Generalized eigen vectors for solving 3 repeated eigen values

Question

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

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

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

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

Answer 1

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 $\frac12$

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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.

Answer 2

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, $t^2\mathbf v_1+2t\mathbf v_2+2\mathbf v_3$.