Dynamical system equilibrium point increment <mstyle displaystyle="true" scriptlevel="0">

Andy Erickson

Andy Erickson

Answered question

2022-05-22

Dynamical system equilibrium point increment
d x 1 d x ( t ) = x 2 ( t )
d x 2 d x ( t ) = k m x 1 ( t ) + f ( t ) m
Then we find the equilibrium point by setting
0 = x 2 ( t )
0 = k m x 1 ( t ) + f ( t ) m
Which gives as result
x e q = [ x 1 x 2 ] = [ f / k 0 ]
Finally, the book defines the increment with respect to the equilibrium point Δ x ( t ) = x ( t ) x e q . Substracting equation ( 1 , 3 ) and ( 2 , 4 ) the result is
d Δ x 1 d x ( t ) = Δ x 2 ( t )
d Δ x 2 d x ( t ) = k m Δ x 1 ( t )
So far so good, but for the next step they get "the general solution, parametrized by the initial state" as
Δ x 1 ( t ) = Δ x 1 ( 0 ) c o s ( ω t ) + Δ x 2 ( 0 ) ω s i n ( ω t )
Δ x 2 ( t ) = Δ x 1 ( 0 ) ω s i n ( ω t ) + Δ x 2 ( 0 ) c o s ( ω t )
This result I don't understand where does it comes from, could someone give me a hint?

Answer & Explanation

aqueritztv

aqueritztv

Beginner2022-05-23Added 10 answers

This is ODE business, the solution of:
d x 1 d t = x 2 d x 2 d t = k m x 1
is of the form you wrote. Let ω 2 = k m for simplification.
A way to prove it is to differentiate the first equation and substitute the second in:
d 2 x 1 d t 2 = d x 2 d t = ω 2 x 1
So the solution for x 1 is of the form:
x 1 ( t ) = A 1 cos ( ω t ) + A 2 sin ( ω t )
Differentiate it to find the solution to x 2 ( t ).
Then substituting the initial conditions you find A 1 = x 1 ( 0 ) and A 2 = x 2 ( 0 ) ω .

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