly
weget 0sin pl npl l
np
Then equation (2) becomes
l
atnc
l
atnc
l
xnctxy
sincossin),( 432 -------------- (3)
Differentiate Partially (3) w.r.to t
),( txt
y
l
atn
l
anc
l
atn
l
anc
l
xnc
cossinsin 432 -------- (4)
Apply the boundary condition (iii) in (4)
0sin)0,( 42
l
anc
l
xncx
t
y
weget 04 c , then equation (3) becomes
l
atn
l
xnc
l
atnc
l
xnctxy
n
cossin
cossin),( 32
where 32cccn
By the superposition principle, the most general solution is
1
cossin),(n
n l
atn
l
xnctxy
----------------(5)
Apply the boundary condition (iv) in (5)
l
x
l
xyl
xc
l
xc
l
xc
l
xc
l
x
l
xy
l
xy
l
xncxy
nn
3sinsin3
4
..........4
sin3
sin2
sinsin
3sinsin3
4
sin1.sin)0,(
0
4321
0
30
1
0..........,4
,0,4
354
032
01 cc
ycc
yc
The equation (5) becomes
l
at
l
xy
l
at
l
xytxy
3cos
3sin
4cossin
4
3),( 00
2. A tightly stretched string of length l has its ends fastened at 0x and lx . The midpoint of the string is then taken to a height h and then released from rest in that position. Obtain an expression for the displacement of the string at any subsequent time.
Sol:
One dimensional wave equation is 2
22
2
2
x
ya
t
y
The conditions are
(i) 0),0( ty (ii) 0),( tly (iii) 0)0,(
xt
y
(iv) )()0,( xfxy
,0l(0,0),0
2
l
,2
lh
X
Y
Consider the interval ),0( 2l , the end points are ),(),0,0( 2 hl
Using two point formula for the straight line
l
hxy
x
h
y
xx
xx
yy
yy
l
2
0
0
0
0
2
21
1
21
1
Consider the interval ),( 2 ll , the end points are )0,(),,( 2 lhl
Again by two point formula for the straight line
)(2
0 2
2
21
1
21
1
xll
hy
l
x
h
hy
xx
xx
yy
yy
l
l
lxxll
h
xl
hx
xfl
l
2
2
),(2
0,2
)(
The correct solution which satisfying the given boundary conditions is
)sincos)(sincos(),( 4321 patcpatcpxcpxctxy -------- (1)
Apply the boundary condition (i) in (1)
),0( ty 0)sincos( 431 patcpatcc
weget 01 c , then equation (1) becomes
)sincos(sin),( 432 patcpatcpxctxy ------- (2)
Apply the boundary condition (ii) in (2)
0)sincos(sin),( 432 patcpatcplctly
weget 0sin pl npl l
np
, then equation (2) becomes
l
atnc
l
atnc
l
xnctxy
sincossin),( 432 ------- (3)
Differentiate Partially (3) w. r. to t
),( txt
y
l
atn
l
anc
l
atn
l
anc
l
xnc
cossinsin 432 ------- (4)
Apply the boundary condition (iii) in (4)
0sin)0,( 42
l
anc
l
xncx
t
y
We get 04 c then equation (3) becomes
l
atn
l
xnc
l
atnc
l
xnctxy
n
cossin
cossin),( 32
where 32cccn
By the superposition principle, the most general solution is
1
cossin),(n
n l
atn
l
xnctxy
(5)
Apply the boundary condition (iv) in (5)
)0,(xy
1
1.sinn
n l
xnc
)(
),(2
0,2
2
2
xf
lxxll
h
xl
hx
l
l
To find the value of nc , expand )(xf in a half range Fourier sine series
We know that half range Fourier sine series of )(xf is
1
sin)(n
n l
xnbxf
(7)
By comparing (6) and (7), we get nn cb
To find nc it is enough to find nb
l
l
n
l
l
dxl
xnxl
l
hdx
l
xn
l
hx
l
dxl
xnxf
lb
2
2
sin)(2
sin22
sin)(2
0
0
n
l
c
n
n
h
n
n
l
l
h
n
n
ln
n
ln
n
ln
n
l
l
h
l
nl
xn
l
nl
xn
xl
l
nl
xn
l
nl
xn
xl
h
l
l
2sin
82
sin8
2sin
2cos
22sin
2cos
2
4
sin)1(
cos)(
sin1
cos4
22
22
2
2
22
22
22
22
2
2
22
02
222
2
2
The equation (5) becomes
1
22cossin
2sin
8),(
n l
atn
l
xnn
n
htxy
3 . The points of trisection of a string are pulled aside through a distance ‘b’ on
opposite sides of the position of equilibrium and the string is released from rest.
