Ecuaciones Diferenciales Parciales Por Expanciones Series de Fourier

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Transcript of Ecuaciones Diferenciales Parciales Por Expanciones Series de Fourier

  • 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

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