One Dimensional Heat Conduction Equation

Let OA be a homogeneous bar of uniform cross – section. Let surface of the rod
be laterally insulated with a material impervious to heat. Let the stream lines
of heat flow be parallel to one another and perpendicular to the cross
sectional area. Let O be the origin and OA as the positive x- axis. Let ρ be the
density (gm/cm3), s be the specific heat (cal./gr.deg) and k the thermal
conductivity (cal./cm.deg.sec). Let u(x,t) be the temperature at a distance x
from O. It can be shown that u(x,t) is governed by

∂2 u 1 ∂u ∂ u 2 ∂2 u
= ∨ =c .................... (1)
∂ x 2 c 2 ∂ t ∂t ∂ x2

2 k
Where c = ρs , and c2 is called the diffusivity of the substance.

The equation (1) is called one dimensional heat flow equation or diffusion
equation.
2
∂ u 1 ∂u
Solution of =
∂ x2 c2 ∂ t
using method of Separation of Variables.

∂2 u 1 ∂u
The one dimensional heat equation is = 2 .................. (1)
∂x c ∂t
2

Let us seek a solution of equation (1) in the form u(x,t) = X(x).T(t)

∂u 1 ∂2 u ∂u 1
= X . T , 2 =X 11 .T and =X .T
∂x ∂x ∂ t

These values substituting in equation (1), we have

1
X 11 .T = 2
X . T1
c

X 11 1 T 1
Hence X = 2 T
c

Since L.H.S is a function of x and R.H.S is a function of t, the equality is possible

if and only if each is equal to the same constant say λ.
 11 1
X 1 T
= =λ ( a real constant)
X c2 T

Here three cases are possible: λ>0, λ=0,λ<0

Case(1): Let λ >0. We can write λ=p2 (p>0). We get

X 11 1 T 1 2
= 2 =p
X c T
11
X (x) 2 11 2
= p ⇨ X ( x )= p X ( x )
X(x)

i.e., X 11 ( x ) −p 2 X ( x )=0
2 2
d X 2 d 2 d
i.e., 2 −p X =0 Let D= dx
,D = 2
dx dx

i.e., D 2 X− p2 X =0 ⇨ ( D2− p2 ) X =0

⸫ The auxiliary equation is f ( m )=m2 −p 2=0

⇨m2= p2 ⇨m=± p

⸫ X ( x )= A1¿ e px + B1¿ e− px

1 T1 2 1 2 2 1 2 2
and also 2 T =p ⇨ T = p c T ⇨ T − p c T =0
c
2 2

⇨T ( t )=C 1 e p c t

Thus u ( x , t )=( A 1¿ e px +B 1¿ e− px ) C 1 e p c t =( A 1 e px + B1 e− px ) e p c t
2 2 2 2

................. (S.1)
¿ ¿
Where constant C1 is absorbed in to A1 and B1 .

Case(2): Let λ =0. We get

X 11 1 T 1
= =0
X c2 T
11
X (x) 11 2
i.e., =0 ⇨ X ( x ) =0 ⇨ D X=0
(
X x )
⸫ The auxiliary equation is f ( m )=m2 =0

⇨m=0,0
¿ ¿
⸫ X ( x ) = A2 + B2 x
1
1 T (t )
and also =0 ⇨ T 1 ( t )=0 ⇨ T =C 2
c 2
T ( t )
¿ ¿
⸫ u ( x ,t )=( A 2 + B2 x ) C 2⇨u ( x , t )= A2 + B2 x .................. (S.2)
¿ ¿
Where C2 is absorbed into A2 ∧B 2

Case(3): Let λ >0. We can write λ=-p2 (p>0). We get

11 1
X 1 T 2
= 2 =− p
X c T
11
X (x) 2 11 2
=− p ⇨ X ( x )=− p X ( x )
(
X x )

i.e., X 11 ( x ) + p2 X ( x )=0
2 2
d X 2 d 2 d
i.e., 2 + p X=0 Let D= dx , D = 2
dx dx

i.e., D 2 X + p2 X=0 ⇨ ( D2+ p2 ) X =0

⸫ The auxiliary equation is f ( m )=m2 + p2=0

⇨m2=− p2 ⇨ m=± √ −p 2 ⇨ m=± pi

¿ ¿
⸫ X ( x )= A3 cos px + B3 sin px

1 T1 2 1 2 2 1 2 2
and also 2 T =− p ⇨T =− p c T ⇨T + p c T =0
c
2 2

⇨T ( t )=C 3 e−p c t
2 2 2 2

Thus u ( x , t )=( A 3¿ cos px + B3¿ sin px ) C 3 e p c t= ( A 3 cos px + B3 sin px ) e− p c t

................. (S.3)
¿
Where constant C3 is absorbed in to A*3and B3 .
Thus there are different forms of solutions of equation (1) is given by

u( x , t)=( A 1 e + B1 e ) e p c t ..................... (S.1)

2 2
px − px

u ( x , t )= A2 + B2 x ............................ (S.2)
2 2

u( x , t)=( A 3 cos px +B 3 sin px ) e− p c t ..................... (S.3)

Depending on the physical conditions of the problem, we choose an

appropriate form of solution. Here as we are dealing with a heat conduction
problem, the solution u(x,t) has to depend on time and it cannot diverge to ∞
as t tends to ∞ . Hence the appropriate form of solution is (S.3)

Problems
1. Find the temperature u(x,t) in a bar OA of length ‘l’ which is perfectly
insulated laterally and whose ends O and A are kept at 0 0C, given that
the initial temperature at any point P of the rod (where OP =x) is given
as u(x,0) = f(x) (0 ≤ x ≤ l¿.

