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Science Forum Index » Math - Numerical Analysis Forum » multistep stability
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| Author |
Message |
| jraul |
 Posted: Thu Mar 22, 2007 12:41 am |
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Guest
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I'm looking at the multistep method:
y_(n+1) = y_(n-1) + h[1/3 f(t_(n-1), y_(n-1)) + 4/3 f(t_n, y_n) + 1/3
f(t_(n+1), y_(n+1))]
and am trying to find the region of stability. I substitute the test
equation f(t,y) = \lambda y
and then form the characteristic polynomial to get (replacing \lambda
by k):
m = 2
b2 = 1/3
b1 = 4/3
b0 = 1/3
a1 = 0
a0 = 1
Q(z, hk) = (1 - hk/3)z^2 - (4hk/3)z - (1+hk/3) = 0
I then solve this for z using the quadratic formula. But from the
resulting expression (involving hk) it is not clear for which values
of hk it has magnitude less than 1. Any advice? The result I got
was:
z_1,2 = [4hk +- sqrt(12h^2k^2 + 36)] / (6-2hk) |
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| Peter Spellucci |
| Posted: Thu Mar 22, 2007 5:27 am |
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In article <1174542111.514463.190280@b75g2000hsg.googlegroups.com>,
"jraul" <jraulinth@yahoo.com> writes:
Quote: I'm looking at the multistep method:
y_(n+1) = y_(n-1) + h[1/3 f(t_(n-1), y_(n-1)) + 4/3 f(t_n, y_n) + 1/3
f(t_(n+1), y_(n+1))]
and am trying to find the region of stability. I substitute the test
equation f(t,y) = \lambda y
and then form the characteristic polynomial to get (replacing \lambda
by k):
m = 2
b2 = 1/3
b1 = 4/3
b0 = 1/3
a1 = 0
a0 = 1
Q(z, hk) = (1 - hk/3)z^2 - (4hk/3)z - (1+hk/3) = 0
k=\lambda
I then solve this for z using the quadratic formula. But from the
resulting expression (involving hk) it is not clear for which values
of hk it has magnitude less than 1. Any advice? The result I got
was:
z_1,2 = [4hk +- sqrt(12h^2k^2 + 36)] / (6-2hk)
this is an "optimal" 2 step method of order 4 and hence no interval
of absolute stability on the negative real axis: use a taylorseries
for the sqrt (cnaceling the 36 abainst the 6, the geometric series for
removing the denominator to get
z_2 = -1 + (1/3)*h*k +O((h*k)^2)
which for k with negative real part is larger than one in magnitude
hth
peter |
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