exame_metodos_2S06_gabarito

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MS650A - Exame Final - 13/12/06
Nome:
RA:
(1) Resolva, usando transformada de Fourier, a equac¸a˜o
y′′ − y = e−α|x|, −∞ < x <∞,
onde α > 0, α 6= 1, e satisfazendo as condic¸o˜es lim
x→±∞
y(x) = lim
x→±∞
y′(x) = 0.
(2) Encontre, usando transformada de Laplace, a soluc¸a˜o do problema{
ux = 2ut + u, x > 0, t > 0,
u(x, 0) = e−x,
que seja limitada para x > 0 e t > 0, e utilizando a fo´rmula complexa de inversa˜o para
efetuar o ca´lculo da transformada inversa.
(3) Resolva, usando o me´todo de separac¸a˜o de varia´veis, o problema
ut = k∇2u, r < 1, 0 ≤ θ ≤ 2pi,
u(1, θ, t) = 0,
u(r, θ, 0) = 1,
|u(r, θ, t)| < M.
OBS: Em coordenadas polares ∇2u = urr + 1rur + 1r2uθθ.
i Valor das questo˜es: (1) 2,5 (2) 3,0 (3) 4,5
K Algumas fo´rmulas eventualmente u´teis:
F [f(x)] = 1√
2pi
∫ ∞
−∞
f(x)eikx dx, F [f ′(x)] = −ikF [f(x)],
(F [f(x)])′ = iF(xf(x)), F [(f ∗ g)(x)] = F [f(x)]F [g(x)]
L[f(t)] =
∫ ∞
−∞
f(t)e−st dt, L[eatf(t)] = F (s− a),
L[f(t− a)H(t− a)] = e−asF (s), L[f ′(t)] = sF (s)− f(0),
L[tf(t)] = −F ′(s), L−1[F (s)] = 1
2pii
∫ γ+i∞
γ−i∞
F (s)est ds
Jν(z) =
∞∑
k=0
(−1)k
k!Γ(k + ν + 1)
(z
2
)2k+ν
, ez(t−1/t)/2 =
+∞∑
n=−∞
Jn(z)tn,
Jn+1(z) = −zn d
dz
(z−nJn(z)), Jn−1(z) = z−n
d
dz
(znJn(z)),
Jn(z) = (−1)nzn
(
1
z
d
dz
)n
J0(z),
∫ a
0
Jν(λνmx/a)Jν(λνnx/a)x dx =
a2
2
[Jν+1(λνn)]2δnm,
an =
1
pi
∫ 2pi
0
f(x) cosnx dx, bn =
1
pi
∫ 2pi
0
f(x) sinnx dx.