Solutions to coupled second order differential equations

Given this system

\begin{align} \ddot{y_1} & = -y_1 + y_2 + 1 - a \\ \ddot{y_2} & = y_1 - 2 y_2 + y_3 + 1 \\ \ddot{y_3} & = y_2 - y_3 + 1 + a \\ \end{align}

putting this into matrix form gives $$\ddot{\textbf{y}} = A \textbf{y} + \textbf{g}$$ wherein $$\textbf{y}, \textbf{g}$$ are the appropriate $$3\times1$$ matrices for the system and $$A$$ is the corresponding coefficient matrix. One problem that I am having is that when the eigenvalues are computed ($$\lambda_1 = 0 \, (\text{multiplicity of 2}), \lambda_2 = \pm i, \lambda_3 = \pm i \sqrt{3}$$), one of the corresponding eigenvectors for $$\lambda_1$$ doesn't exist. As an illustration:

$$\left(\begin{matrix} -1 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1 \\ \end{matrix} \right) \left(\begin{matrix} v_1 \\ v_2 \\ v_3 \\ \end{matrix} \right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right)$$

gives $$\textbf{v} = \left(1 \quad 1 \quad 1 \right)^T$$ as the first eigenvector. Solving for the other eigenvector:

$$\left(\begin{matrix} -1 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1 \\ \end{matrix} \right) \left(\begin{matrix} p_1 \\ p_2 \\ p_3 \\ \end{matrix} \right) = \left(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right)$$

Which leads to this

$$\left(\begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \\ \end{matrix} \right) \left(\begin{matrix} p_1 \\ p_2 \\ p_3 \\ \end{matrix} \right) = \left(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right)$$

which does not make sense. Does this mean that the second eigenvector for $$\lambda_1$$ does not exist? Also as an aside, after solving the homogeneous case, how do I solve for the inhomogeneous case?

• Shouldn't you write those as first order equations first and get a $6~x~6$ from $y' = A y + g$? – Moo Oct 25 '18 at 5:02
• @Moo Yes, however I've read from here that the characteristic equation for the $6 \times 6$ system is just $(-1)^n$ times the characteristic equation for the $3 \times 3$ system. However I do think that I would still have to use the $6 \times 6$ system for the particular solution. – Aldon Oct 25 '18 at 5:22
• You're certainly not going to get 6 eigenvalues and eigenvectors from a $3\times3$ matrix. – Gerry Myerson Oct 25 '18 at 6:45

Given \begin{align} \ddot{y_1} & = -y_1 + y_2 + 1 - a \\ \ddot{y_2} & = y_1 - 2 y_2 + y_3 + 1 \\ \ddot{y_3} & = y_2 - y_3 + 1 + a \\ \end{align}

Writing this as first order equations yields

$$X' = \left(\begin{matrix} x_1' \\ x_2' \\ x_3' \\ x_4' \\x_5'\\x_6' \end{matrix} \right) = AX + g = \left(\begin{matrix} 0 & 1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 1 &0&0&0\\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & -2 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & -1 & 0 \end{matrix} \right) \left(\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\x_5\\x_6 \end{matrix} \right)+ \left(\begin{matrix} 0 \\ 1-a \\ 0 \\ 1 \\ 0\\1+a \end{matrix} \right)$$

The eigenvalues of $$A$$ are

$$\lambda_{1, 2} = \pm~i~\sqrt{3}, ~~\lambda_{3, 4} = \pm~ i, ~~\lambda_{5,6} = 0$$

We have a deficient matrix and need to find a generalized eigenvector for $$\lambda = 0$$. Using Wiki Example $$1$$ as a possible approach, we have

$$[A - \lambda I]v_6 = [A]v_6 = v_5$$

This leads to the augmented matrix $$RREF \left(\begin{array}{rrrrrr|r} 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ -1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & -2 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & -1 & 0 & 0 \\ \end{array}\right) = \left( \begin{array}{rrrrrr|r} 1 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$$

From this, we have the generalized eigenvector

$$v_6 = \left(\begin{matrix} 0 \\ 1 \\ 0 \\ 1 \\0\\1 \end{matrix} \right)$$

For the other four eigenvalues, we can use the typical methods to find linearly independent eigenvectors.

Next, to solve the nonhomogeneous system, refer to the various methods here.

$$A=\pmatrix{-1&1&0\\1&-2&1\\0&1&-1}$$ has characteristic polynomial $$-(q+1)^2(q+2)+2(q+1)=(q+1)(2-2-3q-q^2)=-q(q+1)(q+3)$$ which gives an an eigen-decomposition $$UDU^T$$ with $$D=\pmatrix{-3&0&0\\0&-1&0\\0&0&0}\text{ and } U=\pmatrix{ \frac1{\sqrt6}& \frac1{\sqrt2}&\frac1{\sqrt3}\\ -\frac2{\sqrt6}& 0 &\frac1{\sqrt3}\\ \frac1{\sqrt6}&-\frac1{\sqrt2}&\frac1{\sqrt3} }$$ With the coordinate change to the basis in the columns of $$U$$, the original order 2 system decouples into 2 oscillation equations $$\ddot x_1+3x_1=c_1$$, $$\ddot x_2+x_2=c_2$$ and a simple one of the form $$\ddot x_3=c_3$$.