Picard-Lindelöf Theorem

Definition (Lipschitz Condition).

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A function $F:U\subset X\to Y$ is Lipschitz continuous on $U$ if there exists a real constant $L\ge 0$ such that for all ${\mathbf{x}}_1,{\mathbf{x}}_2\in U$, the inequality $d_Y(F({\mathbf{x}}_1),F({\mathbf{x}}_2))\le L{d}_X({\mathbf{x}}_1,{\mathbf{x}}_2)$ holds. The constant $L$ is called a Lipschitz constant for $F$. For a vector field $F:U\subset {\mathbb{R}}^m\to {\mathbb{R}}^m$, this condition, using the Euclidean norm $\Vert \cdot \Vert$, is $\Vert F({\mathbf{x}}_1)-F({\mathbf{x}}_2)\Vert \le L\Vert {\mathbf{x}}_1-{\mathbf{x}}_2\Vert$.

Theorem (Picard-Lindelöf Existence and Uniqueness).

Let $U\subset {\mathbb{R}}^m$ be an open set and let $F:U\to {\mathbb{R}}^m$ be a Lipschitz continuous vector field. For any initial point ${\mathbf{x}}_0\in U$, the initial value problem (IVP)

$$\{\begin{array}{ll}\frac{d\mathbf{x}}{dt}=F(\mathbf{x}(t)),& \\ \mathbf{x}(0)={\mathbf{x}}_0& \end{array}$$

has a unique solution $\mathbf{x}(t)$ on some time interval $I=[-\alpha ,\alpha ]$ for some $\alpha >0$. Furthermore, this solution depends continuously on the initial condition ${\mathbf{x}}_0$.

Proof. The proof consists of three main parts: 1) reformulation of the IVP as an integral equation and definition of a mapping, 2) application of the Banach Fixed-Point Theorem to show existence and uniqueness of a fixed point, and 3) demonstration of continuous dependence on the initial condition.

Part 1: Integral Equation Formulation. A function $\mathbf{x}(t)$ is a solution to the IVP if and only if it is continuous and satisfies the integral equation:

$$\mathbf{x}(t)={\mathbf{x}}_0+{\int }_0^{t}F(\mathbf{x}(s))ds$$

This equivalence follows from the Fundamental Theorem of Calculus. A solution to this integral equation is a fixed point of the operator $\mathcal{T}$ defined by:

$$(\mathcal{T}\mathbf{y})(t)={\mathbf{x}}_0+{\int }_0^{t}F(\mathbf{y}(s))ds$$

Let’s define a suitable metric space for this operator. Since ${\mathbf{x}}_0\in U$ and $U$ is open, there exists $b>0$ such that the closed ball $\bar{B}({\mathbf{x}}_0,b)=\{\mathbf{y}\in {\mathbb{R}}^m:\Vert \mathbf{y}-{\mathbf{x}}_0\Vert \le b\}$ is contained in $U$. Since $F$ is continuous on the compact set $\bar{B}({\mathbf{x}}_0,b)$, it is bounded there. Let $M={\sup }_{\mathbf{y}\in \bar{B}({\mathbf{x}}_0,b)}\Vert F(\mathbf{y})\Vert$.

Choose $\alpha >0$ such that $\alpha \le b/M$ and $\alpha L<1$, where $L$ is the Lipschitz constant of $F$. Let $I_{\alpha }=[-\alpha ,\alpha ]$. Let $X=C(I_{\alpha },\bar{B}({\mathbf{x}}_0,b))$ be the space of continuous functions from $I_{\alpha }$ to $\bar{B}({\mathbf{x}}_0,b)$. This space, equipped with the supremum norm $\Vert {\mathbf{y}}_1-{\mathbf{y}}_2{\Vert }_{\infty }={\sup }_{t\in I_{\alpha }}\Vert {\mathbf{y}}_1(t)-{\mathbf{y}}_2(t)\Vert$, is a complete metric space.

We show that $\mathcal{T}$ maps $X$ to itself. For any $\mathbf{y}\in X$ and $t\in I_{\alpha }$:

$$\Vert (\mathcal{T}\mathbf{y})(t)-{\mathbf{x}}_0\Vert =\Vert {\int }_0^{t}F(\mathbf{y}(s))ds\Vert \le \left|{\int }_0^t\Vert F(\mathbf{y}(s))\Vert ds\right|\le M|t|\le M\alpha \le b$$

Thus, $(\mathcal{T}\mathbf{y})(t)\in \bar{B}({\mathbf{x}}_0,b)$ for all $t\in I_{\alpha }$. Since the integral of a continuous function is continuous, $\mathcal{T}\mathbf{y}$ is a continuous function. Therefore, $\mathcal{T}:X\to X$.

