# Generalized eigen vectors for solving 3 repeated eigen values

This problem is an extension of the post System of 3 variable differential equations with 3 repeating eigen values

Where now the system is, I need help in identifying why the answer differs from that offered by wolfram alpha

$x' =ax+y$

$y' =ay+z$

$z' =az$

Solving

$det \begin{bmatrix} a-\lambda &1 &0 \\ 0 &a-\lambda &1 \\ 0 &0 &a-\lambda \end{bmatrix} =0 \Rightarrow \lambda_{1,2,3} =a$

Thus using generalized eigen vectors we solve

$\begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 0 &0 &0 \end{bmatrix} \begin{bmatrix} y \\ z \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow v_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, x_{free-variable} =1$

and

$\begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 0 &0 &0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \Rightarrow v_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, x_{free-variable} =0$

and

$\begin{bmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 0 &0 &0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \Rightarrow v_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, x_{free-variable} =0$

then using the formula for the general solutions

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} =c_1v_1e^{λt}+c_2(v_1 te^{λt}+v_2 e^{λt})+c_3(v_1 t^2e^{λt}+v_2 te^{λt}+v_3 e^{λt})$

we get

$x(t)=c_3e^{at}t^2+c_2e^{at}t+c_1e^{at}$

$y(t)=c_2e^{at}+c_3e^{at}$

$z(t)=c_3e^{at}$

and wolfram alpha's answer is this

$x(t)=\frac{1}{2}c_3e^{at}t^2+c_2e^{at}t+c_1e^{at}$

$y(t)=c_2e^{at}+c_3e^{at}$

$z(t)=c_3e^{at}$

Note using the new formula in the linked post would get

$x(t)=c_3e^{at}t^2+c_2e^{at}t+c_1e^{at}$

$y(t)=c_2e^{at}+2c_3e^{at}$

$z(t)=2c_3e^{at}$

• The last two solutions are in fact equivalent.
– amd
Jun 7, 2018 at 8:16
• If you simply work backwards from the equation for $z'$, it’s clear that a factor of $1/2$ will appear in the $t^2$ term of $x(t)$ because integrating $t$ gets you $t^2/2+C$. So, without looking closely at the details of your solution, it’s likely that you’re not computing the generalized eigenvectors correctly.
– amd
Jun 7, 2018 at 8:23
• @amd that is probably it, I got the solution to work with the integrating method, but have been spending 5 hours figuring out how to do it with the generalized eigenvectors. Will try again in another time once I have read and fully understood the material Jun 7, 2018 at 8:26
• I hadn’t looked carefully and thought you were still working on the system in the other question. The coefficient matrix is already in Jordan normal form, so the generalized eigenvectors are just the standard basis vectors, as you’ve computed. The problem is in the formula itself: it’s missing a factor of $1/2$. Where did you get it?
– amd
Jun 7, 2018 at 8:59

It appears my previous method doesn't translate very well to larger systems, and looks to be a bit confusing. I will give a more standard answer below, but be sure to check your text to see what they do.

The coefficient matrix

$$\textbf{A} = \begin{bmatrix} a & 5 & 0 \\ 0 & a & 2 \\ 0 & 0 & a \\ \end{bmatrix}$$

has the following (generalized and regular) eigenvectors for the eigenspace $\lambda = a$

$$\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 0 \\ 1/5 \\ 0 \end{bmatrix}, \quad \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1/10 \end{bmatrix}$$

So we can rewrite $\textbf{A}$ as a Jordan decomposition

$$\textbf{A} = \textbf{V}\textbf{J}\textbf{V}^{-1}$$

where the columns $\textbf{V}$ are the vectors above, and $\textbf{J}$ is the Jordan normal form

$$\textbf{J} = \begin{bmatrix} a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & a \end{bmatrix}$$

