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Section 9.1 Initial Value Problems

By initial value problem we mean a system of first order ordinary differential equations
\begin{equation*} x'(t) = f(t,x(t)),\;\;t\in\Bbb R,\;x(t)\in\Bbb R^m, \end{equation*}
together with a condition
\begin{equation*} x(t_0)=x_0 \end{equation*}
specifying the position of the desired solution at a given time.

Nobody wants to waste computational time looking for solutions that do not exist. This is why the following theorem is of particular importance for us: Example 1. The rhs \(f(t,x)=x\) is smooth for all \(t\) and \(x\text{.}\)

Hence, the ODE
\begin{equation*} \dot x=x, x(t_0)=x_0 \end{equation*}
has a unique solution for all \((t_0,x_0)\text{.}\)

The corresponding solution
\begin{equation*} x(t)=x_0e^{t-t_0} \end{equation*}
in fact is actually defined for all \(t\text{.}\)

Example 2. The rhs \(f(t,x)=x^2\) like in Example 1, is smooth for all \(t\) and \(x\text{.}\)

Hence, the ODE
\begin{equation*} \dot x=x^2, x(t_0)=x_0 \end{equation*}
has a unique solution for all \((t_0,x_0)\text{.}\)

The corresponding solution
\begin{equation*} x(t)=\frac{x_0}{1-x_0(t-t_0)}, \end{equation*}
though, blows up at \(t=t_0+1/x_0\text{.}\)

Namely, as the theorem goes, the solution is defined only "for \(t\) close enough to \(t_0\)".

Example 3 The rhs \(f(t,x)=\sqrt{x}\) is smooth everywhere (in its domain) except at \(x=0\text{,}\) where \(f_x(t,x)=\frac{1}{2\sqrt{x}}\) blows up.

Hence uniqueness is not granted there (for existence, continuity is enough). Indeed, in this case, there are infinitely many solutions to
\begin{equation*} \dot x=\sqrt{x},\,x(0)=0. \end{equation*}
For instance,
\begin{equation*} x_1(t)=t^2/4 \end{equation*}
and
\begin{equation*} x_2(t)=0. \end{equation*}