ordinary differential equations
Express the 2nd order ODE
$$\begin{align}\mathrm d_t^2 u:=\frac{\mathrm d^2 u}{\mathrm dt^2}&=\sin(u)+\cos(\omega t)\qquad \omega \in \mathbb Z /\{0\} \\u(0)&=a\\\mathrm d_t u(0)&=b\end{align}$$
as a system of 1st order ODEs and verify there exists a global solution by invoking the global existence and uniqueness theorems.
I'm not sure how to express second order ODEs as first order ODEs, any tips?
$w$ as defined is a $2$-dimensional vector of functions, the first component of which is the function $u$ and the second component of which is the function $v=\frac{du}{dt}$. When we differentiate, we differentiate componentwise: $$ \frac{dw}{dt}=\begin{bmatrix}\tfrac{du}{dt} \\ \tfrac{dv}{dt}\end{bmatrix}. $$ However, we know both $\frac{du}{dt}$ and $\frac{dv}{dt}$ (you said you understood this part). Just plugging these in, we have that $$ \frac{dw}{dt}=\begin{bmatrix}v \\ 5tu+\sin (v)\end{bmatrix}. $$ Same thing with the initial condition: $$ w(0)=\begin{bmatrix}u(0) \\ v(0)\end{bmatrix}=\begin{bmatrix}1 \\ 0\end{bmatrix}. $$
As for the numerical part, it seems that you should adust your equation $W^{n+1}=W^n+\tau f(t_n,w)$ to read $$ W^{n+1}=W^n+\tau f(t_n,W^n). $$ In any case, when you actually started to work out the problem, it seems as if this is what you did. (Also keep in mind that $t_{n+1}=t_n+\tau$).
The fixed points occur when $\sin(x_1)=C$ and $x_2=0$. When those conditions are met, $x_1'$ and $x_2'$ are both 0. I'm placing a streamline plot with C=0.5 below.
As in Marcus's answer, I get that $U(x(t))= \frac{ x_2^2(t)}2 - C\, x_1(t) - \cos(x_1(t))$ is constant for any solution $x(t)$ of the differential equation (i.e $U(x)= \frac{ x_2^2}2 - C\, x_1 - \cos(x_1)$ is a first integral of the system). The fixed points of the ODE occur at the critical points of $U(x)$ which occur exactly when $\sin(x_1)=C$ and $x_2=0$.
Best Answer
Here's an example to get you started:
$$u^{(3)}(t)+t^3u''(t)+5u'(t)+\sin(t)u=e^{6t}$$
with initial values $u''(0)=1$, $u'(0)=2$, and $u(0)=3$
First, give new names to $u$ and its derivatives (stopping one short of the order of the ODE): $u=x_1$, $u'=x_2$, $u''=x_3$.
Substituting back into the DE (keeping in mind that $u^{(3)}(t)=x'_3(t)$) we get: $$x'_3(t)+t^3x_3(t)+5x_2(t)+\sin(t)x_1(t)=e^{6t}$$
Thus we have the equivalent system:
$$\begin{array}{ccrrrr} x'_1 & = & & x_2 & & \\ x'_2 & = & & & x_3 & \\ x'_3 & = & -\sin(t)x_1 & -5x_2 & -t^3x_3 & +e^{6t} \end{array}$$
Also, $x_1(0)=3$, $x_2(0)=2$, and $x_3(0)=1$.