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 $f:I\times {ℝ}^n\to {ℝ}^n$ be a continuous function, and let

$x^{\prime }(t)=f(t,x(t))$

be the associated ordinary differential equation. If $f$ is Lipschitz continuous in the second argument, then this ODE possesses a unique solution on $[a,a+ϵ]$ for each possible initial value $x(0)=x_0\in {ℝ}^n$, where $ϵ<1/L$, $L$ 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

$\{\begin{array}{cc}{x}^{\prime }(t)=f(t,x(t))& t\in [a,a+ϵ]\\ x(0)=x_0& \end{array}$

are equivalent to the single equation

$\mathrm{\forall }t\in [a,a+ϵ]:x(t)=x_0+\underset{a}{\overset{t}{\int }}f(s,x(s))ds$,

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

$T:𝒞([a,a+ϵ])\to 𝒞([a,a+ϵ]),T(x)(t):=x_0+\underset{a}{\overset{t}{\int }}f(s,x(s))ds$.

Now $T$ satisfies a Lipschitz condition as follows:

$\begin{array}{cc}\Vert T(x)(t)-T(y)(t)\Vert & =\Vert \underset{a}{\overset{t}{\int }}f(s,x(s))ds-\underset{a}{\overset{t}{\int }}f(s,y(s))ds\Vert \\ & \le \underset{a}{\overset{t}{\int }}\Vert f(s,x(s))-f(s,y(s))\Vert ds\\ & \le \underset{a}{\overset{t}{\int }}L\Vert x(s)-y(s)\Vert ds\\ & \le (t-a)L\Vert x-y{\Vert }_{\mathrm{\infty }}\le ϵL\Vert x-y{\Vert }_{\mathrm{\infty }},\end{array}$

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

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 $[a,b]\times {ℝ}^n$.

Picard–Lindelöf theorem (telescopic series version):

Let $f:[a,b]\times \mathrm{\Omega }\to {ℝ}^n$ be a function which is continuous and Lipschitz continuous in the second argument, where $\mathrm{\Omega }\subseteq {ℝ}^n$, and let $(t_0,x_0)\in ℝ\times \mathrm{\Omega }$ with the property that $[t_0-s,t_0+s]\times \overline{B_r(x_0)}\subseteq [a,b]\times \mathrm{\Omega }$ for some $s,r>0$. If in this case $\gamma \le \min \{s,r/M\}$, where $M:=\underset{(t,x)\in [t_0-s,t_0+s]\times B_r(x_0)}{\sup \Vert f(t,x)\Vert }$, then the initial value problem

$\{\begin{array}{cc}{x}^{\prime }(t)=f(t,x(t))& t\in [t_0-\gamma ,t_0+\gamma ]\\ x(t_0)=x_0& \end{array}$

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

$\Vert x(t)-y(t)\Vert =\Vert \underset{t_0}{\overset{t}{\int }}f(t,x(t))-f(t,y(t))dt\Vert \le \underset{t_0}{\overset{t}{\int }}\Vert f(t,x(t))-f(t,y(t))\Vert dt\le 0+\underset{t_0}{\overset{t}{\int }}L\Vert x(t)-y(t)\Vert dt$,

and hence by Gronwall's inequalities

$\Vert x(t)-y(t)\Vert \le 0\cdot e^{\underset{t_0}{\overset{t}{\int }}Ldt}=0$

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),

$x_{n+1}(t):=x_0+\underset{t_0}{\overset{t}{\int }}f(\tau ,x_n(\tau ))d\tau$.

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

$\begin{array}{cc}\Vert x_{n+1}(t)-x_0\Vert & =\Vert \underset{t_0}{\overset{t}{\int }}f(\tau ,x_n(\tau ))d\tau \Vert \\ & \le \underset{t_0}{\overset{t}{\int }}\Vert f(\tau ,x_n(\tau ))\Vert d\tau \\ & \le M\cdot \gamma \\ & \le M\cdot r/M=r.\end{array}$

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 $j\ge 1$,

$\begin{array}{cc}\Vert x_{j+1}(t)-x_j(t)\Vert & =\Vert x_0+\underset{t_0}{\overset{t}{\int }}f(\tau ,x_j(\tau ))d\tau -(x_0+\underset{t_0}{\overset{t}{\int }}f(\tau ,x_{j-1}(\tau ))d\tau )\Vert \\ & \le \underset{t_0}{\overset{t}{\int }}\Vert f(\tau ,x_j(\tau ))-f(\tau ,x_{j-1}(\tau ))\Vert d\tau \\ & \le \underset{t_0}{\overset{t}{\int }}L\Vert x_j(\tau )-x_{j-1}(\tau )\Vert d\tau .\end{array}$

Hence, by induction,

$\Vert x_{j+1}(t)-x_j(t)\Vert \le \underset{t_0}{\overset{t}{\int }}Lr\frac{|\tau -t_0{|}^{j-1}{L}^{j-1}}{(j-1)!}d\tau \le r\frac{|t-t_0{|}^j{L}^j}{j!}\le r\frac{{\gamma }^j{L}^j}{j!}$.

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,

$\sum _{j=1}^n({x_j}^{\prime }(t)-x_{j^{\prime }-1}(t))=\sum _{j=2}^n(f(t,x_{j-1}(t))-f(t,x_{j-2}(t)))+f(t,x_0)=f(t,x_{n-1}(t))$,

which converges to $f(t,x(t))$ for $n\to \mathrm{\infty }$ 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

$x^{\prime }(t)=(\sum _{j=1}^{\mathrm{\infty }}(x_j(t)-x_{j-1}(t)))=\sum _{j=1}^{\mathrm{\infty }}({x_j}^{\prime }(t)-x_{j^{\prime }-1}(t))=f(t,x(t))$

on $[t_0-\gamma ,t_0+\gamma ]$.$◻$