Solution: The temperature distribution u(x,t) is governed by the

equation
2
∂ u 1 ∂u
= 2 ..................... (1)
∂x c ∂t
2

Subject to u(0,t) = 0 for all t ........... (2)

u(l,t) = 0 for all t ............ (3)
u(x,0) =f(x), 0 ≤ x ≤ l ............... (4)
The solution appropriate to the problem is of the form
2 2

u( x , t)=( A 3 cos px +B 3 sin px ) e− p c t .................. (S.3)

Using condition (2)

2 2

u(0,t) = 0 ⇨u ( 0 , t ) =0=( A 3 cos p .0+ B3 sin p.0 ) e− p c t

2 2

⸫ A3 e− p c t =0 for all t
⸫A3 = 0
 2 2
−p c t
u ( 0 , t ) =B 3 sin px e

Now using condition (3)

2 2

u(l,t) = 0 ⇨u ( l ,t )=0=( A3 cos p . l+ B3 sin p .l ) e− p c t

2 2

⇨ B3 sin pl e− p c t=0
⇨sin pl=0 ⇨ pl=nπ where n is a positive integer
nπ
Thus p= l where n = 1,2,3,.......................
Thus a typical solution of equation (1) satisfying conditions (2) and (3)
is given by
2 2 2
n π c
t
nπx l
2
for n = 1,2,3,..............
u ( x , t )=Bn sin e
l

We note that equation (1) is linear. Hence u1(x,t), u2(x,t), u3(x,t)...............

Are solutions of equation (1) satisfying the homogeneous conditions
(2), (3)
The most general solution of equation (1) satisfying (2) and (3) is
∞

∑ u n ( x ,t )
n =1

Hence the most general solution of equation (1) satisfying conditions

2 2 2
∞ n π c

(2) and (3) is ( x ,t )=∑ Bn sin nπx e

t
.............. (5)
2
l

n=1 l
Where Bn’s are arbitrary constants to be determined using condition (4)
Using condition (4) i.e., u(x,0) =f(x). Putting t = 0 in equation (5), we
get
2 2 2
∞ n π c
.0
nπx
u ( x , 0 ) =∑ Bn sin
2
l
e =f (x)
n=1 l
∞
nπx
⇨ ∑ Bn sin =f ( x ) , 0 ≤ x ≤≤ l
n=1 l
l
2 nπx
⸫ Bn = ∫
l 0
f ( x ) sin
l
dx , n = 1,2,3,........ ................... (6)

Hence the solution is given by equations (5) and (6).

 2
∂u 2∂ u
2. Solve the one dimensional heat flow equation ∂ t =c 2 given that u(0,t)
∂x
πx
= 0, u(L,t) = 0, t>0 and u(x,0) = 3 sin L ,0< x < L ( )
Solution: The temperature distribution u(x,t) is governed by the
∂u 2 ∂2 u
equation ∂ t =c 2 .................... (1)
∂x

Let us seek a solution of equation (1) in the form u(x,t) = X(x).T(t)

2
∂u ∂ u ∂u
= X 1 . T , 2 =X 11 .T and =X .T 1
∂x ∂x ∂t

These values substituting in equation (1), we have

11 1 1
X .T = 2
X .T
c
11 1
X 1 T
Hence X = 2 T
c

Since L.H.S is a function of x and R.H.S is a function of t, the equality is possible

if and only if each is equal to the same constant say λ.
11 1
X 1 T
= 2 =λ ( a real constant)
X c T

Here three cases are possible: λ>0, λ=0,λ<0

Case(1): Let λ >0. We can write λ=p2 (p>0). We get

11 1
X 1 T 2
= =p
X c2 T

X 11 ( x )
= p2 ⇨ X 11 ( x )= p2 X ( x )
X(x)

i.e., X 11 ( x ) −p 2 X ( x )=0
2 2
d X 2 d 2 d
i.e., 2
−p X =0 Let D= dx , D =
dx d x2
i.e., D 2 X− p2 X =0 ⇨ ( D2− p2 ) X =0

⸫ The auxiliary equation is f ( m )=m2 −p 2=0

⇨m2= p2 ⇨m=± p

⸫ X ( x )= A1¿ e px + B1¿ e− px

1 T1 2 1 2 2 1 2 2
and also 2 T =p ⇨ T = p c T ⇨ T − p c T =0
c
2 2

⇨T ( t )=C 1 e p c t

Thus u ( x , t )=( A 1¿ e px +B 1¿ e− px ) C 1 e p c t =( A 1 e px + B1 e− px ) e p c t
2 2 2 2

................. (2)
¿ ¿
Where constant C1 is absorbed in to A1 and B1 .