This form, similar to diagonalization, makes it very easy to apply the matrix exponential

$$e^{t\textbf{A}} = \textbf{V}\big(e^{t\textbf{J}}\big)\textbf{V}^{-1} = \textbf{V} \begin{bmatrix} e^{at} & te^{at} & \frac{t^2}{2}e^{at} \\ 0 & e^{at} & te^{at} \\ 0 & 0 & e^{at} \end{bmatrix} \textbf{V}^{-1}$$

A formula for applying a function to the Jordan form is shown here. The general solution, then, is

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = e^{t\textbf{A}} \vec{r}_0 = \textbf{V}\big(e^{t\textbf{J}}\big)\textbf{V}^{-1}\begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix}$$

where the vector $\vec{r}_0$ describes the initial state. Since it is arbitrary (not given), we can assign

$$\textbf{V}^{-1}\begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}$$

as a vector of arbitrary constants. This gives

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{5} & 0\\ 0 & 0 & \frac{1}{10} \end{bmatrix} \begin{bmatrix} e^{at} & te^{at} & \frac{t^2}{2}e^{at} \\ 0 & e^{at} & te^{at} \\ 0 & 0 & e^{at} \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} c_1e^{at} + c_2te^{at} + c_3\frac{t^2}{2}e^{at} \\ \frac{1}{5}\big(c_2e^{at} + c_3e^{at}\big) \\ \frac{1}{10}c_3e^{at} \end{bmatrix}$$

This is probably the answer WolframAlpha gave you, and is more or less equivalent to the one I had before, with the third constant scaled down by a factor of $\frac12$

For the problem in this question, you got lucky because the matrix is already in its Jordan normal form, i.e. $\textbf{A} = \textbf{J}$ and $\textbf{V} = \textbf{I}$, so the solution is a lot less involved.

• You’re solving the system from the preceding question. The one in this question is different.
– amd
Jun 7, 2018 at 11:11
• @amd I know. This was more of a guide. I didn't want to confuse OP by solving two problems two different ways. Jun 7, 2018 at 11:15
• It doesn’t really answer the question at hand, though, which is specifically about the OP getting the wrong answer with the other method.
– amd
Jun 7, 2018 at 11:22
• I gave OP the wrong formula in the first version of my answer. I then made a revision, which lead to the third solution written here. It's really a minor issue that doesn't need any more clarification. I'm sure OP can apply the methods now, since it's just changing a few numbers. Jun 7, 2018 at 11:25
• Yea I got it now, Dylan and amd , thank you both for taking your time with this one. Jun 7, 2018 at 11:30

The discrepancy comes from your having used an incorrect formula. The $t^2$ term is missing a factor of $\frac12$. To see that the formula is incorrect, observe that the term with coefficient $c_3$ must itself satisfy the differential equation. Setting $\mathbf x(t) = e^{at}\left(t^2\mathbf v_1+t\mathbf v_2+v_3\right)$, we have $$\dot{\mathbf x} = ae^{at}\left(t^2\mathbf v_1+t\mathbf v_2+v_3\right)+e^{at}\left(2t\mathbf v_1+\mathbf v_2\right).\tag1$$ On the other hand, by construction $A\mathbf v_1=a\mathbf v_1$, $A\mathbf v_2=a\mathbf v_2+\mathbf v_1$ and $A\mathbf v_3=a\mathbf v_3+\mathbf v_2$, so $$A\mathbf x = ae^{at}\left(t^2\mathbf v_1+t\mathbf v_2+\mathbf v_3\right)+e^{at}\left(t\mathbf v_1+\mathbf v_2\right),$$ which doesn’t match.

The correct term is either $\frac12t^2\mathbf v_1+t\mathbf v_2+\mathbf v_3$ or, if you don’t like the fraction, $t^2\mathbf v_1+2t\mathbf v_2+2\mathbf v_3$.

• Thanks amd, was my fault that I didn't recognize that the second formula Dylan gave me was the right one. Was matching my answer term by term with that in wolframalpha. Jun 7, 2018 at 11:33