Trying to understand eigenvalues with respect to differential

Willie Kelley

Willie Kelley

Answered question

2022-04-10

Trying to understand eigenvalues with respect to differential equations.
I am trying to understand how to find eigenvalues from a matrix consisting of exponential terms, considering a differential equation. The examples I've seen online are ODEs. Without using a vector with exponential terms, here is what I have learned.
ddtx(t)=λx(t)
ddt x1(t) x2(t) x3(t)= λ100 0λ20 00λ3 x1(t=0) x2(t=0) x3(t=0)
Here is what I am trying to understand
Section of a paper with imag eigenvalue
In this paper, the following assumption is made.
J(t)=|J|e-iωt=|Jx|e-iωt|Jy|e-iωt|Jz|e-iωt
They are using partial derivatives (I believe this can be viewed as an ODE then?). Differentiating with respect to time I believe yields the following. Please correct me if I am wrong.
tJ(t)=t|Jx|e-iωt|Jy|e-iωt|Jz|e-iωt= -iω00 0-iω0 00-iω|Jx|e-iωt(t=0)|Jy|e-iωt(t=0)|Jz|e-iωt(t=0)
Or
t|Jx|e-iωt|Jy|e-iωt|Jz|e-iωt= -iω00 0-iω0 00-iω|Jx||Jy||Jz|
Are my assumptions correct? If so, is there a deeper analysis to why this is the case with exponential terms?

Answer & Explanation

strasnihwhge

strasnihwhge

Beginner2022-04-11Added 14 answers

Step 1
Not sure this is what you want to know, but here is a short explanation about the relation between a system of linear ODEs and Eigendecomposition.
Assume a system of first order linear ODEs with constant coefficients. It can be written in matrix form:
dxdt=Mx.
Now assume an exponential solution of the form
x=vezt.
By plugging it in the equation, you get
zvezt=Mvezt
which expresses that z is an Eigenvalue of M, associated to the Eigenvector v:
Mv=zv.
Step 2
So, after finding a diagonalization
M=TΛT1
and setting x=Ty
the system reduces to dydt=Λy
which is a system of three easy independent equations
dykdt=zkyk.
The initial conditions can be dealt with by using y0=T1x0.

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