SIGNALS AND SYSTEMS II

Assignment 1 Solutions

1. Solve the equation

0 = rN_{1}(1 − \frac{N_{1}}{K}) − αN_{1}N_{2}
0 = sN_{2}(1 − \frac{N_{2}}{L}) − βN_{1}N_{2}

We have four equilibrium points

\{ \begin{matrix}N_{1} & = & 0 \\ N_{2} & = & 0 \end{matrix} , \{ \begin{matrix}N_{1} & = & 0 \\ N_{2} & = & L \end{matrix} , \{ \begin{matrix}N_{1} & = & K \\ N_{2} & = & 0 \end{matrix} , \{ \begin{matrix}N_{1} & = & \frac{αKLs−Krs}{αβKL−rs} \\ N_{2} & = & \frac{βKLr−Lrs}{αβKL−rs} \end{matrix} .

The Jacobian matrix (\frac{∂f}{∂x} ) is

[ \begin{matrix}r − \frac{2rN_{1}}{K} − αN_{2} & −αN_{1} \\ −βN_{2} & s − \frac{2sN_{2}}{L} − βN_{1} \end{matrix} ]

Then the linearized equations are

(a) At equilibrium point (0, 0),

( \begin{matrix}ΔṄ_{1} \\ ΔṄ_{2} \end{matrix} ) = ( \begin{matrix}r & 0 \\ 0 & s \end{matrix} ) · ( \begin{matrix}ΔN_{1} \\ ΔN_{2} \end{matrix} ) .

(b) At equilibrium point (0, L),

( \begin{matrix}ΔṄ_{1} \\ ΔṄ_{2} \end{matrix} ) = ( \begin{matrix}−αL + r & 0 \\ −βL & −s \end{matrix} ) · ( \begin{matrix}ΔN_{1} \\ ΔN_{2} \end{matrix} ) .

(c) At equilibrium point (K, 0),

( \begin{matrix}ΔṄ_{1} \\ ΔṄ_{2} \end{matrix} ) = ( \begin{matrix}−r & −αK \\ 0 & −βK + s \end{matrix} ) · ( \begin{matrix}ΔN_{1} \\ ΔN_{2} \end{matrix} ) .

(d) At equilibrium point (\frac{αKLs − Krs}{αβKL − rs}, \frac{βKLr − Lrs}{αβKL − rs}), after some algebra:

[ \begin{matrix}ΔṄ_{1} \\ ΔṄ_{2} \end{matrix} ] = −\frac{1}{αβKL − rs} [ \begin{matrix}rs(αL − r) & αsK(αL − r) \\ βrL(βK − s) & rs(βK − s) \end{matrix} ] [ \begin{matrix}ΔN_{1} \\ ΔN_{2} \end{matrix} ]

2. (a) Solve det(λI − A) = 0, we get two eigenvalues

λ_{1} = σ − iω, λ_{2} = σ + iω.

The eigenvectors are

v_{1} = k_{1} ( \begin{matrix}i \\ 1 \end{matrix} ) , v_{2} = k_{2} ( \begin{matrix}−i \\ 1 \end{matrix} ) , k_{1}, k_{2} ∈ Z.

by solving the equations (λ_{i}I − A)v_{i} = 0, i = 1, 2.

Let T = ( \begin{matrix}i & −i \\ 1 & 1 \end{matrix} ). We have T^{−1} = ( −\frac{i}{2} \frac{1}{2}_{\frac{i}{2} \frac{1}{2}} ).

Since

e^{T^{−1}AT t} = T^{−1}e^{At}T = e^{Λt} = ( \begin{matrix}e^{(σ−iω)t} & 0 \\ 0 & e^{(σ+iω)t} \end{matrix} ) ,

we have

e^{At} = T e^{Λt}T^{−1} = e^{σt} ( \begin{matrix}cos ωt & sin ωt \\ − sin ωt & cos ωt \end{matrix} ) .

(b) We know that

L[e^{At}] = (sI − A)^{−1}
= ( \begin{matrix}\frac{s − σ}{(s − σ)^{2} + ω^{2}} & \frac{ω}{(s − σ)^{2} + ω^{2}} \\ \frac{−ω}{(s − σ)^{2} + ω^{2}} & \frac{s − σ}{(s − σ)^{2} + ω^{2}} \end{matrix} ) .

Then

e^{At} = L^{−1}[(sI − A)^{−1}] = e^{σt} ( \begin{matrix}cos ωt & sin ωt \\ − sin ωt & cos ωt \end{matrix} ) .

(c)

e^{At} = e^{(σI+ωB)t} = e^{σt}e^{ωBt},

where

e^{ωBt} = I + \frac{ωB}{1!} t + \frac{ω^{2}B^{2}}{2!} t^{2} + \frac{ω^{3}B^{3}}{3!} t^{3} + …
= I + ( \begin{matrix}0 & ω \\ −ω & 0 \end{matrix} ) t + ( \begin{matrix}−ω^{2} & 0 \\ 0 & −ω^{2} \end{matrix} ) \frac{t^{2}}{2!} + ( \begin{matrix}0 & −ω^{3} \\ ω^{3} & 0 \end{matrix} ) \frac{t^{3}}{3!} + …
= ( \begin{matrix}1 − \frac{ω^{2}}{2!} t^{2} + \frac{ω^{4}}{4!} t^{4} + … & ω − \frac{ω^{3}}{3!} t^{3} + … \\ −ω + \frac{ω^{3}}{3!} t^{3} + … & 1 − \frac{ω^{2}}{2!} t^{2} + \frac{ω^{4}}{4!} t^{4} + … \end{matrix} )
= ( \begin{matrix}cos ωt & sin ωt \\ − sin ωt & cos ωt \end{matrix} ) .

3. (a) By the definition of e^{At},

e^{At}v = ∑_{k=0}^{∞} \frac{A^{k}t^{k}}{k!} v.

Consider

{ Av = λv ; A^{2}v = A · (Av) = λ(AV) = λ^{2}v ; ⋮ ; A^{k}v = λ^{k}v ,

we have

e^{At}v = ∑_{k=0}^{∞} \frac{λ^{k}t^{k}}{k!} v = e^{λt}v

as required.

(b) Since A has distinct eigenvalues, [v_{1}, v_{2}, …, v_{n}] is a basis of n-dimensional linear space. As to any vector x_{o}, we can write

x_{o} = ∑_{k=1}^{n} c_{k}v_{k}  (1)

Then

x(t) = e^{At} · x_{o}
= ∑_{k=1}^{n} c_{k}e^{At}v_{k}
= ∑_{k=1}^{n} c_{k}e^{λ_{k}t}v_{k}

c_{k} is determined by equation 1.

(c) The eigenvalues are

λ_{1} = 4, λ_{2} = 1.

Their corresponding eigenvectors are

v_{1} = c_{1} ( \begin{matrix}1 \\ 1 \end{matrix} ) , v_{2} = c_{2} ( \begin{matrix}1 \\ −2 \end{matrix} ) .

Then solve equation 1, i.e.,

( \begin{matrix}3 \\ 2 \end{matrix} ) = c_{1} ( \begin{matrix}1 \\ 1 \end{matrix} ) + c_{2} ( \begin{matrix}1 \\ −2 \end{matrix} ) ,

we have

c_{1} = \frac{8}{3}, c_{2} = \frac{1}{3}.

Thus

x(t) = \frac{8}{3}e^{4t} ( \begin{matrix}1 \\ 1 \end{matrix} ) + \frac{1}{3}e^{t} ( \begin{matrix}1 \\ −2 \end{matrix} ) .