MODULE III
PARTIAL DIFFERENTIAL EQUATIONS
An equation involving the partial differential coefficients, independent
variablesand dependent variables is called a partial differential equation.
Egs.
1) x ∂z + y ∂z = z
∂x
∂y
2
2
2) ∂ z + ∂ z = 0
∂x2
∂y2
3) ∂2z 3 -- 3 ∂z 4 +5 ∂z = 5 z
∂x2
∂x
∂x∂y
The order of a partial differential equation is the order of the highest partial
derivative in the equation.
The degree of a partial differential equation is the degree of the highest
partial derivative in the equation.
Thus order of eqn1 is 1 and degree is 1.
Order of eqn2 is 2 and degree is 1.
Order of eqn3 is 2 and degree is 3.
Note:
If z is a function of two independent variables x and y then the different partial
derivatives of z are denoted as p= ∂z , q = ∂z, r =∂2z , s= ∂2z and t = ∂2z
∂x
∂y ∂x2
∂y2
∂x∂y
Formation of partial differential eqns
The p.d.e. can be formed either by the elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation involving 3 or more variables.
If the number of arbitrary constants to be eliminated is equal to the number of
independent variables, the partial differential eqns that arise are of the first order.
If the number of arbitrary constants to be eliminated is more than the
number of independent variables, the partial differential eqns that arise are of the second
or higher order.
If the pde is obtained by the elimination of arbitrary functions then the order of
the pde is, in general, equal to the number of arbitrary functions eliminated.
Egs.
1) Form pde corresponding to z=ax+by
z=ax+by ...........(1)
Differentiating (1) partially with respect to x,
p= ∂z =a
∂x
Differentiating (1) partially with respect to y,
q= ∂z =b
∂y
Substituting the values of a and b in (1),
z=px+qy , which is the required pde.
Soln. of a p.d.e.
The linear p.d.e. of the first order
These are p.d.eqns in which the 1st order partial derivatives p and q occur to the
1st degree only and are not multiplied together
Lagrange’s linear eqn
The p.d.e. of the form Pp+Qq=R , where P and Q are functions of x,y,zis called
Lagrange’s linear eqn.
Method of soln. of Lagrange’s linear eqn
i) Form the auxiliary eqns.
ii) Solve these simultaneous equations either by the method of grouping or by the
method of multipliers or both to get two independent solutins
u=a and v=b
iii) Write the complete solution as Ф(u,v)=0 or u=f(v) or v=g(u)
dx dy dz
Note:- Solution of AE PQ R
1. Method of Grouping
Eg.
dx dy dz
1. Solve z2yz2xy2x
Taking the first two ratios,
dx dy

z2y z2x

dx dy


y
x
x dx – y dy =0
x2 y2 a
 
Integrating ,
2 2 2
ie; x2-y2=a…………..(1)
dy dz
dy dz
Taking the last two ratios z2x  y2x  z2  y2  y2dy-z2dz=0
Integrating ,
y3 z3 b
 
3 3 3
ie; y3-z3=b…………….(2)
Eqns (1) and (2) taken together constitute the required soln. of the given simultaneous
eqns.
2 Method of multipliers
By a proper choice of multipliers l,m,n which are not necessarily constants we
write
dx dy dz ldx
mdy
ndz
  =
P Q R lP
mQ
nRsuch that lP+mQ+nR=0
Then ldx+m dy+n dz=0 can be solved giving the soln. Ф(x,y,z)=c …………(1)
Again find another set of multipliers α,β,γ so that αP+βQ+γR=0 giving
α dx+β dy+γ dz=0 which on integration gives the soln. φ(x,y,z) = C |……………(2)
dx dy dz
Then. (1) and (2) together give the required soln. of PQ R
Note: a p.d.e. may have more than one solns
Eg.
dx dy dz
 2 22 2
1. Solve x
2 2
(
y

z
)
y
(
z

x
)z
(
x

y
)
xdx

ydy

zdz
Taking multpliers as x,y,z; each fraction = x
22 2 22 2 22 2
(
y

z
)

