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In this case the discriminant is nonnegative and we have two real
λ
λ
=
2
r
cos
β
±
±
1
−
4
r
2
sin
2
β
1+2
r
Note that the terms under radical are less than or equal 1 and therefore the numerator is
less than or equal the denominator

λ
≤
2
r
cos
β
+1
1+2
r
≤
1 for all
r
Thus the method is unconditionally stable in this case.
case 2
1
−
4
r
2
(1
−
cos
2
β
)
<
0
Then 4
r
2
(1
−
cos
2
β
)

{z
}
sin
2
β
>
1
In this case the discriminant is negative and we have two complex conjugate
λ
λ
=
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 Fall '08
 BELL,D

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