Case(2): Let λ =0. We get

11 1
X 1 T
= 2 =0
X c T
11
X (x) 11 2
i.e., =0 ⇨ X ( x ) =0 ⇨ D X=0
(
X x )

⸫ The auxiliary equation is f ( m )=m2 =0

⇨m=0,0
¿ ¿
⸫ X ( x ) = A2 + B2 x
1
1 T (t ) 1
and also =0 ⇨ T ( t )=0 ⇨ T =C 2
c
2
T ( t )
¿ ¿
⸫ u ( x ,t )=( A 2 + B2 x ) C 2⇨u ( x , t )= A2 + B2 x .................. (3)
¿ ¿
Where C2 is absorbed into A2 ∧B 2

Case(3): Let λ >0. We can write λ=-p2 (p>0). We get

11 1
X 1 T 2
= =− p
X c2 T
 11
X (x)
=− p2 ⇨ X 11 ( x )=− p 2 X ( x )
X(x)

i.e., X 11 ( x ) + p2 X ( x )=0

d2 X d d2
i.e., 2
+ p2 X=0 Let D= dx , D =
2
2
dx dx

i.e., D 2 X + p2 X=0 ⇨ ( D2+ p2 ) X =0

⸫ The auxiliary equation is f ( m )=m2 + p2=0

⇨m2=− p2 ⇨m=± √ −p 2 ⇨ m=± pi

¿ ¿
⸫ X ( x )= A3 cos px + B3 sin px

1 T1 2 1 2 2 1 2 2
and also 2 T =− p ⇨T =− p c T ⇨T + p c T =0
c
2 2

⇨T ( t )=C 3 e−p c t
2 2 2 2

Thus u ( x , t )=( A 3¿ cos px + B3¿ sin px ) C 3 e p c t= ( A 3 cos px + B3 sin px ) e− p c t

................. (4)
¿
Where constant C3 is absorbed in to A*3and B3 .

Since u(x,t) decreases as time increases, we find that solution (4) only is
consistent with the physical nature of the problem. Hence equation (4) is the
required solution.
2 2

Equation (4) may be written as u ( x , t )=( A cos px+ B sin px ) e− p c t

2 2

Now u(0,t) = 0 ⇨ u ( 0 , t ) =0= ( A cos p .0+ B sin p .0 ) e− p c t

2 2

⇨ A . e− p c t =0 ⇨ A=0
2 2

Hence u ( x , t )=B sin px e− p c t

2 2

Also u(L,t) = 0 ⇨u ( L ,t )=0=( A cos p . L+ B sin p . L ) e− p c t

2 2

B sin p . L e−p c t =0 ⇨ sin pL=0

 nπ
⸫ pL=nπ ⇨ p= L for n = 1,2,3,................
2 2 2
n π c

Hence the solution is u ( x , t )=B sin nπx e ( )

2
t
L
L

Since this is the solution for n = 1,2,3,.........., the sum of these solutions is also a
solution.
2 2 2
n π c

( )
∞
Hence u ( x , t )=∑ Bn sin nπx e
2
t
L

n=1 L

( πx )
Since u(x,0) = 3 sin L , we get
2 2 2
n π c

( ) ( )
∞
πx nπx .0
=¿ ∑ Bn sin
2
L
3 sin e ¿
L n=1 L

( ) ( )
∞
πx nπx
⇨3 sin =¿ ∑ Bn sin ¿
L n=1 L

Comparing coefficients of different terms on both sides,

B1 = 3, B2 = B3 = B4 = ............... =0
2 2
−π c t

( )
Hence u ( x , t )=3 sin πx e , which is the required solution.
2
L
L

3. An insulated rod of length L has its ends A and B maintained at 0 0C and

1000C respectively until steady state conditions prevail. If B is suddenly
reduced to 00C and maintained 00C, find the temperature at a distance x
from A at time t.

Solution: Let u(x,t) be the temperature at P, at a distance x from the end A at

time t.
2
∂u 2∂ u
The equation for conduction of heat is ∂ t =c ................ (1)
∂ x2
First we will find the temperature distribution at any distance x, before the
end B is reduced to 00C. Previous to the temperature change at the end B,
when t = 0, the heat flow was independent of time.( Steady- state condition
means that the temperature at any particular point, no longer varies with
time) When u depends only on x, then equation (1) reduces to

∂2 u d2 u
=0 i .e . , =0 ................. (2) ⸪ D2=0
∂ x2 d x2

⸫ The general solution is u = ax + b ............... (3)

Since u = 0 for x = 0 and u = 100 for x = L

From equation (3), we have b = 0 and 100 = a L⇨ a = 100/L

So equation (3) becomes u = 100x/L which is the temperature at any point

distance x, at time t = 0

i.e., before the temperature at B is reduced.

100 x
Thus the initial condition is expressed by ( x ,0 )= L ................ (4)

Also the boundary conditions for the subsequent unsteady flow are

u(0,t) = 0 for all values of t ............... (5)

and u(L,t) = 0 for all values of t ................ (6)

Thus we have to find a temperature function u(x,t) satisfying in the equation

(1) and the boundary conditions (5) and (6) and the initial condition (4).

Now the solution of equation (1) is of the form

 ................. (7)
2 2
−p c t
u ( x , t )=( A cos px+ B sin px ) e

Substituting equation (5) in (7), we have

2 2

u ( 0 , t ) =0= ( A cos p .0+ B sin p .0 ) e− p c t

⇨ A e− p c t=0, for all values of t

2 2

⇨A=0
2 2

⸫ Equation (7) can be reduces to u ( x , t )=B sin px . e−p c t ........... (8)

From equations (6) and (8), we get

u ( L ,t )=0=B sin pL . e− p c t , for all values of t.

2 2

i.e., B sin pL=0

If B = 0, equation (8) will give a trivial solution u(x,t) = 0 and hence we must
have sin pL=0

i.e., pL=nπ where n is any integer.

nπ
⇨ p= L
2 2 2
−n c π
Hence equation (8) reduces to ( x ,t )=B sin nπx .e
t
............. (9)
2
L
L

Since equation (1) is linear, its most general solution is got by a linear
combination of solutions of the form (9).