y
(
z

x
)

z
(
x

y
)
=
 x dx+y dy+z dz=0
xdx
ydy
zdz
0
2
x
y2 z2 c
  
Integrating ,
2 2 2 2
ie; x2+y2+z2=c …………..(1)
Again taking the multipliers as 1/x,-1/y,-1/z, each fraction= 1/x dx+1/y dy +1/z dz
0
 1/x dx-1/y dy- 1/z dz=0
Integrating, log x- log y- log z= log C’
ie x/yz = C’; or x= C’ yz……………(2)
(1) and (2) together give the required soln.
EXAMPLES
(1)Solve px+qy=z
Soln:
The Lagrange's eqn. Is, Pp+Qq=R
dx dy dz
and the auxilliary eqn. is P = Q = R
dx dy dz
ie; x = y = z …..............(1)
Taking the first two ratios,
dx dy
=
x
y
Integrating, logx=logy+loga
x
ie; y = a …........(2)
Similarly, taking last two ratios of eqn (1),
dy dz
=
y
z
Integrating, logy=logz+logb
y
ie; z = b …............(3)
Eqns (2) and (3) are independent solns of (1).
Hence the complete soln. of the given eqn. is (u,v)=0
x y
ie; ( y , z ) = 0
2)Solve (mz-ny)p+(nx-lz)q=ly-mx
Soln:
dx
dy
dz
Auxilliary eqn. is mz− ny = nx − lz = ly − mx …....................(1)
xdx ydy zdz
Taking the multipliers as x,y,z; each ratio of (1)= x mz− ny y nx− lz z ly − mx
xdx
=
ydy zdz
0
Hence, xdx+yddy+zdz=0
x2 y 2 z 2 a
Integrating, 2 2 2 = 2
ie; x2+y2+z2=a …........(2)
Now choosing l,m,n as multipliers, each ratio of eqn.(1)=
ldx mdy ndz
l mz− ny m nx− lz n ly− mx
ldx mdy ndz
0
=
Hence, ldx+mddy+ndz=0
Integrating, lx+my+nz=b ….............(3)
Eqns (2) and (3) are independent solns of (1).
Complete soln. of the given p.d.e. is, x2+y2+z2=lx+my+nz)
3)Find the general soln. of (y+z)p+(z+x)q=x+y
Soln:
Auxilliary eqn. is
dx
dy
dz
=
=
y z z x x y …...................(1)
Taking the first two ratios,
dx
dy
=
y z z x
dx− dy
= y z − z x
=
dx− dy
y− x
d x− y
= − x− y …...................(3)
dy
dz
dy− dz
Similarly, taking the last two ratios, z x = x y = z x − x y
=
Eqn.(3) – Eqn.(4) 

d y− z
− y− z ….............(4)
− d x− y
− d y− z


x− y
y− z
− d x− y
d y− z
 y− z 
x− y
Integrating, -log(x-y)+log(y-z)= log a
y− z
ie; x− y = a …............(5)
Also from eqn.(2), each ratio =
dx dy dz
y z
z x
x
dx dy dz
=
y
2 x y z …..............(6)
Eqns (3) and (6)

