# Ordinary Differential Equations/The Picard–Lindelöf theorem

In this section, our aim is to prove several closely related results, all of which are occasionally called "Picard-Lindelöf theorem". This type of result is often used when it comes to arguing for the existence and uniqueness of a certain ordinary differential equation, given that some boundary conditions are satisfied.

## Local results

Picard–Lindelöf Theorem (Banach fixed-point theorem version):

Let ${\displaystyle I:=[a,b]}$  be an interval, let ${\displaystyle f:I\times \mathbb {R} ^{n}\to \mathbb {R} ^{n}}$  be a function, and let

${\displaystyle x'(t)=f(t,x(t))}$

be the associated ordinary differential equation. If ${\displaystyle f}$  is Lipschitz continuous in the second argument, then this ODE possesses a unique solution on ${\displaystyle [a,a+\epsilon ]}$  for each possible initial value ${\displaystyle x(0)=x_{0}\in \mathbb {R} ^{n}}$ , where ${\displaystyle \epsilon <1/L}$ , ${\displaystyle L}$  being the Lipschitz constant of the second argument of ${\displaystyle f}$ .

Proof:

We first rewrite the problem as a fixed-point problem. Indeed, using the fundamental theorem of calculus, one can show that the simultaneous equations

${\displaystyle {\begin{cases}x'(t)=f(t,x(t))&t\in [a,a+\epsilon ]\\x(0)=x_{0}&\end{cases}}}$

are equivalent to the single equation

${\displaystyle \forall t\in [a,a+\epsilon ]:x(t)=x_{0}+\int _{a}^{t}f(s,x(s))ds}$ ,

where ${\displaystyle \epsilon }$  is to be determined at a later stage. This means that the function ${\displaystyle x(t)}$  is a fixed point of the function

${\displaystyle T:{\mathcal {C}}([a,a+\epsilon ])\to {\mathcal {C}}([a,a+\epsilon ]),T(x)(t):=x_{0}+\int _{a}^{t}f(s,x(s))ds}$ .

Now ${\displaystyle T}$  satisfies a Lipschitz condition as follows:

{\displaystyle {\begin{aligned}\left\|T(x)(t)-T(y)(t)\right\|&=\left\|\int _{a}^{t}f(s,x(s))ds-\int _{a}^{t}f(s,y(s))ds\right\|\\&\leq \int _{a}^{t}\|f(s,x(s))-f(s,y(s))\|ds\\&\leq \int _{a}^{t}L\|x(s)-y(s)\|ds\\&\leq (t-a)L\|x-y\|_{\infty }\leq \epsilon L\|x-y\|_{\infty },\end{aligned}}}

where we took the norm on ${\displaystyle {\mathcal {C}}([a,a+\epsilon ])}$  to be the supremum norm. If now ${\displaystyle \epsilon <{\frac {1}{L}}}$ , then ${\displaystyle T}$  is a contraction, and hence the Banach fixed-point theorem is applicable, giving us both existence and uniqueness.${\displaystyle \Box }$

Replacing the fixed-point principle by summation techniques, we get a slightly better result in the sense that the domain of definition of the function ${\displaystyle f}$  does not have to be all of ${\displaystyle [a,b]\times \mathbb {R} ^{n}}$ .

Picard–Lindelöf theorem (telescopic series version):

Let ${\displaystyle f:[a,b]\times \Omega \to \mathbb {R} ^{n}}$  be a function which is continuous and Lipschitz continuous in the second argument, where ${\displaystyle \Omega \subseteq \mathbb {R} ^{n}}$ , and let ${\displaystyle (t_{0},x_{0})\in \mathbb {R} \times \Omega }$  with the property that ${\displaystyle [t_{0}-s,t_{0}+s]\times {\overline {B_{r}(x_{0})}}\subseteq [a,b]\times \Omega }$  for some ${\displaystyle s,r>0}$ . If in this case ${\displaystyle \gamma \leq \min\{s,r/M\}}$ , where ${\displaystyle M:={\underset {(t,x)\in [t_{0}-s,t_{0}+s]\times B_{r}(x_{0})}{\sup \|f(t,x)\|}}}$ , then the initial value problem

${\displaystyle {\begin{cases}x'(t)=f(t,x(t))&t\in [t_{0}-\gamma ,t_{0}+\gamma ]\\x(t_{0})=x_{0}&\end{cases}}}$

possesses a unique solution.

Proof:

We first prove uniqueness. To do so, we use Gronwall's inequalities. Suppose ${\displaystyle x,y}$  are both solutions to the problem. Then

${\displaystyle \left\|x(t)-y(t)\right\|=\left\|\int _{t_{0}}^{t}f(t,x(t))-f(t,y(t))dt\right\|\leq \int _{t_{0}}^{t}\left\|f(t,x(t))-f(t,y(t))\right\|dt\leq 0+\int _{t_{0}}^{t}L\left\|x(t)-y(t)\right\|dt}$ ,

and hence by Gronwall's inequalities

${\displaystyle \left\|x(t)-y(t)\right\|\leq 0\cdot e^{\int _{t_{0}}^{t}Ldt}=0}$

for both ${\displaystyle t\in [t_{0},t_{0}+s]}$  (right Gronwall's inequality) and ${\displaystyle t\in [t_{0}-s,t_{0}]}$  (left Gronwall's inequality).

