Solution to Exercise 2.7.20 from „Linear Algebra“

Assume the differential equation has monic auxiliary polynomial

p (t)

of degree

n

. Thus we know that

p (D) (x) = 0

if

x

is a solution. This means that

x^{(k)}

exists for all integer

k \leq n

. We may write

p (t)

as

t^{n} + q (t)

, where

q (t) = p (t) - t^{n}

is a polynomial of degree less than

n

. Thus we have

x^{(n)} = - q (D) (x)

is differentiable since $x^{(n)}$ is a linear combination of lower order terms $x^{(k)}$ with $k \leq n - 1$ . Doing this inductively, we know actualy $x$ is an element in $C^{\infty}$ .

For complex number

c

and

d

, we may write

c = c_{1} + i c_{2}

and

d = d_{1} + i d_{2}

for some real numbers

c_{1}

,

c_{2}

,

d_{1}

, and

d_{2}

. Thus we have

e^{c + d} = e^{(c_{1} + d_{1}) + i (c_{2} + d_{2})} = e^{c_{1}} e^{d_{1}} (\cos (c_{2} + d_{2}) + i \sin (c_{2} + d_{2}))

and

e^{c} e^{d} = e_{1}^{c} e_{1}^{d} (\cos c_{2} + i \sin c_{2}) (\cos d_{2} + i \sin d_{2})

= e_{1}^{c} e_{1}^{d} [(\cos c_{2} \cos d_{2} - \sin c_{2} \sin d_{2}) + i (\sin c_{2} \cos d_{2} + \cos c_{2} \sin d_{2})]

= e^{c_{1}} e^{d_{1}} (\cos (c_{2} + d_{2}) + i \sin (c_{2} + d_{2})) .

This means $e^{c + d} = e^{c} e^{d}$ even if $c$ and $d$ are complex number.¹ For the second equality, we have

1 = e^{0} = e^{c - c} = e^{c} e^{- c} .

So we get

e^{- c} = \frac{1}{e^{c}} .

Let

V

be the set of all solution to the homogeneous linear differential equation with constant coefficient with auxiliary polynomial

p (t)

. Since each solution is an element in

C^{\infty}

, we know that

V \supset N (p (D))

, where

N (p (D))

is the null space of

p (D)

, since

p (D) (x) = 0

means that

x

is a solution. Conversely, if

x

is a solution, then we have

p (D) (x) = 0

and so

x \in N (p (D))

.

Let

c = c_{1} + i c_{2}

for some real numbers

c_{1}

and

c_{2}

. Directly compute that

{(e^{ct})}^{'} = {(e^{c_{1} t + i c_{2} t})}^{'} = {(e^{c_{1} t} (\cos c_{2} t + i \sin c_{2} t))}^{'}

c_{1} e^{c_{1} t} (\cos c_{2} t + i \sin c_{2} t)) + i c_{2} e^{c_{1} t} (\cos c_{2} t + i \sin c_{2} t)

(c_{1} + i c_{2}) e^{c_{1} t} (\cos c_{2} t + i \sin c_{2} t) = c e^{ct} .

Assume that

x = x_{1} + i x_{2}

and

y = y_{1} + i y_{2}

for some

x_{1}

,

x_{2}

,

y_{1}

, and

y_{2}

in

F (ℝ, ℝ)

. Compute that

{(xy)}^{'} = {(x_{1} y_{1} - x_{2} y_{2})}^{'} + i {(x_{1} y_{2} + x_{2} y_{1})}^{'}

= (x_{1}^{'} y_{1} + x_{1} y_{1}^{'} - x_{2}^{'} y_{2} - x_{2} y_{2}^{'}) + i (x_{1}^{'} y_{2} + x_{1} y_{2}^{'} + x_{2}^{'} y_{1} + x_{2} y_{1}^{'})

= (x_{1}^{'} + i x_{2}^{'}) (y_{1} + i y_{2}) + (x_{1} + i x_{2}) (y_{1}^{'} + i y_{2}^{'}) = x^{'} y + x y^{'} .

Assume that

x = x_{1} + i x_{2}

for some

x_{1}

and

x_{2}

in

F (ℝ, ℝ)

. If

x^{'} = x_{1}^{'} + i x_{2}^{'} = 0,

then $x_{1}^{'} = 0$ and $x_{2}^{'} = 0$ since $x_{1}^{'}$ and $x_{2}^{'}$ are real-valued functions. Hence $x_{1}$ and $x_{2}$ are constant in $ℝ$ . Hence $x$ is a constant in $ℂ$ .

Exercise 2.7.20

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Exercise 2.7.20

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