Find an expression for the displacement.
Sol:
One dimensional wave equation is 2
22
2
2
x
ya
t
y
The conditions are
(i) 0),0( ty (ii) 0),( tly (iii) 0)0,(
xt
y
(iv) )()0,( xfxy
Equation of OA:
1 1
1 2 1 2
y y x x y 0 x 0 3bxy in 0, 3
y y x x 0 b 0 3
Equation of AB:
1 1
1 2 1 2
y y x x x 3y b 3by 2x in 3,2 3
y y x x b ( b) 3
Equation of BC:
1 1
1 2 1 2
y y x x y 0 x 0 3by x in 2 3 ,
y y x x 0 b 0 3
3bxin 0, 3
3bf (x) 2x in 3,2 3
3bx in 2 3,
The correct solution which satisfying the given boundary conditions is
)sincos)(sincos(),( 4321 patcpatcpxcpxctxy -------- (1)
Apply the boundary condition (i) in (1)
),0( ty 0)sincos( 431 patcpatcc
weget 01 c , then equation (1) becomes
)sincos(sin),( 432 patcpatcpxctxy ------- (2)
Apply the boundary condition (ii) in (2)
0)sincos(sin),( 432 patcpatcplctly
weget 0sin pl npl l
np
, then equation (2) becomes
l
atnc
l
atnc
l
xnctxy
sincossin),( 432 ------- (3)
Differentiate Partially (3) w. r. to t
),( txt
y
l
atn
l
anc
l
atn
l
anc
l
xnc
cossinsin 432 ------- (4)
Apply the boundary condition (iii) in (4)
0sin)0,( 42
l
anc
l
xncx
t
y
We get 04 c then equation (3) becomes
l
atn
l
xnc
l
atnc
l
xnctxy
n
cossin
cossin),( 32
where 32cccn
By the superposition principle, the most general solution is
1
cossin),(n
n l
atn
l
xnctxy
-------------- (5)
Apply the boundary condition (iv) in (5)
)0,(xy
1
1.sinn
n l
xnc
3bxin 0, 3
3bf (x) 2x in 3,2 3
3bx in 2 3,
To find the value of nc , expand )(xf in a half range Fourier sine series
We know that half range Fourier sine series of )(xf is
1
sin)(n
n l
xnbxf
(7)
By comparing (6) and (7), we get nn cb
To find nc it is enough to find nb
n0
2 n xb f (x)sin dx
2
3 3
n20
3 3
2 3bx n x 3b n x 3b n xb sin dx 2x sin dx x sin dx
nn 2 2
18b nb sin 1 ( 1)
3n
n
2
0 if n is odd
36b nb sin if n is even3n
Substitute nb values in equation (5) we get the required solution
2n 2,4
36b n n x n aty x, t sin sin cos
3n
4. A tightly stretched string with fixed end points 0x and lx is initially at rest
in its equilibrium position. If it is set vibrating by giving each point a
velocity )( xlkx . Find the displacement of the string at any time.