1 d x y z
− d x− y

2 x y z 
x− y
1 d x y z
d x− y
2 x y z  x− y 

1
Integrating, 2 log (x+y+z)+log(x-y)= log b
log x y z +log y= log b
log x y z (x-y)= log b
x y z (x-y)= b
x− y 2 (x+y+z) = b2...................(7)
Eqns (6) and (7) are independent solns of (2).
Complete soln. of the given p.d.e. is, x− y
2
y− z
(x+y+z) = x− y )
Non-linear eqns of first order
A p.d.e. Which involves first order partial derivatives p and q with degree higher
than one and the products of p and q is called a non-linear p.d.e. Of the first order.
Charpit's Method
This is a general method for finding the complete soln. of non-linear p.d.e. Of
first order.
Let the given eqn. be f(x,y,z,p,q)=0..........................(1)
Since z depends on x and y,
dz=z/x dx+ z/y dy
ie; dz= pdx+qdy.......................(2)
Now, if we can find, another relation involving x,y,z,p,q such as
F( x,y,z,p,q )=0.....................(3)
then we can solve eqns (1) and (3) for p and q and substitute in eqn. (2) . This will give
the soln. provided eqn. (2) is integrable.
To determine F
Write the Charpit's auxilliary eqns as
dF
dp
dz
dx
dy
dq
df
df = df
df = − df = − df = 0 
df = − p df
q
p
−q
dy
dz
dx
dz
dy
dq
dq
dq

ny integral of eqn. (4) which involves p or q or both can be taken as the assumed
relation F.
1)Solve (p2+q2)y=qz

Soln:
Given eqn. is p2y+q2y-qz=0.................(1)
Since z is a function of x and y,
dz= pdx+qdy.................(2)
The Charpit's auxilliary eqn. is,
dF
dp
dz
dx
dy
dq
df
df = df
df = − df = − df = 0 
df = − p df
q
p
−q
dy
dz
dx
dz
dy
dq
dq
dq

