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# Séries de Fourier e Equações Diferenciais Parciais

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pi
\u221e\u2211
m=0
(\u22121)m
2m + 1
cos
(2m + 1)pix
L
.
f(x) =
2
pi
\u221e\u2211
m=1
cos mpi
2
\u2212 (\u22121)m
m
sen
mpix
L
=
=
2
pi
\u221e\u2211
m=1
(\u22121)m \u2212 1
2m
sen
2mpix
L
+
2
pi
\u221e\u2211
m=0
1
2m + 1
sen
(2m + 1)pix
L
=
= \u2212 2
pi
\u221e\u2211
m=0
1
2m + 1
sen
(4m + 2)pix
L
+
2
pi
\u221e\u2211
m=0
1
2m + 1
sen
(2m + 1)pix
L
1.2. f(x) =
1
2
+
2
pi
\u221e\u2211
m=1
sen 3mpi
4
\u2212 sen mpi
4
m
cos
mpix
L
.
f(x) =
2
pi
\u221e\u2211
m=1
cos mpi
4
\u2212 cos 3mpi
4
m
sen
mpix
L
1.3. f(x) =
3L
8
+
2L
pi2
\u221e\u2211
m=1
cos mpi \u2212 cos mpi
2
\u2212 mpi
2
sen mpi
2
m2
cos
mpix
L
.
f(x) =
2L
pi2
\u221e\u2211
m=1
mpi
2
cos mpi
2
\u2212mpi cos mpi \u2212 sen mpi
2
m2
sen
mpix
L
1.4. f(x) =
L
4
+
2L
pi2
\u221e\u2211
m=1
2 cos mpi
2
\u2212 1\u2212 (\u22121)m
m2
cos
mpix
L
=
L
4
+
2L
pi2
\u221e\u2211
m=1
2(\u22121)m \u2212 2
4m2
cos
2mpix
L
=
L
4
\u2212 2L
pi2
\u221e\u2211
m=0
1
(2m + 1)2
cos
(4m + 2)pix
L
.
f(x) =
4L
pi2
\u221e\u2211
m=1
sen mpi
2
m2
sen
mpix
L
=
4L
pi2
\u221e\u2211
m=1
(\u22121)m
(2m + 1)2
sen
(2m + 1)pix
L
1.5. f(x) =
3L
16
+
2L
pi2
\u221e\u2211
m=1
cos mpi
4
+ cos 3mpi
4
\u2212 1\u2212 (\u22121)m
m2
cos
mpix
L
.
f(x) =
2L
pi2
\u221e\u2211
m=1
sen mpi
4
+ sen 3mpi
4
m2
sen
mpix
L
Se´ries de Fourier e Equac¸o\u2dces Diferenciais Parciais 22 de novembro de 2007
2.3 Equac¸a\u2dco de Laplace num Reta\u2c6ngulo 57
2. Equac¸o\u2dces Parciais (pa´gina 52)
2.1. Temos que resolver o problema\uf8f1\uf8f4\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f4\uf8f3
\u2202u
\u2202t
=
\u22022u
\u2202x2
u(x, 0) = f(x), 0 < x < 40
u(0, t) = 0, u(40, t) = 0
A soluc¸a\u2dco e´ enta\u2dco
u(x, t) =
\u221e\u2211
n=1
cn sen
npix
40
e\u2212
n
2
pi
2
1600
t
em que cn sa\u2dco os coeficientes da se´rie de senos de f(x), ou seja,
cn =
1
20
\u222b 40
0
f(x) sen(
npix
40
)dx
= 20cn(f
(0)
0,1 )
= \u221220 2
npi
cos s
\u2223\u2223\u2223npi
0
=
40
npi
(1\u2212 cos(npi))
=
40
npi
(1\u2212 (\u22121)n), n = 1, 2, 3 . . .
