We start with the metric space where we will apply the fixed point theorem: the space
\(C\bigl([a,b],\R\bigr)\) of
Example 7.1.8, the space of continuous functions
\(f \colon [a,b] \to \R\) with the metric
\begin{equation*}
d(f,g) \coloneqq \snorm{f-g}_{[a,b]} = \sup_{x \in [a,b]} \abs{f(x)-g(x)} .
\end{equation*}
Convergence in this metric is convergence in uniform norm, or in other words, uniform convergence. Therefore,
\(C\bigl([a,b],\R\bigr)\) is a complete metric space, see
Proposition 7.4.5.
Consider now the ordinary differential equation
\begin{equation*}
\frac{dy}{dx} = F(x,y) .
\end{equation*}
Given some \(x_0, y_0\text{,}\) we desire a function \(y=f(x)\) such that \(f(x_0) = y_0\) and such that
\begin{equation*}
f'(x) = F\bigl(x,f(x)\bigr) .
\end{equation*}
To avoid having to come up with many names, we often simply write \(y' = F(x,y)\) for the equation and \(y(x)\) for the solution.
A subtle issue is how long does the solution exist. Consider the equation
\(y' = y^2\text{,}\) \(y(0)=1\text{.}\) Then
\(y(x) = \frac{1}{1-x}\) is a solution. While
\(F\) is a reasonably “nice” function and in particular it exists for all
\(x\) and
\(y\text{,}\) the solution “blows up” at
\(x=1\text{.}\) For more examples related to Picard’s theorem, see
Section 6.3.
We will look for the solution in \(C\bigl([a,b],\R\bigr)\text{,}\) which may feel strange at first as we are searching for a differentiable function. The explanation is that we consider the corresponding integral equation
\begin{equation*}
f(x)
=
y_0 + \int_{x_0}^x F\bigl(t,f(t)\bigr)\,dt .
\end{equation*}
To solve this integral equation we only need a continuous function, and in some sense our task should be easier—we have more candidate functions to try. This way of thinking is quite typical when solving differential equations.
Theorem 7.6.3. Picard’s theorem on existence and uniqueness.
Let \(I, J \subset \R\) be closed and bounded intervals, let \(I^\circ\) and \(J^\circ\) be their interiors, and let \((x_0,y_0) \in I^\circ \times J^\circ\text{.}\) Suppose \(F \colon I \times J \to \R\) is continuous and Lipschitz in the second variable, that is, there exists an \(L \in \R\) such that
\begin{equation*}
\abs{F(x,y) - F(x,z)} \leq L \abs{y-z}
\qquad \text{for all } y,z \in J, x \in I .
\end{equation*}
Then there exists an \(h > 0\) such that \([x_0-h,x_0+h] \subset I\) and a unique differentiable function \(f \colon [x_0 - h, x_0 + h] \to J \subset \R\) such that
\begin{equation*}
f'(x) = F\bigl(x,f(x)\bigr) \qquad \text{and} \qquad f(x_0) = y_0.
\end{equation*}
Proof.
Without loss of generality, assume \(x_0 =0\) (exercise). As \(I \times J\) is compact and \(F\) is continuous, \(F\) is bounded. So find an \(M > 0\) such that \(\abs{F(x,y)} \leq M\) for all \((x,y) \in I\times J\text{.}\) Pick \(\alpha > 0\) such that \([-\alpha,\alpha] \subset I\) and \([y_0-\alpha, y_0 + \alpha] \subset J\text{.}\) Let
\begin{equation*}
h \coloneqq \min \left\{ \alpha, \frac{\alpha}{M+L\alpha} \right\} .
\end{equation*}
Note \([-h,h] \subset I\text{.}\) Let
\begin{equation*}
Y \coloneqq \bigl\{ f \in C\bigl([-h,h],\R\bigr) : f\bigl([-h,h]\bigr) \subset J \bigr\} .
