MHB Differential equation with eigenvector

[Petrus]

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

Is that correct?

Regards,
$$|\pi\rangle$$

 

[Opalg]

Your third equation should be $x_3=0$. The eigenvector must be a multiple of $$\begin{pmatrix}1\\1\\0\end{pmatrix}.$$

 

[Petrus]

Your third equation should be $x_3=0$. The eigenvector must be a multiple of $$\begin{pmatrix}1\\1\\0\end{pmatrix}.$$

I have hard to understand this

Regards,
$$|\pi\rangle$$

 

[Petrus]

Is this correct?

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

 

[Petrus]

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 $\{(1,0,0)^T,(0,1,0)^T\}.$

- - - Updated - - -
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 $$c_1$$ it should be $$e^{0t}$$
Thanks!
Regards,
$$|\pi\rangle$$

 

[Opalg]

Is this correct?

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

That last formula is partially correct. You wrote $$\begin{pmatrix}f(t) \\g(t) \\h(t)\end{pmatrix} = c_1\begin{pmatrix}1 \\1 \\0\end{pmatrix} + c_2\begin{pmatrix}-1 \\-1 \\1\end{pmatrix}e^t$$. 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 $$\begin{pmatrix}f(t) \\g(t) \\h(t)\end{pmatrix} = c_1\begin{pmatrix}1 \\1 \\0\end{pmatrix} + c_2\begin{pmatrix}-1 \\-1 \\1\end{pmatrix}e^t + c_3\begin{pmatrix}* \\* \\*\end{pmatrix}e^{2t}$$, 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]

Hi,
I forgot that $$\lambda=2$$!

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

Regards,
$$|\pi\rangle$$