Picard–Lindelöf Theorem
Until now, we have been solving (first-order) ODEs and taking the computed answers for granted. A caveat is that we don’t know about the nature of a supposed solution: does it exist to begin with, and is it unique? To address this issue, we will establish the Picard–Lindelöf Theorem which guarantees that a first-order ODE with an initial condition specified will indeed have one and only one solution under certain conditions.
Lipschitz Continuous
To proceed, we need to introduce the notion of Lipschitz continuity, the definition of which is given below.
Definition 1. (Lipschitz continuity) A function \( f(x) \) is called Lipschitz continuous over the interval \( [a,b] \), if there exists a real constant \( K \geq 0 \) such that
\begin{equation}
|f(x_1) -f(x_2)| \leq K |x_1 -x_2| \tag{1}
\end{equation}
for all pairs of \( x_1 \) and \( x_2 \) within \( [a,b] \).
The geometrical meaning is that the absolute value of the gradient along the curve is bounded by the Lipschitz constant (the smallest possible) \( K \). For example, the function \( \sin (3x) \) is Lipschitz continous over \(\mathbb{R}\) with \( K = 3 \). The function \( \ln x \) is Lipschitz continous over \( [1, \infty) \) with \( K = 1 \), but not Lipschitz continous along the entire positive \(x\)-axis \( (0, \infty) \). Finally, for \( f(x) = x^2 \), it is Lipschitz continuous over \( [-1,1] \) with \( K = 2 \), over \( [-5,3] \) with \( K = 10 \), but not Lipschitz continuous along the whole real axis.
Picard’s Successive Approximations
Another technique we will employ is Picard’s method of successive approximations. Given a first-order ODE
\begin{equation}
\frac{dy}{dx} = f(x,y) \tag{2}
\end{equation}
with the initial condition \( y(x_0) = y_0 \), we can integrate both sides to get the equivalent relation of
\begin{align}
[y]_{x_0}^x &= \int_{x_0}^x f(x’, y(x’)) dx’ \\
y(x) &= y_0 + \int_{x_0}^x f(x’, y(x’)) dx’ \tag{3}
\end{align}
where \( x’ \) is a dummy integration variable. This suggests that the sequence of iterations
\begin{align}
\begin{aligned}
y_0(x) &= y_0 \\
y_1(x) &= y_0 + \int_{x_0}^x f(x’, y_0(x’)) dx’ \\
y_2(x) &= y_0 + \int_{x_0}^x f(x’, y_1(x’)) dx’ \\
&\vdots \\
y_n(x) &= y_0 + \int_{x_0}^x f(x’, y_{n-1}(x’)) dx’ \\
&\vdots
\end{aligned} \tag{4}
\end{align}
may converge to the true solution \( y(x) \). Indeed, if \( \lim_{n \to \infty} y_n(x) = y(x) \), then by taking a limit on both sides of (4) and assuming (boldly) the limit can be passed inside the integral and function, we should have
\begin{align}
\lim_{n \to \infty} y_n(x) &= \lim_{n \to \infty} y_0 + \lim_{n \to \infty} \int_{x_0}^x f(x’, y_{n-1}(x’)) dx’ \\
y(x) &= y_0 + \int_{x_0}^x f(x’, \lim_{n \to \infty} y_{n-1}(x’)) dx’ \\
&= y_0 + \int_{x_0}^x f(x’, y(x’)) dx’ \tag{5}
\end{align}
that is consistent with (3). Obviously this is completely heuristic and we are going to prove this rigorously in the following part.
Picard–Lindelöf Theorem
With Lipschitz continuity and Picard’s Iteration explained, we are now ready to derive the main theorem as promised in the beginning.
Theorem 2. (Picard–Lindelöf Theorem) Let \( f(x,y) \) be a continuous function satisfying the Lipschitz condition in \( y \) so that
\begin{equation}
|f(x,y_1) -f(x,y_2)| \leq K|y_1 -y_2| \tag{6}
\end{equation}
over the infinitely long strip \( a \leq x \leq b \) and \( -\infty \leq y \leq \infty \). If \( (x_0, y_0) \) is any point on the strip, then the initial value problem
\begin{align}
\frac{dy}{dx} = f(x,y), & & y(x_0) = y_0 \tag{7}
\end{align}
has one and only one solution \( y = y(x) \) on the interval \( a \leq x \leq b \).
