Prove Picard's Existence and Uniqueness Theorem on Different Levels

IVP and Picard’s

By IVP (Initial Value Problem), we mean the problem about solving

$$\begin{cases} \frac{dy}{dx}=f(x,y) \\ y(x_0)=y_0 \end{cases}$$

Of course we do hope we can solve this problem easily with a simple result. But that won’t happen if the function $f$ is ‘ugly’ enough. Hence the problem is, generally and theoretically, when we can get a unique solution? When does some solution exist? Fortunately Picard ensured that

If $f$ is uniformly Lipschitz continuous in $y$ and continuous in $x$ on $R=\{(x,y):|x-x_0| \leq a,|y-y_0| \leq b\}$, then for some $\varepsilon>0$, there exists a unique solution $y$ to the IVP on the interval $[x_0-\varepsilon,x_0+\varepsilon]$.

Interesting enough, there are several ways to prove it on different levels. This blog post offers two proofs.

Proving Picard’s existence and uniqueness theorem

Some preparation

By uniformly Lipschitz continuous, we mean that for all $y \in R$, $f$ satisfies

$$\left\vert f(x,y_1) - f(x,y_2)\right\vert \leq L\left\vert y_1 - y_2 \right\vert$$

for some $L > 0$. This condition is useful in many different branch of mathematics.

Also, it’s trivial to verify that the IVP is equivalent to

$$y=y_0+\int_{x_0}^{x}f(x,y)dx$$

And yes, our job becomes finding such a $y$ satisfying this equation.

A sketch of the proof in elementary calculus

Honestly the proof is kind of long. I’ll leave the basic steps here. First, we define the Picard sequence by

$$y_{n+1}(x)=y_0+\int_{x_0}^{x}f(x,y_n(x))dx \quad |x-x_0| \leq \varepsilon$$

for $n=0,1,\cdots$. It can be shown by induction that

$$|y_n(x)-y_0| \leq M|x-x_0|$$

where $M = \sup\limits_{(x,y) \in R}|f(x,y)|$ and $\varepsilon = \min\left(a,\frac{b}{M}\right)$. We want to prove that

$$\lim_{n \to \infty}y_n=\sum_{n=1}^{\infty}(y_{n+1}(x)-y_n(x))=y$$

is the solution to the IVP, where Lipschitz condition comes into play by considering

$$|y_{n+1}(x)-y_n(x)| \leq \frac{M}{L}\frac{(L|x-x_0|)^{n+1}}{(n+1)!}$$

Finally, it should be shown that $y$ is the unique solution. To do this, it can be shown that, if we have two solutions, say $y=u(x)$ and $y=v(x)$, then $u(x)-v(x)=0$.

Osgood’s condition ensures uniqueness

Osgood’s condition is much weaker than Lipschitz’s. Under Osgood’s condition, it’s easy to check the uniqueness (existence is not guaranteed), but there is no way to get the result (while Lipschitz’s show you the way) through this condition.

$f(x,y)$ has at most one solution in every point in $R$ if $f$ is continuous and satisfies the Osgood’s condition, namely

$$|f(x,y_1)-f(x,y_2)| \leq F(|y_1-y_2|)$$

in $R$ where $F$ is a continuous function such that $F(t)>0$ for all $r>0$. Also, $F$ is defined in such a way that

$$\int_{0}^{r_1}\frac{1}{F(r)}dr=\infty$$

for some $r_1>0$.

Naturally, if we define $F(r)=Lr$, we have $\int_{0}^{r_1}\frac{1}{Lr}dr=\infty$.

Proof of Osgood’s

We’ll prove this theorem indirectly. That is, if there exists some point $(x_0,y_0) \in R$ such that $f$ has at least two solutions, then $f$ does not satisfy Osgood’s condition, which is equivalent to the statement that if $f$ satisfies Osgood’s, then $f$ has no more than solution.

