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Differential equation with eigenvector

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
solve this system of linear differential equation
\(\displaystyle f'=f-g-h\)
\(\displaystyle g'=-f+g-h\)
\(\displaystyle h'=-f+g+h\)
with boundary conditions \(\displaystyle f(0)=1\), \(\displaystyle g(0)=2\) and \(\displaystyle h(0)=0\)
we get that \(\displaystyle \lambda=1\) and \(\displaystyle \lambda=0\)
now for eigenvector or we can call it basis for eigenvector \(\displaystyle \lambda=0\) i get

Is that correct?

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Your third equation should be $x_3=0$. The eigenvector must be a multiple of \(\displaystyle \begin{pmatrix}1\\1\\0\end{pmatrix}.\)
 

Petrus

Well-known member
Feb 21, 2013
739
Your third equation should be $x_3=0$. The eigenvector must be a multiple of \(\displaystyle \begin{pmatrix}1\\1\\0\end{pmatrix}.\)
I have hard to understand this

Regards,
\(\displaystyle |\pi\rangle\)
 

Petrus

Well-known member
Feb 21, 2013
739
Is this correct?


Edit:Ops on the first one with \(\displaystyle c_1\) it should be \(\displaystyle e^{0t}\)
So basicly we got \(\displaystyle f(t)=c_1-c_2e^t\), \(\displaystyle g(t)=c_1-c_2e^t\) and \(\displaystyle h(t)=c_2e^t\)
Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
The eigenspace associated to $\lambda=0$ is:
$$\ker (A-0I)\equiv\left \{ \begin{matrix} x_1-x_2-x_3=0\\-x_1+x_2-x_3=0\\-x_1+x_2+x_3=0\end{matrix}\right.\sim \ldots \sim x_3=0$$
Then, all solutions of $\ker (A-0I)$ are
$$\begin{pmatrix}{x_1}\\{x_2}\\{x_3}\end{pmatrix}=\begin{pmatrix}{\alpha}\\{\beta}\\{0}\end{pmatrix}=\alpha \begin{pmatrix}{1}\\{0}\\{0}\end{pmatrix}+\beta \begin{pmatrix}{0}\\{1}\\{0}\end{pmatrix}\;,\quad \alpha,\beta \in \mathbb{R}$$
so, a basis of this eigenspace is $\{(1,0,0)^T,(0,1,0)^T\}.$

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The eigenvector is correct.
Im pretty new with the differential equation with eigenvector. I Dont understand how it works with those boundary conditions. Notice that on the \(\displaystyle c_1\) it should be \(\displaystyle e^{0t}\)
Thanks!
Regards,
\(\displaystyle |\pi\rangle\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Is this correct?


Edit:Ops on the first one with \(\displaystyle c_1\) it should be \(\displaystyle e^{0t}\)
So basicly we got \(\displaystyle f(t)=c_1-c_2e^t\), \(\displaystyle g(t)=c_1-c_2e^t\) and \(\displaystyle h(t)=c_2e^t\)
Regards,
\(\displaystyle |\pi\rangle\)
That last formula is partially correct. You wrote \(\displaystyle \begin{pmatrix}f(t) \\g(t) \\h(t)\end{pmatrix} = c_1\begin{pmatrix}1 \\1 \\0\end{pmatrix} + c_2\begin{pmatrix}-1 \\-1 \\1\end{pmatrix}e^t\). That is right in principle. The only thing missing is that a $3\times3$ matrix has three eigenvalues. Your matrix $A$ has a third eigenvalue $\lambda=2$. So your general solution to the system of equations should look like \(\displaystyle \begin{pmatrix}f(t) \\g(t) \\h(t)\end{pmatrix} = c_1\begin{pmatrix}1 \\1 \\0\end{pmatrix} + c_2\begin{pmatrix}-1 \\-1 \\1\end{pmatrix}e^t + c_3\begin{pmatrix}* \\* \\*\end{pmatrix}e^{2t}\), where you have to fill in the stars with the third eigenvector.

When you have done that, you can put $t=0$ in that equation. That will give you a set of three linear equations for the constants $c_1,c_2,c_3$ so that the initial conditions are satisfied.
 

Petrus

Well-known member
Feb 21, 2013
739
Hi,
I forgot that \(\displaystyle \lambda=2\)!


So we get that \(\displaystyle c_1=\frac{3}{2}\), \(\displaystyle c_2=0\) and \(\displaystyle c_3=\frac{1}{2}\) and that should be correct right?

Regards,
\(\displaystyle |\pi\rangle\)