Hence, consider the infinite series

2 2 2
∞ −n c π
t
nπx
u ( x , t )=∑ Bn sin .................. (10)
2
L
.e
n=1 L

Putting t = 0 in equation (10), we have

2 2 2
∞ −n c π ∞
nπx ⇨u ( x , 0 ) =∑ Bn sin nπx
.0
u ( x , 0 ) =∑ Bn sin
2
L
.e
n=1 L n=1 L

Hence in order that the condition (4) may be satisfied, we must have
 ∞

∑ Bn sin nπx
L
=
100 x
L
................... (11)
n =1

100 x
We now expand L
in a half – range Fourier sine series in (0,L).

∞
100 x nπx
=∑ an sin ............... (12)
L n=1 L
L
2 100 x nπx
Where a n= ∫
L 0 L
sin
L
dx

Comparing equations (11) and (12), we have an = Bn

L
200 nπx
Now a n= 2 ∫
x sin dx
L 0 L

[ ( ) ( )]
L
nπx nπx
−cos −sin
200 L L
a n= 2 x −1.
L nπ 2 2
n π
L L
2
0

200
L [
a n= 2 − L .
L
nπ
cos
nπL L2
+
L n2 π 2
sin
nπL 200
L
L2
− 2 0+ 2 2 sin 0
L n π ] [ ]
a n=
[
200 −L2
L2 nπ
cos nπ +
L2
n2 π 2
sin nπ ⇨ an = 2
L]
200 −L2
nπ
cos nπ [ ]
a n=
L
2 [
200 −L2
nπ
n
]
. (−1 ) ⇨ a n=
200
nπ
(−1 )n+1

200 n +1
Hence Bn= nπ (−1 )

⸫ From equation (10), we have

2 2 2
∞ n+1 −n c π
200 (−1 ) nπx .t
∑
2
L
u ( x , t )= sin .e
π n=1 n L
 4. If the ends A and B of a rod 20 cm long have the temperature at 30 0C
and 800C until steady state prevail. If the temperatures of the ends are
changed at 400C and 600C respectively, find the temperature
distribution in the rod at time t.

Solution: For convenience, we shall take l to be the length of the rod and
put l = 20 wherever necessary.
∂u 2 ∂2 u
The equation for the conduction of heat is ∂ t =c 2 ............... (1)
∂x

In order to find the initial temperature distribution in the bar, make use of the
steady- state condition. (Steady- state condition means that the temperature
at any particular point, no longer varies with time) When u depends only on x
and not on t,then equation (1) reduces to
2 2
∂ u d u
2
=0 i .e . , 2 =0 ................. (2) ⸪ D2=0
∂x dx

⸫ The general solution is u = ax + b ............... (3)

To find ‘a’ and ‘b’, we use the boundary conditions

u(0,t) = 30

and u(l,t) = 80 where l is the length of the rod.

From equation (3), we get

30 = a.0 + b ⇨ b = 30

and 80 = a.l + b ⇨80 – 30 =a.l ⇨a.l = 50 ⇨a = 50/l

50
So equation (3) becomes ( x ,0 )= l x+ 30 ................. (4)

Which is the initial temperature distribution in the bar.

When the steady – state condition is reached, the temperatures at the ends A
and B have been changed to 400C and 600C respectively. So the boundary
conditions are
u(0,t) = 40 for all values of t ............ (5)

u(l,t) = 60 for all values of t ............... (6)

Which are non – homogeneous. Therefore assume the solution as

u ( x , t )=u s ( x ) +ut ( x ,t ) ............ (7)

Where u s ( x ) is a solution of equation (1), involving x only and satisfying the

boundary conditions (5) and (6). ut ( x ,t ) is then a function defined by equation
(7) satisfying equation (1). Thus u s ( x ) is then steady – state solution of
equation (1) and ut ( x ,t ) may therefore be regarded as a transient solution
which decreases with increase of t.

d2u
To get u s ( x ), we have to solve the equation =0
d x2

Its general solution is u s ( x )=ax+b ................ (8)

Now u s ( x ) satisfies the boundary conditions (5) and (6).

u s ( 0 )=40∧us ( l )=60

Applying these in equation (3), we get

b = 40 and a = 20 / l
20
Hence u s ( x )= l x +40 .................. (9)

Now from equation (7), we have

u ( x , t )−u s ( x )=ut ( x , t )

⸫ut ( 0 , t )=u ( 0 , t ) −us ( 0 )=40−40=0

20 l
(
and ut ( l , t ) =u (l , t ) −us ( l )=60− l + 40 =60−60=0 )
Also ut ( x , 0 )=u ( x ,0 )−us ( x )

ut ( x , 0 ) =
50
l
x +30−
20 x
l (+40 ) ⸪ From equations (4) and (9)
 50 20 x 30 x
ut ( x , 0 )= x +30− −40= −10
l l l

Hence the boundary conditions relative to the transient solution ut ( x ,t ) are

ut ( 0 , t )=0 ........... (10)

ut ( l , t ) =0 ................ (11)

30 x
and ut ( x , 0 )= l −10 ................. (12)

Since the boundary values are now zero, we use the same procedure to get
ut ( x ,t ) .
2 2

The transient solution is given by ut ( x ,t )= ( A cos px + B sin px ) e− p c t

.................. (13)

Substituting equation (10) in equation (13), we have

2 2
−p c t
ut ( 0 , t )=0=( A cos p .0+B sin p .0 ) e
2 2

⇨ A e− p c t=0 ⇨ A=0
2 2

Now equation (13) reduces to ut ( x ,t )=B sin px e− p c t ............ (14)

Applying the boundary condition (11) in equation (14), we get

2 2

ut ( l , t ) =0=B sin p .l e− p c t ⇨ B sin p . l=0

If B = 0, equation (14) will give the trivial solution ut ( x ,t )=0

nπ
⸫sin pl=0 ⇨ pl=nπ ⇨ p= l where n is any integer.

nπx − p c t 2 2
nπ
Hence equation (14) reduces to ut ( x ,t )=B sin l e where ¿ l ..(15)

The sum of a finite number of terms of the form (15) will also satisfy the p.d.e.
and the boundary conditions (10) and (11) but it will not satisfy the other condition
given by (12).