Part 2: Contraction Mapping. Now we show that $\mathcal{T}$ is a contraction mapping on $X$. Let ${\mathbf{y}}_1,{\mathbf{y}}_2\in X$.

$$\begin{array}{rl}\Vert (\mathcal{T}{\mathbf{y}}_1)(t)-(\mathcal{T}{\mathbf{y}}_2)(t)\Vert & =\Vert {\int }_0^t(F({\mathbf{y}}_1(s))-F({\mathbf{y}}_2(s)))ds\Vert \\ & \le \left|{\int }_0^t\Vert F({\mathbf{y}}_1(s))-F({\mathbf{y}}_2(s))\Vert ds\right|\\ & \le \left|{\int }_0^{t}L\Vert {\mathbf{y}}_1(s)-{\mathbf{y}}_2(s)\Vert ds\right|\\ & \le L\left|{\int }_0^t\Vert {\mathbf{y}}_1-{\mathbf{y}}_2{\Vert }_{\infty }ds\right|\\ & \le L|t|\Vert {\mathbf{y}}_1-{\mathbf{y}}_2{\Vert }_{\infty }\le L\alpha \Vert {\mathbf{y}}_1-{\mathbf{y}}_2{\Vert }_{\infty }\end{array}$$

Taking the supremum over $t\in I_{\alpha }$:

$$\Vert \mathcal{T}{\mathbf{y}}_1-\mathcal{T}{\mathbf{y}}_2{\Vert }_{\infty }\le (L\alpha )\Vert {\mathbf{y}}_1-{\mathbf{y}}_2{\Vert }_{\infty }$$

Since we chose $\alpha$ such that $L\alpha <1$, $\mathcal{T}$ is a contraction on the complete metric space $X$. By the Banach Fixed-Point Theorem, $\mathcal{T}$ has a unique fixed point $\mathbf{x}\in X$. This fixed point is the unique solution to the IVP on the interval $I_{\alpha }$.

Part 3: Continuous Dependence on Initial Conditions. Let $\mathbf{x}(t;{\mathbf{x}}_0)$ be the unique solution for the initial condition ${\mathbf{x}}_0$. Let ${\mathbf{z}}_0$ be another initial condition close to ${\mathbf{x}}_0$, and let $\mathbf{z}(t;{\mathbf{z}}_0)$ be its corresponding solution. Both solutions exist on an interval $[-\alpha ,\alpha ]$ for a sufficiently small $\alpha$.

$$\mathbf{x}(t)={\mathbf{x}}_0+{\int }_0^{t}F(\mathbf{x}(s))ds$$ $$\mathbf{z}(t)={\mathbf{z}}_0+{\int }_0^{t}F(\mathbf{z}(s))ds$$

Subtracting these equations gives:

$$\mathbf{x}(t)-\mathbf{z}(t)=({\mathbf{x}}_0-{\mathbf{z}}_0)+{\int }_0^t(F(\mathbf{x}(s))-F(\mathbf{z}(s)))ds$$

Taking norms and using the triangle inequality and Lipschitz condition:

$$\Vert \mathbf{x}(t)-\mathbf{z}(t)\Vert \le \Vert {\mathbf{x}}_0-{\mathbf{z}}_0\Vert +\left|{\int }_0^t\Vert F(\mathbf{x}(s))-F(\mathbf{z}(s))\Vert ds\right|$$ $$\Vert \mathbf{x}(t)-\mathbf{z}(t)\Vert \le \Vert {\mathbf{x}}_0-{\mathbf{z}}_0\Vert +L\left|{\int }_0^t\Vert \mathbf{x}(s)-\mathbf{z}(s)\Vert ds\right|$$

Let $\varphi (t)=\Vert \mathbf{x}(t)-\mathbf{z}(t)\Vert$. For $t\ge 0$, we have $\varphi (t)\le \Vert {\mathbf{x}}_0-{\mathbf{z}}_0\Vert +L{\int }_0^t\varphi (s)ds$. By Gronwall’s inequality, this implies:

$$\varphi (t)\le \Vert {\mathbf{x}}_0-{\mathbf{z}}_0\Vert e^{Lt}$$

For any $t\in [-\alpha ,\alpha ]$, we have $e^{L|t|}\le e^{L\alpha }$. Therefore:

$$\Vert \mathbf{x}(t;{\mathbf{x}}_0)-\mathbf{z}(t;{\mathbf{z}}_0){\Vert }_{\infty }\le \Vert {\mathbf{x}}_0-{\mathbf{z}}_0\Vert e^{L\alpha }$$

This inequality shows that the solution map ${\mathbf{x}}_0\mapsto \mathbf{x}(t;{\mathbf{x}}_0)$ is continuous (in fact, Lipschitz continuous) on the space of initial conditions. If $\Vert {\mathbf{x}}_0-{\mathbf{z}}_0\Vert \to 0$, then $\Vert \mathbf{x}(t;{\mathbf{x}}_0)-\mathbf{z}(t;{\mathbf{z}}_0){\Vert }_{\infty }\to 0$, which is the definition of continuous dependence. $\square$