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Consider the following ODE, where $x$ is a vector, $M, C, K$ real square matrices and $e_i$ a basis vector (for example $[1\ 0\ \dots 0]^\top$).:

$M\ddot x + C\dot x + Kx = \sin(\omega t)e_i$

with $x(0)=x_0$ and $\dot x(0)=v_0$, $\omega>0$.

It's a common linear second order (matrix) ODE. The one-dimensional case is very common (damped oscillator with harmonic excitation). Is it possible to find a simple solution in the matrix case?

It is easy to derive a particular solution: assume $x(t) = Re\big[x_1 e^{i\omega t}\big]$, then $$x_1 = (-\omega^2 M +i\omega C + K)^{-1} e_i.$$

For the homogeneous solution, as in the one-dimensional case: if $x=e^{\lambda t}x_2$, $$(\lambda^2 M + \lambda C + K)x_2 = 0$$ and there should be a discussion on the value of $\lambda$. But in the general case, $M$, $C$ and $K$ are not simultaneously diagonizable... So, is there a trick?


I managed to solve the ODE using the first-order form, but I am wondering if it is possible to take advantage of the less general form of the original problem to simplify the result:

$$\begin{bmatrix} \dot x \\ \ddot x \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & I \\ -M^{-1}K & -M^{-1}C \end{bmatrix}}_{A} \underbrace{\begin{bmatrix} x \\ \dot x\end{bmatrix} }_{X(t)} + \underbrace{ \sin(\omega t)\begin{bmatrix} O \\ -M^{-1}e_i \end{bmatrix}}_{b(t)}:= \dot X(t) = A X(t) + b(t)$$

The solutions of $X'(t) = AX(t) + b(t)$ are $X = X_h + X_p$ with $X_h = e^{tA}X_1$ and $X_p = e^{tA}\int e^{-sA}b(s) ds$, or $X(t) = e^{tA}\big( X_0 + \int_0^t e^{-sA}b(s)ds\big)$.

$$b(t) =\sin(\omega t) \underbrace{\begin{bmatrix} O \\ M^{-1} e_i \end{bmatrix}}_{H} := \sin(\omega t)H $$ so $$\int_0^t e^{-sA}b(s)ds = Re\left\lbrace \int_0^t e^{-sA+i(\omega s-\pi/2)I}ds\right\rbrace H = Re\left\lbrace e^{-i\pi/2}\int_0^t e^{sA_2}ds \right\rbrace H = Re\left\lbrace e^{-i\pi/2}A_2^{-1}(e^{A_2 t} - I)H \right\rbrace\quad \text{with }A_2 = -A+i\omega I$$ In the end: $$X(t)=e^{tA}\Big(X_0 - Re\left\lbrace i A_2^{-1}(e^{tA_2}-I)\right\rbrace H \Big)\quad \text{with } A_2 = -A + i \omega I$$

and of course, $X_0 = [x_0, \ v_0]^\top$. It does not really look like the solutions of the linear second-order ODE in this form!

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