Ecuaciones diferenciales en derivadas parciales -...

Ecuaciones diferenciales en derivadas parciales

Pedro CorcueraDpto. Matemática Aplicada y Ciencias de la ComputaciónUniversidad de Cantabria

[email protected]

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Ec. derivadas parciales 2

Objetivos

• Revisión de las ecuaciones diferenciales en derivadas parciales y técnicas de solución


Diferenciación

• The mathematical definition of a derivative begins with a difference approximation:

and as ∆x is allowed to approach zero, the difference becomes a derivative:


Diferenciación de funciones continuas

∆y∆x

=f xi + ∆x( )− f xi( )

∆x

dydx

= lim∆x→0

f xi + ∆x( )− f xi( )∆x

• Taylor series expansion can be used to generate high-accuracy formulas for derivatives by using linear algebra to combine the expansion around several points.

• Three categories for the formula include forward finite-difference, backward finite-difference, and centered finite-difference.


Fórmulas de diferenciación de gran exactitud

For a finite


Aproximación por diferencia en adelanto

( ) ( ) ( )x

xfxxfx

xfΔ

Δ0Δ

lim −+→

=′

( ) ( ) ( )x

xfxxfxf∆

−∆+≈′'Δ' x

x x+Δx

f(x)

Graphical Representation of forward difference approximation of first derivative.

• The velocity of a rocket is given by

• Where is given in m/s and is given in seconds.a)Use forward difference approximation of the first derivative

of to calculate the acceleration at . Use a step size of .

b)Find the exact value of the acceleration of the rocket.c) Calculate the absolute relative true error for part (b).


Aproximación por diferencia en adelanto -ejemplo

( ) 300,8.921001014

1014ln2000 4

4

≤≤−

−×

×= tt

ttν

''ν ''t

( )tν st 16=st 2Δ =

• Solution



( ) ( ) ( )t

ttta iii ∆

−≈ + νν 1 16=it 2Δ =t

182161 =+=∆+=+ ttt ii

( ) ( ) ( )2

161816 νν −≈a

( ) ( ) ( )188.91821001014

1014ln200018 4

4

−

−××

=ν m/s02.453=

( ) ( ) ( )168.91621001014

1014ln200016 4

4

−

−××

=ν m/s07.392=

Hence( ) ( ) ( )

2161816 νν −

≈a2

07.39202.453 −≈ 2m/s474.30≈

The exact value of ( )16a can be calculated by differentiating b)

a)

( ) tt

t 8.921001014

1014ln2000 4

4

−

−×

×=ν as ( ) ( )[ ]tν

dtdta =



Knowing that

( )[ ]t

tdtd 1ln = and 2

11ttdt

d−=

( ) 8.921001014

10141014

210010142000 4

4

4

4

−

−×

×

×

−×=

tdtdtta

( ) ( ) ( ) 8.9210021001014

101411014

210010142000 24

4

4

4

−−

−×

×−

×

−×=

tt

tt

32004.294040

+−−−

=

( ) ( )( )163200

164.29404016+−

−−=a 2m/s674.29=

c) The absolute relative true error is

100Value True

Value eApproximat-Value True⋅=∈t 100

674.29474.30674.29 x−

= %6967.2=


Efecto del tamaño de paso en el método de diferencia dividida en adelanto

xexf 49)( =

)2.0(f ′Value of´ using forward difference method.

h )2.0('f aE aε % Significant digits

tE tε %

0.05 88.69336 -8.57389 10.70138 0.025 84.26239 -4.430976 5.258546 0 -4.14291 5.170918 0.0125 82.15626 -2.106121 2.563555 1 -2.03679 2.542193 0.00625 81.12937 -1.0269 1.265756 1 -1.00989 1.260482 0.003125 80.62231 -0.507052 0.628923 1 -0.50284 0.627612 0.001563 80.37037 -0.251944 0.313479 2 -0.25090 0.313152 0.000781 80.24479 -0.125579 0.156494 2 -0.12532 0.156413 0.000391 80.18210 -0.062691 0.078186 2 -0.06263 0.078166 0.000195 80.15078 -0.031321 0.039078 3 -0.03130 0.039073 9.77E-05 80.13512 -0.015654 0.019535 3 -0.01565 0.019534 4.88E-05 80.12730 -0.007826 0.009767 3 -0.00782 0.009766



0

1

2

3

4

5

6

0 1 2 3 4 5 6 7 8 9 10 11 12

Number of times step size halved, n

|Ea|

%

76

80

84

88

92

0 1 2 3 4 5 6 7 8 9 10 11 12


f'(0.

2)

-5

-4

-3

-2

-1

00 2 4 6 8 10 12


E a

0

1

2

3

4

1 2 3 4 5 6 7 8 9 10 11


Leas

t nu

mbe

r of s

igni

fican

t dig

its

corre

ct



-9

-6

-3

00 2 4 6 8 10 12


E t

0

2

4

6

8

10

12

1 2 3 4 5 6 7 8 9 10 11


|Et| %

• Sabemos que

• For a finite ,

• If is chosen as a negative number,


Aproximación por diferencia en atraso

( ) ( ) ( )x

xfxxfx

xfΔ

Δ0Δ

lim −+→

=′

'Δ' x ( ) ( ) ( )x

xfxxfxf∆

−∆+≈′

'Δ' x

( ) ( ) ( )x

xfxxfxf∆−

−∆−≈′ ( ) ( )

xxxfxf

ΔΔ−−

=

• This is a backward difference approximation as youare taking a point backward from x. To find the valueof at , we may choose another pointbehind as . This gives

where


Aproximación por diferencia en atraso

Graphical Representation of backward difference approximation of first derivative.

xx-Δxx

f(x)

( )xf ′ ixx = 'Δ' x1−= ixx

( ) ( ) ( )x

xfxfxf iii ∆

−≈′ −1

( ) ( )1

1

−

−

−−

=ii

ii

xxxfxf

1Δ −−= ii xxx

• Taylor’s theorem says that if you know the value of a function at a point and all its derivatives at that point, provided the derivatives are continuous between and , then

Substituting for convenience


Obtención de la adad a partir de las series de Taylor

f ix

ix 1+ix( ) ( ) ( )( ) ( ) ( ) +−

′′+−′+= +++

2111 !2 ii

iiiiii xxxfxxxfxfxf

ii xxx −= +1Δ( ) ( ) ( ) ( ) ( ) +

′′+′+=+

21 Δ

!2Δ xxfxxfxfxf i

iii

( ) ( ) ( ) ( ) ( ) +∆′′

−∆

−=′ + x

xfx

xfxfxf iii

i !21

( ) ( ) ( ) ( )xOx

xfxfxf iii ∆+

∆−

=′ +1

• The term shows that the error in the approxima-tion is of the order of . It is easy to derive from Taylor series the formula for backward divided difference approximation of the first derivative.

• As shown above, both forward and backward divided difference approximation of the first derivative are accurate on the order of .