This form of the theorem, as well as the following proof, is based on the reference textbook Differential Equations with Applications and Historical Notes by Simmons. Another variant of the theorem is to change \( -\infty \leq y \leq \infty \) to \( c \leq y \leq d \) so the concerned region becomes a closed rectangle: then the existence and uniqueness of the solution will hold within \( x_0 -\varepsilon \leq x \leq x_0 + \varepsilon \) for some small \( \varepsilon > 0 \).
First, define over \( [a,b] \),
\begin{align}
\begin{aligned}
M_0 &= |y_0| \\
M_1 &= \max{|y_1(x)|} \\
M &= M_0 + M_1
\end{aligned}
\tag{8}
\end{align}
Immediately, \( |y_0| = M_0 \leq M \), and by Triangle Inequality
\begin{align}
|y_1(x) -y_0| \leq |y_1(x)| + |y_0| \leq M_1 + M_0 = M \tag{9}
\end{align}
Then as long as \( x_0 \leq x \leq b \), from (4):
\begin{align}
|y_2(x) -y_1(x)| &= \left|\int_{x_0}^x [f(x’, y_{1}(x’)) -f(x’, y_{0}(x’))] dx’\right| \\
&\leq \int_{x_0}^x |f(x’, y_{1}(x’)) -f(x’, y_{0}(x’))| dx’ \\
&\text{(Triangle Inequality for Integrals)} \\
&\leq K \int_{x_0}^x |y_{1}(x’) -y_{0}(x’)| dx’ & & \text{(By (6))} \\
&\leq KM(x -x_0) & & \text{(By (9))} \tag{10}
\end{align}
Similarly,
\begin{align}
|y_3(x) -y_2(x)| &\leq \int_{x_0}^x |f(x’, y_{2}(x’)) -f(x’, y_{1}(x’))| dx’ \\
&\leq K \int_{x_0}^x |y_{2}(x’) -y_{1}(x’)| dx’ & & \text{(By (6))} \\
&\leq K^2M \int_{x_0}^x (x’ -x_0) dx’ & & \text{(By (10))} \\
&= K^2M\frac{(x -x_0)^2}{2} \tag{11}
\end{align}
Going on, we have the general relationship:
\begin{equation}
|y_n(x) -y_{n-1}(x)| \leq K^{n-1}M \frac{(x -x_0)^{n-1}}{(n-1)!} \tag{12}
\end{equation}
The same result is valid for \( a \leq x \leq x_0 \) with \( (x -x_0) \) replaced by the absolute distance \( | x -x_0| \), so
\begin{align}
|y_n(x) -y_{n-1}(x)| &\leq K^{n-1}M \frac{|x -x_0|^{n-1}}{(n-1)!} \\
& \leq K^{n-1}M \frac{(b -a)^{n-1}}{(n-1)!} \tag{13}
\end{align}
for any \( x \) in \( [a,b] \) and all positive integers \( n \). Now consider the infinite series
\begin{align}
\begin{aligned}
&|y_0| + |y_1(x) -y_0| + |y_2(x) -y_1(x)| + \cdots \\
&+ |y_n(x) -y_{n-1}(x)| + \cdots
\end{aligned} \tag{14}
\end{align}
Each of the terms is, by (13), less than or equal to the corresponding term of the series of constants:
\begin{align}
M + M + KM(b-a) + K^2M\frac{(b-a)^2}{2!} + K^3M\frac{(b-a)^3}{3!} + \cdots \tag{15}
\end{align}
which is convergent (readily seen by the Root Test in elementary calculus or recognizing the subsequent terms as \(Me^{K(b-a)}\)). So, for \( x \) in \([a,b]\)
\begin{align}
y_n(x) &= y_0 + \sum_{n=1}^{\infty} [y_n(x) -y_{n-1}(x)] \tag{16}
\end{align}
converges to a function by comparison test, which we will denote by \( y(x) \), uniformly. This means that we can make \( y_n(x) \) as close as needed to \( y(x) \) for all \(x\) in the interval.
