I am interested in bicycle stability and found this article (RG) on it. It seems that they calculate the roots of the characteristic polynom after using an exponential-ansatz (page 1969, (b)).

I fail to see how this has to do something with eigenvalues. Sure it must be the eigenvalue (of some matrix - in the equations of motion) that is calculated to find out if they change some vector (which says something about the movement).

Does anyone have a qualitative explanation for my problem?

In detail there is a ordinary second order matrix differential equation (constant coefficients) just as this one.

$q''A+q'B+qC=0$ where $q=({{q_1},{q_2}})^T$ and $A,B,C$ are matrices. They then go on and use the ansatz $q=q_0e^{\lambda t}$ and find a quadratic characteristic polynom.

My intuition tells me that I am trying to find a $\lambda$ such that $Z(q)=\lambda q$ where $Z$ is a function (including the 2 differential operators). I just can't see how this requires an Ansatz or where the $Z$ is in the equation.

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    $\begingroup$ Take a look at physics.stackexchange.com/q/391989 $\endgroup$ – Bert Barrois Mar 11 at 12:05
  • $\begingroup$ I have a more distinct problem here. In the paper I mentioned they use a special ansatz and therefore get a special matrix - where does the eigenvalue go there? $\endgroup$ – manuel459 Mar 11 at 20:40

The equation of motion for the bike:

$$ \tag{1} \ddot{\vec{q}} \mathbf{A} + \dot{\vec{q}} \mathbf{B} + \vec{q}\mathbf{C} =0 $$

Where the $2\times 2$ constant matrices $ \mathbf{A}$, $ \mathbf{B}$ , and $ \mathbf{C}$ contain the parameter and speed of the bicycle (some of the matrices are not symmetric); and $q = ({{q_1},{q_2}})^T$ the two controlling angles of driving.

This is a constant-coefficient 2nd order matrix differential equation. For constant-coefficient differential equation, there is at least a solution of form:

$$ \tag{2} \vec{q}(t) = \vec{q}_o e^{\lambda t} = \begin{bmatrix} q_{1o} \\ q_{2o} \end{bmatrix} e^{\lambda t}. $$ Where $\vec{q}_o = \left[q_{1o},q_{2o} \right]^T$ is a constant vector, and $\lambda$ a constant exponent. These three parameters will be determined by substitute Eq.(2) into Eq.(1):

$$ \tag{3} \left\{\lambda^2 \mathbf{A} + \lambda \mathbf{B} + \mathbf{C} \right\} \vec{q}_o e^{\lambda t} =0 $$

In Eq.(3), you can see that the function $e^{\lambda t}$ is not changed during differentiations, and thus can be factored out leaving a polynomial equation for $\lambda$. This is the resaon to start with exponential solution for a constant-coefficient differential equation.

We then expand the curry braket part of Eq.(3) in matrix form:

$$\tag{4} \det \begin{bmatrix} A_{11} \lambda^2 + B_{11} \lambda + C_{11} & A_{12} + B_{12} + C_{12} \\ A_{21} + B_{21} + C_{21} & A_{22} \lambda^2 + B_{22} \lambda + C_{22} \end{bmatrix} = 0 $$

Eq. (4) is not an eigne value problem, but a 4th order polynomial equation for $\lambda$. There are 4 roots for Eq. (4). Each root render from Eq.(3) a cooresponding eigen vector $\vec{q}_\lambda = \left[q_{1\lambda},q_{2\lambda} \right]^T$. The resultant solution:

$$ \tag{5} \vec{q}_\lambda(t) = \begin{bmatrix} q_{1\lambda} \\ q_{2\lambda} \end{bmatrix} e^{\lambda t}. $$

  1. For $\lambda$ having a negative real part

    The exponential function in Eq.(5) will converged to zero for large $t$, rendering a stable motion (stable against a small deviation of the driving angles).

  2. For $\lambda$ having a positive real part

    The exponential function in Eq.(5) grows to infinity for large $t$, rendering a unstable motion (it will amplifys small deviations of the driving angles).

  3. For $\lambda$ a pure imaginery number

    The exponential function in Eq.(5) becomes $\cos(\lambda t)$ or$\sin(\lambda t)$, resulting a stable but oscillating motion (it will swing back and forth the small deviations around the equilibrium angle).

  • $\begingroup$ Thank you for this explanation. May I ask you what the vector $\overrightarrow{q_\lambda}$ is an eigenvector from? In linear algebra one usually needs a matrix for this. But then equation (3) must read $\{\lambda^2\textbf{A}+\lambda\textbf{B}+\textbf{C}\}\overrightarrow{q}=\lambda \overrightarrow{q}$ doesn't it? $\endgroup$ – manuel459 Mar 15 at 18:30
  • $\begingroup$ It is not an eigen vector in the form. But is a more general sense: Since there is no other termiology for this kind of relation, may be call it a normal-mode vector. Basically, this is not a linear operator, but a vector solution for this quadratic equation. $\endgroup$ – ytlu Mar 15 at 18:59
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    $\begingroup$ I understand. Thank you for your patience! Could not have thought about that it has a different meaning here. :) $\endgroup$ – manuel459 Mar 15 at 19:07

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