• Can we get better approximations? Yes, another method is called the Central difference approxima-tion of the first derivative.


Obtención de la adad a partir de las series de Taylor

x∆( )xO ∆

( )xO ∆

• From Taylor series(1)

(2)

Subtracting equation (2) from equation (1)


Obtención de la adc a partir de las series de Taylor

( ) ( ) ( ) ( ) ( ) ( ) ( ) +′′′

+′′

+′+=+32

1 Δ!3

Δ!2

Δ xxf

xxf

xxfxfxf iiiii

( ) ( ) ( ) ( ) ( ) ( ) ( ) +′′′

−′′

+′−=−32

1 Δ!3

Δ!2

Δ xxf

xxf

xxfxfxf iiiii

( ) ( ) ( )( ) ( ) ( ) +∆′′′

+∆′=− −+3

11 !322 xxfxxfxfxf i

iii

( ) ( ) ( ) ( ) ( ) +∆′′′

−∆−

=′ −+ 211

!32x

xfx

xfxfxf iii

i

( ) ( ) ( ) ( )211

2xO

xxfxfxf ii

i ∆+∆−

=′ −+

• Hence showing that we have obtained a more accurate formula as the error is of the order of


Obtención de la adc a partir de las series de Taylor

( )2xO ∆

x

f(x)

x-Δx x x+Δx

Graphical Representation of central difference approximation of first derivative

• The velocity of a rocket is given by

• Where is given in m/s and is given in seconds.a)Use central divided difference approximation of the first

derivative of to calculate the acceleration at . Use a step size of .

b)Calculate the absolute relative true error for part (a).


Aproximación por diferencia central -ejemplo

( ) 300,8.921001014

1014ln2000 4

4

≤≤−

−×

×= tt

ttν

''ν ''t

( )tν st 16=st 2Δ =

• Solution


Aproximación por diferencia central -ejemplo

( ) ( ) ( )t

ttta iii ∆

−≈ −+

211 νν 16=it 2Δ =t

142161 =−=∆−=− ttt ii

( ) ( ) ( )22

141816⋅−

≈ννa

( ) ( ) ( )188.91821001014

1014ln200018 4

4

−

−××

=ν m/s02.453=

( ) ( ) ( )148.91421001014

1014ln200014 4

4

−

−××

=ν m/s24.334=

Hence( ) ( ) ( )

4141816 νν −

≈a4

24.33402.453 −≈

2m/s694.29≈

The absolute relative true error is knowing that the exact value atis

b)

a)

st 16=( ) 2m/s674.2916 =a

100674.29

694.29674.29⋅

−=∈t %069157.0=

182161 =+=∆+=+ ttt ii


Efecto del tamaño de paso en el método de diferencia dividida central

xexf 49)( =

)2.0(f ′Value of´ using forward difference method.

h )2.0('f aE aε % Significant digits

tE tε %

0.05 80.65467 -0.53520 0.668001 0.025 80.25307 -0.4016 0.500417 1 -0.13360 0.16675 0.0125 80.15286 -0.100212 0.125026 2 -0.03339 0.041672 0.00625 80.12782 -0.025041 0.031252 3 -0.00835 0.010417 0.003125 80.12156 -0.00626 0.007813 3 -0.00209 0.002604 0.001563 80.12000 -0.001565 0.001953 4 -0.00052 0.000651 0.000781 80.11960 -0.000391 0.000488 5 -0.00013 0.000163 0.000391 80.11951 -9.78E-05 0.000122 5 -0.00003 4.07E-05 0.000195 80.11948 -2.45E-05 3.05E-05 6 -0.00001 1.02E-05 9.77E-05 80.11948 -6.11E-06 7.63E-06 6 0.00000 2.54E-06 4.88E-05 80.11947 -1.53E-06 1.91E-06 7 0.00000 6.36E-07



75

85

95

1 3 5 7 9 11

Number of times the step size is halved, n

f'(0.

2)

-1

01 3 5 7 9 11

Number of steps involved, n

E(a)

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5 6 7 8 9 10 11 12


|E(a

)|,%

0

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8 9 10 11


Leas

t nu

mbe

r of s

igni

fican

t dig

its

corre

ct



-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0.00 2 4 6 8 10 12


E(t)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

1 2 3 4 5 6 7 8 9 10 11

Number of times step size halved,n

|Et|

%

• The results from the three difference approximations are given in the following table.


Comparación de métodos adad, adat, adc

Type ofDifferenceApproximationForwardBackwardCentral

30.47528.91529.695

2.69672.5584

0.069157

Summary of a (16) using different divided difference approximations

( )16a( )2/ sm

%t∈

• In real life, one would not know the exact value of the derivative – so how would one know how accurately they have found the value of the derivative.

• A simple way would be to start with a step size and keep on halving the step size until the absolute relative approximate error is within a pre-specified tolerance.

• Take the example of finding for

at using the backward divided difference scheme.


Obtención del valor de la derivada con una tolerancia específica

( ) tt

t 8.921001014

1014ln2000 4

4

−

−×

×=ν

( )tv′

16=t

• The values obtained using the backward difference approximation method and the corresponding absolute relative approximate errors are given in the following table.


Obtención del valor de la derivada con una tolerancia específica

210.50.250.125

28.91529.28929.48029.57729.625

1.27920.647870.326040.16355

t∆ ( )tv′ %a∈ one can see that the absolute relative approximate error decreases as the step size is reduced. At the absolute relative approximate error is 0.16355%, meaning that at least 2 significant digits are correct in the answer.

125.0=∆t


Fórmulas de diferencia finita en adelanto


Fórmulas de diferencia finita en atraso


Fórmulas de diferencia finita centrada


Sistemas de ecuaciones lineales

• A matrix consists of a rectangular array of elements represented by a single symbol (example: [A]).

• An individual entry of a matrix is an element(example: a23)


Matrices

• Matrices where m=n are called square matrices.• There are a number of special forms of square

matrices:


Matrices especiales

Symmetric

A[ ]=5 1 21 3 72 7 8

Diagonal

A[ ]=a11

a22

a33

Identity

A[ ]=1

11

Upper Triangular

A[ ]=a11 a12 a13

a22 a23

a33

Lower Triangular

A[ ]=a11

a21 a22

a31 a32 a33

Banded

A[ ]=

a11 a12

a21 a22 a23

a32 a33 a34

a43 a44

• The elements in the matrix [C] that results from multiplying matrices [A] and [B] are calculated using:


Multiplicación matricial

cij = aikbkjk=1

n

∑

• Matrices provide a concise notation for representing and solving simultaneous linear equations:


Representación de Algebra Lineal

a11x1 + a12x2 + a13x3 = b1

a21x1 + a22x2 + a23x3 = b2

a31x1 + a32x2 + a33x3 = b3

a11 a12 a13

a21 a22 a23

a31 a32 a33

x1

x2

x3

=

b1

b2

b3

[A]{x} = {b}

• MATLAB provides two direct ways to solve systems of linear algebraic equations [A]{x}={b}:– Left-divisionx = A\b

– Matrix inversionx = inv(A)*b

• The matrix inverse is less efficient than left-division and also only works for square, non-singular systems.


