Btech admission in india

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Partial Differential Equations
Definition
One of the classical partial differential equation of
mathematical physics is the equation describing
the conduction of heat in a solid body (Originated
in the 18th century). And a modern one is the
space vehicle reentry problem: Analysis of transfer
and dissipation of heat generated by the friction
with earth’s atmosphere.
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For example:
Consider a straight bar with uniform cross-section and
homogeneous material. We wish to develop a model for
heat flow through the bar.
Let u(x,t) be the temperature on a cross section
located at x and at time t. We shall follow some basic
principles of physics:
A. The amount of heat per unit time flowing through a
unit of cross-sectional area is proportional to
with constant of proportionality k(x) called the
thermal conductivity of the material.
¶u / ¶x

B. Heat flow is always from points of higher
temperature to points of lower temperature.
C. The amount of heat necessary to raise the
temperature of an object of mass “m” by an amount Du
is a “c(x) m Du”, where c(x) is known as the specific
heat capacity of the material.
Thus to study the amount of heat H(x) flowing from left
to right through a surface A of a cross section during the
time interval Dt can then be given by the formula:
H x k x area t u
admission.edhole=.co-m D ¶
( ) ( )( of A) (x,t)
x
¶

Likewise, at the point x + Dx, we
have
Heat flowing from left to right across the plane during
an time interval Dt is:
+ D = - + D D ¶
( ) ( )( of B) t u (x x,t).
H x x k x x area + D
t
¶
If on the interval [x, x+Dx], during time Dt , additional
heat sources were generated by, say, chemical reactions,
heater, or electric currents, with energy density Q(x,t),
then the total change in the heat DE is given by the
formula:

DE = Heat entering A - Heat leaving B + Heat
generated .
And taking into simplification the principle C above,
DE = c(x) m Du, where m = r(x) DV . After dividing by
(Dx)(Dt), and taking the limits as Dx , and Dt ® 0, we
get:
k x u
x ¶
x t Q x t c x x u
x
¶ ¶
r
( ) ( , ) ( , ) ( ) ( ) (x,t)
é
If we assume k, c, r are constants, then the eq.
Becomes:
t
+ = ¶ úû
ù
êë
¶
¶
u u
+
( , ) 2
2
admission.edhole.¶com b
= ¶
2 p x t
x
t
¶
¶

Boundary and Initial conditions
Remark on boundary conditions and initial condition
on u(x,t).
We thus obtain the mathematical model for the heat
flow in a uniform rod without internal sources (p = 0)
with homogeneous boundary conditions and initial
temperature distribution f(x), the follolwing Initial
Boundary Value Problem:

One Dimensional Heat Equation
2
x t x L t
u
x t u
t
< < >
¶ b
= ¶
( , ) ( , ) , 0 , 0 , 2
x
¶
¶
u t u L t t
= = >
(0, ) ( , ) 0 , 0 ,
u x = f x < x <
L
( ,0) ( ) , 0 .

The method of separation of
variables
Introducing solution of the form
 u(x,t) = X(x) T(t) .
Substituting into the I.V.P, we obtain:
X x T t X x T t x L, t
= b
< < >
( ) '( ) ''( ) ( ) , 0 0.
this leads to the following eq.
Constants. Thus we have
X x
''( )
= =
X x
( )
T t
'( )
T t
( )
b
T t kT t X x kX x
- b
= - =
'( ) ( ) 0 and ''( ) ( ) 0.

Boundary Conditions
Imply that we are looking for a non-trivial solution
X(x), satisfying:
X x kX x
- =
''( ) ( ) 0
X X L
= =
(0) ( ) 0
We shall consider 3 cases:
k = 0, k > 0 and k < 0.

Case (i): k = 0. In this case we have
 X(x) = 0, trivial solution
Case (ii): k > 0. Let k = l2, then the D.E gives X¢ ¢ -
l2 X = 0. The fundamental solution set is: { e lx, e -lx }. A
general solution is given by: X(x) = c1 e lx + c2 e -lx
X(0) = 0 Þ c1 + c2 = 0, and
X(L) = 0 Þ c1 e lL + c2 e -lL = 0 , hence
 c1 (e 2lL -1) = 0 Þ c1 = 0 and so is c2 = 0 .
Again we have trivial solution X(x) º 0 .

Finally Case (iii) when k < 0.
We again let k = - l2 , l > 0. The D.E. becomes:
X ¢ ¢ (x) + l2 X(x) = 0, the auxiliary equation is
r2 + l2 = 0, or r = ± l i . The general solution:
X(x) = c1 e ilx + c2 e -ilx or we prefer to write:
X(x) = c1 cos l x + c2 sin l x. Now the boundary conditions
X(0) = X(L) = 0 imply:
 c1 = 0 and c2 sin l L= 0, for this to happen, we need l L
= np , i.e. l = np /L or k = - (np /L ) 2.
We set Xn(x) = an sin (np /L)x, n = 1, 2, 3, ...

