Lecture Notes in Differential Equations - Bruce E. Shapiro

235
Integrating,
− 2K |t − t 0 | ≤ ln |u(t)/u(t 0 )| ≤ 2K |t − t 0 | (27.59)
Exponentiating,
|u 0 (t)| e −2K|t−t0| ≤ |u(t)| ≤ |u 0 (t)| e 2K|t−t0| (27.60)
By equation (27.39) this last statement is equivalent to the desired result,
which is the fundamental inequality.
Proof. (Theorem 27.3) By the previous theorem, each term in the sum is a
solution of L n y = 0, and hence by the superposition principle, (27.23) is a
solution.
To prove that it is the general solution we must show that every solution
of L n y = 0 has the form (27.23).
Suppose that u(t) is a solution that does not have the form given by (27.23),
i.e., it is not a linear combination of the y ij = e rit t j .
Renumber the y ij to have a single index y 1 , ..., y n , and let u 0 = u(t 0 ), u 1 =
u ′ (t 0 ), ..., u n−1 = u (n−1) (t o ). Then u(t) is a solution of some initial value
problem
L n y = 0, y(t 0 ) = u 0 , y ′ (t 0 ) = u 1 , ..., y (n−1) (t 0 ) = u n−1 (27.61)
and by uniqueness it must be the only solution of (27.61). Let
Differentiating n times,
v = c 1 y 1 + c 2 y 2 + · · · + c n y n (27.62)
v ′ = c 1 y ′ 1 + c 2 y ′ 2 + · · · + c n y ′ n (27.63)
We ask whether there is a solution c 1 , c 2 , ..., c n to the system
v(t 0 ) = c 1 y 1 (t 0 ) + c 2 y 2 (t 0 ) + · · · + c n y n (t 0 ) = u 0 (27.64)
If the matrix
⎛
M = ⎜
⎝
⎞
y 1 (t 0 ) y 2 (t 0 ) · · · y n (t 0 )
y 1(t ′ 0 ) y 2(t ′ 0 ) y n(t ′ 0 )
.
.
.. ⎟ . ⎠
y (n−1)
1 (t 0 ) y (n−1)
2 (t 0 ) y n (n−1) (t 0 )
(27.65)