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 .