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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

Ecuaciones Diferenciales Parciales Series de FourierEcuaciones Diferenciales Parciales Series de Fourier.2