05-The Wave Equation2

8/22/2019 05-The Wave Equation2

1/20

The Wave Equation: 2

Solution Using the Finite Difference

Method


2/20

CM3900 Lecture 4 2

We wish to solve the following problem:

Wave Equation:

Boundary Conditions:

and for all t

Initial Conditions

Basic Problem

2 2

22 2u uct x

=

u(0,t) 0= u(L, t) 0=

t 0

u(x,0) f(x)

u)g(x

t =

=

=


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CM3900 Lecture 4 3

Discretisation MethodWe consider a mesh in one space and one time dimension

xi

t

x

x1x0 x4x3x2t0

t1

t2

t3

tj

t

x

xN = L

i 0x x i x= +

whereand

N x L =j 0t t j t= +

i j i,ju(x ,t ) u=We write

(x i, tj)


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CM3900 Lecture 4 4

Discretisation of the Equation

The power series expansion for a function f(x) at the point

x + h is given by

and about x - h is

Adding these two equations and keeping terms to second

order in h, we obtain

2

hf(x h) f(x) h f (x) f (x)2!

+ = + + +K

2hf(x h) f(x) h f (x) f (x)2! = + +K

2f(x h) f(x h) 2f(x) h f (x)+ + = +


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CM3900 Lecture 4 5


Hence, we see that an approximation to the second

derivative is given by

2

f(x h) 2f(x) f(x h)f (x)

h

+ + =


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CM3900 Lecture 4 6

2i j i j i j

i j2 2

u(x , t t) 2u(x , t ) u(x , t t)u(x , t )

t ( t)

+ + =


We can therefore discretise the second derivatives in the

wave equation using this formula

2i,j 1 i,j i,j 1

i j2 2

ui.e.

2u uu(x , t )

t ( t)

+ + =


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CM3900 Lecture 4 7

2

i j i j i ji j2 2

u(x x, t ) 2u(x , t ) u(x x, t )u (x , t )x ( x)

+ + =


Similarly

2i 1,j i,j i 1,j

i j2 2

u 2u uu(x , t )

x ( x)

+ + =


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CM3900 Lecture 4 8


So the wave equation , becomes2

22

2

2

x

uc

t

u

=

2

j,1ij,ij,1i2

2

1j,ij,i1j,i

)x(

uu2uc

)t(

uu2u

+=

+ ++

( )2

i,j 1 i 1,j i,j i 1,j

i,j 1 i 1,j i,

i,j i,j 1

j i 1,j i,j 1

c( t)u u 2u u 2u u

u u (2 2 )u u u

( x)

+ +

+ +

= + +

= + +

or

i.e.2

x

tc

=

where


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CM3900 Lecture 4 9


ui-1,j ui,j ui+1,j

ui,j+1

ui,j-1

we can construct a diagram to show the relative contribution

of the preceding points to ui,j+1

In terms of the mesh points

t = tj-1

t = tj

t = tj+1

22

-1


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CM3900 Lecture 4 10

To advance the solution from the tj to the tj+1 time-slice, we

use the boundary conditions together with the main equation

There is however a problem with the initial step from t0 to t1.

Because the wave equation is a p.d.e. which is second order

in the time variable, the discretised version approximates thefunction at on the next time-slice by using information from

the preceding, as well as the current time.


i,j 1 i 1,j i,j i 1,j i,j 1u u (2 2 )u u u+ + = + +

for i = 1, 2, N 1


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CM3900 Lecture 4 11


At j = 0, the information for this time-slice is provided by the

initial condition which is discretised to

Unfortunately there is no previous data corresponding to j = -1.

However, we do have initial conditions which give the initial

derivative function

)x(f)0,x(u = )x(fu i0,i =

)x(gtu

0t= =


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CM3900 Lecture 4 12


Discretising this derivative using

setting j = 0, we get:

and so

t2

uu

t2

)tt,x(u)tt,x(u

)t,x(t

u

1j,i1j,i

jiji

ji

=

+=

+

i, 1 i,1 i,1 i

i,j 0

uu u 2( t) u 2 t g(x )

t

=

= =

i,1 i, 1

i,0

u uu

t 2 t

=


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CM3900 Lecture 4 13


To get started therefore, we have

Hence

[ ])x(gt2uu)22()uu(

uu)22()uu(u

i1,i0,i0,1i0,1i

1,i0,i0,1i0,1i1,i

++=

++=

+

+

{ }

[ ] )x(gtu)1(uu2

)x(gt2u)22()uu(

2

1

uu)22()uu(u

i0,i0,1i0,1i

i0,i0,1i0,1i

1,i0,i0,1i0,1i1,i

+++

=

+++=

++=

+

+

+


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CM3900 Lecture 4 14


The diagram for the initial step is therefore

1-/2 /2

t g(xi)

1

t = t0

t = t1


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CM3900 Lecture 4 15

Summary

For the initial step we use:

and for subsequent steps

i,0 i

i,1 i 1,0 i 1,0 i,0 i

j 0 : u f(x )

j 1: u u u (1 )u t g(x )2

+

= =

= = + + +

i,j 1 i 1,j i,j i 1,j i,j 1j 2,3, : u u (2 2 )u u u+ + = = + + K


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CM3900 Lecture 4 16

Example

Numerically solve the 1-D wave equation

with c = 1, subject to

i) for all t

ii)

using x = 0.1 and t = 0.05

i.e. find u(x, t) for x = 0.0 to 1.0 in steps of 0.1 and t = 0.00to 2.00 in steps of 0.05.

2 22

2 2

u uc

t x

=

u(0, t) u(1, t) 0= =

t 0

x 0 x 0.5u(x,0)

1 x 0.5 x 1

u0 0 x 1

t =


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CM3900 Lecture 4 17

Solution

We first calculate and note that g(x) = 0.

Initial step:

4

1

x

tc2

=

=

3/41/8 1/8

0

1

t = t0

t = t1


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CM3900 Lecture 4 18

Solution

Subsequently-1

3/21/4 1/4

1


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CM3900 Lecture 4 19

x

SolutionThe calculation can be set out as a table

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.05 0.000 0.100 0.200 0.300 0.400 0.500 0.400 0.300 0.200 0.100 0.000

0.10 0.000 0.100 0.200 0.300 0.400 0.475 0.400 0.300 0.200 0.100 0.000

0.15 0.000 0.100 0.200 0.300 0.394 0.412 0.394 0.300 0.200 0.100 0.000

0.20 0.000 0.100 0.200 0.298 0.369 0.341 0.369 0.298 0.200 0.100 0.000

0.25 0.000 0.100 0.200 0.290 0.319 0.283 0.319 0.290 0.200 0.100 0.000

0.30 0.000 0.100 0.197 0.266 0.253 0.243 0.253 0.266 0.197 0.100 0.000

0.35 0.000 0.099 0.187 0.222 0.188 0.208 0.188 0.222 0.187 0.099 0.000

0.40 0.000 0.096 0.164 0.160 0.136 0.164 0.136 0.160 0.164 0.096 0.000

0.45 0.000 0.085 0.123 0.094 0.097 0.105 0.097 0.094 0.123 0.085 0.000

0.50 0.000 0.063 0.065 0.035 0.060 0.042 0.060 0.035 0.065 0.063 0.000

etc

t


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CM3900 Lecture 4 20

SolutionThe solution can be graphed for various values of t

0 0.2 0.4 0.6 0.8 1

0.5

0.50.5

0.5

u0 i,

u2 i,

u5 i,

u8 i,

u10 i,

10 xi

05-The Wave Equation2

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