f/p= 2py, f/q= 2qy, f/x=0, f/y= p2+q2 , f/z=-q
Thus, the auxilliary eqns becomes,
dx
dy
dz
dp
dq
=
=
=
= 2 2
− 2py − 2qy − p.2py − q.2qy − pq p q q.− q =F/0..................(3)
dp
dq
eqn(3)  − pq = p2
ie; pdp+qdq=0
p 2 q2 a 2
Integrating, 2 2 = 2
ie; p2+q2=a2..................(4)
Given eqn. is, (p2+q2)y-qz=0
From eqn. (4), a2y-qz=0
2
a y
ie; q= z …...............(5)
Now from eqn.(4), p2=a2-q2
4 2
a y
2
=a - 2
z
2 2
4
a z−a y
=
2
z
2
a2 z2− a2 y2
=
z2
a 2 2 2
Hence, p= z z − a y …..............(6)
2
a 2 2 2
a y
z
−
a
y
Substituting the values of p and q in eqn. (2), dz= z
dx+ z dy
2
2 2
2
ie; zdz= a z − a y dx+ a y dy
2
2 2
2
ie; zdz- a y dy= a z − a y dx
z
2
2
a y
ie; d( 2 − 2
2
2
2 2
) = a z − a y dx
1 d z 2− a 2 y 2
ie; 2
.= a dx
z 2− a 2 y 2
ie;
z 2 − ay 2 =ax+b
ie; z2-a2y2 = (ax+b)2 , which is the required soln.
Home Work
1. Solve using Charpit's method; p xy+ pq+qy=yz
Ans. log(z-ax)+ a log (a+y)= y+b
HOMOGENEOUS LINEAR PARTIAL DIFFERENTIAL
EQUATION WITH CONSTANT COEFFICIENTS
An equation of the form
∂nz + a1 ∂n z + a2 ∂nz + …………………….. + an-1 ∂n z + an ∂n z = F(x,y)
∂xn
∂xn-1 ∂y ∂xn-2 ∂y2
∂x ∂yn-1
∂yn
------------------------------( 1 )
Where a1, a2…………………an-1, an are constants is called a homogeneous linear partial
differential equation of the nth order with constant coefficients.
Putting D = ∂ and D1 = ∂
∂x
∂y
Eqn (1) can be written as
( Dn + a1 Dn-1 D1 + a2 Dn-2 D12+ …………………….. + an-1 D D1(n-1) + an
D1n ) z = F(x, y)
ie f(D,D1)z = F( x, y)
…………………………..(2)
The complete solution of (2) is z = CF +PI
Where CF is the complete solution of the equation f(D,D1)z = 0 and PI is a
particular solution of the equation f(D,D1)z = F( x, y)
PI = 1 F(x,y)
f(D,D1)
AUXILIARY EQUATION
Consider the equation f(D,D1)z = F( x, y). Its symbolic operator equated to
zero, by replacing D by m and D1 by 1.
ie f(D,D1) = 0 is called the AE
AE is f( m,1) =0 .
Thus AE of eqn(1) is
mn + a1 m n-1 +………..+an = 0
RULES FOR FINDING THE CF
1. Write the equation in the symbolic form
2. Write down the auxiliary equation and solve for m.
3. Write the solution as follows
Roots for AE
1. m1 ,m2 ,m3 ,……….
Soln. CF
1. f1 (y+m1x) +f2(y+m2x)+………..
( distinct roots)
2. m1 ,m1 ,m3 ,……….
2. f1 (y+m1x) + x f2(y+m1x)+ f3(y+m3x)
+………..
(2 equal roots)
3. m1 ,m1 ,m1 ,……….
3 f1 (y+m1x) + x f2(y+m1x)+ x2 f3(y+m1x)
+………..
(3 equal roots)
RULES FOR FINDING THE PI
PI =
1 F(x,y)
f(D,D1)
Case 1
When F( x, y) = eax+by
P.I = 1 eax+by
f(D, D1)
= 1 eax+by
Replace D by a and D1by b.
f(a,b)
If f(a,b) = 0 , then the above method fails and we use the following formula
1 eax+by
= x eax+by
f(D – a/bD1)
1!
1 eax+by
f(D – a/bD1)2
= x2 eax+by
2!
1 eax+by
f(D – a/bD1)3
= x3 eax+by
3!
Case 11
When F( x, y) = sin (ax+by) or cos(ax+by) then
1 sin (ax+by)
= 1 sin (ax+by)
f(D2 , D1D ,D12)
f(-a2 , -ab ,-b2)
Replace D2 by –a2 ,DD1by –ab, D12 by –b2
Similar rule holds when F( x, y) = cos(ax+by) .
Case 111
When F( x,y ) = xp yq , where p and q are positive integers.
P.I = 1 xp yq
f(D, D1)
= ( f(D ,D1))-1 xp yq