Solución con Matlab

• Recall that if a matrix [A] is square, there is another matrix [A]-1, called the inverse of [A], for which [A][A]-1=[A]-1[A]=[I]

• The inverse can be computed in a column by column fashion by generating solutions with unit vectors as the right-hand-side constants:


Matriz inversa

A[ ] x1{ }=100

A[ ] x2{ }=

010

A[ ] x3{ }=

001

A[ ]−1 = x1 x2 x3[ ]

• Recall that LU factorization can be used to efficiently evaluate a system for multiple right-hand-side vectors - thus, it is ideal for evaluating the multiple unit vectors needed to compute the inverse.

• Many systems can be modeled as a linear combination of equations, and thus written as a matrix equation:

• The system response can thus be found using the matrix inverse.


Matriz inversa y sistemas estímulo -respuesta

Interactions[ ] response{ }= stimuli{ }

• A norm is a real-valued function that provides a measure of the size or “length” of multi-component mathematical entities such as vectors and matrices.

• Vector norms and matrix norms may be computed differently.


Normas vectoriales y matriciales

• For a vector {X} of size n, the p-norm is:

• Important examples of vector p-norms include:


Normas vectoriales

ini

n

iie

n

ii

xXp

xXXp

xXp

maxmagnitudemaximum:

(length) normEuclidian :2

valuesabsolute theof sum:1

1

1

22

11

≤≤∞

=

=

=−∞=

===

==

∑

∑

X p = xip

i=1

n

∑

1/ p

• Common matrix norms for a matrix [A] include:

• Note - µmax is the largest eigenvalue of [A]T[A].


Normas matriciales

column - sum norm A 1 =1≤ j≤nmax aij

i=1

n

∑

Frobenius norm A f = aij2

j=1

n

∑i=1

n

∑

row - sum norm A∞

=1≤i≤nmax aij

j=1

n

∑spectral norm (2 norm) A

2= µmax( )1/2

• The matrix condition number Cond[A] is obtained by calculating Cond[A]=||A||·||A-1||

• In can be shown that:

• The relative error of the norm of the computed solution can be as large as the relative error of the norm of the coefficients of [A] multiplied by the condition number.

• If the coefficients of [A] are known to t digit precision, the solution [X] may be valid to onlyt-log10(Cond[A]) digits.


Número de condición de una matriz

∆XX

≤ Cond A[ ]∆AA

• MATLAB has built-in functions to compute both norms and condition numbers:– norm(X,p)

• Compute the p norm of vector X, where p can be any number, inf, or ‘fro’ (for the Euclidean norm)

– norm(A,p)

• Compute a norm of matrix A, where p can be 1, 2, inf, or ‘fro’ (for the Frobenius norm)

– cond(X,p) or cond(A,p)• Calculate the condition number of vector X or matrix A using

the norm specified by p.Ec. derivadas parciales 42

Comandos Matlab

• The Gauss-Seidel method is the most commonly used iterative method for solving linear algebraic equations [A]{x}={b}.

• The method solves each equation in a system for a particular variable, and then uses that value in later equations to solve later variables. For a 3x3 system with nonzero elements along the diagonal, for example, the jth iteration values are found from the j-1th iteration using:


Método iterativo: Gauss - Seidel

x1j =

b1 − a12x2j−1 − a13x3

j−1

a11

x2j =

b2 − a21x1j − a23x3

j−1

a22

x3j =

b3 − a31x1j − a32x2

j

a33

• The Jacobi iteration is similar to the Gauss-Seidel method, except the j-1th information is used to update all variables in the jth iteration:

a) Gauss-Seidelb) Jacobi


Iteración de Jacobi

• The convergence of an iterative method can be calculated by determining the relative percent change of each element in {x}. For example, for the ithelement in the jth iteration,

• The method is ended when all elements have converged to a set tolerance.


Convergencia

%1001

, ×−

=−

ji

ji

ji

ia xxxε

• The Gauss-Seidel method may diverge, but if the system is diagonally dominant, it will definitely converge.

• Diagonal dominance means:


Dominancia diagonal

aii > aijj=1j≠i

n

∑

• To enhance convergence, an iterative program can introduce relaxation where the value at a particular iteration is made up of a combination of the old value and the newly calculated value:

where λ is a weighting factor that is assigned a value between 0 and 2.– 0<λ<1: underrelaxation– λ=1: no relaxation– 1<λ≤2: overrelaxation


Relajación

( ) oldnewnew 1 iii xxx λλ −+=


Ecuaciones diferenciales en derivadas parciales

• Ordinary Differential Equations have only one independent variable

• Partial Differential Equations have more than one independent variable

• subject to certain conditions: where u is the dependent variable, and x and y are the independent variables.


Ecuaciones en derivadas parciales

5)0(,353 2 ==+ − yeydxdy x

222

2

2

2

3 yxyu

xu

+=∂∂

+∂∂

• where are functions of , and . is a function of

• can be:Elliptic if B2 – 4AC < 0Parabolic if B2 – 4AC = 0Hyperbolic if B2 – 4AC > 0


Clasificación de EDPs de 2 orden

02

22

2

2

=+∂∂

+∂∂

∂+

∂∂ D

yuC

yxuB

xuA

, , and , .u ux y ux y

∂ ∂∂ ∂

CBA and,, yx andD

• EllipticLaplace equation

• ParabolicHeat equation

• Hyperbolic Wave equation


Ejemplos de EDPs de 2 orden

02

2

2

2

=∂∂

+∂∂

yT

xT

2

2

xTk

tT

∂∂

=∂∂

2

2

22

2 1ty

cxy

∂∂

=∂∂

2

1,0,1c

CBA −===

0,0, === CBkA

1,0,1 === CBA

• Schematic diagram of a plate with specified temperature boundary conditions

• The Laplace equation governs the temperature:


Ejemplo físico de una PDE elíptica

02

2

2

2

=∂∂

+∂∂

yT

xT

bT

lT

tT

rT

L

W

x

y


Discretizando la PDE elíptica

tT

rT

x

y

),( ji ),1( ji +

)1,( −ji

),1( ji −

)1,( +ji

)0,0()0,(m

),0( n

bT

lT ),( ji

x∆

y∆x∆y∆

mLx =∆

nWy =∆

( )22

2 ),(),(2),(),(x

yxxTyxTyxxTyxxT

∆∆−+−∆+

≅∂∂

( )22

2 ),(),(2),(),(y

yyxTyxTyyxTyxyT

∆∆−+−∆+

≅∂∂



tT

rT

x

y

),( ji ),1( ji +

)1,( −ji

),1( ji −

)1,( +ji

)0,0()0,(m

),0( n

bT

lT ),( ji

x∆

y∆x∆y∆

( )2,1,,1

,2

2 2x

TTTxT jijiji

ji ∆

+−≅

∂∂ −+

( )21,,1,

,2

2 2y

TTTyT jijiji

ji ∆

+−≅

∂∂ −+

( )22

2 ),(),(2),(),(x

yxxTyxTyxxTyxxT

∆∆−+−∆+

≅∂∂

( )22

2 ),(),(2),(),(y

yyxTyxTyyxTyxyT

∆∆−+−∆+

≅∂∂

• Substituting these approximations into the Laplace equation yields:

• if,• the Laplace equation can be rewritten as

(Eq. 1)• there are several numerical methods that can be used to solve the

problem:Direct MethodGauss-Seidel MethodLieberman Method



02

2

2

2

=∂∂

+∂∂

yT

xT

( ) ( )0

222

1,,1,2

,1,,1 =∆

+−+

∆

+− −+−+

yTTT

xTTT jijijijijiji

yx ∆=∆

04 ,1,1,,1,1 =−+++ −+−+ jijijijiji TTTTT

• Consider a plate that is subjected to the boundary conditions shown below. Find the temperature at the interior nodes using a square grid with a length of by using the direct method.


Ejemplo 1: Método directo

mm 0.34.2 ×

m6.0

C°50

C°75

C°300

C°100

m4.2

m0.3

x

y

L

W

• We discretize the plate by taking,

• The nodal temperatures at the boundary nodes are given by:



myx 6.0=∆=∆

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4TC°300

C°100

C°50

C°75

3,2,1,3003,2,1,50

4,3,2,1,100

4,3,2,1,75

5,

0,

,4

,0

==

==

==

==

iTiT

jTjT

i

i

j

j

4=∆

=x

Lm 5=∆

=y

Wn

• the equation for the temperature at the node (2,3)

• i=2 and j=3



x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

3,2T

04 3,22,24,23,13,3 =−+++ TTTTT04 3,34,23,22,23,1 =++−+ TTTTT


• We can develop similar equations for every interior node leaving us with an equal number of equations and unknowns.

• For this problem the number of equations generated is 12



• The corner nodal temperature of are not needed• To get the temperature at the interior nodes we have to write Equation 1

for all the combinations of i and j, i=1 and j=1i=1 and j=2i=1 and j=3i=1 and j=4i=2 and j=1i=2 and j=2i=2 and j=3i=2 and j=4i=3 and j=1i=3 and j=2i=3 and j=3i=3 and j=4



0,00,45,45,0 ,,, TTTT

1,....,1;1,....,1 −=−= njmi1254 1,22,11,1 −=++− TTT

754 2,23,12,11,1 −=++− TTTT754 3,24,13,12,1 −=++− TTTT

3754 4,24,13,1 −=+− TTT504 1,32,21,21,1 −=++− TTTT

04 2,33,22,21,22,1 =++−+ TTTTT04 3,34,23,22,23,1 =++−+ TTTTT

3004 4,34,23,24,1 −=+−+ TTTT1504 2,31,31,2 −=+− TTT

1004 3,32,31,32,2 −=+−+ TTTT1004 4,33,32,33,2 −=+−+ TTTT

4004 4,33,34,2 −=−+ TTT

• We can use Excel and matrix operations to solve the linear equations system



T1,1 T1,2 T1,3 T1,4 T2,1 T2,2 T2,3 T2,4 T3,1 T3,2 T3,3 T3,4 RHE-4 1 0 0 1 0 0 0 0 0 0 0 -125 T1,1 74.87191 -4 1 0 0 1 0 0 0 0 0 0 -75 T1,2 95.89590 1 -4 1 0 0 1 0 0 0 0 0 -75 T1,3 127.80360 0 1 -4 1 0 0 1 0 0 0 0 -375 T1,4 196.92881 0 0 0 -4 1 0 0 1 0 0 0 -50 T2,1 78.59170 1 0 0 1 -4 1 0 0 1 0 0 0 T2,2 105.90820 0 1 0 0 1 -4 1 0 0 1 0 0 T2,3 143.38960 0 0 1 0 0 1 -4 0 0 0 1 -300 T2,4 206.32000 0 0 0 1 0 0 0 -4 1 0 0 -150 T3,1 83.58680 0 0 0 0 1 0 0 1 -4 1 0 -100 T3,2 105.75540 0 0 0 0 0 1 0 0 1 -4 1 -100 T3,3 133.52670 0 0 0 0 0 0 1 0 0 1 -4 -400 T3,4 184.9617

300.0 300.0 300.0

75.0 196.9 206.3 185.0 100.0

75.0 127.8 143.4 133.5 100.0

75.0 95.9 105.9 105.8 100.0

75.0 74.9 78.6 83.6 100.0

50.0 50.0 50.0

• Recall the discretized equation

• This can be rewritten as

• For the Gauss-Seidel Method, this equation is solved iteratively for all interior nodes until a pre-specified tolerance is met.


Método Gauss-Seidel


41,1,,1,1

,−+−+ +++

= jijijijiji

TTTTT

• Consider a plate that is subjected to the boundary conditions shown below. Find the temperature at the interior nodes using a square grid with a length of using the Gauss-Siedelmethod. Assume the initial temperature at all interior nodes to be .

Ec. derivadas parciales63

Ejemplo 2: Método Gauss-Seidel

mm 0.34.2 ×

m6.0

C°50

C°75

C°300

C°100

m4.2

m0.3

x

y

L

W

C°0

• Discretizing the plate by taking,

• The nodal temperatures at the boundary nodes are given by:



myx 6.0=∆=∆

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4TC°300

C°100

C°50

C°75

3,2,1,3003,2,1,50

4,3,2,1,100

4,3,2,1,75

5,

0,

,4

,0

==

==

==

==

iTiT

jTjT

i

i

j

j

4=∆

=x

Lm 5=∆

=y

Wn

• Now we can begin to solve for the temperature at each interior node using

• Assume all internal nodes to have an initial temperature of zero.• Iteration 1:

i=1 and j=1 i=2 and j=3 i=1 and j=2 i=2 and j=4 i=1 and j=3 i=3 and j=1i=1 and j=4 i=3 and j=2i=2 and j=1 i=3 and j=3i=2 and j=2 i=3 and j=4



CT º25.311,1 =

5,4,3,2,1;4,3,2,1,4

1,1,,1,1, ==

+++= −+−+ ji

TTTTT jijijiji

ji

CT º5625.262,1 =

CT º3906.253,1 =

CT º098.1004,1 =

CT º3125.201,2 =

CT º7188.112,2 =

CT º27735.93,2 =

CT º344.1024,2 =

CT º5781.421,3 =

CT º5742.382,3 =

CT º9629.363,3 =

CT º827.1344,3 =

• Iteration 2:we take the temperatures from iteration 1 and calculate the approximated error.

