Page 426 - Applied Numerical Methods Using MATLAB

HYPERBOLIC PDE 415
In order for this equation to be solvable, the boundary conditions u(0,t) =
b 0 (t) and u(x f ,t) = b xf (t) as well as the initial conditions u(x, 0) = i 0 (x) and

∂u/∂t| t=0 (x, 0) = i (x) should be provided.
0
9.3.1 The Explicit Central Difference Method
In the same way as with the parabolic PDEs, we replace the second derivatives
on both sides of Eq. (9.3.1) by their three-point central difference approximation
(5.3.1) as
k
k
u k i+1 − 2u + u k i−1 u k+1 − 2u + u k−1 x f T
i
A = i i i with x = , t = (9.3.2)
x 2 t 2 M N
which leads to the explicit central difference method:
t 2
k+1
k−1
k
k
k
u = r(u + u ) + 2(1 − r)u − u with r = A (9.3.3)
i i+1 i−1 i i 2
x
−1 1
Since u = u(x i , − t) is not given, we cannot get u directly from this
i i
formula (9.3.3) with k = 0:
0
1
u = r(u 0 + u 0 ) + 2(1 − r)u − u −1 (9.3.4)
i i+1 i−1 i i
Therefore, we approximate the initial condition on the derivative by the central
difference as −1
1
u − u
i i = i (x i ) (9.3.5)

2 t 0
−1
and make use of this to remove u from Eq. (9.3.3):
i
1
0
1

u = r(u 0 + u 0 ) + 2(1 − r)u − (u − 2i (x i ) t)
i i+1 i−1 i i 0
1
1 0 0 0
u = r(u + u ) + (1 − r)u + i (x i ) t (9.3.6)
i i+1 i−1 i 0
2
1
We use Eq. (9.3.6) together with the initial conditions to get u and then go
i
on with Eq. (9.3.3) for k = 1, 2,.... Note the following facts:
ž We must have r ≤ 1 to guarantee the stability.
ž The accuracy of the solution gets better as r becomes larger so that x
decreases.
It is therefore reasonable to select r = 1.
The stability condition can be obtained by substituting Eq. (9.2.4) into
Eq. (9.3.3) and applying the Jury test [P-3]:
2
−1
λ = 2r cos(π/P ) + 2(1 − r) − λ , λ + 2(r(1 − cos(π/P )) − 1)λ + 1 = 0