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

$x'(t)=f(t,x(t))$

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

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

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

are equivalent to the single equation

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

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

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

Now $T$ satisfies a Lipschitz condition as follows:

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

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

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 $f$ does not have to be all of $[a,b]\times \mathbb {R} ^{n}$.

Picard–Lindelöf theorem (telescopic series version):

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

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

possesses a unique solution.

Proof:

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

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

and hence by Gronwall's inequalities

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

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

Now on to existence. Once again, we inductively define

$x_{0}(t):=x_{0}$ (the constant function),

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

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

For $n=0$, this is trivial.

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

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

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

By the telescopic sum, we have

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

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

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

Hence, by induction,

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

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

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

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

converges uniformly; in particular, $x_{n}$ 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,

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

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

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

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