( ) = -b ( p / )2
, n =
1, 2, 3, ... T t b e
n L t
n n
Thus the function
u x t =
X x T t
( , ) ( ) ( ) satisfies the D.E and
n n n
the boundary conditions. To satisfy the initial
condition, we try :
Finally for T ¢(t) - bkT(t) = 0, k = - l2 .
We rewrite it as: T ¢ + b l2 T = 0. Or T ¢ = - b l2
T . We see the solutions are

u(x,t) = å un(x,t), over all n.
More precisely,
n
p b p
ö çè¥ - æ
u x t c e n
ö çè
å
( , ) sin .
1
We must have :
u x c n
ö çè
¥
( ,0) sin ( ) .
This leads to the question of when it is possible to
represent f(x) by the so 1
called
 Fourier sine series ??
2
x f x
L
x
L
n
t
L
n
= ÷ø
= æ
÷ø
= æ
å
÷ø
p

Jean Baptiste Joseph Fourier
(1768 - 1830)
Developed the equation for heat transmission and
obtained solution under various boundary
conditions (1800 - 1811).
Under Napoleon he went to Egypt as a soldier and
worked with G. Monge as a cultural attache for the
French army.

Example
Solve the following heat flow problem
2
x t
u
< < >
¶
= ¶
¶
p
7 , 0 , 0. 2
x
u
t
¶
u (0, t ) = u ( p
, t) = 0 , t >
0
,
u x = x - x < x <
π
( ,0) 3sin 2 6sin 5 , 0 .
Write 3 sin 2x - 6 sin 5x = å cn sin (np/L)x, and
comparing the coefficients, we see that c2 = 3 , c5 = -6,
and cn = 0 for all other n. And we have u(x,t) =
u2(x,t) + u5(x,t) .

Wave Equation
In the study of vibrating string such as piano wire
or guitar string.
u
= ¶
¶
, 0 , 0 , 2
2
2
u( 0 ,t) = u(L,t), t >
0
,
u(x, ) f(x) , x L ,
u
0 0
(x, ) g(x), x L .
t
x L t
x
t
u
= < <
¶
¶
= < <
< < >
¶
¶
0 0
2
2
a

Example:
f(x) = 6 sin 2x + 9 sin 7x - sin 10x , and
g(x) = 11 sin 9x - 14 sin 15x.
The solution is of the form:
t n x
u x t a n n
p pa pa + =å¥
( , ) [ cos sin ]sin .
1 L
L
t b n
L
n
n
=

Reminder:
TA’s Review session
Date: July 17 (Tuesday, for all students)
Time: 10 - 11:40 am
Room: 304 BH

Final Exam
Date: July 19 (Thursday)
Time: 10:30 - 12:30 pm
Room: LC-C3
Covers: all materials
I will have a review session on Wednesday

Fourier Series
For a piecewise continuous function f on [-T,T], we
have the Fourier series for f:
x b n
ö çè
a n
ö çèæ + »
{ cos sin },


where
2
1 ( ) , and
ò
-
f x n
p
1 ( ) cos ; n 1,2,3,
ö çè
ò
-
f x n
1 ( ) sin ; n 1,2,3,
( )
0
1
0
ö çè
= ÷ø
=
= æ
= ÷ø
= æ
÷ø
æ + ÷ø
ò
å
-
¥
=
T
T
n
T
T
n
T
T
n
n
n
x dx
T
T
b
x dx
T
T
a
f x dx
T
a
x
T
T
a
f x
p
p p

Examples
Compute the Fourier series for
, - π < x <
,
0 0
î í ì < <
x, x
f(x
p
0 .
)
=
g ( x ) = x , - 1 < x <
1
.

Convergence of Fourier Series
Pointwise Convegence
Theorem. If f and f ¢ are piecewise continuous on [ -T,
T ], then for any x in (-T, T), we have
a ¥
a n x
þ ý ü
î í ì
b n x
p p
+ å + f x f x
0 + -
2 n
=
1
where the an
cos sin 1
n n
T
,s and bn
{ ( ) ( } ,
2
= +
T
,s are given by the previous fomulas.
It converges to the average value of the left and right
hand limits of f(x). Remark on x = T, or -T.

Fourier Sine and Cosine series
Consider Even and Odd extensions;
Definition: Let f(x) be piecewise continuous on the
interval [0,T]. The Fourier cosine series of f(x) on [0,T]
is:
a a n T
æ + ò å¥
ö çè
æ = ÷ø
and the Fourier sine series is:
ö çè
xdx
f x n
T
T
x a
T
n
n
n ÷ø
=
cos p , 2 ( ) cos p
0
2 1 0
f x n
b n
sin , 2 ( ) sin ,
æ ö ò =
1 0
n
=
1,2,3,
çè
å¥
ö çè
÷ø
æ = ÷ø
x dx
T
T
x b
T
T
n
n
n
p p

Consider the heat flow problem:
( ) u
= ¶
¶
1 2 0 0 2
( 2 ) u( 0 , t) = u ( , t) , t >
0
,
ì
ï ïî
ïïí
x , < x £
,
£ <
=
< < >
¶
¶
π -x , x π
( ) u(x, )
, x t ,
x
u
t
2
2
0
3 0
2
p
p
p
p

Solution
Since the boundary condition forces us to consider
sine waves, we shall expand f(x) into its Fourier
Sine Series with T = p. Thus
p
p 0
b 2 f (x) sin nxdx n
= ò

With the solution
u ( x , t ) b e sin
nx
0, if n is even,
ì
=
ïî ïí
¥
= å
= -
when n is odd.
4(-1)
n
where
2
(n-1)/2
2
1
2
p
n
n t
n
n
b

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