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Lets say I have a system of equations like this and I need Cauchy's problem for it

$$\begin{align}&\begin{cases} \dot x=-3x+4y-2z ,\\ \dot y=x+z \\ \dot z=6x-6y+5z \\ \end{cases} \\ \end{align}$$ $$x(0)=20$$ $$y(0)=-0.1$$ $$z(0)=-20$$

I am having trouble understanding this whole $x(t)$ concept. How does $t$ play a role here (I don't see it in the system anywhere) and how would Cauchy's problem look for this equation? (don't give solutions (yet))

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    $\begingroup$ $x,y$ and $z$ are all functions that depend on the time $t$. The dot denotes the time derivative. However, if you don't understand the $x(t)$ concept, then you should start to understand simple ODE's first. An elementary example ist $f'(t)=f(t)$ with $f(0)=1$, i.e. a function that coincides with its derivative. $\endgroup$ Jan 31, 2015 at 15:54
  • $\begingroup$ so this would be the same as $x'(t)=-3x(t)+4y(t)-2z(t)$? $\endgroup$ Jan 31, 2015 at 15:58

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Note that all the variables depend on $t$. You have to find $x(t), y(t), z(t)$ such that the system of differential equations holds. Where's the $t$?

For instance you have the first of the equations:

$$\frac{dx}{dt}=-3x(t)+4y(t)-2z(t).$$

All the variables are parameterized by $t$.

Usually what you do is to consider the vector

$$\mathbf X'=\begin{bmatrix}\dot{x}\\\dot{y}\\\dot{z}\end{bmatrix},$$ and the matrix

$$A=\begin{bmatrix}-3&4&-2\\1&0&1\\6&-6&5\end{bmatrix}$$

to solve the system $\mathbf X' =\mathbf{AX}$. One possible way to do this is by supposing that $\mathbf X = \begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}e^{\lambda t}$ is a solution to the differential equations system, then, since $\mathbf X'=\mathbf K\lambda e^{\lambda t}$, we can say that $\mathbf K \lambda e^{\lambda t}=\mathbf {AK}e^{\lambda t}$. This yields $\mathbf{AK}=\lambda \mathbf K$. Therefore you will see that $(\mathbf A-\lambda \mathbf I)\mathbf K =\mathbf 0$. And this turns into a problem of finding eigenvalues (and eigenvectors).

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