i=1, j=1 i=2, j=3 i=1, j=2 i=2, j=4 i=1, j=3 i=3, j=1i=1, j=4 i=3, j=2i=2, j=1 i=3, j=3i=2, j=2 i=3, j=4



CT º9688.421,1 =

CT º7596.382,1 =

CT º7862.553,1 =

CT º283.1334,1 =

CT º8164.361,2 =

CT º8594.302,2 =

CT º4881.563,2 =

CT º150.1564,2 =

CT º3477.561,3 =

%27.271,1=aε %58.83

3,2=aε

%49.312,1=aε

100,

,,,

×−

= presentji

previousji

presentji

jia TTT

ε

%49.543,1=aε

%90.244,1=aε

%83.441,2=aε

%03.622,2=aε

%46.344,2=aε

%44.241,3=aε

%70.312,3=aε

%44.573,3=aε

%12.164,3=aε

CT º0425.562,3 =

CT º8394.863,3 =

CT º747.1604,3 =

67


Node Temperature Distribution in the Plate (°C)Number of Iterations

1 2 10

31.2500 42.9688 73.0239

26.5625 38.7695 91.9585

25.3906 55.7861 119.0976

100.0977 133.2825 172.9755

20.3125 36.8164 76.6127

11.7188 30.8594 102.1577

9.2773 56.4880 137.3802

102.3438 156.1493 198.1055

42.5781 56.3477 82.4837

38.5742 56.0425 103.7757

36.9629 86.8393 130.8056

134.8267 160.7471 182.2278

1,1T

2,1T

3,1T4,1T

1,2T

2,2T

3,2T4,2T

1,3T2,3T

3,3T4,3T

Ec. derivadas parciales

• The numerical solution of Laplace equation at a point is the average of four neighbors

• Example for cell S8: =(S7+S9+R8+T8)/4

• Enter the boundary conditions in the appropriate cells. • Copy and paste to cover the cells where values of the potential are to

be calculated. This calculation contains a "circular reference“.


Ejemplo 2: Método Gauss-Seidel in Excel

41,1,,1,1

,−+−+ +++

= jijijijiji

TTTTT

• To allow circular references and enable iterations:File → Options → FormulasOn the "Calculations options" form select "Enable iterative calculation"

We can increase the Maximum Iterations (100 is the deafult) andreduce the Maximum Change (0.001 is the default). Iterations will stop when the maximum iteration is reached or the change is less than themaximum change.

• F9 to recalculate.Ec. derivadas parciales 69


• Color cell based on valueTo achieve the cell color based on value: Inicio → Estilos → Formato condicional → Escalas de color → Más reglasWe can chose a 3 color scale with blue for mimimum, white or gray formidpoint and red for maximum.



• Plotting the resultsNormally we use the chart type Surface or Contour.




Ejemplo 3: Uso de SolverFinite Difference Solution

8 0 0 0 0 0 0 07 100 51.74 33.90 26.95 25.07 26.95 33.90 51.74 1006 100 73.07 56.89 48.82 46.39 48.82 56.89 73.07 1005 100 83.64 71.80 65.03 62.86 65.03 71.80 83.64 1004 100 89.71 81.62 76.65 75.00 76.65 81.62 89.71 1003 100 93.56 88.31 84.97 83.83 84.97 88.31 93.56 1002 100 96.23 93.11 91.07 90.37 91.07 93.11 96.23 1001 100 98.26 96.80 95.85 95.52 95.85 96.80 98.26 1000 100 100 100 100 100 100 100

x\y 0 1 2 3 4 5 6 7 8

Residuals-squared87 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00006 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00005 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00004 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00003 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00002 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.00000

x\y 0 1 2 3 4 5 6 7 8

sum = 2.7745E-13

=(-4*D5+D4+D6+C5+E5)^2

=SUMA(D17:J23)

• Recall the equation used in the Gauss-Siedel Method,

• Because the Gauss-Siedel Method is guaranteed to converge, we can accelerate the process by using overrelaxation. In this case,

• The λ is known as the “overrelaxation parameter" and is in the range 0 < λ < 2.


Método de Lieberman

41,1,,1,1

,−+−+ +++

= jijijijiji

TTTTT

oldji

newji

relaxedji TTT ,,, )1( λλ −+=

• In the past examples, the boundary conditions on the plate had a specified temperature on each edge. What if the conditions are different ? For example, what if one of the edges of the plate is insulated.

• In this case, the boundary condition would be the derivative of the temperature. Because if the right edge of the plate is insulated, then the temperatures on the right edge nodes also become unknowns.


Condiciones de contorno alternativas

C°50

C°75

C°300

m4.2

m0.3

x

y

Insulated

• The finite difference equation in this case for the right edge for the nodes for

• However the node is not inside the plate. The derivative boundary condition needs to be used to account for these additional unknown nodal temperatures on the right edge. This is done by approximating the derivative at the edge node as



C°50

C°75

C°300

m4.2

m0.3

x

y

Insulated

),( jm 1,..3,2 −= nj04 ,1,1,,1,1 =−+++ +−−+ jmjmjmjmjm TTTTT

),1( jm +

),( jm

)(2,1,1

, xTT

xT jmjm

jm ∆

−≅

∂∂ −+

• Rearranging this approximation gives us,

• We can then substitute this into the original equation gives us,

• Recall that is the edge is insulated then,

• Substituting this again yields,



jmjmjm x

TxTT,

,1,1 )(2∂∂

∆+= −+

04)(22 ,1,1,,

,1 =−++∂∂

∆+ +−− jmjmjmjm

jm TTTxTxT

0,

=∂∂

jmxT

042 ,1,1,,1 =−++ +−− jmjmjmjm TTTT

• The general form for a second order linear PDE with two independent variables and one dependent variable is

• The criteria for an equation of this type to be considered parabolic:

• Examine the heat-conduction equation given by

where thus we can classify this equation as parabolic.


Ecuaciones en derivadas parciales parabólicas

02

22

2

2

=++∂∂

+∂∂

+∂∂

+∂∂

∂+

∂∂ GFu

yuE

xuD

yuC

yxuB

xuA

042 =− ACB

tT

xT

∂∂

=∂∂

2

2

α

0,0,1,0,0,0, ==−===== GFEDCBA α

k = thermal conductivity of rod material,ρ = density of rod material,C = specific heat of the rod material.C

kρ

α =where

• Consider the flow of heat within a metal rod of length L, one end of which is held at a known high temperature, the other end at a lower temperature.– Heat will flow from the hot end to the cooler end. – We'll assume that the rod is perfectly insulated, so that

heat loss through the sides can be neglected.

• We want to calculate the temperature along the length of the rod as a function of time.


