I am interested in which conditions need to be fullfilled in order for a solution of an ODE system to grow algebraicly due to resonance, in best case stronger than just linearly.
Let's assume following two ODEs:
$\frac{\partial^2 u}{\partial x^2}+m^2 u=sin(n x)$
and
$\frac{\partial^2 u}{\partial x^2}+m^2 u=sin(m x)$
The solution of the first ODE reads as:
$u=C_1 sin(mx)+C2 cos(mx)+\frac{sin(nx)}{m^2-n^2}$, while the solution of the second ODE reads as:
$u=C_1 sin(mx)+C2 cos(mx)+\frac{sin(mx)-cos(mx) mx }{2 m^2}$
The second solution if obviously of algebraic growth due to the frequency of the source term (the sin-function) being the resonance frequency of the ODE.
Which conditions need to be fulfilled in order to have an algebraic grwoth which is stronger than linear, hence $x^n$ with $n>1$?
I know that for example the ODE $\frac{\partial^2 u}{\partial x^2}-2 \frac{\partial u}{\partial x}+u=e^x$, leads to $u=C_1 e^x+C_2 x e^x+x^2 \frac{e^x}{2}$, is of algebraic growth (x^2) due to the frequency 1 being a double eigenvalue of the homogeneous ODE $\frac{\partial^2 u}{\partial x^2}-2 \frac{\partial u}{\partial x}+u=0$. However for a trigonometric source terme or a source term of $e^{i x}$, I do not know which conditions an ODE or the source term neead to fulfill in order to have such a solution (x^2).
I would appreciate any help.
In general, for a constant-coefficient linear equation $P(D) u = f(x)$ with forcing term $f(x) = c e^{i \omega x}$ where $P$ is a polynomial, in order for the solution to be a polynomial of degree $d$ times $e^{i\omega x}$ you need $i \omega$ to be a root of $P$ of multiplicity $d$. Since for a real polynomial the complex conjugate of a root is again a root of the same multiplicity, this means if you want a quadratic you'll need at least a polynomial of degree $4$. Thus $i$ is a root of multiplicity $2$ of $(z-i)^2 (z+i)^2 = z^4 + 2 z^2 + 1$, and the differential equation
$$P(D) u = u'''' + 2 u'' + u = \cos(x) $$ has a solution $$ u = \frac{2-x^2}{8} \cos(x) + \frac{3x}{8} \sin(x) $$