Hence the most general solution is given by the infinite series

 2 2 2
∞ −c n π t
nπx
ut ( x ,t )=∑ B n sin ................. (16)
2
l
e
n=1 l

Putting t = 0 in equation (16), we get

∞ ∞
nπx − p c .0 nπx
ut ( x , 0 )=∑ Bn sin ⇨ut ( x , 0 )=∑ Bn sin
2 2

e
n=1 l n=1 l

Hence in order that the boundary condition (12) may be satisfied, we must
have
∞

∑ Bn sin nπx
l
=u t ( x , 0 )=
30 x
l
−10 ................ (17)
n =1

30 x
We now express l −10 in a half – range Fourier sine series in (0,l).

∞
30 x nπx
We know that −10=∑ A n sin ................. (18)
l n=1 l
l
2
Where An = ∫
l 0
30 x
l
−10 sin
nπx
l
dx( )
Comparing equations (17) and (18), we have Bn = An.
l
Bn=
2
l 0 (
∫ 30l x −10 sin nπx
l
dx )

[ ( ) ( )]
l
nπx nπx
−cos −sin
Bn=
2
l ( 30 x
l
−10
nπ )l
−
30
l 2 2
n π
l

l l2 0

[( ) ] [( ) ]
2 2
−2 30 .l l nπ .l 30 l nπ . l 2 30.0 l nπ .0 30 l nπ .0
Bn= −10 cos − sin + −10 cos − sin
l l nπ l l n π
2 2
l l l nπ l l n π
2 2
l

Bn=
−2 20 l
l nπ [ 30 l
. cos nπ − 2 2 sin nπ +
n π
2 −10 l
l nπ
cos 0−0
] [ ]
Bn=
−2 20 l
l nπ [
cos nπ +
2 −10l
l nπ ] [
cos 0 ⇨ Bn=
−40
nπ
20
(−1 )n− .1
nπ ]
 −20
Bn=
nπ
[ 2 (−1 )n +1 ]

Substituting in equation (16), we get

2 2 2
∞ −c n π t
−20 nπx
ut ( x ,t )=∑ [ 2 (−1 ) +1 ] sin
2
n l
e
n=1 nπ l
2 2 2
∞ −c n π t
−20
∑ 1 [ 2 (−1 )n+ 1 ] sin nπx ................ (19)
2
l
ut ( x ,t )= e
π n=1 n l

From equations (7),(9) and (19), we have

2 2 2
∞ −c n π t
20 x 20 1 nπx
+40− ∑ [ 2 (−1 ) +1 ] sin ................ (20)
2
n l
u ( x , t )= e
l π n=1 n l

It is given that length of the rod = 20 cm

⸫ Substituting l = 20 in equation (20), we have

2 2 2
∞ −c n π t
20 x 20 1 nπx
+40− ∑ [ 2 (−1 ) +1 ] sin
2
n 20
u ( x , t )= e
20 π n=1 n 20
2 2 2
∞ −c n π t
20 1 nπx
u ( x , t )=x +40− ∑ [ 2 (−1 ) +1 ] sin
n 400
e
π n=1 n 20

Which is the required solution.

5. A bar 100 cm long, with insulated sides, has its ends kept at 0 0C and
1000C until steady – state conditions prevail. The two ends are then
suddenly insulated and kept so. Find the temperature distribution.

Solution: For convenience, we shall take l to be the length of the bar and put

l = 100 wherever necessary.

The temperature u(x,t) along the bar satisfies the equation

∂u ∂2 u
=c 2 2 ............... (1)
∂t ∂x

Since the ends x = 0 and x = l of the bar are insulated so that no heat can flow
through the ends, therefore the boundary conditions are

( ∂∂ ux )
x=0
=0 for all t

( ∂∂ ux )
x=l
=0 for all t .................. (2)

d2u
The heat equation is steady – state is 2 =0 ⸪ D 2=0
dx

⸫ The general solution is u = ax + b ............... (3)

The boundary conditions in this state are

x =0, u = 0 ⇨ b = 0

and x = 100, u = 100 ⇨100=100a ⇨a = 1

So equation (3) becomes u = x

Thus the initial condition is u(x,0) = x, 0<x<l ................. (4)

Now we have to find u(x,t) by solving the unsteady heat equation given by
equation (1) subject to the conditions (2) and (4).

