ecuaciones diferenciales parciales por expanciones series de fourier
TRANSCRIPT
-
22
2
22
tU
xUa
=
0 x < < U(0,t) = 0 U( ,t) = 0 U(x,0) = 0
Ut t=0
= Sen(x) Como :
( ) ( )U X x T t= U X Tx
=
2
2
U X Tx
=
U XTT
=
2
2
U XTt
=
Tendramos que : 2a X T XT =
2
2
X TX a T
= =
2XX
=
2X X = 2 0X X + =
( )2 2 0D X + = ( )2 2 0D + =
2 2D = D = 2
D = i ( ) 1 2X x c Sen x c Cos x = +
22
Ta T
=
2 2T a T = 2 2 0T a T + =
( )2 2 2 0D a T + = ( )2 2 2 0D a + =
2 2 2D a = D = a2 2
D = ai ( ) 3 4T t c Sena t c Cosa t = +
( ) ( ) ( ),U x t X x T t= ( ),U x t = ( )1 2c Sen x c Cos x + ( )3 4c Sena t c Cosa t +
Como U (0,t) = 0 ( )3 4 0c Sena t c Cosa t +
c1Sen0+ c2Cos0( ) = 0 2 0c =
Como U ( ,t) = 0
U ( ,t) = c1Sen = 0 Para evitar solucin trivial c1 0 Senn = 0( )
Sen = Senn
n =
n = ( ),nU x t = c1n Sen nx( ) c3nSen ant( ) + c4nCos ant( )
U x,t( ) =1n
= Un x,t( )
( ),U x t =1n
= ( )Sen nx BnSen ant( ) + AnCos ant( )
Como U (x,0) = 0 U x,0( ) =1n
= ( )Sen nx BnSen an(0)( ) + AnCos an(0)( )
( ),0U x =1n
= ( )Sen nx Bn(0)+ An(1) = 1n
= ( )Sen nx An = 0 [ ] 0nA =
Como U
t t=0= Sen(x) t
( ),U x t =
Sen nx( ) t
BnSen ant( ) = Sen nx( )BnanCos ant( )
n=1
Ut t=0
= Sen(x)1n
=
= ( )Sen nx ( )0 ( )nB anCos an Sen x=
1n
= Sen nx( ) Bnan = Sen(x)
B1 =
1a
Bn = 0 n 1
( ),U x t = ( )1 Senxa
( )Senat
-
k 2Ux2 =
Ut
0 < x < L t > 0 U(0,t) = 0 U(L,t) = 0 U(x,0) = x(L x) Como :
( ) ( )U X x T t= U X Tx
=
2
2
U X Tx
=
U XTT
=
2
2
U XTt
=
Tendramos que :
k X T = X T
XX
= TkT
= 2
2XX
=
2X X = 2 0X X + =
( )2 2 0D X + = ( )2 2 0D + =
2 2D = D = 2
D = i ( ) 1 2X x c Sen x c Cos x = +
T
kT= 2
T = k
2T T + k2T = 0
D + k 2( )T = 0 D + k 2( ) = 0
D = k2
T t( ) = c3ek
2t
( ) ( ) ( ),U x t X x T t= ( ),U x t = ( )1 2c Sen x c Cos x + c3ek2t
Como U (0,t) = 0
c3e
k2t( ) 0
c1Sen0+ c2Cos0( ) = 0 2 0c = Como U (L,t) = 0
U (L,t) = c1SenL = 0 Para evitar solucin trivial c1 0 Senn = 0( )
SenL = Senn
L = n
= nL
n = 1,2,3......
( ),nU x t = c1n Sen
nL
x
c3nek n
2L2
2
t
= Bn Sen
nL
x
ek n
2L2
2
t
U x,t( ) =1n
= Un x,t( )
( ),U x t =1n
=
Bn Sen
nL
x
ek n
2L2
2
t
Como U (x,0) = x(L x) U x,0( ) =
1n
=
Bn Sen
nL
x
ek n
2L2
2
0
=
1n
=
Bn Sen
nL
x
1 = x(L x)
U x,0( ) = 1n
=
Bn Sen
nL
x
= x(L x)
La cual es una expansin en Series de Fourier de Senos en el semintervalo donde el objetivo es encontrar los coeficientes Bn
Bn =2L x(L x)Sen
nL xdx0
L
=2L xLSen
nL xdx0
L
2L x
2Sen nL xdx0
L
-
U x,t( ) =1n
=
Bn Sen
nL
x
ek n
2L2
2
t
=
1n
=
2L
x(L x)Sen nL
x dx0
L
Sen nL
x
ek n
2L2
2
t
Entonces resolviendo la integral
Bn = 2 xSennL xdx0
L
2L x2Sen nL xdx0
L
v = x dw = Sen nL xdx
dv = dx w = Ln CosnL x
r = x2 ds = Sen nL xdx
dr = 2xdx s = Ln CosnL x
Bn = 2xLn Cos
nL x 0
L
+ 2 Ln CosnL xdx0
L
+ 2L
Ln x
2Cos nL x 0
L
2LLn 2xCos
nL xdx0
L
Bn = 2LLn Cos
nL L + 2(0)
Ln Cos
nL 0 + 2
Ln Cos
nL xdx0
L
+ 2L
Ln L
2Cos nL L 2LLn 0
2Cos nL 0 2LLn 2xCos
nL xdx0
L
Bn = 2L2n (1)
n + 0 + 2 Ln CosnL xdx0
L
+ 2L
2
n (1)n 0 2n 2xCos
nL xdx0
L
Bn = 2Ln Cos
nL xdx0
L
4n xCos
nL xdx0
L
= 2Ln
Ln Cos
nL x
Ln dx0
L
4n xCos
nL xdx0
L
v = x dw = Cos nL xdx
dv = dx w = Ln SennL x
Bn =2L2n2 2 Sen
nL x 0
L
4n
Ln xSen
nL x 0
L
+ 4n SennL xdx0
L
Bn =2L2n2 2 Sen
nL L
2L2n2 2 Sen
nL 0
4nLn LSen
nL L +
4n
Ln 0Sen
nL 0 +
4n Sen
nL xdx0
L
-
Bn =4n Sen
nL xdx =
4Lnn Sen
nL x
nL dx0
L
0
L
= 4Ln2 2 Cos
nL x 0
L
= 4Ln2 2 CosnL L +
4Ln2 2 Cos
nL 0
Bn = 4Ln2 2 Cosn +
4Ln2 2 Cos0 =
4Ln2 2 (1)
n + 4Ln2 2 (1) =4Ln2 2 (1)
n +1 =4Ln2 2 1 (1)
n
Bn =4Ln2 2 1 (1)
n
( ),U x t =1n
=
Bn Sen
nL
x
ek n
2L2
2
t
( ),U x t = 1n
=
4Ln2 2
1 (1)n SennL
x
ek n
2L2
2
t
U x,t( ) =1n
=
4Ln2 2
1 (1)n SennL
x
ek n
2L2
2
t
Ecuaciones Diferenciales Parciales Series de FourierEcuaciones Diferenciales Parciales Series de Fourier.2