\end{equation*}
That is,
\(Y\) is the set of continuous functions on
\([-h,h]\) with values in
\(J\text{,}\) in other words, exactly those functions where
\(F\bigl(x,f(x)\bigr)\) makes sense. It is left as an exercise to show that
\(Y\) is a closed subset of
\(C\bigl([-h,h],\R\bigr)\) (because
\(J\) is closed). The space
\(C\bigl([-h,h],\R\bigr)\) is complete, and a closed subset of a complete metric space is a complete metric space with the subspace metric, see
Proposition 7.4.6. So
\(Y\) with the subspace metric is a complete metric space. We will write
\(d(f,g) = \snorm{f-g}_{[-h,h]}\) for this metric.
Define a mapping \(T \colon Y \to C\bigl([-h,h],\R\bigr)\) by
\begin{equation*}
T(f)(x)
\coloneqq
y_0 + \int_0^x F\bigl(t,f(t)\bigr)\,dt .
\end{equation*}
It is an exercise to check that \(T\) is well-defined, and that for \(f \in Y\text{,}\) \(T(f)\) really is in \(C\bigl([-h,h],\R\bigr)\text{.}\) Let \(f \in Y\) and \(\abs{x} \leq h\text{.}\) As \(F\) is bounded by \(M\text{,}\) we have
\begin{equation*}
\begin{split}
\abs{T(f)(x) - y_0}
&= \abs{\int_0^x F\bigl(t,f(t)\bigr)\,dt} \\
& \leq
\abs{x}M \leq hM \leq \frac{\alpha M}{M+ L\alpha} \leq \alpha .
\end{split}
\end{equation*}
So \(T(f)\bigl([-h,h]\bigr) \subset [y_0-\alpha,y_0+\alpha] \subset J\text{,}\) and \(T(f) \in Y\text{.}\) In other words, \(T(Y) \subset Y\text{.}\) From now on, we consider \(T\) as a mapping of \(Y\) to \(Y\text{.}\)
We claim \(T \colon Y \to Y\) is a contraction. First, for \(x \in [-h,h]\) and \(f,g \in Y\text{,}\) we have
\begin{equation*}
\abs{F\bigl(x,f(x)\bigr) - F\bigl(x,g(x)\bigr)} \leq
L\abs{f(x)- g(x)} \leq L \, d(f,g) .
\end{equation*}
Therefore,
\begin{equation*}
\begin{split}
\abs{T(f)(x) - T(g)(x)}
&= \abs{\int_0^x \Bigl( F\bigl(t,f(t)\bigr) - F\bigl(t,g(t)\bigr)\Bigr)\,dt} \\
& \leq \abs{x} L \, d(f,g)
\leq h L\, d(f,g)
\leq \frac{L\alpha}{M+L\alpha} \, d(f,g) .
\end{split}
\end{equation*}
We chose \(M > 0\) and so \(\frac{L\alpha}{M+L\alpha} < 1\text{.}\) Take supremum over \(x \in [-h,h]\) of the left-hand side above to obtain \(d\bigl(T(f),T(g)\bigr) \leq \frac{L\alpha}{M+L\alpha} \, d(f,g)\text{,}\) that is, \(T\) is a contraction.
The fixed point theorem (
Theorem 7.6.2) gives a unique
\(f \in Y\) such that
\(T(f) = f\text{.}\) In other words,
\begin{equation*}
f(x) = y_0 + \int_0^x F\bigl(t,f(t)\bigr)\,dt .
\end{equation*}
Clearly,
\(f(0) = y_0\text{.}\) By the fundamental theorem of calculus (
Theorem 5.3.3),
\(f\) is differentiable and its derivative is
\(F\bigl(x,f(x)\bigr)\text{.}\) Differentiable functions are continuous, so
\(f\) is the unique differentiable
\(f \colon [-h,h] \to J\) such that
\(f'(x) = F\bigl(x,f(x)\bigr)\) and
\(f(0) = y_0\text{.}\)
