Problemas de ecuacuines diferenciales parciales

PROBLEMAS RESUELTOS:

t − +2t = 0 ,  ∈ (0, 1 ), t > 0
2
(, 0) = 1−2
 (0, t) = (1/ 2, t) = 0

1

Probamos:  =

∞

2
2 2
λn = (2n−1)2 π 2 , n = 1, 2, . . .
→ T = −(2t +λn ) T , Tn = e−t −(2n−1) π t
Xn = {cos(2n−1)π}

conocido:
cn e−t

2

−(2n−1)2 π 2 t

cos(2n−1)π . Debe ser: (, 0) =

n=1

cn =

2
1/ 2

=

1/ 2
0

− cos(2n−1)π

4(1−2)π(2n−1)

=

8
π 2 (2n−1)2

→ (, t) =

∞

cn cos(2n−1)π = 1−2 →

∞

8
π2

8
π(2n−1)

1
(2n−1)2

2

e−t

∞

 +4X +(4+λ)X = 0
X(0) = X(π) = 0

+ c2 e(−2−

3)t

Tn (t) sen n ;

n=1

3) t

(, 0) =

4
cn sen n = e−2 ; (, t) = π

−(2n−1)2 π 2 t

 = XT →

∞
m=1

 +λX = 0
X(0) = X(π) = 0

cos(2n−1)π

, T2 = (c1 +c2 t) e−2t , Tn≥3 =e−2t c1 cos
Tn (0) sen n = 0

e−(2m−1)
2m−1

2t

e−2 sen(2m−1)

λn = n2 , n = 1, 2, . . .
Xn = {sen n}

→ T +4T +n2 T = 0, r = −2±

Tn (0) sen n = sen 2

t (, 0) =

2

λn = n2 , n = 1, 2, . . .
Xn = {e−2 sen n}

n=1

T1 = c1 e(−2+

1/ 2
sen(2n−1)π d
0

T +λT = 0

cn e−n t e−2 sen n ; (, 0) = e−2

tt +4t − = 0,  ∈ [0, π], t ∈ R
(, 0) = sen 2t (, 0) = (0, t) = (1, t) = 0

∞

+

n=1

 = XT →

n=1

=

1/ 2
0

sen(2n−1)π

=

t − −4 −4 = 0,  ∈ (0, π), t > 0
(, 0) = e−2
(0, t) = (π, t) = 0
(, t) =

∞
n=1

1/ 2
(1−2) cos(2n−1)π d
0

8
π 2 (2n−1)2

X + λX = 0
T +(2t +λ)T = 0

= T +2t = −λ →
T

 (0, t) = X (0)T(t) = 0 → X (0) = 0
1
( 2 , t) = X( 1 )T(t) = 0 → X( 1 ) = 02
2

De las condiciones de contorno:
X + λX = 0
1
X (0) = X( 2 ) = 0

X
X

(, t) = X()T(t) →

4−n2 →
n2 −4 t + c2 sen

T2 (0) = c1 = 1
T2 (0) = c2 −2c1 = 0

n2 −4 t .

→  = (1+2t) e−2t sen 2

[La cuerda con rozamiento tiende a pararse].
tt − = 4 sen 6 cos 3,  ∈ [0, π ]
2
(, 0) = t (, 0) = 0, t ∈ R
(0, t) =  ( π , t) = 0
2
∞

 + λX = 0
→
X(0) =X ( π ) = 0
2

[Tn +(2n−1)2 Tn ] sen(2n−1) = 2 sen 3+2 sen 9 ;

n=1

2

∞

2
Yn (y) sen n → Yn −n2 Yn = − π

(−1)n −1
n3

n=1

(1−e−nπ )en −(1−enπ )e−n
enπ −e−nπ

= 0 [ 0 = c1 +c2 ln r ]

0 (1) = 0 (2) = 0

π
0

 = 1 (π −) ,  = − →
2
ch[(2k−1)(y− π )]
2

(2k−1)3 ch[(2k−1) π ]
2

sen n d , Yn = c1 eny +c2 e−ny −

sen(2k −1)

2[(−1)n −1]
→πn3
Yn (0)=Yn (π)=0

− 1 sen n (ambas series deben poder hacerse coincidir).

∞

n = 0, 1, . . . →  = 0 (r)+

r (1, θ) = 0, r (2, θ) = cos 2θ
0
r

→

Θ +λΘ = 0 y 2π-periódica → Θn = {cos nθ, sen nθ} ,

rr + r + θθ = cos θ, 1 < r < 2
r
r2

0 +

(0)=(π)=0

∞
Resolviendo se llega
4
 = 1 (π −)+ π
2
(dice Weimberger) a:
k=1

k=1

2
= π

 = 2(1−cos 3t) sen 3
Tn +(2n−1)2 Tn = 2
9
, n = 2, 5 ,
2
Tn (0) = Tn (0) = 0
+ 81 (1−cos 9t) sen 9

 = −1 ,  = c1 +c2 − 
2

Δ = 0, (0, y) = (π, y) = 0
(, 0) = (, π) = 1 ( − π)
2
∞

Tn (t) sen(2n−1) →

n=1

Convirtiéndolo en homogéneo con una () que cumpla las c.c.:

Δ = −1, (, y) ∈ (0, π)×(0, π)
 = 0 en  = 0,  = π, y = 0, y = π

O bien:  =

∞

[n (r)cos nθ + bn (r) sen nθ]

n=1

,

1 +

1
− 2 = 1 [ 1p = Ar 2 ]
r

1 (1) = 1 (2) = 0

(y n≥3 , bn ≡ 0 ) →  = C +
Δ = y cos , (, y) ∈ (0, π)×(0, 1)
 (0, y) =  (π, y) = 0
y (, 0) = y (, 1) = 0

1
r

=

2

r
3

8
− 14r − 9r cos θ +
9

∞
n=0

Yn (y) cos n →

2

4r
15

,

2 +

2
r

− 42 = 0 [ 2 = c1 r 2 +c2 r −2 ]
r2

2 (1) =0, 2 (2) = 1

4
− 15r 2 cos 2θ

(era de Neumann).

y
Y1 − Y1 = y
−e1−y
→ Y1 = e 1+e −y
Y1 (0) = Y1 (1) = 0

Como es de Neumann aparece (al resolver Y0 = 0 + c.c. ) una C arbitraria:  = C+

ey −e1−y
1+e

−y cos 

Y +λY = 0 , Y (0) = Y (π) = 0 → Yn = {cos ny} , n = 0, 1, . . .

Δ = 2 cos2 y, (, y) ∈ (0, π)×(0, π)
(π, y) = 5+cos y
(0, y) = y (, 0) = y (, π)...