Now we can justify that this \( y(x) \) is the solution of (7) by showing it satisfies (3), i.e.
\begin{align}
y(x) -y_0 -\int_{x_0}^x f(x’, y(x’)) dx’ = 0 \tag{17}
\end{align}
Notice that by rearranging (4):
\begin{align}
y_n(x) -y_0 -\int_{x_0}^x f(x’, y_{n-1}(x’)) dx’ &= 0 \tag{18}
\end{align}
Subtracting the L.H.S. of (18) from that of (17) gives
\begin{align}
& y(x) -y_0 -\int_{x_0}^x f(x’, y(x’)) dx’ \\
={}& y(x) -y_0 -\int_{x_0}^x f(x’, y(x’)) dx’ \\
&- (y_n(x) -y_0 -\int_{x_0}^x f(x’, y_{n-1}(x’)) dx’) \\
={}& y(x) -y_n(x) + \int_{x_0}^x [f(x’, y_{n-1}(x’)) -f(x’, y(x’))] dx’ \tag{19}
\end{align}
and using the familiar Triangle Inequality again, we obtain
\begin{align}
&\left|y(x) -y_0 -\int_{x_0}^x f(x’, y(x’)) dx’\right| \\
\leq{}& |y(x) -y_n(x)| + \left|\int_{x_0}^x [f(x’, y_{n-1}(x’)) -f(x’, y(x’))] dx’\right| \\
\leq{}& |y(x) -y_n(x)| + \int_{x_0}^x |f(x’, y_{n-1}(x’)) -f(x’, y(x’))| dx’ \\
\leq{}& |y(x) -y_n(x)| + K \int_{x_0}^x |y_{n-1}(x’) -y(x’)| dx’ \\
& \text{(By (6))} \\
\leq{}& |y(x) -y_n(x)| + K \max|y_{n-1}(x’) -y(x’)| (x-x_0) \tag{20}
\end{align}
Due to uniform convergence as suggested above, the last line of (20) can be made as small as possible by taking large \( n \), so the first line must be squeezed to zero and hence (17), the existence of a solution, is established.
For the uniqueness part, assume there exists another solution \( \bar{y}(x) \) over the interval. Then we need to show \( y_n(x) \to \bar{y}(x) \) as \( n \to \infty \) for all \(x\) within the interval: but since it is already argued that \( y_n(x) \to y(x) \), it must be \( \bar{y}(x) = y(x) \).
\( \bar{y}(x) \) will also have to obey (3):
\begin{equation}
\bar{y}(x) = y_0 + \int_{x_0}^x f(x’, \bar{y}(x’)) dx’ \tag{21}
\end{equation}
Take \( A = \max|\bar{y}(x) -y_0| \), then similar to the steps from (10) to (12):
\begin{align}
|\bar{y}(x) -y_1(x)| &= \left| \int_{x_0}^x [f(x’, \bar{y}(x’)) -f(x’, y_0(x’))] dx’ \right| \\
&\leq \int_{x_0}^x |f(x’, \bar{y}(x’)) -f(x’, y_0(x’))| dx’ \\
&\leq K \int_{x_0}^x |\bar{y}(x’) -y_0| dx’ \\
&\leq KA(x-x_0) \tag{22}
\end{align}
\begin{align}
|\bar{y}(x) -y_2(x)| &\leq \int_{x_0}^x |f(x’, \bar{y}(x’)) -f(x’, y_1(x’))| dx’ \\
&\leq K \int_{x_0}^x |\bar{y}(x’) -y_1(x’)| dx’ \\
&\leq K^2A \int_{x_0}^x(x’-x_0) dx’ & & \text{(By (22))} \\
&= K^2A\frac{(x-x_0)^2}{2} \tag{23}
\end{align}
By the same logic and considering absolute distance, we have
\begin{equation}
|\bar{y}(x) -y_n(x)| \leq K^nA\frac{|x-x_0|^n}{n!} \leq K^nA\frac{(b-a)^n}{n!} \tag{24}
\end{equation}
The R.H.S. of (24) will eventually tend to zero as \( n \to \infty \) (Factorial grows faster than power), so \( y_n(x) \to \bar{y}(x) \) and we have the uniqueness \( \bar{y}(x) = y(x) \) as just suggested.
A Simple Counter-example
Alice is working on a question about a first-order ODE, where it is simply
\begin{align}
\frac{dy}{dx} = \frac{1}{1-x}
\end{align}
and given \( y(0) = 1 \). She proceeds as follows:
\begin{align}
y = \int\frac{1}{1-x} dx = C -\ln|1-x|
\end{align}
Substituting \( y(0) = 1 \) leads to \( C = 1 \), i.e. \(y = 1 -\ln|1 -x| \). She then concludes that when \(x = 2\), \(y(2) = 1 -\ln|1-2| = 1 -\ln|-1| = 1 -(\ln 1) = 1 -0 = 1\), but the teacher says the answer is wrong (QQ). Explain why.
Answer
The ODE and its solution are discontinuous at \( x = 1 \). Hence, information from the initial condition cannot propagate across the discontinuity.
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