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The Wave Equation: 2
Solution Using the Finite Difference
Method
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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
Discretisation of the Equation
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)
+ + =
Discretisation of the Equation
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)
+ + =
Discretisation of the Equation
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
Discretisation of the Equation
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
Discretisation of the Equation
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.
Discretisation of the Equation
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
Discretisation of the Equation
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
Discretisation of the Equation
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
Discretisation of the Equation
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
Discretisation of the Equation
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
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