I have a system of coupled differential equations (an example )in the form of, x &#x203

juanberrio8a

juanberrio8a

Answered question

2022-06-24

I have a system of coupled differential equations (an example )in the form of,
x + a x + b x c y = 0
y a y + b y c x = 0
The solution to the above system looks like,
x = A e w 1 t + B e w 2 t + C e w 3 t + D e w 4 t
y = E e w 1 t + F e w 2 t + G e w 3 t + H e w 4 t
the frequencies w 1 , w 2 , w 3 and w 4 are functions of a,b and c My goal is to determine that values of a,b and c such that all w are Real. However, my original system of differential equations has nonlinear terms. I managed to derive the solution but in power expansion form or,
x = i = 0 n a i i ! t i
y = i = 0 n a i i ! t i
I need to determine the frequencies so that I can find the values of a,b and c such that all w are Real. Is it possible to derive w from limited power expansion (say 8th term) of x and y?
My first attempt:
I did a test of my method by taking Fourier Sine Transform of analytical solution which give me the answer in form of f ( w ) g ( w ) where w is the frequency. Then solved the g(w) for w which gives me four answers for w which are the frequencies w 1 , w 2 , w 3 and w 4 . However when I attempted with Fourier Sine Transform of series solutions, the answer is different. This is due to the limited power expansion. Is here a way to improve this?
My second attempt:
I linearize the nonlinear terms of differential equations and used the matrix to calculate Determinant of (A-wI) where A is the matrix of the system of differential equation. I managed to calculate the values of a,b and c but they were incorrect because of linear terms I made.
A last method that I am considering is Monte Carlo. Is it possible to get a frequency equation from limited power expansion of differential equation solution? Any ideas for other methods that I missed?

Answer & Explanation

Anika Stevenson

Anika Stevenson

Beginner2022-06-25Added 19 answers

Regarding the linear system of ode's, applying the Laplace transform on
{ x + a x + b x c y = 0 y a y + b y c x = 0
we have
{ ( s 2 + a s + b ) X ( s ) = c Y ( s ) + c 1 x 0 + c 2 x ˙ 0 ( s 2 a s + b ) Y ( s ) = c X ( s ) + c 3 y 0 + c 4 y ˙ 0
and solving for X ( s ) we have
X ( s ) = ( s ( s a ) + b ) ( c 1 x 0 + c 2 x ˙ 0 ) + c ( c 3 y 0 + c 4 y ˙ 0 ) a 2 s 2 ( b + s 2 ) 2 + c 2
now focusing on the denominator, the condition to have only real roots is
a 2 2 b ± a 4 4 a 2 b + 4 c 2 0

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