1
$\begingroup$

If I'm given a system of equation of the form $$\begin{cases} \frac{dx}{dt}= ax+by \\ \frac{dx}{dt}= cx+ey\end{cases}$$

I get the general solution finding the eigenvalues and eigenvectors of the matrix $\left( \begin{array}{ccc} a & b\\ c & e \\ \end{array} \right)$, let's say that I found the eigenvalues $\lambda_1,\lambda_2$ and eigenvectors $v_1,v_2$. Then the general solution have the form $(x,y)=c_1e^{t\lambda_1}+c_2e^{t\lambda_2}$.

What happens if I don't have "enough" eigenvectors to write a solution like the former?

This is what happened when I tried to find the solution for the system

$$\begin{cases} \frac{dx}{dt}= 2x-y \\ \frac{dx}{dt}= x+4y\end{cases}$$

Here the eigenvalues will be given by the roots of $p(\lambda)=(2-\lambda)(4-\lambda)+1=(\lambda-3)^2$. Now if I can only find one eigen vector associated to the double eigenvalue: $$\left( \begin{array}{ccc} 2-3 & -1\\ 1 & 4-1 \\ \end{array} \right)\to\left( \begin{array}{ccc} 1 & 1\\ 0 & 0 \\ \end{array} \right)\implies \left( \begin{array}{ccc} x \\ y \\ \end{array} \right)=\operatorname{gen}\Bigg\{\left( \begin{array}{ccc} 1\\ -1 \end{array} \right) \Bigg\}$$

Then the solution will be simply $\left( \begin{array}{ccc} x \\ y \\ \end{array} \right)= c_1e^{3t}\left( \begin{array}{ccc} 1 \\ -1 \\ \end{array} \right)$ ? or do I need another eigenvector to write the general solution?

$\endgroup$
  • 1
    $\begingroup$ shouldn't be there a $\frac{dy}{dt}$ somewhere? $\endgroup$ – tired Nov 20 '14 at 15:21
  • $\begingroup$ When you don't have enough eigenvectors, you take generalized eigenvectors. See this answer. Edit: The question isn't a duplicate, but my answer answers your question. $\endgroup$ – Git Gud Nov 20 '14 at 15:30
1
$\begingroup$

You need to find a vector $\vec{\rho}$ such that $(A-\lambda I)\vec{\rho} = \vec{\eta}.$
Then, the general solution will be $$c_1e^{\lambda t}\vec{\eta} + c_2(te^{\lambda t}\vec{\eta} + e^{\lambda t}\vec{\rho}).$$ DERIVATION:
Suppose the a solution was of the form $te^{\lambda t}\vec{\eta}$, since that's what you do with second order non-homogeoneous equations. Then, $$\vec{\eta}e^{\lambda t}+\lambda te^{\lambda t}\vec{\eta} = Ate^{\lambda t}\vec{\eta}.$$However, this implies that $$\vec{\eta} = 0.$$But eigenvectors cannot be zero, so this is wrong. Now the problem was that there was a lone term with an exponential in it so let's see if we can correct that. Let's try the guess $$te^{\lambda t}\vec{\eta} + e^{\lambda t}\vec{\rho}.$$Then, $$e^{\lambda t}\vec{\eta}+\lambda te^{\lambda t}\vec{\eta}+\lambda e^{\lambda t}\vec{\rho} = Ate^{\lambda t}\vec{\eta} + Ae^{\lambda t}\vec{\rho}$$ $$\vec{\eta}+\lambda t\vec{\eta}+\lambda\vec{\rho}=At\vec{\eta}+A\vec{\rho}$$ $$\vec{\eta}+\lambda\vec{\rho} = A\vec{\rho}$$ $$\vec{\eta} = A\vec{\rho}-\lambda I\vec{\rho}$$ $$(A-\lambda I)\vec{\rho} = \vec{\eta}.$$

$\endgroup$
  • $\begingroup$ What if no such vector $\rho$ exists? $\endgroup$ – Pancake_Senpai Jan 8 '19 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.