Sol:
One dimensional wave equation 2
22
2
2
x
ya
t
y
The conditions are
(i) 0),0( ty (ii) 0),( tly (iii) 0)0,( xy
(iv) )()0,( xlkxxt
y
The correct solution which satisfying the given boundary conditions is
)sincos)(sincos(),( 4321 patcpatcpxcpxctxy --------(1)
apply the boundary condition (i) in (1)
),0( ty 0)sincos( 431 patcpatcc
weget 01 c then equation (1) becomes
)sincos(sin),( 432 patcpatcpxctxy -----------------(2)
Apply the boundary condition (ii) in (2)
0)sincos(sin),( 432 patcpatcplctly
weget 0sin pl npl l
np
,
then equation (2) becomes
l
atnc
l
atnc
l
xnctxy
sincossin),( 432 --------------(3)
Apply the boundary condition (iii) in (3)
)0,(xy 0sin 32 cl
xnc
here 03 c
then equation (3) becomes
l
atn
l
xnc
l
atnc
l
xnctxy
n
sinsin
sinsin),( 42
where 42cccn
By the superposition principle, the most general solution is
1
sinsin),(n
n l
atn
l
xnctxy
----------------(4)
Differentiate Partially (4) w.r.to t
),( txt
y
l
an
l
atn
l
xnc
nn
1
cossin ------------------------(5)
Apply the boundary condition (iv) in (5)
)0,(xt
y)()(.sin
1
xfxlkxl
an
l
xnc
nn
-------------(6)
To find the value of nc , expand )(xf in a half range Fourier sine series
We know that half range Fourier sine series of )(xf is
1
sin)(n
n l
xnbxf
--------------------------(7)
By comparing (6) and (7), we get nnnn ban
lc
l
ancb
1..200cos.2002
cos2
sin2
cos2
sin)(2
sin)(2
33
3
33
3
03
33
2
222
0
0
n
ln
n
l
l
k
ln
lxn
ln
lxn
xl
ln
lxn
xlxl
k
dxl
xnxlkx
l
dxl
xnxf
lb
l
l
l
n
oddisnifn
kl
evenisnifn
kl
n
l
l
k
n
n
33
2
33
2
33
3
8
0
)1(14
)1(122
44
3
33
2 88
n
kl
n
kl
an
lb
an
lc nn
The equation (4) becomes
5,3,1
44
3
sinsin8
),(n l
atn
l
xn
n
kltxy
4. A rod of length l has its ends A and B kept at c0 and c120 respectively until steady state conditions prevail. If the temperature at B is reduced to c0 and so while that of A is maintained, find the temperature distribution of the rod.
Sol:
One dimensional heat equation 2
22
x
ua
t
u
-----(1)
Steady state equation is 02
2
x
uand the steady state solution is baxxu )( --(2)
The boundary conditions are 120)()(0)0()( luiiui
Apply (i) in (2), 0)0( bu
The equation (2) becomes axxu )( -------- (3)
Apply (ii) in (3)
la
allu
120
120)(
The equation (3) becomesl
xxu
120)(
Now consider the unsteady state condition.
The conditions are
)0,()(0),()(0),0()( xuvtluivtuiiil
x120
The correct solution which satisfying the given boundary conditions is
tpaecpxcpxctxu22
321 )sincos(),( ----------- (4)
Apply the boundary condition (iii) in (4)
0),0(22
31 tpaecctu
weget 01 c
then equation (4) becomes tpaecpxctxu22
32 sin),( ---------- (5)
Apply the boundary condition (iv) in (5)
0sin),(22
32 tpaecplctlu
weget 0sin pl npl l
np
,
The equation (5) becomes
t
l
na
n
tl
na
el
xnc
el
xncctxu
2
222
2
222
sin
sin),( 32
where nccc 32
By the super position principle, the most general solution is
1
2
222
sin),(n
tl
na
n el
xnctxu
------------------ (6)
Apply the boundary condition (v) in (6)
1
)(120
1.sin)0,(n
n xfl
x
l
xncxu
------------------ (7)
To find nc , expand )(xf in half range sine series
We know that half range Fourier sine series of )(xf is
1
sin)(n
n l
xnbxf
------------------ (8)
From the equations (7) & (8) we get nn cb
Now
1
2
02
222
0
0
)1(240
)1(240
cos240
sin)1(
cos240
sin1202
sin)(2
n
n
l
l
l
n
n
ln
l
nn
l
l
l
nl
xn
l
nl
xn
xl
dxl
xn
l
x
l
dxl
xnxf
lb
The required solution is
1
1 2
222
sin)1(240
),(n
tl
nan e
l
xn
n
ltxu
5. The ends A and B of a rod l cm long have their temperatures kept at c30 and c80 , until steady state conditions prevail. The temperature of the end B is suddenly reduced to c60 and that of A is increased to c40 . Find the temperature distribution in the rod after time t.
Sol:
One dimensional heat equation2
22
x
ua
t
u
-----(1)
Steady state equation is 02
2
x
uand the steady state solution is baxxu )( --(2)
the boundary conditions are 80)()(30)0()( luiiui
Apply (i) in (2), 30)0( bu
Then the equation (2) becomes 30)( axxu (3)
Apply (ii) in (3)
la
allu
50
8030)(
Then the equation (3) becomes 3050
)( l
xxu
Now consider the unsteady state condition.
In unsteady state the suitable solution which satisfying the given
boundary conditions is
tpaecpxcpxctxu22
321 )sincos(),( (4)
The boundary conditions are
3050
)()0,()(60),()(40),0()( l
xxuxuvtluivtuiii
Since we have all non zero boundary conditions, we write the temperature
distribution function as ),()(),( txuxutxu ts (5)
)(),(),( xutxutxu st
To find )(xus
The solution is baxxus )( (6)
The boundary conditions are 40)(,60)0( luu ss
,40)0( bus
la
al
ballus
20
6040
60)(
the equation (6) becomes 4020
)( l
xxus (7)
To find ),( txut
Given the boundary conditions are
1030
4020
3050
)()0,()0,()(
06060)(),(),()(
04040)0(),0(),()(
l
x
l
x
l
xxuxuxuviii
lutlutluvii
ututxuvi
st
st
st
In unsteady state, the suitable solution which satisfying the given boundary conditions
is
tpaecpxcpxctxu22
321 )sincos(),( (8)
Apply the boundary condition (vi) in (8)
0),0(22
31 tpat ecctu
here 01 c
the equation (8) becomes tpat ecpxctxu
22
32 sin),( (9)
Apply the boundary condition (vii) in (9)
0sin),(22
32 tpat plecctlu
here
l
np
pl
0sin
the equation (9) becomes t
l
na
t ecxl
xnctxu
2
222
32 sin),(
t
l
na
n
tl
na
t
el
xnc
el
xncctxu
2
222
2
222
sin
sin),( 32
where nccc 32
By the super position principle, the most general solution is
1
2
222
sin),(n
tl
na
nt el
xnctxu
(10)
Apply the boundary condition (viii) in (10)
1
)(1030
1.sin)0,(n
nt xfl
x
l
xncxu
To find nc , expand )(xf in half range sine series
We know that half range Fourier sine series of )(xf is
1
sin)(n
n l
xnbxf
(11)
From the equations (10) & (11) we get nn cb
1.10
cos202
sin30
cos10
302
sin10302
sin)(2
02
22
0
0
n
ln
n
l
l
l
nl
xn
ll
nl
xn
l
x
l
dxl
xn
l
x
l
dxl
xnxf
lb
l
l
l
n
n
n
cn
)1(2120
the equation (10) becomes
1
2
222
sin)1(2120
),(n
tl
nan
t el
xn
ntxu
Then the required temperature distribution function is
1
2
222
sin)1(2120
4020
),(n
tl
nan e
l
xn
nl
xtxu
6. A rectangular plate is bounded by the lines byaxyx ,,0,0 , ab . Its surfaces are insulated. The temperature along 0x and 0y are kept at C0 and
the others at C100 . Find the steady state temperature at any point of the plate.
