2. 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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3. 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
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4. 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
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( ) ( )( of A) (x,t)
x
¶
5. 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:
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6. 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
¶
¶
7. 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:
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8. 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 .
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9. 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
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T t kT t X x kX x
- b
= - =
'( ) ( ) 0 and ''( ) ( ) 0.
10. 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.
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11. 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 .
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12. 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, ...
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13. admission.edhole.com
( ) = -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
14. 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
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15. 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.
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16. 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) .
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17. 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
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18. 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
=
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19. Reminder:
TA’s Review session
Date: July 17 (Tuesday, for all students)
Time: 10 - 11:40 am
Room: 304 BH
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20. 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
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21. 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
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22. 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
.
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23. 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.
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24. 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
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25. 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
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26. 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
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27. 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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