# Differential equation with eigenvector

#### Petrus

##### Well-known member
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,
$$\displaystyle |\pi\rangle$$

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#### Opalg

##### MHB Oldtimer
Staff member
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
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
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
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
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
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,
$$\displaystyle |\pi\rangle$$