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Lecture Notes in Differential Equations - Bruce E. Shapiro

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319<br />

we can replace (31.118) with a set of systems<br />

⎛<br />

⎞<br />

λ i 1 0 0<br />

.<br />

z ′ i = B i z i =<br />

0 λ ..<br />

. i .<br />

⎜<br />

⎝<br />

.<br />

. .. . ⎟ .. 1 ⎠ z i (31.120)<br />

0 · · · 0 λ i<br />

where B i is an m i × m i matrix (m i is the multiplicity of λ i ). Start<strong>in</strong>g with<br />

any block, denote the components of z i as ζ 1 , ζ 2 , ..., ζ m , and (for the time<br />

be<strong>in</strong>g, at least) omit the <strong>in</strong>dex i. Then <strong>in</strong> component form,<br />

ζ ′ 1 =λζ 1 + ζ 2 (31.121)<br />

ζ ′ 2 =λζ 2 + ζ 3 (31.122)<br />

.<br />

ζ n−1 =λζ n−1 + ζ n (31.123)<br />

ζ ′ n =λ i ζ n (31.124)<br />

We can solve this type of system by “back substitution” – that is, start<br />

with the last one, and work backwards. This method works because the<br />

matrix is upper-triangular. The result is<br />

ζ m = a m e λt (31.125)<br />

where a 1 , ..., a m are arbitrary constants. Rearrang<strong>in</strong>g,<br />

⎛<br />

z = ⎜<br />

⎝<br />

ζ 1<br />

ζ 2<br />

.<br />

ζ m<br />

⎞ ⎛<br />

a 1 + a 2 t + · · · + a m<br />

⎟<br />

⎠ = ⎜<br />

.<br />

⎝<br />

a m−1 + a m t<br />

a m<br />

t m−1<br />

(m−1)!<br />

⎞<br />

⎟<br />

⎠ eλt (31.126)<br />

hence<br />

⎛ ⎞<br />

⎛ ⎞ t<br />

⎛<br />

1<br />

z = a 1 e λt 0<br />

1<br />

⎜ ⎟<br />

⎝.<br />

⎠ + a 2e λt 0<br />

+ · · · + a<br />

⎜ ⎟<br />

m e λt ⎜<br />

⎝<br />

⎝.<br />

⎠<br />

0<br />

0<br />

t m−1<br />

(m−1)!<br />

.<br />

t<br />

1<br />

⎞<br />

⎟<br />

⎠<br />

(31.127)<br />

Denot<strong>in</strong>g the standard basis vectors <strong>in</strong> R m as e 1 , ..., e m ,

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