1. ## ODE's help...

Hello, can i get some help in the next ODE?
Consider the oblong O={(t,y):abs(t-t0)<=a,abs(y-y0)<=b} and if f is....(f satisfies the picard theorem anyway..) then show that the initial value problem: y''=f(t,y), y(t0)=y0, y'(t0)=y'0 has oly one solution in the space [t0-σ,t0+σ]...thanx

2. There's the straightforward way: Define the Picard iterations (following the proof of the $Y'=F(t,Y), \ Y(t_0)=Y_0$ initial value problem), show a fixed point exists, and use the Gronwall lemma for uniqueness.

However, who can be bothered to do it? So bypass all of that with a trick.

Define $y'(t)=\omega(t), \ Y(t)=\left(\begin{array}{c}
y(t) \\
\omega(t) \\
\end{array}\right), \ F(t,Y)=\left(\begin{array}{c}
\omega(t) \\
f(t,pr_1Y) \\
\end{array}\right)$
,

where $pr_1$ means projection of the first coordinate of the vector $Y$. What happened here is, that the original problem $y'(t)=f(t,y)$ has been rewritten as $Y'=F(t,Y)$.

Big deal? I mean, we have increased the dimension, is it good? In general, no. But here, we have gained the right to use the (original) Picard theorem!

So, we need to check whether $F$ is Lipschitz for $Y$. For the usual $\ell^1$ norm, we have

$|F(t,Y_1)-F(t,Y_2)|=|f(t,pr_1Y_1)-f(t,pr_1Y_2)|=|f(t,y_1)-f(t,y_2)|\leq L|y_1-y_2|$, where $L$ is the Lipschitz constant for $f$.

Now, by applying the Picard theorem with initial condition $Y(t_0)=\left(\begin{array}{c}
y(t_0) \\
\omega(t_0) \\
\end{array}\right)=\left(\begin{array}{c}
y_0 \\
y'_0 \\
\end{array}\right)$
, we obtain some $\sigma>0$ such that our problem has a unique solution $Y(t)=\left(\begin{array}{c}
y(t) \\
y'(t) \\
\end{array}\right)$
in the interval $[t_0-\sigma,t_0+\sigma]$.

Question: How does $\sigma$ come into play here? What does it signify?