If p<q expand ( f(D ,D1))-1 in powers of D/D1
If p>q expand ( f(D ,D1))-1 in powers of D1/D
Case 1V
Let F( x,y ) be any function of x. and y then
P.I = 1 F( x,y )
f(D, D1)
=
1 F( x,y )
1
(D-m1 D ) (D-m2 D1)………… (D-mn D1)
Where m1, m2,………………mn are roots of AE.
Now 1 F( x,y) = ∫ F(x, c-mnx)dx
D- mnD1
Where c is replaced by y + mnx after integration.
EXAMPLES
∂3z - 3 ∂3 z + 4 ∂3z = ex+2y
∂x3 ∂x2 ∂y
∂y3
1. Solve
Soln:
The symbolic form is ( D3-3D2D1+4D13)z = ex+2y
A.E is m3 – 3m2 +4 =0
m = -1 , 2 ,2
C.F is z = f1(y-x) + f2(y+2x) +x f3(y+2x)
1 ex+2y
D -3D2D1+4D13
=
1 ex+2y
1-3x1x2+4x8
P.I =
3
1 ex+2y
27
The complete solution is
z = f1(y-x) + f2(y+2x) +x f3(y+2x) + 1 ex+2y
27
=
2. Solve (D3-7DD12 - 6D13)z = sin ( x+ 2y) + e 2x+y
Soln:
A.E is m3 – 7m -6 =0
m = -1 , -2 , 3
C.F is z = f1(y-x) + f2(y-2x) + f3(y+3x)
P.I =
1
sin (x+2y) + e2x+y
D -7DD12 - 6D13
3
= 1
sin (x+2y)
D3-7DD12 - 6D13
+
1
e2x+y
D3-7DD12 - 6D13
1
e2x+y
2 -7x2x1 – 6x1
=
1
sin (x+2y)
+
D(D2)-7D(D12 ) - 6D1 (D12 )
=
=
1
sin (x+2y)
- 1
2
2
1
2
D(-1 )-7D(-2 ) - 6D (-2) 12
1
sin (x+2y)
27D+ 24D1
- 1
12
3
e2x+y
e2x+y
=
D sin (x+2y)
27D 2+ 24DD1
=
1 cos(x+2y)
27(-1) 2+ 24(-1x2)
=
- 1 cos(x+2y)
75
e2x+y
- 1
12
-
- 1
12
1
12
e2x+y
e2x+y
The complete solution is
z = f1(y-x) + f2(y-2x) + f3(y+3x) - 1 cos(x+2y) - 1 e2x+y
75
12
3. Solve
∂2z + 3 ∂2 z + 2 ∂2z = x + y
∂x2
∂x ∂y
∂y2
Soln:
The symbolic form is ( D2 + 3DD1+2D12)z = x + y
A.E is m2 + 3m +2 =0
m = -1 , -2
C.F is z = f1(y-x) + f2(y-2x)
P.I = 1 ( x+y)
D2+3DD1+2D12
=
1 ( x+y)
2
D ( 1+3D1+2D12 )
D D2
=
1 ( 1+3D1+2D12 )-1 ( x+y)
D2
D D2
=
1 ( 1- 3D1+……………. ) ( x+y)
D2
D
=
1 ( x+y -3x)
D2
=
1 ( y -2x)
D2
= yx2 – x3
2
3
The complete solution is
z = f1(y-x) + f2(y-2x) + yx2 – x3
2
3
4. Solve
∂2z + ∂2 z -6 ∂2z = y cosx
∂x2
∂x ∂y
∂y2
Soln:
The symbolic form is ( D2 + DD1 -6D12)z = y cosx
A.E is m2 + m -6 =0
m = -3 , 2
C.F is z = f1(y-3x) + f2(y+2x)
P.I =
=
=
1 y cosx
D2+DD1-6 D12
1 y cosx
(D+3D1) (D-2 D1)
∫ (c-2x) cosx dx
1
(D+3D1)
=
1
(D+3D1)
[(c-2x) sinx - ∫ -2sinxdx ]
=
1
(D+3D1)
[(y+ 2x-2x) sinx -2cosx]
=
1
(D+3D1)
[ysinx -2cosx]
= ∫ [(c+3x)sinx – 2cosx ]dx
= [(c+3x) -cosx – ∫ 3(-cosx)dx -2sinx]
= [(y -3x +3x) -cosx + 3sinx -2sinx]
= -ycosx + sinx
The complete solution is
z=
f1(y-3x) + f2(y+2x) - ycosx + sinx
PROBLEMS
1. Solve ∂2z - ∂2 z = sinx cos2y
∂x2
∂x ∂y
2. Solve (2D2-5DD1 +2D12)z = 5sin ( 2x+ y)
3. Solve
∂2z - 4 ∂2 z + 4 ∂2z = e 2x-y
∂x2 ∂x ∂y
∂y2
4. Solve (D2-DD1 +D12)z = 2x+3y
5. Solve (D2-DD1 -2D12)z = (y -1) ex
One dimentional wave eqn
(Vibration of a stretched string)
Consider a uniform elastic string of length l stretched tightly b/w two points O and A, and
displaced slightly from its equilibrium position OA.
When it is released, the string will execute transverse vibrations.
Y
O
x
A
X
Taking the end O as the origin, OA as the X axis and a perpendicular line through O as
the Y axis, we shall find the displacement y as a function of the distance x and time t.
Now the vibration of the string can be represented by an eqn.
2
2 y
2  y

c
where c2=T/m ,( T and m being some constants ) which is called
t 2
x 2
the one dimensional wave eqn.
The boundary conditions which the wave eqn. has to satisfy are
i)
y=0 when x=0
ii)
y=0 when x=l
These should be satisfied for every value of t.
If the sring is made to vibrate by putting it into a curve y=f(x) and then releasing it the
initial conditions are i) y=f(x) when t=0
y
 0 when t=0
ii)
t
Solution of the wave eqn.
2
2 y
2  y

c
The wave eqn. is
……………(1)
t 2
x 2
Let y= XT…………….(2)
Where X is a function of x only and T is a function of t only be the soln. of (1)
y
 XTl
Then
t

2 y
 XTll and
t 2
y
x
=XlT
2 y
 Xll T
.x 2
Sub. These values in (1) XTll=c2 Xll T…………..(3)
Since X is a function of the independent variable x only and T is a function of the
indepaendent variable t only both sides of (3) should be equal to a same constant k.
"
X
1
T
"

kand

k
ie.
2
X
cT
ie X”-kX=0 …….(4) and T”-kc2T=0………….(5)
since we are dealing with a problem on vibrations y must be a periodic function of x
and t.Hence the soln. must involve trigonometric terms. Hence out of 3 different solns
of (4) and (5) we choose the soln. of (4) as X=c1cos px+c2 sin px and that of (5) as
T= c3cos cpt+c4 sin cpt where p is given by the eqn. k= -p2
Thus y= XT=( c1cos px+c2 sin px)( c3cos cpt+c4 sin cpt)………………(6)
Now applying boundary conditions y=0 when x=0 and y=0 when x=l we get

C1( c3cos cpt+c4 sin cpt)=0 ………….(7) and
( c1cos pl+c2 sin pl)( c3cos cpt+c4 sin cpt)=0 ………………(8)
(7)  c1=0 and hence (8) reduces to c2 sin pl ( c3cos cpt+c4 sin cpt)=0 which is
satisfied when sin pl=0 or pl=nπ or p= nπ/l , n=1,2,3…
 From (6) a soln. of the wave eqn. satisfying the boundary conditions is
n

ct
n

ct
nx

c
y=c2(c3cos
4sin ) sin
l
l
l
n

ct
n

ct
nx

b
=(a ncos
where an=c2c3 and bn=c2c4
nsin ) sin
l
l
l
Adding up the solns for different values of n ,we get,

n

ct
n

ct
nx

b
y=  (a ncos
……………….(9)
nsin ) sin
l
l
l
n 1
is also a soln.
Now applying the initial conditions,

y
n
 0 when t=0 , we have f(x)=  an sin
x ……….(10)
y=f(x) and
t
l
n 1
nc
n
x =0 ………………(11)
bn sin
l
l
n 1
Since (10) represents fourier series for f(x) ,
l
2
n

x
f(
x
)
sin
dx
n
we have a
……………..(12)
l0
l

and

From (11) bn  0n

Hence (9) reduces to y=  a ncos
n 1
nct
nx
sin
……………………(13)
l
l
where a n is given by (12) when f(x) . ie, y(x,0) is known.
Problem;
1. A string is stretched and fastened to two points l apart.Motion is started by
x
displacing the string in the form y=a sin
from which it is released at time
l
t=0.Show that the displacement of any point at a distance x from one end at time t
is given by y(x,t)=a sinπx/l cos πct/l
Ans:
Here the boundary conditions are y(0,t)=y(l,t)=0 and the initial conditions are
x
y
 0 when t=0
Y(x,0)=a sin
and
l
t
Now we have the soln. of the wave eqn. as

nct
nx
y=  a ncos
sin
……………(1)
l
l
n 1
2
n

x
f(
x
)
sin
dx
n
where a
………………..(2)
l0
l
Here ,an vanishes for all values of n except 1
l
2a
x
2
sin
dx

l 0
l
Now, a1=
l
2
a1
2

x
(
1

cos
)
dx

l 02
l
l
=
a
l
=
=
2x
sin
x0 2l )
l

l
a
l
{
*0
l l- 2
}
=a
Hence the required soln. y(x,t) = acos
ct
l
x
sin
l
2.A tightly stretched syring with fixed end points x=0 and x=l is initially at rest in its
equilibrium position.If it is set vibrating by giving to each of its points a velocity λx(lx), find the displacement of the string at any distance x from one end at any time t.
2

