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Exercise 7.2.23

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1.

For brevity, we just write

λ

instead of

λ_{i}

. Also denote

u_{k}

to be

{(A - λI)}^{k} u

. So we have

(A - λI) u_{k} = u_{k + 1}

. Let

x = e^{λt} [\sum_{k = 0}^{p - 1} f^{(k)} (t) u_{p - 1 - k}]

be the function vector given in this question. Observe that

(A - λI) x = e^{λt} [\sum_{k = 0}^{p - 1} f^{(k)} (t) u_{p - k}] .

Then compute that

x^{'} = λx + e^{λt} [\sum_{k = 0}^{p - 1} f^{(k + 1)} (t) u_{p - 1 - k}]

= λx + e^{λt} [\sum_{k = 1}^{p - 1} f^{(k)} (t) u_{p - k}]

= λx + (A - λI) x = Ax .

2.

Since

A

is a matrix over

ℂ

, it has the Jordan canonical form

J = Q^{- 1} AQ

for some invertible matrix

Q

. Now the system become

x^{'} = QJ Q^{- 1} x

and so

Q^{- 1} x^{'} = J Q^{- 1} x .

Let $y = Q^{- 1} x$ and so $x = Qy$ . Rewrite the system to be

y^{'} = Jy .

Since the solution of $y$ is the linear combination of the solutions of each Jordan block, we may just assume that $J$ consists only one block. Thus we may solve the system one by one from the last coordinate of $y$ to the first coordinate and get

y = e^{λt} (\begin{matrix} f (t) \\ f^{(1)} (t) \\ ⋮ \\ f^{(p - 1)} (t) \end{matrix}) .

On the other hand, the last column of $Q$ is the end vector $u$ of the cycle. Use the notation in the previous question, we know $Q$ has the form

Q = (\begin{matrix} | & | & \dots & | \\ u_{p - 1} & u_{p - 2} & \dots & u_{0} \\ | & | & \dots & | \end{matrix}) .

Thus the solution must be $x = Qy$ . And the solution coincide the solution given in the previous question. So the general solution is the sum of the solutions given by each end vector $u$ in different cycles.

jephianlin

2011-06-27 00:00

Exercise 7.2.23

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