# Differential equations: Connection between repeated roots of characteristic equation and generalized eigenvectors

My question is about homogeneous linear equations with constant coefficients: $ay''+by'+cy=0$.

When you solve this equation via a characteristic equation (see e.g. here) problems arise when you encounter repeated roots in it. What you would basically do is to combine two chained solutions (principle of superposition and reduction of order).

This equation can also be converted to a system of first order equations (see e.g. here). You then have to diagonalize the resulting matrix. Problems arise when you find that it is not diagonizable because it has to few eigenvectors with repeated eigenvalues. Here again you can remedy the situation by using generalized eigenvectors to find solutions.

My questions:

1. Are both problems and solutions equivalent?
2. Could you please provide a simple example where you show that both methods lead to the same solution.
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You forgot the second $y$ in your equation.

The characteristic equation is the characteristic polynomial of the matrix of the associated first-order system -as you can check. It tells you what the eigenvalues are.

If the eigenvalues are complex they are conjugate because their product is a real number ($c/a$ of your equation) and their sum is $-b$, so your matrix has distinct eigenvalues and is diagonalizable -it has no strictly generalized eigenvector, they are all true eigenvectors.

In case of real eigenvalues (2 real roots of the characteristic equation) if they are different your matrix is diagonalizable. If they are equal then the matrix may be symmetric or not, equivalently, diagonalizable or not. If it is not you do not have 2 linearly independent eigenvectors, you have 1 and you will have strictly generalized eigenvectors.

An example of the last case:

\begin{pmatrix} 1 & 0 \\ \lambda & 1 \\ \end{pmatrix}

is an improper node for $\lambda\ne 0$. One solution of it is $x=e^{\lambda t}$, $y=e^{\lambda t}+te^{\lambda t}$.

In the (phase) plane the solution curves spiral toward ($\lambda<0$) or outward ($\lambda>0$) the origin.

EDIT: I corrected the remark on complex eigenvalues being conjugate: I replaced $a$ by $c/a$ and added the fact that the sum of the roots also needs to be a real number to deduce they are conjugate. The latter implies that they have opposite imaginary parts and then their real part must be equal, by the multiplication formula for complex numbers ($ad+bc=0$ and $b=-d$ imply $a-c=0$). Sorry.

EDIT 2: I should not have used the term spiral as the trajectories do not even make a full turn around the origin. Depending on $\lambda$ and on initial conditions they will bend around it more or less. See pictures here. The solutions for various initial conditions are obtained adding appropriate real coefficients to the exponential summands above, i.e. $x=C_1e^{\lambda t}$, $y=C_2e^{\lambda t}+C_1te^{\lambda t}$. As Gerry Myerson comments the solutions do not cross the eigenline, which is the vertical line here, $x_0=0$ for initial conditions, so $C_1=0$.

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Thank you, corrected the missing y. – vonjd May 11 '12 at 11:24
Spiral? If you had a focus, there would be a spiral, but with a node, the solution curves can't really spiral because they can't cross the "eigenline", right? – Gerry Myerson May 12 '12 at 7:35
@GerryMyerson: True, they do not spiral. They bend once. Google has examples google.com/… For spiraling orbits you would not call the origin a focus, but rather a (spiraling) sink or source -this is the conjugate eigenvalue case, with nonzero real part. – plm May 12 '12 at 9:19
Focus is a word for spiral, according to scholarpedia.org/article/Equilibrium and encyclopedia2.thefreedictionary.com/Phase+Plane+Analysis and any number of sites that turn up when I type "stable focus" into Google. – Gerry Myerson May 12 '12 at 12:43
@GerryMyerson, Ok nomenclature varies. Some call a spiral "focus", some "sink/source", some "spiral sink/spiral source", and I guess there are other terms. I thought the term focus was for the simplest (un)stable node, where the matrix is a multiple of the identity, and all lines are eigenlines. Conclusion: Gerry is right again. :) – plm May 13 '12 at 10:40