Sol:
Two dimensional heat equation is 02
2
2
2
y
u
x
u
Let l be the length of the square plate
Given the boundary conditions are
( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 100
( ) ( , ) 100
i u y ii u x iii u a y
iv u x b
Assume the temperature distribution as
),(),(),( 21 yxuyxuyxu (A)
To find ),(1 yxu
Consider the boundary conditions
1 1 1
1
( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 0
( ) ( , ) 100
v u y vi u x vii u x b
viii u a y
The suitable solution which satisfying the given boundary conditions is
)sincos)((),( 43211 pycpycececyxu pxpx (1)
Apply the boundary condition (v) in (1)
0)sincos)((),0( 43211 pycpycccyu
weget 021 cc
12 cc
then the equation (2) becomes
)sincos)((
)sincos)((),(
431
43111
pycpyceec
pycpycececyxupxpx
pxpx
(2)
Apply the boundary condition (vi) in (2)
0)()0,( 311 ceecxu pxpx
weget 03 c
then the equation (3) becomes
pyceecyxu pxpx sin)(),( 411 (3)
Apply the boundary condition (vii) in (3)
0sin)(),( 411 pbceecbxu pxpx
weget 0sin pb npb b
np
then the equation (4) becomes
b
yn
b
xnc
b
yn
b
xncc
b
ynceecyxu
n
b
xn
b
xn
sinsinh
sinsinh2
sin)(),(
41
411
where 412 cccn
The most general solution is
b
yn
b
xncyxu
nn
sinsinh),(
11
(4)
Apply the boundary condition (viii) in (4)
)(100sinsinh),(1
1 yfb
yn
b
ancyau
nn
(5)
To find nc , expand )(xf in half range sine series
We know that half range Fourier sine series of )(xf is
1
sin)(n
n b
ynbyf
(6)
From the equations (5) & (6) we get
b
anb
cb
ancb n
nnn
sinhsinh
oddisnifn
evenisnifn
nn
b
b
b
nb
yn
b
dyb
yn
b
dyb
ynyf
bb
n
b
b
b
n
400
0
)1(1200
1cos200
cos200
sin1002
sin)(2
0
0
0
nn
cn sinh
400 if n is odd
then the equation (6) becomes
b
yn
b
xn
b
ann
yxun
sinsinhsinh
400),(
5,3,11
To find ),(2 yxu
Consider the boundary conditions
2 2 2
2
( ) (0, ) 0 ( ) ( ,0) 0 ( ) ( , ) 0
( ) ( , ) 100
ix u y x u x xi u a y
xii u x b
and the suitable solution in this case is
))(sincos(),( 87652pypy ececpxcpxcyxu (7)
Apply the boundary condition )(ix in (7)
0)(),0( 8652 pypy ececcyu
weget 05 c
then the equation (8) becomes
)(sin),( 8762pypy ececpxcyxu (8)
Apply the boundary condition )(x in (8)
0)(sin)0,( 8762 ccpxcxu
weget 7887 0 cccc
then the equation (9) becomes
)(sin
)(sin),(
76
7762
pypy
pypy
eepxcc
ececpxcyxu
(9)
Apply the boundary condition )(xi in (9)
0)(sin),( 762 pypy eepaccyau
weget 0sin pa npa a
np
then the equation (10) becomes
a
xn
a
ync
a
xn
a
yncc
eea
xnccyxu
n
a
yn
a
yn
sinsinh
sinsinh2
)(sin),(
76
762
where 762 cccn
The most general solution is
a
xn
a
yncyxu
nn
sinsinh),(
12
(11)
Apply the boundary condition )(xii in (11)
)(100sinsinh),(1
2 xfa
xn
a
bncbxu
nn
To find nc , expand )(xf in half range sine series
Half range Fourier sine series of )(xf is
1
sin)(n
n a
xnbxf
(12)
From the equations (11) & (12) we get
a
bnb
ca
bncb n
nnn
sinhsinh
a
a
a
n
a
na
xn
a
dxa
xn
a
dxa
xnxf
ab
0
0
0
cos200
sin1002
sin)(2
oddisnifn
evenisnifn
nn
a
a
n
400
0
)1(1200
1cos200
a
bnn
cn sinh
400 if n is odd
then the equation (10) becomes
a
xn
a
yn
nnyxu
n
sinsinhsinh
400),(
5,3,12
Then the equation (A) becomes
a
xn
a
yn
nnb
yn
b
xn
nn
yxuyxuyxu
nn
sinsinhsinh
400sinsinh
sinh
400
),(),(),(
5,3,15,3,1
21