Portanto a soluc¸a\u2dco e´ dada por
u(x, t) =
40
pi
\u221e\u2211
n=1
1\u2212 (\u22121)n
n
sen
npix
40
e\u2212
n
2
pi
2
1600
t
=
80
pi
\u221e\u2211
n=0
1
2n + 1
sen(
(2n + 1)pi
40
x)e\u2212
(2n+1)2pi2
1600
t
2.2. Temos que resolver o problema\uf8f1\uf8f4\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f4\uf8f3
\u2202u
\u2202t
=
\u22022u
\u2202x2
u(x, 0) = f(x), 0 < x < 40
u(0, t) = 0, u(40, t) = 60
22 de novembro de 2007 Reginaldo J. Santos
58 2 EQUAC¸O\u2dcES DIFERENCIAIS PARCIAIS
A soluc¸a\u2dco e´ enta\u2dco
u(x, t) =
3x
2
+
\u221e\u2211
n=1
cn sen
npix
40
e\u2212
n
2
pi
2
1600
t
em que cn sa\u2dco os coeficientes da se´rie de senos de
f(x)\u2212 3x
2
= 20\u2212 3x
2
ou seja,
cn = 20cn(f
(0)
0,1 )\u2212
3
2
cn(f
(1)
0,1 )
=
40
npi
cos s
\u2223\u2223\u2223npi
0
\u2212 120
n2pi2
(\u2212s cos s + sen s)
\u2223\u2223\u2223npi
0
=
40
npi
(cos(npi)\u2212 1)\u2212 120
n2pi2
(\u2212npi cos(npi))
=
160((\u22121)n \u2212 1/4)
npi
, n = 1, 2, 3 . . .
Portanto a soluc¸a\u2dco e´ dada por
u(x, t) =
3x
2
+
160
pi
\u221e\u2211
n=1
(\u22121)n \u2212 1/4
n
sen
npix
40
e\u2212
n
2
pi
2
1600
t
Quando t tende a mais infinito a soluc¸a\u2dco tende a soluc¸a\u2dco estaciona´ria v(x, t) =
3x
2
.
2.3. (i) Temos que resolver o problema
\uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3
\u2202u
\u2202t
=
\u22022u
\u2202x2
u(x, 0) = f(x), 0 < x < 40
\u2202u
\u2202t
(0, t) = 0,
\u2202u
\u2202t
(40, t) = 0
A soluc¸a\u2dco e´ enta\u2dco
u(x, t) =
\u221e\u2211
n=0
cn cos
npix
40
e\u2212
n
2
pi
2
1600
t
Se´ries de Fourier e Equac¸o\u2dces Diferenciais Parciais 22 de novembro de 2007
2.3 Equac¸a\u2dco de Laplace num Reta\u2c6ngulo 59
em que cn sa\u2dco os coeficientes da se´rie de cossenos de f(x), ou seja,
c0 =
1
40
\u222b 40
0
f(x)dx = 30,
cn =
1
20
\u222b 40
0
f(x) cos(
npix
40
)dx
= 80
(\u22121)n \u2212 1
n2pi2
, n = 1, 2, 3 . . .
Portanto a soluc¸a\u2dco e´ dada por
u(x, t) = 30 +
80
pi2
\u221e\u2211
n=1
(\u22121)n \u2212 1
n2
cos
npix
40
e\u2212
n
2
pi
2
1600
t
= 30\u2212 160
pi2
\u221e\u2211
n=0
1
(2n + 1)2
cos(
(2n + 1)pi
40
x)e\u2212
(2n+1)2pi2
1600
t
(ii) A soluc¸a\u2dco tende a v(x, t) = 30, quando t tende a mais infinito.
2.4. Temos que resolver o problema\uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3
\u22022u
\u2202t2
= 4
\u22022u
\u2202x2
u(x, 0) = f(x),
\u2202u
\u2202t
(x, 0) = 0, 0 < x < 40
u(0, t) = 0, u(40, t) = 0
A soluc¸a\u2dco e´ enta\u2dco
u(x, t) =
\u221e\u2211
n=1
cn sen
npix
40
cos
npit
20
em que cn sa\u2dco os coeficientes da se´rie de senos de f(x), ou seja,
cn =
1
20
\u222b 40
0
f(x) sen(
npix
40
)dx
=
80
pi2
sen npi
4
+ sen 3npi
4
n2
, n = 1, 2, 3 . . .
Portanto a soluc¸a\u2dco e´ dada por
u(x, t) =
80
pi2
\u221e\u2211
n=1
sen npi
4
+ sen 3npi
4
n2
sen
npix
40
cos
npit
20
22 de novembro de 2007 Reginaldo J. Santos
60 2 EQUAC¸O\u2dcES DIFERENCIAIS PARCIAIS
2.5.
u(x, t) = sen(
pi
20
x) cos(
pi
10
t)
2.6. Temos que resolver o problema\uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3
\u22022u
\u2202t2
= 4
\u22022u
\u2202x2
u(x, 0) = 0,
\u2202u
\u2202t
(x, 0) = g(x), 0 < x < 40
u(0, t) = 0, u(40, t) = 0
A soluc¸a\u2dco e´ enta\u2dco
u(x, t) =
\u221e\u2211
n=1
cn sen
npix
40
sen
npit
20
em que npi
20
cn sa\u2dco os coeficientes da se´rie de senos de g(x), ou seja,
npi
20
cn =
1
20
\u222b 40
0
g(x) sen(
npix
40
)dx
=
80
pi2
sen npi
4
+ sen 3npi
4
n2
n = 1, 2, 3 . . .
cn =
1600
pi3
sen npi
4
+ sen 3npi
4
n3
, n = 1, 2, 3 . . .
