Universidad Simn BolvarDepartamento de Matemticas
Puras y AplicadasEnero - Marzo 2007
Nombre:
Carnet: Seccin:
MA-3111 Segundo parcial (35 %)
1. a) Desarrolle en la serie de Fourier seno la funcin f(x) = x , x [pi , 0].Hacemos la extensin impar g de f al intervalo [0, pi] g(x) = x, x [pi, pi].g(x)
+h=1
bn sennx , bn =2pi
pi0
t sennt dt, x [pi, pi].
bn = 2tpi
(cosnt
n
)|pi0 +
2
pin
pi0
cosnt dt =2
ncosnpi = 2
(1)n+1n
.
f(x) +n=1
2(1)n+1n
sennx , x [pi, 0].
b) a partir de este desarrollo calcule la suma+n=1
1
n2.
g(x) +n=1
2(1)n+1n
sennx , x [pi, pi]Por la igualdad de Parseval-Steklov1
pi
pipi|g(x)|2 =
+n=1
|an|2 =+n=1
4
n2
+1
1
n2=
1
4pi
pipi
x2dx =2
4pi
pi0
x2dx =
=1
2pi pi
3
3=pi2
6. As
+n=1
1
n2=pi2
6.
c) Diga a que converge la serie encontrada en el intervalo [pi,+pi] . x (pi, pi) +n=1
an sennx = g(x), ya que g es diferenciable. Sea x = pi hacemos la extensin2pi peridica g de g.y = g(x) Sn(pi)
ng(pi 0) + g(pi + 0)
2=pi + (pi)
2= 0 ,
Sn(+pi) n
g(pi 0) + g(pi + 0)2
=pi + (pi)
2= 0.
+n=1
(1)n+1 2n
sennx =
{0, si x = pi
x, si x (pi, pi).
DPTO. DE MATEMATICASMA-3111
2
2. Hallar la solucin del siguiente problema:utt = 4uxx , donde u = u(x, t),
u(x, 0) = 0 t 0, x [0, 3] .ut(x, 0) = 8pi sen 4pixu(0, t) = u(3, t) = 0
u(x, t) = X(x)T (t)
u(0, t) = 0 X(0)T (t) = 0 X(0) = 0.u(3, t) = 0 X(3)T (t) = 0 X(3) = 0.
utt = 4uxx T (t)X(x) = 4X (x)T (t) T(t)
4T (t)=X
(x)
X(x)=: .
T (t) + 4T (t) = 0 , X (x) + X(x) = 0.
{
X(x) + X(x) = 0
X(0) = X(3) = 0
1) < 0 X(x) = c1ex + c2e
x
X(0) = 0 c1 + c2 = 0 c2 = c1;X(3) = 0 c1
(e 3 e 3
)= c1 e
3(e2
3 1)= 0 c1 = 0,
c2 = 0 X 0.2) = 0 X (x) = 0 X(x) = c1x+ c2
X(0) = 0 c2 = 0 , X(3) = 0 3c1 = 0 c1 = 0 X 0.3) > 0 X(x) = c1 cos
x+ c2 sen
x
X(0) = 0 c1 = 0 , X(3) = 0 c2 sen3 = 0 3 = pin, n = 1, 2, . . .
= n = (pin/3)2 , X = Xn(x) = cn senn x = cn sen
pin
3x.
T(t) + 4nT (t) = 0 Tn(t) = An cos
4n t+Bn sen
4n t.
u = un(x, t) = Tn(t)Xn(x) = (An cos4n t+Bn sen
4n t)cn sen
n x
un(x, t) = (an cos 2n t+ bn sen 2
n t) sen
n x.
Sea u(x, t) =+n=1
un(x, t) =+n=1
(an cos 2n t+ bn sen 2
n t) sen
nx
u(x, 0) = 0+n=1
an senn x = 0 an = 0.
ut(x, 0) = 8pi sen 4pix+n=1
bn2n sen
n x = 8pi sen 4pix.
+n=1
bn 2 pin3 sen pin
3x = 8pi sen 4pix.
pin
3x = 4pix si n = 12 bn = 0 , n 6= 12.
n = 12 b12 8pi sen 4pix = 8pi sen 4pix b12 = 1.
DPTO. DE MATEMATICASMA-3111
3
u(x, t) = bn sen 2n t sen
n x
n=12 = sen 2pin3 t sen pin3 xn=12
u(x, t) = sen 8pit sen 4pix.
3. Halle la transformada de Fourier de la funcin f(t) = (4 3t)e(t+2)2Solucin:
f = F(4e(t+2)
2 3te(t+2))[]
= 4 F(e(t+2)
2)[] 3F
(te(t+2)
2)[]
= 4F(e(t(2))
2)[] 3 id
dF(e(t(2))
2)[]
= 4ei(2)(et2
)[] 3 idd
(ei(2)F(et2)[]
),
F(et2)[] = F(e2t
2/2)[] =
12pi 2 e
222 =
14pi
e2/4
f() = 4e2i 14pi
e2/4 3 id
d
(e2i 1
4pie
2/4
)=
2e2ipi
e2/4 3i
4pi
(2ie2i e2/4
2e2i e
2/4)
=
(2 + 3 +
3i
4
)e2ie
2/4
pi
=
(5 +
3
4i
)e2ie
2/4
pi
.
4. Halle explcitamente la solucin u = u(x, t) del siguiente problema{ut = 5uxx , donde x R, t 0.
u(x, 0) = 3(x)
DPTO. DE MATEMATICASMA-3111
4
Solucin: ] u(, t) = Fx(u(x, t)).
ut = 5uxx (Fx(u(x, t)))t = 5Fx(uxx(x, t)) ut(, t) = 5(i)2u(, t) ut + 52u = 0.
u(x, 0) = 3(x) Fx(u(x, 0)) = 3Fx((x)) u(, 0) = 3 1
2pi u(, 0) = 3
2pi.
{
ut + 52u = 0 , u = u(, t).
u(, t)|t=0 = 32pi u(, t) = 3
2pie5
2t.
u(x, t) = F1x(u(, t)) =3
2pi
+
e52teixd
= 3 12pi
+
e52t ei(x)d = 3Fx(e52t)[x]
= 3Fx(e10t2/2)[x] = 3 12pi 10t e
x2/(210t)|x
u(x, t) = 320pit
ex2
20t .
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