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I was trying to solve this system of differential equations:

$$\frac{dx}{dt}=3x-y-z$$ $$\frac{dy}{dt}=x+y-z$$ $$\frac{dz}{dt}=x-y+z$$

I found the eigenvalues: $\lambda_1=1,\lambda_2=2$. The last one has multiplicity $2$. I found $\boldsymbol K_1 =\left( \begin{array}{c} 1\\ 1\\ 1 \end{array} \right)$ is an eigenvector, so a solution to the differential equation would be $\boldsymbol X_1=\boldsymbol K_1e^t.$ However when I want to find a solution with an eigenvector $\boldsymbol K_2 = \left( \begin{array}{c} k_1\\ k_2\\ k_3 \end{array} \right)$ associated to $\lambda_2$ I have the equation $k_1-k_2-k_3=0$ three times. From this I know that two eigenvectors are $\left( \begin{array}{c} 1\\ 1\\ 0 \end{array} \right)$ and $\left( \begin{array}{c} 1\\ 0\\ 1 \end{array} \right)$. Now if I choose one of them, let's say the first one, then this will be a second eigenvector $\boldsymbol K_2$, such that $\boldsymbol X_2=\boldsymbol K_2e^{2t}$ solves the system. However if I try to find a third solution (which I already found: $\boldsymbol K_3e^{2t}$ from the third eigenvector; but I want to find it from here) that has the form $\boldsymbol X_3=\boldsymbol K_2te^{2t}+\boldsymbol P e^{2t}$, then I have to solve $(\boldsymbol A-2\boldsymbol I)\boldsymbol P = \boldsymbol K_2$. But this system doesn't have a solution: $$(\boldsymbol A-2\boldsymbol I)\boldsymbol P = \boldsymbol K_2 \Longrightarrow \left(\begin{array}{ccc|c}1 & -1 & -1 & 1\\1 & -1 & -1 & 1\\1 & -1 & -1 & 0\\ \end{array}\right)???$$

Is it possible to obtain a third solution from the second one? I'm trying to construct this solution which should end up with an expression that involves the third eigenvector.

I appreciate any thoughts on this. Thanks.

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  • $\begingroup$ I know that's the general solution... But I don't quite understand why I can't derive a new solution from $\boldsymbol K_2$ or $\boldsymbol K_3$ $\endgroup$ – Vladimir Vargas Nov 29 '14 at 22:30
  • $\begingroup$ @Simon Mmm... How do I know the dimension of the eigenspace? In my example it is quite easy to see, but if I had a more complicated system in which it is not obvious it's dimension. Was it numbers of columns minus the dimension of the nullspace (rank of the matrix)? $\endgroup$ – Vladimir Vargas Nov 29 '14 at 22:45
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We have to look at the matrix $${\bf A} = \begin{bmatrix} 3 & -1 & -1\\ 1 & 1 & -1 \\ 1 & -1 & 1\end{bmatrix},$$ and the system ${\bf X}' = {\bf A}{\bf X}$. I didn't checked, but I'll trust you that the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 2$, they're too nice to be wrong. Let's go find eigenvectors:

  • $\lambda_1 = 1$: Solve $({\bf A}-{\bf I}){\bf K}_1 = {\bf 0}$, that is: $$\begin{bmatrix} 2 & -1 & -1\\ 1 & 0 & -1 \\ 1 & -1 & 0\end{bmatrix}\begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$ The first line is the sum of the other, so it is useless. From the second equation we get $a=c$ and from the third, $a = b$. So far, so good: $${\bf K}_1 = \begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} a \\ a \\ a\end{bmatrix} = a\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix},$$ so we pick the vector $${\bf k}_1 = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix},$$ and get one row for the fundamental matrix of the system: $$e^{\lambda_1t}{\bf k}_1 = e^{1t}\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix}e^t \\ e^t \\ e^t\end{bmatrix}.$$
  • $\lambda_2 = 2$: Solve $({\bf A}-2{\bf I}){\bf K}_2 = {\bf 0}$, that is: $$\begin{bmatrix} 1 & -1 & -1\\ 1 & -1 & -1 \\ 1 & -1 & -1\end{bmatrix}\begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}.$$ Better impossible, we have the relation $a-b-c = 0$, that is, $a = b+c$. Then: $${\bf K}_2 = \begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} b+c \\ b \\ c\end{bmatrix} = b\begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix} + c\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}.$$ Pick the vectors: $${\bf k}_2^{(1)} = \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix} \quad \mbox{and} \quad{\bf k}_2^{(2)} = \begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}.$$ This gives us the more independent rows of the fundamental matrix: $$e^{\lambda_2t}{\bf k}_2^{(1)} = e^{2t}\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix}e^{2t} \\ e^{2t} \\ 0\end{bmatrix} \quad \mbox{and} \quad e^{\lambda_2t}{\bf k}_2^{(2)} = e^{2t}\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix}e^{2t} \\ 0 \\ e^{2t}\end{bmatrix}.$$

So, we get as fundamental matrix for the system, the matrix:

$${\cal A} = \begin{bmatrix} e^t & e^{2t} & e^{2t}\\ e^t & e^{2t} & 0 \\ e^t & 0 & e^{2t}\end{bmatrix},$$ and hence, given $c_1,c_2,c_3 \in \Bbb R$, we get the general solution to the system: $${\bf X} = {\cal A}{\bf C} \implies \begin{bmatrix}x(t) \\ y(t) \\ z(t)\end{bmatrix} = \begin{bmatrix} e^t & e^{2t} & e^{2t}\\ e^t & e^{2t} & 0 \\ e^t & 0 & e^{2t}\end{bmatrix}\begin{bmatrix}c_1 \\ c_2 \\ c_3\end{bmatrix} = \begin{bmatrix}c_1e^t+(c_2+c_3)e^{2t} \\ c_1e^t+c_2e^{2t} \\ c_1e^t+c_3e^{2t}\end{bmatrix}.$$

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  • $\begingroup$ Simon abandoned me in the comments, so I will ask you this question: How do I know the dimension of the eigenspace? In my example it is quite easy to see, but if I had a more complicated system whose dimension is not obvious, how would I know the dimension of the eigenspace?. Was it numbers of columns minus the dimension of the nullspace (rank of the matrix)? $\endgroup$ – Vladimir Vargas Nov 29 '14 at 23:13
  • $\begingroup$ Remember that the eigenspace is $V(\lambda) = \ker ({\bf A} - \lambda{\bf I})$, so it's dimension is the number of columns minus the rank of the matrix. Just use that if ${\bf A}$ is a $n \times n$ matrix, we have: $$n = {\rm rank}({\bf A}-\lambda {\bf I}) + \dim(\ker({\bf A}-\lambda{\bf I})) \implies \dim(V(\lambda)) = n - {\rm rank}({\bf A}-\lambda{\bf I}).$$ $\endgroup$ – Ivo Terek Nov 29 '14 at 23:18
  • $\begingroup$ Sorry, I thought I might be misinterpreting so I deleted my earlier comments. But Ivo has nailed it for you. $\endgroup$ – Simon S Nov 29 '14 at 23:41
  • $\begingroup$ @SimonS don't worry. Thanks to you too $\endgroup$ – Vladimir Vargas Nov 30 '14 at 0:41
  • $\begingroup$ @IvoTerek can you check out a question I've posted on differential equations? No one seems to know the answers and I can see that you clearly are advanced in this subject. math.stackexchange.com/questions/1244304/… $\endgroup$ – Yusha Apr 21 '15 at 15:07

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