I need to solve this differential system :
$ x''+\lambda_1 x' + kx=\alpha + ky$
$y''+\lambda_2 y' + ky = \beta + kx$
with null initial conditions and where $ \lambda_i,k,\alpha,\beta$ are all real parameters but unknown. How would you solve this system? I've tried to transform as a first order system, but the eigenvalues are tedious to compute. Many thanks.
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- $\begingroup$ Consider it as a single system with four variables and solve it. And what do you mean by 'solve' for unknown parameters? Solve symbolically? $\endgroup$Arastas– Arastas2022-12-02 12:26:31 +00:00Commented Dec 2, 2022 at 12:26
- $\begingroup$ @Moo problem with this approach, is that I have to find the roots (one is zero) of a fourth order polynomial with 3 parameters! Same holds for the solution proposed by Arastas. $\endgroup$dodo– dodo2022-12-02 14:47:56 +00:00Commented Dec 2, 2022 at 14:47
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1 Answer
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1 Let $ q_1(t) = x(t) $, $ q_2(t) = \dot{x}(t) $, $ q_3(t) = y(t) $, $ q_4(t) = \dot{y}(t) $, and $ q(t) = \left[ q_1(t), q_2(t), q_3(t), q_4(t) \right]^{T} $. Then, we can rewrite the ordinary differential equation as a first order ODE in the state-space form $$ \begin{align} \dot{q}(t) &= \left[ \begin{matrix} 0 & 1 & 0 & 0 \\ -k & -\lambda_1 & k & 0\\ 0 &0 & 0 & 1 \\ k & 0 & -k & -\lambda_2 \end{matrix} \right]q(t) + \left[ \begin{matrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right] \left[ \begin{matrix} \alpha \\ \beta \end{matrix} \right] \\ &\triangleq A q(t) + B u \end{align} $$ for which the solution is given by $$ q(t) = e^{A(t - t_0)} q(t_0) + \int\limits_{t_0}^{t} e^{A(t-\tau)}Bu d\tau. $$ where $ q(t_0) $ is the initial condition.
- $\begingroup$ As already pointed out, the problem is to compute the eiegenvalues. We would need to find the roots of a parametric third degree polynomial, which is not easy to factorize. $\endgroup$dodo– dodo2022-12-03 08:24:25 +00:00Commented Dec 3, 2022 at 8:24