8
l
1 (
2
m

1
)
ct
(
2
m

1
)
x
sin
sin

Ans: y(x,t)=
4
4
c
(
2
m

1
)
l
l
m

1




One dimensional heat flow
Consider the flow of heat by conduction in a uniform bar.take one end of the bar as
the origin and the direction of flow as the positive X axis.Then the temperature u at
any point of the bar depends on the distance x of the point from one end and the time
u
 2u
2

c
t and is given by the eqn.
which is called the heat eqn.
t
x 2
Soln.
u
 2u
 c2
Consider the heat eqn.
…………….(1)
t
x 2
Let u= XT……………..(2)
where X is a function of x only and T is a function of t only be the soln. of (1)
u
 2u

XT
'
Then
and 2 =X”T
t
x
Sub. These in (1) , we get, XT’= c2 X”T
Separating the variables,
X" 1 T'

……….(3)
X c2 T
Here LHS is a function of x only and RHS is a function of t only and the two
variables x and t are independent.So (3) holds only when both sides of (3) equal to the
same constant k.
X
"
1T
'

kand

k
ie;
2
X
cT
ie; X”-kX=0 ……………..(4)
and T’-k c2T=0 …………….(5)
On solving (4) and (5) ,we get 3 different solns and out of these 3 we choose the soln. of
(4) as X=c1cos px+c2 sin px and that of (5) as
T= e-c2 p2 t .,Which are consistent with the physical nature of the problem.
Thus, from (2), u= ( c1cos px+c2 sin px) e-c2 p2 t
EXAMPLES
1.A rod of length l with insulated sides is initially at a uniform temperature u0.Its ends
are suddenly cooled to 0°C and are kept at that temperature. Find the temperature
function u(x,t)
u
 2u
 c2
Ans. The temperature function u(x,t) satisfies the differential eqn.
. And we
t
x 2
have its soln. as
u(x,t)= ( c1cos px+c2 sin px) e-c2 p2 t ……………..(1)
Here the boundary conditions are u(0,t)=0 and u(l,t)=0 for all t.
Now the initial condition is u(x,0)=u0.
u(0,t)=0  c1 e-c2 p2 t =0
 c1=0
Hence (1) becomes u(x,t)= c2 sin px e-c2 p2 t …………..(2)
U(l,t)=0  c2sin pl e-c2 p2 t =0
 pl=nπ, n being a positive integer
n 2 2
n
2
Now (2) reduces to u(x,t)= c2 sin
x e- c2 l t
l
n 2 2
l2
n
t
x e- c2
l
The most general soln. is obtained by adding all such solns for n=1,2,3…
n 2 2

n
2
ie; u(x,t)=  bn sin
x e- c2 l t …………………….(3)
l
n 1

n
Now u(x,0)= u0.   bn sin
x = u0 which is half range sine series for u0.
l
n 1
=
bn sin
2
n
Hence bn=  u 0 sin xdx
l 0
l
l
n
x
l
n
l
cos
=
2u0
l
=
2
u
0 l
[(

1
)n
1
]
l n

=
0 if n is even
 4u0
if n is odd
n
Hence the temperature function
u(x,t)=
=
1 n

4
u
0
sinx - c2

n l e
t
n

1
,3
,5
,...

0
u
4u0 4

n 2 2
l2
(
2
n

1
)

xe- c2
sin
 l
n

1
(2n1)2 2
t
l2
PROBLEMS
1. .An insulated rod of length l has its ends A and B maintained at 0°C and 100°C
respectively until steady state conditions prevail. If B is suddenly reduced to 0°C and
maintained at 0°C a) find the temperature at a distance x from A at time t.
b)Find also the temperature if the change consists of raising the temperature of A to 20°C
and reducing that of B to 80°C
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