Now the solution of equation (1) is of the form

u ( x , t )=( A cos px+ B sin px ) e− p c t ................ (5)

2 2
Equation (5) partially differentiating with respect to ‘x’,
∂u −p c t 2 2

=(− A p sin px+ Bp cos px ) e .............. (6)

∂x

Putting x = 0 in equation (6), we get

( ∂∂ ux )
2 2

=0=(− A p sin p .0+ Bp cos p .0 ) e− p c t

x=0

2 2

⇨ Bp e−p c t =0 ⇨ B=0

( ∂∂ ux )
x=l
=0=(− A p sin p . l+Bp cos p . l ) e
−p2 c 2 t

2 2

⇨− Ap sin pl e− p c t =0 for all t

⸫ Apsin pl=0 i.e., p ≠ 0, either A = 0 or sin pl=0

If A = 0, equation (5) gives the trivial solution u(x,t) = 0

nπ
⸫sin pl=0 ⇨ pl=nπ ⇨ p= l n = 1,2,3,...............
2 2 2
−c n π t
Hence equation (5) becomes u ( x , t )= A cos nπx e
2
l
l

⸫ The most general solution of equation (1) satisfying the boundary conditions (2)
is
2 2 2
∞ −c n π t
nπx
u ( x , t )=∑ A n cos
2
l
e
n=0 l
2 2 2
∞ −c n π t
nπx
u ( x , t )= A0 + ∑ A n cos .............. (7)
2
l
e
n =1 l

Putting t = 0 in equation (7), we get

2 2 2
∞ −c n π .0
nπx
u ( x , 0 ) =x= A0 + ∑ A n cos
2
l
e
n=1 l
∞
nπx
⇨u ( x , 0 ) =x= A0 + ∑ A n cos ................ (8) by equation (4)
n=1 l
We now expand x into a half – range cosine series in (0,l)
a0 ∞ nπx
Let ¿ + ∑ an cos .................. (9)
2 n=1 l

Comparing equations (8) and (9), we have

a0
A0 = ∧A n=an
2

( ) = 1l l =l
l 2 l
2 2 x
We know that a 0= ∫ x dx=
2
l 0 l 2 0

l
2 nπx
and a n= ∫ x cos dx
l 0 l

[ ( )]
l
nπx nπx
sin −cos
2 l l
a n= x . −1. 2 2
l nπ n π
l l2 0

[ ] [ ]
2 2
2 l nπl l nπl 2 l nπ .0
a n= l. sin + 2 2 cos − 0+ 2 2 cos
l nπ l n π l l n π l

a n=
[
2 1
l nπ
l2
]
2 l2
sin nπ + 2 2 cos nπ − 2 2
n π l n π

2l 2l
a n= 2 2
[ cos nπ−1 ] ⇨a n= 2 2 [ (−1 )n −1 ]
n π n π

l a
Hence A0 = 0 =
2 2

2l
and An =an = 2 2 [ (−1 ) −1 ]
n

n π

{
0 , for n is even
A
⸫ n = −4 l
2 2
, for n is odd
n π

Hence equation (7), we get

 2 2 2
∞ −c n π t
l −4 l nπx
u ( x , t )= + ∑
2
l
2 2
cos e
2 n=1,3,5 …. n π l
2 2 2
∞ −c (2 n−1 ) π t
l 4l 1 (2 n−1) πx
u ( x , t )= − 2 ∑
2
l
cos e
2 π n=1 ( 2 n−1 ) 2
l
2 2 2
∞ −c (2 n−1) π t
100 4.100 1 (2n−1)πx
− 2 ∑
2
100
u ( x , t )= cos e
2 π n=1 ( 2 n−1 ) 2
100
2 2 2
∞ −c (2 n−1 ) π t
400 1 (2 n−1) πx
u ( x , t )=50− 2 ∑ cos e 10000
π n=1 ( 2 n−1 ) 2
100

Which gives the temperature at a point distant x from the left end A.

6. A homogeneous rod of conducting material of length 100 cm has its

ends kept at zero temperature and the temperature initially is
u ( x , 0) = {100−xx ,0,50≤ x≤≤ x50≤ 100
Find the temperature u(x,t) at any time.
2
∂u 2 ∂ u
Solution: The equation for conduction of heat is ∂ t =c 2 ................ (1)
∂x

The boundary conditions are u(0,t) = 0 for all t ............... (2)

u(100,t) = 0 for all t .................... (3)

We are also given the initial temperature distribution i.e., the value of u(x,t)
when t = 0

i.e.,
u ( x , 0)= {100−xx ,0,50≤ x≤≤ x50≤ 100 ................ (4)

Now the solution of equation (1) is of the form

u ( x , t )=( A cos px+ B sin px ) e− p c t ................ (5)

2 2
Substituting equation (2) in (5), we have
2 2

u ( 0 , t ) =0= ( A cos p .0+ B sin p .0 ) e− p c t

=0, for all values of t

2 2
−p c t
⇨ Ae

⇨A=0
2 2

⸫ Equation (5) can be reduces to u ( x , t )=B sin px . e−p c t ........... (6)

Putting x = 100 in equation (6), we have

u ( 100 ,t )=0=B sin p 100 . e− p c t , for all values of t. [⸪From (3)]

2 2

i.e., B sin p 100=0

If B = 0, equation (6) will give a trivial solution u(x,t) = 0 and hence we must
have sin p 100=0

i.e., p .100=nπ where n is any integer.

nπ
⇨ p= 100
2 2 2
−n c π
Hence equation (6) reduces to ( x ,t )=B sin nπx . e
t
10000
............. (7)
100

The most general solution of equation (1) is of the form

2 2 2
∞ −n c π
nπx t
u ( x , t )=∑ Bn sin .e 10000
.................. (8)
n=1 100

Putting t = 0 in equation (8), we get

2 2 2
∞ −n c π
nπx .0
u ( x , 0 ) =∑ Bn sin .e 10000

n=1 100
∞
nπx
⇨u ( x , 0 ) =∑ Bn sin ........... (9)
n=1 100

In order that the boundary condition (4) may be satisfied, we must have
 =u ( x , 0 )={
∞
nπx x , 0 ≤ x ≤50
∑ Bn sin 100 100−x , 50≤ x ≤ 100
.............. (10)
n =1

We now expand u(x,0) given by equation (4) in a half – range sine series in
(0,100).
∞
nπx
We know that ( x ,0 )=∑ An sin ................ (11)
n=1 100
100
2
Where An = ∫ u ( x ,0 ) sin nπx
100 0 100
dx

Comparing equations (9) and (11), we have An = Bn.