Ejemplo de una EDP parabólica

• For a rod of length divided into nodes

• The time is similarly broken into time steps of

• Hence corresponds to the temperature at node ,that is,

and time


Discretización de una EDP Parabólica

L 1+nnLx =∆

t∆

jiT i

( )( )xix ∆= ( )( )tjt ∆=

Schematic diagram showing interior nodes

x

1−i i 1+i

x∆ x∆

• If we define we can then write the finite central divided

difference approximation of the left hand side at a general interior node( ) as where ( ) is the node number along

the time.• The time derivative on the right hand side is approximated by the

forward divided difference method as,


Solución EDP Parabólica: Método explícito

nLx =∆

Schematic diagram showing interior nodes

x

1−i i 1+i

x∆ x∆

i( )2

11

,2

2 2x

TTTxT j

ij

ij

i

ji ∆+−

≅∂∂ −+ j

tTT

tT j

ij

i

ji ∆−

≅∂∂ +1

,

• Substituting these approximations into the governing equation yields

• Solving for the temp at the time node gives

• choosing,

• we can write the equation as,

• we can be solved explicitly: for each internal location node of the rod for time node in terms of the temperature at time node . If we know the temperature at node , and the boundary temperatures, we can find the temperature at the next time step. We continue the process until we reach the time at which we are interested in finding the temperature.


Solución EDP Parabólica: Método explícito

( ) tTT

xTTT j

ij

ij

ij

ij

i

∆−

=∆

+− +−+

1

211 2

α

1+j

( )ji

ji

ji

ji

ji TTT

xtTT 112

1 2)( −+

+ +−∆∆

+= α

2)( xt

∆∆

= αλ

( )ji

ji

ji

ji

ji TTTTT 11

1 2 −++ +−+= λ

1+j j0=j

• Consider a steel rod that is subjected to a temperature of on the left end and on the right end. If the rod is of length ,use the explicit method to find the temperature distribution in the rod from and seconds. Use , .

• Given: , ,

• The initial temperature of the rod is .


Ejemplo 1 EDP Parabólica: Método explícito

C°100C°25 m05.0

0=t9=t mx 01.0=∆ st 3=∆

KmWk−

= 54 37800mkg

=ρKkg

JC−

= 490

C°20

0=i 1 2 3 4 5

m01.0

CT °= 25CT °=100

• Number of time steps

• Recall,

• Then,

• Boundary Conditions

• All internal nodes are at for :



smCk /104129.1

490780054 25−×=×

==ρ

α

33

09=

−=

∆−

=ttt initialfinal

( ) ( )4239.0

01.03104129.1 2

52 =×=

∆∆

= −

xtαλ

3,2,1,0allfor25

100

5

0 =

°=

°=j

CTCT

j

j

C°20 sec0=t 1,2,3,4 allfor ,200 =°= iCTi

CT °=10000

nodesInterior

20

20

20

20

04

03

02

01

°=

°=

°=

°=

CTCTCTCT

CT °= 2505

We can now calculate the temperature at each nodeexplicitly using the equation formulated earlier,

( )ji

ji

ji

ji

ji TTTTT 11

1 2 −++ +−+= λ

• Nodal temperatures vs. Time



sec0=t 0=jCT °=1000

0

nodesInterior

20

20

20

20

04

03

02

01

°=

°=

°=

°=

CTCTCTCT

CT °= 2505

ConditionBoundary10010 −°= CT

CTCTCT

CT

°=

°=

°=

°=

120.22

20

20

912.53

14

13

12

11

ConditionBoundary2515 −°= CT

sec3=t 1=jCT °=1002

0

CTCTCTCT

°=

°=

°=

°=

442.22

889.20

375.34

073.59

24

23

22

21

CT °= 2525

sec6=t 2=j sec9=t 3=jCT °=1003

0

CTCTCTCT

°=

°=

°=

°=

872.22

266.27

132.39

953.65

34

33

32

31

CT °= 2535



Time-dependent Temperature Distribution in a Brass Rod

(Temperature values in bold are constant)

length, cm 10

heat capacity of brass, cal/g/deg 0,09 (hcap)

thermal conductivity of brass, cal/sec/cm/deg 0,26 (k)

density of brass, g/cm3 8,4 (rho)

Coefficient e in general PDE, =k/(hcap*rho) 0,33(e)

∆x 1 (Dx)

∆t 1 (Dt)

f=e*Dt/(Dx^2) 0,33 (f)

Distance x (cm)

0 1 2 3 4 5 6 7 8 9 10

time

t (se

c)

0 100 0 0 0 0 0 0 0 0 0 0

1 100 32,9 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0

2 100 44,2 10,8 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0

3 100 51,6 18,2 3,6 0,0 0,0 0,0 0,0 0,0 0,0 0

4 100 56,5 24,4 7,2 1,2 0,0 0,0 0,0 0,0 0,0 0

5 100 60,3 29,3 10,9 2,8 0,4 0,0 0,0 0,0 0,0 0

6 100 63,2 33,4 14,3 4,7 1,0 0,1 0,0 0,0 0,0 0

7 100 65,5 36,9 17,4 6,6 1,9 0,4 0,0 0,0 0,0 0

8 100 67,5 39,9 20,3 8,6 3,0 0,8 0,1 0,0 0,0 0

9 100 69,1 42,5 22,9 10,6 4,1 1,3 0,3 0,1 0,0 0

10 100 70,5 44,8 25,3 12,5 5,3 1,9 0,5 0,1 0,0 0

11 100 71,8 46,9 27,5 14,4 6,6 2,6 0,9 0,2 0,0 0

12 100 72,9 48,7 29,6 16,1 7,8 3,3 1,2 0,4 0,1 0

13 100 73,8 50,4 31,4 17,8 9,1 4,1 1,6 0,6 0,2 0

14 100 74,7 51,9 33,2 19,4 10,3 4,9 2,1 0,8 0,2 0

15 100 75,5 53,2 34,8 21,0 11,5 5,8 2,6 1,0 0,3 0

16 100 76,2 54,5 36,3 22,4 12,7 6,6 3,1 1,3 0,5 0

17 100 76,9 55,7 37,7 23,8 13,9 7,5 3,7 1,6 0,6 0

18 100 77,5 56,8 39,1 25,1 15,1 8,4 4,3 2,0 0,7 0

19 100 78,1 57,8 40,3 26,4 16,2 9,2 4,9 2,3 0,9 0

20 100 78,6 58,7 41,5 27,6 17,2 10,1 5,5 2,7 1,1 0

21 100 79,1 59,6 42,6 28,8 18,3 10,9 6,1 3,1 1,2 0

22 100 79,6 60,4 43,6 29,9 19,3 11,7 6,7 3,5 1,4 0

23 100 80,0 61,2 44,6 30,9 20,3 12,6 7,3 3,8 1,6 0

24 100 80,4 61,9 45,6 31,9 21,2 13,4 7,9 4,2 1,8 0

25 100 80,8 62,6 46,5 32,9 22,2 14,2 8,5 4,6 2,0 0

0

10

20

30

40

50

0 20 40 60 80 100

Tem

pera

ture

ºC

Time sec.

Temperature at x=5cm.

0

20

40

60

80

100

120

0 2 4 6 8 10

Tem

pera

ture

ºC

Loc ation on rod cm.

