0
$\begingroup$

I have been given the matrix $A = \begin{bmatrix} -3 & -2 & 2 \\ 0 & 2 & 0 \\ -4 & -1 & 3 \\\end{bmatrix}$.

I firstly needed to find the matrix $P$ that diagonalises it, so I found the eigenvalues of $A$, the corresponding eigenvectors and then constructed $P$ with columns being those eigenvectors. So, $P=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 2 & 1 \\\end{bmatrix}$ (with the corresponding diagonal matrix $\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \\\end{bmatrix}$).

I need to use these results to solve the following system of ODEs: $$x_1' =-3x_1 -2x_2 +2x_3$$ $$x_2' =2x_2$$ $$x_3' =-4x_1 -x_2 +3x_3$$ where $x_i'=\frac{dx_i}{dt}$.

Now I recognise that the coefficients in the system of ODEs correspond directly to the values of matrix $A$, however I am not sure how to use diagonalisation to solve it? I guess you could set up an equation where y $=A$x where y is a column vector $[x_1', x_2', x_3']$ and x is the column vector $[x_1, x_2, x_3]$, however I'm not sure how to go from here? I'm guessing I would use $P$?

Any guidance would be greatly appreciated.

$\endgroup$

3 Answers 3

1
$\begingroup$

Using the fact that $A=PDP^{-1}$ we can write the system as x'=$A$x=$PDP^{-1}$x or $(P^{-1}x)^{'}=DP^{-1}x$ so setting $y=P^{-1}x$ we solve $y'=Dy$ (which is easily solved). Then once we know y, use x=$P$y to find x.

$\endgroup$
0
$\begingroup$

$\newcommand{\C}{\mathbb{C}}\newcommand{\R}{\mathbb{R}}$Suppose you have a (time-independent!) matrix $A\in \C^{n\times n}$ and are looking for a solution $x\colon I\to\C^n$ to the linear ODE $$x^\prime(t)=Ax(t), x(0)=x_0.\hspace{3em} (1)$$

Step 1 First, calculate the Jordan Canonical Form $J$ and the transformation matrix $T$, so that $T^{-1}AT=J$.

Step 2 Consider now a solution $z\colon I\to\C^n$ to the linear ODE $$z^\prime(t)=Jz(t), z(0)=z_0.\hspace{3em} (2)$$ The function $x\colon I\to\C^n, t\mapsto Tz(t)$ is then a solution to (1) for inital value $x_0=Tz_0$: $$x^\prime(t)=Tz^\prime(t)=TJz(t)=(TJT^{-1})Tz(t)=Ax(t).$$


In your case the Jordan Canonical Form is diagonal, therefore the system (2) decouples and you can solve the three independent ODEs $$z_1^\prime(t)=-z_1(t), \\ z_2^\prime(t)=z_2(t),\\ z_3^\prime(t)=2z_3(t)$$ to obtain $$z(t)=\begin{pmatrix}e^{-t}&0&0\\0&e^{t}&0\\0&0&e^{2t}\end{pmatrix}z_0=\begin{pmatrix}z_{0,1}e^{-t}\\z_{0,2}e^{t}\\z_{0,3}e^{2t}\end{pmatrix}.$$ Multiply this by the transformation matrix $T$ to obtain the solution to your original ODE.

If the JCF is not diagonal there is a similar formula for the solution to (2).

$\endgroup$
0
$\begingroup$

$$X'=AX$$ where $$A = \begin{pmatrix} -3 & -2 & 2 \\ 0 & 2 & 0 \\ -4 & -1 & 3 \\\end{pmatrix}$$ Since you already have calculated the eignevalues $\lambda_i$ and the eigenvectors $V_i$ then the solution to the system is: $$X(t)=C_1e^{\lambda_1 t}V_1+C_2e^{\lambda_2 t}V_2+C_3e^{\lambda_3 t}V_3$$

$\endgroup$
2
  • $\begingroup$ How did you derive this last equation? $\endgroup$
    – Ruby Pa
    Jun 17, 2020 at 2:36
  • $\begingroup$ @RubyPa It's a well known result for differential system. You have three distincts eigenvalues here. $\endgroup$ Jun 17, 2020 at 2:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .