I'm going through some problem sheets but get confused by a problem as follows
Given a ODE of order n be $$\sum^n_{s=0}\alpha_s\frac{d^s}{dt^s}w=V(w) $$
I am asked to write a first order equation for a vector $u\equiv(w_j)_{j=0,......,n} $ Where $ \frac{dw_j}{dt}=w_{j+1}$
Should be a very fundamental problem but I'm confused by $ w_{j}$ . Is this supposed to be the compnent of $w$ or $$ u=(w_0,\frac{d}{dt}w_0,\frac{d^2}{dt^2}w,......,\frac{d^n}{dt^n}w_0)$$ So if $$ A=(\alpha _0,\alpha_1,\alpha _2,\alpha _3,.......\alpha_n) $$ then $Au=V(w_0) $
Is $u$ supposed to be equivalent as $w$ in first ode equation or else ? Anyhow I am not able to write out the first order expression. Please help!
You are asked to write a first order equation for a vector, which really means a system of first order equations. I'll illustrate the solution for the following third order equation:
$$ a_3 w^{(3)} + a_2 w'' + a_1 w' + w = V(w) $$
We're going to end up with a first order vector DE where the unknown variable is:
$$ \mathbf{u} = (u_1, u_2, u_3) $$
The method is always the same: your first variable is set equal to the original one: $u_1 = w$. Next, you let $u_2 = w'$ and $u_3 = w''$. If the original equation were higher order, you would keep going.
Now you write a differential equation for each one: $u_1'=w'=u_2$, i.e. $u_1'=u_2$. Similarly, $u_2'=u_3$. The DE for $u_3$ is slightly different; $u_3'$ is $w^{(3)}$, which is not one of the new variables. Instead, you can solve for $w^{(3)}$ in the original equation, getting: $w^{(3)}=\left( V(w)-a_2 w'' - a_1 w' \right) / a_3$. Now you replace the $w$s with the $u$s, giving:
$$ u_3'= \left( V(u_1) - a_2 u_3 - a_1 u_2 \right) / a_3 $$
Finally you have the system:
$$ \begin{align} u_1'&=u_2 \\ u_2&'=u_3 \\ u_3' &= \left( V(u_1) - a_2 u_3 - a_1 u_2 \right) / a_3 \end{align} $$
Which can be written as
$$ \mathbf{u}' = \begin{bmatrix} u_1' \\ u_2' \\ u_3' \end{bmatrix} = \begin{bmatrix} u_2 \\ u_3 \\ \left( V(u_1) - a_2 u_3 - a_1 u_2 \right) / a_3 \end{bmatrix}$$
Et voila!