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Suppose now there exists a point $(x_0,y_0) \in R$ such that $y’=f(x,y)$ has two distinct solutions

$$y_1(x)\quad y_2(x)$$

and $y_1 \neq y_2$ for at $x_1 \neq x_0$. W.L.O.G. we suppose that $x_1 > x_0$. Define

$$x_s = \sup_{x \in [x_0,x_1]}\{x:y_1(x)=y_2(x)\}$$

and $r(x)=y_1(x)-y_2(x)$ on $[x_s,x_1]$. The derivative of $r$ is interesting since we have

$$\begin{aligned} r'(x)&=y_1'(x)-y_2'(x) \\ &=f(x,y_1(x))-f(x,y_2(x)) \\ & \leq |f(x,y_1)-f(x,y_2)| \\ & \leq F(|y_1-y_2|)=F(r) \end{aligned}$$

Then the desired improper integral converges since we have

$$\int_{0}^{r_1}\frac{dr}{F(r)}=\int_{x_s}^{x_1}\frac{r'(x)dx}{F(r(x))} \leq x_1-x_s$$

Therefore $f$ does not satisfy Osgood’s condition.

Banach FPT is applied onto Picard’s

What is Banach FPT (Fixed Point Theorem)

A map $T: X \mapsto X$ defined on a complete metric space $(X,d)$ is a contraction if there exists some $k \in [0,1)$ such that

$$d(T(x),T(y)) \leq k(x,y)$$

Banach Fixed Point Theorem states that, $T$ admits a fixed point, namely $T(x)=x$ for some $x \in X$.

The proof is an application of Cauchy sequence (notice that $(X,d)$ is complete). Pick an arbitrary $x_0 \in X$, by defining $T(x_n)=x_{n+1}$, we have

$$d(x_{n},x_{n-1})=d(T(x_{n-1}),T(x_{n-2})) \leq kd(x_{n-1},x_{n-2})$$

which finally goes to

$$d(x_n,x_{n-1}) \leq k^{n-1}d(x_1,x_0)$$

$\{x_n\}$ is Cauchy then since we have

$$\begin{aligned} d(x_m,x_n) &\leq d(x_m,x_{m-1})+d(x_{m-1},x_{m-2})+\cdots+d(x_{n+1},x_n) \\ &\leq (k^{m-1}+k^{m-2}+\cdots+k^{n})d(x_1,x_0) \\ &\leq \frac{q^n}{1-q}d(x_1,x_0) \\ & \to 0 \quad(n \to\infty) \end{aligned}$$

Since $(X,d)$ is complete, we see $\{x_n\}$ converges. Also, $T$ is (uniformly) continuous, therefore

$$\lim_{n \to \infty}x_n=\lim_{n \to \infty}x_{n+1}=x=T(x)$$

Uniqueness of the fixed point follows from the uniqueness of limit.

Proving Picard’s using Banach FPT

Fortunately, Picard’s existence and uniqueness theorem can be directly proved using Banach FPT. All we need is a proper translation.

Complete metric space

Let $\mathcal{C_B}(R)$ be all bounded continuous function on $R$, then $\mathcal{C_B}(R)$ is complete considering the metric by

$$d(f,g)=\sup|f(x)-g(x)|$$

Contraction map

The functional $T:\mathcal{C_B}(R)\mapsto\mathcal{C_B}(R)$ by

$$T(y)=y_0+\int_{x_0}^{x}f(x,y)dx$$

is the contraction we are looking for, if we take uniform Lipschitz’s condition into consideration. Namely, if we have $\varepsilon<\frac{1}{L}$, we can see that

$$\begin{aligned} d(T(r),T(s)) &= \sup\left\vert \int_{x_0}^{x}(f(x,r)-f(x,s))dx \right\vert \\ &\leq\sup\int_{x_0}^{x}|f(x,r)-f(x,s)|dx \\ &\leq L\varepsilon d(r,s) \end{aligned}$$

Therefore by Banach FPT, the functional $T$ has a unique fixed point, which is equivalent to Picard’s existence and uniqueness theorem. Further, the solution can be obtained by taking

$$y_{n+1}=T(y_n)$$

where $y(x_0)=y_0$ comes from the initial value problem. Picard’s follows from FPT.