Differential Equations Lecture Work Solutions 309

Differential Equations Lecture Work Solutions 309 - In this...

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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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