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.
xu ∂∂ /

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 ∆u
is a “c(x) m ∆u”, 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 ∆t can then be given by the formula:
),(A)of)(()( tx
x
u
tareaxkxH
∂
∂
∆−=

Likewise, at the point x + ∆x, we
have
Heat flowing from left to right across the plane during
an time interval ∆t is:
If on the interval [x, x+∆x], during time ∆t , 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 ∆E is given by the
formula:
).,(
u
tB)of)(()( txx
t
areaxxkxxH ∆+
∂
∂
∆∆+−=∆+

∆E = Heat entering A - Heat leaving B + Heat
generated .
And taking into simplification the principle C above,
∆E = c(x) m ∆u, where m = ρ(x) ∆V . After dividing by
(∆x)(∆t), and taking the limits as ∆x , and ∆t → 0, we
get:
If we assume k, c, ρ are constants, then the eq.
Becomes:
),()()(),(),()( tx
t
u
xxctxQtx
x
u
xk
x ∂
∂
=+





∂
∂
∂
∂
ρ
),(2
2
2
txp
x
u
t
u
+
∂
∂
=
∂
∂
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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
.0,)()0,(
,0,0),(),0(
,0,0,),(),( 2
2
Lxxfxu
ttLutu
tLxtx
x
u
tx
t
u
<<=
>==
><<
∂
∂
=
∂
∂
β

Introducing solution of the form
 u(x,t) = X(x) T(t) .
Substituting into the I.V.P, we obtain:
The method of separation of
variables
.0)()(''and0)()('
haveweThusConstants.
)(
)(''
)(
)('
eq.followingthetoleadsthis
.00,)()('')(')(
=−=−
==
><<=
xkXxXtkTtT
xX
xX
tT
tT
L, txtTxXtTxX
β
β
β

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

Case (i): k = 0. In this case we have
 X(x) = 0, trivial solution
Case (ii): k > 0. Let k = λ2
, then the D.E gives X′ ′ -
λ2
X = 0. The fundamental solution set is: { e λx
, e -λx
}. A
general solution is given by: X(x) = c1 e λx
+ c2 e -λx
X(0) = 0 ⇒ c1+c2= 0, and
X(L) = 0 ⇒ c1 e λL
+ c2 e -λL
= 0 , hence
 c1 (e 2λL
-1) = 0 ⇒ c1 = 0 and so is c2= 0 .
Again we have trivial solution X(x) ≡ 0 .

Finally Case (iii) when k < 0.
We again let k = - λ2
, λ > 0. The D.E. becomes:
X ′ ′ (x) + λ2
X(x) = 0, the auxiliary equation is
r2
+ λ2
= 0, or r = ± λ i . The general solution:
X(x) = c1 eiλx
+ c2 e -iλx
or we prefer to write:
X(x) = c1 cos λ x + c2 sin λ x. Now the boundary conditions
X(0) = X(L) = 0 imply:
 c1 = 0 and c2 sin λ L= 0, for this to happen, we need λ L
= nπ , i.e. λ = nπ /L or k = - (nπ /L ) 2
.
We set Xn(x) = an sin (nπ /L)x, n = 1, 2, 3, ...

Finally for T ′(t) - βkT(t) = 0, k = - λ2
.
We rewrite it as: T ′ + β λ2
T = 0. Or T ′ = - β λ2
T . We see the solutions are
: try we condition,
initial e satisfyth To.conditions boundary the
and D.E the satisfies ) ( ) ( ) , (
function the Thus
...3, 2, 1, n , ) (
2
) / (
t T x X t x u
e b t T
n n n
t L n
n n
=
= =−π β

u(x,t) = ∑ un(x,t), over all n.
More precisely,
This leads to the question of when it is possible to
represent f(x) by the so called
 Fourier sine series ??
.)(sin)0,(
:havemustWe
.sin),(
1
1
2
xfx
L
n
cxu
x
L
n
ectxu
n
t
L
n
n
=





=






=
∑
∑
∞






−∞
π
π
π
β

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
Write 3 sin 2x - 6 sin 5x = ∑ cn sin (nπ/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) .
.0,5sin62sin3)0,(
00,(),0(
.0,0,7 2
2
πxxxxu
,, tt)utu
tx
x
u
t
u
<<−=
>==
><<
∂
∂
=
∂
∂
π
π

Wave Equation
In the study of vibrating string such as piano wire
or guitar string.
L .xg(x),)(x,
t
u
L ,xf(x) ,)u(x,
,tu(L,t),,t)u(
tLx
x
u
t
u
<<=
∂
∂
<<=
>=
><<
∂
∂
=
∂
∂
00
00
00
,0,0,2
2
2
2
2
α

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:
Example:
.sin]sincos[),(
1 L
xn
t
L
n
bt
L
n
atxu n
n
n
ππαπα
+= ∑
∞
=

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:


1,2,3,n;sin)(
1
1,2,3,n;cos)(
1
and,)(
1
where
},sincos{
2
)(
0
1
0
=





=
=





=
=






+





+≈
∫
∫
∫
∑
−
−
−
∞
=
T
T
n
T
T
n
T
T
n
n
n
dxx
T
n
xf
T
b
dxx
T
n
xf
T
a
dxxf
T
a
x
T
n
bx
T
n
a
a
xf
π
π
ππ

Examples
Compute the Fourier series for
.x-xxg
xx,
,xπ,
f(x
11,)(
.0
00
)
<<=



<<
<<−
=
π

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
where the an
,
s and bn
,
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.
{ },()(
2
1
sincos
2 1
0 −+
∞
=
+=






++ ∑ xfxf
T
xn
b
T
xn
a
a
n
nn
ππ

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:
and the Fourier sine series is:
Fourier Sine and Cosine series
xdx
T
n
xf
T
ax
T
n
a
a
T
n
n
n 





=





+ ∫∑
∞
=
ππ
cos)(
2
,cos
2 01
0
,3,2,1
,sin)(
2
,sin
01
=






=





∫∑
∞
=
n
dxx
T
n
xf
T
bx
T
n
b
T
n
n
n
ππ

Consider the heat flow problem:






<≤
≤<
=
>=
><<
∂
∂
=
∂
∂
πxπ -x ,
,xx ,
)x,) u((
,, tu, t)) u((
,tx,
x
u
t
u
)(
2
2
0
03
0t),(02
0021 2
2
π
π
π
π

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

With the solution





=
= −
∞
=
∑
odd.isnwhen
n
4(-1)
even,isnif,0
where
sin),(
2
1)/2-(n
2
1
2
π
n
tn
n
n
b
nxebtxu

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