[ ]
50 100
2 nπx nπx
Now Bn= 100
∫ x sin 100
dx+ ∫ ( 100−x ) sin
100
dx
0 50

[{ ( ) ( )} { ( ) ( )} ]
50 100
nπx nπx nπx nπx
−cos −sin −cos −sin
2 100 100 100 100
Bn= x −(−1 ) + ( 100−x ) −(−1 )
100 nπ 2 2
n π nπ 2 2
n π
100 100
2 100 100
2
0 50

B n=
2
100
−50.
nπ[
100
cos
nπ .50 10000
100
− 2 2 sin
n π
nπ .50
100
−
2
100
10000
0− 2 2 sin 0
n π ] [ ]
+2
100 [
−( 100−100 )
100
nπ
cos
nπ .100 10000
100
− 2 2 sin
n π
nπ .100
100
−
2
100
− (100−50 )
100
nπ
cos
100 ] [
nπ .50 10000
− 2 2 sin
n π
nπ .50
100 ]
Bn=
2
100 [
−50.
100
nπ
nπ 10000
cos − 2 2 sin
2 n π
nπ
2
−0+
2
100
10000
0− 2 2 sin nπ −
n π
2
] [
100
−50.
100
nπ
nπ 10000
cos − 2 2 sin
2 n π
nπ
2 ] [ ]
Bn=
2
100 [
−50.
100
nπ
nπ 10000
cos − 2 2 sin
2 n π
nπ
2
−
2
100
−50.
100
nπ
nπ 10000
] [
cos − 2 2 sin
2 n π
nπ
2 ]
2
Bn= ¿
100

Bn=
2
100 [ 10000
2. 2 2 sin
n π
nπ
2
⇨
400
Bn= 2 2 sin
n π
nπ
2]
 {
0 , when n is even
B
Hence n = 400 n
2 2
(−1 ) , when nis odd
n π

From equation (8), we get

2 2 2
∞ n −n c π
400 (−1 ) nπx t
u ( x , t )= 2 ∑ sin .e 10000

π n=1,3,5 …. n
2
100
2 2 2
∞ −(2 n−1) c π
400 (−1 )n (2 n−1) πx t
u ( x , t )= 2 ∑ sin .e 10000
π n=1 (2 n−1) 2
100

7. Find the temperature in a thin metal rod of length L, with both ends
πx
insulated and with initial temperature in the rod is sin L

Solution: The temperature u(x,t) in the rod, satisfies the equation

2
∂u 2∂ u
=c ................... (1)
∂t ∂ x2

If the ends x = 0 and x = L of the rod are insulated so that no heat can flow
through the ends, the boundary conditions are
∂u (0 , t) ∂u( L, t)
=0 , =0 for all t ............... (2)
∂x ∂x

πx
( )
Further, we have ( x ,0 )=sin L for 0 ≤ x ≤ L ............... (3)

The consistent solution of equation (1) satisfying the above conditions is of

the form

................ (4)
2 2
−p c t
u ( x , t )=( A cos px+ B sin px ) e

Equation (5) partially differentiating with respect to ‘x’,

∂u 2 2

=(− A p sin px+ Bp cos px ) e− p c t .............. (5)

∂x

Putting x = 0 in equation (5), we get

( )
∂u 2 2

=0=(− A p sin p .0+ Bp cos p .0 ) e− p c t

∂x x=0

2 2

⇨ Bp e−p c t =0 ⇨ B=0

( )
∂u 2 2

=0=(− A p sin p . L+Bp cos p . L ) e−p c t

∂x x=L

2 2

⇨− Ap sin pL e− p c t =0 for all t

⸫ Apsin pL=0 i.e., p ≠ 0, either A = 0 or sin pL=0

If A = 0, equation (4) gives the trivial solution u(x,t) = 0

nπ
⸫sin pL=0 ⇨ pL=nπ ⇨ p= L n = 1,2,3,...............
2 2 2
−c n π t
Hence equation (4) becomes ( x ,t )= A cos nπx e ............... (6)
2
L
L

Since equation (1) is linear, its most general solution is got by a linear combination
of solutions of the form (5). Hence, consider the infinite series
2 2 2
∞ −c n π t
nπx
u ( x , t )=∑ A n cos
2
L
e
n=0 L
2 2 2
−c n π t
A0 ∞ nπx
u ( x , t )= + ∑ A n cos .............. (7)
2
L
e
2 n=1 L

A0
(Here is written in place of A0)
2

Putting t = 0 in equation (7), we get

2 2 2
−c n π .0

( )
πx A 0
∞
nπx
= + ∑ A n cos
2
L
u ( x , 0 ) =sin e
L 2 n=1 L

( )
πx A0 ∞ nπx
⇨u ( x , 0 ) =sin = + ∑ A n cos ................ (8) by equation (3)
L 2 n=1 L

( πx )
We now expand sin L into a half – range Fourier cosine series in (0,L). We
know that
 ( )
∞
πx nπx
sin =∑ Bn cos .................. (9)
L n=0 l

[ ]
L
πx
L −cos
2 πx 2
Where B0= L ∫ sin L dx ⇨ B0= L
0
π ( )
L

L 0

B0=
−2 L
L π
cos
πL
L [
−cos
π .0 −2
L
=
π ]
[ cos π −cos 0 ]