Temperature distribution along the length of the rod

t = 20 sec.t = 50 sec.t = 100 sec.

• Using the explicit method, we were able to find the temperature at each node, one equation at a time.

• However, the temperature at a specific node was only dependent on the temperature of the neighboring nodes from the previous time step. This is contrary to what we expect from the physical problem.

• The implicit method allows us to solve this and other problems by developing a system of simultaneous linear equations for the temperature at all interior nodes at a particular time.

• The second derivative is approximated by the CDD and the first derivative by the BDD scheme at time level j+1 at node ( i ) as


Solución EDP Parabólica: Método implícito

tT

xT

∂∂

=∂∂

2

2

α ( )2

11

111

1,2

2 2x

TTTxT j

ij

ij

i

ji ∆+−

≈∂∂ +

−++

+

+

tTT

tT j

ij

i

ji ∆−

≈∂∂ +

+

1

1,

• Substituting these approximations into the heat conduction equation yields

• Rearranging yields

given that

• The rearranged equation can be written for every node during each time step. These equations can then be solved as a simultaneous system of linear equations to find the nodal temperatures at a particular time.


Solución EDP Parabólica: Método implícito

tT

xT

∂∂

=∂∂

2

2

α ( ) tTT

xTTT j

ij

ij

ij

ij

i

∆−

=∆

+− ++−

+++

1

2

11

111 2α

ji

ji

ji

ji TTTT =−++− +

+++

−1

111

1 )21( λλλ

( )2xt

∆∆

= αλ

• Consider a steel rod that is subjected to a temperature of on the left end and on the right end. If the rod is of length ,use the implicit method to find the temperature distribution in the rod from and seconds. Use , .

• Given: , ,



Ejemplo 2 EDP Parabólica: Método implícito

C°100C°25 m05.0

0=t9=t mx 01.0=∆ st 3=∆

KmWk−

= 54 37800mkg

=ρKkg

JC−

= 490

C°20

0=i 1 2 3 4 5

m01.0

CT °= 25CT °=100


• Recall,

• Then,





smCk /104129.1

490780054 25−×=×

==ρ

α

33

09=

−=

∆−

=ttt initialfinal

( ) ( )4239.0

01.03104129.1 2

52 =×=

∆∆

= −

xtαλ

3,2,1,0allfor25

100

5

0 =

°=

°=j

CTCT

j

j


CT °=10000

nodesInterior

20

20

20

20

04

03

02

01

°=

°=

°=

°=

CTCTCTCT

CT °= 2505

We can now form the system of equations for the first time step by writing the approximated heat conduction equation for each node

ji

ji

ji

ji TTTT =−++− +

+++

−1

111

1 )21( λλλ

• Nodal temperatures when• For the first time step we can write four such equations with four

unknowns, expressing them in matrix form yields

• The above coefficient matrix is tri-diagonal, so special algorithms (e.g.Thomas’ algorithm) can be used to solve. The solution is given by



sec3=t

=

−−−

−−−

598.302020390.62

8478.14239.0004239.08478.14239.0004239.08478.14239.0004239.08478.1

14

13

12

11

TTTT

=

477.21438.21792.24451.39

14

13

12

11

TTTT

=

25477.21438.21792.24451.39

100

15

14

13

12

11

10

TTTTTT

• Nodal temperatures when:



sec3=t

=

25477.21438.21792.24451.39

100

15

14

13

12

11

10

TTTTTT

sec6=t sec9=t

=

25836.22876.23669.30326.51

100

25

24

23

22

21

20

TTTTTT

=

25243.24809.26292.36043.59

100

35

34

33

32

31

30

TTTTTT

• Using the implicit method our approximation of was of accuracy

, while our approximation of was of accuracy.

• One can achieve similar orders of accuracy by approximating the second derivative, on the left hand side of the heat equation, at the midpoint of the time step. Doing so yields

• The first derivative, on the right hand side of the heat equation, is approximated using the forward divided difference method at time level

,


Solución EDP Parabólica: Método Crank-Nicolson

2

2

xT

∂∂

2)( xO ∆tT

∂∂ )( tO ∆

( ) ( )

∆+−

+∆

+−≈

∂∂ +

−++

+−+2

11

111

211

,2

2 222 x

TTTx

TTTxT j

ij

ij

ij

ij

ij

i

ji

α

1+j

tTT

tT j

ij

i

ji ∆−

≈∂∂ +1

,

• Substituting these approximations into the governing equation for heat conductance yields

• giving

• where

• Having rewritten the equation in this form allows us to discretize the physical problem. We then solve a system of simultaneous linear equations to find the temperature at every node at any point in time.


Solución EDP Parabólica: Método Crank-Nicolson

( ) ( ) tTT

xTTT

xTTT j

ij

ij

ij

ij

ij

ij

ij

i

∆−

=

∆+−

+∆

+− ++−

+++−+

1

2

11

111

211 22

2α

ji

ji

ji

ji

ji

ji TTTTTT 11

11

111 )1(2)1(2 +−

++

++− +−+=−++− λλλλλλ

( )2xt

∆∆

= αλ

• Consider a steel rod that is subjected to a temperature of on the left end and on the right end. If the rod is of length ,use the Crank-Nicolson method to find the temperature distribution in the rod from and seconds. Use , .

• Given: , ,



Ejemplo 3 EDP Parabólica: Método Crank-Nicolson

C°100C°25 m05.0

0=t 9=t mx 01.0=∆ st 3=∆

KmWk−

= 54 37800mkg

=ρKkg

JC−

= 490

C°20

0=i 1 2 3 4 5

m01.0

CT °= 25CT °=100


• Recall,

• Then,





smCk /104129.1

490780054 25−×=×

==ρ

α

33

09=

−=

∆−

=ttt initialfinal

( ) ( )4239.0

01.03104129.1 2

52 =×=

∆∆

= −

xtαλ

3,2,1,0allfor25

100

5

0 =

°=

°=j

CTCT

j

j


CT °=10000

nodesInterior

20

20

20

20

04

03

02

01

°=

°=

°=

°=

CTCTCTCT

CT °= 2505

We can now form the system of equations for the first time step by writing the approximated heat conduction equation for each node

ji

ji

ji

ji

ji

ji TTTTTT 11

11

111 )1(2)1(2 +−

++

++− +−+=−++− λλλλλλ

• Nodal temperatures when• For the first time step we can write four such equations with four

unknowns, expressing them in matrix form yields

• The above coefficient matrix is tri-diagonal, so special algorithms (e.g.Thomas’ algorithm) can be used to solve. The solution is given by



sec3=t

=

−−−

−−−

718.52000.40000.4030.116

8478.24239.0004239.08478.24239.0004239.08478.24239.0004239.08478.2

14

13

12

11

TTTT

=

607.21797.20746.23372.44

14

13

12

11

TTTT

=

25607.21797.20746.23372.44

100

15

14

13

12

11

10

TTTTTT

• Nodal temperatures when:



sec3=t sec6=t sec9=t

=

25607.21797.20746.23372.44

100

15

14

13

12

11

10

TTTTTT

=

25730.22174.23075.31883.55

100

25

24

23

22

21

20

TTTTTT

=

25042.24562.26613.37604.62

100

35

34

33

32

31

30

TTTTTT

• The table below allows you to compare the results from all three methods discussed in juxtaposition with the analytical solution.


