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These lecture slides are delivered at The LNM Institute of Information Technology by Dr. Sham Thakur for subject of Mathematical Modeling and Simulation. Its main points are: Partial, Differential, Equations, PDE, Models, Hyperbolic, Wave, Dimensional, Boundary, Conditions
Typology: Slides
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Topic: Wave Equation
Some PDE Models
( , ) ( , )
1 ( , ) 2
2
2
2
r t g r t t
r t
c
PDE: Wave Equation
2
2
2
2
2
2 2
2 2
z
u x y z t
y
u x y z t
x
u x y z t
t
u x y z t
u r t t
u x y z t
In two dimensional plane, it reduces to
2
2
2
2 ( , , ) 2 ( , , ) ( , , )
y
u x y t
x
u x y t
t
u x y t
This will be further simplified in one-dimensional wave as:
2
2 ( , ) 2 ( , )
x
u x y
t
u x y
One Dimensional Wave Equation
Finite Difference Method
The first step is to choose n and maxk integers.
Define step sizes h and k by
h = (b – a)/n ;
Δt = (d – c)/maxk
This means partitioning the interval (0, L) into n equal parts of width h and time interval (0, T) into maxk equal parts of width Δt.
x0 x1 x2 x3 x4 xn
0
1
2
3
m
This generates a grid.
Finite Difference Method
1 , 2 , ,max.
1 , 2 , ,.
y c j t for all j k
x a ih for all i n
i
i
0 1 2 3 4 n
0
1
2
3
m
4
2 4
2
2
4
2 4
2
2
2 2 2
2
x
h u
h
u i j u i j u i j
t
k u
k
u i j u i j u i j
x
u x t
t
u xi tj i j
Using above two formulas, the Poisson equation at point (xi, yj) is
For all i = 1, 2, 3,... , (n - 1); and j = 1, 2, 3,... , (m - 1)
(^00 1 2 3 4) n
1
2
3
m
2
2
2
Neglecting error and simplifying we get
For all i = 1, 2, 3,... , (n - 1); and j = 1, 2, 3,... , (maxk - 1)
(^00 1 2 3 4) n
1
2
3
m
2 2 2
Consider a problem of determining the wave distribution in a thin metal wire. The zero boundaries are along x- and y-axis and problem equation is
( , ) 0 1 , 0 max
( , 0 ) sin ; 0 1
R x t x t t
x t
u x
u x x x
u t u t t
2
2
2
2
Subjected to following conditions:
If n = m = 4, this problem has a grid shown in the figure.
The boundary conditions and mesh points are shown here.
Red mesh points need calculations. These are nine unknowns.
MATLAB Code wave1d.m -- continued
%%%%%%%%% initialize for t = 0 and t = 1 for i=2:nx w(i,1) = sin(pix(i)); w(i,2)= (1-xlamxlam)sin(pix(i)) ...
1D wave Equation: Results
alpha = 1.0; L = 1.0; T = 2.; maxk = 100; dt = T/maxk; nx = 20.; h = L/nx; dx = h; xlam = dtalpha/h*
The shape of wave at different time or k-values.
Stability:
xlam = 0.4 <
stable
Two Dimensional Wave Equation
( , ) 0 , 0 max
R x t x L t t
g x x L t
u x
u x f x x L
u y t u L y t y W t
0 , 0 ; 0 ; 0.
( , , ) ( , , ) ( , , ) 2
2 2
2 2 2
2
^
x L y W t y
u x y t x
u x y t t
u x y t
In two dimensional case, we consider the following wave equation:
It is subjected to following conditions:
Two Dimensional Wave Equation
0 , 1 , 2 , , max; , 1 , 2 , , 1.
u x y f x g y for all i j n
for all k t for all i j n
u y t u L y t u x t u x W t
i j i j
j k j k i k i k
where t h
u i j k u i j k u i j k u i j k u i j k
u i j k u i j k
/
( 1 , , ) ( 1 , , ) ( , 1 , ) ( , 1 , ) ( , , 1 )
( , , 1 ) 2 ( 1 ) ( , , ) 2
2
For all i ,j= 1, 2, 3,... , (n - 1); and k = 1, 2, 3,... , (maxk - 1).
Δx = Δy