1. 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.
2. 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 ∂∂ /
3. • 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
∂
∂
∆−=
4. 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 ∆+
∂
∂
∆∆+−=∆+
5. ∆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
+
∂
∂
=
∂
∂
β
6. 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:
8. • 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
9. • 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
10. • 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 .
11. 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, ...
12. 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
13. 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
14. 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.
15. 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
16. 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
17. • 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
+= ∑
∞
=
18. Reminder:
• TA’s Review session
• Date: July 17 (Tuesday, for all students)
• Time: 10 - 11:40 am
• Room: 304 BH
19. 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
20. 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
21. Examples
• Compute the Fourier series for
.x-xxg
xx,
,xπ,
f(x
11,)(
.0
00
)
<<=
<<
<<−
=
π
22. 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
23. • 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
25. 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