Comparación de métodos: temperaturas en 9 seg.

Node Explicit Implicit Crank-Nicolson Analytical

34

33

32

31

TTTT

042.24562.26613.37604.62

243.24809.26292.36043.59

872.22266.27132.39953.65

610.23844.25084.37510.62

• The general form for a second order linear PDE with two independent variables and one dependent variable is

• The criteria for an equation of this type to be considered hyperbolic:

• The wave equation (oscillatory systems) given by

where thus we can classify this equation as hyperbolic.


Ecuaciones en derivadas parciales hiperbólicas

02

22

2

2

=++∂∂

+∂∂

+∂∂

+∂∂

∂+

∂∂ GFu

yuE

xuD

yuC

yxuB

xuA

042 >− ACB

2

2

2

2

xyk

ty

∂∂

=∂∂

1,0,1 −=== CBA

T = tension,g = gravitational constant,w = weight/unit =W/L ,W=weight, L=lengthw

Tgk =where

• A string of certain length and weight is under a fixed tension. Initially the mid-point of the string is displaced some distance from its equilibrium position and released.

• We want to calculate the displacement as a function of time at fixed intervals along the length of the string.


Ejemplo de una EDP hiperbólica

• Once again, we can solve the problem by replacing derivatives by finite differences.

• which, when rearranged, yields

• If we set , the above equation is simplified to

• When employing the simplified equation, the value of is determined by the expression .To begin the calculations (value at t1),

it is required values of the function at t0 = 0 and also a value at t-1. We can get a value for the function at t-1 by making use of the fact that the function is periodic. We can use for the first row.


Solución EDP hiperbólica: Método explícito

( ) ( )

∆

+−=

∆+− −+

−+

211

2

11 22x

TTTw

Tgt

TTT ji

ji

ji

ji

ji

ji

( )( )

( )( )

ji

ji

ji

ji

ji T

xt

wTgTTT

xt

wTgT

∆∆

−+−+∆∆

= −−+

+2

21

112

21 12)(

( ) ( ) 122 =∆∆ xwtTg1

111 −

−++ −+= j

ij

ij

ij

i TTTTt∆

wTgxt/

∆=∆

2

01

011 −+ +

= iii

TTT

• A string 50 cm long and weighing 0.5 g is under a tension of 33 kg. Initially the mid-point of the string is displaced 0.5 cm from its equilibrium position and released. We want to calculate the displacement as a function of time at 5 cm intervals along the length of the string, using equation

• From equation the must be 8.8 x 10-5 seconds.


Ejemplo 1 EDP hiperbólica

111

1 −−+

+ −+= ji

ji

ji

ji TTTT

wTgxt/

∆=∆ t∆


Ejemplo 1 EDP hiperbólicaThe Wave Equation:Vibration of a String

length, cm 50 (L)tension, g 33000 (T)weight,g 0,5 (Wt)weight per unit length, g/cm 0,01 (w)

gravitational constant, cm/sec2 980 (g)Dx 5 (Dx)Dt 8,79E-05 (Dt)

Distance x (cm)0 5 10 15 20 25 30 35 40 45 50

time

t (se

c)

0 0 0,1 0,2 0,3 0,4 0,5 0,4 0,3 0,2 0,1 08,8E-05 0 0,1 0,2 0,3 0,4 0,4 0,4 0,3 0,2 0,1 01,8E-04 0 0,1 0,2 0,3 0,3 0,3 0,3 0,3 0,2 0,1 02,6E-04 0 0,1 0,2 0,2 0,2 0,2 0,2 0,2 0,2 0,1 03,5E-04 0 0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1 04,4E-04 0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 05,3E-04 0 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 06,2E-04 0 -0,1 -0,2 -0,2 -0,2 -0,2 -0,2 -0,2 -0,2 -0,1 07,0E-04 0 -0,1 -0,2 -0,3 -0,3 -0,3 -0,3 -0,3 -0,2 -0,1 07,9E-04 0 -0,1 -0,2 -0,3 -0,4 -0,4 -0,4 -0,3 -0,2 -0,1 08,8E-04 0 -0,1 -0,2 -0,3 -0,4 -0,5 -0,4 -0,3 -0,2 -0,1 09,7E-04 0 -0,1 -0,2 -0,3 -0,4 -0,4 -0,4 -0,3 -0,2 -0,1 01,1E-03 0 -0,1 -0,2 -0,3 -0,3 -0,3 -0,3 -0,3 -0,2 -0,1 01,1E-03 0 -0,1 -0,2 -0,2 -0,2 -0,2 -0,2 -0,2 -0,2 -0,1 01,2E-03 0 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 01,3E-03 0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 01,4E-03 0 0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1 01,5E-03 0 0,1 0,2 0,2 0,2 0,2 0,2 0,2 0,2 0,1 01,6E-03 0 0,1 0,2 0,3 0,3 0,3 0,3 0,3 0,2 0,1 01,7E-03 0 0,1 0,2 0,3 0,4 0,4 0,4 0,3 0,2 0,1 01,8E-03 0 0,1 0,2 0,3 0,4 0,5 0,4 0,3 0,2 0,1 01,8E-03 0 0,1 0,2 0,3 0,4 0,4 0,4 0,3 0,2 0,1 01,9E-03 0 0,1 0,2 0,3 0,3 0,3 0,3 0,3 0,2 0,1 02,0E-03 0 0,1 0,2 0,2 0,2 0,2 0,2 0,2 0,2 0,1 02,1E-03 0 0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1 02,2E-03 0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 0,0 02,3E-03 0 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 -0,1 02,4E-03 0 -0,1 -0,2 -0,2 -0,2 -0,2 -0,2 -0,2 -0,2 -0,1 02,5E-03 0 -0,1 -0,2 -0,3 -0,3 -0,3 -0,3 -0,3 -0,2 -0,1 02,5E-03 0 -0,1 -0,2 -0,3 -0,4 -0,4 -0,4 -0,3 -0,2 -0,1 02,6E-03 0 -0,1 -0,2 -0,3 -0,4 -0,5 -0,4 -0,3 -0,2 -0,1 02,7E-03 0 -0,1 -0,2 -0,3 -0,4 -0,4 -0,4 -0,3 -0,2 -0,1 02,8E-03 0 -0,1 -0,2 -0,3 -0,3 -0,3 -0,3 -0,3 -0,2 -0,1 0

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 0.0005 0.001 0.0015 0.002 0.0025 0.003

Disp

lacem

ent (

cm.)

Time (sec.)

Displacement evolution at 25 cm

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