−2 4
B0= [ −1−1 ] =
π π
L

and Bn=
2
L0
∫ sin πx
L ( )
cos
nπx
L
dx

[ ]
L
1 πx πx
Bn= ∫ sin ( 1+n ) + sin ( 1−n ) dx
L0 L L

[ ]
L
πx πx
−cos ( 1+n ) cos ( 1−n )
1 L L
Bn= −
L π π
(1+ n) (1−n)
L L 0

1
Bn= ¿
L

Bn=
1
[L
L (1+n) π
cos ( 1+n ) π−
L
(1−n) π
1
cos ( 1−n ) π − ¿
L ]
Bn=
π [
1 cos ( 1+n ) π cos ( 1−n ) π
1+n
−
1−n
−
1
−
1
1+n 1−n ( )]
2
Bn= [ (−1 )n+1−1 ] ( n≠ 1)
π ( n −1 )
2

Comparing equations (8) and (9), we have A0 = B0 and An = Bn.

L L
2 πx πx 1 2 πx
Now, A1= ∫ sin cos dx= ∫ sin dx=0
L0 L L L0 L

Also An =0 if n is odd.
From equation (7), we have
2 2 2
∞ −c n π t
2 2
u ( x , t )= + ∑ [ (−1 )n+1−1 ] cos nπx
2
L
e
π n=2,4,6 … π ( n2−1 ) L
2 2 2
∞ −c n π t
2 4 1 nπx
∑
2
L
u ( x , t )= − cos e
π π n=2,4,6 … ( n −1 )
2
L

8. Derive the complete solution for the one dimensional heat equation
with zero boundary conditions problem with initial temperature
u(x,0)=x(L-x) in the interval (0,L).

Solution: The initial boundary value problem consists of

2
∂u 2 ∂ u
(i) Heat equation: ∂ t =c 2
∂x
(ii) Zero boundary conditions : u(x,0)=0,u(L,0)=0, for any t
(iii) Initial condition: u(x,0)=x(L-x),0<x<L

Thus we have to find a temperature function u(x,t) satisfying the differential

equation (i) subject to the boundary conditions (ii) and the initial condition
(iii).

Now the solution of (i) is of the form

u ( x , t )=( A cos px+ B sin px ) e− p c t ............. (1)

2 2

By u(0,t) = 0, we have
2 2

u ( 0 , t ) =0= ( A cos p .0+ B sin p .0 ) e− p c t

2 2
−p c t
Ae =0 ⇨ A=0
2 2

Now equation (1) reduces to ( x ,t )=( B sin px ) e− p c t ........... (2)

By u(L,t)=0, we have
2 2
−p c t
u ( L ,t )=0=( A cos p . L+ B sin p . L ) e

=0, for all values of t

2 2
−p c t
B sin p . L e
⇨sin p . L=0 , B ≠ 0
nπ
⇨ pL=nπ ⇨ p= L , where n is any integer
2 2 2
−c n π t
Hence equation (2) reduces to u ( x , t )=B sin nπx e
2
L
L

Since equation (i) is linear, its most general solution is got by a linear combination
of solutions of the form (2). Hence, consider the infinite series
2 2 2
∞ −c n π t
nπx
u ( x , t )=∑ Bn sin ............... (3)
2
L
e
n=1 L
2 2 2
∞ −c n π .0

Putting t = 0, u ( x , 0 ) =∑ Bn sin nπx e

2
L

n=1 L
∞
nπx
u ( x , 0 ) =∑ Bn sin .............. (4)
n=1 L

In order that the initial condition (iii) may be satisfied, (iii) and (4) must be same.
This requires the expansion of x(L-x) as a half – range Fourier sine series in
(0,L). Thus
∞ L
nπx 2 nπx
x ( L−x )=∑ b n sin where b n= ∫ ( Lx−x 2) sin dx
n=1 L L 0 L

[ ]
L
nπx nπx nπx
−cos −sin cos
2 L L L
b n= ( Lx−x 2 ) . −( L−2 x ) + (−2 ) 3 3
L nπ 2 2
n π n π
L L 2
L3 0

[ ] [
2
2 L3
b n= ( L . L−L2) . −L cos nπL −( L−2 L ) −L sin
nπL
−2 cos
nπL 2
− ( L .0−02 )
−L
cos
nπ .0 −L
−( L−2.0 ) 2
L nπ L 2 2
n π L 3 3
n π L L nπ L n

b n=
2
L [ −L2
n π
L3
n π
2 −L2 L3
] [
0−(−L ) 2 2 sin nπ−2 3 3 cos nπ − 0−L . 2 2 sin 0−2 3 3 cos 0
L n π n π ]
[ ] [ ]
3 3
2 L 2 L
b n= −2 3 3 cos n π + 2 3 3 .1
L n π L n π
 3 2
2 2L
b n= [ −cos nπ + 1 ] ⇨b n= 43L 3 [−(−1 )n +1 ]
L n3 π3 n π

{
0 ,if nis even
⸫ x ( L−x ) = 8L
2

3 3
, if n is odd
n π

Hence from equation (3) gives

2 2 2
−c n π t

( )
2 ∞
8L
∑ n13 sin nπx
2
L
u ( x , t )= 3 e
π n=1,3,5… L
2 2 2
−c (2 n−1) π t

( )
2 ∞
8L 1 (2n−1)πx
∑ (2 n−1)
2
L
u ( x , t )= 3 sin e
π n=1
3
L