I am new with eigenfrequency analysis and I am stuck to understand some physical facts with the mathematical theory.
Eigenfrequency analysis
I have a physical system (e.g. a beam, a membrane, acoustics in a fluid... etc) whose behavior is described by a system of second order ODEs:
$$ \ddot{u} = Au + b $$
where $A$ comes from the discretization of some linear operator (e.g. laplacian) and $b$ represents an excitation of the system.
As far as I understand, eigenfrequency analysis looks for solutions of the system without any excitation: $b = 0$, written under the form:
$$ u(t) = u_\omega e^{i \omega t} $$
where $\omega$ is some pulsation and $u_\omega$ is the spatial mode associated to this pulsation, which gives the (space-discretized) Helmholtz equation:
$$ (A + \omega^2 I) u_\omega = 0 $$
Pairs $(\omega, u_\omega)$ are called "eigenfrequencies" and "eigenmodes".
Misunderstood facts
Resonance
Here is my mathematical "understanding" of the "resonance" behavior: when the system is excited exactly at some eigenfrequency:
$$ \ddot{u} = A u + b e^{i \omega t} $$
there are no solutions of the form $u = u_\omega e^{i \omega t}$, since:
$$ (A + \omega^2 I) u_\omega = b $$
has no solution because $det(A + \omega^2 I) = 0$.
Why does it mean that somehow, the solution blows up?
Behavior with an excitation near an eigenfrequency
Assume that the system is excited at $\omega_0$ a frequency near an eigenfrequency $\omega$:
$$ \ddot{u} = A u + b e^{i \omega_0 t} $$
Experience shows that the system vibrates more and more at some specific mode (so $u$ is continuous against $\omega_0$), as soon as $\omega_0$ is near $\omega$.
Is the solution $u$ near the eigenmode $u_\omega$ for excitation of frequency $\omega_0$ near eigenfrequency $\omega$?
Behavior without excitation from some initial condition
When I play the guitar, regardless of the attack of the string, it vibrates at some eigenfrequency.
Is this fact true? If it is, how to mathematically justify that for "any" initial condition and no excitation $b=0$, that the system will finally vibrates at a specific eigenfrequency?
You label some of your statements "misunderstandings" but I don't think anything is actually incorrect, just incomplete.
"Why does it mean that somehow, the solution blows up?"
If the determinant of the matrix $A + \omega^2 I$ vanishes, that means it has a mode $u$ with eigenvalue zero. If you choose to set the drive $b$ to this zero mode the equation can't be satisfied because the eigenmodes span the system. The ones with zero eigenvalues will be destroyed by the matrix and the ones with nonzero eigenvalues can't be turned into the drive term with eigenvalue zero.
But, you might say, you still have your original differential equation. What happens if you actually apply the dynamics? Something must happen, but it won't take the oscillatory form you assume. Consider the simplest example: $b=1$ is a scalar and $A=0$, so that the equation of motion is
$$\ddot{u} = 0$$.
This is the dynamics of a free particle that continues on at a constant velocity, the displacement growing linearly in time. Similarly, even the amplitude a mode at some nonzero frequency would grow linearly in time if driven at this resonant frequency.
Physically, an example of the latter would be pushing someone on a swing set. They swing back and forth at some resonant frequency (approximately) and you push them at that same frequency. You keep putting energy into the system, so they keep swinging higher and higher. Eventually, with swing sets and pretty much everything else, dissipation becomes important and a drag term pointing in the opposite direction of velocity stops the system amplitude from growing without limit.
"Is the solution u near the eigenmode uω for excitation of frequency ω0 near eigenfrequency ω?"
Sometimes. We can think of the change of frequency as a small perturbation to the matrix. Per nondegenerate perturbation theory, a small change in the matrix should yield a small change in the mode. But per degenerate perturbation theory, such a small perturbation could cause a large change within a degenerate mode space.
"When I play the guitar, regardless of the attack of the string, it vibrates at some eigenfrequency.
Is this fact true? If it is, how to mathematically justify that for "any" initial condition and no excitation b=0, that the system will finally vibrates at a specific eigenfrequency?"
By linearity, you can decompose the initial conditions of the guitar string into a linear superposition of eigenmodes. Each of them evolves forward in time according to a harmonic oscillator equation. That means that they will oscillate and decay, with the decay occurring because of being off-resonance and because of a drag term. The decay rate will be at a minimum on resonance, so if you wait long enough the modes near resonance will be exponentially larger than those far from resonance.