Portanto a soluc¸a\u2dco e´ dada por
u(x, t) =
1600
pi3
\u221e\u2211
n=1
sen npi
4
+ sen 3npi
4
n3
sen
npix
40
sen
npit
20
2.7.
u(x, t) =
10
pi
sen(
pi
20
x) sen(
pi
10
t)
2.8. Temos que resolver o problema\uf8f1\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f4\uf8f4\uf8f4\uf8f3
\u22022u
\u2202t2
= 4
\u22022u
\u2202x2
u(x, 0) = f(x),
\u2202u
\u2202t
(x, 0) = g(x), 0 < x < 40
u(0, t) = 0, u(40, t) = 0
Se´ries de Fourier e Equac¸o\u2dces Diferenciais Parciais 22 de novembro de 2007
2.3 Equac¸a\u2dco de Laplace num Reta\u2c6ngulo 61
A soluc¸a\u2dco e´ a soma das soluc¸o\u2dces dos problemas com apenas uma das func¸o\u2dces f(x)
e g(x) na\u2dco nulas.
u(x, t) =
\u221e\u2211
n=1
cn sen
npix
L
cos
anpit
L
+
\u221e\u2211
n=1
dn sen
npix
L
sen
anpit
L
em que cn e
npi
20
dn sa\u2dco os coeficientes da se´rie de senos de f(x) e de g(x), respectiva-
mente, ou seja,
cn =
1
20
\u222b 40
0
f(x) sen(
npix
40
)dx
=
80
pi2
sen npi
4
+ sen 3npi
4
n2
, n = 1, 2, 3 . . .
npi
20
dn =
1
20
\u222b 40
0
g(x) sen(
npix
40
)dx
=
80
pi2
sen npi
4
+ sen 3npi
4
n2
n = 1, 2, 3 . . .
dn =
1600
pi3
sen npi
4
+ sen 3npi
4
n3
, n = 1, 2, 3 . . .
Portanto a soluc¸a\u2dco e´ dada por
u(x, t) =
80
pi2
\u221e\u2211
n=1
sen npi
4
+ sen 3npi
4
n2
sen
npix
40
cos
npit
20
+
1600
pi3
\u221e\u2211
n=1
sen npi
4
+ sen 3npi
4
n3
sen
npix
40
sen
npit
20
2.9. A soluc¸a\u2dco e´ enta\u2dco
u(x, y) =
\u221e\u2211
n=1
cn sen
npiy
2
senh
npix
3
em que cn senh(
3npi
2
) sa\u2dco os coeficientes da se´rie de senos de k(y), ou seja,
cn senh(
3npi
2
) =
\u222b 2
0
k(y) sen(
npiy
2
)dx
=
4
pi2
sen npi
4
+ sen 3npi
4
n2
, n = 1, 2, 3 . . .
22 de novembro de 2007 Reginaldo J. Santos
62 2 EQUAC¸O\u2dcES DIFERENCIAIS PARCIAIS
cn =
4
pi2
sen npi
4
+ sen 3npi
4
n2 senh(3npi
2
)
, n = 1, 2, 3 . . .
Portanto a soluc¸a\u2dco e´ dada por
u(x, y) =
4
pi2
\u221e\u2211
n=1
sen npi
4
+ sen 3npi
4
n2 senh(3npi
2
)
sen
npiy
2
senh
npix
3
=
8
pi2
\u221e\u2211
n=0
(\u22121)n
senh(3(2n+1)pi
2
)(2n + 1)2
sen
(2n + 1)piy
2
senh
(2n + 1)pix
3
2.10. A soluc¸a\u2dco e´ enta\u2dco
u(x, y) =
\u221e\u2211
n=1
cn sen
npiy
2
senh(
npi
3
(3\u2212 x))
em que cn senh(
3npi
2
) sa\u2dco os coeficientes da se´rie de senos de h(y), ou seja,
cn senh(
3npi
2
) =
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