# Solving a system of ODEs with a zero eigenvalue and non-zero initial velocity

Consider a system of two second-order linear ODEs for which I have found a solution:

$$\pmatrix{y_1 \\ y_2} = (A_1\cos\omega t + B_1\sin\omega t )\pmatrix{a \\ b} + \pmatrix{C_1 \\ C_2}$$

the constant term being due to an eigenvalue of zero for the matrix $$A$$ in the equation:

$$\ddot{\boldsymbol{y}} = -A\cdot \boldsymbol{y}$$

Say I am now given an initial condition of $$\dot{\boldsymbol{y}} = (v, 0)^T$$ at $$t = 0$$. Differentiating my solution, I find:

$$\dot{\boldsymbol{y}} = \pmatrix{v \\ 0} = B_1\omega\pmatrix{a \\ b}$$

Clearly there is no constant $$B_1$$ which satisfies this condition. How is it physically possible that my system cannot have an initial velocity? I'm really struggling to see what I might be doing wrong here, and can provide more context for how I arrived at my original equation if necessary. Sorry if this is confusing.

Original ODE:

$$M\cdot\ddot{\boldsymbol{y}} = -\begin{bmatrix} k & -k \\ -k & k \end{bmatrix}\boldsymbol{y}$$

where $$M = \mathrm{diag}(m_1, m_2)$$, two known constants. $$k$$ is also known.

• Could you show us the original ODE? Feb 17, 2021 at 19:11
• @Cesareo apologies for the wait. I have edited the question. Feb 17, 2021 at 20:06
• The system's determinant of the matrix is zero, giving a degenerated case. No mystery: as a consequence, solutions (trajectories) are straight lines Feb 17, 2021 at 20:51
• Ok, thank you. But why does this mean that I can't set an initial velocity for the system? Feb 17, 2021 at 21:31

As $$\left( \begin{array}{cc} k & -k \\ -k & k \\ \end{array} \right) = T^{-1}\Lambda T$$ with $$T = \left( \begin{array}{cc} -1 & 1 \\ 1 & 1 \\ \end{array} \right)$$ and $$\Lambda = \left( \begin{array}{cc} 2 k & 0 \\ 0 & 0 \\ \end{array} \right)$$ we have

$$m\ddot y + T^{-1}\Lambda Ty=0\ \ \Rightarrow mT\ddot y + \Lambda T y = 0$$

so calling $$z = T y$$ we have the decoupled linear system

$$m\ddot z + \Lambda z = 0$$

with solution

$$\cases{ z_1 = c_1\cos\left(\sqrt{\frac{2k}{m}}t\right)+ c_2\sin\left(\sqrt{\frac{2k}{m}}t\right)\\ z_2 = c_3+c_4 t }$$

so we have four independent constants to configure with the initial/boundary conditions. From $$z$$ to $$y$$ we proceed as $$y = T^{-1}z$$. Concluding, we are free to configure those initial/boundary conditions as needed.