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# Picard's theorem.
We will state a common version of uniqueness and existence theorem for a local solution to a first order ODE with an initial point specified.
> **Picard's theorem.**
> Let $f(x,y)$ be a slope field such that on some rectangle $R = [a,b] \times [c,d]$ both $f$ and $\partial f / \partial y$ are continuous. Then for any interior point $(x_{0}, y_{0})$ of $R$ (so not on the boundary), there **exists** a **unique** **local** solution to the initial value problem (IVP) $$
y'(x) = f(x,y) , \quad y(x_{0}) = y_{0}.
$$ In other words, there exists some $h > 0$ such that above IVP has a unique solution $y=y(x)$ on the interval $[x_{0} - h, x_{0} +h]$.
Note Picard's theorem cannot tell us about global behaviors of these solutions, just that locally they exists and are unique, provided that the hypotheses about the slope field are satisfied.
$\blacktriangleright$ Proof.
We will prove this in several steps. First we restate the problem in an equivalent integral form. Next, we construct a candidate local solution. Then we show that indeed this candidate solution is indeed a local solution, and that, lastly, it is a unique local solution.
Step 1. Re-expression of the IVP as an integral equation.
Observe that by the fundamental theorem of calculus, the differential equation is equivalent to seeking a function $y = y(x)$ such that $$
y(x) = y_{0} + \int_{x_{0}}^{x} f(t,y(t)) dt
$$So we will be looking for a local solution to this integral equation instead.
Step 2. A candidate solution.
Notice the integral equation has a "recursive nature", that putting $y(x)$ into the integrand gives $y(x)$ back. This motivates the following sequence of functions: $$
\begin{align*}
y_{0}(x) & = y_{0} \\
y_{1}(x) & = y_{0} + \int_{x_{0}}^{x} f(t,y_{0}(t)) dt \\
y_{2}(x) & = y_{0} + \int_{x_{0}}^{x} f(t,y_{1}(t)) dt \\
& \vdots \\
y_{n+1}(x) & = y_{0} + \int_{x_{0}}^{x} f(t,y_{n}(t)) dt \\
& \vdots
\end{align*}
$$ on some yet-to-be-determined small interval $[x_{0}- h , x_{0} + h]$, for some $h > 0$.
We will eventually claim that this sequence of functions $(y_{n}(x))_{n=0}^{\infty}$ converges to some continuous function $y(x)$ on the interval $[x_{0} - h, x_{0} + h]$, and that $y(x)$ solves the integral equation uniquely on this small interval.
Ok. What $h > 0$ should we even pick? We will pretend we already have picked one, and we will see what value of $h$ would be appropriate for us.
Since the slope field $f(x,y)$ is continuous on the rectangle $R$, we know it is bounded, so there is some $M$ where $$
|f(x,y)| \le M
$$for every $(x,y) \in R$. Similarly, as $\partial f / \partial y$ is continuous on $R$ as well, we also have some $K$ where $$
\left|\frac{\partial f}{\partial y} (x,y)\right| \le K
$$ for all $(x,y) \in R$.
Also, by mean value theorem, for any two points $(x,y_{1})$ and $(x,y_{2})$ in this rectangle $R$, we have $$
|f(x,y_{1}) - f(x,y_{2})| = \left| \frac{\partial f}{\partial y} (x,y^{\ast}) \right| |y_{1} - y_{2}|
$$for some $y^{\ast}$ between $y_{1}$ and $y_{2}$. And since this point $(x,y^{\ast}) \in R$, we conclude that $$
|f(x,y_{1}) - f(x,y_{2})| \le K |y_{1} - y_{2}|
$$whenever both $(x,y_{1})$ and $(x,y_{2})$ are in $R$.
Now, supposed we have picked some $h > 0$ and some $\ell >0$ such that the small rectangle $R' = [x_{0} - h , x_{0} + h] \times [y_{0} - \ell , y_{0} + \ell]$ would be inside the rectangle $R$.
To show the sequence $(y_{n}(x))_{n=0}^{\infty}$ converges on $[x_{0} - h, x_{0} + h]$, we can show the series $$
y_{0}(x) + \sum_{k=0}^{\infty} y_{k+1}(x) - y_{k}(x)
$$converges. We will show it converges absolutely by using comparison test, on a suitable choice of $h > 0$.
Note on the interval $x\in [x_{0} - h, x_{0} + h]$, we have $$
|y_{1}(x) - y_{0}(x) | = \left| \int_{x_{0}}^{x} f(t,y_{0}(t)) dt \right| \le M h
$$since the points $(t,y_{0}(t))$ are always in $R$.
At this point we will choose $h$ such that $y_{0} \pm Mh$ stays inside the rectangle $R$. So $\ell = Mh$. This ensures the graph of $y_{1}(x)$ stays in $R'$.
Next, again on the same interval, we have $$
|y_{2}(x) - y_{0}(x) | = \left| \int_{x_{0}}^{x} f(t,y_{1}(t)) dt \right| \le M h
$$
Again, the graph of $y_{2}$ stays in the rectangle $R'$.