Now on to existence. Once again, we inductively define

${\displaystyle x_{0}(t):=x_{0}}$  (the constant function),
${\displaystyle x_{n+1}(t):=x_{0}+\int _{t_{0}}^{t}f(\tau ,x_{n}(\tau ))d\tau }$ .

Since ${\displaystyle f}$  is not necessarily defined on any larger set than ${\displaystyle [t_{0}-s,t_{0}+s]\times B_{r}(x_{0})}$ , we have to prove that this definition always makes sense, i.e. that ${\displaystyle f(t,x_{n}(t))}$  is defined for all ${\displaystyle n}$  and ${\displaystyle t\in [t_{0}-\gamma ,t_{0}+\gamma ]}$ , that is, ${\displaystyle x_{n}(t)\in B_{r}(x_{0})}$  for ${\displaystyle t\in [t_{0}-\gamma ,t_{0}+\gamma ]}$ . We prove this by induction.

For ${\displaystyle n=0}$ , this is trivial.

Assume now that ${\displaystyle x_{n}(t)\in B_{r}(x_{0})}$  for ${\displaystyle t\in [t_{0},t_{0}+\gamma ]}$ . Then

{\displaystyle {\begin{aligned}\|x_{n+1}(t)-x_{0}\|&=\left\|\int _{t_{0}}^{t}f(\tau ,x_{n}(\tau ))d\tau \right\|\\&\leq \int _{t_{0}}^{t}\|f(\tau ,x_{n}(\tau ))\|d\tau \\&\leq M\cdot \gamma \\&\leq M\cdot r/M=r.\end{aligned}}}

For ${\displaystyle t\in [t_{0}-\gamma ,t_{0}]}$  we obtain an analogous bound.

By the telescopic sum, we have

${\displaystyle x_{n}(t)-x_{0}=\sum _{j=1}^{n}(x_{j}(t)-x_{j-1}(t))}$ .

Furthermore, for ${\displaystyle t\in [t_{0},t_{0}+\gamma ]}$  and ${\displaystyle j\geq 1}$ ,

{\displaystyle {\begin{aligned}\|x_{j+1}(t)-x_{j}(t)\|&=\left\|x_{0}+\int _{t_{0}}^{t}f(\tau ,x_{j}(\tau ))d\tau -\left(x_{0}+\int _{t_{0}}^{t}f(\tau ,x_{j-1}(\tau ))d\tau \right)\right\|\\&\leq \int _{t_{0}}^{t}\|f(\tau ,x_{j}(\tau ))-f(\tau ,x_{j-1}(\tau ))\|d\tau \\&\leq \int _{t_{0}}^{t}L\|x_{j}(\tau )-x_{j-1}(\tau )\|d\tau .\end{aligned}}}

Hence, by induction,

${\displaystyle \|x_{j+1}(t)-x_{j}(t)\|\leq \int _{t_{0}}^{t}Lr{\frac {|\tau -t_{0}|^{j-1}L^{j-1}}{(j-1)!}}d\tau \leq r{\frac {|t-t_{0}|^{j}L^{j}}{j!}}\leq r{\frac {\gamma ^{j}L^{j}}{j!}}}$ .

Again, by the very same argument, an analogous bound holds for ${\displaystyle t\in [t_{0}-\gamma ,t_{0}]}$ .

Thus, by the Weierstraß M-test, the telescopic sum

${\displaystyle x_{n}(t)-x_{0}=\sum _{j=1}^{n}(x_{j}(t)-x_{j-1}(t))}$

converges uniformly; in particular, ${\displaystyle x_{n}}$  converges.

It is now possible to interchange differentiation and summation in the latter sum; for, on the one hand, we are uniformly convergent, and on the other hand,

${\displaystyle \sum _{j=1}^{n}(x_{j}'(t)-x_{j-1}'(t))=\sum _{j=2}^{n}(f(t,x_{j-1}(t))-f(t,x_{j-2}(t)))+f(t,x_{0})=f(t,x_{n-1}(t))}$ ,

which converges to ${\displaystyle f(t,x(t))}$  for ${\displaystyle n\to \infty }$  due to theorem 2.5 and the convergence of ${\displaystyle x_{n}}$ ; note that the image of each ${\displaystyle x_{n}}$  is contained within the compact set ${\displaystyle {\overline {B_{r}(x_{0})}}}$ , the closure of ${\displaystyle B_{r}(x_{0})}$ . Hence indeed

${\displaystyle x'(t)=\left(\sum _{j=1}^{\infty }(x_{j}(t)-x_{j-1}(t))\right)'=\sum _{j=1}^{\infty }(x_{j}'(t)-x_{j-1}'(t))=f(t,x(t))}$

on ${\displaystyle [t_{0}-\gamma ,t_{0}+\gamma ]}$ .${\displaystyle \Box }$